BUREAU’S HIGHER SECONDARY (+2) CHEMISTRY VOL. - I Dr. A. K. Das Dr. H. K. Patnaik Former Chairman, CHSE, Odisha Former Principal, Govt. College, Bhubaneswar. Rourkela. Dr. S. Behera Dr. G. C. Dash Former Principal, D. D. College, Former Principal, Keonjhar. S.C.S. (Autonomous) College, Puri. Dr. A. K. Panigrahi Dr. Hrushikesh Mohanty Former Director, Odisha State Bureau of Associate Professor & HOD, Chemistry, Textbook Preparation and Production, B.J.B. Autonomous College, Bhaubaneswar. Bhubaneswar. Dr. Santosini Patra Dr. B. C. Singh Associate Professor in Chemistry Former Professor of Chemistry, R.D. Women's University, Bhubaneswar Ravenshaw College, Cuttack. Dr. Panchanan Goud Dr. B. K. Mohapatra Associate Professor & H.O.D., Chemistry, Former Chairman, CHSE, Odisha, Khalikote University, Berhampur. Bhubaneswar. Dr. I. B. Mohanty Dr. J. N. Kar Deputy Director Former Principal, Govt. College, Phulbani. Department of Higher Education, Odisha Dr. Sakuntala Jena Asst. Professor, Department of Chemistry, Govt. Women's College, Dhenkanal. PUBLISHED BY THE ODISHA STATE BUREAU OF TEXTBOOK PREPARATION AND PRODUCTION, PUSTAK BHAVAN, BHUBANESWAR.

Published by : THE ODISHA STATE BUREAU OF TEXTBOOK PREPARATION AND PRODUCTION, Pustak Bhavan, Bhubaneswar, Odisha, India. First Edition : 2000 / 7000 Revised and Enlarged Edition : 2002 / 3000 Revised Edition : 2004 / 1000 Revised Edition : 2006 / 1000 Revised Edition : 2008/2000 Reprint - 2010/2000 Revised Edition : 2013/2000 New Edition : 2016-17 / 5000 Publication No : 176 ISBN : 978-81-8005-349-8 © Reserved by the Odisha State Bureau of Textbook Preparation and Production, Bhubaneswar. No part of this publication may be reproduced in any form without the prior written permission of the publisher. Typesetting and Diagram by : PRINT-TECH OFFSET PVT. LTD., Bhubaneswar Printed at : PRINT-TECH OFFSET PVT. LTD., Bhubaneswar Price : ` 360/-

FOREWORD (New Edition - 2016) The Council of Higher Secondary Education, Odisha has revised the Courses of Studies in Chemistry for its examination, 2017 and onwards. It is really heartening to know that the Chemistry - Vol-I is an exclusive textbook of CHSE and its new edition is being published by the Odisha State Bureau of Text Book Preparation and Production, Bhubaneswar. I acknowledge with thanks to the Board of Writers and Reviewers who have worked hard in writing the chapters of the book and setting new pattern of questions as per the requirement of the new syllabus of CHSE. My special thanks due to Dr. Akhil Krishna Panigrahi, Dr. Gobinda Chandra Dash, Dr. Hrushikesh Mohanty, Dr. Panchanan Goud, Dr Santosini Patra & Dr. Sakuntala Jena for taking pain and strain of doing arduous work in preparing the book within the frame work of the new syllabus of CHSE. I also take this opportunity to convey my heartfelt thanks to Dr. Jibanananda Kar who have gone through the proofs of the entire book before final printing. The new edition of the book could see the light of the day in record time due to sincere effort of Sri Biraja Bhusan Mohanty, Deputy Director of the Bureau. My earnest gratitude is also due to the Commissioner Cum Secretary, Department of Higher Education, Odisha and Chairman, Council of Higher Secondary Education, Odisha in entrusting the Bureau the task of publishing the book for Higher Secondary students of the state. Improvement has no limit specially when one aims at excellence. The Bureau is always alive to any constructive suggestions from the students as well as the teachers to make the book more purposeful. 11.3.2016 Dr. Geetika Patnaik Director Odisha State Bureau of Text Book Preparation and Production, Pustak Bhavan, Bhubaneswar, Odisha

PREFACE India is on the verge of a great leap into the global scientific and technological advancement in the New Millennium. Our Universities and Council of Higher Secondary Education have taken up the upgrading of science curriculum as a challenge. Society is becoming largely knowledge based. To prepare our young students to achieve the goal, Council of Higher Secondary Education, Odisha has revised the syllabus of all science subjects and has taken up the challenge to arm our students with advanced scientific education. The biggest challenge in the present times in the field of scientific education is the preparation of textbooks suited to the needs of the students. Some of the most experienced, learned and brilliant teachers of the State have made attempts towards fulfilling the national need of providing a good textbook in Chemistry for +2 students. As a result the book titled +2 Chemistry has been prepared in accordance with the new syllabus of C.H.S.E. Odisha which will be effective for the students who will be admitted in 2016 and onwards. This book has many special features, the salient ones of which may be enumerated as follows : (i) The text book has been prepared keeping in view the type of questions set in the +2 as well as entrance examinations. (ii) The subject matter has been put in a lucid manner and in a simple language to be easily followed by the students. (iii) Large numbers of numerical problems have been worked out. (iv) Neat diagrams are given to provide suitable explanation of the texts. (v) Large number of questions of very short answer type, short answer type, multiple choice type and long answer type including questions of H. S. Examinations are given in each chapter. (vi) At the end of every chapter, summary of the topics dealt in the chapter is given under Chapter at a glance . The authors express their gratitude to the authorities of C.H.S.E. for accepting them as the members of Board of Editors and to the ODISHA STATE BUREAU OF TEXTBOOK PREPARATION AND PRODUCTION for publishing the book. The authors sincerely hope that their endeavour would fulfil the need of students. There may be minor errors of omissions and commissions in the book. The authors welcome constructive criticism and suggestions for the improvement of the book. Bhubaneswar Board of Writers 14.3.2016

Unit - I COURSES OF STUDIES IN CHEMISTRY (THEORY) FOR HIGHER SECONDARY EXAMINATION (Effective from 2016 Admission Batch) FIRST YEAR : Some Basic Concepts of Chemistry General Introduction : Importance and scope of chemistry. Nature of matter, laws of chemical combination, Dalton's atomic theory: concept of elements, atoms and molecules. Atomic and molecular masses. and equivalent mass of elements, acid, base, salt, oxidants and reductants, mole concept and molar mass, percentage composition, empirical and molecular formula, chemical reactions, stoichiometry and calculations based on stoichiometry. Unit - II : Structure of Atom Discovery of Electron, Proton and Neutron, atomic number, isotopes and isobars. Thomson's model and its limitations. Rutherford's model and its limitations, Bohr's relationship, Heisenberg uncertainity principle, concept of orbitals, quantum numbers, shape of s, p and d orbitals, rules for filling electrons in orbitals - Aufbau principle, Pauli's exclusion principle and Hund's rule, electronic configuration of atoms, stability of half filled and completely filled orbitals. Unit -III : Classification of Elements and Periodicity in Properties Significance of classification, brief history of the development of periodic table, modern periodic law and the present form of periodic table, periodic trends in properties of elements - atomic radii, ionic radii, inert gas radii. Ionization enthalpy, electron gain enthalpy, electronegativity, valency and oxidation state, Nomenclature of elements with atomic number greater than 100. Unit - IV : Chemical Bonding and Molecular Structure Valence electrons, ionic bond, covalent bond; bond paramaters, Lewis structure, polar character of covalent bond, covalent character of ionic bond, valence bond theory, resonance, geometry of covalent molecules, VSEPR theory, concept of hybridization, involving s, p and d orbitals and shapes of some simple molecules, molecular orbital theory of homonuclear diatomic molecules (qualitative idea only), hydrogen bond.

Unit - V : State of Matter : Gases and Liquids Three states of matter, intermolecular interactions, types of bonding, melting and boiling points, role of gas laws in elucidating the concept of the molecule, Boyle's law, Charle's law, Gay Lussac's law, Avogadro's law, ideal behaviour, empirical derivation of gas equation. Avogadro's number, ideal gas equation, Deviation from ideal behaviour, liquefaction of gases, critical temperature, kinetic energy and molecular speeds (elementary idea). Liquid state: vapour pressure, viscocity and surface tension (qualitave idea only, no mathematical derivations.) Unit - VI : Chemical Thermodynamics Concept of System and types of systems, surroundings, work, heat, energy, extensive and intensive properties, state functions. First law of thermodynamics - internal energy and enthalpy, heat capacity and specific heat, measurement of ΔU and ΔH, Hess's law of constant heat summation, enthalpy of bond dissociation, combustion, formation, atomization, sublimation phase transition, ionization, solution and dilution. Second law of Thermodynamics (brief introduction) Introduction of entropy as a state function, Gibb's energy change for spontaneous and non-spontaneous process, criteria for equilibrium. Third law of thermodynamics (brief introduction) Unit - VII : Equilibrium Equilibrium in physical and chemical processes, dynamic nature of equilibrium, law of mass action, equilibrium constant, Kc, Kp and Kx and their relationship factors affecting equilibrium. Le Chatelier's principle, ionic equilibrium-ionization of acids and bases strong and weak electrolytes, degree of ionization, ionization of poly basic acids, acid strength, concept of pH, Henderson Equation. hydrolysis of salts (elementary idea), buffer solution, solubility product, common ion effect (with illustrative examples), numerical problems. Unit - VIII : Redox Reaction Concept of oxidation and reduction, redox reactions, oxidation number, balancing redox reactions, in terms of loss and gain of electrons and change in oxidation number, applications of redox reactions.

Unit - IX : Hydrogen Position of hydrogen in periodic table, occurrence, isotopes, preparation, properties and uses of hydrogen, hydrides-ionic, covalent and interstitial; physical and chemical properties of water, heavy water, hydrogen peroxide - preparation, reactions and structure and use, hydrogen as a fuel. Unit - X : s-Block Elements (Alkali and Alkaline Earth Metals) Group 1 and Group 2 Elements General introduction, electronic configuration, occurrence, anomalous properties of the first element of each group, diagonal relationship, trends in the variation of properties (such as ionization enthalpy, atomic and ionic radii), trends in chemical reactivity with oxygen, water, hydrogen and halogens, uses. Preparation and Properties of some important Compounds : Sodium Carbonate, Sodium Chloride, Sodium Hydroxide and Sodium Hydrogencarbonate, Biological importance of Sodium and Potassium. Calcium Oxide and Calcium Carbonate and their industrial uses, biological importance of Magnesium and Calcium. Unit - XI : Some p-Block Elements General Introduction to p-Block Elements Group 13 Elements : General introduction, electronic configuration, occurrence, variation of properties, oxidation states, trends in chemical reactivity, anomalous properties of first element of the group, Boron-physical and chemical properties, some important compounds, Borax, Boric acid Boron Hydrides, Aluminium: Reactions with acids and alkalis, uses. Group 14 Elements : General introduction, electronic configuration, occurrence, variation of properties, oxidation states, trends in chemical reactivity, anomalous behaviour of first elements. Carbon-catenation, allotropic forms, physical and chemical properties, use of some important compouds: oxides. Important compounds of Silicon and a few uses: Silicon Tetrachloride. Silicones, Silicates and Zeolites, their uses. Unit - XII : Organic Chemistry - Some Basic Principles and Technique General introduction, methods of purification, qualitative and quantitave analysis, classification and IUPAC nomenclature of organic compounds. Electronic

displacements in a covalent bond: inductive effect, electromeric effect. resonance and hyperconjugation, homolytic and heterolytic fission of a covalent bond free radicals, carbocations, carbanions, electrophiles and nucleophiles, types of organic reactions. Unit - XIII : Hydrocarbons Classification of Hydrocarbons Aliphatic Hydrocarbons Alkanes - Nomenclature, isomerism, conformation (ethane only), methods of preparation, physical properties, chemical reactions including free radical mechanism of halogenation, combustion and pyrolysis. Alkenes - Nomenclature, structure of double bond (ethene) geometrical isomerism, physical properties, methods of preparation, chemical reactions: addition of hydrogen, halogen, water, hydrogen halides (Markownikov's addition and peroxide effect), ozonolysis, oxidation, mechanism of electrophilic addition. Alkynes - Nomenclature, structure of triple bond (ethyne), physical properties, methods of preparation, chemical reactions: acidic character of alkynes, addition reaction of - hydrogen, halogens, hydrogen halides and water. Aromatic Hydrocarbons : Introduction, IUPAC nomenclature, benzene: resonance, aromaticity chemical properties: mechanism of electrophilic substitution: nitration, sulphonation, halogenation Friedel Craft's alkylation and acylation, directive influence of functional group in monosubstituted benzene. Carcinogenicity and toxicity. Unit - XIV : Environmental Chemistry Environmental pollution - air, water and soil polution, chemical reactions in atmosphere, smog, major atmospheric pollutants, acid rain, ozone and its reactions, effects of depletion of ozone layer, green house effect and global warming-pollution due to industrial wastes, green chemistry as an alternative tool in reducing pollution, strategies for controal of environmental pollution. .....

Chapter CONTENTS Pages Subjects UNIT - I CHAPTER - 1 : SOME BASIC CONCEPTS OF CHEMISTRY : 1 - 33 Introduction - 1.1. Important contributions of chemistry, 1.2. Units of measurement, system of units, 1.3. Matter, 1.4. Atoms and molecules, 1.5. States of matter, 1.6. Symbol, 1.7. Valency, 1.8. Radicals, 1.9. Formula, 1.10. Naming of compounds, 1.11. Chemical equation. CHAPTER - 2 : BASIC CONCEPTS - ATOMS & MOLECULES 34 - 113 2.1. Laws of chemical combinations, 2.2. Dalton's atomic theory, 2.3. Dalton's atomic theory and the law of chemical combination, 2.4. Atomic mass, 2.5. Determination of relative atomic mass by Dulong and Petit's method, 2.6. Molecular mass, 2.7. Equivalent mass, 2.8. Equivalent masses of acids, bases and salts, 2.9. Percentage composition, molecular formula and empirical formula, 2.10. Avogadro's hypothesis and the mole concept, 2.11. Stoichiometry (calculations based on chemical equation). UNIT - II CHAPTER - 3 : STRUCTURE OF ATOM : 114 - 154 3.1. Fundamental particles and their properties, 3.2. Thomson's model for the structure of atom, 3.3. Rutherford's experiment- scattering of alpha particles, 3.4. Characteristics of radiation and Planck's theory, 3.5. Bohr's model of atom, 3.6. Bohr's equation for energy of electron in Hydrogen atom, 3.7. Solar spectrum, 3.8. Bohr's theory and hydrogen spectrum, 3.9. Dual nature of electron, 3.10. Quantum numbers, 3.11. Pauli's exclusion principle, 3.12. Heisenberg's uncertainty principle and the concept of probability, 3.13. Aufbau principle, 3.14. Hund's rule, 3.15. Electronic configuration of elements, 3.16. Extra stability of half filled and completely filled orbitals. UNIT - III CHAPTER - 4 : CLASSIFICATION OF ELEMENTS AND 155 - 175 PERIODICITY IN PROPERTIES Introduction - 4.1. Brief history of Periodic table, 4.2. Long form of Periodic table, 4.3. Cause of periodicity and magic numbers, 4.4. Classification with respect to s, p, d, f blocks, 4.5. Periodicity

Chapter Subjects Pages in properties of elements 4.6. Nomenclature of elements with atomic number more than 100. UNIT - IV CHAPTER - 5 : CHEMICAL BONDING AND MOLECULAR 176 - 232 STRUCTURE Introduction - 5.1. Why do atoms combine?, 5.2. Valence electrons, 5.3. Types of chemical bond, 5.4. Bond parameters, 5.5. Lewis structure, 5.6. Polar character of covalent bonds, 5.7. Covalent character of Ionic bonds. 5.8. Valence bond theory, 5.9. Resonance, 5.10. VSEPR Theory-Geometry of covalent molecules, 5.11. Hybridisation, 5.12. Molecular Orbital Theory (MOT), 5.13. Hydrogen bond. UNIT - V CHAPTER - 6 : STATES OF MATTER : GASES AND LIQUIDS 233 - 296 General Introduction - Gaseous state : 6.1. Characteristic properties of Gases, 6.2. State variable of gases, 6.3. Gas Laws- Boyle's law, Charle's law, Gay-Lussac's law, combined gas equation, Ideal gas equation, Dalton's law of partial pressure, Diffusion of gases, Graham's law of diffusion, 6.4. Kinetic theory of gases : 6.4.1. Postulates of Kinetic theory of gases, 6.4.2. Derivation of Kinetic gas equation, 6.4.3. Relationship between average kinetic energy and absolute temparature, 6.4.4. Explanation of Gas Laws in the light of Kinetic Molecular Theory, 6.4.5. Deduction of Gas laws from Kinetic Gas equation, 6.4.6. Molecular velocity and their calculation - Behaviour of real gases, deviation of Real gases, causes of deviation. 6.4.7. Equation of state for real gases, 6.4.8. Maxwell's law of distribution of molecular velocities - Liquification of Gases - Joule - Thomson effect, Critical temperature. CHAPTER - 7 : LIQUID STATE 297 - 305 7.1. Characteristic properties of the liquids 7.2. Viscocity, 7.3. Surface tension 7.4. Vapour pressure. UNIT - VI CHAPTER - 8 : THERMODYNAMICS 306 - 357 Introduction- 8.1. Some basic terms used in Chemical Energetic, 8.2. Concept of Internal Energy, 8.3. Concept of Enthalpy, 8.4. First law of Thermodynamics (Conservation of Energy),

Chapter Subjects Pages 8.5. Heat capacity and specific heat of a system, 8.6. Thermochemistry and Thermochemical reactions, 8.7. Enthalpy changes in chemical reaction (Heat of reaction) 8.8. Types of Enthalpies of reactions, 8.9. Enthalpy of phase changes, 8.10. Hess's law of constant heat summation, 8.11. Bond Energy, 8.12. Second law of Thermodynamics-Limitations of First Law, 8.13. Scope of Second law, 8.14. Statement, 8.15. Spontaneous or Irreversible process. 8.16. Entropy(S). 8.17. Free Energy(G). 8.18. Gibb's Free energy and Equilibrium constant, 8.19. Spontaneity of a process, 8.20. Third law of Thermodynamics. UNIT - VII CHAPTER - 9 : EQUILIBRIA 358 - 418 9.1. Irreversible and Reversible reactions, 9.2. Equilibria in physical process. Chemical equilibria: 9.3. State of Equilibrium, 9.4. Law of Mass Action, 9.5. Le-Chatelier principle. Ionic Equilibria: 9.6. Theories of Acids and Bases, 9.7. Ionisation of weak acids and bases, 9.8. Ionisation of Polybasic acids, 9.9. Factors affecting acid strength, 9.10. Ionisation of water, 9.11. Hydrogen ion exponent: pH, 9.12. Buffer solution, 9.13. Hydrolysis of salts, 9.14. Solubility product, 9.15. Common Ion Effect. UNIT - VIII CHAPTER - 10 : REDOX REACTION 419 - 459 10.1. Redox reactions, 10.2. Competative electron transfer reactions, 10.3. Oxidation number or Oxidation state, 10.4. Types of Redox reactions, 10.5. Uses of oxidation number, 10.6. Applications of Redox reactions: 10.6.1. Quantitative analysis, 10.6.2 Redox reaction and extraction of metals, 10.6.4. Redox reaction and supply of energy 10.6.5. Redox reaction and Photosynthesis, 10.7. Theroretical aspects of Acid- Base titration-volumetric analysis. UNIT - IX CHAPTER - 11 : HYDROGEN 460 - 487 11.1. Position of Hydrogen in the periodic table, 11.2. Occurrence, 11.3. Isotopes of hydrogen, 11.4. Dihydrogen, 11.4.1. Laboratory methods of preparation, 11.4.2. Commercial methods of preparation, 11.4.3. Properties, 11.4.4. Uses

Chapter Subjects Pages 11.5. Hydrides, 11.5.1. Ionic or saline or saltlike hydrides, 11.5.2. Covalent or molecular hydrides, 11.5.3. Metallic or Non- stoichiometric or Interstitial hydrides, 11.6. Water, 11.6.1. Physical properties of water, 11.6.2. Chemical properties of water, 11.6.3. Hard and soft water, 11.7. Heavy water, 11.8. Hydrogen peroxide, 11.8.1. Preparation, 11.8.2. Manufacture of Hydrogen peroxide, 11.8.3. Concentration of Hydrogen peroxide solution, 11.8.4. Properties of Hydrogen peroxide, 11.9. Hydrogen as a fuel. UNIT - X CHAPTER - 12 : S-BLOCK ELEMENTS: ALKALI AND 488 - 510 ALKALINE EARTH METALS. ALKALI METALS : (GROUP - 1) 12.1. Introduction, 12.2. Electronic configuration, 12.3. General characteristics of Alkali metals, 12.4. Chemical characteristics, 12.5. Preparation and properties of some important compounds: Sodium carbonate, Sodium chloride, Sodium hydroxide (Caustic Soda), Sodium hydrogen carbonate (Baking soda), 12.6. Analytical tests for sodium, 12.7. Biological importance of sodium and potassium. CHAPTER - 13 : ALKALINE EARTH METALS : (GROUP - 2) 511 - 525 13.1. Introduction, 13.2. Electronic configurations, 13.3. General characteristics of Alkaline earth metals (Group-2), 13.4. Some important compounds of calcium: Calcium oxide (Quick lime) and Calcium carbonate, 13.5. Biological importance of magnesium and calcium. UNIT - XI THE P-BLOCK ELEMENTS CHAPTER - 14 : GROUP - 13 ELEMENTS 526 - 555 Introduction. 14.1. Group-13 elements : The Boron family, 14.2. General characteristics of p-Block elements, 14.3. Occurrence, 14.4. General characteristics of the elements. 14.5. Boron; Boranes, 14.6. Borax, 14.7. Boric acid, 14.8. Aluminium : Reaction with acids and alkalis, uses, 14.9. Alum. CHAPTER - 15 : THE CARBON FAMILY (GROUP-14 ELEMENTS) 556 - 580 15.1. Electronic configurations, 15.2. Occurrence, 15.3. General characteristics of the elements, 15.4. Oxidation states, 15.5. General trends in the chemical properties of the elements, 15.6. Anomalous character of carbon, 15.7. Carbon: catenation,

Chapter Subjects Pages 15.8. Allotropes of carbon, 15.8.1. Diamond structure, 15.8.2 Graphite 15.8.3. Recently discovered allotrope of carbon: Fullerenes, 15.9. Some important compounds of carbon; 15.9.1. Carbon monoxide, 15.9.2. Carbond dioxide, 15.9.3. Carbon tetrachloride, 15.9.4. Carbides, 15.10. Compounds of silicon: 15.10.1. Silicon dioxide, 15.10.2. Silicates and zeolites, 15.10.3. Silicon carbide, 15.10.4. Silicon tetrachloride, 15.10.5. Silicones. UNIT - XII ORGANIC CHEMISTRY CHAPTER - 16 : SOME BASIC PRINCIPLES AND TECHNIQUES 581 - 710 16.1. General introduction, 16.2. Basic concepts of organic compounds, 16.3. Characteristics of carbon atom, 16.4. Hybridisation and tetracovalency of carbon, 16.5. Methods of purification of organic compounds, 16.6. Qualitative analysis of organic compounds, 16.7. Quantitative analysis, 16.8. Classification of organic compounds, 16.9. Nomenclature of organic compounds, 16.10. Isomerism, 16.11. Electronic displacements in a covalent bond: Inductive effect, Inductomeric effect, Electromeric effect, Resonance, Hyperconjugation, 16.12. Types of organic reaction, 16.13. Fission of a covalent bond, 16.14. Electrophiles and Nucleophiles. UNIT - XIII 711 - 713 HYDROCARBONS CHAPTER - 17 : ALIPHATIC HYDROCARBONS : SATURATED 714 - 744 HYDROCARBONS-ALKANES 17.1. Saturated hydrocarbons (Paraffins) or Alkanes, 17.2. General formula, 17.3. Types of carbon atom in alkanes, 17.4. IUPAC rules for naming alkanes, 17.5. Structure of alkanes, 17.6. Isomerism in alkanes, 17.7. Occurrence 17.8. Methods of preparation, 17.9. General properties. CHAPTER - 18 : UNSATURATED HYDROCARBONS- ALKENES 745 - 775 18.1. Alkenes (Olefins), 18.2. Nomenclature of alkenes, 18.3. Structure of alkenes, 18.4. Isomerism, 18.5. Stability of alkenes, 18.6. Methods of preparation, 18.7. Properties. CHAPTER - 19 : UNSATURATED HYDROCARBONS- ALKYNES 776 - 796 19.1. Nomenclature, 19.2. Structure of alkynes, 19.3. Isomerism, 19.4. Methods o preparation, 19.5. Properties, 19.6. Uses of acetylene, 19.7. Distinction of ethane, ethene and ethyne.

Chapter Subjects Pages CHAPTER - 20 : AROMATIC COMPOUNDS: AROMATIC 797 - 841 HYDROCARBONS 20.1. Introduction, 20.2. Aromatic hydrocarbons or Arenes, 20.3. Meaning of aromaticity, 20.4. Molecular orbital structure of Benzene, 20.5. Nomenclature of aromatic compounds, 20.6. Methods of preparation of Benzene, 20.7. Properties of Benzene- General mechanism for electrophilic substitution- Halogenation, Nitration, Sulphonation, Friedel Crafts alkylation and acylation; Addition and oxidation reaction in Benzene, 20.8. Directive influence of substituents; Activating and deactivating influence of substituents, 20.9. Carcinogenicity and Toxicity. UNIT - XIV CHAPTER - 21 : ENVIRONMENTAL CHEMISTRY 842 - 881 21.1. Introduction, 21.2. Environmental pollution, 22.3. Air pollution, 21.3.1. Sources of air pollution, 21.3.2. Harmful effects of air pollutants, 21.3.3. Units of measurements of air pollutants 21.3.4. Control of air pollution, 21.4. Water pollution. 21.4.1. Sources of pollution 21.4.2. Effect of water pollution, 21.4.3. Prevention and control of water pollution, 21.4.4. International standards for drinking water, 21.5. Soil pollution, 21.5.1. Sources of soil pollution, 21.5.2. Effect of soil pollution 21.5.3. Control of soil pollution, 21.6. Chemical reactions in atmosphere 21.7. Smog, 21.8. Acid rain, 21.9. Ozone layer-Formation and Depletion of ozone layer, ozone hole, effect of depletion and protection of ozone layer. 21.10. Green House effect and Global warming, 21.11. Pollution due to industrial wastes, 21.12. Green Chemistry as an alternative tool for reducing pollution, 21.13. Strategy for control of environmental pollution, 21.14. Waste management. .....

UNIT – I SOME BASIC CONCEPTS OF CHEMISTRY CHAPTER - 1 INTRODUCTION Chemistry is one of the major disciplines of science. It deals with the study of the composition, structure and properties of matter along with the changes involved in matter. In other words, chemistry is defined as the science of matter and its transformations. Chemistry has been further classified into following main different branches. 1. Inorganic chemistry : It deals with the study of all the elements and their compounds except the covalently bonded compounds of carbon. 2. Organic chemistry : It deals with the study of the carbon compounds derived from living organisms, except oxides of carbon, carbides, carbonates and bicarbonates. In other words, organic chemistry is defined as the study of hydrocarbons and their derivatives. 3. Physical chemistry : It deals with the fundamental principles governing various chemical phenomena. 4. Analytical chemistry : It deals with the identification, separation and qualitative determination of the composition of different substances. 5. Industrial chemistry : It deals with the chemistry involved in various chemical industry. 6. Biochemistry : It deals with the study of chemical changes, which take place in living organisms. 7. Nuclear chemistry : It deals with the study of structure and processes involved in nuclei of atoms. 8. Polymer chemistry : It deals with the study of structure, synthesis and reactions of macromolecules. 1.1 IMPORTANT CONTRIBUTIONS OF CHEMISTRY : The present day chemistry is providing man with more comforts for a better, healthier and happier life. It is of great importance in industry, medicines, synthetic fibres, food stuffs, fertilizers and generation of power. The important contributions of chemisty to the modern world are : (i) In foodstuff : Food enriched with vitamins helps us for the improvement of our health. The modern methods of processing food and selection of diets have solved malnutrition problem. The satisfactory production and protection of food grains have been achieved by the application of chemical fertilizers and pesticides such as D.D.T, Gammaxene etc.

2 +2 CHEMISTRY (VOL. - I) (ii) In housing : Chemical substances, like cement, paints, varnishes, plastic, metals and their alloys are useful in building the modern house. (iii) In medicine : A number of medicines like quinine, antibiotics, insulin, streptomycin etc. have helped human beings in leading healthy life. Now radioactive isotopes are used in diagnosis of diseases. (iv) In industry : Chemical industries like steel, fertilizers, artificial fibres, synthetic rubber, petrochemicals, cement etc. have helped us to achieve self sufficiency in the respective fields. (v) In warfare : Powerful explosives like T.N.T, R.D.X and poisonous gases like phosgene are the contributions of chemistry. Atom bomb and hydrogen bomb can destroy the whole mankind. However, these are now used for peaceful purposes like generation of power and utilisation in telecommunications. (vi) Pollution control : The effluents are chemically treated before mixing with air and water to prevent pollution. Besides the above, chemistry will have a wide range of applications in the 21st century for modern living and better future. 1.2 UNITS OF MEASUREMENT SYSTEM OF UNITS : The system of units commonly used are : (1) French system or C.G.S. System, (2) British system or F.P.S. System, (3) Metre-Kilogram Second or M.K.S. System (4) International system of units or S.I.System. 1. C.G.S. System : In this system the units of length, mass and time are centimetre, gram and second respectively. The CGS System is widely used because its small and large units are multiples of ten. However, now a days S.I. units are preferred. The units in C.G.S. system for expressing mass, volume and length are explained in following table. Units for mass, volume and length. Physical(Quantity) Kilo Hecto Deca Deci Centi Milli 0.001 Mass (in gm) 1000 100 10 0.1 0.01 0.001(1ml) Volume (in litres) 1000 100 10 0.1 0.01 Length (in metres) 1000 100 10 0.1 0.01 0.001 (1cm) 2. F.P.S. System : In this system the units of length, mass and time are foot, pound and second respectively. 3. M.K.S. System : The unit of mass in this system is kilogram, the unit of length is metre and the unit of time is second. 4. International System of Units (S.I.Units) : This is the recent version of the metric system (M.K.S) and is supposed to be used throughout the world. This system consists of seven fundamental units with the help of which the derived units for all other possible quantities can be determined.

INTRODUCTION 3 (a) Fundamental Units : There are seven fundamental units. Fundamental quantity and units Physical quantity Unit Symbol Length Metre m Mass Kilogram Kg Time Second S Electric current Ampere A Temperature Kelvin K Luminous intensity Candela Cd Amount of substance Mole mol (b) Derived Units : There are many derived units. Only few of them which are mostly useful in the study of chemistry are : (i) Pressure: It is defined as the force per unit area. Hence pressure is expressed in Nm2. (ii) Volume : Volume of a solid is expressed in m3, whereas the volume of a liquid is generally expressed in litres. 1 litre = 10–3 m3 Names and Symbols for some S.I.derived units. Physical quantity Symbol Units Special Symbol name of for Area A m2 S.I.Units S.I.Units Density P Kg.m–3 Force F Kg.m.S–2 Newton N Pressure P Kg.m–1.S–2 Pascal Pa Energy E Kg.m2.S–2 Joule J Viscosity coefficient m Kg.m-1.S–1 Surface tension r Kg.S–2 Electric potential V Kg.m2.S-2A–1 Volt V Electric resistance R Kg.m2.S-2A–2 Ohm W Electric change Coulomb C Frequency I— Hertz Hz u S–1 (cycles per second)

4 +2 CHEMISTRY (VOL. - I) Multiples of SI Units : Knowledge of some units which are not S.I. Units but are commonly used and are fractions or multiples of S.I.Units is useful in various fields of chemistry. Such units and their relation with S.I.Units is described in the following table. Some Units and their relation with S.I.Units Physical quantity Unit Symbol Relation with S.I.Unit Length Angstrom AO 10–10 m Length Micron m 10–6 m Length Picometre pm 10–12 m Force dyne dyne 10–5N Pressure bar bar 105Nm–2 Energy erg erg 10–7J Area barn b 10–25m2 Volume litre l 10–3m3 = 1dm3 Temperature measurement : Temperature is measured with the help of thermometre. Three scales are used for measuring temperature. (i) Centigrade scale : The thermometer with centigrade scale has a zero mark for the temperature of ice at one atmospheric pressure. The mark 1000C indicates the temperature of boiling water at one atmospheric pressure. The space between any two consecutive divisions (representing 10C) is further divided into 10 subdivisions. (ii) Fahrenheit scale : In this scale, the temperature of ice at one atmospheric pressure has a mark 320F while that of boiling water is 2120F. Centigrade and Fahrenheit scales are related as follows. F= 9 C + 32 5 Example : Convert 250C into a temperature on Fahrenheit scale. Solution : Temperature on centigrade scale = 250C Hence, temperature on Fahrenheit scale = 9 x 25 + 32 5 = 77 0F (iii) Absolute scale : It is also known as Kelvin scale. On this scale, the temperature of ice at one atmospheric pressure is 2730A or 273K. This scale is related to centigrade scale in the following way; Temperature on centigrade scale + 273 = Temperature on absolute or kelvin scale. Hence, O0C = 2730A or 273 K. and – 2730C = O0A or Absolute zero.

INTRODUCTION 5 1.3 MATTER Matter : Matter is defined as anything that has definite mass and that occupies space. Examples : Water, air, gases, wood, solutions of salts etc. are examples of matter. All these substances possess definite mass and occupy some space. Classification of matter : Matter is classified into the following two categories. 1. Homogeneous matter : Matter that has uniform composition and identical properties throughout the phase is called homogeneous matter. It consists of only one phase. Examples : Water, sulphur, sugar, oxygen, copper etc. 2. Heterogeneous matter : Matter, which is made of two or more components, that are physically distinct is called heterogeneous matter. It has neither uniform composition nor identical properties. It consists of two or more number of phases. Example : (i) Mixture of salt and sand. (ii) Mixture of water and benzene. Elements, Compounds and Mixtures : 1. Element : An element is the simplest form of matter which can not be broken down into still simpler form by any physical or chemical means. Modern concept of element : According to the modern concept, an element can be broken down into its simplest form called atom in which all the properties of element are observed. For example, if a piece of iron is divided and subdivided continuously a stage will be reached when further subdivision will not be possible. The last portion will however exhibit all the properties of iron, that is, it will be attracted by a magnet. So far 116 elements have been discovered. Classification of elements : Elements have been classified into the following three categories depending upon their physical properties. (i) Metals : All metals are solids except mercury, which is a liquid. Metals are good conductors of heat and electricity and possess high melting and boiling points. Examples : copper, silver, gold etc. (ii) Non-metals : Non-metals exist in all the three states of matter. They are bad conductors. Examples : Solid non-metal – Sulphur, iodine, carbon Liquid non-metal – Bromine Gaseous non-metal – Chlorine, oxygen, nitrogen etc. (iii) Metalloids : Elements which have the properties of metals as well as non-metals are called semi-metals or metalloids.

6 +2 CHEMISTRY (VOL. - I) Example : Bismuth, arsenic, tin etc. Compound : A compound may be defined as a substance formed by the chemical union of two or more substances in certain fixed proportion by mass. The compound so formed differs entirely in properties from those of its constituents. Example : Water, carbondioxide, sodium chloride etc. A compound is always homogeneous and on decomposition it produces the substances from which it is obtained. Mixture : A mixture is formed by simply mixing two or more elements or compounds in any proportion. Here, the properties of elements or compounds forming the mixture do not change at all. Mixture can be homogeneous or heterogeneous. Examples of homogeneous mixture : (i) Mixture of two miscible liquids (water and alcohol) (ii) Mixture of a solid and a liquid (solution of any salt) (iii) Mixture of two or more gases (air) (iv) Mixture of two or more solids (steel, brass etc.) Examples of heterogeneous mixture : (i) Gun powder is a mixture of nitre, sulphur and charcoal (ii) Salt and sand 1.4 ATOMS AND MOLECULES : We know that matter is composed of very minute particles, called molecules. But it has been found that, a molecule is not the ultimate particle but is made up of two or more particles, called atoms. So, an atom is the ultimate particle of an element while a molecule is the ultimate particle of a compound. Atom : An atom is defined as the smallest particle of an element which may or may not have independent existence. Example : (i) Atoms of copper, iron, silver etc. have independent existence. (ii) Atoms of hydrogen, oxygen etc. do not have independent existence. However, now a days atom is not considered as indivisible, rather, it can be further sub-divided into more simpler substances, called the fundamental particles. The fundamental particles present in an atom are electron, proton and neutron. Molecule : A molecule is defined as the smallest particle of an element or a compound which has independent existence.

INTRODUCTION 7 Molecules can be classified into two main types : (a) Homoatomic molecules : In this case, two or more atoms of the same element combine to form molecules. Examples : (i) Monoatomic molecules : Monoatomic molecules contain single atoms of the same element. Noble gas elements, He, Ne, Ar are monoatomic molecules. (ii) Diatomic molecules : Diatomic molecules contain two atoms of the same element. Oxygen and nitrogen are diatomic molecules (iii) Triatomic molecules : Triatomic molecules contain three atoms of the same element. Ozone (O3) is a triatomic molecule. (b) Heteroatomic molecules : In this case, two or more atoms of the different elements combine to form molecules. Examples : (i) Molecule of water (H2O) (ii) Molecule of sulphuric acid (H2SO4) 1.5 STATES OF MATTER : Matter exists in any one of the three states, such as solid, liquid and gas, depending upon temperature, pressure and its nature. Solid ¾h¾e¾at® Liquid ¾h¾e¾at® Gas ¬¾co¾ol¾ ¬¾co¾ol¾ (i) Solid state : Matter in the solid state has definite shape and volume. Since solids are rigid and hard, they show very small change in volume by the application of pressure and temperature. However, on strong heating, solid melts. Solids also possess high densities due to the strong electrostatic force of attraction between the atoms in a molecule. Example : Silver, copper, sugar etc. (ii) Liquid state : Matter in the liquid state has definite volume but no definite shape. It acquires the shape of the container in which it is kept. By the application of pressure and temperature, the volume of the liquid changes appreciably. The forces of attraction between the atoms here is comparatively less than solid. Examples : Water, alcohol etc. (iii) Gaseous state : Matter in the gaseous state has neither definite shape nor definite volume. The gases occupy the entire available space and also acquire the shape of the vessel.

8 +2 CHEMISTRY (VOL. - I) Temperature and pressure have a marked effect on the volume of a gas. By the application of pressure a gas can be compressed or expanded. They possess low values of densities as compared to solids and liquids due to very weak forces of attraction. Distinction between solid, liquid and gaseous state The main distinction between the three states of matter. Property Solid Liquid Gas 1.State of aggregation The molecules are 1. The molecules are 1. The molecules are at closely packed. loosely packed a sufficient distance apart. 2. Intermolecular forces The forces of attraction are 2. The forces of 2. Very weak forces maximum attraction are less than of attraction. those in case of solids 3. Molecular motion The molecules possess 3. Molecules have 3. Molecules have 4. Density vibratory motion. translatory as well as random motion. They vibratory motion. possess translatory, The density is high vibratory and rotatory 4. The denisity is less motion. than those of solids 4. Very small value of density. 5. Shape and volume It has definite shape and 5. It has definite volume 5. It has neither definite shape nor volume but no definite shape definite volume. 6. Compressibility Solids are least 6. The compressibility 6. Gases are highly 7. Diffusion compressible of liquids is more than compressible. soilds. Solids do not diffuse 7. Liquids have a small 7. Gases diffuse tendency for diffusion. easily. 1.6 SYMBOL Symbol is a chemical notation to represent the name of an element in an abbreviated form. It also represents one atom of the element or one mole of the element. For example, the symbol 'C' represents the element Carbon, 1 atom of carbon or 1 mole of carbon atoms.

INTRODUCTION 9 Name Elements and their symbols Argon Symbol Name Symbol Aluminium Arsenic Ar Hydrogen H Antimony (Stibium) Astatine Al Helium He Actinium Americium As Holmium Ho Beryllium Boron Sb Hafnium Hf Bromine Barium At Hahnium Ha Bismuth Berkelium Ac Iron (ferrum) Fe Carbon Chlorine Am Iodine I Calcium Chromium Be Indium In Cobalt Copper (Cuprum) B Iridium Ir Cadmium Cesium Br Krypton Kr Cerium Curium Ba Kurchatovium Ku Californium Dysprosium Bi Lithium Li Europium Erbium Bk Lanthanum La Einsteinium Fluorine C Lutetium Lu Francium Fermium Cl Lead Pb Gallium Germanium Ca Lawrencium Lr Gold (Aurum) Gadolinium Cr Magnesium Mg Co Manganese Mn Cu Molybdenum Mo Cd Mercury (Hydrargyrum) Hg Cs Mendelevium Md Ce Nitrogen N Cm Neon Ne Cf Nickel Ni Dy Niobium Nb Eu Neodymium Nd Er Neptunium Np Es Nobelium No F Oxygen O Fr Osmium Os Fm Phosphorus P Ga Potassium (Kalium) K Ge Palladium Pd Au Praseodymium Pr Gd Promethium Pm

10 +2 CHEMISTRY (VOL. - I) Name Symbol Name Symbol Platinum Pt Titanium Ti Polonium Po Technetium Tc Protactinium Pa Tin (Stannum) Sn Plutonium Pu Tellurium Te Rubidium Rb Terbium Tb Ruthenium Ru Thulium Tm Rhodium Rh Tantalum Ta Rhenium Re Tungsten (wolfram) W Radon Rn Thallium Tl Radium Ra Thorium Th Sodium (Natrium) Na Uranium U Silicon Si Vanadium V Sulphur S Xenon Xe Scandium Sc Yttrium Y Selenium Se Ytterbium Yb Strontium Sr Zinc Zn Silver (Argentum) Ag Zirconium Zr Samarium Sm As is evident from the symbols of different elements. the initial letter of the name of the element in capital usually represents the element. For example, H stands for hydrogen, C stands for carbon, O stands for oxygen, etc. The initial letter C above cannot represent the elements Chlorine, Calcium, Chromium, Cobalt, Copper, Cadmium, etc. The initial letter C being used for carbon, for above elements, the initial letter in capital along with a small letter from the name of the elements represents the element. So the symbols Cl, Ca, Cr, Co, Cu, Cd are used respectively to represent them. For some metals, the latin names of the elements (written in parentheses against names of elements in table above) are used as symbol. Certain elements have been named after the scientists, important laboratories, cities, countries and planets. Given below is the list of elements whose symbols are derived from the names of scientists, laboratories, etc.

INTRODUCTION 11 Elements after the names of Scientists, Laboratories etc. Element Scientist Symbol Curium Madam Curie Cm Einsteinium Alfred Einstein Es Fermium Enrico Fermi Fm Mendelevium Mendeleef Md Nobelium Alfred Nobel No Californium LABORATORY Cf Berkelium University of California Bk CITY Berkeley COUNTRY Americium America Am Germanium Gallium Ga Polonium Galia (old name of France) Ge Germany Po Uranium Poland Neptunium U Plutonium PLANET Np Pu Uranus Neptune Pluto 1.7 VALENCY According to the old concept, Valency of an element denotes the combining capacity of its atoms to form compounds. Hydrogen being the lightest element, its valency is chosen as one and the reference. The valencies of other elements are compared with that of hydrogen. For example, one atom of chlorine combines with one atom of hydrogen to form one molecule of hydrogen chloride, HCl. So the valency of chlorine is one. One atom of oxygen combines with two atoms of hydrogen to form one molecule of water, H2O. Hence the valency of oxygen is two. One atom of nitrogen combines with three atoms of hydrogen to form one molecule of ammonia, NH3. Thus the valency of nitrogen is three.

12 +2 CHEMISTRY (VOL. - I) Valencies of some elements Element Valency Element Valency Hydrogen 1 Magnesium 2 Chlorine 1 Calcium 2 Sodium 1 Nitrogen 3 Potassium 1 Aluminium 3 Oxygen 2 Carbon 4 When elements have valencies 1, 2, 3, 4, 5, 6, etc. they are said to be monovalent or univalent, divalent, trivalent, tetravalent, pentavalent, hexavalent respectively. Thus, valency is a number that describes the combining capacity of an element. It does not carry any plus or minus sign. Valency is helpful in writing the formula of the compounds, In any compound, the valency of an element is always a whole number. Formulae of organic compounds cannot be written and explained on the basis of this concept of valency. The modern concept of valency is related to the number of electrons lost, gained or shared with an atom of an element in order to attain the stable configuration of the nearest inert gas element. For example, in sodium chloride NaCl, sodium loses one electron to attain the stable configuration of neon, Ne. Chlorine gains one electron and attains the stable configuration of inert gas argon, Ar. Therefore, in sodium chloride valency of sodium is one and that of chlorine is one. In hydrogen chloride HCl, hydrogen shares its only electron with one electron of chlorine. After sharing hydrogen attains the stable configuration of inert gas helium, He and chlorine attains the stable configuration of argon, Ar. Therefore the valency of hydrogen is one and that of chlorine is one. Variable valency : Certain elements having more than one valency are said to exhibit variable valencies. Variable valencies of some elements Element Valency Element Valency Arsenic 3,5 Manganese 2, 3, 4, 6, 7 Antimony 3,5 Mercury 1, 2 Chlorine 1, 3, 5, 7 Nitrogen 1, 2, 3, 4, 5 Copper 1, 2 Phosphorus 3, 5 Cobalt 2,3 Platinum 2, 4, 6 Gold 1, 3 Sulphur 2, 4, 6 Iron 2,3 Tin 2, 4

INTRODUCTION 13 While naming a particular compound the lower valency is referred to as –ous and higher valency as –ic. The following metallic elements combine with chlorine to form metal chlorides, exhibiting their variable valencies. Some metallic elements with variable valencies in their chlorides Metal chloride Systematic name Formula Valency of the metal Ferrous chloride Iron (II) chloride 2 Ferric chloride Iron (III) chloride FeCl2 3 Cuprous chloride Copper (I) chloride FeCl3 1 Cupric chloride Copper (II)chloride CuCl or Cu2Cl2 2 Mercurous chloride Mercury (I) chloride CuCl2 1 Mercuric chloride Mercury (II) chloride HgCl or Hg2Cl2 2 Stannous chloride Tin (II) chloride HgCI2 2 Stannic chloride Tin (IV) chloride SnCl2 4 SnCl4 Non-metallic elements such as nitrogen, N exhibits variable valencies of 1, 2, 3, 4, 5 in its oxides like N2O, NO, N2O3, NO2, N2O5 respectively. 1.8 RADICALS A radical in the molecule of a compound behaves as a single unit with a cluster of atoms. Radicals possess electrical charge. A simple radical contains one or more atoms of the same element. For example, oxide O2– and peroxide, O2– 2. Valencies of some simple radicals Radical Valency Radical Valency Hydrogen, H+ 1 Chloride, Cl– 1 Sodium, Na+ 1 Bromide, Br– 1 Potassium, K+ 1 Oxide O2– 2 Magnesium Mg2+ 2 Sulphide S2– 2 Calcium Ca2+ 2 Nitride N3– 3 Aluminium Al3+ 3 Phosphide P3– 3 Hydride, H– 1 Carbide C4– 4 A compound radical consists of two or more atoms of different elements. For example ammonium NH4+, nitrate NO3–, sulphate SO42–, phosphate PO43–, etc. When radicals contain oxygen atoms, the suffixes, –ite and –ate are used. The names of radicals ending in 'ite' contains less number of oxygen atoms than radicals ending in –ate. For example, nitrite NO– 2 and nitrate NO3–.

14 +2 CHEMISTRY (VOL. - I) Radical Valencies of some compound radicals Valency Radical Valency Ammonium, NH+4 1 Manganate, MnO24– 2 Hydroxide, OH– 1 Chromate, CrO24– 2 2 Dichromate, Cr2O72– 2 Carbonate, CO32– 1 Arsenate, AsO43– 3 Bicarbonate HCO3– 1 Arsenite, AsO33– 3 Bisulphide, HS– 2 Antimonate, SbO43– 3 1 Pyroantimonate, H2Sb2O27– 2 Sulphite, SO32– 1 Borate, BO33– 3 Bisulphite, HSO3– 1 Metaborate, BO2– 1 Nitrite, NO2– 2 Meta aluminate, AlO2– 1 Nitrate, NO3– 1 Aluminate, AlO33– 3 Sulphate, SO24– 3 Zincate, ZnO22– 2 Bisulphate, HSO4– 3 Hypochlorite, ClO– 1 Phosphate, PO34– 1 1 Phosphite, PO33– 4 Chlorite, ClO2– 1 Metaphosphate, PO3– 2 Chlorate, ClO3– 1 Pyrophosphate, P2O47– 1 Perchlorate, ClO4– Molybdate, MoO24– Permanganate, MnO4– Radical bearing one, two, three, four units of charges are called univalent, divalent, trivalent, tetra or quadrivalent radicals respectively. A salt has usually two parts or radicals. The electropositive part or basic radical is the cation, where as the electronegative part or acid radical is the anion. For example in common salt i.e. sodium chloride the basic radical is Na+ and the acid radical is Cl– . 1.9 FORMULA The formula of a compound represents the number of atoms of each element present in one molecule of the compound. The formula of the compound sodium chloride is NaCl, which means that one molecule of sodium chloride contains one atom of sodium and one atom of chlorine. A chemical formula also states the name of a compound as in the case of symbols. For some elements, the molecule of the element exists. In such case, the formula of an element denotes the number of atoms occurring in one molecule of the element. The formula H2, O3, P4, etc, indicate that two atoms of hydrogen are present in a molecule of the element hydrogen, three atoms of oxygen are present in a molecule of ozone, four atoms of phosphorus are present in a molecule of phosphorus respectively.

INTRODUCTION 15 If we know the valency of each element in a compound we can easily write its formula. The symbol of the positive part is written to the left hand side and the symbol of the negative part to its right. The valency of each element is placed as subscript by the side of the symbol of other element as shown below. Subscript 1 is not often written. Example - 1 What is the formula of calcium chloride ? Solution : The valency of calcium is 2 and that of chlorine is 1. Criss - crossing the valency number and using them as subscripts, we have the formula : Ca Cl º CaCl2 12 Example - 2 What is the formula of sodium sulphide ? Solution : The valency of sodium is 1. and that of sulphur is 2. Criss - Crossing the valency number, we can write the formula : Na S º Na2S 21 Example - 3 A compound consists of aluminium and oxygen. What is its formula ? Solution : Aluminium has a valency number of 3. oxygen has a valency number of 2. Criss - crossing the valency numbers and using them as subscript, we have the formula : Al O º Al2O3 23 Example - 4 What is the formula of magnesium sulphide ? Solution : Magnesium has a valency number of 2, and that of sulphur is 2. Therefore the formula of magnesium sulphide is MgS as explained below : Mg S º Mg2S2 22 This formula tells us that the ratio of magnesium to sulphur atoms is 2 : 2. This is, of course, is the same as a ratio of 1 : 1 and the formula becomes : Mg2S2 º MgS Example - 5 To write the formula of calcium carbonate. Solution : The valency number of calcium is 2 and that of carbonate radical is 2. Criss- crossing the valency numbers, we write the formula :

16 +2 CHEMISTRY (VOL. - I) Ca (CO3) º Ca2(CO3)2 º Ca1(CO3)1 º CaCO3 22 Example - 6 What is the formula of ammonium phosphate ? Solution : Ammonium has the valency Number 1 and that of phosphate radical is 3. Criss- crossing the valency numbers the formula of the compound may be written : (NH4) (PO4) 3 º (NH4)3 (PO4)1 º (NH4)3PO4 1 Correctness of a formula : Since a compound contains two parts that is the basic and the acid part, the correctness of its formula may be tested as follows. (i) The number of the basic part is multiplied with its valency. (ii) The number of the acid part is multiplied with its valency. (iii) If both the values of the products are equal, then the formula is found to be correct. Testing the correctness of a formula. Formula No. of basic No. of acid Remark on parts x valency parts x valency correctness. CaCl2 Na2PO4 1x2=2 2x1=2 correct Al2(SO4)3 wrong (NH4)3SO4 2x1=2 1x3=3 correct. Mg(NO3)2 wrong 2x3=6 3x2=6 correct 3x1=3 1x2=2 1x2=2 2x1=2 Structural formula : Sometimes, the formulae of compounds are written graphically indicating how the atoms in a molecule are linked with each other. The structural formula may also be called as graphic formula. In a compound, atoms of elements present are assumed to possess one or more bonds which are represented by small straight lines. A single bond is represented by a small straight line. Two straight lines mean a double, bond three lines, a triple bond etc.

INTRODUCTION 17 Structural or graphic formulae of some compounds Compound Molecular formula Structural formula Hydrogen chloride HCl H – Cl H Methane CH4 l Carbon dioxide CO2 H–C–H C2H4 l Ethylene HCN H C2H2 O=C=O Hydrogen cyanide Acetylene HH 1.10 NAMING OF COMPOUNDS C=C HH H–CºN H–CºC–H The way of naming is called nomenclature. A binary compound contains only two elements. Out of the two elements, one may be a metal and the other, a nonmetal or both may be non-metals. The following sequence is followed to name a binary compound that contains a metal and a nonmetal. 1. The name of the metallic part is mentioned first. 2. The stem of the name of the second element i.e. the nonmetal ends with \"ide\". 3. If the binary compound contains metal of variable valency, the metal is followed by a Roman numeral in bracket to indicate its valency state. In common method of naming, the lower valency state of the metal is indicated by suffix \"–ous\" and its higher valency state by suffix \"–ic\". 4. Although the metal hydroxides and ammonium chloride contain three elements and they are not binary compounds, they end in \"–ide\", e.g, sodium hydroxide NaOH, ammonium chloride NH4Cl. Name of some binary compounds containing metals of fixed valency states. Formula Name of the Stem name of the second Name of the compound first element (metal) element ending with '–ide' NaCl Sodium Chlor + (–ide) Sodium chloride KBr Potassium Brom +(–ide) Potassium bromide. CaO Calcium Ox + (–ide) Calcium oxide MgS Magnesium Sulph +(–ide) Magnesium sulphide AlN Aluminium Nitr + (–ide) Aluminium nitride.

18 +2 CHEMISTRY (VOL. - I) Name of some binary compounds containing metals of variable valency states. Formula Valency of Conventional name Systematic name the metal Lead(II) chloride PbCl2 2 Plumbous chloride Lead (IV) chloride PbCl4 4 Plumbic chloride Arsenic (III) sulphide As2S3 3 Arsenous sulphide Arsenic(V) sulphide As2S5 5 Arsenic sulphide Antimony (III) oxide Sb2O3 or Sb4O6 3 Antimonous oxide Antimony (V) oxide Sb2O5 5 Antimonic oxide Manganese (II) oxide MnO 2 Manganous oxide Manganese (III) oxide Gold (I) chloride Mn2O3 3 Manganic oxide Gold (III) chloride AuCl 1 Aurous chloride AuCl3 3 Auric chloride In order to name a binary compound containing two nonmetals, number of each of the atoms are indicated by Greek prefixes. The prefix hemi, mono, sesqui, di, tri, tetra etc are used for ½, 1, 1 ½, 2, 3, 4 atoms respectively. The prefix \"mono\" is usually not mentioned when naming the compound. There are some exceptions to the name of binary compounds such as water H2O, ammonia NH3, etc. Name of binary compounds composed of nonmetals. Formula Name of first Greek prefix plus the Name of the nonmetal with stem name of the second compound Greek prefix nonmetal ending with '–ide' BCl3 Boron tri-chlor-ide Boron trichloride CO2 Carbon di-ox-ide Carbon dioxide CCl4 Carbon tetra-chlor-ide Carbon tetrachloride N2O Di-nitrogen oxide Dinitrogen oxide N2O3 Di-nitrogen tri-ox-ide Dinitrogen trioxide N2O5 Di-nitrogen penta-ox-ide Dinitrogen pentoxide PCl3 Phosphorus tri-chlor-ide Phosphorus trichloride PCl5 Phosphorus penta-chlor-ide Phosphorus pentachloride SO2 Sulphur di-ox-ide Sulphur dioxide SO3 Sulphur tri-ox-ide Sulphur trioxide

INTRODUCTION 19 The above procedure is followed to name binary compounds containing hydrogen and nonmetals. Such a compound existing in gaseous or liquid state remains soluble in water and the resulting aqueous solution is acidic in nature. To name its acid the prefix hydro is used followed by the stem of the name of the non- metal with the suffix \"–ic\" and the word acid. Name of some binary compounds of hydrogen and nonmetals Formula Name of the compound Name of the compound in the gaseous or liquid state in aqueous state i.e. acid. HCl Hydrogen chloride Hydrochloric acid HBr Hydrogen bromide Hydrobromic acid H2S Hydrogen sulphide Hydrosulphuric acid H2Se Hydrogen selenide Hydroselenic acid Inorganic oxy-acids contain hydrogen, oxygen and another element, called a parent element. The name of the oxy-acid is derived from the parent element ending in -ous or - ic. The acid with less number of oxygen ends with -ous and the acid with more number of oxygen ends with - ic. The oxy-acid in which the valency of the parent element is higher than the - ic acid and contains one more oxygen is called per-ic acid. An oxy-acid containing the parent element of lower valency than the -ous acid is called hypo -ous acid. Name of some oxyacids and their sodium salts Oxy-acid Sodium salt HNO2 Nitrous acid NaNO2 Sodium nitrite HNO3 Nitric acid NaNO3 Sodium nitrate H2SO3 Sulphurous acid Na2SO3 Sodium sulphite H2SO4 Sulphuric acid Na2SO4 Sodium sulphate HOCl Hypochlorous acid NaOCl Sodium hypochlorite Chlorous acid Sodium chlorite HClO2 Chloric acid NaClO2 Sodium chlorate HClO3 Perchloric acid NaClO3 Sodium perchlorate HClO4 NaClO4 The name of the salts is based upon the name of the acid form which it is derived. The salt derived from the acid ending in –ous ends in –ite. A salt derived from an acid ending in –ic, ends in –ate. The salt of a hypo –ous acid is named as a hypo –ite salt.

20 +2 CHEMISTRY (VOL. - I) 1.11 CHEMICAL EQUATION A chemical equation states a chemical reaction represented by symbols and chemical formulae. When sulphur burns in air it combines with oxygen in air to produce sulphur dioxide due to the chemical combination of sulphur and oxygen. S + O2 = SO2 Reactants Products The chemical substances on the left hand side and right hand side of an equation are called reactants and products respectively. The above chemical equation also expresses that one atom of sulphur reacts with one molecule of oxygen to form one molecule of sulphur dioxide. Again the equation indicates that one mole of sulphur reacts with one mole of oxygen to produce one mole of sulphur dioxide. On examining this equation we observe that the total mass of the reactants is equal to the total mass of the products i.e the total mass is conserved as per law of conservation of matter. Let us look at another reaction. Magnesium burns in air to produce magnesium oxide and is represented by the equation : Mg + O2 ® MgO : Skeleton equation In this reaction there is loss of oxygen from the product side which is not in accordance with law of conservation of matter. The equation is called a skeleton equation and needs to be balanced. The equation would be balanced if we could place number 2 before MgO and number 2 before Mg. Thus, 2 Mg + O2 = 2MgO is a balanced equation. A skeleton equation is balanced by placing numbers, also called coefficients before the formulae of the substances in the reaction. In the left hand side there are two magnesium atoms and two oxygen atoms and in the right hand side there are also same number of magnesium and oxygen atoms. In a balanced equation therefore, the number of atoms of different elements appearing as reactants are equal to number of atoms of the same elements present as products. Let us look at one more example. The following chemical equation N2(g) + 3H2(g) = 2NH3(g) says that 1 mole of gaseous nitrogen combines with 3 moles of gaseous hydrogen to produce 2 moles of gaseous ammonia. This also says that 1 volume of gaseous nitrogen combines with

INTRODUCTION 21 3 volumes of gaseous hydrogen to form 2 volumes of gaseous ammonia. The equation may be elaborately written as : N2(g) + 3H2(g) 2NH3(g), DH = – 92 KJ The reaction is a reversible one and represented by the sign which indicates that the reaction may proceed both in forward or reverse direction.The reversibility of a reaction is represented by the sign . The above reaction evolves heat and we call it a exothermic reaction (the heat changes i.e. DH has a –ve value). An endothermic reaction absorb heat and its DH value is +ve. However, the heat changes (DH) in reactions are mentioned in a thermochemical equation whenever necessary. The gaseous, liquid, solid states and the aqueous solution are given the symbol (g), (l), (s) and (aq) respectively For example the reaction of sodium chloride and silver nitrate is represented as follows : NaCl(aq) + AgNO3(aq) = AgCl(s) + NaNO3(aq) In the laboratory carbon dioxide gas is prepared by the action of dilute hydrochloric acid on calcium carbonate. The reaction is represented by the equation : CaCO3(s) + 2HCl(aq) = CaCl2(aq) + CO2(g) + H2O(l) When gas is evolved in a reaction, a vertical arrow - is sometimes used. In a reaction when a precipitate is formed, a downward arrow ¯ is often used. The required conditions of temperature, pressure, catalyst etc, are often mentioned in the equation as shown below. N2(g) + 3H2(g) 200 atms. 2NH3(g) -, DH = – 92 kJ 673 K, Fe catalyst 'Mo promoter' Limitations of a chemical equation The conditions such as temperature, pressure and the catalyst those bring about a chemical reaction are sometimes mentioned in the chemical equation. The concentration of the reactants and products are never mentioned in an equation. An equation cannot furnish informations about all experimental conditions required for the reaction. The heat evolved or heat absorbed in a reaction are not ordinarily stated. A chemical equation does not speak about the speed with which the reaction proceeds and also the time needed for completion of the reaction. Balancing of chemical equations Any chemical equation must be written in the balanced form, since the number atoms on the reactant side must be equal to the number of atoms on the product side due to conservation of atoms. There are five different methods for balancing a chemical equation.

22 +2 CHEMISTRY (VOL. - I) (a) Hit and trial method (b) Partial equation method (c) Algebraic method (d) Oxidation number method (e) Ion electron method (a) Hit and trial method Simple equations are balanced by hit and trial method. The following steps may be employed to balance the equation in order to equalize atoms of each kind on both sides of the skeleton equation. (i) The skeleton equation be written to indicate the reactants and products. (ii) The biggest formula is selected and the different atoms present in it be balanced with the same atoms on the other side. (iii) Elementary gaseous reactants be written in atomic state, N, O etc. and they should be balanced at last. (iv) Atoms of elements which occur in maximum number of places be balanced then. (v) The balanced equation in the atomic state is changed into the molecular form and all the reactants and products are expressed in whole number multiples. While balancing the equations, methods of addition, subtraction, multiplication and division are used. Example 1 Potassium chlorate when heated decomposes to potassium chloride and oxygen gas is liberated. Step 1 Skeleton equation is written as KClO3 ¾¾D® KCl + O2 Step 2 Biggest formula here is KClO3 and it is balanced. 2KClO3 ® 2KCl + 3O2 (Balanced equation) Example 2 Iron pyrite burns in oxygen to produce ferric oxide and sulphur dioxide Step 1 FeS2 + O2 ® Fe2O3 + SO2 (Skeleton equation) Step 2 (iron balanced) 2FeS2 + O2 ® Fe2O3 + SO2 (sulphur balanced) Step 3 2FeS2 + O2 ® Fe2O3 + 4SO2 (atomic state of equation) Step 4 2FeS2 + O ® Fe2O3 + 4SO2 (oxygen balanced) Step 5 2FeS2 + 11O ® Fe2O3 + 4SO2 (balanced equation in molecular form) 4FeS2 +11O2 ® 2Fe2O3 + 8SO2

INTRODUCTION 23 Example 3 When steam is passed over red-hot iron, ferroso ferric oxide and hydrogen are produced. Step 1 Fe + H2O ® Fe3O4 + H2 (Skeleton equation) Step 2 3Fe + 4H2O ® Fe3O4 + 4H2 (Fe in Fe3O4 when balanced, the equation is balanced for oxygen and hydrogen respectively) Example 4 Balance the skeleton equation KMnO4 + HCl ® KCl + MnCl2 + H2O + Cl2 Step 1 KMnO4 + HCl ® KCl + MnCl2 + 4H2O + Cl2 (KMnO4 balanced) Step 2 KMnO4 + HCl ® KCl + MnCl2 + 4H2O + Cl (Equation is written in atomic form) Step 3 KMnO4 + 8HCl ® KCl + MnCl2 + 4H2O + Cl (hydrogen balanced and oxygen balanced) Step 4 KMnO4 + 8HCl ® KCl + MnCl2 + 4H2O + 5Cl (chlorine balanced) Step 5 2KMnO4 + 16HCl ® 2KCl + 2MnCl2 + 8H2O + 5Cl2 (Balanced equation in molecular form) Partial equation method In this method the reaction is thought to take place in several steps and each step is represented by partial equation. The partial equations are balanced by hit and trial method and may be multiplied by suitable integers to omit the products which do not appear in the final equation. Finally the partial equations are added to have the balanced equation. Example 1 Copper and sulphuric acid react to produce sulphur dioxide gas, copper sulphate and water. Balance the reaction by partial equation method. Solution : The skeleton equation may be written as : Cu + H2SO4 = CuSO4 + SO2 + H2O The above reaction is supposed to break up as follows. H2SO4 ® SO2 + H2O + O Cu + O ® CuO Example 2 CuO + H2SO4 ® CuSO4 + H2O ———————————————— Cu + 2H2SO4 ® CuSO4 + SO2 + 2H2O. Metallic copper reacts with moderately concentrated nitric acid to produce cupric nitrate, nitric oxide and water. Balance the reaction by partial equation method.

24 +2 CHEMISTRY (VOL. - I) Solution : The skeleton equation may be written as : Cu + HNO3 ® Cu(NO3)2 + NO + H2O The above reaction is supposed to break up as follows : 2HNO3 ® 2NO + H2O + 3O (Cu + O ® CuO) x 3 (CuO + 2HNO3 ® Cu(NO3)2 + H2O) x 3 ——————————————————— 3Cu + 8HNO3 ® 3Cu(NO3)2 + 2NO + 4H2O Example 4 Hydrogen peroxide reacts with lead sulphide to produce lead sulphate and water. Balance the equation by partial equation method. Solution : The skeleton equation is written as : PbS + H2O2 ® PbSO4 + H2O The above reaction is supposed to break up as follows : ( H2O2 ® H2O + O ) x 4 PbS + 4O ® PbSO4 ————————————— PbS + 4H2O2 ® PbSO4 + 4H2O Example 4 Balance the following equation by partial equation method. Solution : P + HNO3 ® H3PO4 + NO2 + H2O The above reaction is supposed to occur as follows : (2HNO3 ® 2NO2 + H2O + O) x 5 2P + 5O ® P2O5 P2O5 + 3H2O ® 2H3PO4 ——————————————————— 2P + 10HNO3 ® 2H3PO4 + 10NO2 + 2H2O Some important and supposed break up reactions. The following reactions may be written to be used as first reactions while balancing equations by partial equation method. (1) O3 ® O2 + O (2) H2O2 ® H2O + O (3) H2O2 ® O2 + 2H

INTRODUCTION 25 (4) X2+ H2O ® 2HX + O (X = Cl, Br or I) (5) Cl2 + H2O ® HCl + HOCl (6) 2HNO3 ® 2NO2 + H2O + O (concentrated HNO3) (7) 2HNO3 ® 2NO + H2O +3O (Moderate HNO3) (8) 2HNO3 ® N2O + H2O + 4O (dilute HNO3) (9) H2SO4 ® SO2 + H2O + O (10) MnO2 ® MnO + O (11) MnO2 + 2HCl ® MnCl2 + H2O + O (12) MnO2 + H2SO4 ® MnSO4 + H2O + O (13) 2KMnO4 ® K2O + 2MnO + 5O (acidic medium) (14) K2Cr2O7 ® K2O + Cr2O3 + 3O (acidic medium) (15) 2KMnO4 + 3H2SO4 ® K2SO4 + 2MnSO4 + 3H2O + 5O (16) K2Cr2O7 + 4H2SO4 ® K2SO4 + Cr2(SO)3 + 4H2O + 3O Algebraic method This method is useful to balance a complicated equation. Example 1 Solution : Balance the equation for the combustion of benzene. The skeleton equation for the combustion of benzene is C6H6 + O2 ® CO2 + H2O Let the balance equation be a C6H6 + b O2 ® c CO2 + d H2O According to the conservation of atoms, the number of carbon atoms on the reactant side must be equal to that on the product side. 6 a = c ...................(1) Similarly, considering hydrogen atoms 6 a = 2 d .................(2) and taking oxygen atoms into account 2 b = 2 c + d .............(3) The above equations can be reduced to c=6a d=3a b = 15 a 2 Assuming the smallest value a = 1

26 +2 CHEMISTRY (VOL. - I) C = 6, d = 3 and b = 15 2 The balanced equation is thus C6H6 + 15 O2 = 6CO2 + 3H2O 2 or, 2C6H6 + 15O2 = 12CO2 + 6H2O Example 2 Write the balanced equation for the oxidation reaction of phosphorus by nitric acid. Solution : The skeleton equation for the reaction is written as P4 + HNO3 ® NO2 + H3PO4 + H2O Let the balanced equation be a P4 + b HNO3 ® c NO2 + d H3PO4 + e H2O Phosphorus 4a = d ................................................(1) Nitrogen b = c ..................................................(2) Hydrogen b = 3d + 2e .......................................(3) Oxygen 3b = 2c + 4d + e ..............................(4) The above equations reduce to d = 4a ; c = b, b = 12a + 2e, b = 16a + e Therefore, e = 4a, b = 20, c = b and d = 4a Assuming a = 1, then b = 20, c = 20, d = 4 and e = 4 Thus the balanced equation is P4 + 20 HNO3 u 20 NO2 + 4H3PO4 + 4H2O Oxidation number method and ion-electron method Chemical equations corresponding to redox chemical reactions (involving both oxidation and reduction) in aqueous solutions are balanced by oxidation number method. Such reactions can also be balanced by ion electron methods. Both willl be discussed later . Ionic equations Metallic zinc reacts with dilute hydrochloric acid to liberate hydrogen gas for which the complete equation is : Zn(s) + 2HCl(aq) ® ZnCl2(aq) + H2(g) The ionic equation for the above reaction is written as : Zn(s) + 2H+(aq) ® Zn2+(aq) + H2(g) Silver nitrate solution when added to aqueous solution of sodium chloride, silver chloride solid is precipitated.

INTRODUCTION 27 The chemical equation for the precipitation reaction is given below. NaCl(aq) + AgNO3(aq) ® AgCl(s) + NaNO3(aq) The substances NaCl, AgNO3 and NaNO3 almost completely dissociate into their ions in aqueous solution. As a result the number of Na+(aq) and NO–3(aq) on both the reactant and product side equal and the equation is written as : Na+(aq) + Cl–(aq) + Ag+(aq) + NO3- (aq) ® AgCl(s) + Na+(aq) + NO3- (aq) and hence, the ionic equation becomes Ag+(aq) + Cl–(aq) ® AgCl(s) CHAPTER (1) AT A GLANCE Symbol : Symbol is a chemical notation to represent the name of an element, one atom of the element or one mole of the element. Valency : According to the old concept valency of an element denotes the combining capacity of its atoms to form compounds. According to modern concept, valency is related to the number electrons lost, gained or shared with an atom of an element in order to attain the stable configuration of the nearest inert gas element. Variable valency : Certain elements having more than one valency are said to show variable valencies. Radical : A radical in the molecule of a compound behaves as a single unit with a cluster of atoms. Radical possesses electrical charge. A simple radical contains one or more atoms of the same element. Cation and Anion : A salt has usually two parts or radicals. The electropositive part or basic radical is called the cation. The electronegative part or acid radical is called the anion. Formula : The formula of a compound represents the number of atoms of each element present in one molecule of the compound. The formula of an element denotes the number of atoms present in one molecule of the element. A formula also represents a quantity of the compound equal in mass to its molecular mass. Structural or Graphic formula : The structural formula or graphic formula of compounds are written graphically indicating how the atoms in a molecule are linked with each other. Chemical equation : A chemical equation states a chemical reaction represented by symbols and chemical formulae. Balanced equation : A skeleton equation is balanced by placing numbers called coefficients before the formulae of the reactants and products mentioned in the reaction. In a balanced equation the number of atoms of different elements appearing as reactants are equal to the number of atoms of the same elements present as products.

28 +2 CHEMISTRY (VOL. - I) QUESTIONS (A) Very short answer type (1 mark each) Write the formulae of the following compounds 1. Sodium phosphate 29. Nitrosyl chloride 2. Silver nitrate 30. Sodium zincate 3. Aluminium nitride 31. Perchloric acid 4. Aluminium phosphate 32. Stannous chloride 5. Sulphurous acid 33. Potassium chromate 6. Nitrogen trioxide 34. Sodium silicate 7. Potassium permanganate 35. Barium peroxide 8. Mercurous chloride 36. Chloric acid 9. Sodium bicarbonate 37. Mercuric chloride 10. Carbonic acid 38. Stannic sulphide 11. Manganese dioxide 39. Bismuth oxychloride 12. Silicon fluoride 40. Calcium carbide 13. Sodium thiosulphate 41. Potassium ferrocyanide 14. Boric acid 42. Potassium ferricyanide 15. Antimonous sulphide 43. Ferric ferricyanide 16. Mercurous nitrate 44. Diammine silver chloride 17. Nitrogen pentoxide 45. Zinc sulphide 18. Hydrogen peroxide 46. Magnesium nitride 19. Manganous sulphate 47. Ferric sulphate 20. Cuprous chloride 48. Lead acetate 21. Potassium pyroantimonate 49. Sodium meta-aluminate 22. Nitrous acid 50. Chromium sulphate 23. Lead dioxide 51. Potassium cobaltinitrite. 24. Ammonium dichromate 52. Strontium nitrate 25. Sodium tetraborate 53. Lead iodide 26. Auric chloride 54. Beryllium chloride 27. Potassium manganate 55. Lithium oxide. 28. Phosphoric acid 56. Magnesium nitride

INTRODUCTION 29 (B) Short answer type (2 marks each) (i) Define the following terms 1. Symbol. 2. Valency 3. formula 4. Radical (ii) Explain the following terms 1. Variable valency 2. Compound radical 3. Molecular formula 4. Empirical formula 5. Structural formula. (iii) Write down the formulae and chemical names of the following 1. Dry ice 25. slaked lime 2. Sand 26. lime water 3. Oil of vitriol 27. chalk 4. Baking soda 28. marble 5. Caustic soda 29. Gypsum 6. soda lime 30. Plaster of paris 7. Soda ash 31. Bleaching powder 8. Washing soda 32. Sal-ammoniac 9. Common salt 33. Smelling salt 10. Table salt 34. Borax 11. Chili salt petre 35. Alumina 12. Nitre cake. 36. Potash alum 13. Salt cake 37. Green vitriol 14. Glauber's salt 38. Mohr's salt 15. Sodamide 39. Blue vitriol 16. Water glass 40. White vitriol 17. Hypo 41. Lunar caustic 18. Micro cosmic salt 42. Quick silver 19. Caustic potash 43. Calomel 20. Nitre 44. Corrosive sublimate 21. Potash or pearl ash 45. Litharge 22. Milk of magnesia 46 Red lead 23. Epsom salt 47. Sugar of lead 24. lime or quick lime 48. White lead.

30 +2 CHEMISTRY (VOL. - I) (iv) Balance the following equations by Hit and trial method 1. Na + H2O ® NaOH + H2 2. Al + NaOH + H2O ® NaAlO2 + H2 3. Si + NaOH + H2O ® Na2SiO3 + H2 4. Na2CO3 + HCl ® NaCl + CO2 + H2O 5. Na2CO3 + CO2 + H2O ® NaHCO3 6. Mg + HNO3 ® Mg(NO3)2 + H2 7. Fe3O4 + CO ® FeO + CO2 8. NH4NO2 ® N2 + H2O 9. NH4NO3 ® N2O + H2O 10. Mg3N2 + H2O ® Mg(OH)2 + NH3 11. NH3 + O2 ® NO + H2O 12. NH3 + Cl2 ® NH4Cl + N2 13. NH3 + Cl2 (excess) ® NCl3 + HCl 14. CaOCl2 + NH3 ® CaCl2 + H2O + N2 15. Pb(NO3)2 ® PbO + NO2 + O2 16. Hg2(NO3)2 ® Hg + NO2 + O2 17. Na + NH3 ® NaNH2 + H2 18. HNO3 + H2S ® NO2 + H2O + S 19. HNO3 + HI ® NO + I2 + H2O 20. H2S + SO2 ® H2O + S 21. P4 + H2O + NaOH ® NaH2PO2 + PH3 22. FeSO4 ® Fe2O3 + SO2 + SO3 23. CuSO4 + PH3 ® Cu3P2 + H2SO4 24. KMnO4 ® K2MnO4 + MnO2 + O2 25. NaOH + F2 ® NaF + H2O + O2 26. Cu + HNO3 ® Cu(NO3)2 + H2O + NO 27. Al + Fe3O4 ® Al2O3 + Fe 28. NaCl + H2SO4 ® Na2SO4 + HCl

INTRODUCTION 31 ANSWERS A. 1. Na3PO4 2. AgNO3 3. AlN 4. AlPO4 5. H2SO3 6. N2O3 7. KMnO4 8. Hg2Cl2 9. NaHCO3 10. H2CO3 11. MnO2 12. SiF4 13. Na2S2O3 14. H3BO3 15. Sb2S3 16. Hg2(NO3)2 17. N2O5 18. H2O2 19. MnSO4 20. Cu2Cl2 23. PbO2 24. (NH4)2Cr2O7 21. K2H2Sb2O7 22. HNO2 27. K2MnO4 28. H3PO4 31. HClO4 32. SnCl2 25. Na2B4O7 26. AuCl3 35. BaO2 36. HClO3 39. BiOCl 40. CaC2 29. NOCl 30. Na2ZnO2 44. Ag(NH3)2Cl 43. Fe[Fe(CN)6] 48. Pb(CH3COO)2 33. K2CrO4 34. Na2SiO3 47. Fe2(SO4)3 52. Sr(NO3)2 51 K3[Co(NO2)6] 56. Mg3N2 37. HgCl2 38. SnS2 55. Li2O. 41. K4[Fe(CN)6] 42. K3[Fe(CN)6] 45. ZnS 46. Mg3N2 49. NaAlO2 50. Cr2(SO4)3 53. PbI2 54. BeCl2 B.(iii) 1. CO2 , Carbon dioxide 2. SiO2, Silicon dioxide 3. H2SO4, Sulphuric acid 4. NaHCO3, Sodium bicarbonate 5. NaOH, Sodium hydroxide 6. NaOH + Ca(OH)2, Mixture of soidum hydroxide and calcium hydroxide 7. Na2CO3, Sodium carbonate 8. Na2CO3, Sodium carbonate 9. NaCl, Sodium chloride 10. NaCl, Sodium chloride 11. NaNO3, Sodium nitrate 12. NaHSO4, Sodium bisulphate 13. Na2SO4, Sodium sulphate 14. Na2SO4. 10H2O, Sodium sulphate decahydrate. 15. NaNH2, Sodamide. 16. Na2SiO3, Sodium silicate 17. Na2S2O3, Sodium thiosulphate 18. Na(NH4)HPO4. 4H2O, Sodium ammonium hydrogen phosphate tetrahydrate. 19. KOH, Potassium hydroxide 20. KNO3, Potassium nitrate

32 +2 CHEMISTRY (VOL. - I) 21. K2CO3, Potassium carbonate 22. Mg(OH)2, Magnesium hydroxide. 23. MgSO4. 7H2O, Magnesium sulphate heptahydrate 24. CaO, Quick lime 25. 26. Ca(OH)2, Calcium hydroxide suspension 27. 28. Ca(OH)2, Dissolved calcium hydroxide 29. 30. CaCO3, Calcium carbonate 31. 32. CaCO3, Calcium carbonate 33. 34. CaSO4.2H2O, Calcium sulphate dihydrate 35. 1 36. 2CaSO4.H2O or CaSO4. 2 H2O, Calcium sulphate half hydrate. 37. 38. CaOCl2, Calcium chlorohypochlorite. 39. 40. NH4Cl, Ammonium chloride 41. 42. (NH4)2CO3, Ammonium carbonate 43. 44. Na2B4O7.10H2O, Sodium tetraborate decahydrate 45. 46. Al2O3, Aluminium oxide 47. 48. K2SO4.Al2 (SO4)3.24H2O, Hydrated potassium aluminium sulphate . B.(iv) 1. FeSO4.7H2O, Ferrous sulphate heptahydrate 2. 3. FeSO4.(NH4)2SO4.6H2O, Ferrous ammonium sulphate hexahydrate 4. 5. CuSO4. 5H2O, Copper sulphate pentahydrate ZnSO4 . 7H2O, Zinc sulphate heptahydrate AgNO3, Silver nitrate Hg, Metallic mercury Hg2Cl2 , Mercurous chloride HgCl2, Mercuric chloride PbO, Lead oxide (yellow) Pb3O4, Lead oxide(Red) Pb(CH3COO)2, Lead acetate 2PbCO3. Pb(OH)2, Basic lead carbonate. 2Na + 2H2O ® 2NaOH + H2 2Al + 2NaOH + 2H2O ® 2NaAlO2+ 3H2 Si + 2NaOH + H2O ® Na2SiO3 + H2 Na2CO3 + 2HCl ® 2NaCl + CO2 + H2O Na2CO3 + CO2 + H2O ® 2NaHCO3

INTRODUCTION 33 6. Mg + 2HNO3 ® Mg(NO3)2 + H2 7. Fe3O4 + CO ® 3FeO + CO2 8. NH4NO2 ® N2 + 2H2O 9. NH4NO3 ® N2O + 2H2O 10. Mg3N2 + 6H2O ® 3Mg(OH)2 + 2NH3 11. 4NH3 + 5O2 ® 4NO + 6H2O 12. 8NH3 + 3Cl2 ® 6NH4Cl + N2 13. NH3 + 3Cl2(excess) ® NCl3 + 3HCl 14. 3CaOCl2 + 2NH3 ® 3CaCl2 + 3H2O + N2 15. 2Pb(NO3)2 ® 2PbO + 4NO2 + O2 16. Hg2(NO3)2 ® 2Hg + 2NO2 + O2 17. 2Na + 2NH3 ® 2NaNH2 + H2 18. 2HNO3 + H2S ® 2NO2 + 2H2O + S 19. 2HNO3 + 6HI ® 2NO + 3I2 + 4H2O 20. 2H2S + SO2 ® 2H2O + 3S 21. P4 + 3NaOH + 3H2O ® 3NaH2PO2 + PH3 22. 2FeSO4 ® Fe2O3 + SO2 + SO3 23. 3CuSO4 + 2PH3 ® Cu3P2 + 3H2SO4 24. 2KMnO4 ® K2MnO4 + MnO2 + O2 25. 4NaOH + 2F2 ® 4NaF + 2H2O + O2 26. 3Cu + 8HNO3 ® 3Cu(NO3)2 + 4H2O + 2NO 27. 8Al + 3Fe3O4 ® 4Al2O3 + 9Fe qqq

34 +2 CHEMISTRY (VOL. - I) CHAPTER - 2 BASIC CONCEPTS ATOMS & MOLECULES 2.1 LAWS OF CHEMICAL COMBINATIONS The laws of chemical combinations helped in the development of the atomic theory and later the molecular theory. These laws govern chemical reactions. Law of conservation of mass Matter is neither created nor destroyed during any transformation of matter. In any chemical reaction the total mass of the reactants is equal to the total mass of the products. This law has been verified by repeated experiments on chemical changes using delicate balances to measure the weights of reactants and products. In 1774, scientist Lavoisier formulated the law. This law is also applicable in case of a physical change. Landolt’s experiment Landolt conducted many experiments to verify the law of conservation of mass. One of the experiments is discussed below. In a specially designed H-shaped glass tube (Fig 2.1) aqueous solution of silver nitrate and sodium chloride were taken as shown. The open ends of the tube were sealed and the tube was weighed. AgNO3 solution NaCl solution Fig. 2.1 Landolt’s experiment

ATOMS & MOLECULES 35 Then the two solutions were thoroughly shaken and mixed. During mixing, the reactants combine to form white AgCl precipitate and NaN03 solution. NaCl + AgNO3 ® AgCl ¯ + NaNO3 Now the tube was weighed again and it was observed that the weight remained un­changed. Thus, the mass was found to be the same as before and the law was verified. Example : 0.22 g of a hydrocarbon containing carbon and hydrogen only, on complete combustion with oxygen resulted 0.9 gm of water and 0.44 gm of carbon dioxide. Show that these results are in accordance with the law of conservation of mass (C = 12, H = 1, O = 16) The molecular mass of carbon dioxide CO2 and water H,0 calculated from their respective formula. Molecular mass of CO2 = 12 + 32 = 44 Molecular mass of H2O = 2 + 16 = 18 Now mass of carbon in 0.44 g of CO2 = 12 × 0.44 = 0.12g. and 44 mass of hydrogen in 0.9 g. of H2O = 2 × 0.9 = 0.1 Og. 18 Hence, the total mass of C and H after combustion = 0.12 + 0.10 = 0.22g This total mass 0.22 g of C and H becomes equal to the same mass 0.22g of hydrocarbon used in the combustion. Thus, the results are in accordance with the law of conservation of mass. Law of definite or constant proportions According to Proust, the law states that : A given compound always contains the same elements combined in the same proportions by mass. When a pure sample of water collected from any source is analysed, it is found that one part by mass of water contains 1 part i.e. 11.1 % hydrogen and 8 part i.e. 88.9 % oxygen. The 9 9 ratio of mass of the two elments hydrogen and oxygen is always 1:8. Example : 2.75 g of cupric oxide was reduced by hydrogen and the weight of copper so obtained was 2.20 g. In another experiment 2.36 g of copper was dissolved in nitric acid and the resulting copper nitrate was converted to cupric oxide which weighed 2.95 g by ignition. Show that these results illustrate the law of definite proportion. 2.20 g of copper was obtained from 2.75 g of cupric oxide. Therfore. the weight of oxygen that combined with copper was (2.75 – 2.20 =) 0.55 g. Hence, 2.20 g of copper combined with 0.55 g of oxygen. So 1 g of copper combines with 0.55 = 0.25 g of oxygen. 2.20

36 +2 CHEMISTRY (VOL. - I) Again the other data reveal that, 2.36 g of copper combines with (2.95 – 2.36 =) 0.59 g of oxygen So 1 g of copper combines with 0.59 = 0.25 g of oxygen. 2.36 Thus from whatever sources cupric oxide may be obtained, lg of copper always combines with 0.25 g of oxygen to form cupric oxide. Therefore, in the formation of cupric oxide, the ratio of masses of copper to oxygen is always 4:1. Law of multiple proportions The law was formulated by John Dalton. Whenever two elements combine to form more than one compound the masses of the elements that combine with a constant mass of the other element are in a ratio of simple whole numbers. Hydrogen combines with oxygen to form water and hydrogen peroxide separately. On analysis it is evident that : In water, 1 part by mass of hydrogen combines with 8 parts by mass of oxygen. In hydrogen peroxide 1 part by mass of hydrogen combines with 16 parts by mass of oxygen. So in H2O and H2O2, one part by mass of hydrogen combines with 8 and 16 parts by-mass of oxygen respectively. It means that the masses of oxygen that combine with constant mass of hydrogen (in both the compounds) are in a ratio 8 : 16 or 1 : 2. Example 1 1.5 g of each of two oxides of a metal on reduction resulted 1.331 g and 1.196 g of metal. Show that the data illustrate the law of multiple proportions. Weight of oxygen in two oxides are 1.5 g - 1.331g = 0.169g and 1.5 g - 1.196g = 0.304g In the first oxide, 1.33 lg of metal combines with 0.169g of oxygen. So lg of metal combines with 0.169 = 0.127 g of oxygen. 1.331 In the second oxide, 1.196 g of metal combines with 0.304g of oxygen So lg of the metal combines with 0.304 = 0.253g of oxygen 1.196 The mass of oxygen that combines with a fixed or constant mass of the metal (here lg) are in the ratio of 0.127 : 0.253 and that is 1 : 2. Thus the results are in conformity with the law of multiple proportion.

ATOMS & MOLECULES 37 Example 2 Two chlorides of a metal contains 34% and 51% chlorine respectively. Show that the results are in accordance with law of multiple proportions. The first chloride contains 34% CI Hence, the metal present in it = 100 - 34 = 66% Therefore, 66 g of metal combine with 34 g of chlorine. So, 1 g of metal combines with 34 = 0.515 g of chlorine. 66 The second chloride contains 51% CI Hence, the metal present in it = 100 - 51 = 49%. Therefore, 49 g of metal combine with 51 g of chlorine So, 1 g of metal combines with 51 = 1.041 g of chlorine. Now the varying masses of 49 chlorine which combine with fixed mass of the metal are in a ratio of 0.515 : 1.041 = 1 : 2 Law of equivalent or reciprocal proportions The masses of two or more different elements which separately combine with a definite mass of another element are either the same as or simple multiple of the masses of different elements when they combine amongst themselves. Let two different elements-A and B separately combine with another element C. According to this law, the ratio in which the masses of A and B combine with definite mass of C will be the same or be a simple multiple of the ratio in which A and B combine with each other. For example, sulphur and oxygen separately combine with hydrogen to form hydrogen sulphide and water respectively. In hydrogen sulphide : 2 grams of hydrogen combine with 32 grams of sulphur. In water : 2 grams of hydrogen combine with 16 grams of oxygen. Thus, the masses of sulphur and oxygen which combine separately with a definite mass of hydrogen that is 2 grams of hydrogen bear a simple ratio of 32 : 16 or 2 : 1. When sulphur and oxygen combine together to form the compound sulphur dioxide, according to the law of reciprocal proportions, the ratio of their masses should be the same i.e. 2 : 1 or a simple multiple of this ratio. In sulphur dioxide we see that 32 grams of sulphur combine with 32 grams of oxygen and the ratio of the masses of sulphur and oxygen is 1 : 1. Thus the two ratios that is 2 : 1 and 1 : 1 are simple multiple of each other being 2:1. Example 1 Water and sulphur dioxide contain 88.9% and 50% oxygen respectively. Hydrogen sulphide contains 94.1% sulphur. Show that these figures are in agreement with the law of reciprocal proportions.

38 +2 CHEMISTRY (VOL. - I) In H2O, 88.9 % oxygen is present. Hence, 100 – 88.9 = 11.1 % hydrogen is present. So, the mass ratio of H : O = 111 : 88.9 =1:8 In H2S, 94.1 % sulphur is present. Hence, 100 - 94.1 = 5.9 % hydrogen is present. So, the mass ratio of H : S = 5.9 : 94.1 = 1 : 16 Now the mass ratio of sulphur and oxygen in H,S and H,0 when combining with fixed mass of hydrogen becomes 16 : 8 i.e. 2:1. Now, SO2 contains 50% oxygen. Hence, 100 - 50 = 50% sulphur is present. So the mass ratio of S : O = 50 : 50 = 1 : 1 Thus, the above two ratios that is 2 : 1 and 1 : 1 are simple multiple of each other being 2 : 1 Hence, the data are in agreement with law of reciprocal proportions. Example 2 Carbon dioxide contains 27.3% of carbon. Carbon disulphide contains 15.8% of carbon and sulphur dioxide contain 50% of sulphur. Show that these results illustrate the law of reciprocal proportions. In CO2, 27.3% carbon is present. Hence, 100 - 27.3 = 72.7% oxygen is present. So, the mass ratio of C : O = 27.3 : 72.7 = 1 : 2.66 In CS,, 15.8% carbon is present. Hence, 100 - 15.8 = 84.2% sulphur is present. So, the mass ratio of C : S = 15.8 : 84.2 = I : 5.33. Now, the mass ratio of sulphur and oxygen in CS, and CO, when combining with fixed mass of carbon becomes 5.33 : 2.66 = 2:1 Now, SO2, contains 50% sulphur Hence, 100 - 50 = 50% oxygen is present. So the mass ratio of S : O = 50 : 50 = 1 : 1 Thus, the above ratios that is 2 : 1 and 1 : 1 are simple multiple of each other being 2 : 1 :: 1 : 1 = 2:1 Hence, the results illustrate the law of reciprocal proportions. Law of gaseous volumes Gases react with one another in simple ratio by volume and if the product of the reaction is gaseous then the volume of the product also bears a simple ratio to the volumes of the combining gases under similar conditions of temperature and pressure. Gay-Lussac, a French scientist stated the law of gaseous volume.

ATOMS & MOLECULES 39 Gay-Lussac’s law of combining volumes When gases combine they do so in simple ratio by volume to each other and to the gaseous products, all the volumes being measured at the same temperature and pressure. In the synthesis of ammonia 1 volume of nitrogen combines with 3 volumes of hydrogen to form 2 volumes of ammonia. Thus, the” volume ratio is nitrogen : hydrogen : ammonia :: 1 : 3 : 2. 2.2 DALTON’S ATOMIC THEORY According to Greek philosophers matter can be divided again and again till the smallest possible piece of matter, i. e. atom, is reached, The word ‘atom’ in Greek means (a - tomos) for not cut. In the begining of 19th century, the English chemist John Dalton put forth the atomic theory of matter (1808). This theory was based on the first four laws of chemical combinations as already described. Dalton’s theory, deals with the microscopic nature of matter. Postulates (1) Matter is composed of tiny particles called atoms. Atoms are indestructible i.e. they can neither be created nor destroyed by any chemical reaction. Atoms cannot be further divided during a chemical reaction. Chemical reactions can only involve rearrangement of atoms leaving the total mass unchanged. (Dalton named the smallest part of an element as simple atom and that of a compound as compound atom. However, atom is the smallest particle of an element and molecule, that of a compound. Thus, an element consists of only one kind of atoms.) (2) Element is composed of one kind of atoms. Atoms of the same element are all alike in shape, size etc. and differ from atoms of another element. (3) Atom of an element possesses a definite mass. Atoms of different elements are different in mass and exhibit different properties. (The atomic mass of carbon and oxygen have atomic masses 12 and 16 respectively. Further, these two elements are entirely different in their properties.) (4) When two or more different elements chemically combine, they do so by combination of atoms to form molecule. (The word molecule has been used by replacing “compound atom” as suggested by Dalton) The combination of atoms of different elements to form molecules occurs in simple ratios. (5) The combining ratio of the masses of the elements represents the combining ratio of the masses of the atoms. Atoms can not be considered indestructible. In subsequent chapters while studying Atomic structure and Nuclear reactions, it will be shown that atom is further composed of still more microscopic particles. Again it will be shown

40 +2 CHEMISTRY (VOL. - I) in the study of isotopes that it is possible to distinguish atoms of the same element. Isotopes are atoms of the same element having different masses. 2.3 DALTON’S ATOMIC THEORY AND THE LAWS OF CHEMICAL COMBINATIONS Law of conservation of mass Since atoms are indestructible that is they cannot be created or destroyed in a chemical reaction, all the atoms present in the reactants must be present at the end of the chemical reaction. Thus chemical reactions are only due to rearrangement of atoms. Again atom of an element having fixed mass, the total mass of the reactants must be equal to the total mass of the products. Law of definite proportions According to the atomic theory atoms of different elements combine to form molecules. Molecules are the smallest particles of compounds. Let atoms of element A combine with atoms of element B to form molecules of the compound AB. This is possible if equal number of atoms of element A and B combine. This shows that the compound AB always has a diffinite composition of element A and B i.e. the ratio in this case is 1:1. Again in compounds of elements A and B the molecules may be of the type A2B, AB,, AB3, etc. Since the combination of indivisible atoms must occur in whole numbers the ratio of elements of A and B in the above compounds are 2 : 1, 1 : 2, 1 : 3 etc. Fig 2.2 Combination of atoms to form molecules Since atom of an element has a fixed mass the constant composition of elements in a compound leads to a definite ratio of masses of the elements in a compound. Law of multiple proportions When the elements A and B combine to form a compound AB, the combining ratio of the atoms A and B is 1 : 1. In the compound AB2, similarly, the combining atomic ratio is 1 : 2. In a molecule of the compound AB2 there are present 1 atom of A and 2 atoms of B. Since atom of an element has fixed mass, the mass B that combines with a certain mass of A in the compounds AB and AB2 should be in the ratio 1:2.

ATOMS & MOLECULES 41 SOLVED PROBLEMS Problem 1 Different samples of sodium chloride obtained from sea, lake and rock gave the following results. a. Sample A : 5.85g of NaCl gave 3.54g of chlorine b. Sample B : 2.92g of NaCl gave 1.77g of chlorine c. Sample C : 1.95g of NaCl gave 1.18g of chlorine Show that the results agree with law of definite proportions Solution Sample A : 5.85g of NaCl contain 3.54g of CI So, % of chlorine = 3.54 × 100 = 60.51 5.85 Hence, % of Na = 100 - 60.51 = 39.49 Sample B : 2.92g of NaCl contain 1.77g of CI So, % of chlorine = 1.77 × 100 = 60.61 2.92 Hence, % of Na = 100 – 60.61 = 39.39 Sample C : 1.95g of NaCl contain 1.18g of CI So, % of CI = 1.18 × 100 = 60.51 Hence, % of Na = 100 - 60.51 = 39.49 1.95 From the above results it is found that the percentage of Na and CI are nearly constant in all the three samples. Therefore, the results agree with the law of definite proportion. Problem 2 Three oxides of manganese contain respectively 30.4%, 36.8% and 40% of oxygen. Show that these results illustrate the law of multiple proportions. Solution In the first compound, 30.4g of oxygen combine with 100 – 30.4 = 69.6g of manganese So, lg of oxygen combines with 69.6 = 2.29g of manganese. 30.4 In the second compound, 36.8 g of oxygen combines with 100 – 36.8 = 63.2g of manganese So, lg of oxygen combines with 63.2 = 1.72g of manganese 36.8

42 +2 CHEMISTRY (VOL. - I) In the third compound, 46.6g of oxygen combines with 100 – 46.6 = 53.4 g of manganese. So, lg of oxygen combines with 53.4 = 1.15g of manganese 46.6 Thus, in the three compounds lg of oxygen that combines with 2.29g, 1.72g and 1.15g of manganese is 2 : 1.5 : 1 and that is 4 : 3 : 2. This is a ratio of simple whole numbers. Thus, the results are in conformilty with the law of multiple proportion. Problem 3 Water contains 11.11% hydrogen, ammonia contains 82.35% nitrogen and nitrogen trioxide contains 63.15% oxygen. Show that these results can be explained with the help of the law of reciprocal proportions. Solution In H2O, 11.11% hydrogen is present Hence, 100 – 11.11 = 88.89% oxygen is present Thus the mass ratio of H : O = 11.11 : 88.89 = 1:8 In NH3, 82.35% of nitrogen is present Hence, 100 - 82.35 = 17.65% hydrogen present. Thus the mass ratio of H : N = 17.65 : 82.35 = 1 : 4.67 In the two compounds H20 and NH3 the weight of nitrogen and oxygen that combine with same wt. that is 1 part by wt. of hydrogen, are in the ratio of 4.67 : 8 = 1 : 1.71 In nitrogen trioxide, 63.15% oxygen is present Hence, 100 - 63.15 = 36.85% nitrogen is present. Thus, the mass ratio of N : O = 36.85 : 63.15 = 1 : 1.71 In both of the above cases the mass ratio of N : O becomes the same i.e 1 : 1.71 which illustrates the validity of the law of reciprocal proportions.

ATOMS & MOLECULES 43 CHAPTER (2.1 - 2.3) AT A GLANCE Law of conservations of mass Matter is neither created nor destroyed during any transformation of matter. In any chemical reaction the total mass of the reactants is equal to the total mass of the products. Law of definite proporations or constant proportions A given compound always contains the same elements combined in the same proportion by mass. Law of multiple proportions Whenever two elements combine to form more than one compound, the masses of the elements that combine with a constant mass of the other element are in a ratio of simple whole numbers. Law of equivalent or reciprocal proportions The masses of two or more different elements which separately combine with a definite mass of another element are either the same as or simple multiple of the masses of different elements when they combine amongst themselves. Gay-Lussac’s law of combining volumes When gases combine, they do so in simple ratio by volume to each other and to the gaseous products, all the volumes being measured at the same temperature and pressure. QUESTIONS A. Very short answer type (1 mark each) State true or false 1. Atoms can be divided. 2. Copper oxide is formed from metallic copper with a net increase in mass. 3. Mass and energy are interconvertible. 4. The law of reciprocal proportions is also called law of equivalent proportions. 5. Law of conservation of mass was formulated by Lavoisier. 6. Atoms of one element can be changed into another element. 7. Gay-Lussac’s law of combining volumes is applicable to the formation of SOy from sulphur. 8. Matter is destructible. 9. Atoms of different elements are different in all respects. 10. Atoms of the same element are similar in all respects. 11. The law of definite proportions is also called the law of constant proportions. 12. The law of reciprocal proportions was enunciated by Richter. 13. John Dalton formulated the law of multiple proportions. 14. Atoms of different elements always combine in simple ratio to form their compounds.

44 +2 CHEMISTRY (VOL. - I) 15. Proust formulated the law of definite proportions. 16. According to the law of definite proportions, a chemical compound obtained from different sources has different composition. B. Short answer type (2 marks each) 1. State the law of multiple proportions. [CHSE 1991, 1994,2001 IR] 2. State the law of reciprocal proportions [CHSE 1997] 3. Define law of conservation of mass 4. State the Gay-Lussac’s law of combining volumes. 5. State the law of definite proportions. C. Long answer type (10 marks each) (Statement with explanation 7 marks, Problem 3 marks) 1. State the postulates of Dalton’s atomic theory. How this theory explains the laws of chemical combinations ? 2. State the law of reciprocal proportions. Water and sulphur dioxide contain 88.9% and 50% oxygen respectively. Hydrogen sulphide contains 94.1% sulphur. Show that these figures are in agreement with the law-of reciprocal proportions. 3. State and explain the law of multiple proportions. 1.5 grams of each of the two oxides of an element on reduction produced 1.331 grams and 1.196 grams of the elements respectively. Show that the results are in agreement with the law of multiple proportions. 4. State and explain the law of conservation of mass. 6.3 grams of sodium bicrabonate are added to 15.0 grams of acetic acid solution and the residue is found to weigh 18.0 grams. What mass of carbon dioxide is released during the reaction ? [Hints : NaHCO3 + CH3COOH ® CH3COONa + H2O + CO2. In the reaction mass of the residue = mass of CH3COONa + mass of H2O] 5. State and explain the law of definite proportions. 0.09g of carbon burns in oxygen to give 0.33g of carbon dioxide. Carbon dioxide is also obtained by burnig 0.225g of carbon in 0.6g of oxygen. Show that the above data illustrate the law of constant proportions. 6. State Gay-Lussac’s law of gaseous volumes and give two examples of the reactions which illustrate the law. D. NUMERICAL PROBLEMS 1. How much silver nitrate would be required to react with 1. 16gms of sodium chloride to produce 1.70 gms of sodium nitrate and 2.87 gms of silver chloride ? Which law of chemical combinations holds good in this case ? [Hint : mass of AgNO3 + mass of NaCl = mass of AgCl + mass of NaNO3] 2. Weight of copper oxide obtained by treating 2.16g of metallic copper with nitric acid and subsequent ignition was 2.7g. In another experiment, 1.15g of copper oxide on reduction produced 0.92g of copper. Show that these results illustate the law of constant proportions.

ATOMS & MOLECULES 45 3. In an experiment 0.6g of carbon when heated with excess of oxygen produced 1.12 litres of carbon monoxide at NTP. In another experiment 2.24 litres of carbon dioxide upon reduction with 1.2g of carbon produced 4.48 litres of carbon monoxide at NTP. Show that the results illustate the law of definite proportions. [Hint : 2C + O2 = 2CO = 2 × 22.4 litres of CO = 2 × 28 = 56 g. CO2 + C = 2CO The calculation will be based on the concepts of mole and molar volume which is dicussed in chapter-6] 4. Carbon and oxygen are known to form two compounds. The carbon content in one of these is 42.9% while in the other is 27.3%. Show that the data are in agreement with the law of multiple proportions. 5. Three oxides of a metal M are found to contain 24.26, 29.95 and 49.95 percent of oxygen. Show that these results are in agreement with law of multiple proportions. 6. Phosphorus chloride contains 22.57% phosphorus, phophine contains 8.82% hydrogen and hydrogen chloride contains 97.23% chlorine. Show that these figures agree with the law of reciprocal proportions. 7. KI contains 23.6% potassium. KC1 contains 52% potassium. IC1 contains 77.8% iodine. Show that these data are in agreement with law of reciprocal proportions. ANSWERS A. 1. False 2. True. 3. True 4. True 5. True 6. True 7. False 8. False 9. False 10. False 11. True 12.’True 13. True 14. False 15. True 16. False. C. 2. Reference : Example 1 of law of reciprocal proportions . 3. Reference : Example 1 of the law of multiple proportions. 4. 3.3g. 5. C : O mass ratio 1 : 2.67. D. 1. 3.4lg , The law of conservation of mass. 2. C = 80% 0 = 20% 3. In both the experiments C : O ratio = 3:4. 4. The weights oxygen combining with fixed weight of carbon are in a simple whole number ratio of 1 : 2. 5. The ratio of weights of the metal M that combines with one part by weight of oxygen is 3:2:1. 6. The ratio weights of CI and H which combine with fixed weight of phosphoais is 35.5 : 1 and the ratio weights of CI and H in HC1 is also 35.5 : 1. 7. The ratio weights of I : CI which combine with fixed weight of K is 3.52 : 1, where as the ratio weights of I : CI is 3.50 : 1. Both the values of I : CI ratio- weights are within the limits of experimental error.

46 +2 CHEMISTRY (VOL. - I) 2.4 ATOMIC MASS It is analysed that water contains 88.9% mass of oxygen and 11.1% mass of hydrogen. Thus the mass of hydrogen and oxygen is about 1 : 8. The molecular formula of water is H2O. It is evident that there are 2 atoms of hydrogen and one atom of oxygen in a molecule of water. It leads to the fact that 1 atom of oxygen is 16 times as heavy as that of 1 atom of hydrogen. Since hydrogen is the lightest element, if the mass of a hydrogen atom is regarded as 1 then the mass of oxygen atom is found more correctly as 15.88. Considering the mass of hydrogen atom as the reference, the relative atomic masses of other elements may be expressed as mentioned below. The atomic mass or the relative atomic mass of an element is a number which expresses how many times the mass of one atom of the element is heavier than the mass of one atom of hydrogen. Atomic mass of an element = Mass of 1 atom of the element Mass of 1 atom of hydrogen. The atomic mass of nitrogen is 14. This means that one atom of nitrogen is 14 times heavier than one atom of hydrogen. If, however, an oxygen atom was chosen as the reference and assigned the value 16.000, then the atomic masses of majority of elements become whole numbers or near whole numbers. Thus, oxygen was taken as the accepted standard. The atomic mass or relative atomic mass of an element is the ratio of the mass of an 1 atom to 16 th of the mass of oxygen atom. These masses are expressed in atomic mass units (amu). The atomic mass of hydrogen is 1.008 amu in this scale. It has been stated earlier that isotopes of an element are atoms of the same element having different masses. Natural oxygen is known to consist of three isotopes of masses, 16, 17 and 18. Hence chemical atomic mass scale based upon the mass of natural oxygen as 16 is of less validity from the standpoint of determining isotopic masses accurately. Thus, a physical atomic scale based upon the oxygen –16(16O) was devised. The physical atomic mass of an element is the ratio of the mass of one atom of the 1 element to 16 of the mass of one atom of 16O. On this scale the atomic mass of natural oxygen is 16.0043 amu. The atomic masses of most elements are very accurately determined in a mass spectrometer by using volatile compounds of carbon as reference. Thus the mass of carbon –12 isotope (12C)

ATOMS & MOLECULES 47 was accepted as the standard by IUPAC (International Union of Pure and Applied Chemistry) in 1961. The physical atomic mass scale based on 12C as a reference is expressed as Mass of 1 atom of the element 1 12 th of the mass of 12C. Carbon has two other isotopes of masses 13 and 14. Thus the present scale is based on 12C equal to 12. This scale is used for both chemical atomic masses and isotopic masses. The atomic mass or the relative atomic mass of an element is a number which expresses how many times the mass of one atom of the element is heavier than 1 th 12 of the mass of carbon –12 (12C). Atomic mass of an element = Mass of 1 atom of the element 1 12 th of the mass of 12C. Relative atomic mass is expressed in the unit (a.m.u), atomic mass unit. Gram atomic mass The gram atomic mass or the gram - atom of an element is its relative atomic mass expressed in grams. For example, the relative atomic mass of helium is 4.003. Hence its gram - atomic mass is 4.003g. 1 g - atom of helium represents 4.003 g of helium. Table 2.1 Relative atomic masses of elements based upon carbon –12, 12C = 12.000 amu. Element Relative atomic mass (amu) Element Relative atomic mass (amu) Hydrogen 1.007967 Magnesium 24.312 Helium 4.003 Aluminium 26.98 Lithium 6.939 Silicon 28.086 Beryllium 9.102 Phosphorus 30.974 Boron 10.81 Sulphur 32.064 Carbon 12.01115 Chlorine 35.453 Nitrogen 14.007 Argon 39.948 Oxygen 15.99943 Potassium 39.102 Fluorine 19.00 Calcium 40.08 Neon 20.183 Scandium 44.956 Sodium 22.9898 Titanium 47.90

48 +2 CHEMISTRY (VOL. - I) Element Relative atomic mass (amu) Element Relative atomic mass (amu) Vanadium 50.942 Gadolinium 157.25 Terbium 158.924 Chromium 52.00 Dysprosium 162.50 Manganese 54.94 Holmium 164.93 Iron 55.85 Erbium 167.26 Cobalt 58.933 Thulium 168.934 Nickel 58.71 Ytterbium 173.04 Copper 63.54 Lutetium 174.97 Zinc 65.37 Hafnium 178.49 Gallium 69.72 Tantalum 180.948 Germanium 72.59 Tungsten 183.85 Arsenic 74.92 Rhenium 186.23 Selenium 78.96 Osmium 190.20 Bromine 79.909 Iridium 192.20 Krypton 83.80 Platinum 195.09 Rubidium 85.47 Gold 196.97 Strontium 87.62 Mercury 200.59 Yttrium 88.905 Thallium 204.37 Zirconium 91.22 Lead 207.19 Niobium 92.91 Bismuth 208.98 Molybdenum 95.94 Polonium 210† Technetium 99† Astatine 210† Ruthenium 101.07 Radon 222† Rhodium 102.905 Francium 223† Palladium 106.4 Radium 226† Silver 107.87 Actinium 227 Cadmium 112.40 Thorium 232.038 Indium 114.82 Protactinium 231† Tin 118.69 Uranium 238.03 Antimony 121.75 Neptunium 237† Tellurium 127.60 Plutonium 242† Iodine 126.90 Americium 243† Xenon 131.30 Curium 247† Caesium 132.91 Berkelium 249† Barium 137.34 Californium 251† Lanthanum 138.91 Einsteinium 254† Cerium 140.12 Fermium 253† Praseodymium 140.907 Mendelevium 256† Neodymium 144.24 Nobelium 254† Promethium 147† Samarium 150.35 Lawrencium 257† Europium 151.96 † symbol indicates the mass number of the longest lived or best known isotope.

ATOMS & MOLECULES 49 Absolute mass of an atom The atom is so small that its exact weight cannot be measured accurately. The absolute mass of an atom is the mass of one atom in gram. Atomic mass in gram Absolute mass of an atom in gram = Avogadro number, (6.023 x 1023 ). Table 2.2 Atomic mass and absolute atomic mass of atoms of some elements Element Atomic mass (amu) Absolute mass of an atom (gm) H 1.008 1.008 = 1.67 x 10–24 C 12 6.023 x 1023 N 14 O 16 12 = 1.99 x 10–23 6.023 x 1023 14 = 2.32 x 10–23 6.023 x 1023 16 = 2.66 x 10–23 6.023 x 1023 Atomic mass, isotopes and isobars F. W. Aston, an English physicist determined the relative mass of atoms in the mass spectrograph. It is found that some naturally occuring elements contain atoms that are chemically identical but have different masses. Atoms of the same element having different masses are called isotopes. In case of existing isotopes of an atom, it is significant to find the average atomic mass of the atom. Chlorine has two isotopes of atomic mass 35 and 37 occuring in the ratio 3 : 1. Therefore, the average mass of chlorine is (3 x 35) + (1 x 37) = 35.5 3+1 Atoms of different elements having the same atomic mass are called isobars. Argon has one isotope having atomic mass 40 and one of the isotopes of potassium has atomic mass 40. These two atoms of different elements argon and potassium are called isobars. 2.5 DETERMINATION OF RELATIVE ATOMIC MASS BY DULONG AND PETIT'S METHOD The French Scientists Dulong and Petit studied the relationship between specific heat and atomic mass of elements. Dulong and Petit's law states that the product of the relative atomic mass of an element and its specific heat is approximately equal to 6.4 for all solid elements at

50 +2 CHEMISTRY (VOL. - I) ordinary temperature (except carbon silicon, boron & beryllium). This product is called the atomic heat. Mathematically, it can be stated : Relative atomic mass x Specific heat ~ 6.4 The specific heat is usually expressed in calorie per degree per gram of the element. If however, the specific heat is expressed in joule, then relative atomic mass x specific heat ~ 26.8. Dulong and Petit's law can be applied to determine the approximate relative atomic mass of an element. In order to find out the correct relative atomic mass of an element, the following steps are adoptd. (1) The approximate relative atomic mass of the element is calculated from the known specific heat of the element with the help of Dulong and Petit's law. Approximate relative atomic mass = 6.4 Specific heat (2) The equivalent mass of the element is determined from the experiment. (3) The approximate relative atomic mass is divided by the equivalent mass to get an approximate valency. Relative atomic mass (approx) = Equivalent mass x Valency (approx) or Valency (approx) = Relative atomic mass (approx) Equivalent mass (4) The approximate valency is converted to the nearest whole number. (5) The correct relative atomic mass of the element is calculated from the product of equivalent mass and the valency (whole number) Relative atomic mass = Equivalent mass x Valency NUMERICAL PROBLEMS Example 1 The equivalent mass of a metal is 32.2. Its specific heat is 0.095. Determine the relative atomic mass of the metal. Solution The approximate relative atomic mass of the metal = 6.4 = 67.4 0.095 The approximate valency of the metal is 67.4 = 2.1 32.2 When converted to nearest whole number becomes 2. Hence the relative atomic mass of the metal = Eq.mass x Valency = 32.2 x 2 = 64.4

ATOMS & MOLECULES 51 Example 2 The equivalent mass and specific heat of a metal is 9.0 and 0.214 respectively. Calculate the atomic mass of the metal. Solution The approximate atomic mass of the metal. = 6.4 =29.9 0.214 The approximate valency of the metal = Atomic mass of the metal Eq.mass of the metal = 29.9 = 3.32 9 When converted to the nearest whole number becomes 3. Hence, the atomic mass of the metal = Eq.mass x Valency = 9.0 x 3 = 27 Determination of atomic masses from physical measurement The atomic masses of elements are determined and the existence of isotopes is proved by experiments performed in a Mass Spectrograph. In 1919 the instrument mass spectrograph was devised by scientists Thomson and Aston who studied the deflection of positive ions by simultaneously applying electric and magnetic fields. In this method, atoms of gaseous elements are converted into ions in a discharge tube. The positive rays thus formed are passed through two slits S1S2 and then subjected to the influence of electric (E1E2) and magnetic fields M (Fig 2.3) Photographic plate, P S1 S2 E1 Electric Positive S2 field + rays E2 Magnetic S1 field, M Fig.2.3 Mass spectrograph. The narrow beam of positive ions with different masses deflect into different extents. Ions of heavier mass suffer less deflection. The impressions of the deflections are recorded on a photographic plate, P. Thomson produced positive ions by taking neon gas in a discharge tube and analysed it in the mass spectrogram. Thomson found a very faint trace due to singly charged neon of atomic mass 22 along with parabolic trace due to singly charged neon of atomic mass 20. He concluded that neon has two isotopes of atomic masses 20 and 22.

52 +2 CHEMISTRY (VOL. - I) 18 20 22 24 26 28 30 32 34 36 38 Fig 2.4 Spectrogram From the position of the lines, the relative masses of the positive ions are calculated (Fig..2.4). Several lines may be recorded for the elements consisting of isotopes. An element recording lines at mass numbrs A1 and A2 and thickness of lines remaining in the ratio of n1 and n2 the atomic mass becomes. n1A1 + n2A 2 n1 + n2 CHAPTER (2.4) AT A GLANCE The atomic mass or the relative atomic mass of an element The atomic mass or the relative atomic mass of an element is a number which expresses 1 how many times the mass of one atom of the element is higher than 12 th of the mass of carbon –12(12C). Atomic mass unit (a.m.u.) The relative atomic mass is expressed with the unit a.m.u, that is atomic mass unit. The gram atomic mass or the gram-atom The gram atomic mass or the gram-atom of an element is its relative atomic mass expressed in grams. Absolute mass of an atom The absolute mass of an atom is the mass of one atom in grams. Atomic mass in gram Absolute mass of an atom in grams = Avogadro number (6.023 x 1023 ). Isotopes are atoms of the same element having different masses. Average atomic mass It can be calculated by taking into account of the number of isotopes of an atom and the isotopic occurrence of its mass ratio.

ATOMS & MOLECULES 53 Relationship among atomic mass, equivalent mass and valency of an element Atomic mass = Equivalent mass x Valency Dulong and Petit's law Dulong and Petit's law states that the relative atomic mass of an element and its specific heat is approximately equal to 6.4 (atomic heat) for all elements at ordincry temperature (except Be, B, C, Si) Relative atomic mass x Specific heat P 6.4 (Atomic heat) (approx) Mass Spectrograph The atomic masses of elements can be determined in a mass spectrograph. Solved Problems Problem 1 0.875 g of a metal produced 1.095 g of its oxide. The specific heat of the metal is 0.09. Determine the exact atomic mass of the metal. Solution : The mass of oxygen that combined with 0.875 g of metal = 1.095 – 0.875 = 0.22 g Hence 0.22 g of oxygen combined with 0.875 g of metal 0.875 or 8 g of oxygen combines with 0.22 x 8 = 31.82, which is the equivalent mass of the metal. Now the approximate atomic mass of the metal = 6.4 = 71.1 0.09 71.1 The approximate valency of the metal = 31.2 = 2.23 When converted to the nearest whole number becomes 2. Hence, the atomic mass of the metal = Eq.mass x Valency = 31.82 x 2 = 63.64 Problem 2 The specific heat of a metal M is 0.25 and its equivalent mass is 12. What is its correct atomic mass ? Mention the formula of its chloride. Solution : The approximate atomic mass of the 2m152.e6tal==2.1063..245an=d 25.6 the nearest whole The approximate valency of the metal = therefore number valency becomes 2. Hence the correct atomic mass of the metal = .12 x 2 = 24. Since the valency of the metal is 2, the formula of its chloride becomes MCl2. Problem 3 Specific heat of an element is 0.057. Its equivalent mass is 37.8. What is its exact atomic mass ? (CHSE 1994 A)

54 +2 CHEMISTRY (VOL. - I) Solution : The approximate atomic mass of the metal = 6.4 = 112.28 0.057 The approximate valency of the metal = Approx. atomic mass of element = 112.28 Eq.mass of the element 37.8 = 2.97 and converted to the nearest whole number valency 3. Now the exact atomic mass of the element = Equivalent mass x Valency = 37.8 x 3 = 113.4 Problem 4 The chloride of a metal M contains 23.2% of metal. Its specific heat is 0.224. What is the accurate atomic mass of the metal ? (CHSE 1992) Solution : In 100g of metal chloride the mass of metal = 23.2 g So, the mass of chlorine = 100 – 23.2 = 76.8 g 76.8g of chlorine combines with 23.2 g of the metal Hence, equivalent mass of the metal = 23.2 x 35.45 = 10.71 76.8 Specific heat of the metal = 0.224 The approximate atomic mass of the metal = 6.4 = 28.57 0.224 28.57 The approximate valency = 10.71 = 2.67 and making the valency a whole number it becomes 3. Now the accurate atomic mass of the metal = Eq.mass x Valency = 10.71 x 3 = 32.13 QUESTIONS A. Very short answer type (1 mark each) 1. Atomic mass of which element is taken as standard when defining the relative atomic mass of other elements. 2. Define isotopes. 3. What value has the relative atomic mass of oxygen ? 4. What is the mass of 1 atom of carbon ? 5. The equivalent mass of a trivalent metal is 9. What will its relative atomic mass ? 6. What is the value of atomic heat ? 7. What is the unit of specific heat ? 8. Does the specific heat change with temperature ? 9. Mention the relationship among atomic mass, molecular mass and atomicity.

ATOMS & MOLECULES 55 10. How many atoms are present in 1 gram atom of the element ? 11. State Dulong and Petit's law (CHSE 2001 ER) B. Short answer type (2 marks each) 1. An oxide of a bivalent metal contains 40% oxygen. Calculate its atomic man. 2. What is the difference between relative atomic mass and gram atomic mass ? 3. What is the mass of 3 atoms of nitrogen ? 4. What is the relationship among equivalent mass, atomic mass and valency ? 5. Explain atomic mass unit. 6. The specific heat of an element is 0.214 cal/gm. What is its approximate atomic weight ? 7. Explain average atomic mass. 8. Differntiate between atomic mass and absolute mass of an atom. 9. \"Equivalent mass of an element may vary\" – Explain with example. 10. State Dulong and Petit's law. How can it be used to calculate the exact atomic mass of a solid element. C. Long answer type (7 marks each) 1. State and explain Dulong and Petit's law. Define atomic mass. Specific heat of an element is 0.057. Its equivalent mass is 37.8. What is its exact atomic mass ? 2. State Dulong Petit's law. Describe how the atomic mass of an element is determined by the use of the law. A certain metal oxide contains 60% of oxygen. The specific heat of the metal is 0.401. Calculate the exact atomic mass, equivalent mass and valency of the metal. 3. State and explain Dulong and Petit's law. The choloride of a metal 'M' contains 23.2% of metal. Its specific heat is 0.224. What is the accurate atomic mass of the metal ? D. Numerical Problems 1. A certain metal forms an oxide containing 60% of oxygen. The specific heat of the metal is 0.401. Calculate the valency of the metal. 2. 0.1g of a metal M, when dissolved in dil HCl evolved 124.2 ml of hydrogen at NTP. The specific heat of the metal is 0.214. Calculate the atomic mass of the metal. What are the formulae of the oxide and chloride of the metal ? 3. An element forms an oxide containing 47% oxygen. The specific heat of the element is 0.22. Calculate the relative atomic mass of the element.

56 +2 CHEMISTRY (VOL. - I) 4. 0.108g of a metal when treated with excess of dilute H2SO4 liberated 40.2 ml of hydrogen gas at 170C and 770 mm pressure. Aqueous tension at 170C is 14.4 mm pressure. The specific heat of the metal is 0.095. Determine the relative atomic mass of the metal. 5. The chloride of a metal contains 79.8% of chlorine. The specific heat of the metal is 0.224. Calculate the correct relative atomic mass of the metal. ANSWER A. 1. Carbon, 12C. 3.16 amu 4. 1.99 x 10–23g 5. 27 6. 6.4 approx. 7. cal/gm/0C. 8. Yes. 9. Molecular mass = At.mass x Atomicity 10. 6.023 x 1023 atoms. B. 3. 6.96 x 10–23g 4. Atomic mass = Equivalent mass x Valency 6. 29.9 C. 1. 113.4 (Ref : solved problem no 3. in this Chapter) D. 1. 3 2. At.mass = 27,M2O3 and MCl3 3. 27.06 4. 64.4 5. 26.94 2.6 MOLECULAR MASS : The molecular mass or the relative molecular mass of an element or a compound is expressed as the mass of one molecule of the element or compound 1 as compared to 12 th of the mass of one atom of carbon -12 (12C = 12.000) in the physical atomic mass unit scale. The relative molecular mass of an element or compound = Mass of 1 molecule of the element or compound 1 12 th of the mass of 12C. The gram-molecular mass or the gram-mole or the molar mass of a substance is its relative molecular mass expressed in grams. For example, the relative molecular mass of oxygen is 32. So, 1 gram-molecule of oxygen represents 32 g.

ATOMS & MOLECULES 57 The relative molecular mass of a compound is the sum of the relative atomic masses of all constituent elements. Based on the carbon–12 standard in the physical atomic mass scale, the mass of oxygen atom is 15.99943 » 16. The relative molecular mass of oxygen (O2) is 2 x 16 = 32 amu. Thus, one gram-molecule of oxygen represents 32 g. The relative molecular masses of some compounds have already been calculated in Chapter – 1 of this book. Table 2.3 Relative molecular masses and gram molecular masses of some elements and compounds. Molecule/ Formula Relative molecular Gram molecular No.of molecules. compound mass mass Hydrogen H2 2 a.m.u 2 g 6.023 x 1023 Nitrogen N2 28 a.m.u 28 g 6.023 x 1023 71 a.m.u 71 g 6.023 x 1023 Chlorine Cl2 18 a.m.u 18 g 6.023 x 1023 44 a.m.u 44 g 6.023 x 1023 Water H2O 34 a.m.u 34 g 6.023 x 1023 40 a.m.u 40 g 6.023 x 1023 Carbon dioxide CO2 Hydrogen peroxide H2O2 Sodium hydroxide NaOH Determination of Relative molecular mass Various methods are used to determine the relative molecular masses. DETERMINATION OF RELATIVE MOLECULAR MASS OF VOLATILE LIQUIDS BY VICTOR MEYER'S METHOD. A known weight of a volatile liquid is vapourised which displaces an equal volume of air. From this volume of collected air and weight of liquid, the vapour density and then the relative molecular mass of the liquid is calculated. Victor Meyer's apparatus consists of a long cylindrical glass vessel having a bulb at the bottom , a side tube and a funnel fitted with a cork at the top.

58 +2 CHEMISTRY (VOL. - I) Victor Meyer's tube Displaced air Graduated tube Outer jacket tube (copper) Hofmann's bottle Liquid of higher boiling point Fig: 2.5 Victor Meyer's apparatus In the bulb at the bottom small amount of asbestos is placed so that the bottom of the cylindrical glass vessel does not crack when the Hofmanns' bottle is dropped. This cylindrical glass vessel is kept inside an outer jacket-tube made of copper ending in a bulb at the lower end. The glass vessel is kept in position by a cork fitted to the outer jacket tube. A bent tube is also fitted to the cork for the exit of the vapours of boiling liquid in the outer jacket. A liquid whose boiling point is about 25–300C higher than the boiling point of the volatile liquid (whose relative molecular mass is to be determined) is taken in the outer jacket-tube. For the determination of relative molecular mass of low boiling liquid, usually water is taken in the outer jacket-tube. The top of the funnel of the cylindrical glass vessel is closed with a cork. To start with the experiment the liquid in the outer jacket-tube is heated to boiling. Now the side tube of the cylindrical glass vessel is dipped under water in a trough. When no more bubles appear, the apparatus has attained a steady temperature. Then a graduated tube filled with water is inverted over the end of the side tube. In the mean while a known weight of the volatile liquid (about 0.1 to 0.2 g) is taken in a small bottle fitted with a loose stopper known as Hofmann's bottle. This bottle is quickly dropped into the cylindrical glass vessel and the mouth of the funnel of the glass vessel is closed with the cork. Since the Hofmann's bottle is loosely stoppered, the stopper opens and the volatile liquid vapourises quickly. The vapour so produced displaces the same volume of air that is collected in the graduated tubes at the room temperature and pressure. When the water level in the graduated tube remains constant for quite sometime, the graduated tube is removed with the thumb at its open end with care so that the water level is constant. Then the graduated tube is carefully lowered in a water jar and the volume of air is noted by making the level of water inside and outside the tube the same. The room temperature and the atmospheric pressure is also noted. Now since the volume of the vapour from a known weight of a volatile liquid at a particular temperature and pressure is known, the vapour density and the molecular mass of the volatile liquid can be calculated as follows :

ATOMS & MOLECULES 59 Calculation Let mass of the volatile liquid = m g Volume of air collected = V1cc = The volume the vapour of the volatile liquid. Atmospheric pressure = P1mm of mercury Room temperature = t10, C = (t1 + 273)0 A = T1 Aqueous tension at t01C = f mm So, pressure of dry air = (P1 – f) mm In order to calculate the vapour density of the vapour from the volatile liquid, the volume of vapour of the volatile liquid V1 be converted to V i.e. the volume of vapour at NTP. P = 760 mm T = 2730A V = The volume of vapour of the volatile liquid at NTP = ? From the combined gas equation PV = P1V1 T T1 V= P1V1 x T = (P1 - f ) x V1 x 273 = Volume of vapour of the volatile liquid. T1 P (t1 + 273) x 760 Method- 1 Direct calculation of relative molecular mass from the idea of molar volume. Now, weight of Vcc of vapour at NTP = m g So, weight of 22,400 cc of the vapour at NTP = M x 22,400 g V = The relative molecular mass of the volatile liquid. Method- 2 The relative molecular mass = 2 x V.D. The vapour density (V.D) of the vapour is given by V.D = Weight of Vcc of the vapour at NTP Weight of the same Vcc of the hydrogen gas at NTP = m (Since weight of 1cc of hydrogen gas at NTP is 0.000089 g) V x 0.000089 Now the relative molecular mass M = 2.016 x V.D. = 2 x V.D (Since 2.016 » 2, we consider) Method- 3 g The relative molecular mass can be calculated from the gas equation PV = M RT

60 +2 CHEMISTRY (VOL. - I) Since the entire amount of volatile liquid is converted to its vapour, the mass of the vapour is also the mass of the volatile liquid. From the gas equation, PV = g RT M where g is the mass of vapour whose relative molecular mass is M. In this case, Pressure = P1 mm = (P1 – f) mm = P1 - f atms. 760 V1 Volume = V1cc = 1000 litres. R = 0.082 litre–atm. degree–1 mol–1. T1 = (t1 + 273)0 A g g M M Substituting these values in PV = RT, Here P1V1 = RT1 or, M = g.RT1 = m x 0.082 x (273 + t1) x 760 x 1000 P1V1 (P1 - f )V1. = The relative molecular mass of the volatile liquid. NUMERICAL PROBLEMS Example 1 : In an experiment 0.61 g. of a volatile liquid displaces 122.9 ml of moist air at 200C and 757.5 mm Hg pressure. If aqueous tension at 200C is 17.4 mm, calculate the vapour density and molecular mass of the substance. Solution Given, mass of volatile liquid m = 0.61 g Volume of moist air collected = Say V1 = 122.9 ml Atmospheric pressure = 757.5 mm of mercury Temperature = t1 = 200C = (20 + 273)0A = 2930A. Aqueous tension at 200C = 17.4 mm. Thus, the pressure of dry air = P1 = (757.5 – 17.4) = 740.1 mm P = 760 mm T = 2730A and V = Volume of the vapour of the volatile liquid at NTP = ? Applying the combined gas equation, PV = P1V1 T T1 V= P1V1 x T = 740.1 x 122.9 x 273 = 111.51 ml T1 P 293 x 760 Now vapour density, V.D = Wt. of Vml. of the vapour at NTP Wt. of same volume of hydrogen at NTP = 0.61 = 61 111.51 x 0.000089 and therefore, the molecular mass = 2 x V.D = 2 x 61 = 122

ATOMS & MOLECULES 61 Example 2 : 0.309 g of a compound is found to occupy 488 ml at 250C and 737 ml pressure. What is the molecular mass of the compund ? Solution Given, the mass of the compound = 0.309 g Volume of the gas occupied, V1 = 488ml. Temperature t10 c = (25 + 273)0A = 2980A = T1 P1 = 737 mm(since no aqueous tension is given, the pressure of the gas is to be considered as pressure of dry gas) Normal pressure = P = 760 mm, Normal temperature = T = 2730A. From combined gas equation PV = P1V1 ,V= P1V1 x T T T1 T1 P Now, the volume of gas occupied at NTP = V So, V = P1V1 x T = 737 x 488 x 273 = 433.53 ml. T1 P 298 x 760 Now, 433.53 ml of the gas at NTP is liberated from 0.309 g of the compound. Hence 22,400 ml of the gas at NTP = 0.309 x 22,400 = 16 433.53 Therefore, the molecular mass of the compound is 16. Example 3 In Victor Meyer's experiment 0.235 g of a liquid is vapourised and the volume of air displaced measured in a gas burette is 42cc at 230C and 730 mm pressure. The vapour pressure of water at 230C is 21 mm. Calculate the molecular weight of the substance. Solution Pressure of dry air = The pressure exerted by vapour = (730 – 21) = 709 mm = 709 atm. 760 Volume of the air displaced = Volume of the vapour = 42cc = 42 litres 1000 Temperature = (23 + 273) = 2960A Mass of the vapour = mass of the liquid = 0.235 g Applying the gas equation PV = g RT M Now, the molecular weight M= gRT PV = 0.235 ´ 0.082 ´ 760 ´ 1000 ´ 296 = 145.6. 709 ´ 42 (Since R = 0.082 litre–atm. degree–1 mol–1 Therefore, the molecular weight of the substance is 145.6

62 +2 CHEMISTRY (VOL. - I) CHAPTER (2.6) AT A GLANCE Molecular mass : The molecular mass or the relative molecular mass of an element or a compound is the mass of one molecule of the element or the compound compared with 1 th of the mass of one 12 atom of carbon–12 (12C = 12.000). Gram - molecular mass : The gram–molecular mass or the gram–mole or the molar mass of a substance is its relative molecular mass expressed in grams. The relative molecular mass is the sum of the relative atomic masses of all the constituent elements. The unit of the relative molecular mass is a.m.u. (atomic mass unit) The relative molecular masses of volatile liquids can be determined by Victor Meyer's Experiment. The following important relationships are used to calculate the relative molecular masses. 1. Mass of 22,400 ml of vapour at NTP = The relative molecular mass of the volatile liquid. 2. The relative molecular mass = 2 x V.D (Vapour density) 3. The relative molecular mass can be calculated by applying gas equation PV = g RT (where g = mass of the substance, M = molecular mass) M Solved Problems : Problem 1 0.1 gm of a volatile substance on vapourisation in Victor Meyer's experiment displaced 123 ml of moist air at 150C and 740 mm pressure. Calculate the molecular mass of the substance. (Aqueous tension at 150C = 12.7 mm. (CHSE 1997 A) Solution Given : Mass of the substance = 0.1 g Measured : Pressure, P1 = (740 – 12.7) = 727.3 mm Volume V = ? Temperature, T1 = (15 + 273) = 2880 A Volume V1 = 123 ml. At NTP, pressure, P = 760 mm Temperature, T = 2730A Applying combined gas equation PV = P1V1 or, V= P1V1 x T = 727.3 x 123 x 273 = 111.5 ml = Volume at NTP. T T1 T1 P 288 x 760

ATOMS & MOLECULES 63 Now, 111.5 ml of the vapour is liberated from 0.1 g of the substance Hence, 22,400 ml of the vapour will be liberated from : 0.1 x 22,400 ml = 20.1 111.5 Therefore, the molecular mass of the volatile substance is 20.1 Problem 2 In a Victor Meyer's experiment 0.6 g of a volatile substance displaced 115 ml of air at 200C and 756 mm pressure. Calculate the molecular weight of the substance (Aqueous tension at 200C = 17.4 mm). (CHSE 1989 A) Solution Mass of volatile substance = 0.6 g Observed : Volume V1 = 115 ml , pressure P1 = (756 – 17.4) = 738.6 mm Temperature T1 = 20 + 273 = 2930A At NTP, P = 760 mm T = 2730A V = ? Applying the combined gas laws PV = P1V1 T T1 or, V = P1V1 x T = 738.6 ´ 115 ´ 273 = 104.13 ml T1 P 293 ´ 760 Now, 104.13 ml of vapour at NTP weigh 0.6 g. So, 22,400 ml of at NTP weigh = 0.6 ´ 22,400 = 128.07 g. 104.13 Therefore, the molecular weight of the substance is 128.07 Problem 3 One litre of hydrogen at NTP weighs 0.09 g. 100ml of another gas at NTP weigh 0.0765 g. Calculate the vapour density and the relative molecular mass of the gas. Solution Weight of 100 ml of gas at NTP = 0.0765 g Therefore, weight of 1000 ml or 1 litre of the gas at NTP = 0.765 g Weight of 1 litre of hydrogen at NTP = 0.09 g. Vapour density of the gas = Wt. of 1 litre of gas = 0.765 = 8.5 Wt of 1 litre of H 2 0.09 Hence, the relative molecular mass = 2 x V.D = 2 x 8.5 = 17. Problem 4 A certain gas occupies 0.418 litre at 270C and 740 mm of mercury. If the same weighs 3.0 g , what is the molecular mass of the gas ? Solution P = 740 mm = 740 atm V = 0.418 l 760 T = 27 + 273 = 3000A, g = 3.0g, R = 0.082 litre – atm. per degree per mole. Applying the relation PV = g RT or M = gRT M PV and incorporating the above values M = 3 ´ 0.082 ´ 300 ´ 760 = 181.3 740 ´ 0.418 Hence, the relative molecular mass of the gas is 181.3

64 +2 CHEMISTRY (VOL. - I) QUESTIONS A. Very short answer type (1 mark each) 1. One gram mol. of oxygen will weigh how many grams ? 2. What unit is used for vapour density ? 3. Which method is suitable for determining the molecular mass of volatile substances ? 4. How many molecules are present in one gram molar volume of a gas ? 5. Why is asbestos or glass wool placed in the Victor Meyer's cylindrical glass vessel ? 6. What is the relationship between molecular mass and vapour density of a gas or vapour ? 7. Why should the Victor Meyer's tube be dried before start of the experiment ? 8. After inserting Hofmann's bottle, why should the Victor Meyer's tube is corked at once ? 9. What is the vapour density of N2O ? 10. Which will weigh more, one gram mol of oxygen or 24 gms of nitrogen ? 11. What is the vapour density of ozone ? B. Short answer type (2 marks each) 1. Define molecular mass. 2. Explain gram molecular mass. 3. What is the difference between normal density and relative vapour density ? 4. Is there any difference between the molecular weight and mass of one molecule of a gas ? Explain. 5. Explain why some gases show abnormal density. C. Long answer type (10 marks each) 1. Describe Victor Meyer's method for determining the molecular mass of a substance. 0.1 gm of a volatile substance on vapourisation in Victor Meyer's experiment displaced 123 ml moist air at 150C and 740 mm pressure. Calculate the molecular mass of the substance. (Aqueous tension at 150C = 12.7 mm) (CHSE 1997 A) 2. Define vapour density. How will you determine the vapour density of a volatile substance ? 3. Derive the relationship between molecular mass and vapour density of a gas from Avogadro's hypothesis. 0.796 g of a metal oxide was heated in a cuvvent of dry hydrogen when. 0.18 g of water was formed. Calculate the equivalant mass of the metal. (CHSE, 2001 ER)

ATOMS & MOLECULES 65 ANSWERS A. 1. 30g 2. No unit and V.D. is a number 3. Victor Meyer's method 4. 6.023 x 1023 molecules. 5. The glass vessel does not crack when Hofmann's bottle is dropped into it. 6. Molecular mass = 2 x V.D. 7. Because there will be error in the volume of vapour collected, if some moisture remains in the tube. 8. Because the vapours of volatile liquid will immediately escape. 9. 22 10. 1 gm mole of oxygen 11. 24gms of nitrogen. C. 1. 20.1 (Reference : Solved problem No.1 in this chapter) NUMERICAL PROBLEMS 1. 3.01 g of a gas occupies 1.3 litres at 00C and 760 mm pressure. Calculate the molecular mass of the gas. (Ans. 51.44) 2. In the Victor Meyer's experiment, 0.0926 g of a volatile liquid resulted 28 ml of its vapour at 160C and 753.5mm pressure. Find out the vapour density and molecular weight of the liquid (Aqueous tension at 160C = 13.5 mm) (Ans : 79.8) 3. In the Victor Meyer's apparatus 0.23 gm of a volatile substance displaced air which measured 112 ml at STP. Calculate the vapour density and molecular weight of the substance. (Ans : V.D = 23, Mol. wt = 46) 4. 33.6 cc of phosphorus vapour weighs 0.062 gm at 5460C and 760 mm pressure. The atomic weight of phosphorus is 31. What is the molecular weight of phosphorus ? How many atoms are present in one molecule of phosphorus ? (Ans : Mol. wt. 124, 4 atoms) 2.7 EQUIVALENT MASS The equivalent mass of an element is a number which expresses how many parts by mass of the element combines with or displaces from a compound 1.008 parts by mass of hydrogen or 8 parts by mass of oxygen or 35.45 parts by mass of chlorine. One equivalent of any substance reacts with one equivalent of the other substance to produce one equivalent of each of the products. Thus, it is evident that all substances combine or appear on the basis of an equivalence. The equivalent mass or equivalent weight expressed in gram is called gram- equivalent mass or gram-equivalent weight.

66 +2 CHEMISTRY (VOL. - I) Many metals such as zinc, magnesium, aluminium etc. react with acids liberating hydrogen gas. Now, one mole of hydrogen gas occupies 22.4 litres at NTP, and weighs 2.016g. Hence, 11.2 litres of hydrogen gas at NTP would weigh 1.008g. Therefore, in the displacement method, that mass of the metal in grams which would liberate 11.2 litres of hydrogen at NTP would be the gram equivalent mass of the metal. Relation between equivalent mass, valency and relative atomic mass of an element. Table 2.4 Relation of relative atomic mass and equivalent mass with valency Element Relative atomic mass(A) Equivalent mass(B) (A)/(B) Valency Sodium 22.997 22.997 1 Magnesium 24.32 Zinc 65.38 12.16 2 Aluminium 26.97 Hydrogen 1.008 32.69 2 Oxygen 16.000 8.99 3 1.008 1 8.000 2 From the table 8.1, it is clear that the ratio of the relative atomic mass to the equivalent mass of an element is equal to its valency. Therefore, the relative atomic mass of the element = the equivalent mass of the element x valency Variable equivalent mass Some metals such as copper,iron show variable valency. Copper has valencies 1 and 2. Iron shows valencies 2 and 3. Elements showing variable valency have equivalent masses corresponding to each of its valency. For such elements the equivalent mass of an element depends upon its valency shown by the element in a particular compound. For example, the element iron forms two oxides, ferrous oxide FeO and ferric oxide Fe2O3. The valencies of iron in these two compunds are 2 and 3 respectively. The relative atomic mass of iron is 55.85. And hence the equivalent masses of iron in FeO is 55.85/2 = 27.92 and in Fe2O3 is 55.85/3 = 18.62. DETERMINATION OF EQUIVALENT MASS OF METALS : Hydrogen displacement method Metals such as zinc, mangnesium, aluminium, iron, tin etc, displace hydrogen gas from dilute acids. A known amount of the metal is allowed to react with dilute acid and the volume

ATOMS & MOLECULES 67 of evolved hydrogen gas, measured under atmospheric pressure and at room temperature, is converted to the volume at NTP. It has been mentioned earlier that the gram-equivalent mass of a metal is that weight of the metal in grams which would liberate 11.2 litres of hydrogen gas at NTP. EQUIVALENT MASS OF ZINC BY HYDROGEN DISPLACEMENT METHOD Experiment A piece of AR grade (analytical reagent) zinc (< 0.1g) of about 0.08g in weight placed in a watch glass is weighed out accurately. This piece of zinc, placed on a piece of and in contact with platinum foil is kept immersed in a beaker under water. The zinc metal along with the platinum foil is covered with an inverted small funnel such that the stem of the funnel remained immersed under water. An eudiometer tube (graduated tube) is filled with bench sulphuric acid and then the open end of the eudiometer tube closed with thumb is inverted over the funnel so that the stem of the funnel remained inside the tube. A little concentrated sulphuric acid is added to the water in the beaker. The zinc metal reacted with sulphuric acid liberating hydrogen gas. The gas is collected by downward displacement of acid in the eudiometer tube. When the metal is completely dissolved in the acid and no more gas evolved, the tube is removed by closing the mouth of the tube with the thumb. The tube, the open end of which is closed with the thumb is then plunged inside a tall jar of water in an inverted position. The thumb is slowly released and the tube is held with a piece of paper in position. The tube is not held with hand, since the body temperrature will expand the gas. The tube is allowed to remain for sometime in the tall jar so that the gas in the tube is to attain the room temperature i.e. the temperature of water. Then the volume of the hydrogen gas collected is read out by adjusting the tube so that the surface of water level both inside and outside the tube became the same. The room temperature and the atmospheric pressure are noted. Since the gas is collected over water, the pressure exerted by dry hydrogen gas is less than the atomospheric pressure by an amount, the aqueous tension of water vapour. This aqueous tension of water vapour is also noted at room temperature. Reaction : Zn (s) + 2H2SO4 (aq) ® ZnSO4 (aq) + H2 (g) eudiometer tube Fig 2.6 Determination of equivalent mass of zinc

68 +2 CHEMISTRY (VOL. - I) (a) Calculation of Volume of hydrogen at NTP Let the weight of zinc taken be Wg. The volume of hydrogen gas collected at room temperature t0C and atmospheric pressure Pmm of mercury be V1cc. The aqueous tension due to water vapour at t0C is f mm of mercury. Therefore, the pressure exerted by dry hydrogen gas is (P – f) mm of mercury. Thus, V1cc of dry hydrogen gas is collected at t0C and under (P – f) mm of mercury pressure. This volume is converted to the volume Vcc at NTP. Here, P1 = (P – f), T1 = (t + 273), V1 = measured volume of gas or, PV =PT1PV1T11V1 1x T or, T P V= V= (P - f) x V1 x 273 (t + 273) x760 (b) Calculating equivalent mass : Method 1 Now Vcc of dry hydrogen gas at NTP is liberated from Wg of zinc. So, 11.2 litres i.e 11,200cc of hydrogen gas at NTP is liberated from = W x 11,200 g V of zinc. Hence, the equivalent weight of zinc = W x 11,200 V Method 2 1cc of hydrogen at NTP weighs 0.000089 g. So, Vcc of hydrogen at NTP weigh V x 0.000089 g. Now V x 0.000089 g of hydrogen is displaced by Wg of metal Hence, 1.008 g of hydrogen displaced by W x 1.008 g V x 0.000089 So, equivalent mass of metal = W x 1.008 V x 0.000089 NUMERICAL PROBLEMS Example 1 0.27 g of a metal on treatment with dilute H2SO4 produced 112 ml of hydrogen at NTP. What is the equivalent mass of the metal ?

ATOMS & MOLECULES 69 Solution : 112 ml of hydrogen gas at NTP is liberated from 0.27g of the metal. 11,200 ml of hydrogen gas is liberated from 0.27 x 11,200 = 27 112 Example 2 0.425g of a metal liberated 436.5 ml of moist hydrogen at 170C and 745.4 mm when reacted with HCl. Determine the equivalent mass of the metal. ( Aqueous tension at 170C is 14.4mm) Solution : Volume of moist hydrogen V1 = 436.5 ml Pressure P1 = pressure observed – aqueous tension = 745.4 – 14.4 = 731 mm Temperature T1 = 17 + 273 = 2900 A At NTP : volume = V, pressure, P = 760 mm, Temperature T = 2730 A Now, PV = P1V1 or, V= P1V1T T T1 T1P = 731 x 436.5 x 273 = 395.2 ml 290 x 760 1 ml of H2 at NTP weighs 0.000089 gm So, 395.2 ml of H2 at NTP weighs 0.000089 x 395.2 = 0.035 gm Now 0.035 gm of H2 displaced by 0.425 g of the metal. Therefore 1.008 gm of H2 displaced by 0.425 x 1.008 = 12.24 gm of metal. 0.035 Hence, The equivalent mass of the metal is 12.24. Determination of equivalent mass of a metal by indirect oxidation method. A metal is oxidised to its oxide either by heating the metal in air or by an indirect way. If a known amount of a metal is converted to its oxide, then the amount of the metal which was combined with 8g of oxygen can be calculated and that weight of metal is its equivalent mass. EQUIVALENT MASS OF COPPER BY INDIRECT OXIDATION METHOD The metal copper does not readily and quantitatively produce its oxide when heated directly in excess of air. Hence, a known amount of copper is treated with nitric acid to dissolve the metal. The copper nitrate solution so produced is heated to dryness. Thereafter this dry copper nitrate is strongly heated to decompose to cupric oxide. From the weight of the oxide, as has been said, the equivalent mass of copper is calculated. Experiment A porcelain crucible with its lid is cleaned, dried, heated for sometime, and then cooled in a desiccator. The weight of the crucible with its lid is determined. The procedure

70 +2 CHEMISTRY (VOL. - I) of heating, cooling and then weighing the crucible along with the lid is repeated till they weigh constant weight. Then a piece of copper wire weighing about 1g is placed in the crucible and weighed. Thus the weight of copper can be known. The crucible is kept nearly covered when concentrated nitric acid is added dropwise to react with copper wire. Brown fumes are evolved and reaction appears to be vigorous. Addition of nitric acid dropwise is continued until the metal dissolved completely. When no more brown fumes appear, the reaction has apparently ceased. Fig 2.7 Determination of equivalent mass by oxide formation method. Now the solution is slowly evaporated on water bath to dryness. Care is taken so that there is no loss of solution due to spurting. The residue so obtained, green in colour, is copper nitrate. Then the crucible with its lid is strongly heated on a clay pipe triangle when copper nitrate completely decomposes to cupic oxide, black in colour. Heating is then discontinued, the crucible is cooled in a desiccator, and its weight is determined. The procedure of heating, cooling and weighing along with cupric oxide is repeated till a constant weight is obtained. Reaction Cu(s) + 4 HNO3(aq) ® Cu(NO3)2 (aq) + 2NO2(g) + 2H2O(l) 2Cu(NO3)2(aq) ® 2CuO(s) + 4NO2(aq) + O2(g) Calculation Weight of crucible with lid = W1g (constant weight) (constant weight). Weight of crucible with lid and copper wire = W2g Weight of crucible with lid and black cupric oxide = W3g Thus, the weight of copper = (W2 – W1)g, the weight of cupric oxide = (W3 – W1)g

ATOMS & MOLECULES 71 Hence, the weight of oxygen = (W3 – W1) – (W2 – W1) = (W3 – W2) g So, (W3 – W2)g of oxygen combines with (W2 – W1)g of copper Hence, 8g of oxygen will combine with W2-W1 x 8g of copper. W3-W2 Therefore, equivalent weight of copper is 8 W2-W1 W3-W2 Relationship among atomic mass, equivalent mass and valency of an element. Consider an element X with its atomic mass A, equivalent mass E and valency V. Hydrogen being lightest its atomic mass is 1.008. Now, A parts by mass of the element will combine with V x 1.008 parts by mass of hydrogen. So, 1.008 parts by mass of hydrogen will combine with A/V parts by mass of the element and by defination this is equal to equivalent mass E of the element. Hence A = E or A= Ex V V Therefore Atomic mass = Equivalent mass x Valency NUMERICAL PROBLEMS Example 1 3.5 g of copper was dissolved in excess concentrated nitric acid and the solution was evaporated to dryness. The copper nitrate so formed was ignited to its oxide which weighed 4.4 g. Calculate the equivalent mass of copper. Solution : Weight of copper = 3.5 g Weight of copper oxide = 4.4 g Therefore, weight of oxygen = (4.4 – 3.5) = 0.9 g Hence, 0.9 g of oxygen combines with 3.5 g of copper So, 8 g of oxygen combines with 3.5 x 8 = 31.1g 0.9 Thus, Equivalent mass of copper is 31.1. Example 2 0.84g of a divalent metal on burning in air produced 1.4g of metal oxide. Solution : Calculate the equivalent mass and the relative atomic mass of the element. Weight of the metal = 0.84g Weight of the metal oxide = 1.4g Therefore, the weight of oxygen = (1.4 – 0.84) = 0.56g 0.56g of oxygen combines with 0.84g of the metal. So, 8g of oxygen will combine with 0.84 x 8 = 12g 0.56 Hence the equivalent mass of the metal = 12. Since the metal is divalent its valency is 2, and its relative atomic mass becomes 12 x 2 = 24.

72 +2 CHEMISTRY (VOL. - I) Equivalent mass of a compound Equivalent mass of a compound is a number which expresses what mass of the compound contain 1.008 parts by mass of hydrogen or 8.0 parts by mass of oxygen or 35.45 parts by mass of chlorine. Thus, 98 parts by mass of H2SO4 contain 2.016 parts by mass of hydrogen. So, 49 parts by mass of H2SO4 contain 1.008 parts by mass of hydrogen. Therefore, equivalent mass of H2SO4 is 49. In the case of NaCl, 58.45 parts by mass NaCl contain 35.45 parts by mass of chlorine. Therefore, the equivalent mass of NaCl is 58.45. Equivalent mass of a compound may be calculated by the expression, Molecular mass of the compound The total number of charges on the cations or anions NUMERICAL PROBLEMS Example 1 Calculate the equivalent mass of sodium carbonate. Solution : Molecular mass of Na2CO3 = 2 x 23 + 12 + 3 x 16 = 46 + 12 + 48 = 106g Total number of charges of cation i.e. sodium (two numbers of sodium ions each having one charge) = 2 Now, equivalent mass of Na2CO3 = 106 = 53. 2 Example 2 Calculate the equivalent mass of barium chloride. Solution : Molecular mass of BaCl2 = 137 + 2 x 35.45 = 207.9 » 208 There are one barium cation having charges 2 which becomes the total charges. Now, equivalent mass of BaCl2 = 208 = 104 2 Equivalent mass of metals 1. Chloride formation method A metal may directly combine with chlorine to form metal chloride. For example, silver reacts with chlorine to form silver chloride

ATOMS & MOLECULES 73 2Ag + Cl2 ® 2AgCl In this case the mass of metal and the mass of metal chloride are accurately determined and the equivalent mass of the metal is calculated as below. Equivalent mass of metal = Mass of metal x 35.45 Mass of chlorine Equivalent mass of silver can also be determined by completely dissolving silver metal in dilute nitric acid. The silver nitrate thus formed is treated with dil HCl to precipitate silver chloride. The weight of silver and weight of silver chloride are accurately measured and then the equivalent mass of silver is calculated. Weight of silver = x gram Weight of silver chloride = y gram Therefore, weight of chlorine = (y – x) g Thus, (y – x) g of chlorine combines with xg of silver Hence, 35.45 g of chlorine will combine with x x 35.45 g y-x Equivalent weight of silver = 35.45 x y-x Example 1 2.7 g of silver when heated with excess of chlorine resulted 3.586 g of silver chloride. Calculate the equivalent mass of silver. Solution : The mass of chlorine that combined with silver = 3.586 – 2.7 = 0.886 g So, Equivalent mass of silver = Mass of silver x 35.45 Example 2 Mass of chlorine = 2.7 x 35.45 = 108.03 0.886 1.62 g of silver was dissolved in nitric acid. The solution was then treated with hydrochloric acid to precipitate 2.15 g of silver chloride. Determine the equivalent weight of silver. Solution : Weight of silver = 1.62g Weight of silver chloride = 2.15g Therefore, weight of chlorine = (2.15 – 1.62) = 0.53 g 1.62 So, equivalent weight of silver = 0.53 x× 35.45 = 108.36 2. Metal to metal displacement method More electropositive metals like Zn, Fe, Al etc. displace less electropositive metals like Cu, Ag, Au etc, from their salt solutions.

74 +2 CHEMISTRY (VOL. - I) Zn + CuSO4 ® ZnSO4 + Cu In such methods, the masses of displacing metal and displaced metal take place in equivalent amounts and are in the ratio of their equivalent masses. Here, Mass of Zn = Equivalent mass of Zn Mass of Cu Equivalent Mass of Cu Example : 2.06 g of zinc displace 2.0 g of copper from copper sulphate solution. If the equivalent mass of Zn is 32.5, find out the equivalent mass of copper. Solution Mass of Zn = Equivalent mass of Zn Mass of Cu Equivalent mass of Cu or, 2.06 g = 32.5 2.00 g Equivalent mass of Cu So, Equivalent mass of Cu = 32.5 x 2.00 = 31.55 2.06 3. Double decomposition method In this method, a known amount of a water soluble salt such as NaCl or KCl solution is acidified with dilute nitric acid and treated with silver nitrate solution to precipitate AgCl. The weight of AgCl is accurately determined. NaCl(aq) + AgNO3(aq) ® AgCl(s) + NaNO3(aq) The following expression is used to calculate the equivalent mass of the metal. Weight of metal chloride = Eq.mass of metal + Eq.mass of chlorine Weight of silver chloride Eq.mass of silver + Eq.mass of chlorine Example : 0.2 g of a metal chloride dissolved in water was treated with excess of silver nitrate solution to precipitate 0.5g of silver chloride. Calculate the equivalent mass of the metal. (Equivalent mass of silver and that of chlorine are 108 and 35.45 respectively). Solution : Let the equivalent mass of the metal be x. Weight of metal chloride = 0.2 g Weight of silver chloride = 0.5 g Equivalent mass of silver = 108 Equivalent mass of chlorine = 35.45 Therefore, Equivalent mass of silver chloride is (108 + 35.45) = 143.45 Weight of metal chloride = Eq.mass of metal + Eq.mass of chlorine Weight of silver chloride Eq.mass of silver + Eq.mass of chlorine

ATOMS & MOLECULES 75 or, 0.2 = x + 35.45 = x + 35.45 0.5 108 + 35.45 1 4 3 .4 5 or, x + 35.45 = 143.45 x 0.2 = 57.4 0.5 or, x = 57.4 – 35.45 = 21.95 Hence the equivalent mass of the metal is 21.95 4. Reduction of the oxide method A known weight of a metal oxide is heated in a current of hydrogen gas to the corresponding metal. The equivalent mass of the metal can be known from the weight of metal oxide, weight of the metal and the weight of oxygen. CuO(s) + H2(g) ® Cu + H2O(g) 5. Decomposition of a metal carbonate into its oxide Metal carbonates such as NiCO3 and CuCO3 on heating produce their oxides. CuCO3(s) ® CuO(s) + CO2(g) The equivalent mass of such metals can be calculated with the help of the following expression. Weight of the metal carbonate = Equivalent mass of the metal carbonate Weight of the metal oxide Equivalent mass of the metal oxide = Equivalent mass of the metal + 30 Equivalent mass of the metal + 8 Equivalent mass of oxygen is 8 and that of carbonate ( CO 2- ) 3 = 12+16 X 3 = 30 2 Example : 1.05g of a metallic carbonate on heating produced 0.5g of the oxide. Calculate the equivalent mass of the metal. Solution : Weight of the metal carbonate = 1.05g Weight of the metal oxide = 0.5g Let the equivalent mass of the metal be x, Therefore, Weight of the metal carbonate = Equivalent mass of the metal carbonate or, Weight of the metal oxide Equivalent mass of the metal oxide 1.05 = x + 30 0.5 x+8

76 +2 CHEMISTRY (VOL. - I) or, x + 30 = 2.1 (x + 8) or, 1.1x = 13.2 or, x = 12 Hence, the equivalent mass of the metal is 12. Equivalent mass of nonmetals In determination of the equivalent mass of nonmetals like carbon and sulphur, a known amount of the nonmetal is burnt in excess of air. C + O2 ® CO2 The oxide of the nonmetal so formed is absorbed in an suitable solution. Here, CO2 is absorbed in causitc potash solution. CO2 + 2KOH ® K2CO3 + H2O There would be an increase in weight of the solution, and from this increase in weight, the weight of the nonmetal oxide can be determined. The equivalent mass of the nonmetal can be calculated from the known weight of the nonmetal, the weight of the metal oxide and the weight of oxygen. In the case of nonmetals such as chlorine, bromine etc, which react with hydrogen easily, the equivalent mass of the nonmetal can be calculated from the percentage composition of the compound of the nonmetal and hydrogen. Nonmetals like phosphorus which chemically combine with chlorine, the equivalent mass of the nonmetal can be determined from the percentage composition of the nonmetal chloride. CHAPTER (2.7) AT A GLANCE Equivalent mass The equivalent mass of an element is a number which expresses how many parts by mass of the element combines with or displaces 1.008 parts by mass of hydrogen or 8 parts by mass of oxygen or 35.45 parts by mass of chlorine. Equivalents One equivalent of any substance reacts with one equivalent of the other substance to produce one equivalent of each of the products. All substances combine or appear on the basis of an equivalent. Gram equivalent mass The equivalent mass expressed in grams is called gram equivalent mass.

ATOMS & MOLECULES 77 Solved Problems Problem 1 0.08 g of a metal produces 22.4ml of dry hydrogen at NTP. What is its equivalent mass ? Solution 22.4 ml of dry hydrogen at NTP was obtained from 0.08 g of metal. So 11,200ml of hydrogen gas is produced from 0.08 x 11,200 = 40 g 22.4 Therefore, the equivalent mass of the metal = 40 Problem 2 0.205 g of a metal on treatment with an acid gave 106.60 ml of moist hydrogen Solution collected at 755 mm pressure and 170C. Calculate the equivalent mass of the metal (Aqueous tension at 170C = 14.4 mm). Problem 3 Solution Volume of moist hydrogen 106.60 ml = V1 (say) Problem 4 pressure P1 = (755 – 14.4) = 740.6 Solution Temperature T1 = 17 + 273 = 2900A PV = P1V1 or V = P1V1T = 740.6 x 106.6 x 273 = 97.79 ml T T1 T1P 290 x 760 Now, 97.79 ml of hydrogen gas at NTP is liberated from acid by 0.205 g of metal. So, 11,200 ml of hydrogen gas at NTP is liberated by, 0.205 x 11,200 = 23.48 g of metal 97.79 Therefore, the equivalent mass of the metal is 23.48. 0.325 gm. of a metal when dissolved in dil H2SO4 liberated 0.01 gm of hydrogen. What is the equivalent weight of the metal ? (CHSE 1996 A) 0.01 gm of hydrogen is equivalent to 0.325 gm of the metal. Therefore 1.008 gm of hydrogen will be equivalent to 0.325 x 1.008 = 32.76 0.01 gm of the metal. Hence, the equivalent weight of the metal is 32.76. 1.62 g of a metal was dissolved in nitric acid to prepare its nitrate. On strong heating of the nitrate, 2.02 g of oxide was obtained. Find out the equivalent weight of the metal. Weight of the metal = 1.62 g Weight of its oxide = 2.02 g Hence, the weight of oxygen that combined with the metal = (2.02 – 1.62) = 0.40 g Now, 0.40 g of oxygen combined with 1.62 g of the metal So, 8.0 g of oxygen combines with 1.62 x 8 = 32.4 g of the metal. 0.40 Therefore, equivalent weight of the metal is 32.4.

78 +2 CHEMISTRY (VOL. - I) Problem 5 Calculate the equivalent mass of the metal whose oxide contains 35.2% oxygen. Solution Mass of oxide of the metal = 100 g Problem 6 Solution Mass of oxygen = 35.2 g Hence, the mass of the metal = (100 – 35.2) = 64.8 g. Now, 35.2 g of oxygen combines with 64.8 g of the metal. Therefore 8.0 g of oxygen will combine with 64.8 x 8 = 14.72 g of the metal. 35.2 So the equivalent mass of the metal is 14.72. 0.51 gm of a metal combines with oxygen to give 0.68 gm of metal oxide. Calculate the equivalent weight of the metal. (CHSE 1989) Weight of the metal = 0.51 gm Weight of the metal oxide = 0.68 gm The weight of oxygen = 0.68 – 0.51 = 0.17 gm The equivalent weight of the metal = 0.51 x 8 = 24 0.17 Problem 7 0.54 g. of a metal gave 0.7175 g of the metal chloride. Determine the equivalent Solution weight of the metal. Problem 8 The weight of combined chlorine = 0.7175 – 0.54 = 0.1775 g. Hence, equivalent weight of the metal = 0.54 x 35.5 = 108 0.1775 1.08 g of copper displace 3.67 g of silver from silver nitrate solution. find the equivalent mass of copper, that of silver being 108. Solution Here, Eq.mass of copper = Weight of copper Eq. mass of silver Weight of silver displaced Problem 9 Solution or, Eq.mass of Cu = 1.08 108 3.67 or, Eq.mass of copper = 1.08 x 108 = 31.78. 3.67 1.52 g of a metal hydroxide gave 0.995 g. of its oxide. Calculate the equivalent mass of the metal. Eq. mass of metal + Eq.mass of OH = 1.52 Eq. mass of metal + Eq.mass of O 0.995 (Eq mass of OH = 17 = 17 and eq.mass of O = 16 = 8) 1 2

ATOMS & MOLECULES 79 So, Eq. mass of metal + 17 = E + 17 = 1.52 Eq. mass of metal + 8 E+8 0.995 Solving , E = 9.06, and that is the equivalent mass of the metal. Problem 10 0.5 g of a metal combines with 142 ml of oxygen at NTP. Calculate the equivalent mass of the metal. Solution 22,400 ml of oxygen weigh 32 g Hence, 142 ml of oxygen weigh 32 x 142 = 0.20 g 22,400 Thus 0.20 g of oxygen combines with 0.5 g metal. Hence 8 g of oxygen combine with 0.5 x 8 = 20 g of the metal. 0.2 Therefore, the equivalent mass of the metal is 20. QUESTIONS A. Very short answer type (1 mark each) 1. Which method is used to determine the equivalent mass of zinc. ? 2. What mass of an element combines with 8 parts by mass of oxygen ? 3. How many gram-equivalents will be 40 g of Ca. ? 4. Which method is used to determine the equivalent mass of copper ? 5. What is the equivalent weight of NaOH ? 6. Does the equivalent mass of an element change with temperature ? 7. What is the equivalent mass of hydrogen ? 8. Atomic weight of a metal is 52 and its valency is 3. What is the equivalent weight of the metal ? 9. What is the equivalent mass of OH– ? 10. What equivalent of oxygen will measure 5.6 litres of oxygen at NTP ? B. Short answer type (2 marks each) 1. Two metals A and B have the same equivalent mass but the atomic mass of A is twice that of B. How can it be explained ? 2. What is the relationship between atomic mass and equivalent mass of oxygen ? 3. Define equivalent mass. 4. Explain gram equivalent mass.

80 +2 CHEMISTRY (VOL. - I) 5. What is the equivalent mass of sodium carbonate ? 6. Why cannot the equivalent mass of copper be determined by hydrogen displacement method ? 7. 74.5 g of a metal chloride contains 35.5 g of chlorine. What is the equivalent mass of the metal ? 8. Explain why the equivalent mass of copper is different in Cu2O and CuO. 9. X g of an element combines with oxygen to give Y g of its oxide. What is the equivalent mass of the element ? 10. Explain that iron shows variable equivalent weight in its compounds. C. Long answer type (7 marks each) 1. Define equivalent weight. How will you determine the equivalent weight of zinc by hydrogen displacement method ? 0.325 g of a metal when dissolved in dil H2SO4 liberated 0.01 g of hydrogen. What is the equivalent weight of the metal ? 2. Define equivalent mass of an element. How would you determine the equivalent mass of copper ? 3. Define equivalent mass. What is the difference between equivalent mass and gram equivalent ? 4. How will you determine the equivalent mass of zinc by hydrogen displacement method ? D. Numerical problems : 1. If 1.2 g of a metal displaces 1.12 litres of hydrogen at NTP, what would be the equivalent weight of the metal ? 2. 0.01 g of a metal when dissolved in HCl liberates 10.65 ml of hydrogen collected over water at 270C. The barometric pressure is 760 mm Hg. Aqueous tension at 270C is 25 mm of Hg. Calculate the equivalent weight of the metal. 3 2.943 g of a metal when treated with excess dil H2SO4 the volume of hydrogen collected was 1083.6 ml at 150C and 752.5 mm of pressure. Calculate the equivalent mass of the metal. (Aqueous tension at 150C = 12.5 mm) 4. 0.54 g of a metal gave 0.90 g of its oxide. Calculate the equivalent mass of the metal. 5. An oxide of a metal contains 52.9% metal. Calculate its equivalent mass. 6. 4.74 g of cupric oxide on being heated in a current of hydrogen gave 3.78 g of the metal. Calculate the equivalent mass of copper. 7. The chloride of a metal contained 52.85% of metal. What is the equivalent mass of the metal ?

ATOMS & MOLECULES 81 8. 0.13 g of aluminium displaces 0.47 g of copper from copper sulphate solution. The equivalent mass of copper is 31.75. Calculate the equivalent mass of aluminium. 9. 0.5054 g of a metal nitrate is converted into 0.373 g of its chloride. The equivalent weight of nitrate group (NO - ) is 62. Calculate the equivalent weight of the metal. 3 10. What mass of the metal of equivalent mass 12 will give 0.475 g of its chloride ? ANSWERS A. 1. Hydrogen displacement method 2. Equivalent mass 3. 2 4. Oxide formation method 5. 40 6. No 7. 1.008 8. 17.33 9. 17 10. 1 B. 1. Valency of metal A is double that of valency of B. 2. Atomic mass of oxygen = Equivalent mass of oxygen x Valency of oxygen (i.e.2) 5. 53. 6. Copper cannot displace H2 from an acid. 7. 39 9. 8x y-x C. 1. 32.76 (Ref : solved problem No.3) D. 1. 12 2. 12 3. 32.2 4. 12 5. 8.98 6. 31.5 7. 39.78 8. 8.8 9. 39.1 10. 0.12 g 2.8 EQUIVALENT MASSES OF ACIDS, BASES AND SALTS (a) Equivalent mass of acid : Equivalent mass of an acid is a number which shows how many parts by mass of the acid contain one part by mass of replaceable hydrogen. Molecular mass of HCl = 36.5 (At. mass of H=1, Cl = 35.5). Thus, 36.5 parts by mass of HCl contain 1 part by mass of replaceable hydrogen. Hence, equivalent mass of HCl is 36.5. Molecular mass of H2SO4 = 98 (At mass of H = 1, S = 32, O = 16). Thus, 98 parts by m=as9s28of=H429S.O4 contain 2 parts by mass of replaceable hydrogen. Hence, equivalent mass of H2SO4 Basicity of the acid is the number of replaceable hydrogen atoms it contains. Therefore, Equivalent mass of the acid = Molecular mass of the acid Basicity of the acid [Note : Acetic acid (CH3COOH) contains four hydrogen atoms in all, but the number of replaceable hydrogen in it is only one. Therefore the equivalent mass of acetic acid is its molecular mass ].

82 +2 CHEMISTRY (VOL. - I) Table 2.5 Equivalent mass of some acids. Acid Formula Basicity Molecular mass Equivalent mass. Hydrochloric HCl 1 36.5 36.5 = 36.5 Nitric 1 63.0 1 Sulphuric HNO3 2 98.0 Phosphoric 3 98.0 63.0 = 63.0 Acetic H2SO4 1 60.0 1 Oxalic (anhydrous) H3PO4 2 90.0 98.0 = 49.0 2 CH3COOH COOH 98.0 = 32.66 I 3 COOH 60.0 = 60.0 1 90.0 = 45.0 2 Oxalic COOH. 2 126.0 126.0 = 63.0 (hydrated) 2 I .2H2O COOH (b) Equivalent mass of base : Equivalent mass of a base is a number which shows how many parts by mass of a base can just be neutralised by one equivalent mass of an acid. Consider the reaction NaOH + HCl ® NaCl + H2O (23+16+1) (1+35.5) Thus, 40 parts by mass of NaOH is completely neutralised by 36.5 parts by mass of HCl i.e. one equivalent mass of HCl. Therefore, the equivalent mass of NaOH is 40. Na2CO3 + H2SO4 ® Na2SO4 + H2O + CO2 2+32+(16×4) (23×2)+12+(16×3) Thus, 106 parts by mass of Na2CO3 is completely neutralised by 98 parts by mass of i.e. two equivalent mass of H2SO4. Therefore, equivalent mass of 106 H2SO4 Na2CO3 = 2 = 53. The number of moles of monobasic acid which can completely neutralise one mole of the base determines the acidity of the base. Acidity of the base can also be defined as the number of replaceable hydroxyl group present in one molecule of the base. NaOH + HCl ® NaCl + H2O (Acidity of NaOH = 1) (1 mole) (1 mole) Ba(OH)2 + 2HCl ® BaCl2 + 2H2O (Acidity of Ba(OH)2 = 2) (1 mole) (2 moles) Na2CO3 + 2HCl ® 2 NaCl + CO2 + H2O (Acidity of Na2CO3 = 2) (1 mole) (2 moles)

ATOMS & MOLECULES 83 Hence, equivalent mass of the base = Molecular mass of the base Acidity of the base Table 2.6 Equivalent mass of some bases. Base Formula Acidity Molecular mass Equivalent mass NaOH 1 40 Sodium 40 = 40 hydroxide 1 Potassium KOH 1 56 56 = 56 hydroxide 1 Calcium Ca(OH)2 2 74 74 = 37 hydroxide 2 Barium Ba(OH)2 2 171 171 = 85.5 hydroxide 2 Aluminium Al(OH)3 3 78 78 = 26 hydroxide 3 (c) Equivalent mass of salt :- Equivalent mass of a salt is a number which shows how many parts by mass of the salt contain one part by mass of hydrogen equivalent of an active element of a positive group. NaCl is a salt whose molecular mass 58.5, 58.5 parts by mass of NaCl contain 23 parts by mass of Na which in turn is equivalent to one part by mass of hydrogen. So, 58.5 parts by mass of NaCl contain one equivalent mass of the metal and equivalent mass of NaCl is 58.5. BaCl2 is another salt whose molecular mass is 208. 208 parts by mass of BaCl2 contain 137 parts by mass of barium which in turn is equivalent of two parts by mass of hydrogen. So, 208 parts by mass of BaCl2 contain two equivalent mass of the metal and equivalent mass of BaCl2 = 208 2 =104. Molecular mass of the salt In general, Equivalent mass of the salt = Total valency of the metal ion. Table : 2.7 Equivalent mass of some simple salts. Salt Formula Mol. mass Total Valency Equivalent of the salt of the metal ion mass Sodium sulphate Na2SO4 142 2 71 Ferrous chloride FeCl2 127 2 63.5 Ferric chloride FeCl3 162.5 3 54.16 Aluminium chloride AlCl3 133.5 3 44.5

84 +2 CHEMISTRY (VOL. - I) Gram-Equivalent Mass : Equivalent mass expressed in grams is called gram - equivalent mass, written as gm - equivalent. Thus, one gram - equivalent of HCl = 36.5g, one gram - equivalent of Na2CO3 is 53g, one gram - equivalent of H2SO4 is 49 g. and so on. CHAPTER (2.8) AT A GLANCE 1. Equivalent mass of acid : How many parts by mass of the acid contain one part by mass of replaceable hydrogen. Number of repalceable hydrogen = Basicity. 2. Equivalent mass of the base : How many parts by mass of a base can just be neutralised by one equivalent mass of an acid. Number of moles of monobasic acid neutralising one mole of the base completely = Acidity = Number of replaceable hydroxyl group present in one molecule of the base. 3. Equivalent mass of the salt : How many parts by mass of the salt contain one part by mass of hydrogen equivalent of an active element of a positive group. Also molecular mass of the salt divided by the total valency of the metal ions. 4. Gram - equivalent mass. Equivalent mass expressed in grams. 2.9 PERCENTAGE COMPOSITION, MOLECULAR FORMULA AND EMPIRICAL FORMULA Calculation of percentage composition Percentage of composition is denoted as the member of parts by mass of each element present in hundred parts by mass of a compound. Percentage of the element = Number of parts by mass of the element x 100 Molecular mass Example - 1 Calculate the percentage composition of calcium carbonate. (At mass : Ca = 40, C = 12, O = 16) Solution : Formula of calcium carbonate is CaCO3 Therefore its molecular mass = 1 x 40 + 12 + 3 x 16 = 100 Hence, % of Ca = 40 x 100 = 40 100 % of C = 12 x 100 = 12 100 % of O = 48 x 100 = 48 100

ATOMS & MOLECULES 85 Example - 2 Calculate the percentage composition of ferrous sulphate crystal. (Fe = 56, S = 32, O = 16, H = 1) Formula of ferrous sulphate crystal is FeSO4 7H2O. Solution : Formula of ferrous sulphate crystal is FeSO4 7H2O. Therefore, its molecular mass = 56 + 32 + 4 x 16 + 7(1 x 2 + 16) = 56 + 32 + 64 + 126 = 278 Hence , % of Fe = 56 x 100 = 20.14 278 % of S = 32 x 100 = 11.51 278 % of O = 64 x 100 = 23.02 278 % of H2O = 126 x 100 = 45.33 278 Example - 3 Calculate the percentage of composition of sodium phosphate. (Na = 23, P = 31, 0=16) Solution : Formula of sodium phosphate is Na3PO4 Therefore its molecular mass is 3 x 23 + 31 + 4 x 16 = 69 + 31 + 64 = 164 164 parts of Na3PO4 contains 69 parts of Na, Hence % of Na = 69 x 100 = 42.07 164 % of P = 31 x 100 = 18.90 164 % of O = 64 x 100 = 39.02 164 Calculation of mass of element present in a given mass of subtance : Example - 1 How many grams of sodium are present in 5.85 grams of sodium chloride ? (Na = 23, Cl = 35.5) Solution : Molecular mass of sodium chloride NaCl = 23 + 35.5 = 58.5 58.5g of NaCl contain 23g of sodium So 5.85gm of NaCl will contain 23 x 5.85 = 2.3g of sodium 58.5

86 +2 CHEMISTRY (VOL. - I) Example - 2 How many grams of sulphur are present in 6.0 g of magnesium sulphate ? (Mg = 24, S = 32, O =16) Solution : Molecular mass of magnesium sulphate, MgSO4 = 24 + 32 + (4 x 16) = 120 120g of MgSO4 Contain 32g of sulphur Hence, 6.0g of MgSO4 contain 32 x 6.0 = 1.6g of sulphur 120 MOLECULAR FORMULA AND EMPIRICAL FORMULA A formula is a ratio of number of atoms of each element present in the compound. Molecular formula : The molecular formula shows the exact number of atoms of each element present in one molecule of the compound. Empirical formula : The empirical formula shows the simplest whole number ratio of atoms of each element present in one molecule of the compound. It is the simplest formula which gives the atomic ratio in terms of smallest whole numbers of each type of atom present. If the percentage composition of a compound is determined, then its simplest formula can be computed. Relationship between molecular formula and empirical formula. The molecular formula of hydrogen peroxide is H2O2. Its empirical formula is HO. The molecular formula of acetic acid is CH3COOH or C2H4O2 and its empirical formula is CH2O. The molecular formula of glucose is C6H12O6 and its empirical formula is CH2O. From the above examples, we arrive at a conclusion for a relationship between molecular formula and empirical formula and that is Molecular formula = (Empirical formula)n. Where, the value of n is a whole number. For H2O2 the value of n = 2, for CH3COOH, n = 2, and for glucose C6H12O6, n = 6. For some compounds the molecular formula is the same as its empirical fomula. Both formulae for hydrogen chloride is HCl and have the value of n = 1. Determination of empirical formula and molecular formula. This is determined from percentage composition. The following steps are followed in determining the empirical formula. 1. The percentage composition of each element is divided by relative atomic mass to get the relative number of atoms present in the compound. 2. The quotients from (1) are divided by the smallest quotient to get the simplest ratio of atoms of different elements constituting the molecule of a compound. If the above ratio of atoms is not a whole number ratio, then this ratio is multiplied by an integer to convert the same to a whole number ratio.

ATOMS & MOLECULES 87 3. Once the whole number ratio is determined it is possible to write the empirical formula. The molecular formula sometimes may be same as the emipirical formula or it may be a simple multiple of the empirical formula. Molecular formula = (Empirical formula)n. The value of n is a whole number and is determined by dividing the molecular mass of the compound by its empirical formula mass. n= Molecular mass Empirical formula mass Example - 1 Calculate the empirical formula of a compound that contains 22.22% nitrogen, 1.59 % hydrogen and the rest oxygen. Solution : The % of oxygen = [100 – (22.2 + 1.59)] = 76.19 Element Percentage Atomic mass Relative Simplest Whole number composition number of atoms ratio ratio. H 1.59 1.008 or 1 1.59/1 1.59 1 N 22.22 14 22.22/14 1.59 1 O 76.0 16 76.19/16 4.76 3 So, the compound containing H, N, and O atoms with whole number ratio 1, 1 and 3 respectively, its empirical formula is HNO3. As we know the molecular formula of nitric acid is HNO3, so here the value of n = 1 and the molecular formula and empirical formula of nitric acid is the same, that is HNO3. Example - 2 A gaseous compound of carbon and nitrogen, containing 53.8% by weight of nitrogen was found to have vapour density of 25.8. What is the molecular formula of the compound ? (IIT. 1971) Solution : Element Percentage Atomic Relative number Simplest Composition mass of atoms ratio N 53.8 14 53.8 = 3.84 3.84 =1 14 3.84 C 100 – 53.8 = 46.2 12 46.2 = 3.85 3.85 =1 12 3.84 Hence the empirical fomula of the gaseous compound is CN. Now the empirical formula mass is 12 + 14 = 26. We shall learn later in this chapter that molecular mass of a compound is equal to twice its vapour density. Since the vapour density of the above compound is 25.8, the molecular mass is 2 x 25.8 = 51.6. We Know, n = Molecular mass = 51.6 = 1.98 » 2 Empirical formula mass 26

88 +2 CHEMISTRY (VOL. - I) Thus the molecular formula of the gaseous compound is (CN)2 or C2N2. Percentage of oxygen in an organic compound In the analysis of elements of an organic compound, the elements carbon, hydrogen, nitrogen, sulphur and halogen can be estimated by chemical methods. But the element oxygen cannot be estimated directly. So the percentage of oxygen is obtained by subtracting the sum of percentage of other elements from 100. Therefore, percentage of oxygen = 100 – (Sum of percentage of other elements) Example - 3 An organic compound contains 40% carbon and 6.66% hydrogen and rest oxygen. The vapour density of the compound is 30. Calculate its empirical and molecular formula. Solution : Element Percentage Atomic Relative number of Simplest ratio Composition mass atoms C 40 12 40 = 3.33 3.33 = 1 12 3.33 H 6.66 1.008 6.66 = 6.61 6.61 = 1.98 » 2 1.008 3.33 O 100 –(40 + 6.66) 16 53.34 = 3.33 3.33 = 1 16 3.33 = 53. 34 Hence the empirical formula is CH2O. The molecular formula would be thus (CH2O)n. The empirical formula mass = 12 + 2 x 1.008 + 16 = 30.016 » 30. Given the vapour density of the compound 30. So the moleculr mass is 2 x 30 = 60 Therefore, n= 60 =2 30 So the molecular formula of the compound is (CH2O)2 or C2H4O2. Example - 4 A compound on analysis was found to contain the following percentage composition Fe = 20.14% ; S = 11.51% ; O = 63.3% and H = 5.08%. The molecular mass of the substance is 278. Determine the molecular formula assuming that all the hydrogen atoms are present with oxygen as water of cystallisation. (At. mass Fe = 56, S =32, O = 16, H = 1.008).

ATOMS & MOLECULES 89 Element Percentage Atomic Relative Simplest Composition mass number of atoms ratio Fe 20.14 56 20.14 = 0.36 0.36 =1 56 0.36 S 11.51 32 11.51 = 0.36 0.36 =1 32 0.36 O 63.30 16 63.30 = 3.96 3.96 = 11 16 0.36 5.08 5.03 H 5.08 1.008 1.008 = 5.03 0.36 = 14 Thus, the empirical formula of the compound is FeSO11H14 The empirical formula mass = 56 + 32 + (11 x 16) + (14 x 1) = 278 Given, molecular mass = 278 So n= Molecular mass = 278 =1 Empirical formula mass 278 Hence, the molecular formula is FeSO11H14 In this case we know, all the fourteen hydrogen atoms combine with seven oxygen atoms to form seven molecules of water. So, the molecular formula is to be written as FeSO4. 7H2O. CHAPTER (2.9) AT A GLANCE 1. Percentage composition : Percentage of composition is denoted as the number of parts of each element present in hundred parts by mass of a compound. 2. Molecular formula : The molecular formula shows the exact number of atoms of each element present in one molecule of the compound. 3. Empirical formula : The empirical formula shows the simplest whole number ratio of atoms of each element present in one molecule of the compound. QUESTIONS Long answer type 1. What do you understand by empirical and molecular formula of a compound ? How are they related ? An organic compound contains C = 62.15% and H = 10.34%. Its vapour density is 29. Find out the empirical and molecular formula of the compound. (Ans :CH2, C4H8) 2. What do you understand by empirical and molecular formulae of a compound ? How are they related ? A compound contains carbon, hydrogen and nitrogen in the ratio 18:2:7. Calculate its empirical formula. If the molecular mass of the compund in 108, what is its molecular formula ? (Ans : C3H4N, C6H8N2)

90 +2 CHEMISTRY (VOL. - I) NUMERICAL PROBLEMS 1. What is the molecular mass of : (a) Hydrochloric acid (b) Sodium hydroxide (c) Sodium carbonate (d) BCaC=l212. ,2OH2O= (e) Barium sulphate. (f) Aluminium oxide. ( At mass, Cl = 35.5, Na = 23, 16, Ba = 137, S = 32, Al = 27) 2. Calculate the percentage composition of : (a) sodium chloride (b) Sodium sulphate 3. A given sample of chalk contains 55% by mass of calcium carbonate. Calcualte the percentage of calcium, carbon and oxygen in the sample. 4. A compound on analysis gave the following percentage composition. Na = 14.31%, S = 9.97%, H = 6.22%, O = 69.5%. The molecular mass of the substance is 322. Find its molecular formula assuming that all the hydrogen atoms are present in combination with oxygen as water of crystallisation (At. mass Na = 23, H = 1, S = 32, O = 16) 5. An organic compound contains 52.18% carbon and 13.04% hydrogen, the remaing being oxygen. Vapour density of the compound is 23. Calculate the molecular formula. 6. In case of a hydrocarbon, it was found that the hydrogen content is 1 of that of carbon by weight. What is the empirical formula of the compound ? 12 7. A compound with molecular weight 108, contained 88.89% C and 11.11% H. Calculate its molecular formula. ANSWERS TO NUMERICAL PROBLEMS 1. (a) 36.5 (b) 40 (c) 106 (d) 244 (e) 233 (f) 102 2. (a) Na = 39.3%, Cl = 60.7% (b) Na = 32.4%, S = 22.5% O = 45.1% 3. Ca = 22%, C = 6.6%, O = 26.4% 4. Na2SO4.10H2O 5. C2H6O 6. CH 7. C8H12 2.10 AVOGADRO'S HYPOTHESIS AND THE MOLE CONCEPT According to Dalton, when elements combine, they do so in simple ratios by atoms. We also know from Gay Lussac's law that when gases combine, they do so in simple ratios by volume. J. J. Berzelius, a Swedish chemist attempted to correlate Dalton's atomic theory and Gay Lussac's law of gaseous volumes with the idea that there must be some relationship between the volume of a gas and the number of atoms present. Berzelius put forward a relationship called as Berzelius hypothesis.

ATOMS & MOLECULES 91 Berzelius hypothesis : Equal volumes of all gases under similar conditions of temperature and pressure contain equal number of atoms. Berzelius hypothesis when applied to some gaseous reactions failed to explain Dalton's atomic theory. Example : It was observed experimentally that for a reaction, H2 + Cl2 ® 2HCl 1 vol 1 vol 2vol 1 volume of hydrogen combines with 1 volume of chlorine to produce 2 volumes of hydrogen chloride, the volumes measured under similar conditions of temperature and pressure. Let 1 volume of a gas contain 'n' atoms and applying Berzelius hypothesis, n atoms of hydrogen combine with n atoms of chlorine to form 2n atoms of hydrogen chloride. or 1 atom of hydrogen combines with 1 atom of chlorine to form 2 atoms of hydrogen chlorinde. or, 1 atom of hydrogen chloride contains ½ atom of hydrogen and ½ atom of chlorine. The above statement contradicts Dallton's atomic theory that says atoms is indivisible. In 1811 Italian chemist, Amedeo Avogadro modified Berzelius hypothesis by distinguishing atoms and molecules. Atom : An atom is the smallest particle of an element which takes part in a chemical change. An atom may or may not have independent existence. Molecule : A molecule is defined as the smallest particle of a compound which contains atoms of different kind. In case of an element, the smallest particle may be a molecule composed of atoms of the same kind. The molecule is capable of existing in the free state and retaining the properties of matter. In a chemical reaction between molecules reorganisation of atoms occurs. The reactions between gases may be understood in terms of Avogadro's hypothesis. AVOGADRO'S HYPOTHESIS Equal volumes of all gases at the same temperature and pressure contain equal number of molecules. Avogadro's hypothesis explains correctly the reaction between hydrogen and oxygen forming water vapour. 2H2 + O2 ® 2H2O 2vol 1vol 2vol

92 +2 CHEMISTRY (VOL. - I) 2 volumes of hydrogen gas combine with 1 volume of oxygen gas to form 2 volumes of water vapour measured under same condition of temperature and pressure. Let there be n number of molecules present in 1 volume of hydrogen. Thus there are 2n molecules present in 2 volumes of hydrogen. According to Avogadro's hypothesis, there would be n molecules of oxygen in 1 volume of oxygen gas and 2n molecules of water in 2 volumes of water vapour. Thus, 2n molecules of hydrogen combine with n molecules of oxygen to form 2n molecules of water vapour. 2 molecules of hydrogen combine with 1 molecule of oxygen to form 2 molecules of water vapour. Therefore, 1 molecule of hydrogen combines with 1/2 molecule of oxygen to form 1 molecule of water vapour. According to Dalton's atomic theory atoms connot be divided in a chemical reaction, but molecules are divisible. If ½ molecule of oxygen is regarded as 1 atom of oxygen making oxygen atom diatomic, then 1 molecule of hydrogen combines with 1 atom of oxygen to form 1 molecule of water vapour. This deduction from Avogadro's hypothesis does not contradict Dalton's atomic theory. Advantages of Avogadro's hypothesis 1. It could explain the difference between an atom and a molecule. 2. It proved that molecules are the ultimate particles of the elementary gases. 3. It explained correctly the law of gaseous volumes. APPLICATIONS OF AVOGADRO'S HYPOTHESIS : 1. Explanation of Gay Lussac's law of gaseous volumes In the chemical combination of gases A and B to form a gaseous compound AB, Let 'a' molecules of gas A combine with 'b' molecules of gas B (a & b are simple integers). Under the same condition of temperature and pressure, If 1 cc of gas A contains n molecules then 1 cc of gas B would also contain the same number of n molecules. This is as per Avogadro's hypothesis. Since 1 cc of the gas A contains n molecules then 'a' molecules of the gas A would be present in ( an ) cc. And similarly 'b' molecules of the gas 'B' would be present in ( nb ) cc. Thus, the volumes of gases A and B will combine in the ratio : i.e. a : b (a simple ratio, since a and b are whole members), when the volumes are measured at the same temperature and pressure. Avogadro's hypothesis thus, explains the law of gaseous volumes.

ATOMS & MOLECULES 93 2. Determination of atomicity of elementary gases Atomicity means the number of atoms present in a molecule. (a) Atomicity of oxygen molecule The fact that a molecule of oxygen is diatomic will be deduced from Avogadro's hypothesis. It has been already stated that 2 volumes of hydrogen combine with 1 volume of oxygen to produce 2 volumes of water vapour at the same temperature and pressure. It is also deduced by considering n molecules being present in 1 volume of hydrogen that 2 molecules of hydrogen combine with 1 molecule of oxygen to form 2 molecules of water vapour. And hence, 1 molecule of hydrogen combines with ½ molecule of oxygen to form 1 molecule of water vapour. Since water is a compound of hydrogen and oxygen only, each molecule of water must contain at least 1 atom of hydrogen and 1 atom of oxygen. This 1 atom of oxygen in water must have come from ½ molecule of oxygen gas. Now since 1 atom of oxygen is present in ½ molecule of oxygen, the oxygen molecule should be diatomic. (b) Atomicity of hydrogen and chlorine molecules Similarly from the gaseous combination of hydrogen and chlorine, 1 volume of hydrogen combines with 1 volume of chlorine to give 2 volumes of hydrogen chloride gas, volumes measured at the same temperature and pressure. It can be deduced with the help of Avogadro's hypothesis that hydrogen and chlorine molecules are diatomic. (3) Determination of the relationship between molecular mass and vapour density The vapour density of a gas or vapour is defined as the ratio of the mass of a definite volume of the gas or vapour to the mass of the same volume of hydrogen, both the volumes measured at the same temperature and pressure. Vapour density of a gas or vapour = Mass of a definite volume of gas or vapour (at the same temperature and pressure) Mass of the same volume of hydrogen The vapour density of a gas or vapour can be measured experimentally. Let Vcc of a gas or vapour contain n molcules.

94 +2 CHEMISTRY (VOL. - I) Vapour density (V.D) = Mass of Vcc of the gas or vapour Mass of Vcc of hydrogen (at the same temperature and pressure) According to Avogadro's hypothesis, Vcc of hydrogen gas under the same conditions of temperature and pressure would also contain n molecules of hydrogen. Therefore, V. D. = Mass of n molcules of the gas or vapour Mass of n molecules of hydrogen = Mass of 1 molcule of the gas or vapour Mass of 1 molecule of hydrogen Since hydrogen molecule is diatomic (this fact has also been deduced from Avogadro's hypothesis), V. D. = Mass of 1 molecule of the gas or vapour Mass of 2 atoms of hydrogen So, V. D. x 2 = Mass of 1 molecule of the gas or vapour Mass of 1 atom of hydrogen = Relative molecular mass of the gas or vapour. Thus, the relative molecular mass of a gas or vapour is twice its vapour density. The relative molecular mass of gas or vapour = 2 x V.D. The atomic mass of hydrogen is 1.008 on the basis of atomic mass scale. In order to calculate the exact relation, the relative molecular mass of a gas or vapour is equal to 2.016 times its vapour density, i.e., Relative molecular mass of a gas or vapour = 2.016 x Vapour density of the gas or vapour. 4. Determination of Gram molecular mass or Molar mass and Gram molecular volume or Molar volume. Gram molecular mass of any substance is the molecular mass expressed in grams. According to Avogadro's hypothesis, Vapour density = Mass of 1 litre of gas at NTP Mass of 1 litre of hydrogen at NTP = Mass of 1 litre of gas at NTP 0.089 gram (Since 1 litre of hydrogen weighs 0.089 gram)

ATOMS & MOLECULES 95 Now, Vapour density = Mass of 1 litre of gas at NTP 0.089 gram Since, molecular mass or gram molecular mass=2 x V.D = 2 x mass of 1 litre of gas at NTP 0.089 gram = Mass of 22.4 litres of the gas at NTP. Thus, gram molecular mass of any gas or vapour occupies 22.4 litres at NTP which is called the gram molecular volume or molar volume. 5. Determination of molecular formulae of gases Molecular formula of a gas can be determined from its volumetric composition. For example, it has been found from experiment that. 1 volume of nitrous oxide contains 1 volume of nitrogen gas. If n number of molecules occupy 1 volume of nitrous oxide, according to Avogadro's hypothesis, 1 volume of nitrogen will also contain n molecules. Thus, n molecules of nitrous oxide contain n molecules of nitrogen. So, 1 molecule of nitrous oxide will contain 1 molecule of nitrogen. (It can be deduced from Avogadro's hypothesis that nitrogen molecule is diatomic and hence its atomicity is 2) Therefore, 1 molecule of nitrous oxide contains 2 atoms of nitrogen. Thus, its formula may be written as N2Ox , x being the number of oxygen atoms present in a molecule of nitrous oxide. The vapour density of nitrous oxide has been determined from experiment to be 22. Again from Avogadro's hypothesis, the relative molecular mass of a gas or vapour = 2 x Vapour density Hence, the relative molecular mass of nitrous oxide gas is 2 x 22 = 44. Again, the relative molecular mass of a compound is the sum of the relative atomic masses of all constituent elements. So, (2 x 14 + 16x) = 44 (The relative atomic masses of nitrogen and oxygen are 14 and 16 respectively) Therefore, x = 1 and the molecular formula of nitrous oxide is N2O. 6. Determination of atomic mass of elementary gases Avogadro's hypothesis can be applied to determine the atomic mass of elementary gases Molecular mass = 2 x Vapour density

96 +2 CHEMISTRY (VOL. - I) Again , Molecular mass = atomic mass x atomicity Therefore, atomic mass x atomicity = 2 x Vapour density and hence, atomic mass = 2 x Vapour density atomicity 7. Explanation of Dalton's atomic theory According to Avogadro's hypothesis molecules are divisible whereas atoms are not. This distinguishes between atoms and molecules and explains Dalton's atomic theory. AVOGADRO'S NUMBER AND MOLE CONCEPT The basic unit of substances in different types of chemical problems is an atom, ion or a molecule. While dealing with a definite and macroscopic quantity of substance the number of atoms, ions or molecules which compose the substance will have to be counted. Chemists have devised the unit mole to express the number of atoms, ions or molecules. A mole is the quantity of a substance that contains the same number of atoms, ions or molecules as are contained in 12g of carbon - 12 (12C). The number of atoms of carbon in 12g of carbon -12 has been found to be 6.023 x 1023. Thus, 1 mole of any substance is that quantity of the substance which contains 6.023 x 1023 particles or units of that substance. This number is called Avogadro's number (N) or (L). One gram-atom of an element is the mass of 6.023 x 1023 atoms of the element. For example, the gram-atomic mass of sodium is 22.9898 g. The mass of Avogadro number of sodium atoms is 22.9898g. Hence, the mass of 1 atom of sodium = 22.9898 = 3.82 x 10–23g 6.023 x 1023 Thus, mass of one atom of the element = Gram-atomic mass N Similarly, the gram-molecular mass of hydrogen is 2.016 g. The mass of 1 mole of hydrogen molecule is 2.016g and that contains Avogadro number of hydrogen molecules. Hence, the mass of 1 molecule of hydrogen = 2.016 = 3.85 x 10–24g 6.023 x 1023 Thus, mass of one molecule of the substance = Gram molecular mass N The quantity 'mole' is written in an abbreviated form as 'mol'.

ATOMS & MOLECULES 97 Again, the number atoms and number molecules in a given mass of element can be calculated. Number of atoms = Mass of substance ´ N Gram atomic mass Number of molecules = Mass of substance x N Gram molecular mass Molar Volume The gram-molecular volume or molar volume of all gases is the same at the same temperature and pressure and is 22.4 litres at NTP. (I litre = 1000 cm3 = (10cm)3 = 1 dm3) It is known from experiment that 1 litre of hydrogen gas at NTP weighs 0.089g, the molar mass of hydrogen being 2.016g, 2.016 g of hydrogen occupy 2.016 = 22.4 litres at NTP. This volume is called 0.0899 gram – molecular volume or molar volume of hydrogen. This means that one mole of hydrogen molecules occupies 22.4 litres at NTP. According to Avogadro's hypothesis , if we consider any other gas such as oxygen or carbon dioxide, 22.4 litres at NTP of any gas would contain the same number of molecules as are present in 22.4 litres of hydrogen gas at NTP. Now we know that one mole of molecules of hydrogen occupy 22.4 litres at NTP. Hence I mole of molecules of any gas occupies 22.4 litres at NTP. Again the molar volume of any gas weighs its molar mass. Thus, the molar volume or gram - molecular volume of any gas is the volume occupied by 1 mole of molecules of the gas at NTP and equals to 22.4 litres. The molar volume weighs the molar mass. Hence, O2 N2 CO2 Cl2 Molar mass (g) 32 28 44 71 Molar volume (litres at NTP) 22.4 22.4 22.4 22.4 NUMERICAL PROBLEMS Example 1 Calculate the number of molecules of carbon dioxide present in 500 ml at NTP. Solution The number of molecules present in a given volume of a gas at NTP = Volume of gas in litres x N 22.4 litres Volume of gas = 500 ml. = 0.5l

98 +2 CHEMISTRY (VOL. - I) Therefore, the number of molecules = 0.5l x 6.023 x 1023 = 1.344 x 1022 molecules. 22.4l Example 2 2.8 litres of a gas at NTP weighs 4 grams. Calculate the molecular mass of the gas. Solution Molecular mass of the gas = Mass of the gas ing (W) x 22.4l = W x 22.4 Example 3 V Volume of gas in litres (V) Hence, The molecular mass = W x 22.4 = 4 x 22.4 = 32 V 2.8 Calculate the density of chlorine gas at NTP. Solution Density of the gas = Gram molecular mass of the gas Molar volume (22.4l) Gram molecular mass of chlorine = 71 g. Hence, Density of chlorine gas = 71 = 3.17 gm/litre = 0.00317 gm/ml. 22.4 Mole concept and chemical equations The chemical equation for the reaction of hydrogen and oxygen forming water may also be interpreted in terms of moles of reactants and products. 2H2(g) + O2(g) ® 2H2O(l) This equation also represents that 2 moles of hydrogen react with 1 mole of oxygen to form 2 moles of water, i.e., 2(6.023 x 1023) molecules of hydrogen combine with (6.023 x 1023) molecules of oxygen to produce 2(6.023 x 1023) molecules of water. Since the weight of 1 mole in grams is the molar mass, the above fact implies that 4g of hydrogen reacts with 32g of oxygen to produce 36g of water. 2H2(g) + O2(g) ® 2H2O(l) 2mol 1mol 2mol 2(6.023 x 1023) 6.023 x 1023 2(6.023 x 1023) Molecules Molecules Molecules 2 x 2g 32g 2 x 18g The chemical equation for the preparation of carbon dioxide gas in the laboratory is as follows. CaCO3(s) + 2HCl ® CaCl2(aq) + CO2(g) + H2O(l) 1mol 2mol 1Mol 1mol 1mol 100g 2 x 36.5g 111g 44g 18g

ATOMS & MOLECULES 99 The above equation represents that 1 mole of calcium carbonate reacts with 2 moles of hydrochloric acid to produce 1 mole of carbon dioxide gas, 1 mole of calcium chloride and 1 mole of water. Here, we are mainly concerned with calcium carbonate, hydrochloric acid and carbon dioxide gas. The relative molecular mass of calcium carbonate is 100. That is the sum of relative atomic masses of calcium, carbon and oxygen. Relative molecular mass of CaCO3 = (40 + 12 + 3 x 16) = 100 Now 1 Mole of carbon dioxide gas occupies 22.4l at NTP. Thus, 100g of CaCO3 reacts with 1 mole of HCl to liberate 22.4l of carbon dioxide at NTP. From the above examples it is evident that mole concept is very useful in the calculations involving stoichiometric relationship. NUMERICAL PROBLEM Example : Calculate the volume of oxygen gas at NTP that can be produced by heating 4.9 grams of potassium chlorate (K = 39, Cl = 35.5, O = 16) Solution : The chemical equation for this reaction is 2KClO3(s) u 2KCl(s) + 3O2(g) 2mol 2mol 3mol Thus, 2 moles of KClO3 liberate three moles of oxygen. The relative molecular mass of KClO3 = 39 + 35.5 + (3 x 16) = 122.5 and 1 mole of oxygen gas occupies 22.4 litres at NTP. Hence, from the equation 2 x 122.5 = 245 g of KClO3 produce 3 x 22.4l = 67.2l of oxygen gas at NTP. Therefore, 4.9g of KClO3 will produce 67.2 x 4.9 = 1.344 l of oxygen at NTP. 245 CHAPTER (2.10) AT A GLANCE Atom An atom is the smallest particle of an element which takes part in a chemical change. An atom may or may not have independent existence. Molecule A molecule is the smallest particle of a compound that contains atoms of different kinds. In case of an element, the smallest particle may be a molecule composed of atoms of the same kind. The molecule is capable of existing in the free state and retaining the properties of matter. Berzelius hypothesis Equal volumes of all gases under similar conditions of temperature and pressure contain equal number of atoms.

100 +2 CHEMISTRY (VOL. - I) Avogadro's hypothesis Equal volumes of all gases at the same temperature and pressure contain equal number of molecules. Atomicity of elementary gases Atomicity means the number of atoms in elementary gases present in a molecule. Relation between molecular mass and vapour density The relative molecular mass of a gas or vapour is twice its vapour density. Relative Molecular mass = 2 x V. D. Mole and Avogadro's number One mole of any substance is that quantity of substance which contains 6.023 x 1023 particles or units of that substance. This number is Avogadro's number (N). Gram-atom and Gram atomic mass One gram -atom or one mole-atom of an element contains 6.023 x 1023 atoms of the element. Gram-atomic mass of an element is the atomic mass expressed in grams. Gram-molecule and Gram molecular mass One gram-molecule or one mole-molecule of a substance contains 6.023 x 1023 molecules of that substance. Gram molecular mass of any substance is the molecular mass expressed in grams. Molar volume or Gram molecular volume Molar volume or Gram molecular volume is the volume occupied by one gram molecular mass of any gas or vapour at NTP. Molar volume of any gas or vapour occupies 22.4 litres at NTP. Molar volume of all gases is the same at the same temperature and pressure and is 22.4 litres at NTP. Solved problems Problem 1 Calculate the mass of 1 atom of carbon –12 isotope and hence find out the value of 1 amu in grams. Evaluate the mass of one molecule of oxygen. Solution Mass of 1 atom of C – 12 = gram atomic mass of C -12 N 12 = 6.023 x 1023 = 2 x 10–23g. 1 atomic mass unit (amu) = 1 th of mass of one atom of carbon – 12 = 12 g 1 12 x 2 x 10–23 = 1.66 x 10–24 The relative molecular mass of oxygen is 32. Therefore, the mass of an oxygen molecule is 32 amu which is equal to 32 x 1.66 x 10–24 = 5.31 x 10–23 g.

ATOMS & MOLECULES 101 Problem 2 What is the mass of half mole of oxygen atom ? (C.H.S.E. 1991(A)) Solution Number of moles = Mass of atoms Problem 3 gram atomic mass Solution or, mass of atoms = number of moles x gram.. atomic mass So, mass of oxygen atom = 0.5 x 16 = 8g. Calculate the number of moles in 25 gm. of CaCO3. (C.H.S.E. 1989(S)) Number of moles = Mass of substance gram molecular mass of substance Here, Number of moles = 25 = 0.25 mole (since molecular mass of CaCO3 = 100) 100 Problem 4 Calculate the number of molecules present in 1.6 gm of methane. (C.H.S.E. 1994(S)) Solution Number of molecules = gm. Mass of substance x N molecular mass of the substance Here, molecular mass of methane, CH4 = 16 and number of molecules = 1.6 X 6.023 X 1023 = 6.023 x 1022 molecules. 16 Problem 5 Calculate the number of molecules of sodium hydroxide present in 4 gms of the sample. (C.H.S.E. 1991(S)) Solution Number of molecules of NaOH = Mass of NaOH X N gm. molecular mass of NaOH = 4 X 6.023 X 1023 = 6.023 x 1022 molecules. 40 Problem 6 Write the number of atoms present in 1 mole of sulphuric acid.(O. J. E. E 1993) Solution Total number of atoms in the substance =Number of moles x total number of atoms x N Problem 7 Here, number atoms in one molecule of H2SO4 = 7 Solution and therefore total number of atoms present = 1 x 7 x 6.023 x 1023 = 4.216 x1022 atoms In 93 g of phosphorus calculate the number of molecules. Phosphorus molecule is tetra atomic, P4 Hence its molecular mass = 124 g Number of molecules = Mass of substance x N gram molecular mass = 93 x 6.023 x 1023 = 4.517 x 1023 molecules. 124 Problem 8 Calculate the total number of electrons present in 1.6 gms of methane. Solution Molecular formula of methane = CH4 and hence, Molecular mass of methane = 16 So, number of molecules = Mass of substance x N gram molecular mass

102 +2 CHEMISTRY (VOL. - I) = 1.6 x 6.023 x 1023 = 6.023 x 1022 molecules 16 One molecule of methane contains 10 electrons. Hence 6.023 x 1022 molecules of methane contain 6.023 x 1022 x 10 = 6.023 x 1023 electrons. Problem 9 How many atoms of each of the constituent elements are there in 25g. of CaCO3 ? (Ca = 40, C = 12, O = 16). Solution Gram molecular mass of CaCO3 = 40 + 12 + 3 x 16 = 100g. Thus, 25 g of CaCO3 = 25 = 0.25 moles of CaCO3. 100 Now, 1 mole of CaCO3 contains 6.023 x 1023 atoms of Ca, 6.023 x 1023 atoms of C and 3 x 6.023 x 1023 atoms of O. Hence, the number of atoms of the constituent elements in 25g i.e. 0.25 moles of CaCO3 is equal to Ca = 0.25 x 6.023 x 1023 = 1.506 x 1023 atoms C = 0.25 x 6.023 x 1023 = 1.506 x 1023 atoms O = 0.25 x 3 x 6.023 x 1023 = 4.517 x 1023 atoms Problem 10 How many gram atoms of oxygen and hydrogen are there in 0.2 moles of water ? What are the number of atoms of each element present in this quantity of water ? Solution 1 mole of H2O or 18g of H2O contains 2 gram atomic mass of hydrogen and 1 gram atomic mass of oxygen. So, 0.2 moles of water will contain 2 x 0.2 = 0.40 gram atomic mass of hydrogen and 1 x 0.2 = 0.20 gram atomic mass of oxygen Now, the number of hydrogen atoms = 0.40 x 6.023 x 1023 = 2.409 x 1023 atoms and the number of oxygen atoms = 0.20 x 6.023 x 1023 = 1.205 x 1023 atoms Problem 11 What will be number of molecules of a gas occupying 280 ml at NTP ? (C.H.S.E 1987 (A)) Solution 280 ml of the gas = 0.28 litres of the gas So number of molecules = 0.28 x 6.023 x 1023 = 7.529 x 1021 molecules 22.4 Problem 12 A tetra-atomic gas occupies 1.4 litres at 00C and 760 mm pressure. Find the number of atoms in the gas. Solution Number of molecules of the gas present = 1.4 x 6.023 x 1023 molecules. 22.4 Since the gas is tetra-atomic, one molecule contains 4 atoms. Therefore, the number of atoms present in the gas = 1.4 x 6.023 x 1023 x 4 22.4 = 1.505 x 1023 atoms.

ATOMS & MOLECULES 103 Problem 13 Density of water at room temperature is 1.0 g./ml. How many molecules are Solution there in a drop of water if the volume is about 0.05 ml ? Weight of 1 drop or 0.05 ml of H2O = 0.05 = 0.05 g 1 Since molecular mass of water is 18, number of moles of H2O in 0.05g of it = 0.05 18 Each mole of water contain 6.023 x 1023 molecules. Hence, The number of water molecules in one drop = 0.05 x 6.023 x 1023 = 1.673 x 1021 molecules. 18 Problem 14 Calculate the weight of 60% sulphuric acid required to decompose 25 g of chalk, calcium carbonate. (Ca = 40, C = 12, O = 16, S = 32) Solution Chemical reaction : CaCO3 + H2SO4 = CaSO4 + CO2V + H2O 100g 98g From the above equation it is known that 98g of H2SO4 required to decompose 100g of chalk (CaCO3). So, weight of H2SO4 required to decompose 25g of chalk = 98 x 25 = 24.5g. 100 Since, 60g of H2SO4 are contained in 100 gm of 60% H2SO4, 24.5 g of 100% H2SO4 = 100 x 24.5 = 40.83g of 60% H2SO4. 60 Q U E S T I ON S A. Very short answer type (1 mark each) State true or false : 1. Avogadro number has value 6.023 x 1023. 2. Equal volumes of all gases contain equal number of molecules at all temperature and pressure. 3. Half a molecule of oxygen is equal to one atom of oxygen. 4. Avogadro's hypothesis explains the law of gaseous volume. 5. Atomicity is equal to valency of an element. 6. One mole of hydrogen contains 6.023 x 1023 atoms. 7. 6.023 molecules of hydrogen weigh 1 gram.

104 +2 CHEMISTRY (VOL. - I) 8. 2 grams molecular mass of a substance contain 2 moles. 9. 1 mole of chlorine occupies 22.4 ml at NTP. 10. 1 mole of electrons contains 6.023 x 1022 electrons. 11. One mole of KClO3 contains 18.069 x 1023 atoms of oxygen. 12. Molecular mass = 22.4 litres at 270C. 13. Gram molecular volume of any substance at NTP is 22.4 litres. 14. One mole of oxygen contains 2 gm atoms. B. Short answer type (1 mark) 1. How many atoms are present in 10 gms of calcium carbonate ? 2. What is the mass of 224ml of oxygen at NTP ? 3. What is the volume of 17.75 g of chlorine at NTP ? 4. What is the mass of 2.8 litres of CO2 at NTP ? 5. How many electrons are their in ammonium ion ? 6. How many atoms of oxygen are present in 4.4 gms of CO2 at NTP ? 7. How many gms of H2O2 are present in 1 litre of 20 volumes of H2O2 solution ? C. Short answer type (2marks each) 1. What is Avogadro's number ? 2. State Berzelius hypothesis 3. State Avogadro's law. 4. What is meant by atomicity of a gas ? 5. What is vapour density ? 6. What is the relationship between vapour density and molecular mass ? 7. Define gram molecular volume. 8. Define gram molecular mass. 9. Write the merits of Avogadro's hypothesis. D. Long Answer Type (7 marks each) 1. State Avogadro's hypothesis. What are its applications ? The vapour density of a gas is 11.2. Calculate the volume occupied by 1.6 grams of the gas at STP. 2. State and explain Avogadro's hypothesis. Derive the relations between molecular mass and vapour density of a gas. 3. (i) Determine the atomicity of oxygen by applying Avogadro's law, (ii) Derive from the Avogadro's law that molecular mass is equal to mass of 22.4 litres of gas or vapour at NTP. 4. State and explain Avogadro's law. How does it explain Gay-Lussac's law of gaseous volumes. How it could discard Berzelius hypothesis ? 5. State and explain Avogadro's hypothesis. How is the hypothesis applied to determine the gram molecular mass and molecular formulae of gases ?

ATOMS & MOLECULES 105 6. Derive the relationship between molecular mass and vapour density of a gas from Avogadro's hypothesis. 0.796 g of a metal oxide was heated in a current of dry hydrogen when 0.18g of water was formed. Calculate the equivalent mass of the metal. 7. What is Avogadro's hypothesis ? How can it be derived from combined gas equation ? 1.40 litre of a gas at 270C and 900 mm pressure weighs 3.5g. Calculate the molecular mass of this gas. E. Numerical Problems : (3 marks) 1. (a) How many molecules will be present in one gram molecular mass of hydrogen gas ? (b) How many atoms of hydrogen are present in two moles of H3PO4 ? 2. How many atoms of oxygen are present in one gram of oxygen ? 3. How many calcium atoms are present in 10 g of CaCO3 ? 4. (a) How many moles of hydrogen molecules are present in one mole of hydrogen peroxide ? (b) How many molecules of oxygen are present in one mole of sulphuric acid ? 5. How many molecules are there in 90 gms of H2O ? 6. How many atoms are present in 49 gms of sulphuric acid ? 7. What is mass of half mole of nitrogen atom ? 8. Avogadro's number of helium atoms weigh how many grams ? 9. How many moles do 3g of water represent ? 10. How many moles do 2 x 1022 molecules of hydrogen represent ? 11. A piece of copper weighs 0.635 g. How many atoms of copper does it contain ? 12. Between 1 gram of water and 1 gram of methanol, which has more number of molecules ? 13. What will be the weight of 1 x 1022 molecules of ammonia ? 14. What is the mass of 2.5 x 1023 atoms of nitrogen ? 15. What is the mass of a carbon dioxide molecule ? 16. Calculate the number of molecules present in carbon dioxide gas weighing 4.4g. 17. Calculate the number of moles in 30 gms of sodium hydroxide. 18. Calculate the mass of 1 ml. of hydrogen at NTP. 19. Find the volume occupied by 6.023 x 1023 molecules of oxygen at NTP. 20. 5.6 litres of gas at NTP weighs 8.25 gms. Calculate its vapour density. 21. Calculate the density of nitrogen at NTP. 22. Calculate the number of electrons present in 0.3 mole of electrons. 23. How many moles of water should be electrolysed to produce 80g of oxygen ? 24. Find out the amounts of magnesium and oxygen consumed to prepare 60g of magnesium oxide. 25. What would be the volume of 44 grams of CO2 at NTP ? 26. What volume of CO2 is liberated at NTP from 0.1 mole of CaCO3 ?

106 +2 CHEMISTRY (VOL. - I) 27. What weight of water will be formed by the combination of 11.2 litres of oxygen at NTP with excess of hydrogen ? 28. 6 gms of carbon was completely burnt in oxygen. What would be the volume of carbon dioxide produced at NTP and how many molecules will be present in that gas ? 29. Calculate the volume of oxygen at STP required to completely burn 30ml of acetylene at STP. 30. Calculate the total number of electrons present in 2.24 litres of methane at NTP. 31. What is the mass of one mole of electrons ? 32. How many atoms of fluorine are there in 1.9 x 10–6 gms of Fluorine ? (F = 19 amu) ANSWERS A. 1. True 2. False 3. True 4. True 5. False 6. False 7. False 8. True. 9. False 10. False 11. True 12. False 13. True. 14. True. C. 1. 1.6 litres. E. 1. (a) 6.023 x 1023 molecules (b) 3 x 2 x 6.023 x 1023 atoms 2. 3.76 x 1023 atoms 3. 6 x 1022 atoms 4. (a) 1 mole (b) 2 x 6.023 x 1023 molecules 5. 3.0115 x 1024 molecules 6. 2.108 x 1024 atoms 7. 7g 8. 4g 9. 0.167 mole 10. 0.333 mole 11. 6.023 x 1021 atoms. 12. 1g of water 13. 0.282 g 14.5.81g 15.7.3 x 10–23g 16. 6.023 x 1022 molecules 17. 0.75 mole 18. 0.00009g 19. 22.4 litres 20. 16.5 21. 0.00125 gm/ml 22. 1.807 x 1023 electrons 23. 5moles 24. Mg 36g & oxygen24g 25. 22.4 litres 26. 2.24 litres 27.18g 28.11.2 litres,3.01 x 1023 molecules 29. 75 ml. 30. 6.023 x 1023 electrons. 31. 9.1 x 6.023 x 10–5 g. 32. 6.023 x 1016 atoms. 2.11 STOICHIOMETRY (CALCULATIONS BASED ON CHEMICAL EQUATION) A chemical change is represented both qualitatively and quantitatively by a chemical equation. Since a symbol or a formula has a quantitative meaning, a balanced chemical equation gives quantitative informations regarding reactants and products. Calculations based on chemical equations are known as stoichiometric calculations. They take into account the molecule, molecular mass, mole concept, gram molar volume etc. in their calculation. Let us consider the combustion of ethane (C2H6). The balanced equation of the chemical change is represented by

ATOMS & MOLECULES 107 2C2H6(g) + 7O2(g) 4CO2 (g) + 6H2O(g) 2moles 7moles 4 moles 6 moles 2[24 + 6]g 7[32]g 4[12+32]g 6[2+16]g 2×22.4L 7×22.4L 4×22.4L 6×22.4L (at N.T.P) The above chemical equation means- 2moles of C2H6(g) reacts with 7 moles of O2(g) to give 4 moles of CO2(g) and 6 moles of H2O(g) 60g of C2H6(g) reacts with 224g of O2(g) to gives 176g of CO2 and 108 g of H2O(g) 44.8L of C2H6(g) reacts with 156.8 L of O2(g) to give 89.6L of CO2(g) and 134.4L of H2O(g) (all measured under N.T.P) From this, it is evident that mole, molecular mass and volume are all correlated. The calculations based on chemical equations are of three types : (i) calculations based on mass-mass relationship (ii) calculations based on mass-volume relationship (iii) calculations based on volume-volume relationship Calculations based on mass-mass relationship In general, the following steps are followed for the calculation. (a) First write down the balanced chemical equation. (b) Write down the molecular masses along with their respective coefficients multiplied with molecular masses (written below the respective species). These represent the theoretical amount of the reactants and products. (c) Then establish the mass-mass relationship and find out the mass of the required spe- cies by unitary method. Example -1. Calculate the mass of O2 required for completely burning 30g of ethane to produce CO2. Also calculate the mass of H2O produced during the reaction. Solution : The balanced chemical equation is 2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O (g) 6 ×18 2× 30 7×32 For burning 60g of ethane, O2 required in 224g 224× 30 g= 112g Thus, for burning 30g of ethane, O2 required will be 60 On burning 60g of ethane, H2O produced is 108g

108 +2 CHEMISTRY (VOL. - I) Thus, on burning 30g of ethane, H2O produced will be 108× 30g= 54g 60 Example -2 Calculate the mass of lime required to remove the hardness of 10,00,000 L of water con- taining 16.2g of calcium bicarbonate per litre. Solution : Ca(HCO3)2 in water causes hardness. Lime (CaO) reacts with Ca(HCO3)2 present in hard water precipitating CaCO3 and thus removes hardness. Ca (HCO3)2 + CaO 2CaCO3 + H2O 162 56 Total amount of Ca(HCO3)2 present in 10,00,000 L of water = 16.2g L–1 × 10,00,000L = 16200 kg 162 g of Ca(HOO3)2 requires 56g of lime (as per equation) 16200 kg of Ca (HCO3)2 will 56g require 162g × 16200 kg = 5600 kg of lime. Hence, 5600 kg of lime will be required for removing hardness. Limiting reagent : Often, reactions are carried out when the reactants are not present in the amounts required by a balanced chemical reaction. The reactant which is present in the lesser amount gets consumed and no further reaction takes place eventhough the other reactant is present. Thus, the reactant present in lesser amount limits the amount of product formed and is, therefore named as limiting reagent. In stoichiometric calculations, this fact should be borne in mind. Example - 3 100g sample of calcium carbonate is allowed to react with 75g of orthophosphoric acid. Calculate the mass of calcium phosphate that could be produced and the mass of excess reagent that will remain unreacted. Solution : The balanced equation is represented as 3CaCO3 + 2H3PO4 Ca3 (PO4)2+3CO2 + 3H2O 1 mole 3 mole 2mole 3[40 + 12 + 48] 2[3 + 31 + 64] [3 × 40 + 2 × 95] 300g 196g 310g 300g of CaCO3 produce Ca3 (P31O04×)2 = 310 g or 1 mole mole of Ca3(PO4)2 100g of CaCO3 will produce 300 100 = 103 g or 0.33 or, 196g of wHil3lPpOr4odpurocdeucCea3C(Pa3O(4P)2O=4)213=91603×107g5 or 1 mole or 0.38 mole 75g of H3PO4 = 118.6g

ATOMS & MOLECULES 109 The values show that CaCO3 is the limiting reagent. Therefore, Ca3 (PO4)2 formed is 0.33 mole or 103 g. = 196× 103 = 65.12g For producing 103g of Ca3 (PO4)2, H3PO4 required will be 310 Therefore, mass of remaining H3PO4 = (75 – 65.12)g = 9.88g Calculations based on mass-volume relationship These calculations are based on the fact that 1 mole or 1 gram molecular mass of the substance occupies 22.4 litres or 22400 mL at N.T.P. The following example shows tha mass- volume relationship. MgCO3 + 2 HCl MgCl2 + H2O + CO2 1 mole 2 mole 1 mole 84 g 73 g 44g 22.4L at N.T.P. Volume of a gas at any temperature and pressure can be converted to mass or vice-versa using the equation W PV= M RT where W = mass of the gas with molecular mass M and R is molar gas constent. P,V & T have their usual meaning. Example -1 What volume of chlorine gas at N.T.P. will be produced when 8.7g of MnO2 reacts with excess of conc. HCl ? (Mn = 55, O = 16, Cl = 35.5) Solution : The balanced reaction is MnO2 + 4HCl MnCl2 + Cl2 + 2H2O 55+32 71 1 mole 1 mole One mole i.e. 87g of MnO2 produce 22.4L of Cl2 at N.T.P. Hence, 8.7g of MnO2 will produce 2.24L of Cl2 at N.T.P. Example -2 10g of an impure sample of potassium chlorate on complete decomposition gave 1.12 litre of oxygen at N.T.P. What is the percentage purity of the sample of potassium chlorate ? Solution : The balanced equation for decomposition is 2KClO3 2KCl + 3O2 2mole 3 mole 245g 3 × 22.4L at N.T.P.

110 +2 CHEMISTRY (VOL. - I) 3 × 22.4L of O2 is obtained from KClO3 = 245g 245g 1.12 L of O2 will be obtained from KClO3 = 3× 22.4 L × 1.12 L = 4.08g =P10e4rg1c.00oe8gnfgtsa×agm1e0pp0lue=rcit4oy0n.ot8afitnhseKsaCmlOp3le==4.A08mAgmouonutnotfoifmppuurreesasmamplpele× 100 Calculations based on volume - volume relationship These calculations are based on two laws : (i) Avogadro’s law & (ii) Gay-Lussac’s law For example, 2CO + O2 2CO2 (Avogadro’s law) 2 mole 1 mole 2 mole 2 × 22.4L 22.4 L 2 × 22.4L (Under similar conditions of temperature and pressure, equal moles of gases occupy equal volumes) or, 2CO + O2 2CO2 (Gay - Lussac’s law) 2vol 1 vol 2 vol (Under similar conditions, gases react in simple ratio of their volumes) It may be remembered that the law holds good only for gaseous reactants & products, which do not liquify on cooling. Example -1 What volume of oxygen at N.T.P. is necessary for complete combustion of propane (C3H8) measured at 270C and 760 mm of Hg. Solution : The balanced chemical equation is C3H8 + 5O2 3CO2 + 4H2O 22.4L 5× 22.4L Since the volume of propane is at 270C & 760 mm of Hg pressure, it has to be converted to N.T.P. condition applying gas law. Pressure remains constant at 760 mm of Hg. Therefore, ac- cording to Charle’s law, 273K 273 300 Volume at NTP = Given volume × (27+ 273)K = 20 L × = 18.2 L Hence, 18.2 L of propane will undergo combustion at N.T.P. 22.4 L of C3H8 requires O2 = 5×22.4L = 112L 18.2 L of C3H8 will requires 112L 22.4L × 18.2L = 91.0 L Therefore, 91 L of O2 at N.T.P. is necessary for complete combustion of C3H8

ATOMS & MOLECULES 111 Example -2 One litre of a mixture of carbon monoxide and carbondioxide is passed through a tube containing red hot charcoal. The volume now becomes 1.6 litre. Volumes are measured under similar conditions. Find the composition of gaseous mixture. [I.I.T. 1980] Solution : When the mixture is passed through red hot charcoal, only CO2 is reduced to carbon mon- oxide resulting in the increase in volume according to the equation CO2 + C 2CO 1 vol 2 vol Let the volume of CO2 in the mixture = x L volume of CO in the mixture will be = (1 – x) L x vol. of CO2 produce 2x vol. of CO Total volume of CO = 2x + 1 –x = (1 + x) vol = 1.6L (Given) Therefore, x = 0.6L = volume of CO2 in the mixture. and volume of CO in the mixture = (1 – 0.6)L = 0.4L Example -3 A gas mixture of 3 litres of propane (C3H8) and butane (C4H10) on complete combustion of gas at 250C produced 10 litres of CO2. Find out the composition of the mixture. Solution : Let x litre of C3H8 and y litre of C4H10 be present in the mixture. x + y = 3 (given)................ (i) C3H8 + 5O2 3CO2 + 4H2O 3 vol 1 vol 13 2 4CO2 + 5H2O C4H10 + O2 4 vol 1 vol CO2 formed by C2H8 = 3x CO2 formed by C4H10 = 4y 3x + 4y = 10 (given)........... (ii) Solving equations (i) and (ii) x = 2 litre = volume of C3H8 y = 1 litre = volume of C4 H10

112 +2 CHEMISTRY (VOL. - I) QUESTIONS (Stoichiometric Calculations) 1. 1.5g of an impure sample of Na2CO3 is dissolved in water. On adding excess amount of CaCl2 solution, a white precipitate was obtained which was filtered, dried and found to weigh 1g. Calculate the percentage purity of the Na2CO3 sample. 2. Find out the amout of litharge (PbO) produced by heating 132.4g of lead nitrate of 50% purity. (Pb = 207.2) (Ans. 44.6g) 3. A 2g sample containing Na2CO3 and NaHCO3 loses 0.248g when heated to 3000C, the temperature at which NaHCO3 decomposes to Na2CO3, CO2 and H2O. What is the percent- age of Na2CO3 in the mixture ? [Ans. 66.4%] 4. 2g of a mixture of NaCl and KCl was dissolved in water and the resulting solution was treated with excess of AgNO3 solution. AgCl when weighed after drying was found to be 4.483g. Calculate the percentage composition of the mixture. [Ans. NaCl = 59.85%, KCl = 40.15%] 5. How many litres of oxygen at NTP is required to burn completely 2.2 g of propane (C3H8) ? [Ans. 5.6L] 6. How much KClO3 is needed to produce oxygen enough for burning 10g of coke ? (Ans. 68.05g) 7. 1.84 g of a mixture of CaCO3 and MgCO3 is strongly heated till no further loss of weight takes place. The residue weighs 0.96g. Calculate the percentage composition of the mix- ture. (Ca=40, Mg = 24) [Ans. 54.3% : 45.7%] 8. What volume of CO2 at STP can be obtained from 500 g of marble containing 80% CaCO3 ? [Ans : 89.6L] 9. 8.7g pure MnO2 is heated with excess of HCl and the gas evolved is passed through a solution of KI. Calculate the mass of iodine liberated (Mn = 55, Cl = 35.5, & I = 127) [Ans. 25.4g] 10. How many grams of potassium permanganate completely oxidise 4.5g of oxalic acid present in an acid solution ? (K = 39, Mn = 54.94, C = 12) [ Ans. 3.16g] 11. 5g of a natural gas consisting of methane and ethylene was burnt in excess of oxygen producing 14.5g of carbon dioxide and some amount of water as products. Find out the percentage of ethylene in the natural gas. (Ans. 40.4%) Multiple choice questions Select the correct answer for the following. 1. The number of grams of oxygen which will completely react with 27g of aluminium is (a) 8 g (b) 16g (c) 24 g (d) 32g [Ans. (c)]

ATOMS & MOLECULES 113 2. 2.76 g of silver carbonate on being strongly heated yield a residue weighing (a) 2.48g (b) 2.16g (c) 2.32g (d) 2.64g [Ans. (b)] 3. The mass of CO2 produced by heating 21g of NaHCO3 is [Ans. (a)] (a) 5.5g (b) 11.0g (c) 7.5g (d) 5.9g 4. The amount of zinc (65) required to produce 224 mL of H2 at S.T.P on treatment with dil. H2SO4 will be (a) 0.65g (b) 6.5g (c) 65g (d) 0.065g [Ans. (a)] 5. 56g of CaO has been mixed with 63g of HNO3. The amount of calcium nitrate formed will be (a) 4g (b) 3.28g (c) 164 g (d) 82g [Ans.(d)] 6. 10mL of gaseous hydrocarbon on combustion gives 40mL of CO2(g) and 50mL of H2O (vap). The hydrocarbon is (a) C4H5 (b) C8H10 (c) C4H8 (d) C4H10 [Ans. (d)] 7. 4L of CO at NTP will require for complete oxidation, the volume of O2 at NTP (a) 4L (b) 8L (c) 2L (d) 1L [Ans (c)] 8. The minimum quantity of H2S needed (in g) to precipitate 63.5g of Cu2+ will be nearly (a) 63.5g (b) 31.75g (c) 34g (d) 20g [Ans. (c)] 9. Which of the following wll have largest number of atoms ? (a) 1gAu(s) (b) 1gNa(s) (c) 1gLi(s) (d) 1g of cl2(g) [Ans. (b)] qqq

114 +2 CHEMISTRY (VOL. - I) UNIT – II CHAPTER - 3 STRUCTURE OF ATOM 3.1 FUNDAMENTAL PARTICLES AND THEIR PROPERTIES John Dalton (1809) developed famous \"atomic theory\". He considered the atom as a hard, dense and smallest indivisible particle of matter. According to the theory each element consists of a particular kind of atoms differing in mass. But the mass difference of atoms of different elements could not be explained by him. Subsequent investigators were of the opinion that there should be some material difference for making atoms of different elements. But the experimental evidence in support of it was obtained during the last part of nineteenth century. The passage of electricity through gases at very low pressure and discovery of radioactivity showed the presence of three fundamental particles i.e. electron, proton and neutron in the atoms. CONDUCTION OF ELECTRICITY THROUGH GASES - CATHODE RAYS DISCOVERY OF ELECTRON Normally gases are poor conductors of electricity. But under the influence of high voltage (20,000 - 30,000 volts) gases conduct electricity. Even very interesting results are obtained at low pressure. Experiments were conducted in discharge tubes of hard glass of 60 cm length and 4 cm in diameter closed at both ends. (Fig. 3.1) Two metal electrodes 'C' and 'A' were fused into the tube and a side tube was connected to the vacuum pump 'V'. cA Cathode Rays. + TO VACUUM PUMP Fig. 3.1 Production of Cathode Rays.

STRUCTURE OF ATOM 115 When pressure was reduced in side the tube to 0.06 - 0.03 mm by operating the vacuum pump and a high potential of 20,000 volts was applied across the terminals of the tube, the tube began to glow and a stream of faint greenish light was seen to travel from Cathode. These were known as Cathode rays. Properties of Cathode Rays : 1. These rays travel in straight lines with velocity approaching the velocity of light and cast shadows of objects placed in their path. 2. These are considered to be possessing energy as they rotate a small paddle wheel placed in their path. 3. These rays penetrate through, thin metal sheets, produce phosphorescence, affect photographic plates, ionise gases and produce X-rays on striking a metal target. 4. These rays are deflected in electric and magnetic fields and consist of negatively charged particles as found by its deflection in an electric field (Fig 3.2). On applying electric field across the tube, the rays are bent towards the positive plate. + Cathode Rays. + + Fig 3.2 Deflection of cathode rays in electric field. Therefore, Cathode rays consist of negatively charged particles. These negatively charged particles are called electrons. J. J. Thompson carried out a number of experiments to find out charge and mass of the electron. Charge of the electron was found to be 1.602 x 10–19 coulomb. Mass of the electron was found to be 9.1 x 10–28 g. ( )charge em = 1.7599 x 108 coulombs per gram. mass of electron Thompson carried out a number of measurements on the value of e m of Cathode rays in which he varied the metal forming the Cathode as well as the gas in the vacuum tube. In every case the ratio of charge to mass (e m) was the same. Millikan found that charge (e) on each electron was 1.602 x 10–19 coulomb, irrespective of the gas taken inside the tube. From these observations, it was concluded that electrons are universal constituent of matter. Discovery of Protons : Positive Rays : Since electron is an essential constituent of atom and atom as a whole is electrically neutral, it follows that an equal magnitude of positive electricity must also be present in the atom. In 1886, M. Goldstein experimented with discharge tubes containing perforated cathode and showed the presence of another type of radiation that passed through the holes in the cathode and carried positive charge. These rays were called positive rays or anode rays. (Fig. 3.3)

116 +2 CHEMISTRY (VOL. - I) + He H+ H e H+ Anode H e Positive Hrays + Perforated Cathode Fig. 3.3 Positive Rays or Anode Rays. Properties of Anode Rays : 1. These travel in a straight line. 2. These rays cause phosphorescence, affect photographic plate, and penetrate through very thin aluminium foils. 3. These rays are deflected by electric and magnetic field, but deflection takes place in opposite direction to that of the electrons. Thus, these rays are found to be positively charged. The ratio of charge to mass (e m) of positive rays gases was determined by taking different in the discharge tube. Hydrogen gave the highest value for e m . This shows that if charge (e) is the same in the case of each gas, the mass (m) of positive ray particle produced in the hydrogen tube is the lowest. It was concluded that the positive particle given out by hydrogen is one of the fundamental particles of matter. Its charge was found to be equal in magnitude but opposite in sign to that of the electron. Mass was found to be 1.672 x 10– 24 gm. This lightest positive particle is known as proton. Existence of charges on cathode and anode rays indicates that atom consists of one or more electrons and a positive residue. Since electrons have negligible mass, it follows that almost the entire mass of atom is associated with the positive residue. Discovery of Neutron : In 1932, James Chadwick bombarded thin sheets of light elements such as Beryllium, Boron etc with a– particles. It was observed that a stream of highly penetrating rays were produced. Thses rays were found to consist of neutral particles. These neutral particles were found to have a mass of 1.675 x 10–24 gm. These particles were called neutrons. Neutron is present in all atoms excepting hydrogen atom. 3.2 THOMSON’S MODEL FOR THE STRUCTURE OF ATOM After the discovery of the sub-atomic particles, the electrons, protons and neutrons, several models for the structure of atom were proposed. The first model of the atom was proposed by Sir J.J. Thomson in 1898. He assumed that the basic body of an atom is a spherical object containing electrons confined in homogeneous jelly like but relatively massive positive charge distribution whose total charge cancels that of the number of electrons present. The schematic drawing of the model is shown in fig. Thomson’s model is sometimes called a plum pudding model. This model was soon discarded and is only of historical importance. + + + ++ + + + ++ + + + + + + + Fig. 3.4 Thomson’s model of atom

STRUCTURE OF ATOM 117 Defects : Thomson’s model of the atom could not explain the large angle scattering of a- particles. Further, it could not explain the experimental observations of the atomic spectra of elements. 3.3 RUTHERFORD'S EXPERIMENT - SCATTERING OF a - PARTICLES. In 1911 E.Rutherford performed experiments with a-particles which gave insight into the structure of atom. A radioactive substance spontaneously emits radiations consisting of a– particles which are positively charged, b -particles which are negatively charged and l-rays which are neutral. In fact a-particles are doubly charged helium nucleus. It carries two units of positive charge and has mass equal to four times that of hydrogen atom. Rutherford produced a-particles from a radioactive source (polonium) and bombarded it with thin (0.00004 cm thick) sheet of silver or gold metal. After passing through the metal sheet, a– particles struck a zinc sulphide screen. It was found that most of the a–particles passed through the metal sheet with no change in their path. Only very few were deflected from the original path through wider angles. Only 1 out of 20,000 a–particles was deflected backwards retracing the path. (Fig 3.5) a –particles NUCLEUS Fig 3.5 Scattering of a –particles by the atom. Rutherford after carrying out a series of experiments concluded that (1) Since most of the a particles pass through the metal sheet, without being deflected, it is concluded that the atom consists of empty space predominantly. (2) Since a few of the a–particles (having appreciable mass and positive charge) get deflected through wide angles or even backwards, there must be present in the atom a heavy positively charged mass. For such large scattering of a–particles, these must have come close to even collide with the positively charged body of heavy mass. (3) Since only very few a–particles suffer such deflection, the volume occupied by this heavy positively charged mass is a very small fraction of the volume of the atom and present in the centre of the atom. RUTHERFORD'S NUCLEAR MODEL Based on his observations and conclusions derived from it Rutherford proposed a nuclear model of atom. According to this model, the atom consists of a minute but heavy

118 +2 CHEMISTRY (VOL. - I) positively charged body at its centre, called the nucleus. Since atom is largely empty and electrons have negligible mass, the entire mass of the atom is due to the protons present in the nucleus. Positive charge of the nucleus is due to the presence of the protons. Electrons are referred as planetary electrons which move round the nucleus in different orbits or path just as planets move round the sun. Thus, atom was compared to a mini solar system. Since the atom is electrically neutral, the number of planetary electrons is equal to the number of protons present in the nucleus. From the Scattering experiments, Rutherford also calculated that the nucleus has a diameter of the order of 10–13cm Where as the atom has a diameter of the order of 10–8cm. To justify the revolving of the electrons around the nucleus without falling into the nucleus, Rutherford suggested that the attraction forces acting between the electrons and the nucleus acting towards the nucleus is balanced by the centrifugal force due to movement of electrons acting away from the nucleus. MOSLEY'S EXPERIMENT AND CONCEPT OF ATOMIC NUMBER In 1913, H.G.J. Mosley, a young British physicist working in Rutherford's laboratory, devised a method to find out the exact charge on the nucleus. He produced X-rays by bombarding different elements (used as anode in discharge tubes)with cathode rays. He suggested that the frequency of X-rays produced in this manner was related to the charge present in the nucleus of an atom of the element used as anode. Data obtained by him fitted to the equation. n = a (Z–b) where 'n' is the frequencey of X-ray, 'Z' is the nuclear charge and 'a' and 'b' are constants. Mosley found that the nuclear charge increases by one unit in passing from one element to the next arranged by Mendeleeff in the order of increasing atomic weight. Increase in positive charge on the nucleus was taken to imply an increase in the number of protons in the nucleus. The number of unit positive charge carried by the nucleus of an atom is termed as the atomic number of the element. Therefore, atomic number of an element (Z) = Number of unit positive charges present in the nucleus. = Number of protons present in the nucleus. = Number of electrons present outside the nucleus of the atom . Thus, atomic number is taken as a fundamental property of an element. The elements are now arranged in the periodic table in the order of increasing atomic numbers. Mass number : The sum of the number of protons and neutrons is called the mass number. Since each proton and each neutron has a mass approximately equal to 1, on the atomic weight scale, the atomic weight of an element is approximatley equal to its mass number (A). Knowing the atomic number (Z) and mass number (A) of an element, the number of protons, neutrons and electrons can be calculated. If the mass number of an elements X is ‘A’ and its mass number is ‘Z’, then the element can be represented as A X . Then, the number of protons = Z, number of electrons = Z Z, and number of neutrons = A – Z

STRUCTURE OF ATOM 119 For example, in case of sodium atom, Z = 11, A = 23. Then, number of protons = 11, number of electrons = 11 and number of neutrons = 23 – 11 = 12. ISOTOPES, ISOBARS AND ISOTONES : Isotopes : Isotopes are defined as the atoms of the same element which has the same atomic number but different mass numbers. Isotopes have same physical and chemical properties, same atomic number, same position in the periodic table, but different number of neutron and different atomic weights, density, atomic volume, melting and boiling points. For example, hydrogen has three isotopes. Protium (ordinary hydrogen) 1 H 1 Deuterium 2 H (mass number – 2) 1 Tritium 3 H (mass number – 3) 1 Isobars : Isobars are the atoms of different elements having the same mass number but differ in their atomic numbers. These have different physical and chemical properties, different atomic numbers, different position in the periodic table. For example, Argon = 40 Ar 18 Calcium = 40 Ca 20 Isotones : Atoms of different elements which possess the same number of neutrons are called isotones. For example, 76 Ge and 77 As (contain 44 neutrons each). 32 33 DEFECT OF RUTHERFORD'S MODEL In view of Clark Maxwell's classical electromagnetic theory Niels Bohr pointed out that Rutherford's atom should be highly unstable. Clark Maxwell had shown that whenever an electric charge is subjected to acceleration, it emits radiation and loses energy. Bohr argued that an electron (charged particle) moving round the nucleus in an orbit will be subjected to acceleration due to continuous change in its direction of motion. Therefore, the electron will continuously emit radiation and lose energy, as a result of which the curvature of the path (orbit) of the electron will decrease and ultimately it will fall into the nucleus (Fig.6.6). Bohr modified the defect in the Rutherford's model and provided the Bohr's model of atom. Fig 3.6 Decrease in path of the electron.

120 +2 CHEMISTRY (VOL. - I) 3.4 CHARACTERISTICS OF RADIATION AND PLANCK'S QUANTUM THEORY Light, X - rays, infrared and ultraviolet rays are some examples of radiant energy. Clark Maxwell proposed that radiant energy has wave character. He called radiant energy as electromagnetic radiations or electromagnetic waves. The term electromagnetic was used because similar waves can be produced by moving a charged body or a magnet to and fro in an oscillating manner. Electromagnetic waves do not need any medium for propagation. There are three fundamental characteristics of wave motion i.e.wavelength, frequency velocity and amplitude. l Fig 3.7 Wave motion Wave length —The distance between two nearest crests or troughs. It is denoted by ''l \" (lambda) and is expressed in centimetres or Angstrom (A0) units 1 A0 = 10–8cm. Amplitude – Height of crest or depth of trough of a wave Frequency – The number of times a wave passes through a given point in one second. It is denoted by 'u' (nu) Velocity – The distance travelled by the wave in one second. It is denoted by 'C'. The product of wavelength and frequency gives the velocity. Thus, C = u l or u= C l All types of electromagnetic radiations travel with the same velocity but their wavelength and frequency are different. Planck's quantum theory — In 1901, Max Planck studied the radiations from black (hot) bodies and proposed a theory which goes by the name of Planck's quantum theory. The postulates of his theory are :- (1) Radiant energy, such as light and heat is emitted or absorbed not continuously but discontinuously in the form of small packets or bundles. (2) The packet or bundle of energy is called quantum and in case of light the quantum is called photon. Photon is a massless bundle of energy. (3) The energy of the photon is given by the expression E = hu where 'u ' is the frequency of radiation and 'h' is the Planck's constant. Its value is equal to 6.62 x 10–27 erg. second.

STRUCTURE OF ATOM 121 (4) A body can emit or absorb either one quantum of energy (hu) or any whole number multiples of this unit i.e. 2hu, 3hu.....nhu. E = nhu, where n = 1,2,3................. 3.5 BOHR'S MODEL OF ATOM Danish physicist Niels Bohr gave a model of atom in 1913. He accepted the provisions of Rutherford's model such as the positively charged nucleus at the centre contained the protons and neutrons. He also agreed to the fact that the electrons are revolving round the nucleus, but modified the model to meet the objections of Clark Maxwell's theory. On the basis of Planck's quantum theory Bohr gave a number of postulates. (1) The electrons move round the nucleus in stationary orbits without losing energy, because energy in fractions of a quantum cannot be lost. This means that energy of an electron remains constant as long as it remains in the same orbit. Thus, electron in each orbit is associated with definite amount of energy or a definite whole number quantum of energy. The orbits are therefore, called as energy levels or energy shells and are given numbers 1, 2, 3, 4 etc. starting with the nearest orbit to the nucleus. These numbers are now called principal quantum numbers and the orbits are designated as K, L, M, N etc. (Fig 3.8) Nearest orbit to the nucleus has the least energy and increases on going away from the nucleus. As long as the electron is in a particular energy level, it neither emits nor absorbs energy. n=4 n=3 n=2 n=1 E4 E3 E2 E1 KLMN Fig. 3.8 Energy levels of the electron (2) The angular momentum of the electron moving round the nucleus is quantised. An electron moving in a circular orbit has an angular momentum given by 'mvr' where 'm' is the mass, 'v' is the velocity of the electron and 'r' is the radius of the orbit. According to Bohr, angular momentum of the electron is quantised or has definite values given by the following expression :

122 +2 CHEMISTRY (VOL. - I) m22phom,en23Atphunmg..u..lo.a.f.r...am.no2nmpehleen.cttTurmhoins=icsamntverrbm=eedn2hpa2hsp , where n = 1, 2, 3 ......... etc. Thus, the angular , such as or whole number multiples of h 2p quantisation of angular momentum. (3) Energy is emitted or absorbed by an atom when electron moves from one energy level to another. Thus, if energy associated with the electron in orbits 1, 2, 3, 4 etc. are E1, E2, E3, E4 etc. respectively, where E1 < E2 < E3 < E4 , the change in energy on movement of electron from one orbit to the other is given by D E. When electron jumps from a lower energy orbit (E1) to a higher energy orbit (E2), the amount of energy absorbed is given by D E = E2 – E1. Similarly, this amount of energy is emitted when electron jumps from a higher energy orbit (E2) to a lower energy orbit (E1). 3.6 BOHR’S EQUATION FOR ENERGY OF ELECTRON IN HYDROGEN ATOM The expression for energy of an electron in hydrogen atom or hydrogen like atom was calculated by Bohr basing on his postulates. According to Bohr the electron continues to stay in a particular orbit so long as it is associated with a definite amount of energy. The electrostatic force of attraction exerted by the nucleus is exactly balanced by the centrifugal force resulting from its circular motion. A hydrogen atom has one proton in the nucleus and one electron in the extra nuclear part of the atom. Let Z = Atomic number of hydrogen like species (one electron system) e = charge in the electron revolving round the nucleus. For hydrogen atom Z = 1 For hydrogen like atoms such as He+, Li 2+ , Z = 2 and 3 respectively Hence, the charge on the nucleus = Ze Suppose, r ® Distance between the revolving electron and the nucleus i.e. the radius of the orbit in which the electron is revolving. m ® mass of the electron v ® Tangential velocity of the revolving electron. Now two types of forces are seen to play simultaneously. (i) Electrostatic force of attraction (Centripetal force) This force exists between the revolving electron and the nucleus. It tends to attract the electron towards the nucleus. > Ze2 The centripetal force = Ze ´ e = r2 Nucleus > r2 + (ii) Repulsive force (Centrifugal force) e This force tends to keep the electron away from the nucleus The centrifugal force acting on the electron = mv 2 r Both the above forces act in opposite direction. In order to keep the electron revolving round the nucleus in its orbit, Bohr assumed that these forces must be equal.

STRUCTURE OF ATOM 123 Hence Ze2 mv2 .............. (1) or r2 = r v2= Ze 2 ...................(2) mr According to Bohr’s postulate the angular momentum is an integral multiple of h 2p i.e. mvr = nh ....................... (3) 2p or v= nh ........................(4) 2pmr Squaring equation (4), v2= n2h2 ............... (5) 4p2m2r 2 From equation (2) and (5) Ze2 = n2h2 ................. (6) mr 4p2m 2r 2 n2h2 or Ze2 = 4p2mr ....................................... (7) or r≡ rn = n2h2 .................................(8) 4p2Zme2 For hydrogen atom, Z = 1, hence rn = n2h2 ............................................(9) 4p2me2 In the ground state n = 1 So, r1 = Bohr’s radius = a0 = h2 .................. (10) 4p2me2 The total energy ‘E’ of the electron revolving in the nth orbit is given by the sum of kinetic energy and potential energy. Kinetic energy = 1 mv2 ............. (11) 2 - Ze2 Potential energy = rn .............(12) \\ E = ½ mv 2 - Ze 2 ................. (13) rn From equation (2), mv 2 = Ze 2 ............. (14) rn Substituting the value of mv2 from equation (14) in equation (13) , we have Ze2 Ze 2 – Ze 2 2rn rn 2rn ............... (15) E= = Substituting the value of rn from equation (8) in equation (15), we have

124 +2 CHEMISTRY (VOL. - I) E= - Ze2 ´ 1 = - Ze2 ´ 1 2 rn 2 n2h2 4p2 Zme 2 = - Ze2 ´ 4p2Zme2 2 n2h2 = – 2p2Z2me4 n2h2 So, energy of electron in the nth orbit En is given by En = – 2p2Z2me4 ................. (16) n2h2 Equation (16) is the Bohr’s equation for energy of electron in nth orbit. This equation is appli- cable to both hydrogen and hydrogen like atoms (He+, Li2+ Be3+, etc) Calculation of rn is CGS unit rn = n2h2 4p2me2Z Here, h = 6.624 × 10–27 ergs.sec p = 3.14 m = 9.108 × 10–28 g e = 4.8 × 10–10 esu. ( ( ) ( ) )rn= 4× n2× 6.624× 10 27 erg . sec 2 Z (3.14)2× 9.108× 10 28 g × 4.8× 10 10 esu 2× ( ( ) ( ) )= 4× n2× 6.624× 10 27 2 erg2. s2 Z 4.8× 10 10 2 g. esu 2× (3.14)2× 9.108× 10 28 × Use the relation erg = g cm2s–2 esu = g½cm3/2s 1 (( ) )rn= 0.529× 10 8× n2 × g cm2s 2 s2 Z g. g½cm3/2s 1 2 = 0.529× 10 8× n2 × g2 cm4s 4 . s2 Z g. g .cm3.s 2 = 0.529× 10 8× n2 = 0.529× n2 A0 ZZ Calculation of rn is SI units ( )rn = n 2h 2 4p ∈ 0 4p 2 Zme 2 The factor 4pe0 is known as Permittivity factor

STRUCTURE OF ATOM 125 h = 6.624 × 10–34 Js ∈ 0= 8.85× 10 12 J 1c2m 1 p = 3.14 m = 9.108 × 10–31 kg e = 1.602 × 10–19c Use the relation J = kg m2s–2 ( ( ) ) ( )rn= n2× 6.624× 10 34 Js 2× 8.85× 10 12 J 1c2m 1 3.14× 9.108× 10 31 Kg × 1.602× 10 19 c 2× Z n 2× 5.29× 10 11× J2s2J 1c2m 1 Kg c2× Z = n 2× 5.29× 10 11× Js2m 1 Kg× Z = 5.29× 10 11 × n 2× Kg m2s 2s2m 1 Kg× Z = = 0.529 ´10-10 ´n2 m Z Bohr’s radius (r1)H or rH or a0 (i) in CGS unit rH = 0.529A0 (ii) in SI unit rH = 0.529 × 10–10m Calculation of En in CGS unit En = 2p 2 Z 2 me 4 n2h2 Putting the values of p, m, e and h p = 3.14 m = 9.108 × 10–28g e = 4.8 × 10–10 esu h = 6.624 × 10–27erg . s ( ( ) ( ) )En= 2× (3.14)2× 9.108× 10 28 g × 4.8× 1010 esu 4 Z2 n2× 6.624× 10 27 erg . s 2 = 2.18× 10 11 × Z2 × g(esu)4 n2 (erg. s)2 Utilising the relations erg = g cm2s–2 esu = g½ cm3/2 s–1 We have, (( ) )En= 2.18× 10 11× Z2 × g g½cm3/2s 1 4 n2 g cm2s 2s 2

126 +2 CHEMISTRY (VOL. - I) 2.18× 10 11× Z2 × g× g2cm6s 4 n2 g2cm4s 2 = =– 2.18× 10 11× Z2 g cm2s 2 n2 =– 2.18× 10 11× Z2 erg atom 1 n2 Value of En in ev. atom–1 Since 1 erg = 6.2419 × 1011 ev En = –2.18 × 10–11 × 6.2419 × 1011 × Z2 ev . atom–1 n2 = – 13.6× Z2 ev . atom 1 n2 Calculation of En in SI units 2p 2 Z 2 me 4 4p ∈ 0 2 n 2h 2 ( )En = where 4pÎ0 is the permittivity factor. or, En = me 4 Z 2 8∈ 2 n2h2 0 Putting m = 9.109 × 10–31kg e = 1.602 × 10–19 c ∈ 0= 8.85×10 12 J 1c2m 1 h = 6.626 × 10–34Js En = 2.18 ×10–18 × Z2 J atom–1 n2 = 13.1279× 105× Z2 J mole 1 n2 = 13.1279× 102× z2 kJ mole 1 n2 Energy of an Electron in the 1st orbit of H - atom (E1) H Here n = 1, Z = 1 (i) In CGS unit 2p 2 me 4 Z2 h2 n2 (E1) H = × = – 2.18 × 1011 erg atom–1 = – 13.1279 × 1010 erg mol–1 = –13.6 ev. atom–1

STRUCTURE OF ATOM 127 (ii) In SI units (E1 )H = me 4 × Z2 n2 8 ∈ 2 h 2 0 = – 2.18 × 10–18J atom–1 = –13.1279 × 102 kJ mol–1 To find out the value of (E1)H, (E2)H, (E3)H etc (E1)H, (E2)H, (E3)H are the values of energy of H atom moving in 1st, 2nd and 3rd orbit respec- tively. (E1)H = – 2.18 × 10–11 × (1)2 erg = – 2.18 × 10–11erg (1)2 (E2)H = – 2.18 × 10–11 × (1)2 erg = – 0.5448 × 10–11erg (2)2 (1)2 (E3)H = – 2.18 × 10–11 × (3)2 erg = – 0.2421 × 10–11erg Thus, (E1)H < (E2)H < (E3)H............ Thus the energy of electrons goes on increasing as it moves away from the nucleus. Why is Energy of an electron negative ? Consider an electron situated at infinite distance from the nucleus. There is no interaction between the electron and the nucleus. As the electron approaches the nucleus, it does some work and some energy is released due to attractive forces xrating between the nucleus and the elec- tron. Thus the energy of the electron decreases. The energy of electron increases i.e (becomes less and less –ve) as it moves away from the nucleus. Expression for velocity of an electron From eqn (2), v2 = Ze 2 mr From eqn (4), v = nh 2pmr On dividing eqn (2) by eqn (4) we have v2 = Ze2 × 2pmr = 2pZe2 or vn = 2pZe2 v mr nh nh nh (i) Value in CGS unit Putting p = 3.14 e = 4.8×10–10 esu h = 6.624 × 10–27 erg.s and using the relation erg = gcm2s2 esu = g½ cm3\\2s–1 We get, vn = Velocity of electron in nth orbit of H atom ( )= 10 2 2× 3.14× Z× 4.8× 10 1 n× 6.624× 10 27 cm s

128 +2 CHEMISTRY (VOL. - I) = 21.8368× 107 × Z cm s 1 n = 2.1836× 108× Z cm s 1 n = 0.2136× 109× Z cm s 1 n (ii) Value of SI unit vn = 2pZe2 , where 4p Î0 Permittivity factor. 4p ∈ 0 .nh Ze2 = 2 Î0 .nh Putting the value of e, Î0 , h etc. vn = 0.2185× 107× Z ms 1 n Number of revolutions made by an electron moving with a velocity vn in nth orbit is given by vn 2π rn Substituting the value of vn & rn in the above expression, Number of revolutions = 65.711´1014 ´ Z 2 s -1 n3 Problems (i) Find the expression for energy of an electron in nth orbit of He+, Li2+ and Be3+ ions. Solution For He+, Li2+ and Be3+ , Z = 2, 3 and 4 respectively. And all are 1 electron species. EHe+ = 2p2me4Z2 = 2p2me4 (2)2 = 8p2me4 n2h2 n2h2 n2h2 ELi2+ = 2p2me4 (3)2 = 18p2me4 n2h2 n2h2 2p2me4 (4) 2 32p2me4 n2h2 n2h2 EBe3+ = = (ii) Compare the frequencies and wavelengths of radiation emitted by H atom with those emitted by He+, Li2+ and Be3+ ions. Solution We know that E = h u or un = En n

STRUCTURE OF ATOM 129 Again Q En = 2p 2 me 4 Z 2 , un = 2p2me4Z2 n2h2 n2h3 For H atom, Z = 1, so uH = - 2p2me4 n2h3 uHe+ =- 8p2me4 , uLi+ = 18p2me4 , uBe3+ = 32p2me4 n2h3 n2h3 n2h3 So , uH = 2p2me4 × n2h3 = 1 uHe+ n2h3 8p2me4 4 uH = 2p2me4 × n2h3 = 1 uLi2+ n2h3 18p2me4 9 uH = 2p2me4 × n2h3 = 1 uBe3+ n2h3 32p2me4 16 Comparison of wave length We know that, E = h u = hc l So , l = hc E lH = hEc= 2p hc = n2h3c 2me4 2p2me4 n 2h 2 l He+ = n2h3c , n2h3c - 8p2me4 lLi+ = -18p2me4 n2h3c lBe3+ = - 32p2me4 Thus, lH = 4 , lH = 9 , lH = 16 lHe+ 1 lLi2+ 1 lBe3+ 1 3.7 SOLAR SPECTRUM When sunlight or any other white light is passed through a prism, it is separated into different colours. Assembly of these colours is called spectrum . Different colours correspond to waves of different frequencies and wavelengths. The wavelengths of various radiations constituting the visible spectrum are shown below.

130 +2 CHEMISTRY (VOL. - I) WAVELENGTH (A0) 7 X 103 6 X 103 5 X 103 4 X 103 Low Infrared Red Green Yellow Ultra High energy Violet Violet energy Green Blue 4 X 10–5 7 X 10–5 6 X 10–5 5 X 10–5 WAVE LENGTH (CM) Fig. 3.9 Wavelengths of visible spectrum The wavelengths of some important electromagnetic radiations are given below. (Fig.3.10) VISIBLE 3RADIO WAVE MICRO WAVE U3V X-R3AYS GAMMA COSMIC LOW 3 3INFRARED RA3YS RA3YS HIGH ENERGY ENERGY ® 2x104 30 5x10–2 1.5x10–6 10–9 10–10 5x10–12 WAVE LENGTH (CM) Fig. 3.10 Wavelengths of electromagnetic radiations. ATOMIC SPECTRA If sunlight is passed through a prism, it gets separated into a continuous spectrum extending from red at one end to violet at the other. But when a crystal of sodium chloride or any other sodium salt is shown to the bunsen flame, a bright yellow light is produced. If this light is passed through a prism and then allowed to fall on a photographic plate (or examined through a spectroscope) it is seen to consist of two isolated yellow lines separated by dark space. The wavelengths of these two lfionuensdatrhea5t8e9a0ch(Del1e)manednt5e8m96itsAi0ts(Do2w).nScuhcahraacstpeericsttriucmlinise known as line spectrum. It has been spectrum which is different from that of any other element. Therefore, the atoms of the elements produce line spectra. Hence, line spectra is called atomic spectra. Absorption and Emission spectrum — When an element is exposed to a strong source of light (light from arc lamp), one of the electrons present in the atom of the element may absorb energy and pass from a lower energy level to a higher energy level. This is in ct'aEhoc2cer'roetsrhspdepeanocnntcrdeEui2nwmg–i.tthEToB1heio=nshedrrDg'asyrEkthEwel2ioni–lrelyE.bc1Ioefwntasihbltelistobuerelteebmcsetdrai.sobsnAsinosjgurampatnprieodssnfucrlosotnpmoseefceqttnruheueirsnmgtytl.hylWeeavshdepelaen'crEtka1r'naltlioanlteiaonmwleeivolleloflsafeoprsefpqeeeunnaeeernrrggcinyyy so that the spectral line corresponding to Eco2 n–stEit1u=tesD E of specific wavelenght will be emitted. This will appear as a bright line which the emission spectrum. For example, in case odfarskodliinuems ,atphpeeatwr oinlitnheesircpolrarcees.pTonhdeisnegtwtoowdaavrkelleinngetshcsoonfstDit1utaendtheD2abasroerfpotiuonnd missing. Two spectrum of sodium.

STRUCTURE OF ATOM 131 3.8 BOHR'S THEORY AND HYDROGEN SPECTRUM In the hydrogen atom, where there is one proton and one electron, the energy of the electron is given by the expression : -2p 2me4 = - 313.5 k.cal per mole................ (1) n2h2 n2 where m = mass of the electron, e = charge of electron, h = planck's constant = 6.624 x 10–27 erg.sec.and n = 1, 2, 3, 4 etc. Substituting the value of 'n', energy of electron in various energy levels can be easily calculated . Negative value of energy of electron is accounted for by the following explanation. When the electron is at infinite distance from the nucleus, there is no electrical interaction between the two and energy of the electron is taken as zero. When electron moves closer to the nucleus, work is done by the electron due to electrical attraction. Consequently, the energy of the electron decreases i.e. it becomes negative. It becomes more and more negative as the electron moves nearer and nearer to the nucleus. Hydrogen atom contains only one electron, but its spectrum consists of a large number of lines. This is explained by the fact that a sample of hydrogen contains very large number of atoms. When energy is supplied to the sample of hydrogen, different atoms absorb different amount of energy. The single electron in different atoms will absorb different amount of energy and shift to different energy levels such as second, third, fourth or even higher energy levels. The electrons then tend to fall back to one or other of the lower energy levels. The various possibilities by which electrons fall back from the various excited states to lower energy levels are shown in (Fig. 3.11). During each transition, energy is released which appears in the form of radiations of specific frequency and specific wavelength. PFUND SERIES n1= 5, n2 = 6,7,8.... BRACKET SERIES n1= 4, n2 = 5,6,7.... PASCHEN SERIES n1= 3, n2 = 4, 5, 6.... BALMER SERIES n1= 2, n2 = 3, 4, 5, 6 .... LYMAN SERIES n1= 1, n2 = 2, 3, 4.... Fig.3.11 Origin of hydrogen spectrum.

132 +2 CHEMISTRY (VOL. - I) Suppose the electron lies in the energy level n2 where its energy is E2 and electron jumps to lower energy level n1 where its energy is E1. Then according to the equation of electron energy, -2p 2me4 E2 = -2p 2me4 and, E1 = 2n 2 h2n22 h 1 Difference in energy, D E = E2 – E1 = hu or, hu = -2p 2me4 + 2p 2me4 h2 2 h 2 n12 n 2 2 p 2 me 4 1 1 h3 n12 n22 ( )or, u = - 2p 2me4 1 1 ..............................(2) h3c n12 n22 ( )or, u = u = - c Emission of various series of spectral lines observed in the emission spectra of hydrogen can be explained with the help of equation (2). Lyman series is produced when electron jumps from second, third, fourth or higher energy levels to the first energy level. Frequencies of various lines of this series are obtained by substituting n1= 1 and n2= 2 , 3 , 4 etc in equation (2). Similarly, frequencies for lines of Balmer series lines are obtained by putting n1= 2, n2= 3, 4, 5, 6 etc. In the same way for Paschen series, n1= 3, n2= 4, 5, 6, 7 etc. Bracket series n1= 4, n2= 5, 6, 7, 8 etc and Pfund series n1= 5, n2= 6, 7, 8 etc. The frequencies of the spectral lines of different series calculated by equation (2) almost agree with the experimentally determined values. This has offered a strong support in favour of the Bohr's theory of hydrogen atom. Ritz combination principle — Equation (2) given above is similar to Ritz equation ( )u=1 = u = R = 1 - 1 . Here R is called the Rydberg constant, which is given by l c x2 y2 2p 2me4 R= h3c . The value of 'R' is found to be 109737 per cm which is comparable to the experimental value of 109678 per cm obtained by Ritz . Equation for the various series of spectral lines of hydrogen atom is similar to the Ritz equation. This gives further support to the Bohr's theory of atom. Limitations of Bohr's model — Using spectroscopes of high resolving power the line spectra of hydrogen was found to consist of a group of very fine lines. To explain this Sommerfeld put forward the idea of elliptical orbits. Thus. the circular orbit of Bohr is a special case of elliptical orbits. Hence, Bohr's theory is referred as Bohr - Sommerfeld theory . Following are the limitations of the theory. 1. Spectra of many electron atoms can not be explained by Bohr's theory. 2. Bohr's orbits are planar, but according to the modern concept electron moves in three dimensional orbits. 3. It does not accept the dual (wave and particle) nature of electron. 4. According to Bohr's therory the position and momentum of electron can be determined which is objected by Heisenberg's uncertainity principle.

STRUCTURE OF ATOM 133 3.9 DUAL NATURE OF ELECTRON Einstein in 1905 suggested that light has a dual character, particle as well as wave. In 1923, French physicist Louis de Broglie suggested that matter has also a dual character, particle and wave. Therefore, according to him electron has also the dual character. He derived an expression for calculating the wavelength ' l ' of a particle of mass 'm' moving with velocity h 'u' according to which l = mu . The equation was derived in the following manner. According to Einstein, E = mc2, also E = hu. Equating, hu = mc2 . Since u = C l , substituting the value of u, we have hc = mc2 l or, l = mhc Substituting 'c' with velocity of electron 'u', the equation becomes l = mhu . or, l = ph where p = momentum = mu. (de Broglie equation) 3.10 QUANTUM NUMBERS Quantum numbers are identification marks for the electrons or in other words they give the addresses of the electrons. In an atom. there can be many electrons which are alike. So they can only be identified by the quantum numbers. There are four quantum numbers. These are : 1. Principal quantum number 'n'. 2. Azimuthal quantum number 'l' 3. Magnetic quantum number 'm'. 4. Spin quantum number 's'. All the quantum numbers are discussed below. 1. Principal quantum number (n) — It determines the shell or energy level or orbit of the electron. It has integral values 1, 2, 3, 4 etc which indicates the shells or energy levels K, L, M, N etc respectively. It also determines the total energy of the electron in an energy level. En = - 2p 2 me 4 , where En = Energy of the electron in nth orbit and p2h2

134 +2 CHEMISTRY (VOL. - I) p m, e and h have their usual meaning. From the wave mechanical concepts, principal quantum number gives the size of electron wave. 2. Azimuthal quantum number (l) — It is also called secondary quantum number or angular momentum quantum number or orbital quantum number. It determines the sub-energy level of the electron. It also determines the energy contribution due to the angular momentum or orbital motion of the electron. From the wave mechanical concept it determines the shape of electron wave. 'l' carries 'n' number of values. It has values from 0 to n –1. Depending upon the value of 'l' the sub-shells of the electron are indicated. Value of 'l' Sub-Shell 0s 1p 2d 3f When n = 1, the electron is in K-shell and 'l' value is 'o' indicating that the electron is present is s-subshell or orbital. When n = 2 electron is in L - shell, 'l' has values 0 and 1, which indicates that in M - Shell, s and p - sub shells are present. When n = 3, electron is in M-Shell, the 'l' values are 0, 1, 2, which indicates that in M-Shell, s, p, and d - sub shell are present. Similarly, when n = 4 for N-shell, 'l' values are 0, 1, 2, 3 which indicates that in N-shell, s, p, d and f- subshells are present. 3. Magnetic quantum number (m) : Motion of the electron in an orbit is comparable to flow of electric current in a loop. The flow of current generates a magnetic. field which interacts with external electric or magnetic field and as a result the electrons in a given energy sub-level orient themselves in certain specific region of space around the nucleus. These regions of space are called orbitals. Thus, magnetic quantum number determines the orientation of orbital in space or the number of orbitals of a particular type in a shell or energy level. 'm' can have values from +1 through 'o' to –1. The number of values it carries is (2l + 1). For example when l = 0, m has only one value 'o' which means in a shell there can be only one 's' orbital. When l = 1, m = –1, 0 and + 1, thus there can be three p-orbitals in a shell i.e. p- orbitals are three in number or it has three orientations. When l = 2, m = – 2, – 1, 0, + 1, + 2, five values, thus d-orbitals will be five in number or it has five orientaitons. Similarly, when l = 3, m = – 3, – 2, – 1, 0, + 1, + 2, + 3, seven values. Thus, 'f' orbital is seven in number or it has seven orientations.

STRUCTURE OF ATOM 135 4. Spin quantum number (s) — Electron during its motion around the nucleus also rotates or spins about its own axis. A charged particle which spins about its own axis behaves as a small magnet. Spin of the electron adds to the angular momentum and contributes to the total energy. The electron can spin in clockwise or anticlockwise direction and accordingly spin quantum number 's' has two values i.e. + ½ and –½ respectively. Thus there can be maxium two electrons with opposite spins in an orbital. 3.11 PAULI'S EXCLUSION PRINCIPLE Position of an electron in an atom can be ascertained by the four quantum numbers. These quantum numbers give idea about the position of electron in the main energy level (n), sub-energy level (l), orientation of sub-energy level (m) and the direction of spin (s). It is possible to identify the electron by means of the quantum numbers and these serve as address for the electron. But identification of electrons is posiible only on the basis of restriction imposed by the Pauli's exclusion principle. It states that no two electrons in a given atom can have all the four quantum numbers same Maximum three quantum numbers can be same, but the fourth quantum number must be different, otherwise it will not be possible to identify the electron. For example, suppose there are two electrons in the K-shell of an atom. Let us write down the quantum numbers for the two electrons. nl ms Ist electron 1 0 0 + 1 2 2nd electron 1 0 0 – 1 2 Thus, three quantum numbers are same, but the fourth one is different. In any orbital there can be maximum two electrons of opposite spin. since these two electrons are in K-shell, they are in the s-orbital of K-shell. 3.123HEISENBERG'S UNCERTAINTY PRINCIPLE AND THE CONCEPT OF PROBABILITY The principle states– \"It is not possible to determine exactly both position and momentum of a small particle like electron simultaneously. According to Bohr's model, the electron in an orbit is at a fixed distance from the nucleus and its position and velocity can be determined. But according to uncertainty principle it is not possible to determine both position and momentum of the electron. The physical significance of the principle can be known from the following. In order to observe something, light photons must strike the object. Electron is a small particle like photon. During the observation due to the impact of photon on the electron, the position and

136 +2 CHEMISTRY (VOL. - I) velocity of the electron will suffer a change. Hence, the position and velocity can not be determined simultaneously. Uncertainty principle is expressed mathematically as, ( D x) ( D p) » h 4p where ' Dx' is the uncertainty in determining position and ' D p' is the uncertainty in determining momentum. If ' D x' becomes very large, D P , becomes very small and momentum can be determined. When D p is very large, D x becomes very small and position can be determined. Thus, Bohr's concept of definite orbit or difinite position of the electron becomes vague. Only it is possible to predict the probability of locating an electron of definite energy in a given region in space. CONCEPT OF ORBITAL While Bohr's model restricts the electron to a definite orbit at a fixed distance from the nucleus, the wave meachanical model gives merely the probability of locating the electron at a given distance from the nucleus. Thus, the orbital is the region in space around the nucleus where there is maximum probability of finding the electron. Difference between Orbit and Orbital — Orbit Orbital 1. It is the circular path around the 1. It is the three dimensional space around the nucleus in which electron moves nucleus where there is maximum, probability of finding the electron. 2. It has a circular shape 2. The shape of different orbitals are different 3. In it electron moves in one plane 3. In the orbital electron moves in three dimensional space. 4. In the orbit position and momentum 4. In the orbital position and momentum of of the electron can be determined. the electron can not be determined simultancously. 5. In the orbit the maximum number of 5. In an orbital there can be maximum two electrons is determinded by 2n2 rule. electrons of opposite spin. SHAPES OF ORBITALS s-orbital - Since s-orbital has only one orientation, it is spherically symmetrical about the nucleus and is undirectional. As the size of the orbital depends on the value of the principal quantum number, 2s-orbital is larger than 1s orbital (Fig. 3.12). Each s-orbital is symmetrical along the three axes, x, y and z. S - orbital Fig. 3.12

STRUCTURE OF ATOM 137 p - orbitals : There are three p-orbitals (px, py and pz). Since they are of equal energy, they are oriented along the three axes x, y and z which are mutually at right angles to each other. Each orbital looks like two pears attached at their narrow ends. It looks like a dumb- bell or it has the shape of a flattened sphere. The lobes of px, py and pz orbitals are directed along the three axes x, y and z respectively (Fig. 3.13 – a, b, c) px - orbital py - orbital pz - orbital Fig. 3.13 (a) Fig. 3.13 (b) Fig. 3.13 (c) d - orbitals : There are five d-orbitals. These are of equal energy, but differ only in their orientations in three dimensional space. Three of these orbitals (dxy , dyz , dzx) project in between the coordinate axes x–y, y–z and z–x respectively. The other two orbitals (dx2- y2 , dz2 ) lie along the coordinate axes. (Fig. 6.14) Y XX YZ Z dxy X dxz Z dyz Y d x2 - y2 d z2 Fig.3.14 Shapesofd - orbitals 3.13 AUFBAU PRINCIPLE The meaning of \"aufbau\" is building up. It states that electrons enter various orbitals in the increasing order of energies. It means that the orbital of lowest energy is filled up first and then electrons go to the orbitals with increasing order of energies. The sequence of filling up of orbitals is shown below (Fig 3.15). The following rules are taken into account during filling up of the orbitals. (i) (n+l) Rule - The orbital with lower (n+l) value possesses lower energy therefore is filled up first. e.g. 4s orbital is filled up prior to 3d orbital, n+l = 3 + 2 = 5

138 +2 CHEMISTRY (VOL. - I) (ii) If the (n+l) value is same for both the orbitals, then the orbital having lower 'n' value is filled up first . e.g. 3p orbitals are to be filledup prior to 4s orbital For 3p, n+l = 3+1 = 4 and for 4s , n+l = 4+0 = 4. Energy 5s A new electron enters the orbital where 4p (n+l) value is minimum. In case the 3d values of (n+l) are same, the new electron 4s enters the orbital where n value is 3p minimum. 3s 2p 1s 2s 1s 2s 2p 03s 3p 3d 0 04s 4p 5s Fig. 3.15 - Sequence of filling of the orbitals. 3.14 HUND'S RULE Hund's rule of maximum multiplicity states that \"electron pairing in p, d and f- orbitals take place only after all the orbitals of a given set have one electron each.\" On the basis of application of Hund's rule electronic configurations of some elements are shown below (Fig. 12.14) Nitrogen Is 2s 2p (atomic number-7) Oxygen (atomic number-8) Fluorine (atomic number-9) Neon (atomic number-10) (Fig. 3.16) 3.15 ELECTRONIC CONFIGURATION OF ELEMENTS Distribution of electrons in atoms of elements is guided by the following rules. 1. 2n2 rule – The maximum capacity of the different energy levels is given by this rule.

STRUCTURE OF ATOM 139 Energy level. n Maximum number of electrons. K1 2 x 12 = 2 L2 2 x 22 = 8 M3 2 x 32 = 18 N4 2 x 42 = 32 2. Aufbau principle - Electrons fill the orbitals in the increasing order of energies. 3. Pauli's exclusion principle - An orbital can have maximum two electrons of opposite spin. 4. Hund's rule of maximum multiplicity - Electron pairing in p,d,f orbitals does not take place till all the orbitals of the given set have one electron each. Electronic configuration of atoms can be presented by box diagram method or according to the formula nlx. Box -diagram method : In this case each orbital is represented by a box and electrons are indicated by arrow in the box. For example – Carbon (atomic number-6) -¯ -¯ -- Nitrogen (atomic number-7) -¯ -¯ - - - Oxygen (atomic number-8) -¯ -¯ -¯ - - Electronic configuration by the formula nlx. n = 1, 2, 3, 4 etc, the main energy levels or shells l = orbital x = number of electrons in the orbital For example – Hydrogen (atomic number = 1) = 1s1 It has a single electron in the K-shell or first shell and s-orbital. Helium (atomic number = 2) = 1s2 Lithium (atomic number = 3) = 1s22s1

140 +2 CHEMISTRY (VOL. - I) Electronic configuration of different atoms upto atomic number 30 is given below. Element Symbol Atomic number Electronic Configuration Hydrogen H 1 1s1 Helium He 2 Lithium Li 3 1s2 Beryllium Be 4 Boron B 5 1s2 2s1 Carbon C 6 Nitrogen N 7 1s2 2s2 Oxygen O 8 Florine F 9 1s2 , 2s2 2px1 Neon Ne 10 1s2 , 2s2 , 2px1 , 2py1 Sodium Na 11 1s2 , 2s2 2px1 2py1 2pz1 Magnesium Mg 12 1s2 , 2s2 2px2 2py1 2pz1 Aluminium Al 13 1s2 , 2s2 2px2 2py2 2pz1 Silicon Si 14 1s2 , 2s2 2px2 2py2 2pz2 Phosphorus P 15 1s2 , 2s2 2p6 3s1 Sulphur S 16 Chlorine Cl 17 1s2 , 2s2 2p6 3s2 Argon Ar 18 *Potassium K 19 1s2 , 2s2 2p6 3s2 3px1 Calcium Ca 20 1s2 , 2s2 2p6 3s2 3px1 3py1 Scandium Sc 21 1s2 , 2s2 2p6 3s2 3px1 3py1 3pz1 Titanium Ti 22 1s2 , 2s2 2p6 3s2 3px2 3py1 3pz1 Vanadium V 23 1s2 , 2s2 2p6 3s2 3px2 3py2 3pz1 *Chromium Cr 24 1s2 , 2s2 2p6 3s2 3px2 3py2 3pz2 Manganese Mn 25 1s2 , 2s2 2p6 3s2 3p6 4s1 Iron Fe 26 1s2 , 2s2 2p6 3s2 3p6 4s2 Cobalt Co 27 Nickel Ni 28 1s2 , 2s2 2p6 3s2 3p6 4s2 3d1 *Copper Cu 29 Zinc Zn 30 1s2 , 2s2 2p6 3s2 3p6 4s2 3d2 1s2 , 2s2 2p6 3s2 3p6 4s2 3d3 1s2 , 2s2 2p6 3s2 3p6 4s1 3d5 1s2 , 2s2 2p6 3s2 3p6 4s2 3d5 1s2 , 2s2 2p6 3s2 3p6 4s2 3d6 1s2 , 2s2 2p6 3s2 3p6 4s2 3d7 1s2 , 2s2 2p6 3s2 3p6 4s2 3d8 1s2 , 2s2 2p6 3s2 3p6 4s1 3d10 1s2 , 2s2 2p6 3s2 3p6 4s2 3d10 *Chromium should have the configuration 4s23d4, but actually it has the configuration 4s13d5 in the valence shell because 3d5 i.e half filled configuration is a stable configuration. Similarly, Copper should have the configuration 4s23d9, but actually it has the configuration 4s13d10, because 3d10 (completely filled) configuration is a stable configuration.

STRUCTURE OF ATOM 141 3.16 EXTRASTABILITY OF HALF FILLED AND COMPLETELY FILLED ORBITALS There is a difference between electronic configuration and electronic arrangement. Electronic configuration tells us that how many electrons are present in a set of orbitals. But electronic arrangment tells us the manner in which the electrons are placed is a given set of orbitals. Let us take the case of 3d2 configuration in Ti2+ ion. These two electrons can be placed in a set of five 3d orbitals in as many as 45 different ways. Some of the arrangements are shown below. Thus a given electronic configuration may lead to a large number of electronic arrange- ments. When the two electrons are placed in same or different orbitals they are subjected to mutual repulsion. The more the repulsion the greater is the energy and less is the stability. A stable electronic configuration is always concerned with stable electronic arrangement of that configuration . Hund’s rule of maximum multiplicity predicts the most stable electronic arrange- ment belonging to a given electronic configuration. Hund’s rule and its application According to Pauli’s exclusion principle ‘‘No two electrons in an atom can have an identi- cal set of quantum numbers’’. Hence, if two electrons have the same spin i.e the same spin quantium number they must have to occupy different orbitals since they can not have the same value of n, l and m. On the other hand, if the two electrons have opposite spin they can occupy the same orbital i.e. they have the same value of n, l and m. According to Hund’s Rule The most stable electronic arrangement is one in which there is maximum number of unpaired electrons. The degenerate orbitals are orbitals of equal energy. This is due to the fact that the degen- erate electronic arrangements have the same inter electronic repulsions and hence may have the same energy. Let us take 3 degenerate p-orbitals and find out the most stable electronic arrange- ment for a p2 configuration. There are 15 ways of arranging these two electrons in a set of 3 p- orbitals. (1) (2) (3) In arrangement (1) the two electrons are placed in the same orbital. The electrostatic repul- sion is more, so energy is more and this arrangement is least stable. The stability of arrangement of (2) or (3) is decided on the basis of concept of exchange energy. Exchange energy If the positions is space of the two electrons having parallel spins are exchanged, there is no change in electronic arrangement and this leads to decrease in energy. The pair of electrons is known as exchange pair and the decrease in energy per exchange pair of electrons is known as exchange energy. Exchange energy carries a –ve sign.

142 +2 CHEMISTRY (VOL. - I) There is one exchange pair of electrons in arrangement (2) and none in arrangement (3). The arrangement (2) in thus associated with lower energy than arrangement (3). Thus arrange- ment (2) is more stable than arrangement (3). Pairing Energy - The energy required for placing two electrons with antiparallel spins in the same orbital is known as pairing energy. This pairing energy is having +ve sign. It tends to increase the energy of the system i.e. to destabilise it. Consider the following arrangement for p3 configuration. If E - Exchange energy (–ve sign) stabilising P - Pairing energy (+ve sign) destabilising Arrangement (1) (–E +P), since one exchange pair of electrons (a, b) and one pair of electron (a). Arrangement (2) (–3E + 0), since three exchange pair of electrons (a,b) , (a,c) and (b,c) and no pair of electrons. Arrangement (3) (–E + 0), since one exchange pair of electrons (a,b) and no pair of electron. Here, order of stability is 2 > 3 > 1 Thus, the electronic arrangement containing a set of exactly half filled degenerate orbitals is found to be most stable. Stability of completely filled orbitals : Consider the electronic arrangement of p4, p5, p6 configurations p6 p4 p5 (1) (2) (3) The relative stabilities of the three arrangements can not thus be predicted on the basis of aggregates of exchange energy and pairing energy. Experimentally it has been found that ar- rangement (3) is the most stable arrangement. Hence there most be some other stabilising factor in favour of arrangement (3). The factor is symmetrical distribution of charge. Configurations having uniform or sym- metrical distribution of charge in all directions are associated with lower energy and hence higher stability. Thus the configurations having exactly half filled orbitals or completely filled orbitals have symmetrical distribution of charge and hence are considered to be more stable. Solved Problems Example 1. Calculate the number of protons, neutrons and electrons in an isotope of the element having atomic number 9 and atomic weight 19. Solution Atomic number = number of protons = number of extranuclear electrons = 9. Number of neutrons = Atomic mass - atomic number = 19 – 9 = 10 Threfore, There are 9 protons, 9 electrons and 10 neutrons. Example 2 An oxide of nitrogen has a molecular weight of 30 Find the total number of electrons in one molecule of the compound (Atomic numbers, N = 7, O = 8)

STRUCTURE OF ATOM 143 Solution Molecular formula weight of the compound = 30 The oxide would have the formula = NO Atomic number of 'N' and 'O' are 7 and 8 respectively, So total number of electrons = 7 + 8 = 15. Example 3 A neutral atom of an element has 2K, 8L and 5M electrons. Find out the Solution following : (a) Atomic number (b) Total number of s-electrons (c) Total number of p-electrons (d) Number of protons (e) valency of the element. Atomic number = no. of protons = no. of electrons = 2+8+5=15 Total no. of p-electrons = 2 in K-shell + 2 in L-shell + 2 in M-Shell = 6 Total no. of p-electrons = 6 in L-shell + 3 in M - shell = 9 No. of protons = 15 Electronic configuraion = 1s2, 2s22p6, 3s2 3p3. Since there are 3-electrons in p-orbital, valency = 3. Example 4 An atom with an atomic mass of 28 has a ground state electronic configuration 1s2, 2s2 2p6, 3s2, 3p2. Find the atomic number of element, number of electrons in valence shell, number of neutrons. Solution Atomic number = 14, valency = 4 Number of neutrons = 28 – 14 = 14 Example 5 An electron has mass = 9.1 x 10–28g. and is moving with a velocity of 105 cm/ Solution sec. Calculate its kinetic energy and wavelength where, h = Planck's constant = 6.625 x 10–27 erg. sec. K.E = 1 mv2 = 1 x 9.1 10–28 x 105 x 105 = 4.55 x 10–18 ergs. 2 2 l = h = 6.625 x 10-27 = 7.28 x 10–5 cm. mv 9.1 x 10-28 x 10-28 Example 6 Calculate the de Broglie wavelength associated with a particle having an energy of 7.7 x 10–13 J and a mass of 6.6 x 10–24 g. ( h = 6.6 x 10–34 J. ) Solution Energy = E = 1 mv2 = 7.7 x 10–13 J. Example 7 2 v= 7.7 x 10-13 x 2 = 1.525 x 105 cm. sec–1. 6.6 x 10-24 6.6 x 10-34 l = mhv = 6.6 x 10-24 x 1.525 x 10-5 = 6.56 x 10–16 cm . Calculate the wavelength associated with an electron (mass = 9.1 x 10–31 g.) moving with a velocity of 103 m/sec. ( h = 6.6 x 10–34 kg. m2sec–1 ) Solution l = mhv = 6.6 x 10-34 9.1 x 10-31 x 103

144 +2 CHEMISTRY (VOL. - I) Example 8 The energy of the electron in the second and third orbit of hydrogen atom is – 5.42 x 10–12 erg. and –2.41 x 10–12 erg. Calculate the wavelength emitted when Solution the electron drops from the third to the second orbit. (h = 6.626 x 10–27 erg. sec.) DE = E2 – E1 = –2.41 x 10–12 – (–5.42 x 10–12 ) ergs = 3.01 x 10–12 erg. DE = hc or, l = hc = 6.626 x 10-27 x 3 x 1010 erg.sec l DE 3.01 x 10-12 erg. = 6.63 x 10–5 cm or 6630 A0 Example 9 What is the mass of photon of sodium light ? Solution l = 5894 A0 = 5.894 x 10–7 metres, sec–1 V = 3 x 108 metres, h=6.6 x 10–34 kg.m2. sec–1) l = h or m = h = 6.6 x 10-24 x 103 g.m2 sec-1 =3.73 x 10–33 g. mv lv 5.894 x 10-7 x 3 x 103 m2 sec-1 Example 10 Wavelength of blue light is 4800 A0. Calculate the frequency and wave-number of light. Solution l = 4800 A0 = 4800 x 10–8 cm. C = 3.0 x 1010 cm. u = c = 3.0 x 1010 = 6.25 x 1014 sec–1. l 4800 x 10-8 Wave number = u = 1 = 1 = 2.08 x 10–1 cm–1. l 4800 x 10-8 Example 11 Calculate the frequency of radiation of visible light having wavelength 750cm. Solution C= ul or u = c Here C = 3 x 108 ms–1, l = 750 nm. l = 750 x 10–9 m. u = c = 3 x 108 ms-1 = 4 x 1014 s–1 = 4 x 1014 Hz. l 7.5 x 10-7 m Example 12 Energy difference between two electronic states of hydrogen atom is 245.9 kJ mol–1. Calculate the wavelength of light emitted when an electron drops from the higher to the lower state. ( h = 3.99 x 10–13 kJ sec. mol–1 ) Solution DE = hu = hc or l = hc = 3.99 x 10-13 kJ Sec mol-1 x 3 x 108 m sec-1 l DE 245.9 kJ mol-1. = 4.867 x 10–7 m = 4.867 x 102 nm. Example 13 Calculate the wavelength of the third line in the Brackett series of hydrogen. ( R = 10.97 x 106 m–1 )

STRUCTURE OF ATOM 145 Solution ( )Rydberg equation,1 x 1 1 l = 10.97 106 m–1 n12 – n22 Example 14 Solution For the Brackett series n1= 4 and n2=7 . substituting the values of n1 and n2 in the above equation. ( )1 1 1 42 72 l = 10.97 x 106 m–1 – – 10.97 x 106 m–1 (0.0420918) = 461747 m–1 l = 2.1 x 10–6 m . Calculate the wavelength of spectral line when electron in a hydrogen atom undergoes a transition from an energy level with n=4 to n=2. What is the colour of radiation ? 1 1 1 1 1 n12 n22 22 42 ( ) ( )l = 10.97 x 106 m–1 – = 10.97 x 106 m–1 – = 10.97 x 106 m–1 (0.25 – 0.25) = 10.97 x 106 m–1 x 0.1875 l = 4.86 x 10–7 m = 486 nm., colour of light = bluish green Example 15 Calculate the momentum of a particle which has a de Broglie wavelength Solution of 0.1 nm. l = 0.1 nm = 0.1 x 10–10 m , = 1010 m. l= h , p = momentum p p = h = 6.62 x 10-34 kg m2s-1 = 6.62 x 10–24 Kg. m. s–1. l 10-10 m Example 16 Write the values of 'l' and 'm' for an electron in 3d orbital Solution For a d – orbital, l = 2 When l = 2, m = any one of –2, –1, 0 +1, +2. Example 17 For an atom, the maximum value of 'm' obtained from experiment is +2. What are the possible values of n, l, and m) Solution l = 0 to n–1 m = –l through zero to +l. When, m = +2, l = 2, n = 3 (since l = n–1) and m = –2, –1, 0, +1, +2. Example 18 An electron is present in 4s - sub shell. What are the possible values of n, l and m. Solution n = 4, l = 0, m = 0 Example 19 An electron is present in 4f subshell. What are the possible values of quantum numbers. Solution n = 4, l = 3, m = any one of –3, –2, –1, 0, 2, 3, s = + 1 and – 1 2 2

146 +2 CHEMISTRY (VOL. - I) Example 20 Quantum numbers of an electron are given below. In which orbital the electron will be present. Solution (i) n = 3, l = 1, the orbital is 3p (ii) n = 4, l = 0 , the orbital is 4s Example 21 Which of the following sets of quantum numbers are not permissible ? Solution (i) n = 2 , l = 1, m = 0, s = – 1 2 (ii) n = 2, l = 2, m = –1, s = – 1 2 (iii) n = 3, l = 2, m = 0, s = – 1 2 (iv) n = 3, l = 2, m = +1, s = 0 Solution (ii) is not allowed as when n = 2, l can not be 2 (should be n – 1) (iv) is not allowed. Here s can not be zero. CHAPTER (3) AT A GALANCE 1. Electron, proton and neutron are the three fundamental particles of an atom. 2. Charge of electron is 1.602 x 10–19 coulombs and mass is 9.1 x 10–28 Kg. 3. Electrons are universal constituent of matter. 4. Mass of proton is 1.67 x 10–27 Kg where as that of neutron is 1.67 x 10–27 Kg. 5. Atomic number : The number of unit positive charge carried by the nucleus of an atom or the number of electrons present outside the nucleus of an atoms. 6. Mass number : The sum of the number of protons and neutrons present in the nuclus of an atom. 7. Isotopes : Atoms of same element having same atomic number but different mass number. 8. Isobars : Atoms of different elements having same mass number but different atomic numbers. 9. Isotones : Atoms of different elements which possess the same number of neutrons. 10. Dual nature of electron was suggested by de Broglie. According to him electron behaves both as a particle and also as a wave. 11. Quantum numbers are identification marks for the electrons in an atom. Position of an electron in an atom can be ascertained by these quantum nos. The four quantum numbers are Principal, azimuthal, magnetic and spin quantum number.

STRUCTURE OF ATOM 147 12. Paulis Exclusion principle : No two electrons in an atom can have the four quantum numbers same. Heisenberg's uncertainty principle : It is not possible to determine exactly the position and momentum of an electron simultaneously. 14. Orbit is the circular path around the nucleus in which the electron moves where as orbital is the region of space around the nucleus where the probability of finding an electron is maximum. 15. There is only one 's' orbital, three p – orbitals (px, py & pz), five d-orbitals (dxy, dxz, dyz, d x2 -y2 , d z2 ) and seven f-orbitals. 16. Hund's Rule of maximum multiplicity : States that electron pairing in p, d and f orbitals takes place only after all the orbitals of a given set has one electron each. 17. Aufbau Principle : Electrons enter the various orbitals in the increasing order of energy. 18. Some important relations : (a) Planck's constant h = 6.62 x 10–27 erg. sec. = 6.62 x 10–34 J. sec. = 6.62 x 10–34 Kg. m2. sec–1. (b) Rydberg constant 'R' = 109678 cm–1. (c) Wave number 'y ' = 1 R 1 1 l n12 n22 (d) Energy 'E' = hy = h c = hcy l (e) de-Broglie wavelength l = h = h . p mv (f) Heisenberg's Uncertainty Principl Dx ´ Dp » h 4p QUESTIONS (A) Very short answer type questions. (1 marks) 1. State Pauli's exclusion principle. 2. Write the electronic contiguration of Cu+ ion. 3. Give the no. of electrons present in P3– ion. 4. Which quantum no. specifies the shape of an orbital in an atom ? 5. List the quantum numbers (n & l) of electrons for 3d orbitals.

148 +2 CHEMISTRY (VOL. - I) (B) Short questions (2 marks) 1. State Aufbau Principle. 2. An atom has its K and L shells completely filled and five electrons in M - shell. Find out the total number of p electrons. 3. Which quantum number has different values for the two electrons of helium atom ? 4. Electronic configuration of an atom is 1s22s22p23p63s2. Determine its atomic number. How many neutrons are present in the nucleus if its atomic wt is 24 ? 5. State the number of neutrons and protons present in C12 and C14 6. Give the quantum numbers of electron in hydrogen atom. 7. Calculate the energy of photon having frequency of 1.0 x 1045 sec–1 (planck's constant h = 6.63 x 10–34 J.S) 8. Name the different spectra of H-atom. 9. Write de Broglie equation. 10. Which is more stable configuration and why ? 4s23d9 and 4s1 3d10 11. What is the electronic configuration of an element of atomic number 25 ? 12. Give the value of n and l for each of the subshell 2p, 4s, 4d, 6f. 13. Write the quantum number for a 3d electron. 14. State Hund's rule. 15. What are the various quantum numbers of the valence electron of Na(11). 16. What was the objections to the Rutherford's model ? What it was overcome by Niels Bohr ? 17. Write down the value of the quantum numbers for the valency electron of sodium. 18. What is the maximum numbers that may be present in all the atomic orbitals with principal quantum numbers and azimuthal quantum number 2 ? 19. Write the n, l, m and s-values for an electron in 3d - orbital 20. How many protons & neutrons are present in P31 ? 15 (C) Short questions (3 marks) 1. Write brief notes on : (each carry 3 marks) (a) Heisenberg's uncertainty principle (b) Aufbau principle (c) Pauli's exclusion principle (d) Hund's rule 2. State the postulates for Bohr's atomic model. 3. What are the limitations of Bohr's atomic model ? 4. Find out the number of electrons, protons and neutrons in K, Na+ and O2–. 5. A nuetral element has mass number 39. It has got one neutron more than protons. Find out the number of electrons, protons and neutrons in this element. 6. The molecular mass of a saturated hydrocarbon is 58. Find out the total number of electrons in one molecule of the hydrocarbon.

STRUCTURE OF ATOM 149 7. Find out the total number of electrons in 1.6 gm of methane. 8. Explain why half-filled and completely filled orbital have extra stability. 9. Give the electron configuration in orbitals of the atoms of the elements Nitrogen, Chlorine and Sodium. 10. Write low the values of all the quantum numbers for the valency electrons of chlorine. 11. Represent mathematically : (a) Energy of one quantum of light. (b) Electron's angular momentum as postulated by Bohr. (c) Relationship between energy and mass. (D) Long questions 1. Give an account of Rutherford's nuclear atom. How did Bohr improve upon Rutherford's model of the atom ? 2. Derive de Broglie equation and state its significance. 3. What are quantum numbers ? Briefly describe the four quantum numbers. 4. Write a note on Pauli's exclusion principle. 5. What are the fundamental particles ? Write their properties. 6. Write a note on Aufbau Principle 7. Give an account of hydrogen spectrum. 8. Write notes on : (a) shapes of s, p, d orbitals (b) Hund's rule. 9. Write note on Aufbau Principle. 10. Discuss about the four quantum numbers and their significance. 11. Discuss Bohr's theory of atom. What are the defects of the theory ? How an orbital differs forms that of an orbit ? 12. What are the various postulates of Bohr's model of atom ? Discuss its drawbacks. 13. What are quantum numbers ? Describe briefly the four quantum numbers. ADDITIONAL QUESTIONS Long answer questions : 1. What are cathode rays ? Discuss its characteristics. 2. What observations of a-ray scattering led Rutherford to propose the model of atom ? 3. Give essential features of Bohr's model atom. 4. Discuss the nuclear model of atom. 5. What was the objection to Rutherford's model. 6. How do you explain the existence of various lines in the hydrogen spectum ? 7. What are the defects of Bohr's model.

150 +2 CHEMISTRY (VOL. - I) 8. What is the significance of quantum numbers ? Describe the four quantum numbers. 9. Write short notes on :- (a) Atomic number (b) Hund's rule (c) Pauli's exclusion principle. (d) Aufbau principle (e) Dual nature of matter and de Broglie equation (f) Planck's quantum theory (g) Concept of orbital. 10. How will you justify that electron is universal constituent of matter ? 11. Describe the shapes of s, p and d-orbital. 12. What are the differences between orbit and orbital. SHORT ANSWER QUESTION (A) Indicate true or false statements in the following : (i) Neutrons are produced when a-rays bombarded a metal sheet. (ii) X-rays are produced when cathode rays hit a metal anticathode. (iii) Energy of the electron in an orbit is negative as electron is negatively charged. (iv) Cathode rays can ionise gases. (v) Charge of the electron is same as that of proton with opposite sign. (vi) Bohr's theory can explain the spectra of all elements . (vii) Anode rays affect photographic plate. (viii) Isotopes of an element have same position in the periodic table. (ix) The energy of shell decreases with increase in the value of n. (x) The neutron was discovered by J.J. Thomson. (xi) The number of unpaired electrons in oxygen is 2. (xii) The shape of an orbital is given by azimuthal quantum number. (xiii) An electron radation energy spontaneously while in stationary orbit. (xiv) When azimuthal quantum number ‘l’ is one (1) it refers to ‘s’ subshell. (B) Give the correct answer : (i) What is the mass of an electron ? (ii) What is the charge of an electron ? (iii) What is the mass of a proton ? (iv) What is the electronic configuration of element of atomic number 24 ? (v) Which rule states that \"no two electrons in atom will have all the four quantum numbers same\"? (vi) What is the approximate size of nucleus ? (vii) Write the electronic configuration of the element of atomic number 29. (viii) What is the rule which gives the sequence in which electrons fill the orbital ? (ix) Name the rule of maximum multiplicity. (x) Draw the shape of dxy orbital.

STRUCTURE OF ATOM 151 (C) Fill in the blanks (one mark) : (i) Principal quantum number gives the —— of electron wave. (ii) Azimuthal quantum number gives the —— of electron wave. (iii) Magnetic quantum number gives —— of orbitals in space. (iv) d - orbitals can accomodate maximum —— number of electrons. (v) 1s2 2s2 2p6 3s2 3p6 4s1 is the electronic configuration of —— atom. (vi) An orbital represents —— motion of the electron around the nucleus. (vii) —— number of unpaired electrons are there in oxygen atom. (viii) Isotopes of the same element differ in their ——. (ix) Isotones are atoms of different elements having —— number of neutrons. (x) Atoms of different elements having same mass number but differing in atomic number are called ——. (xi) The maximum number of electrons on any arbit is ——. (xii) The atomic number of elements having maximum number of unpaired 3p electrons is ——. (xiii) If n = 3, then l can have values from —— to —— (xiv) The number of unpaired electrons in the ground state of carbon is —— . (D) Multiple choice questions (one mark) : (1) Which of the following electronic configuration is permissible ? (a) 1s2 2s2 2px1 2py1 (b) 1s2 2s2 2px2 (c) 1s2 2s2 3s2 (d) 1s2 2s2 2py2 (2) The number of neutrons in Zn+2 ion with mass number is : (a) 34 (b) 36 (c) 38 (d) 40 (3) The radius of atomic nucleus is of the order of (a) 10–10 cm (b) 10–13 cm (c) 10–12 cm (d) 10–15 cm (4) Electmagnetic radiation of maximum wavelength is : (a) UV (b) Radio Waves (c) X-rays (d) Infrared (5) Which of the following sets of quantum numbers is not permissible ? (a) 2, 1, 0, + ½ (b) 2, 2, –1, + ½ (c) 2, 1, +1, –½ (d) 3, 2, 0, – ½ (6) The wavelength of spectral line for an electronic transition is inversely related to – (a) number of electrons undergoing transition . (b) nuclear charge of the atom.

152 +2 CHEMISTRY (VOL. - I) (c) Difference in energy of the energy levels involved in transition (d) velocity of the electron undergoing transition. (7) Energy of the electron in an orbit is determined by : (a) Principal quantum number (b) Azimuthal quantum number (c) Spin quantum number (d) Magnetic quantum number (8) Number of electrons present in an element having 2K, 8L and 5M electrons is : (a) 15 (b) 8 (c) 6 (d) 5 (9) Bohr's model can explain (a) spectrum of hydrogen atom only (b) spectrum of an atom or ion having one electron only (c) spectrum of hydrogen molecule (d) solar spectrum (10) The triad of nuclei that is isotonic is : (a) 6C14 , 7N15 , 9C17 (b) 6C12 , 7N14 , 9C19 (c) 6C14 , 7N14 , 9C17 (d) 6C14 , 7N14 , 9C19 (11) Nitrogen is having three unpaired electrons according to (a) Hund's Rule (b) Aufbau Principle (c) Heisenberg's Principle (d) None of the above. (12) In an atom no two electrons can have the same value for all the quantum numbers. This was proposed by (a) Hund (b) Pauli (c) Dalton (d) Avogadro. (13) Correct set of four quantum numbers for valence electrons of rubidium (Z = 37) is (a) 5, 0, 0, + ½ (b) 5, 1, 0, + ½ (c) 5, 1, 1, + ½ (d) 6, 0, 0, + ½ (14) For azimuthal quantum number l = 3, the maximum number of electrons will be (a) 2 (b) 6 (c) 0 (d) 14

STRUCTURE OF ATOM 153 (15) Krypton has the electronic configuration [Ar] 4s2 3d10 4p6, the 37th electron will go into which of the subshells ? (a) 4f (b) 4d (c) 3p (d) 5s (16) If uncertainty in position of an electron is zero, the uncertainty in its momentum would be : h 2p (a) Zero (b) h (c) p (d) Infinite. (17) The magnetic quantum number for valence electron of sodium is (a) 3 (b) 2 (c) 1 (d) 0 (18) Which of the following transition metal cations has maximum unpaired electrons ? (a) Mn (b) Ni (c) Co (d) Fe. (19) A particular element has configuration 1s2 2s2 2p5. In its chemical reaction, it is most likely to (a) gain one electron (b) Lose one electron (c) Lose three electrons (d) gain three electron. (20) Number of electrons in the nucleus of an element having atomic number 14 is (a) 14 (b) 20 (b) 0 (d) 7 (21) The nucleus of tritium consists of (a) 1 Proton +1 neutron (b) 1 proton +3 neutrons (c) 1 proton + no neutron (d) 1 proton + 2 neutrons (22) The species isoelectromic with CN– are (a) CO (b) O+2 (c) F2– (d) Si (23) Which of the following relates to proton both as wave motion and stream of particles ? (a) Interference (b) E = mc2 (c) E = h (d) Diffraction (24) The preference of three unpaired electrons in the nitrogen atom can be explained by (a) Pauli's exclusion principle (b) Aufbau principle (c) Uncertainty principle (d) Hund's rule

154 +2 CHEMISTRY (VOL. - I) (25) Electromagnetic radiation with maximum wavelength is (a) UV (b) Radio wave (c) X-ray (d) IR (26) 2p orbital have : (a) n = 1, l = 2 (b) n = 2, l = 1 (c) n = 1, l = 0 (d) n = 2, l = 0 (27) Correct set of four quantum numbers for the valence (outermost) electron of rubidium (Z = 37) is : (a) 5, 0, 0, + 1/2 (b) 5, 1, 1, + 1/2 (c) 5, 1, 0, + 1/2 (d) 6, 0, 0, + 1/2 (28) The azimuthal quantum number of the 17th electron of chlorine atom is : (a) one (b) two (c) three (d) zero ANSWERS C. Fill in the Blanks (vi) Three dimensional (xi) 2n2 (i) Size (vii) Two (xii) 15 (ii) Shape (viii) Mass number (xiii) 0 to 2 (iii) Orientation (ix) Same (xiv) 2 (iv) Ten (x) Isobars (v) K D. Multiple Choice type 1. a 4. b 7. a 10. a 13. a 16. d 19. a 22. a 25. b 28. a 2. d 5. b 8. a 11. a 14. d 17. d 20. c 23. b 26. c 3. b 6. c 9. b 12. b 15. d 18. a 21. d 24. d 27. a qqq

UNIT – III CHAPTER - 4 CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES TOPICS DISCUSSED : l Significance of classification l Brief histroy of periodic table l Modern periodic table l Present form of periodic table l Periodic trends in properties of elements l Nomenclature of elements with atomic number greater than 100. INTRODUCTION : There are about 116 elements discovered till today. It is difficult to remember all the properties separately. For systematic study, it was necessary to classify them into different groups on the basis of their characteristics. Once the general behaviour is known, one can always go into details of specific behaviour of each element. The method of classifying the elements having somewhat similar physical and chemical properties is known as periodic classification of elements. The table which contains the elements so classified is called the Periodic table. 4.1 BRIEF HISTORY OF PERIODIC TABLE In 1829, German chemist Johann Dobereiner suggested the relationship between atomic masses of elements with their properties. He noticed a similarity in properties of several groups consisting of three elements, called Triads. In each of the triads the average of the atomic masses of first and third element would be equal to the atomic mass of the second element. For example, in the Triad of Li (7), Na (23) and K (39), the average of the atomic masses of Li and K (7+39) would be equal to the atomic mass of the middle element Na (23). This relationship is called Debereiner's law of Triad. In 1864, Newlands observed that when the elements were arranged in the order of increasing atomic masses, elements with similar properties were repeated at intervals of seven i.e. every eighth element was similar in properties to the first one. He suggested this as law of octaves. e.g. lithium and sodium were considered to be one octave apart as in musical system.

156 +2 CHEMISTRY (VOL. - I) Lother Meyer in 1869 plotted atomic volumes with atomic masses and found that similar elements occupied similar positions in the plot. In 1869, a Russian scientist Dmitri Mendeleev made a significant contribution to develop the periodic table. Only 63 elements were known at the time and he studied the properties of these elements and gave a law of Periodic table which was stated as : \"The physical and chemical properties of the elements are periodic functions of their atomic masses\". After the study of atomic structure, atomic number was considered as fundamental properties of an element. Mosley in 1913 proved that the properties of the elements depend on the atomic numbers rather than atomic masses. Mendeleev realised that the best way to arrange the elements should not be according to atomic masses, but according to atomic numbers. Mendeleev's periodic table has the follwoing advantages and drawbacks. ADVANTAGES OF MENDELEEV'S PERIODIC TABLE There were only 63 elements known when Mendeleev published his periodic table. Within one year of this, 23 new elements were discovered. Some of the advantages of Periodic table are : (1) Prediction of new elements :In Mendeleev's table there were some vacant places. Mendeleev predicted the properties of the elements that would occupy the vacant places. Actually after a few years of his prediction, these elements were discovered and fitted into the vacant places in the periodic table. These elements were scandium, germanium etc. (2) Correction of atomic mass : Mendeleev placed Be in Gr II along with Mg, Ca etc. These have valency two. The atomic mass of Be was found experimentally to be 13.5. Hence, it should be placed after 'C' whose atomic mass is 12. Mendeleev determined its equivalent mass and multiplying with 2, found its atomic mass to be 9.1. (3) Atomic structure : Mendeleev's periodic table helped in the study of electronic configuration of elements. Drawbacks of Mendeleev's periodic table. Mendeleev's periodic table had many limitations. 1. Anomalous position of some elements : The atomic masses of K, Ni and I are less than those of Ar, Co and Te respectively. In Mendeleev's table K follows Ar, Ni follows Co and I follows Te. 2. Position of Triads : In Mendeleev's table, the space for one element is occupied by three elements in the triads like Fe, Co and Ni. This is against the periodic law. 3. Position of dissimilar elements : Alkali metals and coinage metals differ in their physical and chemical properties. But these are placed in one group. 4. Position of noble gases, lanthanides and actinides : The noble gases, lanthanides and actinides were not known. Noble gases were placed in a separate group called zero group. Lanthanides and actinides have been given one position each which goes against periodic law. 5. Position of hydrogen : Hydrogen is similar to alkali metals and to halogens. The position of hydrogen is thus anomalous which can belong to Gr IA and Group VII A. 6. Position of isotopes : Mendeleev's periodic table does not speak about the positions of isotopes.

CLASSIFICATION OF ELEMENTS AND PERIODICTY IN PROPERTIES 157 Modern Periodic Law : The modern periodic law states that \"the physical and chemical properties of elements are periodic functions of their atomic numbers\". The improvised and modified periodic table has seven horizontal rows called periods. The first period consists of two elements hydrogen and helium. The second period starts with lithium and is completed with neon. There are eight elements in this period. Similarly the third period starts with sodium and is completed with argon. This period also contains eight elements. The fourth period starts with potassium and ends with krypton. The fifth period is similar to fourth one. These are called long periods containing eighteen elements in each. However, the sixth period has 32 elements. The seventh period is incomplete. Table 4.1 The old form of the Periodic Table. GROUPS A I B A I B A I B A I B A I B A I B A I B A I B ZERO 1st Period 1 2 H He 2nd Period 3 45 6 7 89 10 Li Be B C N OF Ne 3rd Period 11 12 13 14 15 16 17 18 Na Mg Al Si P S Cl Ar 19 20 21 22 23 24 25 26 27 28 K 29 Ca 30 31 Sc 32 Ti 33 V 34 Cr 35 Mn36 Fe Co Ni 4th Period Cu Zn Ga Ge As 44 45 46 36 5th Period Se Br Fe RURH Pb Kr 37 38 39 40 41 42 43 54 Rb Sr Y Zr Nb Mo Tc Xe 47 48 49 50 51 Ag Cd in Sn Sb 52 53 Te I 55 56 57* 72 73 74 75 76 77 78 Cs Ba La Hf Ta W Re Os Ir Pt 6th Period 79 80 81 82 83 84 85 86 Au Hg Tl Pb Bi Po At Rn 7th Period 87 88 89# 104 105 Fr Ra Ac Ku Ha THE RARE EARTHS *Lanthanide 58 59 60 61 62 63 64 65 66 67 68 69 70 71 series Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu #* 90 91 92 93 94 95 96 97 98 99 100 101 102 103 Actinide series Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr The vertical columns are known as groups. There are nine groups starting from I to VIII and zero. All the groups except zero and VIII are divided into two subgroups A and B. The elements placed to the left in the vertical column form subgroup A and those placed to the right form subgroup B. Zero group consists of noble gas elements. Group VIII consists of transition elements.

158 +2 CHEMISTRY (VOL. - I) 4.2 LONG FORM OF PERIODIC TABLE In the long form of periodic table the elements are arranged in the order of increasing atomic number. This table has seven horizontal rows called periods and eighteen vertical columns known as groups. Period : The seven periods are represented as period 1 to 7. First period contains two elements hydrogen (1s1) and helium (1s2). In these case s, the first shell (K) is completed. The second period starts with n = 2, the first member of which is lithium (2s1) and the shell (L) is completed at neon (2s22p6). PERIODIC CHART OF THE ELEMENTS Table 4.2 Long form of Periodic Table

CLASSIFICATION OF ELEMENTS AND PERIODICTY IN PROPERTIES 159 There are eight elements in this period. Similarly the third period (n = 3) starts with sodium (3s1) and is completed at argon (3s23p6). This period also contains eight elements. The fourth period (n = 4) starts with the filling up of the 4s orbital and ends with the completed 4p orbital. This period which starts with potassium (4s1) and ends with krypton (4s24p6) has ten more elements than the earlier one (eighteen elements in all). This is due to the elements in which the filling up of electrons in the 3d orbitals takes place after 4s orbital but before 4p orbitals. The fifth period (n = 5) is similar to the fourth one. However, the sixth period (n = 6) has 32 elements in which the filling up of electrons takes place in 6s, 4f, 5d and 6p orbitals in that order. The seventh period (n = 7) would have been similar to the sixth period, but it is incomplete. Group : The eighteen groups are designated as Gr IA to VIIA, Gr IB to VIIB, Group Zero and Gr VIII (three vertical columns). Group IA consists of elements hydrogen (Is1), lithium (2s1), sodium (3s1) potassium (4s1) etc. All these elements have the common outermost electronic configuration ns1. The elements of Group IIA have ns2 configuration. Similarly elements from Group IIIA to group zero have common electronic configuration varying from ns2np1 to ns2 np6. In addition to the above group ten more columns are present between group IIA and IIIA. The elements present in them have common electronic configuration (n – 1) d 1–10ns 0–2.The first of these columns is called group III B which consists of elements scandium (3d14s2), yttrium (4d105s2), lanthanum (5d1 6s2) and actinum (6d1 7s2) and the last one IIB having zinc (3d10 4s2); cadmium (4d105s2) and mercury (5d106s2). The elements present in columns 6, 7 and 8 are grouped as one group and called group VIII. In group III B there are additional 14 elements along with lanthanum and also with actinum. These elements have common electronic configuration (n – 2) f 1–14 (n – 1) d 0–1 ns2 and have been shown separately in two rows below the periodic table usually called the lanthanide and actinide series. The other groups are known as IVB, VB, VIB, VIIB, VIII and IB. 4.3 CAUSE OF PERIODICITY AND MAGIC NUMBERS From the above discussion it is quite evident that when the elements are arranged in the increasing order of their atomic numbers similar elements are repeated at regular intervals. These intervals are 2, 8, 8, 18, 18, 32 and are known as magic numbers. The elements belonging to a particular group have similar properties and they have similar electronic configuration. By adding the magic number to the atomic number of an element belonging to a particular group the atomic number of the subsequent elements may be found out. For example, in Gr IA the elements are H (1), Li(3), Na(11), K(19), Rb(37), Cs(55) and Fr(87), the atomic numbers are 2, 8, 8, 18, 18, and 32 units apart. Since all these elements have same number of valence electrons we may regard \"the repetition of similar electronic configuration after regular intervals\" as the cause of periodicity of elements.

CLASSIFICATION OF ELEMENTS AND PERIODICTY IN PROPERTIES 161 4.5 PERIODICITY IN PROPERTIES OF ELEMENTS : ATOMIC AND IONIC RADII The radius of an atom is the distance between the centre of the nucleus of an atom and its outermost electron shell. This determines the approximate size of the atom. In general atomic radius increases while going down the group of the periodic table. This is due to successive addition of new shells. Atomic size decreases across a period due to increase in nuclear charge. This increases nuclear attraction for the electrons. Therefore, atomic size decreases. Table 4.4 Atomic radii in a group Elements : Li (3) Na (11) K(19) Rb(37) Cs(55) 1.57 2.03 2.16 2.35 Atomic radius (A0) : 1.23 Table 4.5 Atomic radii in a period Elements : Li (3) Be(4) B(5) C(6) N(7) O(8) F(9) 0.74 0.74 0.72 Atomic radius (A0) : 1.23 0.89 0.80 0.77 Several types of atomic radii such as covalent radii, vander Waal's radii and ionic radii are known. Fig 4.1(a) Variation of atomic radius with Fig 4.1(b) Variation of atomic radius with atomic number across the second period atomic number for alkali metals and halogens

162 +2 CHEMISTRY (VOL. - I) (a) Covalent radius : It is difined as one half the distance between the nuclei of the two atoms forming a covalent bond. In a diatomic molecule, the internuclear distance is also called bond length. Example 1 : Internuclear distance in a hydrogen molecule is 0.74AO, Covalent radius of hydrogen = 0.37A0 Diagram 12.1 Example 2 : Internuclear distance in a chlorine molecule is 1.98A0 Covalent radius of chlorine = 0.99A0. (b) van der Waal's radius : It is defined as one half of the distance between the nucleus of two neighbouring nonbonded atoms. Example 1: van der Waal's radius of hydrogen = 1.2A0. Example 2 : van der Waal's radius of chlorine = 1.8A0. van der Waal's radius is greater than covalent radius. (c) Ionic radius : It is the effective distance from the nucleus of an ion up to the point where its influence over the electron cloud ceases. A positive ion (cation) is always smaller than the parent atom and a negative ion (anion) is always larger than its parent atom When an atom loses one or more electrons positively charged ion or cation is formed. M – e ® M+ (Atom) (Cation) The number of electrons in a cation is always less than the number of protons. Effective nuclear charge increases. Sometimes, the formation of a cation involves removal of the outermost shell. Table 4.6 Radii of atoms and cations Atom radius (A0) Cation radius(A0) Li 1.23 Li+ 0.60 Na 1.57 Na+ 0.95 Mg 1.36 Mg2+ 0.60 Al 1.18 Al3+ 0.50 When an atom gains one or more electrons, negatively charged ion, an anion is formed. A + e ® A– (Atom) (Anion) The number of electrons in an anion is always greater than the number of protons. Effective nuclear charge decreases. The electron cloud expands.

CLASSIFICATION OF ELEMENTS AND PERIODICTY IN PROPERTIES 163 Table : 4.7 Radii of atoms and anions Atom radius(A0) Anion radius (A0) F. 0.72 F– 1.36 Cl. 0.97 Cl– 1.81 Br 1.13 Br– 1.95 I 1.35 I– 2.16 ATOMIC VOLUME Atomic volume is defined as the volume in c.c. occupied by one gram atom of the element in the solid state and commonly called gram atomic volume. It is obtained by dividing the atomic mass of the element by its density. Atomic volume = Atomic mass Density Thus, the atomic volume is the volume in c.c. occupied by 6.023 x 1023 atoms of an element and the volume of a single atom may be obtained by dividing gram atomic volume with Avogadro's number of atoms. Atomic volume increases more or less regularly in going down a group due to the increase in the number of shells. The larger the number of shells, the bigger is the atomic volume. In going from left to right in a period, it deceases at first, becomes minimum in the middle and then increases. The variation is influenced by (i) nuclear charge (The increased nuclear charge attracts each electron more strongly towards the nucleus resulting in the decrease in the volume) and (ii) number of valence electrons (Towards the end of a period, the number of valence-electrons increases and therefore the volume of the atom increases so that it may accomodate all the electrons). These two factors combine to give the above observation. Table 4.8. Atomic volumes of some elements Group ® IA IIA IIIA IVA VA VIA VIIA Zero ¯Period 1. H He 14.1 31.8 2. Li Be B C N O F Ne 13.1 5.0 4.6 5.3 17.3 14.0 17.1 16.8 3. Na Mg Al Si P S Cl Ar 23.7 14.0 10.0 12.1 17.0 15.5 18.7 24.4 4. K Ca Ga Ge As Se Br Kr 45.3 29.9 11.8 13.6 13.1 16.5 23.5 32.2

164 +2 CHEMISTRY (VOL. - I) IONISATION ENTHALPY OR IONISATION ENERGY OR IONISATION POTENTIAL Ionisation enthalpy or ionisation energy of an element is defined as the energy required to remove an electron from the isolated neutral gaseous atom. It is designated as DiH. Na(g) + I.E ® Na+(g) + e This energy is measured in units of electron volt per atom or kJ mole–1. One electron volt per atom is equivalent to 23 kilocalories or 96.3 kilojoules per mole. (Avogadro number of atoms) Fig 4.2 Variation of first ionizaion enthalpies (DiH) with atomic number for elements with Z = 1 to 60 Atomic number(Z) Factors affecting Ionisation Enthalpy. (i) Size of atom : The larger the size of the atom, the smaller is the ionisation energy. With the increase in atomic size the outer electrons lie further away from the nucleus and can be easily removed at the expense of lower amount of energy. (ii) Effect of Nuclear charge : With the increase in nuclear charge the electrostatic attraction between the nucleus and the outer electrons increases. Thus, the removal of outer electron becomes difficult leading to higher ionisation energy. (iii) Number of electrons in the inner shell The inner electrons tend to shield the outer electrons from the pull of the nucleus as a result of which the outer electron can be easily knocked out. This effect is known as screening effect. This results in decrease of ionisation energy. (iv) Half-filled or completely filled orbital According to Hund's rule exactly half-filled or completely filled orbitals are more stable. So more energy is required to detach an electron from atoms having such orbitals and therefore more is the ionisation energy. (v) Nature of electron to be removed : The 's' electrons experience more attraction than the 'p' electrons since the 's' orbital is closer to the nucleus than the 'p' orbital of the same orbit. Thus removal of 's' electron involves more IE than that– 'p' electron. In general the IE follows the order s > p > d > f orbital of the same orbit.

CLASSIFICATION OF ELEMENTS AND PERIODICTY IN PROPERTIES 165 Ionization energy increases form Li to Ne with exception of B and O. Table 4.9 (IE of elements in second period) Element Li(3) Be(4) B(5) C(6) N(7) O(8) F(9) IE (kJ. mole–1) 520 900. 800 1086 1403 1314 1681 In general there is an increasing trend of IE across a period. This is due to increase in nuclear charge as we go across a period. The electrons are more strongly held by the attractive force between the nucleus and the electrons. In B(5) and O(8) removal of electron requires lower energy since the resultant ions have stable half-filled or full-filled orbitals. On the otherhand IE decreases as we go down the group. The outer electron is farther from the nucleus and the effect of nuclear charge is decreased by the presence of electronic shells. 4.3 (a) 4.3 (b) Fig 4.3 (a) First ionization enthalpies (DiH) of elements of the second period as a function of atomic number (Z) and Fig. 4.3 (b) DiH of alkali metals as a function of Z. Properties of elements and compounds as predicted by ionisation energy (i) Lower value of ionisation energy indicates the greater reducing power of an element, hence greater reactivity. (ii) Basic character of an element can be roughly estimated. (iii) The value of ionisation energy provides us indication regarding the number of valence electrons. Abnormally high value of ionisation energy indicates the removal of electron other than valence electron. Table : 4.10 (I.E of alkali metals) Elements : Li Na K Rb Cs IE (kJ mole–1) 520 495 418 403 374

166 +2 CHEMISTRY (VOL. - I) The ionisation energy to remove the second electron (IE2) is higher than ioni sation energy to remove the first electron (IE1). The successive values of ionisation energy of an element show an increasing trend. The second electron has to be removed from a positive ion and this requires more energy. ELECTRON GAIN ENTHALPY OR ELECTRON AFFINITY The energy released when a neutral gaseous atom gains an electron is called electron affinity or electron gain enthalpy. Cl(g) + e ® Cl–(g) + energy Electron affinity is a measure of tightness of binding of an extra electron to an atom. Electron affinity has also been renamed as electron gain enthalpy (Deg H0). The difference comes in the representation of values with sign only. Suppose the electron affinity of an element is 20kJ mol–1, its Deg H0 value will be –20kJ mol–1, because it is an exothermic process. Factors affecting Electron affinity (i) Atomic size : Electron affinity decreases with increase in atomic size. (ii) Nuclear charge : More the nuclear charge stronger will be the attractive force for incoming electron. Thus more energy is released as a result of this attraction and more will be the electron affinity. (iii) Electronic configuration : An atom having stable electronic configuration has zero electron affinity. For example, the noble gases have ns2np6 configuration. These atoms have no tendency to gain electron, thus have zero value of EA. On the other hand the halogens have electronic configuration ns2np5. In order to acquire stable configuration ns2np6, halogens have maximum tendency to gain electron and therefore, are associated with high value of EA. Elements of Group VIIA (halogens) have high electron affinity. This is because by gaining an electron these attain stable noble gas configuration. The electron affinity decreases as we go down the group because the attractive force of the nucleus decreases with increase in size. Fluorine has slightly lower electron affinity than chlorine probably due to compact 2p orbital in fluorine. Table : 4.11 (Electron affinity of halogens) Elements : F(9) Cl(17) Br(35) I(53) 348 324 295 Electron affinity (kJ mole–1) 322 Along a period, the nuclear hold on the outermost electron increases as we move from left to right and hence electron affinity increases. Properties of elements and their compounds as predicted by Electron affinity (i) The oxidising power of an element can be predicted from the value of E.A. More the value of E.A more is the tendency to gain electron and greater is the oxidising power.

CLASSIFICATION OF ELEMENTS AND PERIODICTY IN PROPERTIES 167 (ii) The ionic character of the bond between two atoms can be predicted. By knowing the relative values of ionisation energy and electron affinity of the two elements the nature of bond between them can be predicted. Increase in difference of these two values decreases the ionic character. ELECTRONEGATIVITY When two dissimilar atoms i.e. H and Cl form a covalent bond, the electrons forming the bond are not equally shared. The electron pair is shifted to one of the atoms which becomes partially negatively charged. Thus, chlorine has greater electronegativity than hydrogen. The electronegativities of various elements increase in a period of the periodic table. It is due to increase of nuclear charge which results in decrease in size. Electronegativity decreases in a group. Although atoms of chlorine and sulphur have the same size, chlorine has higher electronegativity than sulphur because of greater nuclear charge. Both EA and EN measure the electron attracting power of an atom. But the former is concerned with an isolated gaseous atom whereas the latter, with an atom in a molecule. Electronegativity may be expressed in the following three scales. (i) Mulliken Scale : According to Mulliken, electronegativity is taken as the average value of Ionisation energy and electron affinity of an atom. EN = IE + EA 2 (ii) Alfred Rochow Scale : Alfred and Rochow regarded electronegativity as the electrostatic force operating between the nucleus and the valence electrons. They expressed it as EN = 0.359 Z + 0.744 r2 where, Z is Effective nuclear charge r is Covalent radius of atom (in A0), (iii) Pauling Scale : The most widely used scale was that predicted by Pauling. It is based on excess bond energies. If XAand XB be the electronegativities of two atoms A and B XA – XB = 0.208 D where D = Actual bond energy – (EA-A x EB-B The factor 0.208 arises due to conversion of kcal to electron volt ( 1ev = 23.01 kcal/ mole). Properties of elements as predicted by Electronegativity (i) Nature of bond between two atoms can be predicted. If the difference in value of EN between two atoms is more, more is the ionic character of the bond formed between them. If the difference is zero, this indicates the presence of a covalent band.

168 +2 CHEMISTRY (VOL. - I) (ii) Greater the value of electronegativity of an element more is its tendency to gain electron and therefore more is its oxidising power. This also indicates that the element is more nonmetallic in nature. OXIDATION STATES Mendeleev defined groups of elements on the basis of their valencies or oxidation states. The formulae of compounds formed by an element depends on the oxidation states. Valencies of the representative elements (s and p block elements) of Gr I A to VII A are generally given by the following simple formula. If G = group number of the representative elements, valency = G and (8 - G) These are some exceptions to this rule. Transition elements show variable valency but all of them have a common valency of two. It is of interest to know the connection between valency and electronic configuration. The outermost group of orbitals are known as valence orbitals. These are involved in chemical bonding. The electrons occupying the valence orbitals are referred to as valence electrons. The valency is determined by the valence electrons. Group 1 Table 4.12 Valency and Group 5 Examples: HCl 2 34 NH3 H2O BeCl2 BCl3 CH4 N2O5 LiCl CaCl2 Al2O3 CO2 PCl3 Li2O CaO AlCl3 SiO2 PCl5 NaCl TlCl3 SnO2 SrO BaO 4.6 NOMENCLATURE OF ELEMENTS WITH ATOMIC NUMBER MORE THAN 100 : About the nomenclature of elements with atomic number greater than 100 new controversies came up in recent years. Usually the naming of the new elements had been traditionally done after the name of its discoverer or discoverers and the suggested name was ratified by the IUPAC. But sometimes it so happened that scientists before collecting the reliable data on the new element got tempted to claim for its discovery. For example, both american and Soviet Union scientists claimed credit for discovering elements 104. The Americans named it Rutherfordium while the Soviets named it Kurchatovium. To avoid such confusion, the IUPAC has made recommendation that until the discovery of a new element is proved and its name is officially recognised, a systematic nomenclature be derived directly from the atomic number of the element using the numerical roots for zero and numbers one to nine as

CLASSIFICATION OF ELEMENTS AND PERIODICTY IN PROPERTIES 169 shown in Table 4.14. The roots are put together in order of digits which makeup the atomic number and \"ium\" is added at the end. The IUPAC names of elements with atomic number more than 100 are shown in 4.13. Table 4.13 Nomenclature or elements with atomic number more than 100 Atomic Name Symbol IUPAC IUPAC Number Offical Name Symbol Unnilunium Unu 101 Unnilbium Unb Mendelevium Md 102 Unniltrium Unt Nobelium No 103 Unnilquadium Unq Lawrencium Lr 104 Unnilpentium Unp Rutherfordium Rf 105 Unnilhexium Unh Dubnium Db 106 Unnilseptium Uns Seborgium Sg 107 Unniloctium Uno Bohrium Bh 108 Unnilennium Une Hassnium Hs 109 Ununnillium Uun Meitnerium Mt 110 Unununnium Uuu Darmstadtium Ds 111 Ununbium Uub Rontgenium Rg 112 Ununtrium Uut Copernicium Cn 113 Ununquadium Uuq * 114 Ununpentium Uup Flerovium Fl 115 Ununhexium Uuh + 116 Ununseptium Uus Livermorium Lv 117 Ununoctium Uuo + 118 + * Offical IUPAC name yet to be announced + Elements yet to be discovered. Digit Name Abbreviation 0 nil n 1 un u 2 bi b 3 tri t 4 quad q 5 pent p 6 hex h 7 sept s 8 oct o 9 enn e Table 4.14 Notation for IUPAC Nomenclature of Elements with Atomic number more than 100.

170 +2 CHEMISTRY (VOL. - I) CHAPTER (4) AT A GLANCE 1. Atomic volume : It is defined as the volume occupied by one gram atom of the element in the solid state at its melting point. 2. Mendeleev's periodic law : According to this law, the properties of elements are periodic function of their atomic masses. 3. Modern periodic law : The physical and chemical properties of elements are the periodic function of their atomic numbers. 4. Covalent radius : The covalent radius of an atom is one half the distance between the nuclei of the two atoms forming the covalent bond. 5. van der Waal's radius : It is defined as one half the distance between the two nearest nonbonded atoms. 6. Ionic radius : It is defined as the distance between the centre of the nucleus up to which the electron cloud is extended in an ion. 7. Ionisation enthalpy or Ionisation energy : It is defined as the amount of energy required to remove an electron from the isolated neutral gaseous atom. 8. Electron gain enthalpy or Electron affinity : It is defined as the amount of energy released when an electron is added to the neutral gaseous atom. If electron affinity is 20 kJmol–1, then electron gain enthalpy will be –20kJ mol–1. 9. Electronegativity : It is defined as the power of an atom in a molecule to draw the shared pair to electron towards itself. Summary : The physical and chemical properties of the elements are periodic function of atomic numbers. Vertical columns are called groups. The horizontal rows are known as periods. Long form periodic table has sub groups with eighteen vertical columns. It has s,p,d and f blocks. A number of physical and chemical preperties of elements vary periodically with atomic number. These properties include atomic and ionic radius, ionisation energy, electron affinity, electronegativity and oxidation states.

CLASSIFICATION OF ELEMENTS AND PERIODICTY IN PROPERTIES 171 QUESTIONS GROUP - A I. Multiple choice question : (One mark) 1. Long form of periodic table is based on (a) Atomic mass (b) Atomic number (c) Atomic size (c) Electron affinity 2. Which pair of elements is chemically most similar ? (a) Na, Al (b) Cu, S (c) Ti, Zr (d) Zr, Hf 3. Elements of 1 give colour in Bunsen burner due to (a) Low IP (b) Low MP (c) Softness (d) one electron in the outermost shell. 4. Elements of same vertical group of the periodic table have (a) Same atomic size (b) Same electronic configuration (c) Same number of electrons in the outermost shell of their atoms. (d) Same number of atoms. 5. The element with highest value of 1st ionisation potential is (a) B (b) C (c) N (d) O 6. The ion having highest radius is (a) Al3+ (b) N3- (c) Na+ (d) F- 7. Atomic radii of Fluorine and Neon in AO units are respectively given by (a) 0.72, 1.60 (b) 1.60, 1.60 (c) 0.72, 0.72 (d) None of these 8. Which of the following electronic configuration represents an inert gas ? (a) 2, 8, 1 (b) 2, 8, 8 (c) 2, 1, 7 (d) 2, 8, 2 9. The maximum tendency to form unipositive ion is for the element with the following electronic configuration. (a) 1s2 2s2 2p6 3s1 (b) 1s2 2s2 2p6 3s2 3p1 (c) 1s2 2s2 2p6 3s2 3p2 (d) 1s2 2s2 2p6 3s2 3p3 10. Characteristic of transition element is (a) Incomplete d-orbital (b) Incomplete f - orbital (c) Incomplete s- orbital (d) Incomplete p- orbital.

172 +2 CHEMISTRY (VOL. - I) 11. Which of the following has the largest ionic radius ? (a) Be2+ (b) Mg2+ (c) Ca2+ (d) Sr2+ 12. The first ionisation energy of lithium is (a) greater than Be (b) less than Be (c) equal to Na (d) equal to F 13. The electronegativity of the following elements increases in the order (a) C, N, Si, P (b) N, Si, C, P (c) Si, P, C, N (d) P, Si, N, C 14. The 1st transition series contains elements having atomic numbers from (a) 22 to 30 (b) 21 to 30 (c) 21 to 31 (d) 21 to 29 15. Which is the lightest metal in the periodic table ? (a) H (b) Mg (c) Ca (d) Li 16. A newly discovered element 'X' is placed in the group 1 of the Periodic table because it forms (a) an oxide which is acidic (b) A volatile chloride having formula XCl. (c) An ionic chloride having formula XCl (d) An insoluble XCO4 17. The correct order of metallic character of the elements B, Al, Mg and K (a) B > Al > Mg > K (b) Al > Mg > B > K (c) Mg > Al > K > B (d) K > Mg > Al > B 18. The correct order of the chemical reactivity of the elements F, Cl, O and N is (a) F > O > Cl > N (b) F > Cl > O > N (c) O > F > N > Cl (d) Cl > F > O > N 19. The correct order of non-metallic character of the elements B, C, N and F is (a) F > N > C > B (b) C > B > N > F (c) B > C > N > F (d) F > N > C > B 20. Size of the isoelectronic species F–, Ne and Na+ is affected by (a) Nuclear charge (b) Valence principal quantum number (c) Electron-electron interaction in the outer orbitals (d) None of the above 21. In periodic table, on moving along a period, the ionisation potential (a) Increases from left to right (b) Remains unchanged (c) First increases then decreases (d) Decreases from left to right

CLASSIFICATION OF ELEMENTS AND PERIODICTY IN PROPERTIES 173 22. Which of the following atoms possesses the smallest volume? (a) S (b) Si (c) P (d) He 23. Ionisation potential of an element does not depend upon (a) Electrical neutrality (b) Nuclear charge (c) Penetrating effect (d) Shielding effect 24. Zero group was introduced by (a) Mendeleev (b) Ramsay (c) Lothar Meyer (d) Johnson GROUP - B II. Very short answer type questions (one mark) 1. Which of the following has the smallest size ? Na+, Mg+2, Al+3 2. Which of the following has the smallest electron affinity ? Na, O, C, F 3. Which element has the electronic configuration ? Is2 2s2 2p6 3s2 3p6 4s2 4. Name any two elements of Group 14 of the periodic table. 5. Name any two elements of Group 16 of the periodic table 6. Name the group of the periodic table in which the element having atomic number 6 is placed. 7. Between iodine and iodide ion which has the larger size ? 8. Which of the following has the largest size ? Na, Na+, N, F 9. Write the electronic configuration of the second element of Gruop 16 of the periodic table. 10. Between lithium and sodium which is more electropositive ? 11. Which of the following has the lowest electron affinity ? K, B, C, F 12. Name two elements of group 5 of the periodic table. 13. Which of the following elements shows variable valency ? Iron, sodium, calcium, strontium, 14. Name any two elements of group 2 of the periodic table. 15. Which one of the following elements has the lowest electronegativity ? P,N,O,F 16. Which of the following has the highest ionisation potential ? Na, He, Ca, Mg, P 17. Arrange the following in order of increasing size : Cl, Cl+, Cl– 18. Name the elements in the group 1 of the periodic table.

174 +2 CHEMISTRY (VOL. - I) 19. Which two elements of the second period of the periodic table have both positive and negative oxidation number ? 20. Write the electronic configuration of an element present in 4th period and 16 group. 21. Why is ionisation energy of Mg greater than Al ? 22. Which of the following has smallest ionisation energy ? F, Na, Cl, Mg, Cs III. Short answer type questions : (Two marks ) 1. State Mendeleev's periodic law. 2. The ionic radii of alkaline earth metals are smaller than those of the nearest alkali metals. Why ? 3. What is electron affinity ? How does it vary in a period. 4. Why are elements of group I and 2 of the periodic table called s- block elements. 5. The electronic configuration of the element is 1s2 2s2 2p6 3s2 3p6. In which group of the periodic table will it be placed and why ? 6. Justify the position of carbon and lead in the second group. 7. Explain with reasons : (a) Alkali metals do not form dipositive ions. (b) Electronegativity values of inert gases are zero. 8. How does the metallic character of elements generally vary in periods and groups of the periodic table ? 9. How does ionisation energy vary in a period and in a group ? Give reasons. 10. Mg2+ ion in smaller than O2– ion although both have the same electronic structure. Explain. IV. Short answer type questions : (Three marks ) 1. Write two characteristics of d-block elements. 2. Name the group of elements classified as s, p and d blocks. 3. Why ionisation energy of Nitrogen is greater than that of oxygen? 4. Between Na and Mg which has higher second ionisation energy and why? 5. Third period has eight and not eighteen elements. Explain. 6. Alkali metals donot form dipositive ions, why ? 7. Why first ionisation energy of Mg is more than that of Al? 8. The ionic radii of alkaline earth metals are smaller than those of the nearest alkali metals? Why? 9. Give four characteristic properties of group 14 elements. 10. Justify the position of carbon and lead in the periodic table on the basis of electronic configuration.

CLASSIFICATION OF ELEMENTS AND PERIODICTY IN PROPERTIES 175 V. Long answer type questions : (Seven marks) 1. Describe the long form of periodic table. What advantages it has over the Mendeleev's periodic table ? 2. What is periodic law ? Discuss how the properties of the elements vary in periodic table. 3. How the properties of the elements vary in each group and each period of the periodic table ? 4. What are electron affinity and ionisation energy ? Explain. Arrange nitrogen, oxygen and fluorine in the increasing order of their ionisation energy giving reasons. 5. What is long form of periodic table ? How do the properties of elements vary in groups and periods of the periodic table ? 6. Describe long form of periodic Table. How and why ionisation energy vary in groups and periods ? 7. Write notes on – (5 marks) Variation of size of atoms in Periods & Groups. 8. Write notes on s,p,d,f blocks of elements of the periodic table. ANSWERS TO MULTIPLE CHOICE TYPE QUESTIONS 1. (b) 6. (b) 11. (d) 16. (c) 21. (a) 2. (c) 7. (a) 12. (b) 17. (d) 22. (d) 3. (a) 8. (b) 13. (c) 18. (b) 23. (a) 4. (c) 9. (a) 14. (b) 19. (d) 24. (b) 5. (d) 10. (a) 15. (d) 20. (a) qqq

176 +2 CHEMISTRY (VOL. - I) UNIT – IV CHAPTER - 5 CHEMICAL BONDING AND MOLECULAR STRUCTURE INTRODUCTION : Atom is the smallest particle of an element, which may or may not have free existence. But most of the substances exist in the form of clusters or aggregates of atoms. Any such cluster, in which atoms of same or different elements combine together is called molecule. The molecule is electrically neutral and have free or independent existence. For example, hydrogen, oxygen, nitrogen, chlorine etc. exist as diatomic molecules. Here, these diatomic molecules are more stable than their constituent atoms. Hence, a chemical bond is defined as the force of attraction which holds the constituent atoms in a molecule. It is generally represented by a dash (–), which is used to link the constituent atoms. But, the stability of monoatomic gaseous molecules of noble gases, like He, Ne, Ar etc. led the scientists to raise the following questions: (i) Why do certain elements combine to form molecule, while certain other elements like He, Ne, Ar etc. exist as monoatomic gases and do not form molecules ? (ii) What is the nature of force that binds the atoms together ? 5.1 WHY DO ATOMS COMBINE ? The elements like, He, Ne, Ar, Kr, Xe and Rn are placed at the zero group in the periodic table. These are called inert gases. They do not combine with other elements and also they do not form even diatomic molecules. However, recent researches have shown that, they do enter into chemical combinations with some specific elements under certain specific conditions, forming chemical compounds. The word, inert, therefore, is no longer considered appropriate, so they are called noble gases. The word noble, signifying that they enter into very few chemical reactions.

CHEMICAL BONDING AND MOLECULAR STRUCTURE 177 The outer electronic configurations of noble gases are given below ; Element Atomic number Outer electronic configuration (Valence shell electrons) Helium 2 1s2 Neon 10 2s2p6 Argon 18 3s2p6 Krypton 36 4s2p6 Xenon 54 5s2p6 Radon 86 6s2p6 Atoms of noble gases are considered to be most stable. Therefore, s2p6 configuration (8 electrons) in the outer energy level (valence shell) constitutes a structure of maximum stability, except He, where 2 electrons are present. Following points are required to be understood for the causes of combination of atoms. 1. Electronic structure : Noble gas elements are stable. They contain 8 electrons (except He) in their valence shell. Hence, elements having less than 8 electrons in their valence shell enter into chemical combination. The valence electrons in two or more combining atoms rearrange to form molecules. 2. Net attractive forces between atoms : When two atoms approach each other, the following two types of forces operate between them. i. Attractive forces between the electrons of one atom and nucleus of the other. ii. Repulsive forces between the electrons and nuclei of the two atoms. . .e- R e- A +R A + AB Fig: 5.1 Attractive (A) and Repulsive (R) forces between two atoms. These two types of interactions counteract each other. When the resultant force is attractive, the two atoms combine. But when the resultant force is repulsive, the atoms do not combine. 3. Lowering of energy : For chemical combination, the resultant force is attractive, which results with decrease of energy. Because, during their approach, some work has been done by the system. Hence, it is clear that atoms combine with the net decrease in energy.

178 +2 CHEMISTRY (VOL. - I) 0 Repulsion When the two isolated atoms are at infinite attraction distance apart from each other the potential Potential energy of the system is taken to be zero. As Energy the distance between the atoms decreases the potential. energy decreases. At a certain point r0 x 'x' the potential energy becomes minimum. Fig : 5.2 Potential energy curve. This point is attributed to the formation of molecule. The distance between the two nuclei is the bond length r0. With further decrease in internuclear distance, net repulsion takes place and the potential energy suddenly increases. Thus, right hand portion of curve is attributed to net attraction whereas the left hand portion to net repulsion. 4. Octet rule : Kossel and Lewis (1916) observed that, atoms of noble gases have 8 electrons in the valence shell. These are stable and enter into few chemical reactions. Thus, atoms having less than 8 electrons in the valence shall are reactive and capable of chemical combination. In order to complete their octet, the atoms of such elements mutually share or transfer one or more valence electrons. Thus, each combining atom has a tendency to attain the nearest noble gas configuration of maximum stability and minimum energy, in order to form a compound. 5.2 VALENCE ELECTRONS Kossel and Lewis (1916) proposed ‘Electronic Theory of Valency’ and their generalisations are as follows; (i) The presence of 8 electrons in the valence shell of an atom, like that of noble gases (except He, where the no. of such electrons is 2) constitutes the stable electronic structure of an atom. (ii) The capacity of an atom to take part in chemical combination is determined by the number of valence electrons that is, electrons present in the outermost orbit of an atom. (iii) During chemical combination, the transfer of one or more valence electrons from one atom to another or mutual sharing of valence electrons amongst themselves takes place. (iv) By the process of transference or by the mutual sharing of valence electrons, each combining atom acquires the nearest stable noble gas configuration. (v) The number of electrons, which an atom loses or gains or mutually shares to attain noble gas configuration is called its valency.

CHEMICAL BONDING AND MOLECULAR STRUCTURE 179 5.3 TYPES OF CHEMICAL BOND The following types of chemical bonds are possible, depending upon the process of valence electron rearrangement between the constituent atoms. (i) Ionic or electrovalent bond. (ii) Covalent bond. (iii) Co-ordinate covalent bond. and (iv) Hydrogen bond. I. IONIC OR ELECTROVALENT BOND : Definition : Ionic bond may be defined as the union of two or more atoms, through the redistribution of valence shell electrons by the process of transference of electrons from one atom to another, so that all the atoms acquire the stable noble gas configuration of minimum energy. By the process of transference of one or more valence electrons, the atom which loses electrons, acquires positive charge and becomes positive ion or cation. Similarly, the atom which gains the electrons, acquires negative charge and becomes negative ion or anion. In this way, the two atoms become oppositely charged ions. The two oppositely charged ions acquire the electronic configuration of nearest noble gas. The electrostatic attraction between the oppositely charged ions is called the ionic or electrovalent bond. The electrostatic attraction always tends to decrease the potential energy. Hence, the potential energy of the system is much less now than it was before the formation of the ionic bond. This type of bond is also known as heteropolar bond. The compounds which contain electrovalent bond are called electrovalent or ionic compounds. Examples of ionic compounds : (a) Formation of sodium chloride : Here, the constituent atoms sodium (At. No. 11) and chlorine (At. No.17) have the following electronic configuration. 11Na = 1s2, 2s2p6, 3s1 17Cl = 1s2, 2s2p6, 3s2p5 There is only one electron in the valence shell of sodium atom and seven valence shell electrons in case of chlorine atom. Hence, sodium has a tendency to lose the outermost electron to chlorine atom, in order to acquire the stable configuration of Neon. Similarly, chlorine atom has a tendency to gain one electron in order to acquire the nearest noble gas configuration of argon. Fig : 5.3 Formation of sodium chloride.

180 +2 CHEMISTRY (VOL. - I) Finally, the positive and negatively charged ions get attracted by the electrostatic forces of attraction. Ionic bond is strong as it is formed by a large decrease of energy. Here, each ion is a charged particle. It is surrounded by an electricfield. This field is non-directional, that is, it is uniformly distributed about the ion. Ionic bond is, therefore, considered as non-directional. (b) Formation of Calcium Fluoride : Here the constituent atoms calcium (At.No. 20) and fluorine (At.No. 9) have the following electronic configuration. 20Ca = 1s2, 2s2p6, 3s2p6, 4s2 , 9F = 1s2, 2s2p5 Calcium has two valence shell electrons. It can lose two electrons and attain the nearest noble gas configuration of argon. Similarly, two fluorine atoms can gain one electron each and attain the noble gas configuration of neon. x ...F.... Ca (2, 7) x (2, 8, 8, 2) (.(..22...FF,.....7) 2+ [ .....F....... - (2, 8) F- Ca++ F- ] [Ca] (2, 8, 8) [ F ] (2, 8) ... - .....[ F ] (2, 8) [ F ] (2, 8) Finally, the oppositely charged ions get attracted by electrostatic forces of attraction. Ca2+ + 2F- ® CaF2 (c) Formation of magnesium sulphide : Here the electronic configuration of constitutent atoms, magnesium (At. no.12) and sulphur (At. no.16) are given below. 12Mg = 1s2, 2s2p6, 3s2 16S = 1s2, 2s2p6, 3s2p4 . Magnesium has two valence shell electrons. It can lose two electrons and attain the nearest noble gas configuration of neon. Similarly, sulphur atom has a tendency to gain two electrons and acquire the nearest noble gas configuration of argon. Finally, the two oppositely charged ions get attracted by electrostatic forces of attraction. MMgxMxgxxg .S...: MMgg2-+ S2- MgS (2,8,2) ¯ (2,8,6) [Mg]22+ [:S¯....: ] 2– (2,8) (2,8,8)

CHEMICAL BONDING AND MOLECULAR STRUCTURE 181 Difference between atoms and ions : It must be remembered that ions differ from the corresponding atoms in several respects. These are given below. i. The ions are charged particles, may be positive or negative, where as atoms are perfectly neutral. ii. Ions have the stable s2p6 configuration in the outer energy level. But all atoms have not the stable configuration. Therefore, ions are more stable than the corresponding atoms. iii. Generally, a positive ion is invariably smaller in size and a negative ion is invariably larger than the corresponding atom. Na + Cl Na+ + Cl- Atom Atom Ion Ion Born - Haber cycle : Lattice energy of ionic solids can be calculated by applying Born-Haber cycle. Lattice energy is defined as the energy released during the formation of one mole of a crystal of the ionic solid from the constituent gaseous ions. Let us consider the formation of one mole of crystalline sodium chloride (NaCl) from sodium solid and chlorine gas. The heat change involved in the reaction is –94.6 kcal per mole. The reaction is represented as, Na(S) + 1/2Cl2(g) NaCl(S), DHf = – 94.6 kcal/mole. DHf = heat of formation of one mole of sodium chloride. The above reaction may be considered to take place through the following three hypothetical steps. (i) The first step involves the sublimation of one mole of solid sodium into gaseous state and dissociation of half a mole of chlorine gas into gaseous chlorine atom. a. Na(S) Na(g) DH1 = 26.0 kcal ½b. Cl2(g) Cl(g) DH2 = 28.9 kcal These processes are involved the absorption of energy equal to 26.0 and 28.9 kcal and are called heat of sublimation and heat of dissociation respectively. (ii) In the second step. the gaseous atoms of sodium and chlorine are converted to gaseous ions. The energy absorbed for the formation of gaseous sodium ion from gaseous sodium atom is called ionisation energy. Also the energy released for the formation of gaseous chlorine atom is called electron affinity.

182 +2 CHEMISTRY (VOL. - I) (c) Na(g) Na+(g) + e– , DH3 = 119.0 kcal (d) Cl(g) + e– Cl–(g), DH4 = – 86.5 kcal. (iii) The final step involves combination of gaseous sodium ion and chloride ion to give one mole of sodium chloride crystal lattice. The energy released in this process is called lattice energy. Na+(g) + Cl–(g) [Na+] [Cl–] + Lattice Energy (U) Solid According to Hess's law, the heat of formaiton of one mole of sodium chloride will be the same irrespective of the fact whether it takes place directly or through the three hypothetical steps involved. Hence, DHf = DH1 + DH2 + DH3 + DH4 + U Hence, – 94.6 = 26.0 + 28.9 + 119.0 – 86.5 + Lattice Energy. \\ Lattice Energy = – 182.0 kcal/ mole. The negative sign indicates that the energy is released or evolved. Since, the formation of ionic bond is possible only if there is a net decrease in the potential energy of the system, the lattice energy is responsible for the formation of the ionic bond. The Born - Haber Cycle explained above may be represented as below. Na(s) + ½ Cl2(g) NaCl(s) DH1 DH2 U (Lattice Na(g) Cl(g) Energy) DH3 DH4 Na+(g) Cl–(g) Factors influencing formation of ionic bond : From the above discussion, we may conclude that the following three factors favours the formation of ionic bonds. between metals and non-metals.

CHEMICAL BONDING AND MOLECULAR STRUCTURE 183 (i) The ionisation energy of the metal atoms should be low. (ii) The electron affinity of the non-metal atoms should be high. (iii) The lattice energy of the ionic compound formed should be high. (i) Low ionisation energy : Ionisation energy is the amount of energy required to remove an electron from the neutral gaseous atom and form a cation. For example, Na+(g) + e– Na(g) + Ionisation Energy (= 119.0 kcal/mole) We know, for the formation of ionic bond, the overall energy of the system should be minimum. Since, energy is absorbed during ionisation, therefore, lesser the ionisation energy, greater is the tendency of an atom to change itself into a cation. That is why, alkali metals form ionic bonds more readily than alkaline earth metals, which have high ionisation energy. (ii) High electron affinity : The second atom gain electron and therefore, an anion is formed. This process is accompanied by the release of energy, known as electron affinity. For example Cl(g) + e– Cl–(g) + Electron affinity (= –86.5 kcal /mole) Since, energy is released during the formation of an anion, therefore, greater the value of electron affinity, greater will be the ease of formation of anion.. That is why halogens (with high electron affinity) form ionic bonds more readily than chalcogens (O, S, etc). (iii) High lattice energy : Lattice energy is–the amount of energy released to form one mole of crystalline ionic solid. For example, [Na+] [Cl ] + Lattice energy Na+(g) + Cl–(g) (= – 182.0 kcal/mole) Since, energy is released, therefore, greater the lattice energy, stronger will be the ionic bond. We know that the electrostatic forces of attraction between the oppositely charged ions will be high, at high lattice energy. q1 . q2 But, according to Coulomb's law, the electrostatic forces of attraction, F = d2 where, q1 and q2 are the respective charges of ions and d is the distance between the ions. So, if the size of the ions are small, the inter-ionic attraction will be increased. Hence, greater will be the strength of the ionic bond. Characteristics of Ionic compounds : The important characteristics of ionic compounds are as follows : (i) Physical state : Due to the strong electrostatic forces of attraction between the oppositely charged ions in ionic solids, ionic compounds are hard and rigid.

184 +2 CHEMISTRY (VOL. - I) (ii) High density : The strong electrostatic forces of attraction between the oppositely charged ions in ionic compounds bring them very close to one another. Consequently, the volume decreases and its density becomes high. (iii) High melting and boiling points : Ionic compounds have high melting and boiling points due to the strong inter-ionic forces of attraction. A considerable amount of heat energy is required to overcome this force and cause fusion or vapourisation of ionic compounds. (iv) Electrical conductivity : Ionic compounds do not conduct electricity in solid state. But they conduct electricity in the molten state. The anions and cations in ionic compounds remain intact occupying fixed positions in the crystal lattice, due to electrostatic forces of attraction. Therefore, the ions are unable to move when an electric field is applied. Hence, no current flows. But in the molten state, since the cations and anions are mobile, they conduct electric current. (v) Solubility in water : Ionic compounds are highly soluble in polar solvents like water and insoluble in non-polar solvent like benzene, chloroform etc. This arises due to the high dielectric constant of polar solvents. These compounds follow the principle of like dissolves like, that is, polar compounds are soluble in polar solvents. Electrovalency : The number of electrons gained or lost by an atom to form an ionic bond is called its electrovalency. Hence, electrovalency of sodium is 1 and that of calcium is 2. But, iron, cobalt and nickel show variable electrovalency. They form divalent as well as trivalent ions. Metal Configuration of atoms Ion Configuration of ions Fe 2, 8, 3s2p6d6, 4s2 Fe++ 2,8, 3s2p6d6 Co 2, 8, 3s2p6d7, 4s2 Co++ 2,8, 3s2p6d7 Ni 2, 8, 3s2p6d8, 4s2 Ni++ 2,8, 3s2p6d8 The outer electronic configuration of above divalent ions are not as stable as the s2p6 configuration. Hence, these ions can easily lose one of the d-electron producing trivalent ions, like, Fe3+, Co3+ and Ni3+ ions. These ions are relatively more stable, since the effective positive charge on the nucleus is now higher and it prevents the elimination of any more electron.

CHEMICAL BONDING AND MOLECULAR STRUCTURE 185 Hence, electrovalency of iron, cobalt and nickel may be 2 or 3. Similarly, copper shows variable electrovalency as 1 or 2. II COVALENT BOND : Introduction : G.N. Lewis in 1916, first suggested that atoms may combine with one another by mutual sharing of electrons in their valence shells. The shared electrons are counted towards the stability of both the atoms and hence these atoms attain the nearest noble gas configurations. This type of bond is called covalent bond. The compounds formed by this process are called covalent compounds. Hence, covalent bond is formed by the process of equal sharing and equal distribution of valence shell electrons between the atoms. Definition : Covalent bond is defined as the union of two or more atoms through the redistribution of valence shell electrons by the process of mutual sharing amongst themselves, so that all the atoms acquire the stable noble gas configuration of minimum energy. Examples : A. Formation of single bond : 1. Formation of hydrogen molecule : Hydrogen (At.No.1) has electronic configuration, 1s1. When two hydrogen atoms approach each other, each contributes one electron to the common pair which is then shared equally by both the atoms and thereby acquiring the stable configuration helium atom. Thus, .H .H x H or H – H Hx 2. Formation of chlorine molecule : In a chlorine molecule two chlorine atoms are combined together by means of a covalent bond. Chlorine (At.No.17) has electronic configuration 1s2 , 2s2p6 , 3s2p5. So chlorine has seven electrons in its valence shell. Here each chlorine atom contributes one electron to form a common pair and the common pair of electron is then shared equally by the two chlorine atoms. Thus, both the chlorine atoms attain stable electronic configuration of argon. Thus, ...C..l.. xx. ......xx Cl xxCxxlxxx xx x Cl or Cl – Cl - (2, 8, 7) (2, 8, 7) (2, 8, 8) (2, 8, 8)

186 +2 CHEMISTRY (VOL. - I) 3. Formation of methane : In methane the central atom carbon is linked with four hydrogen atoms by means of four single covalent bonds as shown below. Carbon (At No. 6) has the electronic configuration : 1s2 , 2s2 p2. Hydrogen (At. No.1) has the electronic configuration : 1s1. H. H l x or H – C – H l . .H x C x H H .x H B. Formation of double bonds : When two electron pairs are shared between the two atoms, double bond results. It is denoted by two dashes (=). 1. Formation of oxygen molecule : The atomic number of oxygen is 8. Its electronic configuration is 1s2 2s2p4. Hence each oxygen atom has six valence electrons. So it is deficient of two electrons. For the formation of an oxygen molecule, each oxygen atom contributes two electrons. Thus, two electron pairs so formed are distributed equally between them and attain the stable noble gas configuration of neon. o ox x .. . . .. o oxxxx.. . ... o ox x . . .. xx + xx xx = xx (2, 6) (2, 6) (2, 8) (2, 8) Double bond 2. Formation of ethylene molecule (C2H4) : The formation of ethylene molecule is represented as below. HH . .. . . . . .. . .H xHH x C x x H2 x x C x H x xx x Cx H C H ¯H II H–C=C–H - Double bond

CHEMICAL BONDING AND MOLECULAR STRUCTURE 187 C. Formation of triple bonds : When three electron pairs are shared between two atoms a triple bond results. It is denoted by three dashes (º). 1. Formation of nitrogen molecule : The nitrogen (At.No.7) has the electronic configuration 1s22s22p3. Hence each nitrogen atom has five valence electrons. In order to form a nitrogen molecule, each nitrogen atom contributes three electrons. Thus, three electron pairs formed are then shared equally between the atoms in order to attain the noble gas configuration of neon. .. ...: + xxx xx .. ... xxx xx .. xx N N N N N N (2, 5) (2, 5) (2, 8) (2, 8) 2. Formation of acetylene (C2H2) : The formation of acetylene molecule can be represented as given below. . . . . . .x H C xxx C x H ® H x C xxxC x H ® H – C º C – H. Factors influencing the formation of covalent bond : The following factors influence the formation of covalent bond. (a) High ionisation energy : Atoms having high ionisation energy are incapable of forming cations. Therefore, they favour covalent bond formation. (b) Equal electron affinities : Atoms having equal electron affinities favour covalent bond formation, since covalent bond is formed due to equal sharing of electrons. (c) Electronegativity : Atoms having either little difference or no difference in their electronegativities favour covalent bond formation. This is because, in such cases, the transfer of electron from one atom to the another is hindered. Covalency and variable covalency : The number of electrons contributed by an atom for sharing, to establish a covalent bond is called its covalency. The covalency of hydrogen is one, while oxygen, nitrogen, and carbon have covalencies 2, 3, 4 respectively. But the covalency of phosphorus in PCl3 is three and five in PCl5.

188 +2 CHEMISTRY (VOL. - I) Characteristics of covalent compounds : The characteristics of covalent compounds are described below. (i) Physical state : Most of the covalent compounds exist either in gaseous or liquid state. This is due to the weak force of attraction between the atoms constituting the covalent bond. (ii) Low melting and boiling points : The force of attraction between the constituent atoms in covalent compound is weak in comparision to ionic compound. Hence, the melting and boiling points of covalent compounds are low. (iii) Bad conductors : Covalent compounds are bad conductors of electricity and carry no current, because they do not ionise by passing electric current through them. (iv) Solubility : Covalent compounds are usually insoluble in polar solvents like water. However, they dissolve in non-polar or organic solvents, like, benzene, carbon tetrachloride etc. (v) Slow rates of reaction : Since covalent compounds show molecular reactions, their rates of reaction are slow. (vi) Isomerism : Covalent bonds are rigid and directional. So they can give rise to different arrangements of atoms in space. Therefore, covalent compounds can appear in different structures in space with different properties. This property is called isomerism, which is exhibited by covalent compounds. 5.4 BOND PARAMETERS : 1. Bond Length : The equilibrium distance between the nuclei of two bonded atoms in a molecule is known as bond length (also called bond distance or inter-nuclear distance). Bond length of a molecule can be determined by various methods like electro-negativity method, electron diffraction method, neutron diffraction method and molecular spectra method. As it is possible to measure the radius of an isolated atom or ion, the following concepts are used to define atomic and ionic radius. These are covalent radius, vanderWaals radius and ionic radius. The covalent radius is defined as half of the distance between the nuclei of two similar atoms, covalently bonded to form molecule. rA = d A- A 2 vander Waals radius of an element can be defined as half of the inter nuclear distance between the nuclei of the adjacent atoms, belonging to two neighbouring molecules of that element, in the solid state. Ionic radius is the distance between the nucleus of the ion and the point upto which the nucleus exerts its attractive force on the electron cloud of the ion.

CHEMICAL BONDING AND MOLECULAR STRUCTURE 189 2. Bond Order : Bond order (also called as bond multiplicity) is defined as the number of covalent bonds formed between two atoms in a given molecule. For example, bond order in H2 is 1, in O2 is 2 and in N2 is 3, which also indicates the number of shared pairs of electrons between the atoms. It is observed that with the decrease of bond order between the two atoms, the bond length between the atoms increases. Thus, the relation between bond order and bond length is expressed as, Bond Order µa Bond 1 Length It is also a fact that with increase in bond order, bond energy increases; i.e. Bond order µ Bond energy. 3. Bond Angle : The angle between the bonding electron pairs in a given molecule or an ion is called bond angle. For example, HCH bond angle in CH4 molecule is equal to 109.5°. It is very helpful in understanding the distribution of orbitals around the central atom in a molecule or an ion and also the shape of the molecule or ion. The mangitude of bond angles is affected by the following factors. (i) Number of lone pairs of electrons on the central atom of a given molecule or an ion. (ii) Electronegativity and size of central atom and other atoms in a molecule. (iii) Presence of multiple bonds in a molecule. 4. Bond Energy : In the formation of a bond some amount of energy is released. This energy is called ‘bond formation energy’ or simply ‘bond energy’ or ‘bond enthalpy’. A+B AB Two atoms AB molecule + Bond formation energy (released) } } }} Likewise, in the dissociation of a bond, some amount of energy is required, which is called as ‘bond dissociation energy’. AB A+B AB molecule + Bond dissociation energy (required) Two atoms Bond energy is defined as the amound of energy required to break one mole (one Avogadro's number) of bonds, formed between the constituent atoms (in gaseous state). For example, H2(g) ® H(g) + H(g). DH = 435.8 kJmol–1 Similarly, bond energy of HCl molecule is 431 kJmol–1 HCL(g) ® H(g) + Cl(g) DH = 431 kJmol–1 The most important application of bond energy is that this can be used to evaluate the heat of reaction of a given process.

190 +2 CHEMISTRY (VOL. - I) 5.5 LEWIS STRUCTURE : The concept of electron dot structure was given by Lewis. The electron dot structure will enable us to understand the formation of a bond between the constituent atoms. The essential points for representation of an atom are as follows : (i) The symbol for an element represents the nucleus and the electrons in the inner energy levels, which are not involved in bond formation. (ii) The dots on the symbol represent the number of valence electrons, that is electrons present in the outermost energy level. (iii) When combination takes place between two similar atoms, valence electrons on two atoms are represented differently. Dots are put on one atom and crosses on the other. Example :1.(a) The atomic number of chlorine is 17. Its electronic configuration is 1s2, 2s2p6, 3s2p5 The outer electronic configuration is 3s23p5 ....It is writen as, : Cl. (i) Cl represents the nucleus of chlorine and 10 (2, 8) inner electrons. (ii) Seven dots on it represent valence electrons. (b) The formation of chlorine molecule is represented as, ....: Cl. ClXXX X XX X (c) The formation of chloride ion is represented as, [:..C....l...:] – Here, one electron is added to the chlorine atom. Hence, chloride ion attains the electronic configuration of argon, with the electronic configuration 2, 8, 8. 2. In the Lewis structure of a molecule, a covalent bond between atoms is ordinarily shown as straight line between atoms, Unshared electron pairs, belonging entirely to one atom are shown as dots. The Lewis structures for the molecules formed by hydrogen with C, N, O and F are CH4 NH3 H2O HF H HCH HNH HOHH F HH

CHEMICAL BONDING AND MOLECULAR STRUCTURE 191 In each case the central atom (C, N, O, F) is surrounded by eight valence electrons and a single electron pair is shared between two bonded atoms. These bonds are called single bonds. Depending upon the number of shared pairs of electrons, double bonds and triple bonds occur. (see later). Many polyatomic ions can be assigned simple Lewis structures. For example, the OH– and NH + ions can be shown as : 4 + H O H and HNH H With OH– ion, this is one more than the number contributed by the neutral atoms (6 + 1 = 7). The extra electron is accounted for by the –1 charge of the ion. With the NH + ions, 4 4 hydrogen atoms and a nitrogen atom supply 9 (4 + 5 = 9) valence electrons. One of these is missing in the NH + ion, accounting for its +1 charge. 4 Electron Dot (Lewis) structure of a few simple molecules and ions. (a) Phosphine 3H x + P = Hx P xH or HPH Hx H (b) Hydrogen sulphide 2H x + S = S S S H H xx xx HH HH (c) Sulphur dioxide x xx x 2O= x x S + (d) Sulphur trioxide OO xx S xx S xx + 3 O = O

192 +2 CHEMISTRY (VOL. - I) (e) C arbonate ion C O – – 3 O 2 O C O (f) Sulphate ion SO42– O2 OS O O (g) Nitrate ion NO3– OO N O (h) Phosphate ion PO 3- 4 O 3 OP O O 5.6 POLAR CHARACTER OF COVALENT BONDS : A covalent bond is formed by equal contribution and equal sharing of valence shell electrons between the constituent atoms in a molecule.But, we know, electronegativity of an atom in a melecule is its relative power or tendency to attract the shared pair of electrons towards itself. The electronegativity values of some of the elements are given in the following Table-5.1.

CHEMICAL BONDING AND MOLECULAR STRUCTURE 193 Table: 5.1 Electronegativity of some common elements. Name of the element Electronegativity value Hydrogen 2.1 Boron 2.0 Carbon 2.5 Nitrogen 3.0 Oxygen 3.5 Fluorine 4.0 Chlorine 3.0 Bromine 2.8 Iodine 2.5 Polar covalent bonds : From the above discussion, we may conclude that, if in a molecule the values of electronegativity of constituent atoms differ much, then the shared pair of electron is shifted slightly towards more electronegative atom. As a result of which, there occurs charge separation. Thus, the more electronegative atom acquires a partial negative charge and less electronegative atom gets a positive charge. These are written as d+ and d– . ...... .......H x Cl ® ®d+ d– d+ d– H (a) HCl x Cl Od– H – Cl (b) Water, d+ (c) Ammonia, d– d+ H N d+ H d+H d+ H H Greater the difference in the electronegativities of the atoms, greater is the charge separation.Since , here dipole results, these bonds are called polar covalent bonds. Non-polar covalent bonds : When a molecule is formed by two similar atoms or by the atoms which differ very slightly in their electronegativities, then such a molecule is said to be non-polar. In such molecules, the shared electron pair remains exactly midway between the two nuclei and no charge separation takes place. Examples : H2, Cl2, F2, O2, N2, CH4 etc.

194 +2 CHEMISTRY (VOL. - I) Ionic character of bonds and polar molecules (Dipole moments) : Heteronuclear molecules are made up of different kinds of atoms. More electronegative atom withdraws the shared pair of electrons more towards itself and acquires partial negative charge. The other atoms acquires the same amount of positive charge. Such molecules having positive and negative charges, separated by some distance form dipoles and thus possess dipole moments. Dipole moment : The degree of polarity of a bond or a molecule is usually expressed in terms of dipole moment. The dipole moment is defined as the product of the magnitude of the charges and the distance between them. Mathematically, Dipole moment (m) = e x d Here, e = magnitude of charges or net +ve or –ve charge and is of the order of 10–10 e.s.u. d = distance between the charges, that is, bond length and is of the order of 10–8cm. Thus, the values for dipole moments of molecules are of the order of 10–18 e.s.u–cm. This quantity is called a Debye and is represented as D. For example, dipole moment of HCl is 1.03 x 10–18 e.s.u – cm and is expressed as 1.03 D. Representation : The dipole moment is a vector quantity having magnitude as well as direction. The direction of dipole moment is indicated with the help of an arrow head (+ ®) pointing towards the more electronegative element. For example, the dipole moment of HF is shown as, +® m H —— F ( m = 1.98 D). Dipole moment in diatomic molecules : A diatomic molecule has two atoms bonded to each other by a covalent bond. The dipole moment of a diatomic molecule with one polar bond is equal to the dipole moment of its individual bond. For example, HF has one polar bond. Its dipole moment is 1.98 D, which is the same as that of single H – F bond. Dipole moment in molecules having more than one polar bond The dipole moment of a molecule with more than one polar bond is equal to the resultant of dipole moments of all the indivisual bonds. It also depends upon the orientations of the bonds in a molecule in space. For example, carbon dioxide and water are both triatomic molecules having two polar bonds. But dipole moment of carbon dioxide is zero, whereas that of water is 1.84 D. This difference in dipole moments is due to the difference in their structures.

CHEMICAL BONDING AND MOLECULAR STRUCTURE 195 Carbon dioxide is a linear molecule. The dipole moment of each C = O bond is 2.3 D. But, the resultant dipole moment is zero due to its linear structure. Hence, CO2 is a non- polar molecule. But, water is a polar molecule having dipole moment 1.84 D. It is due to the bent structure of water. Due to the bent structure of water, the dipole moment of each O – H bond can not cancel out and shows a resultant dipole moment. Similarly, the dipole moment of ammonia molecule is 1.49D, which is the resultant of three polar N – H bonds oriented in space at an angle of 1070 with respect to each other. ®d– d+ d– O=C = O (m = 0) o d– d– N d+ d+ d+ d+ H H H H m = 1.84 D d+ H m = 1.49 D. Similarly, BF3, CH4 and CCl4 have no resultant dipole moment, because of their geometry. Hd+ Cld– F d– B d+ C d– C d+ Fd– Fd– d+ H d+ d+ Cld– Cld– m=0 H H Cldm– O. m=O m=O = m=0 m=0. Applications of dipole moment : Some of the applications of dipole moment are described below : (i) Shape of molecules : Dipole moment values can help in predicting the shape of molecules. For example, water can have a linear or an angular structure. d+ d– d+ Od– HO H d+ d+ H H m=0 m = 1.84 D (Linear structure) (Angular structure)

196 +2 CHEMISTRY (VOL. - I) If water will have linear structure, its dipole moment should be zero. But experimental data shows that water has a dipole moment of 1.84 D, which is possible with angular structure. Thus, water molecule can not be linear, but must have an angular shape. Similarly, since the net dipole moment value of carbon dioxide is zero, it must have a linear structure. (ii) Distinction between polar and non-polar molecules : The dipole moment values can also be used to predict the polar and non-polar nature of molecules. If a molecule has some specific dipole moment, it is a polar molecule. For example, water ( m = 1.84 D), and NH3( m = 1.49 D) are polar molecules. But, if a molecule has zero dipole moment, it is a non-polar molecule. For example, H2, N2, Cl2, BF3, CH4 and CCl4 (where, m = 0) are non-polar molecules. We have observed that covalent compounds containing non-polar bonds are non-polar in nature. But a covalent compound containing polar bonds may be polar or non- polar in nature. This can be seen from the Table 5.2. Table 5.2 : Dipole moment of some common molecules. Molecule Nature of bond Dipole moment Nature of molecule H2, O2, Cl2 Non-polar 0 Non-polar CO2, BF3 CH4, CCl4 Polar 0 Non-polar HF Polar 1.98 D Polar H2O Polar 1.84 D Polar NH3 Polar 1.49 D Polar SO2 Polar 1.60 D Polar HCl Polar 1.03 D Polar Polar 1.02 D Polar CHCl3 (iii) Degree of polarity : The degree of polarity in molecules can be predicted by the dipole moment values. For example, HF (m = 1.98 D ) has greater dipole moment than HCl ( m = 1.03D). Hence, HF is more polar than HCl . CALCULATION OF PERCENTAGE OF IONIC CHARACTER FROM DIPOLE MOMENT Dipole moment (m) is defined as the product of charge (e) and the internuclear distance i.e the distance of separation between two nuclei. m = e × d, Charge is expressed in esu and distance ‘d’ in cm, so m is expressed in esu. cm or in terms of debye (D). 1D = 1 × 10–18 esu. cm. Consider the case of HX where ‘X’ is a halogen atom.

CHEMICAL BONDING AND MOLECULAR STRUCTURE 197 (i) If the bond is ionic i.e H + X–, there is complete transfer of electrons from H to X, the bond has an appreciable value of dipole moment. Dipole moment is determined by the product of electronic charge and the internuclear distance betwen H and X : (ii) If the bond is covalent, the value of dipole moment would be negligibly small. (iii) If the bond is polar covalent, the value of dipole moment lies inbetween the value for a pure covalent bond and a pure ionic bond. The value of dipole moment in this case is given by the product of charge separated due to polarisation and the internuclear distance. Thus the value of m depends upon the magnitude of ionic character developed in the bond. Percentage of ionic character = Actual dipole moment of the bond ´ 100 Dipole moment of pure ionic bond For example : (a) In case of HF molecule Internuclear distance ‘d’ = 0.92A0 If the molecule is ionic H+F– m=e×d = 4.8 × 10–10 esu × 0.92 × 10–8 cm = 4.42D. The actual dipole moment of HF = 1.98D So, percent ionic character = Actual DM ´100 = 1.98 D ´100 DM of pure ionic bond 4.42D = 44.8 ~ 45 (b) In case of HCl Internuclear distance = 127 pm = 127 × 10–12m Actual Dipole moment = 3.44 × 10–30 coulombs metre. Dipole moment of HCl considering it to be purely ionic compound = e × d = 1.6 × 10–9 coulombs × 127 × 10–12 m = 2.03 × 10–29 cm Percent ionic character = 3.44 ´10-30 cm ´100 = 16.9 ~ 17 2.03´10-29 cm

198 +2 CHEMISTRY (VOL. - I) 5.7 COVALENT CHARACTER OF IONIC BOND : Ionic bonds possess some covalent character due to polarisation of ions. During the formation of an ionic bond by two oppositely charged ions coming very close to each other, the cation attracts the electron charge cloud of the outer-most shell of the anion towards itself and hence the symmetrical shape of the anion. The ability of a cation to polarise an anion is called its ‘polarising power’ and the tendency of an anion to get polarised by a cation is called its ‘polarisability’. The cation, due to its smaller size and its electron cloud being strongly held to the nucleus, gets very less polarised by the anion. Greater is the polarising power of a cation, greater will be the amount of covalent character produced in the ionic molecule. Fajans’ rule explains the partial covalent character of ionic bonds with following conditions. 1. Higher is the positive charge on the cation, higher will be its polarising power and hence greater will be the magnitude of covalent character of an ionic bond. 2. Smaller is the size of the cation, greater is its polarising power and hence higher is the degree of covalent character in the ionic bond. 3. A cation with ns2p6d10 configuration posseses higher polarising power than the cation having ns2p6 configuration and hence imparts more covalent character to the ionic bond. 4. Greater is the negative charge on the anion, more it gets polarised by the cation and hence more is the covalent character of ionic bond. 5. Larger is the size of the anion, more strongly it is polarised by the cation and consequently the covalent character increases. 5.8 VALENCE BOND THEORY : Theis theory was proposed by Heitler and London (1927) and later extended by Pauling and Slater (1931). The main features of the theory are : (i) Valency shell atomic orbitals of th two atoms overlap to form a covalent bond. (ii) Only half filled atomic orbitals, i.e. orbitals with single electron can overlap and the resulting bond, thus, aquires a pair of electrons with opposite spins. (iii) Maximum electron density is found somewhere in the overlapped region. The force that binds the atoms in a covalent bond is due to the electrostatic attraction between the nuclei and the accumulated electron cloud between them. (iv) More the extent of overlapping, greater is the strength of the covalent bond. The amount of energy released per mole of bonds formed is called bond energy. (v) Covalent bonds show directional properties i.e. the direction of covalent bond is considered to be in the region of maximum electron destity in overlapping region. (vi) It so happens sometimes that paired electrons of the valency shell get unpaired and promoted to vacant orbitals of the same valency shell with slightly higher energy,

CHEMICAL BONDING AND MOLECULAR STRUCTURE 199 which can only exaplain boron's trivalency, carbon's tetravalency, phosphorus pentavalency etc. Application of VB Therory : Valency Bond Theory can be exaplained in simple way by considering the formation of hydrogen molecule. Consider two hydrogen atoms A and B approaching each other with nuclei NA and NB, while their electrons are denoted by eA and eB. When these atoms A and B are brought closer to each other, force of attraction as well as replusion sets in between the atoms. While attraction betwen nucleus of one atom and its electron, NA – eA and NB – eB, along with nucleus of one atoms and electron of other atom, NA – eB and NB – eA, accounts for the attractive force, repulsive force is caused by the repulsion between electrons of two atoms, eA – eB, and nuclei of two atoms, NA – NB. It is the attractive force which brings the two atoms close to each other whereas the repulsive force tends to push them away. When atoms A and B approach each other, a stage is reached where the attractive force balances the repulsive force and the two hydrogen atoms get bonded to form a hydrogen molecule. During the formation of the bond, energy is released, which is called ‘Bond Energy’ or ‘Bond Enthalpy’. It is the same amount of energy which is needed to dissociate one mole of hydrogen molecule. H2(g) + 435.8 kJmol–1 ® H(g) + H(g) ORBITAL CONCEPT OF COVALENCY We know, an orbital can accomodate maximum two electrons having opposite spins. But such atomic orbitals are unable to go for chemical combination. However, an atomic orbital containing a single electron in the valence shell have a tendency to combine with another atom having a single electron with opposite spin. This results in the formation of covalent bond between the atoms. Thus, according to orbital theory a covalent bond is formed between two atoms by simply coupling of electrons with opposite spins present in the atomic orbitals of valence shells of the two atoms. Here the atomic orbitals overlap with each other, so that the electron pair belongs to each orbital. By doing so, the potential energy of the system decreases. Example : Formation of hydrogen molecule : The formation of a hydrogen molecule as a result of the overlapping of atomic orbitals of two hydrogen atoms is explained below. The point at which the potential energy is minimum the s-orbitals of two atoms overlap and form a covalent bond. ++ + ® ++ Hydrogen atom Hydrogen atom. Hydrogen molecule.

200 +2 CHEMISTRY (VOL. - I) Types of covalent bonds : The covalent bonds are of two types, depending on the nature of overlapping of various half - filled atomic orbitals. They are 1. sigma bond (s - bond) 2. Pi bond (p - bond). 1. Sigma bond : A sigma bond is formed by the overlapping of half-filled atomic orbitals of two atoms along their internuclear axis. This type of overlapping is also called head on overlapping or end-to-end overlapping. The sigma bond is fairly strong, as the extent of overlapping in it is maximum. Hence, s - s, s - p and p - p sigma bond formation is possible. Examples : (a) Formation of hydrogen molecule : The sigma bond formed due to the s - s overlapping of atomic orbital is found in the hydrogen molecule. In this molecule, the half filled 1s orbital of one atom overlaps with the other atom along their axes and form a sigma molecular orbital. + ++® + + ® H–H Hydrogen atom(1s1) Hydrogen atom (1s1) s – s sigma bond. (b) Formation of Hydrogen fluoride molecule : The electronic configuration of fluorine (Atomic No.9) is 1s2 2s2 2pX2 2pY2 2pZ1 . It has one half filled 2pZ atomic orbital. Hydrogen atom has one half-filled 1s1 atomic orbital. When 1s atomic orbital of hydrogen and 2pZ orbital of fluorine overlap along their axes, a sigma bond is formed. Here, the nature of the sigma bond is s - p type. H 1s F 2s 2px 2py 2pz 1s + H —- F s – p sigma bond.

CHEMICAL BONDING AND MOLECULAR STRUCTURE 201 (c) Formation of fluorine molecule : 9F = 9F = 1s 2s 2px 2py 2pz Here, half - filled 2pZ atomic orbital of one fluorine atom overlap with 2pZ orbital of another fluorine atom along their axes to form p - p type sigma bond. pz pz F—F p – p molecular orbital. p – p sig-ma bond 2. Pi (p) bond : A pi (p) bond is formed by the lateral or sidewise overlapping of half- filled atomic orbitals present in the valence shells of two atoms. Here, the overlapping occurs above and below the internuclear axis. The Pi ( p ) bond is a weak bond, as the extent of overlapping is very small. The formation of p –bond restricts free rotation between the two atoms. Examples : (a) Formation of oxygen molecule : The electronic configuration of oxygen (At.No. 8) is Is2 2s2 2pX2 2pY2 2p1z . Since, two p - orbitals are half filled, two oxygen atoms get attached by a double bond to form oxygen molecule. 8O 8O 2s z 2pX 2pY 2pz. z pz Is 2 z *17 17 (oxygen atom) y s O ——— O Oy *+O y (oxygen atom) (oxygen atom) p o p o o o2 s Here, 2py atomic orbital of each oxygen atom overlap along their internuclear axes, to form a sigma bond. But, 2pz orbital is at right angle to 2py orbital. Thus, 2pz orbital of each oxygen atom overlap sidewise to form a p – bond.

202 +2 CHEMISTRY (VOL. - I) (b) Formation of nitrogen Molecule : The electronic configuration of nitrogen (at. no. 7) is 1s2 2s2 2pX1 2p1Y 2pz1. Since, here, three p - orbitals are half-filled, two nitrogen atoms get attached by a triple bond to form nitrogen molecule. 7N = 7N = 2s 2pX 2pY 2pZ. 1s z y y z x +x (nitrogen atom) (nitrogen atom) ®x y py z x 2 p _2 s p p2 2 NN .. .. NN So, one nitrogen molecule consists of a s – bond and two p–bonds between the atoms.

CHEMICAL BONDING AND MOLECULAR STRUCTURE 203 Comparison between sigma (s) and pi (p) bonds Sigma bond Pi bond i. A sigma bond is formed by the i. A Pi bond is formed by the sidewise overlapping of half-filled atomic overlapping of half-filled p–atomic orbitals along their internuclear axes. orbitals. ii. A sigma bond is strong, since the extent ii. A pi-bond is weak, since here the of overlapping in it is greater. atomic orbitals overlap to a small extent. iii. A sigma molecular orbital consists of a single electron cloud, which is iii. Here, the overlapping of orbitals occur symmetrical along the internuclear above and below the internuclear axis. axis. Thus, p -molecular orbitals consists of two electron cloud. iv. About the sigma bond, free rotation of atoms are possible. iv. The p - bond restricts the free rotation of atoms. 5.9 RESONANCE : It is observed that certain molecules and ions can be represented by more than one valence bond structures but none of these structures can correctly account for the properties like bond length, bond energy etc. of the molecules or ions. According to this concept, if two or more alternate valence bond structures can be writeen for a molecule, the actual structure is said to be a resonance or mesomeric hybrid of all these alternate structures. For example, CO2 molecule can be represented by the following three structures : O=C=O + º C – O O – C º + I O O II III While the calculated values of C = O and C º O bond lengths are 1.22A° and 1.10A° respectivlely, the experimental value of bond length between carbon and oxygen in CO2 is 1.15A°, which suggests that none of the above structures is the correct structure of CO2. Thus, the correct structure of CO2 molecule is a resonance hybrid of the above three structures. The resonating structures are also called resonance forms or canonical forms or mesomeric formrs. ‘‘When a molecule or ion can be respresented by more than one Lewis structures, none of these structures is able to explain the properties of the molecule or ion and the true structure of the molecule/ion is a resonance hybrid of these various structures, the phenomenon is called resonance.’’ Rules for writing resonating structures : 1. The relative position of all the atoms in each of the resonating structures should be the same and it is the position of electrons which differ. 2. There should not be much difference in energy between the canonical forms.

204 +2 CHEMISTRY (VOL. - I) 3. Each of the canonical forms must have same number of unpaired and paired electrons. 4. It is mostly preferred that in the resonating structures, negative charge resides on more electronegative atom. Characteristics of resonating structures: 1. None of the canonical forms can fully describe all the porperties of the molecule or ion. 2. All the structures contribute to the actual structure of the molecule or ion. 3. In canonical forms, similar charges should not reside on adjacent atoms while dissimilar charges should reside on the adjacent atoms. 4. Greater is the number of covalent bonds in a resonating structure, greater is the stability of that structure. 5. More is the number of canonical forms, greater is the stability of the molecule. Resonance energy : Resonance energy of a molecule is defined as the difference between the energy of its most stable mesomeric form (i.e. form with the lowest energy) and the energy of the structure determined experimentally. For example, resonance energy of CO2 is 126 kJ/mol, which implies that the energy of CO2 molecule is less than that of the most stable resonating structure of this molecule by an amount equal to 126 kJ. Resonance energy is considered as a measure of the stability associated with any molecule. Resonance structures of some molecules and ions (i) Nitrogen molecule : Its actual structure is supposed to be a resonance hybrid of the following resonating structures. ::Nº N: :N N N N: : : : : : : : : : (ii) Nitrogen dioxide molecule : It is a resonance hybrid of the following structures : x x xx xx NN N N : O: : O: : O : O: : O. : O: : O : O. : (I) (II) (III) (IV) (iii) Carbonate ion : Following resonance structures are written for the ion. OO O CC C O O OO OO (I) (II) (III)

CHEMICAL BONDING AND MOLECULAR STRUCTURE 205 5.10 VSEPR THEORY - GEOMETRY OF COVALENT MOLECULES : This theory is based on the repulsive interactions of pairs of valence electrons. The geometry of covalent molecules are determined basing on this theory. The main postulates of valence shell electron pair repulsion theory are as follows. (i) The orbitals are oriented in space, so that the distance between them is maximum and repulsion is minimum. (ii) The lone pair-lone pair repulsion is greater than the lone pair - bond pair repulsion, which, in turn is greater than the bond pair - bond pair repulsion. That is, Lone pair - Lone pair repulsion > Lone pair - bond pair repulsion > Bond pair - bond pair repulsion. Shapes of other molecules : 1. Shape of ethylene (C2H4) molecule : In one ethylene molecule, two carbon atoms combine with four hydrogen atoms. We know, the electronic configuration of carbon atom in the excited state is , 1s2 2s1 2px1 2py1 2pz1 Here, 2s, 2px and 2py orbitals of carbon atom after excitation hybridised to form three sp2-hybridised orbitals. Each sp2-hybrid orbital contain one half-filled electron. First of all one sp2-hybrid orbital of each carbon atom overlap along the internuclear axis to form a sigma bond. Remaining two sp2-hybrid orbitals of each carbon atom overlap with s-orbital of hydrogen atom to form sigma bonds. The unhybridised 2pz orbital of each carbon atom then overlaps laterally to form p -bond. Thus in between two carbon atoms, one sigma and one Pi-bond is formed. The shape of ethylene molecule is trigonal or plannar and the bond angle is 1200. pz - orbital pz - orbital _ HH sp2 sp2 HH Hs H ps Cs C 1200 Hs sH (Trigonal or Planar structure of ethylene)

206 +2 CHEMISTRY (VOL. - I) 2. Shape of acetylene (C2H2) molecule : In one acetylene molecule, two carbon atoms combine with two hydrogen atoms. We know, the electronic configuration of carbon in excited state is, 1s2 2s2 2px1 2p1y 2p1z Here, 2s and 2px orbitals of carbon atom after excitation hybridised to form two sp- hybrid orbitals. One sp-orbital of each carbon atom overlap along their internuclear axes to form a sigma bond. The other sp-orbital of each carbon atom overlaps with 1s-orbital of hydrogen atom. Then the unhybridised 2py and 2pz orbitals on each carbon atom overlap laterally to form two p -bonds. Hence, the shape of acetylene molecule is linear and the bond angle is 1800. py pz py pz HH 1800 p H —C s C — H sp (Linear structure of acetylene) 3. Shape of Ammonia molecule : The electronic configuration of nitrogen 1s2 2s1 2px1 2py1 2pz1 . Here, the central atom, nitrogen shows sp3-hybridisation and forms four sp3-hybrid orbitals. Each of the three half-filled sp3-hybrid orbitals overlaps with 1s-orbital of hydrogen atom to form sigma bonds. One lone pair of electron is present in the fourth sp3-orbital. Here, due to lone pair-bond pair repulsion, according to VSEPR theory, the regular tetrahedral geometry is distorted. The shape of NH3 molecule becomes pyramidal and the bond angle is reduced to 1070 .. H .. H N H 1070 H H H (Pyramidal structure of NH3).

CHEMICAL BONDING AND MOLECULAR STRUCTURE 207 4. Shape of water molecule : In water, the central atom oxygen, has the following electronic configuration in the ground state. 1s 2s 2px 2py 2pz -- a. Oxygen (Ground state). -- b. sp3 - hybridisation. Oxygen contains two unpaired electrons in its valence shell even in the ground state. So this can form two covalent bonds with hydrogen atoms, without involving hybridisation. But, in that case, the water molecule, involving only two orbitals should have been linear and non- polar, like BeCl2. But, this actually is not true. The water molecule is known to have a bent structure and it is highly polar. The reason for the bent structure is the central atom oxygen in water undergoes sp3-hybridisation. Out of the four sp3-orbital two contain lone pair electrons. Other two hybrid orbital containing half filled electron overlap with hydrogen atom forming two sigma bonds. Although the central atom shows sp-hybridi sation, the shape of water is not tetrahedral. This is because of the greater lone pair - lone pair repulsion. Hence, the tetrahedral geometry of water distorted to 'V' shaped or bent. Thus, the geometry of water is 'V'-shaped and the bond angle is 1040 27I or 104.50 . Lone pair .. electrons H o.... o .. 104.50 1040 27I HH HH H 5.11 HYBRIDISATION : Introduction : The overlapping concept of half - filled atomic orbitals explains the formation of covalent bonds. According to this concept, the number of half-filled orbitals in an atom represents the number of covalency of that element.. For example : 1H = 1s1 7N = 1s2 2s2 2px1 2py1 2pz1. 8O = 1s2 2s2 2px2 2py1 2p1z .

208 +2 CHEMISTRY (VOL. - I) From the electronic configuration of above elements we know, the covalency of hydrogen is one. But the covalency of oxygen is two and that of nitrogen is three. Hence, hydrogen is monovalent, oxygen is divalent and nitrogen is trivalent. But beryllium, boron and carbon furnish interesting cases of compound formations. The electronic configuration of these elements are, 4Be = 1s2 2s2 5B = 1s2 2s2 2p1X 2pY0 2pz0. 6C = 1s2 2s2 2pX1 2pY1 2p0z. According to orbital theory, beryllium should behave as a noble gas, since it has no half-filled orbital. Since boron and carbon has one and two half-filled orbitals, boron should be monovalent and carbon should be divalent. But this is not actually the case. Beryllium forms a number compounds like BeH2, BeF2 etc. in which it is divalent. Boron forms a number of compounds like BH3, BCl3 etc. in which it is trivalent. Carbon forms large number of compounds in which it is tetravalent. The simple examples are CH4, CCl4 etc. In order to explain these anomalies, we make use of a new concept, called hybridisation. The phenomenon of hybridisation involves the following steps. (i) Excitation : If an atom in its valence shell has vacant orbitals, there is possibility of electron promotion or excitation from the paired orbitals to these vacant orbitals. The promotion of electrons to the vacant orbitals of the same energy level is possible, because the energy difference between various orbitals of the same level is not much. (ii) Re - orientation : The atomic orbitals on the same atom containing half-filled electron now have identical energy. These atomic orbitals have an unusual ability to merge with one another in an additive way, forming the same number of new orbitals, called hybrid orbitals.The phenomenon is called hybridisation. Definition : The phenomenon of redistribution of energy in different orbitals belonging to the valence shell of an atom to give new orbitals of equivalent energy is called hybridisation. Conditions of hybridisation : The conditions for hybridisation are as follows : (i) The orbitals of one and the same atom participate in hybridisation. Only the orbitals and not electrons get hybridised. (ii) The energy difference between the hybridising orbitals should be small. Characteristics of hybridisation : The characteristics of hybridisation are as follows : (i) The number of hybrid orbitals is equal to the number of hybridising orbitals. (ii) Like the atomic orbitals, a hybrid orbital can not have more than two electrons. These two electrons must also have opposite spins.

CHEMICAL BONDING AND MOLECULAR STRUCTURE 209 (iii) The number of electrons in hybridising orbitals remain the same after hybridisation. (iv) A hybrid orbital which has an unpaired (single) electron can only take part in overlapping, that is bond formation. (v) The shape of the hybrid orbital is influenced by the most dominating orbital taking part in the process of hybridisation. (vi) The hybrid orbitals of a particular type are all similar in shape and energy, but they differ from one another largely in orientation in space. (vii) Due to the electronic repulsions in hybrid orbitals they try to remain away from one another as far as possible. Types of hybridisation : Hybridisation is classified as different types, depending upon the number and the nature of orbitals taking part in the process. The process of hybridisation involving the merger of s and p-orbitals are of following types. (i) sp - hybridisation. (ii) sp2 - hybridisation. (iii) sp3 - hybridisation. (i) sp - hybridisation : The process of hybridisation involving the merger of one s and one p-orbital is termed as sp-hybridisation. In this process, p orbital being the dominant one, the shape of the hybrid orbital is influenced by it. The number of hybrid orbitals become equal to two and these orbitals arrange themselves in a linear fashion at an angle of 1800 between the axes of the two orbitals. In sp - hybrid orbitals the percentage of s and p - character is 50% each. Illustration : .+ One s-orbital One p-orbital Two sp-hybrid orbital. 1800 Linear arrangement of hybrid orbitals

210 +2 CHEMISTRY (VOL. - I) Examples : Formation of BeCl2 molecule : - - orbitals 2p - sp - hybrid 2s - 1s Be (in ground state) Be (in excited state) Be (in hybridised state) The electronic configuration of chlorine is , 1s2 2s2p6 3s2 3p2X 3p2Y 3p1z . First of all with the approach of chlorine atoms, the beryllium atom gets excited by promoting one electron from 2s-orbital to 2px orbital. Then one 2s and one 2px orbital intermixed to produce two sp-hybrid orbitals. Each sp-hybrid orbital overlaps with the 3pz orbital of chlorine atoms and forms BeCl2. Two sp-hybrid orbitals overlap with two 3pz orbitals of chlorine atoms in the head on fashion and form two sigma bonds. Since the shape of sp-hybrid orbital is linear, the shape of BeCl2 is linear and the angle between them is 1800. z z Be 3pz -orbital of sp-hybrid chlorine atom orbital 1800 Cl Be Cl 3pz - orbital of chlorine atom Sp-hybrid orbital (ii) sp2- hybridisation : The process of hybridisation involving the merger of one s and two p-orbitals is termed as sp2- hybridisation. Each sp2- orbitals has 33% s - character and 66% p- character. The number of sp2 - hybrid orbitals produced is equal to three. These hybrid orbitals arrange themselves in space making an angle of 1200 with each other and the orientation is trigonal.

CHEMICAL BONDING AND MOLECULAR STRUCTURE 211 Illustration : .+ One s - orbital Two p - orbitals Three sp2 - hybrid orbitals 1200 Three sp2 hybrid orbitals 120 Three Sp2 hybrid orbitals (Trigonal orientati(oTnr)igonal orientation) Example : Formation of BCl3 molecule : Boron, the central atom in BCl3 molecule shows following electronic configuration in ground state and also in excited state. 2p - -- -- - sp2 - hybrid 2s - orbitals 1s Boron(in excited state) Boron(in hybridised state) Boron (in ground state) In boron trichloride, one boron atom combines with three chlorine atoms. The electronic configurations of chlorine is 1s2, 2s2p6, 3s2 3px2 3py2 3pz1 .

212 +2 CHEMISTRY (VOL. - I) The sp2 - hybrid orbitals of boron atom are directed in space making an angle of 1200 with each other. Each hybrid orbital contain one half - filled electron. Now three 3pZ orbitals of chlorine with one electron in each overlap with three sp2-hybrid orbitals of boron, separately. As a result, three sigma bonds are formed. Therefore, the shape of BCl3 is trigonal and the bond angle is 1200. pz Pz Cl Cl B 1200 Cl Trigonal shape of BCl3. Pz Trigonal shape of BCl3. (iii) sp3-hybridisation : The hybridisation arising out of the merger of one s and three p-orbitals is known as sp3-hybridisation. The total number of hybrid orbitals obtained in this case is four. These four sp3.-hybrid orbitals arrange themselves in space with their lobes directed towards the corners of a regular tetrahedron making an angle of 1090 28. Illustration : .+ (one s-orbital) (three p-orbitals) four sp3-hybrid orbitals.

CHEMICAL BONDING AND MOLECULAR STRUCTURE 213 1090 28 sp3-hybrid orbitals (Regular tetrahedron) Example : Formation of methane (CH4) molecule. Carbon, the central atom in CH4 molecule, shows following electronic configuration in ground state and also in the excited state. 2p - - - - -- -- -- 2s - sp3 - hybrid orbitals 1s Carbon atom Carbon atom Carbon atom (in ground state) (in excited state) (hybridised state) In methane one carbon atom combines with four hydrogen atoms. In CH4, the central atom carbon, shows sp3-hybridisation and produces four sp3 hybrid orbitals. These sp3-hybrid orbitals are arranged in a regular tetrahedral fashion. Since each sp3 - hybrid orbital contain half-filled electron, four hydrogen atoms (each containing half-filled electron) overlap with it. As a result of which four sigma bonds result. Therefore, the shape of CH4 molecule is tetrahedral and the bond angle is 1090 28l. (Regular tetrahedral structure of CH4 molecule).

214 +2 CHEMISTRY (VOL. - I) (iv) COMMON TYPE OF HYBRIDISATION AND GEOMETRY OF COMPLEXES. Co-ordination Hybridisation Geometry Example Number (CN) dsp2 Square planar [ ]Pt(NH3 )4 2+ , [Pt Cl4 ] 2 4 dsp3 Trigonal bipyramidal d2sp3 [Fe(CO)5 ] 5 Octahedral [Fe(CN)6 ]4 (Inner d complex) 6 Geometry of [Ni(CN)4 ]2 NC CN Ni NC CN Square planar structure of [Ni (CN )4 ] 2 CN– is a strong ligand. The unpaired electons in Ni2+ are paired making one inner d-orbital vacant for the ligand. Ni2+ º [Ar] 4s 4p Ni2+in[Ni(CN)4] 2 ≡ [Ar] 4s 4p [Ni(CN)4 ] 2 ≡ [Ar] xx XX XX XX dsp2 electron pair of ligand Nature of hybridisation - dsp2 Nature of complex - Inner d-orbital complex Magnetic character - Diamagnetic Geometry / Shape - Square planar Bond Angle = 900 Geometry of [Fe(CO )5 ] Oxidation number of Fe = 0 C.N. of Fe = 5 Fe º [Ar] 4s 4p Fe in [Fe(CO) 5 ] ≡ [Ar] 4s 4p

CHEMICAL BONDING AND MOLECULAR STRUCTURE 215 Unpaired electrons are paired and outer 4s electrons are accomodated in 3d [Fe(CO)5 ] ≡ [Ar] xx xx xx xx xx dsp3 CO i.e carbonyl is a strong ligand, hence electronic configuration is affected. The unpaired elec- trons are paired. Nature of hybridisation – dsp3 Nature of complex – inner d-complex Magnetic character - Diamagnetic Geometry / Shape - Trigonal bipyramidal Band angle - 1200 and 900 Geometry of [Fe(CN)6 ]4 Oxidation number of Fe = +2 C.N. of Fe = 6 4s2 4p Fe º [Ar] Fe2+ º [Ar] 4s 4p Fe2+ in [ ]Fe2+(CN)6 4 ≡ [Ar] xx xx xx xx xx xx d2sp3 CN– is a strong ligand. It makes all the unpaired electrons paired. The six empty hybrid orbitals are required by six CN– ligands. Nature of hybridsation - d2sp3 Nature of complex - Inner d-complex Magnetic character - Diamagnetic Geometrical shape - Octahedral Bond angle - 900 Magnetic moment = 0 (zero spin complex or spin paired complex)

216 +2 CHEMISTRY (VOL. - I) 5.12 MOLECULAR ORBITAL THEORY (MOT) : This theory was proposed by Hund and Mulliken. So this theory is known as Hund- Mulliken theory. According to this theory, (a) The atomic orbitals combine and form an orbital which is known as the molecular orbital. (b) Orbitals of same energy level of each bonded atom involved in molecule formation lose their identity and merge together to give rise to an equivalent number of new molecular orbitals. (c) All the electrons pertaining to both the bonded atoms are considered to be moving along the entire molecule under the influence of all the nuclei. (d) Electrons while filling the molecular orbitals follow Pauli's exclusion principle as well as Aufbau's principle. (e) The nomenclature, s, p and d used for atomic orbitals is replaced by s , p and d for molecular orbitals. (d) The number of molecular orbitals formed is equal to the number of the atomic orbitals involved in their formation. Thus two atomic orbitals after interaction will produce two molecular orbitals. Of these two molecular orbitals, one is of lower energy than either of the two atomic orbitals and the other is of higher energy. The orbital having lower energy is known as bonding molecular orbital and orbital having higher energy is known as antibonding molecular orbital. The electrons present in these orbitals are called bonding and antibonding electrons respectively. (g) Electrons which are present inside the shell of atoms and do not take part in bond formation are called non-bonding electrons. (h) The antibonding moleculer orbital is represented by a superscript asterisk (*). (i) The bonding molecular orbitals are stable and the antibonding molecular orbitals are unstable. Antibonding M.O. (s * or p* ) Atomic orbital Atomic orbital Energy Bonding M.O ( s or p ) Fig 5.4. Formation of molecular orbitals by the combination of two atomic orbitals.

CHEMICAL BONDING AND MOLECULAR STRUCTURE 217 Table 5.3 Comparison between bonding and antibonding molecular orbitals. Bonding molecular orbitals s or p Anti-bonding molecular orbitals s * or p* 1. These are formed by linear 1. These are formed by linear combination combination of two atomic orbitals of two atomic orbitals when their wave when their wave functions are added. functions are subtracted. That is, yb = y1 + y2 that is, ya = y1 – y2 2. It is formed when atomic orbitals with 2. It is formed when two atomic orbitals same signs of their lobes combine. with opposite signs of their lobes combine. 3. Their energy is less than the combining 3. Their energy is more than the combining atomic orbitals. atomic orbitals. 4. They increase the electron density 4. They decrease the electron density between the nuclei and therfore between nuclei and therefore destabilise stabilise the molecule. the molecule. Rules of filling-up electrons in molecular orbitals : Following rules are followed during the filling-up of the molecular orbitals with the available electrons. (i) The molecular orbitals are filled-up in the order of increasing energy. The molecular orbital with lowest energy is filled first (Aufbau's principle) (ii) Maximum capacity of electrons in a molecular orbital is two. (iii) If there are two or more molecular orbitals of same energy, these are first singly filled and then pairing starts. (Hund's rule) (iv) Whenever bonding of atoms is to take place, there should be an excess of bonding electrons over antibonding electrons. Thus, no bonding occurs, if the number of antibonding electrons is equal to or more than the number of the bonding electrons.

218 s *2pz +2 CHEMISTRY (VOL. - I) 2pz 2py2px p*2px p*2py 2px2py2pz 2s p2px p2py 2s s 2pz s 2*S Increasing energy s 2S s 1*S 1s s2*S 1s s2S s 1S Atomic orbitals Molecular orbitals. Atomic orbitals Fig : 5.5 Energy levels of different molecular orbitals. Order of energy of molecular orbitals: The molecular orbitals in order of their increasing energies, upto nitrogen molecule are, s 1S < s *1S < s 2S < s *2S < p2px = p2py < s 2pz < p*2px = < p*2py < s *2pz But beyond nitrogen, the order is as follows ; s 1S < s *1S < s 2S < s *2S < s 2pz < p2px = p2py < p*2px = p*2py< s *2pz

CHEMICAL BONDING AND MOLECULAR STRUCTURE 219 Electronic configuration and molecular behaviour : The behaviour and nature of a molecule depends upon its electronic configuration. The electronic configuration can be used to predict the bond order, stability and magnetic behaviour of the molecule. These are described below. (i) Bond order : It is defined as the number of covalent bonds in a molecule. 1 Bond order = 2 (Nb – Na) , where Nb is the number of bonding electrons and Na is the number of antibonding electrons. (ii) Stability of the molecule : We know that the electrons in bonding molecular orbitals contribute towards the lowering of energy (stability), whereas the electrons in antibonding molecular orbital contribute towards increase of energy (instability). Thus, IfTIfhNNebbs>t=abNNilaiat;y; tthohefeammmoololeeclceuucluleeleiisscsautnanbsaltelasbo:leIbf. eNebxp<reNssae;d the molecule is unstable : in terms of bond order (a) If bond order is positive, then molecule is stable. (b) If bond order is negative or zero, the molecule is unstable. Higher the bond order, more is the stability of the molecule. Bond order and 1 bond length are also related to each other as follows : Bond order µ Bond length. Bond order of nitrogen is three (N º N) and that of oxygen is two (O = O). Hence, N2 is more stable than oxygen. Stability of a molecule is directly proportional to the bond strength and inversely proportional to the bond length. (iii) Magnetic character : The molecule in which there are no unpaired electrons are diamagnetic. But the presence of one or more unpaired electrons is the cause of paramagnetism of the molecule. Greater the number of unpaired electrons, more is the paramagnetic behaviour. Molecular orbitals in some simple molecules : (i) Hydrogen dmiaotolemciuclehy(dHro2)ge:n Each hydrogen atom has one electron in its 1s-orbital. Therefore, molecule has two electrons and two nuclei. Two 1s atomic orbitals thus, combine to form two molecular orbitals, like, s1S and s*1S . According to Pauli's exclusion principle, both the electrons are accomodated in bonding molecular orbital, which has lower energy. Therefore, electronic configuration of H2 molecule is, H2 = (s1s)2. 1 Hence, Nb = 2 and Na = zero. \\ Bond order = 2 (Nb – Na) = 1. Thus, two atoms are bonded together through one covalent bond. It is diamagnetic as there is no unpaired electron. The energy diagram is given below. Energy s *1S 1S 1sS1*S Energy* s 1S 1s- s 1S - 1s Atomic orbitalMolecule orbital Atomic orbital Atomic orbital Molecular orbital Atomic orbital Fig. 5.6 Molecular orbital energy diagram for H2 molecule.

220 +2 CHEMISTRY (VOL. - I) (ii) Nitrogen molecule (N2) : Each nitrogen atom has seven electrons. Therefore, there are fourteen electrons in nitrogen molecule. These are filled in the increasing order of their energies. Therefore, electronic configuration of nitrogen molecule is N2 = (s 1S)2 (s * )2 (s 2S)2 (s *2S)2 (p2px )2 (p2py)2 (s 2pz)2 . 1S --- s *2pz --- 2pz 2py 2px p*2px p*2py 2px 2py 2pz s 2pz p2px p2py Increasing energy s 2*S 2s 2s s 2S s 1*S 1s 1s s 1S Fig.5.7 Molecular orbital energy diagram for N2 molecule.

CHEMICAL BONDING AND MOLECULAR STRUCTURE 221 Here, Nb = 10 Na = 4 \\ Bond order = 1 (Nb – Na) = 1 (10 – 4) = 3. 2 2 Hence, nitrogen consists of a triple bond between the two atoms. It is diamagnetic since no unpaired electron is present. (iii) Oxygen molecule (O2) : Each oxygen atom has eight electrons. therefore, there are sixteen electrons in oxygen molecule. These are filled in the increasing order of their energies. Therefore, electronic configuration of oxygen molecule is O2 = (s 1S)2 (s *1S)2 (s 2S)2 (s *2S)2 (s 2pz)2 (p2px )2 = (p2py )2, (p*2px )1 (p*2py )1. Here, Nb = 10, Na = 6 \\ Bond order = 1 (Nb – Na) = 1 (10 – 6) = 2 2 2 Hence, there is a double bond in oxygen molecule. Due to the presence of two unpaired electron, it is paramagnetic. The energy diagram is given below : Increasing energy s 2*S s 2S s 1*S s 1S Fig. 5.8 M.O diagram for oxygen molecule.

222 +2 CHEMISTRY (VOL. - I) 5.13 HYDROGEN BOND : We know HF is a polar compound. Due to the higher electronegativity of fluorine the electron pair shared between them lies far away from the hydrogen atom. Thus, hydrogen atoms are highly electropositive with respect to fluorine atom. The phenomenon of charge separation between hydrogen and fluorine atom in HF molecule is represented as, Hd+ — Fd . Thus, due to the charge separation, the molecule behaves as a dipole. The electrostatic forces of attraction between such molecules are very strong. Thus, two or more molecules of HF may associate together to form large cluster of molecules as shown below. H H H F FF F FF H H H Hydrogen bond is defined as the force of attraction which binds hydrogen atom of one molecule with highly electronegative atom of another molecule of the same substance. The hydrogen bond is generally represented by dotted lines, as shown above. It may be noted that, hydrogen atom is bonded to fluorine atom by a covalent bond in one molecule and by hydrogen bond to the fluorine atom in the adjacent molecule. Thus, hydrogen atom is seen to act as a bridge between the two fluorine atoms. Conditions for formation of hydrogen bond : (i) The molecule should be polar covalent. (ii) One of the atoms in the molecule should be highly electronegative. (iii) The highly electronegative atom must be small in size. So, fluorine, oxygen and nitrogen atoms satisfy the above conditions. Hence, all compounds containing hydrogen and one of these electronegative atoms show the property of hydrogen bonding. A hydrogen bond is very much weaker than a covalent bond. Types of hydrogen bonding : Hydrogen bonding are of two types. (i) Intermolecular hydrogen bonding (ii) Intramolecular hydrogen bonding. (i) Intermolecular hydrogen bonding : The intermolecular hydrogen bonding results from the electrostatic forces of attraction between the positive and negative poles of different molecules of the same substance. The following compounds show this type of hydrogen bonding. (a) In water : Oxygen being more electronegative than hydrogen, in the water molecule, intermolecular hydrogen bonding is established involving the oxygen atom of one with the hydrogen atom of the other molecule.

CHEMICAL BONDING AND MOLECULAR STRUCTURE 223 d+ O d..–...... d+ d– d+ d– .......H — H— O ............ H— O............... Hd+ Hd+ Hd+ (b) In ammonia : Due to the polarity in the molecule, a hydrogen bond is established using the nitrogen atoms of one and the hydrogen atom of the neighbouring molecule. Hd+Hd.+.......... ....... Nd– Nd– Hd+Hd..+.......... Nd– Hd+Hd.+..... Hd+ Hd+ Hd+ (c) In alcohol (R – OH) : Association in alcohol molecules takes place due to the intermolecular hydrogen bonding. RR R ll l .......Od–— Hd+........ Od–— Hd+............ Od– — Hd+ ............... (ii) Intramolecular hydrogen bonding : The intramolecular hydrogen bonding involve the electrostatic forces of attraction betwen hydrogen and an electronegative element both present in the same molecule. For example, ortho-nitrophenol shows intramolecular hydrogen bonding. O H O N (o - nitrOophenol) (O - nitrophenol) Strength of hydrogen bond : A hydrogen bond is a much weaker bond and requires only 5 - 10 kcal per mole of energy to break. The bond strength depends on the electronegativity of the element with which a hydrogen bond is established. The decreasing strength of a hydrogen bond takes up the following pattern. F — H ......... F > O — H ......... O > N — H ........ N > Cl — H ........ Cl.

224 +2 CHEMISTRY (VOL. - I) A chlorine atom has approximately the same electronegativity (3.0) value as that of a nitrogen atom , but due to its smaller size the nitrogen atom shows a stronger hydrogen bonding than the chlorine atom. Hydrogen bond strength of some molecules are given below. Hydrogen bond Strength (kcal/mole) H..........F 10.0 H..........O 6.0 H..........N 3.0 Consequences of hydrogen bonding : (i) Compounds showing the property of hydrogen bonding have high melting and boiling points. (ii) Compounds showing the property of hydrogen bonding are highly soluble in water. Alcohol and ammonia are soluble in water, as these molecules form hydrogen bonding with water molecule.

CHEMICAL BONDING AND MOLECULAR STRUCTURE 225 CHAPTER (5) AT A GLANCE 1. Atomic orbital : It is a region around the nucleus in a single atom where the probability of finding the electron is maximum. 2. Chemical bond : It is defined as the force of attraction which holds the constituent atoms in a molecule. 3. Co-ordinate bond : This type of bond is formed by one side contribution and equal sharing of valence shell electron between the constituent atoms. 4. Covalent bond : It is defined as the the force of attraction which arises by the process of mutual sharing of valence electrons between the constituent atoms. 5. Dipole moment : It is defined as the product of the net positive or negative charge on one of the atoms and the distance between the two charges. 6. Hybridisation : It is the intermixing of the orbitals of an atom having nearly the same energy which result in the redistribution of energy to form an equal number of new orbitals of equivalent energy. 7. Hydrogen bond :It is defined as the force of attraction which binds hydrogen atom of one molecule with highly electronegative atom of another molecule of the same or different substance. 8. Ionic bond : It is formed by the transference of one or more electrons from one atom to the other resulting in the formation of ions which get attracted by the force of attraction. 9. Molecular orbital : They exist around the nuclei of all the bonded atoms in a molecule. 10. Sigma bond : This bond is formed by the overlapping of half-filled atomic orbitals of two atoms along their internuclear axis. 11. Pi-bond : This bond is formed by the lateral or sidewise overlapping of half-filled atomic orbitals above and below the internuclear axis. QUESTIONS Very short Answer Questions (1 mark each) 1. State whether SiO2 is an ionic or a covalent compound. 2. Between KF and CHCl3 which is likely to have a higher melting point ? 3. Write the structural formula of nitric acid. 4. Name a monoatomic gas. What is its valency ? 5. Why ammonia gas is not collected over water ? 6. (a) What is the shape of C2H2 molecule ? (b) What type of hybridisation is found in acetybne ?

226 +2 CHEMISTRY (VOL. - I) 7. How. many s and p bonds are there in a molecule of ethylene ? 8. What is the angle between the C – H bonds of methane ? 9. Why is CCl4 immiscible in water ? 10. What type of hybridisation exists in the central atom of BCl3 molecule ? 11. What is the shape of ammonia molecule ? 12. Find out the total number of electrons in a water molecule. 13. What is the shape of BCl3 molecule ? 14. What is the bond angle between O – H bonds of a water molecule ? 15. What type of hybridisation takes place in carbon atom for the formation of graphite ? 16. Between sigma and pi-bond, which bond is weaker ? 17. How do you account for the fact that H2O is a liquid and H2S is a gas at room temperature ? 18. Name the type of overlapping and type of bond in forming HF molecule. 19. (a) What is the shape of NH3 molecule ? (b) What is the shape of NH3 molecule ? What is the reason for it ? 20. What type of hybridisation takes place in carbon atom for the formation of diamond ? 21. Which of the halogens forms hydrogen bond ? 22. Why do H2O molecules remain associated in liquid water ? 23. Why is sigma bond stronger than pi-bond ? 24. Which of the following contains a lone pair of electrons in the central atom ? NH3, CH4, CCl4. 25. What is the shape of CO2 molecule ? 26. What is the bond angle between two hybrid bonds in sp2 hybridisation ? 27. How many sigma and pi-bonds are there in a molecule of acetylene ? 28. Among the compounds NH3, HF and CH4, in which hydrogen bonding is most prominent and why ? 29. What is the maximum number of electrons that can margin in a molecular orbital ? 30. In which of the following types of hybridisation the structure formed has maximum bond angle ? sp3, sp2, sp. 31. What is the shape of methane molecule ? 32. What is the bond angle in NH4+ ion ?

CHEMICAL BONDING AND MOLECULAR STRUCTURE 227 33. What are the overlapping orbitals of carbon and oxygen in CO ? 34. Mention the hybrid state of sulphur in H2S molecule. 35. What is the hybridisation of carbon in CO2 ? Short Answer type questions. (2 marks each) 1. Give reasons, why ethanol is completely soluble in water but benzene is not. 2. Explain why H2O is a polar molecule but CO2 is not. 3. Give one example each of molecules with sp2 and sp3 hybrid bonds. 4. Write the electronic structures of CO–3 – and NH+4 ions with lines and arrows. 5. Why covalent molecules have definite shape ? 6. Show the hybridisation process for the four valence electrons of carbon. 7. Why hydrofluoric acid is the weakest of all halogen acids ? 8. Describe two characteristics properties of electrovalent compounds. 9. What do you understand by hydrogen bond ? 10. What is co-ordinate bond ? Give one example. 11. Explain by an example what do you mean by polarity of a covalent bond. 12. The covalent compounds possess low melting and boiling point, why is it so ? 13. What are the conditions of hybridisation ? 14. Which of the following is an ionic compoud ? CO2, KCl, CH4, NH3. Short Answer type questions. (3 marks each) 1. Why three p-orbitals of each of two atoms connot form more than one sigma bond ? 2. Explain the formation of co-ordinate bond in ammonium ion. 3. When is HF a liquid where as other hydrides of halogens are gases ? 4. Why CO2 is non-polar but SO2 is polar ? 5. Differentiate between orbit and orbital. 6. Explain why oxygen molecule is paramagnetic 7. From the molecular orbital diagram of N2, find out its bond order. 8. HF is a liquid, whereas HCl is a gas. Explain 9. Draw the orbital diagram of CO2 and indicate the orbitals used by the elements. 10. What is the order of increasing bond angle (HxH) of the following ? What is the theory involved ? CH4, H2O, NH3

228 +2 CHEMISTRY (VOL. - I) 11. Why boron trifluride is called electron deficient compound ? Explain 12. Why is sigma bond stronger than pi bond 13. Write the electron dot structure for hydronium ion. 14. Write the molecular orbital configuration of O2. Long type questions. (7 marks each) 1. Explain the term hybridisatione What are the shapes of sp, sp2 and sp3 hybrid orbitals ? Give one example of a compounds of each. 2. State four physical properties that can be used to distinguish between covalent and ionic compounds. Discuss with examples. 3. Discuss with examples the directional properties of covalent bonds. 4. Explain with examples the term electrovalency, covalency and co-ordinate valency. 5. What is covalent bond ? Explain why the covalent bonds between oxygen and hydrogen atoms in water molecule are polar. 6. What is hydrogen bond ? How does it influence the properties of compounds ? Explain with two examples. 7. Write notes on Co-ordinate bond 8. What is hybridisation ? How does it explain the shapes of NH3 and H2O molecules ? 9. Explain the term hybridisation. Discuss the shape of CO2 and PH3 molecules. 10. Write a note on metallic bond 11. Write notes on (any two) (a) Electrovalent bond (b) Covalent bond (c) Hydrogen bond [2002 (A), CHSE] ADDITIONAL QUESTIONS 1. Define an electrovalent linkage. What are the necessary conditions for the formation of ionic bond ? Give two examples of such compounds. 2. What is lattice energy ? How the lattice energy of solid NaCl can be calculated. 3. Write short notes on : (a) hydrogen bond. (b) sigma and p-bond. 4. Explain why oxygen molecule is paramagnetic and nitrogen molecule is diamagnetic. Draw the molecular orbital energy level diagram of nitrogen molecule. (OJEE - 1991). 5. Using valence shell electron pair repulsion (VSEPR) theory predict the structure of the following compounds and write whether the bond angles are likely to be distorted from theoretical values. (a) CH4 (b) H2O (c) NH3.

CHEMICAL BONDING AND MOLECULAR STRUCTURE 229 Multiple Choice Questions 1. The polar molecule among the following is : (a) CCl4 (b) CO2 (c) CH2Cl2 (d) CH2 = CH2 2. How many s (sigma) and p (pi) bonds are there in tetracyanoethylene [C2(CN)4] molecule ? (a) Five s and nine p bonds (b) Nine s and nine p bonds (c) Six s and eight p bonds (d) Nine s and seven p bonds. 3. Which of the following gases does not have eight electrons in the outermost orbit ? (a) Kr (b) Ne (c) He (d) Ar 4. On hybridisation of one s and one p-orbital, we get, (a) Two mutually perpendicular orbitals (b) Two orbitals at 1800 (c) Four orbitals directed tetrahedrally (d) Three orbitals in a plane 5. Which of the following compounds has zero dipole moment. (a) CCl4, (b) CHCl3 (c) HF (d) NH3. 6. The species in which the central atom uses sp2 hybrid orbitals in its bonding is (a) PH3 (b) NH3 (c) SO2 (d) SiO2 7. The molecule that has linear structure is : (a) CO2 (b) NO2 (c) SO2 (d) SiO2. 8. Which is planar molecule ? (a) H2O (b) NH3 (c) BF3 (d) CH2Cl2. 9. The nature of hybridisation in carbon atoms in the ethylene (CH2 = CH2) molecule is, (a) sp, and sp (b) sp and sp2 (c) sp and sp3 (d) sp2 and sp2 10. Octet rule is not followed in (a) CO2 (b) H2O (c) O2 (d) CO

230 +2 CHEMISTRY (VOL. - I) 11. CO2 is isostructural with (a) HgCl2 (b) SnCl2 (c) C2H2 (d) NO2 12. The hybridisation of sulphur atom in SO2 is (a) sp, (b) sp3 (c) sp2 (d) dsp2 13. Which one of the following has a dipolemoment ? (a) CH4 (b) BCl3 (c) CO2 (d) H2O 14. A species which is formed by co-ordinate covalency is (a) NH+4 (b) NH3 (c) BF3 (d) PCl5 15. The hydrogen bond is strongest in (a) O – H ......... S (b) F – H ......... F (c) S – H ...........O (d) F – H ......... O 16. Out of the following which is least ionic (a) P - F (b) F - F (c) S – F (d) Cl - H 17. An example of a co-ordinate compound is (a) NH3 (b) HCl (c) CaCl2 (d) NaNO3 18. The angle between two odjancent sp2 hybridized orbitial is (a) 900 (b) 104.30 (c) 1200 (d) 109. 28 19. Covalent bond formation is favoured by (a) Small cation and large anion (b) Small cation and small anion. (c) Large anion and large cation (d) Large cation and small anion 20. Which of the following liquid is completely miscible with water ? (a) CHCl3 (b) CCl4 (c) CH3OH (d) CS2

CHEMICAL BONDING AND MOLECULAR STRUCTURE 231 21. Which of the following is most soluble in water. (a) Na2S (b) CuS (c) FeS (d) Ag2S 22. The outermost electronic configuration of the most electronegative element is (a) ns2np3 (b) ns2np4 (c) ns2np5 (d) ns2np6 23. The type of hybridisation in H2O is (a) sp3 (b) sp2 (c) sp (d) d2sp3 24. Hydridisation is the (a) removal of two electrons (b) adding electron to neutral atom (c) Separation of atomic orbitals (d) mixing of atomic orbitals 25. The molecule having one unpaired electron is (a) NO (b) CO (c) CN– (d) O2 26. The types of bond present in CuSO4. 5H2O are only (a) electrovalent and covalent (b) electrovalent and co-ordinate covalent (c) electrovalent, covalent and co-ordinate covalent (d) covalent and co-ordinate covalent. 27. If a molecule MX3 has zero dipole moment, the sigma bonding orbitals used by M (atomic number less than 21) are, (a) Pure p (b) sp2 hybrid (c) sp3 hybrid (d) sp-hybrid 28. Structure of ammonia is (a) trigonal (b) octahedral (c) trigonal bipyramidal (d) pyramidal 29. A molecule of acetylene (C2H2) consists of (a) Three s and two p bond (b) Two s and one p bond (c) s bonds only (d) p bonds only 30. The bond angle in CH4 is approximately (a) 200 (b) 1200 (c) 1090 (d) 1800

232 +2 CHEMISTRY (VOL. - I) 31. The bond angle in NH4+ ion is (OJEE 1990) (a) 1070 (b) 109.280 (c) 1040 (d) 900 32. Which of the following is paramagnetic ? (a) H2O (b) NO (c) CO2 (d) SO2 ANSWERS TO MULTIPLE QUESTIONS 1. c 7. a 13. d 19. a 25. c 31. b 2. b 8. c 14. a 20. c 26. c 32. b 3. c 9. d 15. b 21. a 27. b 4. b 10. b 16. b 22. c 28. d 5. a 11. a 17. d 23. a 29. a 6. c 12. c 18. c 24. d 30. c qqq

UNIT – V STATES OF MATTER : GASES AND LIQUIDS GENERAL INTRODUCTION : Three States of Matter : Matter is defined as something which has mass and occupies some space. Examples are iron, copper, chair, table, ice, water, air etc. Matter exists in three states: solid liquid and gas. The chemical properties of a substance do not change with the change of its physical state, but the rate of chemical reactions depends upon the physical state. This is because of the physical laws which govern the behaviour of matter in different states. Therefore, it is necessary to know about the nature of intermolecular forces, molecular interactions and effect of heat energy on the motion of the particles. Solid state : Matter in the solid state has definite shape and volume. The constituent atoms, ions or molecules are held together by strong force of attraction. They have high melting, boiling points and density. Examples are copper, iron, gold, sand, sugar etc. Liquid state : Matter in the liquid state has no definite shape but have definite volume. They have melting points, boiling points less than those of solids. Since the constituent atoms or molecules are held together by weak forces of interaction, they can flow from higher level to lower level. Their densities are also less than those of solids. Gaseous state : Gases have neither definite shape or definite volume. They assume the shape of the container and spread throughout. Weak forces of interaction exist between the constituent atoms or molecules. Due to this reason gases can be compressed. The volume of gases are greatly affected by change of temperature and pressure. The above three states of matter are interconvertible. For example, by increasing temperature and pressure solid ice can be converted into liquid water. In turn by further increasing temperature at constant pressure liquid water can be converted to water vapour. H2O (s) l H2O (l) l H2O (vap.) Similarly by going in the reverse way i.e. by decreasing temperature and increasing pressure water vapour can be converted to liquid water. By further lowering the temperature liquid water can be converted into solid ice. Intermolecular Interactions : In matter forces of attraction and repulsion exist among the constituent atoms and molecules. Attractive intermolecular forces are known as van der Waals forces. These forces

234 +2 CHEMISTRY (VOL. - I) vary considerably in magnitude and include dispersion forces or London forces, dipole-dipole forces and dipole-induced dipole forces. Hydrogen bonding is a particular type of strong dipole-dipole interaction. All these forces are responsible for the different physical states of matter. The constituent atoms or molecules of matter also experience repulsive forces among one another. When two particles are brought into close contact with each other the repulsion between the electron clouds and that between the nuclei produce repulsive force. The magnitude of the repulsion rises very rapidly as the distance of separation between the nuclei produce repulsive force. The magnitude of the repulsion rises very rapidly as the distance of separation between the particles decreases. Due to this reason liquids and solids do not chage their volume on compression, but gases do. In gases the distance of separation between the approaching particles is large so that they experience less repulsive force at a given temperature and pressure. Types of Bonds : As discussed above, in case of solids the constituent atoms or molecules experience different types of interactions as compared to liquids and gases. Metals except Mercury are solids and the bond type is metallic bond. Electron pool theory suggests that positively charged metallic kernels are fixed in the pool of valence electrons. In liquids the molecules are held together by van der Waals forces of attraction, dipole-dipole forces and hydrogen bond. Hydrogen bond is due to the coulombic force of attraction between the lone pair of electrons of the electronegative atom of one molecule and the hydrogen atom of same or other atom. The former is intramolecular hydrogen bond and latter is intermolecular hydrogen bond. In case of gases the molecules are mostly held together by van der Waals force of attraction. When molecular attractions are very weak, molecules do not come closer on application of pressure unless thermal energy is reduced by lowering the temperature. Gases do not liquify on compression only unless the thermal energy of molecules is not reduced. Melting point and Boiling point : The melting point and boiling point of solid, liquid and gas depend upon the type of interactions involved among their constituent atoms and molecules. Solids have high melting points due to the strong forces of interaction among the constituent particles. The amount of energy required to separate them is high. In case of liquids the forces of interaction between the constituent atoms or molecules are less than that of the solids, hence less amount of energy is required to overcome these forces. They boil at temperatures at which the vapour pressure of the liquid become equal to the atmospheric pressure.

GASEOUS STATE 235 CHAPTER - 6 THE GASEOUS STATE Physical classification indicates that matter exists in three different states namely solid, liquid and gas. This classification is more obvious since in our day-to-day life we have come across solid ice, liquid water and gaseous steam. Under suitable conditions of temperature and pressure almost all substances exist in three different states. The gaseous state is the simplest of all the three states of matter. 6.1 CHARACTERISTIC PROPERTIES OF GASES 1. No definite shape or volume : A gas has neither definite volume nor definite shape. It has no bounding surface due to random motion of the molecules. It acquires the shape and volume of the container in which it is filled. 2. Expansibility : A gas can occupy all the available space. It spreads uniformly throughout the container. It can be expanded to a desired volume by lowering the pressure or by raising the temperature. 3. Compressibility : A gas can be compressed to any desired volume by the application of pressure. 4. Diffusion : When two or more gases are introduced in to the same vessel, they intermix with each other and form a homogeneous mixture. This property is known as diffusion. The pleasant smell of kitchen and the pungent smell of laboratory are the common examples of diffusion. 5. Density : A gas has low density. This is because the intermolecular distance between the molecules of the gas is very large and the number of molecules per unit volume is lower than that in case of solids and liquids. For example, density of air is 0.00120 gm/cm3. 6. Pressure : A gas when enclosed in a vessel exerts pressure. The pressure is due to the collision of gas molecules on the walls of the container. 7. Homogeneity : When a gas is introduced into a container it fills the container uniformly. This property is known as homogeneity. 8. Liquefaction : A gas can be liquefied when subjected to high pressure below its critical temperature. 9. Colour : All gases are transparent. Most of the gases are colourless. Only a few are coloured e.g. chlorine is greenish yellow whereas bromine is reddish brown and iodine is violet. 6.2 STATE VARIABLES OF GASES There are four measurable properties of a gas. These are : (i) mass , (ii) volume , (iii) pressure and (iv) temperature. (i) Mass : Mass of the gas is expressed in grams and kilograms. Also the mass is related to the number of moles by the equation

236 +2 CHEMISTRY (VOL. - I) n = m where n = no. of moles M m = mass of the gas in grams M = molecular mass of the gas. (ii) Volume : The volume of the gas is equal to that of the container in which the gas is kept. The units in which volume is expressed are litre (l), cubic metre (m3). The smaller units are cubic decimetre (dm3) and cubic centimetre (cm3).The conversion factors are lm3 = 103dm3 = 106 cm3. However the units of litre(l) and millilitre(ml) are more commonly used. Litre is defined as volume of 1kg. of pure water at 1 atmospheric pressure and 3.980C. For all practical purposes cm3 and ml are taken to be equal. 1l = 1dm3 = 103 cm3 (iii) Pressure : A gas when enclosed in a vessel exerts pressure. A gas consists of small particles known as molecules. The molecules of a gas are always in a state of constant rapid zigzag motion in all possible directions with different velocities. During motion the molecules may collide among themselves or with the walls of the container. As a result of collisions on the walls, the molecules exert force on the walls. The total force exerted by the collisions of molecules on the walls of the container per unit area determines the gas pressure. Pressure = force area The common units in which pressure is expressed are (a) atmosphere (b) cm of Hg (c) mm of Hg (d) Torr (Torricellie) 1 atmosphere = 76 cm of Hg = 760 mm Hg. 1 mm Hg = 1 torr The SI Unit of pressure is pascal. It is defined as the pressure exerted when a force of 1N (Newton) acts on an 1m2 area. 1 atm = 101.325 KPa For all practical purposes 1 atm = 102 K Pa = 105 Pa. One standard atmospheric pressure is the pressure exerted on one cm2 area by 76 cm Hg at 00C and standard gravity (g = 980.665cm/sec2) (iv) Temperature : It indicates the intensity of heat or hotness with respect to a standard. It is an indicator of the average kinetic energy possessed by molecule. With increase in temperature the kinetic energy increases. The units to represent temperature are (a) celsius temperature (oC) (b) Kelvin temp (K) [ K = (oC) + 273 ] In the celsius or centigrade scale the freezing point of water (0oC) and boiling point of water (100oC) at one atmospheric pressure are taken as the reference points. 6.3 GAS LAWS BOYLE'S LAW (Relationship between Pressure and Volume) Robert Boyle in 1662 proposed a relationship between pressure of the gas and the volume occupied by it. He found that at constant temperature the volume of a given mass of gas goes on decreasing with increase in pressure. The behaviour was generalised and was named Boyle's law after his name. Statement : Temperature remaining constant, the volume of a given mass of gas is inversely proportional to its pressure.

GASEOUS STATE 237 Derivation : Let a given mass of gas occupy volume 'V' at pressure 'P' and constant temperature 'T'. According to Boyle's law, Vµ 1 [T is constant] P 1 Þ V= Const. x P Þ PV = Const. ('K' say) 'K' is the proportionality constant. It depends upon the mass of the gas and the temperature at which the measurements are made. If the volume of the same mass of gas be changed to V1, then pressure will change to P1 and under such condition P1V1 = Const. ( K1 say ) If the pressure is changed to P2 and volume to V2 then P2V2 = Const. ( K2 ) at the same temperature 'T' Since the constants K, K1 and K2 depend upon the mass of the gas and temperature and here both are kept constant, we can conclude and hence K = K1 = K2 PV = P1V1 = P2V2 = K (Constant) Another statement of Boyle's law \"The product of pressure and volume of a given mass of gas at constant temperature is constant.\" If pressure is doubled, volume will be reduced to half and vice versa. Similarly when pressure becomes three times its original value, volume is reduced to one third of its value. Validity of Boyle's law (i) It is possible to demonstrate the validity of the law by determining the values of volume of a given mass of gas at different pressures and at constant temperature. The product PV remains constant in all the cases as given in table 6.1 Table 6.1 Changes in volume of a fixed mass of gas at different pressure and constant temperature. Expt.No Press. 'P' in atm Vol. 'V' in litres PV (lit-atm) 1. 0.20 112.0 22.4 2. 0.25 89.2 22.3 3. 0.35 64.2 22.47 4. 0.40 56.25 22.50 5. 0.60 37.40 22.44 6. 0.80 28.1 22.48 7. 1.00 22.4 22.40 (ii) The validity can be tested by plotting a graph between volume in cm3 and pressure in atmosphere of a given mass of gas at constant temperature.

(in cm3)238 +2 CHEMISTRY (VOL. - I) V1 (P1, V1) V2 (P2, V2) P1 P2 ¾(P¾irnesa¾stu¾mre.)® Fig. 6.1 Plot of 'V' in cm3 against 'P' in atm. (iii) Validity may also be tested by plotting a graph between PV in lit-atm against 'P' in atmosphere. We get a horizontal line parallel to pressure axis. PV T3 Lit-atm T2 T1 T3 > T2 > T1 P (atm) Fig. 6.2 Plot of 'PV' versus 'P' (iv) Boyle's law can be illustrated by plotting 'P' in atmosphere against 1 in lit–1. A straight line passing through the origin is formed. V P .. . in atm. 1 Lit -1 V Fig. 6.3 Plot of 'P' versus 1 V

GASEOUS STATE 239 Utility of the law : It helps in determining (a) The volume of the gas at a given pressure if its volume at some other pressure and at constant temperature is known. (b) The value of pressure corresponding to any volume if the pressure at some other volume and at constant temperature is known. NUMERICAL PROBLEMS Example 1. 400 ml of air at 700 mm pressure were compressed to 200 ml. What will be the new pressure, if the temperature remains constant ? Solution :- Given conditions Final conditions applPVy1i1n==g 700 mm PV22 = ? ml. 400 ml = 200 Boyle's law By P701V01x=40P02V=2 P2 x 200 Þ P2 = 700 x 400 = 1400 mm or 140 cm. 200 Example 2. What is the volume of a sample of oxygen at a pressure of 3 atmosphere and 270C, if its volume is 4.5 litres at 1 atmosphere and same temperature. Solution :- Given conditions Final conditions P1 = 1 atm P2 = 3 atm ApplyinVg1B=o4y.l5e'slitlraews V2 = ? P1V1 = P2V2 1 x 4.5 = 3 x V2 Þ V2 = 1 x 4.5 = 4.5 = 1.5 litres. 3 3 Example 3. A certain volume of the gas was found to be at a pressure of 1000 mm of mercury. When the pressure was decreased by 500 mm, the gas occupied a volume of 2000 ccmon3s. tCanatlcteumlapteertahteurien.itial volume occupied by the gas if the change was done at a Solution :- Initial conditions Final conditions P1 = 1000 mm P2 = 1000 – 500 = 500 mm V1 = ?? V2 = 2000 cm3 Applying Boyle's law P1V1 = P2V2 1000 x V1 = 500 x 2000 Þ V1 = 500 x 2000 = 1000 cm3 . 1000

240 +2 CHEMISTRY (VOL. - I) Example 4. For use in a certain industrial process the volume of a particular gas had to be reduced by 50% of its present volume. The original pressure of the gas was measured and found to be 5 atmospheres. Find the final pressure of the gas if the temperature of the gas was constant throughout the process. Solution :- Initial conditions Final conditions Vol = V1 Volume V2 = V1 – 0.5V1 = 0.5 V1 Pressure = P1 = 5 atm. Pressure P2 = ?? Applying Boyle's law P1V1 = P2V2 5 x V1 = P2 x 0.5 V1 Þ P2 = 5xV1 = 5 = 10 atm. 0.5V1 0.5 CHARLE'S LAW (Relationship between Volume & Temperature) Jacques Charles in 1787 studied the relationship between temperature of a gas and the volume occupied by it at constant pressure. He observed that at constant pressure the volume o2n173coonffiitrsmveadlubeyatJo0s0eCphfoGr eavye-Lryudsseagcr.ee of a given mass of gas expands or contracts by celsius rise or fall in the temperature. This was later Statement :- \"Pressure remaining constant the volume of a given mass of gas increases or 1 decreases by 273 of its value at 0 0C for each 10C rise or fall in temperature.\" Derivation : Let 'V'O be the volume of a given mass of gas at 0 0C and at constant pressure. ( )Volume of the gas at 10C = VO + Vo x1 = Vo + Vo = Vo 1 + 1 273 273 273 ( )Volume 2Vo 2 of the gas at 20C = VO + 273 = VO 1 + 273 ( )Volume of the gas at t 0C = VO + tVo = VO 1 + t 273 273 When t = –2730C ( )V–273 = VO1- 273 = VO x 0 = 0 (Zero) 273 Thus, gases have no volume at –2730C. This temperature is known as Absolute zero of temperature. Absolute zero : The temperature at which a gas ceases to exist or at which volume of the gas becomes zero is called Absolute Zero. Thus, –2730C is the absolute zero of temperature. This temperature has no physical significance. Reduction of volume of a gas to zero cc is only a theoretical concept. Practically all gases get liquefied or solidified before reaching this temperature.

GASEOUS STATE 241 Absolute scale of Temperature : The scale of temperature with –2730C as its Zero (0K) is called Absolute scale of temperature. Thus, –2730C = 0K or (–273 + 273)0C = (0 + 273) K Þ 00C = 273 K Þ t0C = (t + 273) K. The effect of temperature on volume of gas can be graphically represented as -2 2 Volume (in lit) 1 –2730C –1000C 00C 1000C 2730C 0K 173K 273K 373K 546K –2730c –1000c 00c 1000c 2730c 0K 173K Te2m73pK®373K 546K Temp 2 Fig. 6.4 Plot of V against T Again let us assume that V1 ml & V2 ml be the volume of a given mass of gas at t10C and t20C respectively and at constant pressure. V0 x t1 t1 273 + t1 V0 T1 273 273 273 273 ( ) ( )V1 V0 1+ = V0 + = = V0 = V0 t2 t2 273 + t2 V0T2 273 273 273 273 ( ) ( )V2 = V0 + 1+ = V0 = V0 = where T1 & T2 are the temperatures in the absolute scale. Dividing V1 by V2 we have, V1 = V0 T1 x 273 = T1 , V2 273 V0 T2 T2 Þ V1 = V2 T1 T2 Þ V = constant, at constant pressure. T Thus, V µ T , at constant pressure.

242 +2 CHEMISTRY (VOL. - I) Another definition of CHARLE'S LAW : \"Pressure remaining constant, the volume of a given mass of gas is directly proportional to absolute temperature.\" Validity of the law : (i) It is possible to demonstrate the validity of the law by determining the volume of a given mass of gas at different temperatures and at constant pressure. The constancy of value V in each case proves the validity as is evident from the following table 6.2. T Table-6.2 Changes in volume of a given mass of gas with temperature at constant pressure. No. of Temp in Temp in Volume V Kelvin (K) T Experiments deg. centigrade (0C) in cm3 1. – 50 223 223 1 2. – 0 273 273 1 3. 50 323 323 1 4. 100 373 373 1 5. 150 423 423 1 (ii) The validity of the law can also be tested by plotting a graph (Fig 6.5) between volume and temperature of a given mass of gas at constant pressure. A straight line is formed which shows that V T is constant. At different pressures the straight lines formed in the graph meet at the same point '0' upon extension. This point refers to zero volume and absolute zero of temperature. Volume P1 (in litre) P2 P3 O0 (–2730C) Temp. P1 < P2 < P3 Fig 6.5 Volume - Temp. relationship Utility of the law : The law can be used in determining the volume of a given mass of gas at a certain temperature provided the volume at some other temperature is known keeping the pressure constant. GAY - LUSSAC'S LAW Statement : At constant volume, the pressure of the given mass of gas is directly proportional to its absolute temperature. i.e. P µ T at constant volume.

GASEOUS STATE 243 If P1 & P2 be the pressure of the gas at temperature T1 & T2 respectively, Then, P1 = T1 at constant volume. P2 T2 NUMERICAL PROBLEMS Example 1. A certain volume of gas is kept at 670C. When the temperature is decreased by 170C the gas occupies a volume of 800 ml. What was the initial volume of the gas ? Assume that the pressure of the gas remains constant. Solution : Initial condition Final condition V1 = ? V2 = 800 ml T1 = 273 + 67 T2 = 273 + (67 – 17) = 340 K. = 273 + 50 = 323 K. According to Charle's Law V1 = V2 T1 T2 Þ V1 = 800 340 323 Þ V1 = 800 x 340 = 842.10 ml 323 Example 2. A sample of hydrogen gas is found to occupy 900 cm3 at 370c . Calculate the temperature at which it will occupy 500 cm3. Solution : Given V1 = 900 cm3 T1 = 37 0C = (273 + 37) = 310 K V2 = 500 cm3 T2 = ?? V1 = V2 Applying Charle's Law T1 T2 Þ 900 = 500 310 T2 Þ T2 = 500 x 310 = 172 K (or – 1010 C) 900 Example 3. 500 ml of Oxygen is collected at 270C. If the volume is reduced to 1 th its 4 original volume, find the temperature to which the gas has to be cooled ? Solution : Given V1 = 500 ml T1 = 273 + 27 = 300 K V2 = 1 V1 = 1 x 500 = 125 ml 4 4 T2 = ??

244 +2 CHEMISTRY (VOL. - I) Applying Charle's Law V1 = V2 T1 T2 Þ 500 = 125 300 T2 Þ T2 = 300 x 125 = 75 K (or – 1980 C) 500 Example 4. 3000cc of a gas is heated from 270C to 1270C. What will be the new volume of the gas at constant pressure ? Solution : Given V1 = 3000 cc V2 = ? ? T1 = 273 + 27 = 300 K T2 = 127 + 273 = 400 K Applying Charle's Law, V1 = V2 T1 T2 Þ 3000 = V2 300 400 Þ V2 = 3000 x 400 = 4000 cc. 300 Example 5. A given mass of oxygen gas occupies 300 cm3 at 270C. What is the volume at o0C, the pressure remains constant ? Solution : Given V1 = 300 cm3 T1 = 270C = 273 + 27 = 300 K V2 = ?? T2 = 00C = 273 K Applying Charle's Law, V1 = V2 T1 T2 Þ 300 = V2 300 273 Þ V2 = 300 x 273 = 273 cm3 300 COMBINED GAS EQUATION The gas equation deals with the simultaneous effect of change of temperature and pressure on the volume of a given mass of gas. It is derived on the combination of both Boyle's law and Charle's law. Derivation : Let us suppose that a given mass of gas occupies a volume 'V1' at a pressure P1 and temperature T1 (i) Apply Boyle's law : Keep Temp 'T1' to be constant. When pressure changes from P1 to P2, let the volume change from V1 to u .

GASEOUS STATE 245 So, V1 µ 1 and uµ 1 P1 P2 Thus, P1V1 = P2 y Þ u= P1 V 1 ............................... (1) P2 (ii) Apply Charle's law : Keep Pressure 'P2' constant. When temperature changes from T1 to T2 , let the volume change from y to V2. So y µ T1 and V2 µ T2 Thus u = V2 T1 T2 Þ u= V 2 T1 ............................... (2) T2 Comparing (1) & (2) we have P1 V1 = V 2T1 P2 T2 Þ P1 V1 = P2 V 2 = K [constant] T1 T2 Þ PV = constant (K) ® Gas Equation. T Utility of the gas equation : The volume of a given mass of gas at a certain temperature and pressure can be determined provided its value at some other temperature and pressure is known. Standard Temperature and Pressure (STP or NTP) : The volume of a given mass of gas varies with temperature and pressure. Hence a standard reference condition is chosen to compare the measurable properties P, V, T of gases. Standard Temperatures = 00C = 273 K. Standard Pressure = 760 mm Hg. = 76 cm Hg. = 1 atmos. pressure. NUMERICAL PROBLEMS Example 1. A flask contains 300 ml of gas at 170C and 50 cm of Hg. Find the final pressure when it is transferred to another flask of 100 ml. capacity at a temperature of 370C. Solution : Final condition Initial condition P1 = 50 cm Hg P2 = ?

246 +2 CHEMISTRY (VOL. - I) T1 = 273 + 17 = 290 K T2 = 273 + 37 = 310 K V1 = 300 ml V2 = 100 ml Using the gas equation, P1 V1 = P2 V 2 T1 T2 Substituting the values, 50 x 300 = P2 x 100 290 310 Þ P2 = 50 x 300 x 310 = 160.34 cm. 290 x 100 Example 2. A gas occupies 500 ml at STP. Find the volume of the gas when its pressure is 200 mm of Hg and its temp 200C. Solution : Initial condition (STP) Final Condition P1 = 760 mm Hg P2 = 200 mm Hg V1 = 500 ml T1 = 273K V2 = ? Applying gas equation, T2 = 273 + 20 = 293 K P1 V1 = P2 V 2 T1 T2 Substituting the values, 760 x 500 = 200 x V2 273 293 Þ V2 = 760 x 500 x 293 = 2039.19 ml. 273 x 200 Example 3. At a given temperature the pressure of the gas reduces to 60 % of its initial value and volume increases by 45% of its original value. Find this temperature if the initial temperature was –200C. Solution : Initial condition Final condition P1 = P1 P2 = 60 P1 = 3P1 100 5 V1 = V1 ( )V2 = V1 + 45V1 = 145 V1 100 100 T1 = –20 + 273 = 253 K T2 = ? Applying gas equation, P1 V1 = P2 V 2 T1 T2 Þ P1 V1 = 3P1x 145V1 253 5 x 100 x T1

GASEOUS STATE 247 Þ T2 = 3 x 145 x 253 = 220.11 K. 5 x 100 Example 4. A spherical balloon of 21 cm diameter is to be filled up with hydrogen at NTP from a cylinder containing the gas at 20 atm and 270C. If the cylinder can hold 2.82 litres of water, calculate the number of balloons that can be filled up. Solution : Initial condition Final condition V1 = 2.82 l V2 = ? T1 = 270C = 300 K T2 = 273 K P1 = 20 atm P2 = 1 atm Applying gas equation P1 V1 = P2 V 2 T1 T2 Þ 20 x 2.82 = 1 x V2 300 273 Þ V2 = 20x 2.82x 273 = 51.324 l = 51324 ml. 300 The volume of spherical balloon ( )=4 3 4 x x 21 3 ml. 3 pr = 3 3.14 2 = 4847 The capacity of the cylinder = 2.82 l = 2820 ml. The amount of gas to be filled into balloons = 51324 – 2820 = 48504 ml. L No of balloons = Vol. of gas transferred Vol. of balloon = 48504 @ 10 4847 Example 5. If the density of a gas at 270C and 1 atmospheric pressure is 1.5 gms / litre, find the density at 1270C and 4 atm. pressure Solution : According to combined gas equation, P1 V1 = P2 V 2 T1 T2 V µ 1 (at constant temp) d P1 P2 The equation may be expressed as d 1T1 = d 2T2 P1 = 1 atm P2 = 4 atm d1 = 1.5 gms/lit d2 = ?? T1 = 27 + 273 = 300 K T2 = 127 + 273 = 400 K

248 +2 CHEMISTRY (VOL. - I) From the above equation d2 = P2d1T1 = 4 x 1.5 x 300 = 4.5 gms/ lit. P1T2 1 x 400 Example 6. The pressure of a given mass of gas becomes half the original pressure and simultaneously the volume becomes four times the original volume. What should have happened to the temperature ? Solution : Let the initial pressure and volume be P1 & V1 respectively. Final pressure P2 = P1 2 Final volume V2 = 4 V1 We know P1 V1 = P2 V 2 T1 T2 Þ P1 V1 = P1 x4V1 = 2 P1 V1 T1 2 T2 T2 Þ T2 = 2 T1 i.e. the temperature is doubled. IDEAL GAS EQUATION Ideal gas equation can be derived on the basis of Boyle's law, Charle's law and Avogadro's law. The equation gives the general relationship between pressure, volume, temperature and no. of moles of the gas. According to Boyle's Law : V µ 1 when temp 'T' and no. of moles 'n' are kept constant. P According to Charle's Law : V µ T when pressure 'P' and no. of moles 'n' are kept constant. According to Avogadro's Law : V µ n when press. 'P' and Temp. 'T' are kept constant. [Equal volumes of all gases under similar conditions of temperature and pressure contain equal no. of molecules] Combining all the three laws, Vµnx 1 xT P nT or, V = const x P or, PV = n RT ® Ideal gas equation. where R is a constant known as Universal gas constant.

GASEOUS STATE 249 If W ® Mass of the gas M ® Molecular mass n= W M Then, PV = n RT = W RT M or M= WRT = d RT (where d = density of gas = W ) PV P V SIGNIFICANCE OF R The ideal gas equation is PV = n RT So, R = PxV nxT = Press. xVolume No. of moles x Abs. temp. = (Force/Area) xVolume No. of moles x Abs. temp. Force x (Length)3 (Length)2 = No. of moles x Abs. temp. = (Force x Length) No. of moles x Abs. temp. = Work No. of moles x Abs. temp. Thus, 'R' may be represented in the units of work per mole per degree. NUMERICAL VALUE OF 'R' (i) In litre - atmosphere / cm3 – atm/lit – mm When Pressure is in atmosphere and volume is in litres, 'R' is represented in litre atmosphere For 1 mole of gas at NTP, we have P = 1 atm, T = 273 K V = 22.4 litres \\ R= PxV = 1 x 22.4 = 0.0821 lit. atm deg–1 mole–1. nxT 1 x 273 R = 82.1 cm3. atm K–1 mole–1 (When 'P' is in atm. and 'V' is in cm3) = 62.39 litre. mm K–1 mole–1 (When 'P' is in mm and V is in litre.)

250 +2 CHEMISTRY (VOL. - I) (ii) In ergs / calorie : When Pressure is expressed in dynes per square centimetre and volume in cubic centimetre, 'R' is represented in terms of ergs/Joules /Calories. For 1 mole of gas at STP P = 1 atm = 76 x 13.6 x 981 dynes/cm2 V = 22.4 litres = 22,400 cm3 T = 273 K \\ R= PxV = 76 x 13.6 x 981 x 22,400 nxT 1 x 273 = 8.31 x 107 ergs K–1 mole–1 Since 1 Joule = 107 ergs R = 8.31 Joules K–1 mol –-1 (SI Unit) Again we know, 1 calorie = 4.182 x 107 ergs So, R= 8.314 x 107 = 1.99 Cal K–1 mol–1 4.182 x 107 (iii) (SI Unit of Work) R in Joule Pressure is expressed in pascals or Nm–2 and volume in m3 For 1 mole of gas at STP P = 1 atm = 1.013 x 105 pa = 1.013 x 105 Nm–2 V = 22.4 litre = 22.4 x 10–3 m3 T = 273 K \\ R= PV = 1.013 x 105 x Nm-2 x 22.4 x 10-3 (m3 ) nT 1 x 273 = 8.31 Nm K–1 mol–1 = 8.31 pa m3 K–1 mole–1 = 8.31 K Pa Dm3 K–1 mole–1 = 8.31 Joule K–1 mole–1 ( Q 1N m = 1J ) NUMERICAL PROBLEMS Example 1. Calculate the volume occupied by 7 gms of N2 at 270C and 750 mm Hg. Solution : Given W = 7 gms, T = 300 K, P = 750 atm , M = 28 760

GASEOUS STATE 251 PV = nRT = W RT = 7 x 0.82 x 300 M 28 Þ 750 x V = 7 x 0.082 x 300 760 28 \\ V= 760 x 7 x 0.082 x 300 = 6.239 ltres. 750 x 28 Example 2. Calculate the the weight of methane in a 9 litre cylinder at 16 atm and 270C (R = 0.08 l atm.K–1) Solution : P = 16 atm Given V = 9 litres T = 273 + 27 = 300 K M = 12 + 4 = 16 We know PV = nRT = W RT M Substituting the values, 16 x 9 = W x 0.08 x 300 16 Þ W= 16 x 9 x 16 = 96 gms. 0.08 x 300 Example 3. Calculate the temperature at which 28 gms of N2 occupies a volume of 10 litres at 2.46 atm. Solution : W = 28 gms Given V = 10 litres P = 2.46 atm M = 28 The ideal gas equation is PV = nRT = W RT M Substituting the values, 2.46 x 10 = 28 x 0.082 x T 28 Þ T= 2.46 x 10 = 299.64 K. 0.082 Example 4. O2 is present in one litre flask at a pressure of 7.6 x 10–10 mm of Hg. Calculate the no. of O2 molecules at O0C Solution : Given P = 7.6 x 10–10mm = 7.6 x 10-10 atm . 760 V = 1 litre, T = 273 K

252 +2 CHEMISTRY (VOL. - I) The ideal gas equation is PV = nRT Substituting the values, 7.6 x 10-10 x1 = n x 0.0821 x 273 760 Þ n= 7.6 x 10-10 = 4.46 x 10 –14 moles. 760 x 0.0821 x 273 1 mole of O2 contains Avogadro no. (6.02 x 1023) of molecules \\ 4.46 x 10–14 moles contain 4.46 x 10–14 x 6.02 x 1023 = 2.68 x 1010 no. of molecules. Example 5. Calculate the density of CO2 at 1000C and 800 mm Hg pressure. Solution : P= 800 atm . Given 760 T = 273 + 100 = 373 K The ideal gas equation is PV = nRT or, PV = W RT M or, P = W RT MxV or, P = d RT ( l W = mass of CO2 = density ) M V volume Substituting the values 800 = d x 0.082 x 373 ( l mol mass of CO2 = 12 + 32 = 44) 760 44 Þ d= 800 x 44 = 1.5124 gl –1 760 x 0.082 x 373 DALTON'S LAW OF PARTIAL PRESSURE The behaviour of a mixture of gases that do not react chemically with each other is frequently of interest. A law dealing with such behaviour was stated by John Dalton in 1801. Statement : The total pressure of the mixture of gases that do not react chemically with each other is equal to the sum of partial pressures of all the gases present. If the total pressure be PTotal and partial pressures of gases present be P, P etc. A B PTotal = P + P + ......... A B

GASEOUS STATE 253 Partial Pressure : It is defined as the pressure which each gas would exert if it is separately confined in the whole volume occupied by the mixture of gases Partial Pressure = mole fraction x Total Pressure. Mathematical deduction : Let there be two chemically non- reacting gases A and B. Let us suppose that nA moles of gas A are mixed with nB moles of gas B. Total no. of moles in the reaction mixture = ( n A + n B ) The ratio of no. of moles of A to the total no. of moles present is known as the molefraction of 'A'. It is represented by XA . Thus, XA = (nAn+AnB) Similarly, mole fraction of 'B' , XB = (nAn+BnB) Let PTotal be the total pressure exerted by the mixture and PA, B be partial pressures of A and B respectively. Partial pressure of gas 'A' will be its mole fraction times the total pressure . ie. PA = XA PTotal = (nAn+AnB) x PTotal Similarly, partial pressure of gas 'B' will be its molefraction times the total pressure ie. PB = XB PTotal = (nAn+AnB) x PTotal Thus P +P = (nAn+AnB) PTotal + (nAn+BnB) PTotal AB = PTotal [ (nAn+AnB) + (nAn+BnB) ] = PTotal [ nnAA + nnBB ] + = PTotal i.e. PTotal = P +P AB So, if a mixture contains 1 mole of A and 4 moles of B, then total no. of moles is (1 + 4) = 5 . The mole fraction of A = 1 and that of B = 4 . The Partial pressure of 'A' 5 5 1 4 is 5 th of total Pressure whereas that of B is 5 th of total pressure. APPLICATIONS OF DALTON'S LAW 1. Determination of Total Pressure : The law is used to determine the total pressure exerted by a mixture of chemically non- reacting gases. The total pressure is the sum of partial pressures of individual gases.

254 +2 CHEMISTRY (VOL. - I) 2. Determination of Pressure of dry gas : When a gas is collected over water it is associated with some water vapour since at all temperatures there is a conversion of water into its vapour. So the pressure, we measure of a moist gas is the sum of the partial pressure of dry gas and the pressure of water vapour at that temperature. Thus, P = P + Pmoist gas dry gas water vapour The partial pressure of water vapour is known as \"Aqueous Tension\". So, Pmoist gas = Pdry gas + Aqueous Tension or, Pdry gas = Pmoist gas – Aqueous Tension So, by knowing the aqueous tension at that particular temperature, the pressure of dry gas can be found out. NUMERICAL PROBLEMS Example 1. O2 is collected over water at 280C. The pressure inside the gas is 745 mm of Hg. What is the pressure due to O2 alone if vapour pressure of H2O is 15 mm at 280C ? Solution : Pmoist O2 = 745mm, PH2O = 15mm or, Pmoist O2 = PO+2 PH2O or, 745 = PO2 + 15 or, PO2 = 745 – 15 = 730 mm. Example 2. A container of 2.461 litre at 270C has a mixture of 0.3 moles of N2, 0.5 moles of He and 6.2 moles of O2. What will be the partial pressure of gases ? Solution : Given V = 2.461 litres, T = 273 + 27 = 300 K No. of moles of N2 = 0.3 No. of moles of He = 0.5 No. of moles of O2 = 6.2 Total No. of moles = 0.3 + 0.5 + 6.2 = 7 Apply the equation PV = nRT P= nRT = 7 x 0.082 x 300 = 70.06 atm. V 2.461 We know that partial press. = molefraction x Total Press. molefraction of N2 = 0.3 7 0.5 molefraction of He = 7 molefraction of O2 = 6.2 7 0.3 partial press of N2 = 7 x 70.06 = 3.003 atm. partial press. of He = 0.5 x 70.06 = 5.004 atm. 7

GASEOUS STATE 255 partial press. of O2 = 6.2 x 70.06 = 62.053 atm. 7 Example 3. 300 ml of nitrogen at a pressure of 740 mm and 350 ml of oxygen at a pressure of 600 mm are put together in a one litre flask. If the temp. is kept constant what will be the total pressure ? Solution : (i) Calculate Partial pressure of N2 as follows : V1 = Vol. of Nitrogen = 300 ml P1 = Press. of Nitrogen = 740 mm V2 = 1 litre = 1000 ml P2 = ?? Apply P1V1 = P2V2 ( Q temp. is kept constant ) 740 x 300 = P2 x 1000 \\ P2 = 740 x 300 = 222 mm. 1000 \\ Partial pressure of N2 i.e. PN2 = 222 mm. (ii) Calculate Partial pressure of O2 : V1 = 350 ml P1 = 600 ml V1 = 1000 ml P2 = ?? Applying P1V1 = P2V2 600 x 350 = P2 x 1000 \\ P2 = 600 x 350 = 210 mm. 1000 i.e. Partial pressure of O2 , Po2 = 210 mm. \\ Total pressure = PN2 + Po2 = 222 mm. + 210 mm = 432 mm. Example 4. Two gases A and B having molecular mass 60 and 45 respectively are enclosed in a vessel. The weight of A is 0.50gms and that of B is 0.2gms. The total pressure of mixture is 750 mm. Calculate the partial pressure of each gas. Solution : wt of A = 0.50 gms Given molecular mass of A = 60 \\ No. of moles of A = 0.5 60

256 +2 CHEMISTRY (VOL. - I) wt. of B = 0.2 gms mol. mass of B = 45 0.2 45 \\ No. of moles of B = Total Pressure = 750 mm. Total no. of moles = ( 0.5 + 0.2 ) 60 45 0.5/ 60 ( )Partial pressure of A = 0.5 + 0.2 60 45 Applying the formula Partial pressure = Total pressure x molefraction 0.5/ 60 ( )Partial pressure of A = 750 x = 489.23 mm 0.5 0.2 60 + 45 Partial pressure of B = 750 – 489.23 = 260.77 mm. Example 5. A mixture of gases in a gas cylinder at 760 mm pressure contains 70% nitrogen, 20% oxygen and 10% CO2 by volume. What is the partial pressure of each gas in mm. ? Solution : Combining Dalton's law and Avogadro's law we may conclude that % of each gas = Partial Pressure x 100 Total Pressure i.e. volume fraction will correspond to mole fraction. So, Partial pressure = Total Press x % of each gas 100 760 x 70 Partial pressure of N2 = 100 = 532 mm Partial pressure of O2 = 760 x 20 = 152 mm 100 760 x 10 Partial pressure of CO2 = 100 = 76 mm DIFFUSION OF GASES Gases possess the property of diffusion ie. they can intermix with each other irrespective of their densities. This is due to the fact that there are intermolecular spaces or voids between the molecules and the molecules of a gas are always in a state of constant rapid motion. The molecules of a gas can be easily accomodated within the spaces between the molecules of other gas. For example, (i) In a closed room if a bottle of conc. NH4OH is opened at one corner, the smell of NH3 can be perceived at the other end. (ii) If a jar full of air is inverted over a jar of bromine gas, it is observed that bromine gas rises up, mixes with air forming a homogeneous mixture. This is revealed from the colour (i.e. reddish brown). Bromine is a heavier gas. Yet it rises up against the forces of gravity. This is possible only due to random perpetual motion of gas molecules. This property, by virtue of which gases can intermix with each other forming a homogeneous mixture irrespective of gravitational force is known as diffusion.

GASEOUS STATE 257 GRAHAM'S LAW OF DIFFUSION In 1869, Thomas Graham studied the effect of rates of diffusion of gases on their densities and expressed his experimental findings in the form of a law known as Graham's Law. Statement : Under similar conditions of temperature and pressure, the rates of diffusion of gases are inversely proportional to the square root of their densities. Mathematically it can be represented as follows : Let us take two gases with rates of diffusion r1 and r2 and densities d1 and d2. According to Graham's law r1 µ 1 , r2 µ 1 d1 d2 So, r1 = d 2 (at same temp. and pressure) r2 d1 Again we know that molecular mass is twice the vapour density. Therefore, the above expression may be written as : r1 = 2d 2 = M2 where M1 and M2 are the molecular masses r2 2d 1 M1 of gases having densities d1 and d2 respectively. Another statement of the Law : From the above expression the law may also be stated as : \" The rates of diffusion of gases are inversely proportional to the square root of their molecular masses.\" Again the rate of diffusion is the volume of gas that diffuses in unit time. ie. Rate of diffusion 'r' = Vol. of gas diffused 'v' Time taken for diffusion 't' If V cc of two gases take t1 and t2 seconds respectively to diffuse then r1 = V and r2 = V t1 t2 So, r1 = V/ t1 = t2 r2 V/t2 t1 i.e. Rate of diffusion of gas is inversely related to time of diffusion. If different volumes of two gases diffuse in the same time, the relationship between volume and density of the gas can be derived. Let V1 & V2 be the volumes of two different gases that diffuse in the same time i.e.'t' secs. Then r1 = Rate of diffusion of 1st gas = V1 t r2 = Rate of diffusion of 2nd gas = V2 t

258 +2 CHEMISTRY (VOL. - I) and r1 = V1 / t = V1 x t = V1 , Thus, r1 = V1 = d2 r2 V2 /t t V2 V2 r2 V2 d1 Thus the volumes of gases that diffuse in the same period of time are inversely proportional to the square root of their densities under similar conditions of temperature and pressure. From the above discussion we may write r1 = V1 = t2 = d2 = M2 r2 V2 t1 d1 M1 So, we may conclude that a gas with lower molecular mass will diffuse faster than that with higher molecular mass. Applications of Gaseous Diffusion : 1. When there are two gases having different rates of diffusion it can be applied to separate these two gases. 2. A gas detector, which works on the principle of diffusion, can be used in mines for detecting the presence of poisonous gases. 3. Law of diffusion helps us in calculating the molecular mass of gases and their relative densities. 4. The effect of poisonous or foul gases can be diluted by diffusion into the air. 5. The homogeneity of atmospheric air is maintained due to diffusion of its constituent gases. Effusion : When a gas under pressure is allowed to pass through a small orifice into a region of low pressure, the process is called Effusion. So effusion is a special case of diffusion. Graham's law also holds good in case of Effusion. The law may be stated as \"The relative rates of effusion of different gases are in the inverse ratio of square root of their densities.\" NUMERICAL PROBLEMS Example 1. 150 ml of a certain gas diffuse in the same time as 125 ml of Chlorine under the same condition. Calculate the molecular mass of the gas. Solution : Given Volume of gas (V1) = 150 ml Volume of Chlorine (V2) = 125 ml Molecular mass of chlorine,M2 = 71 Molecular mass of the gas,M1 = ?? By Graham's law, V1 = d2 = M2 V2 d1 M1

GASEOUS STATE 259 Substituting the values we have 150 = 71 125 M1 Þ 71 = 150 x 150 M1 125 x 125 Þ M1 = 71 x 125 x 125 = 49.30 150 x 150 Example 2. 130 volumes of Hydrogen take 25 minutes to diffuse out of a vessel. How long will 100 volumes of Oxygen take to diffuse out from the same vessel under the same condition ? Solution : r1 = Rate of diffusion of H2 = 130 25 100 r2 = Rate of diffusion of O2 = t Mol. wt of H2 = 2 Mol. wt of O2 = 32 By Graham's law, r1 = M2 r2 M1 or, 130 x t = 32 = 16 = 4 25 100 2 or, 13t = 4 250 or, t= 250 x 4 = 76.92 minutes 13 Example 3. Calculate the molecular mass of the gas X which diffuses twice as fast as another gas Y, which in turn diffuses thrice as fast as another gas Z. Given, molecular mass of the gas Z is 144. Solution : Let the rate of diffusion of gas Z(rz) be 1. So rate of diffusion of gas Y (ry) be 3 and that of gas x (rx) be 6 According to Graham's law ry = Mz = 144 rz My My or, 3 = 144 1 My Þ 3 x 3 = 144 1 x 1 My Þ My = 144 = 16 3x3

260 +2 CHEMISTRY (VOL. - I) Again ry = Mx rx My or, or, 3 = Mx or, 6 My Mx = 3x 3 16 6x 6 MX = 16 x 3 x 3 =4 6x6 Example 4. The rates of diffusion of an unknown gas and chlorine are in the ratio of 6:5. Assuming the density of chlorine to be 36, calculate the molecular wt of unknown gas. Solution : According to Graham's law r1 = d2 when r1 2 rate of diffusion of unknown gas, r2 d1 r2 ® rate of diffusion of chlorine, d1 ® density of unknown gas, d2 ® density of chlorine. \\ 6 = 36 5 d1 Þ 36 = 36 25 d1 Þ d1 = 25 \\ Molecular mass = 2 x V.D = 2 x 25 = 50 Example 5. The relative ratio of diffusion of two gases A & B are found to be 0.3 and 0.2 respectively. If the density of 'A' is 4, find the relative density of B. Solution : According to Graham's law r1 = dB r2 dA or 0.3 = dB 0.2 4 or dB = 9 4 4 or dB = 9

GASEOUS STATE 261 Example 6. 180 ml of a hydrocarbon diffuses through a porous membrane in 15 minutes, while 120 ml of SO2 under identical condition diffuse in 20 minutes. What is the molecular wt of hydrocarbon ? Solution : Volume of hydrocarbon diffuse = 180 ml Time taken for diffusion = 15 minutes \\ Rate of diffusion (r1) = 180 15 Similarly rate of diffusion of SO2 (r2) = 120 20 Molecular mass of hydrocarbon (M1) = ?? Molecular mass of SO2 (M2) = 64 From Graham's law r1 = M2 r2 M1 Substituting the values 18015 = 64 120 20 M1 or 180 x 20 = 64 15 120 M1 or 64 =2 M1 or 64 =4 M1 or M1 = 64 = 16 4 i.e. Molecular wt of hydrocarbon is 16. Example 7. A straight glass tube has two inlets X and Y at the two ends.The length of the tube is 200 cm. HCl gas through the inlet tube X and NH3 gas through the inlet tube Y are allowed to enter into the tube at the same time. White fumes appeared at a point P. Find the distance of 'P' from X. Solution : .HCl X P Y NH33 x (200 - x) Let the distance of 'P' from 'X' be x cm. So, Py= (200 – x) Time taken by HCl to move x cm. must be equal to time taken by NH3 to move (200 – x) cm.

262 +2 CHEMISTRY (VOL. - I) Rate of diffusion of HCl = distance Px = M NH3 Rate of diffusion of NH3 distance Py M HCl \\ x = 17 = 0.68 200 - x 36.5 or, x = 0.68 ( 200 – x ) or, x = 80 cm. 6.4 KINETIC THEORY OF GASES In all the three states of matter the molecules exhibit some sort of motion. In solid state only vibratory motion, in liquid state both vibratory and rotatory and in gaseous state all the three types, namely vibratory, rotatory and translatory motions are conspicuous. By virtue of these different types of motion the molecules acquire some energy. Such energy is called energy of motion or kinetic energy. Thus the theory that explains the behaviour of gases is known as kinetic theory of gases. This theory was putforth by Bernoulli in 1728 and later on developed by Clausius, Kronig, Maxwell, Boltzmann and Waterson. The theory is applicable to perfect or an ideal gas. 6.4.1 POSTULATES OF KINETIC THEORY OF GASES : 1. Every gas consists of minute particles called molecules. The molecules have free existence. 2. The molecules of a gas are similar in shape, size and mass, but they are very much different from those of the other gases. 3. The actual volume occupied by the molecules is negligible compared to the total volume of the gas. 4. The molecules of a gas are always in a state of constant, rapid, zig-zag motion in all possible directions with different velocities. They travel in straight lines. The direction of motion is changed either on collision with other molecules or with the walls of the containing vessel. 5. The period during which the collision occurs is negligible compared to the time taken to traverse the distance between two successive collisions. 6. The molecules are supposed to be spherical in shape and perfectly elastic. As a result no energy is lost during their collisions with one another or with the walls of the containing vessel. 7. Unequal bombardments of molecules on the walls of the container give rise to pressure of the gas. The pressure is the average force per unit area which the molecules exert during their impacts on the walls per unit time. More the number of collisions, more will be the pressure. 8. At relatively low pressure the average distance between the molecules is large compared to molecular diametre. The forces of attraction between the molecules are therefore negligible. The molecules can then be considered as point masses. 9. The gravitational force on the molecules is negligible. 10. The average kinetic energy of gas molecules is a direct measure of absolute temperature. i.e. Average K.E µ Absolute temperature.

GASEOUS STATE 263 Basing on the above postulates of Kinetic theory of gases it is possible to derive an expression for the pressure of an ideal gas which is known as \"Kinetic gas equation\". 6.4.2 DERIVATION OF KINETIC GAS EQUATION Consider a certain quantity of ideal gas to be enclosed in a cubical vessel A B C D E F G H . H G lcm A B mcx Face CY C – mcx BGFC lcm CX FX E CZ D lcm C Z Fig. 6.7 Transfer of momentum to the wall of the vessel Fig. 6.6 Cubical vessel and resolution of molecular velocity Let us suppose :- l cm ® Length of each side of the cube. n ® Number of molecules present. m ® Mass of each molecule. Consider a molecule moving with a velocity 'C' cm/sec. This velocity vector can be resolved into three components Cx, Cy and Cz along three axes x, y and z respectively. These components are at right angles to each other and parallel to three sides of the cube. The components will be related to velocity 'C' by the relation. C2 = C2x + C2y + C2z ................. (1) Consider the motion of the molecule along X-axis striking the face BGFC. Its momentum will be m.Cx (mass X velocity). As the molecule is perfectly elastic, on striking the wall it will rebound with same velocity but opposite sign. Principle of conservation of momentum is assumed to hold good. Thus the momentum of the molecule after striking the wall in opposite direction = – mcx \\ Change in momentum for one impact = m Cx – ( – m Cx) = 2 m Cx After striking the face BGFC the molecule moves 'l' cm towards left and then 'l' cm towards right to strike the same face again. In other words the molecule has to traverse a distance of 2l cm for each successive collision on the same face. Since Cx is the velocity of the molecule along X-axis, the time taken to traverse a distance of 2l cm = 2l secs. Cx

264 +2 CHEMISTRY (VOL. - I) So the time taken by a molecule for one collision is 2l secs. Cx Cx and the no. of impacts per sec on the same face = 2l For one impact the change of momentum = 2 mCx . 2 So for Cx impacts the change of momentum = 2 mCx x Cx = m C x . 2l 2l l The cube has two opposite faces on the X-axis. Hence the total change in momentum per second due to impact of molecules on the two opposite faces of the cube along 2 X-axis = 2mC x l Similarly the total change in momentum per second on the two opposite faces along Y-axis and Z-axis are 2 m C 2 and 2mC 2 respectively. y z l l Thus the total change of momentum of a molecule on all the six faces of a cube per second = 2mC 2 + 2 m C 2 + 2m C 2 x y z l 2m l l l = ( Cx2 + Cy2 + Cz2 ) = 2mC2 (Q C2 = Cx2 + Cy2 + Cz2 ) ................... (2) l If there are 'n' molecules having velocities C1 , C2 , C3 .....................Cn , the total change in momentum per second due to 'n' molecules = 2m ( C12 + C22 + ......................Cn2 ) l = 2mn (C12 +C22 +...........+ C 2 ) ........................................(3) l n n (multiplying both numerator and denominator by n) But (C12 +C22 +...........+Cn2 ) is mean square velocity and let it be represented Cby2 Thus the total change innmomentum per second due to 'n' number of molecules =2mnc 2 .........(4) l Again according to Newton's second law, the change in momentum per second or the rate of change of momentum is the Force. force = 2mnc 2.........................(5) 2mnc 2 × 1 mnc 2 mnc 2 l 6l 2 3l 3 3V \\ l = = Gas pressure 'P' = Force = Area ( l The area of six faces of cubical vessel = 6l2 ) Thus PV = 1 m nc 2 , where C = Root mean square velocity ............. (6) 3 This equation (6) is called the kinetic gas equation. The equation is valid for a vessel of any shape since such vessels may be considered to be made up of a large number of small cubes.

GASEOUS STATE 265 6.4.3 RELATIONSHIP BETWEEN AVERAGE KINETIC ENERGY AND ABSOLUTE TEMPERATURE The kinetic gas equation is PV = 1 mn 2 3 C where, P ® Pressure of the gas V ® Volume occupied by the gas m ® Mass of each molecule n ® Number of molecules present C ® Root mean square velocity (RMS velocity) Suppose one mole of gas is under consideration. One mole of gas contains Avogadro number i.e. 'N' number of molecules. Hence , PV = 1 mN C2 3 Comparing this equation with ideal gas equation PV = RT, we have 1 mN C2 = RT 3 or, 2 x 1 mN C2 = RT 3 2 or, 1 mN C2 = 3 RT 2 2 3 or, Kinetic energy = 2 RT (Q 1 m N C2 = K.E of one mole of ideal gas) 2 \\ Kinetic energy per molecule = 3 R T 2 N or Average Kinetic Energy = 3 KT ................. (7) 2 where , K = Boltzmann Constant = R = 8.314 = 1.38 x 10 –23 J deg–1. mole–1 . N 6.02 x 1023 or Average K.E µ T Thus, Average Kinetic Energy of every gas is directly proportional to the absolute temperature. From equation (7) it is evident that average kinetic energy of all gases is the same at the same temperatures. From equation (7) we can write T= 2 Av . KE. 3 K It follows that at absolute zero, the Kinetic energy is zero i.e. the molecules cease to move at absolute zero of temperature.

266 +2 CHEMISTRY (VOL. - I) 6.4.4 EXPLANATION OF GAS LAWS IN THE LIGHT OF KINETIC MOLECULAR THEORY 1. Boyle's law : According to Kinetic molecular theory, the pressure of a gas is due to the collision of gas molecules on the walls of the containing vessel. The total force exerted by the molecules on the walls per unit area gives the pressure of the gas. At constant temperature there is a definite no. of collisions made by the gas molecules on the walls of the containing vessel. If the volume of the container is increased the no. of collisions decrease since the molecules get more space for movement. So the pressure exerted by the gas decreases. On the other hand when the volume decreases the no. of collisions made on the walls increases and therefore pressure increases. Thus, keeping temperature constant the pressure and volume become inversely related. This is what is Boyle's law. 2. Charle's law : According to Kinetic molecular theory the average kinetic energy is a direct measure of absolute temperature. So when temperature increases, the molecules gain kinetic energy. As a result the molecular velocity increases and the molecules strike the walls harder. Thus the pressure of the gas becomes more. So in order to keep the pressure constant, the volume has to increase. In other words it can be concluded that with increase of temperature there will be proportionate increase in volume if pressure is kept constant. This is how Charle's law can be explained. 3. Avogadro's law : Consider equal volumes of two gases at the same temperature and pressure. The pressure of the gas depends upon two factors (i) The no. of molecules present per unit volume. (ii) Average kinetic energy Since the temperature remains same, the average kinetic energy of two gases remains constant. The two gases are also having the same pressure. This is possible if they possess same no. of molecules. In otherwords it can be said that equal volumes of all gases at the same temperature and pressure contain equal no. of molecules. This is how Avogadro's law can be explained. 4. Dalton's law of Partial pressure : According to Kinetic theory, the intermolecular forces of attraction are almost negligible in case of a gas. So if a mixture of chemically non-reacting gases is taken in a container, then their molecules do not exert any force of attraction on each other. The pressure of each gas (partial pressure) is due to hits recorded by its molecules on the walls of the container. Since the total pressure is due to the total no. of collisions made by the molecules of all the gases present we may conclude that the total pressure is the sum of the partial pressures. This is how Dalton's law can be explained.

GASEOUS STATE 267 6.4.5 DEDUCTION OF GAS LAWS FROM KINETIC GAS EQUATION 1. Boyle's Law : The Kinetic gas equation is PV = 1 mn C 2 = 2 x 1 mn C 2 3 3 2 According to Kinetic molecular theory the average kinetic energy remains constant 1 mC 2 at constant temperature i.e. at constant temperature KE, 2 is constant. Further, if a definite mass of gas is considered, the no. of molecules 'n' will remain m nC 2 is constant at constant temperature. unchanged and hence 1 2 So, 2 x 1 mn C 2 is a constant at constant temperature. 3 2 or, PV = Constant (at constant temperature). which is Boyle's law. 2. Charle's Law : The Kinetic gas equation is PV = 1 m n C 2= 2 x 1 mn C 2 3 3 2 But 1 mn C 2 µT 2 Þ 1 mnC 2 = KT, where K = Proportionality constant 2 So, PV = 2 KT 3 V 2 K or, T = 3 P .......................(8) If pressure is kept constant, the R.H.S of above equation (8) in constant Þ V = Constant at constant pressure , which is Charle's law. T 3. Avogadro's law : Let us consider two gases I and II For gas I, the kinetic gas equation is P1V1 = 1 m1n1 C 2 For gas II, the kinetic gas equation is P2V2 = 31 1 3 m2n2 C 2 2 If the two gases are at the same pressure and they occupy the same volume i.e. P1 = P2 , and V1 = V2 Then, P1 V1 = P2 V2 Þ 1 m1n1 C 2 = 1 m2n2 C 2 3 1 3 2 or, or, 2 x 1 m1n1 C 2 = 2 x 1 m2n2 C 2 3 2 1 3 2 2 1 m1n1 C12 = 1 m2n2 C.22................(9) 2 2

268 +2 CHEMISTRY (VOL. - I) If both the gases are at same temperature, then mean K.E. of both the gases are same. i.e. 1 m1C 2 = 1 m2 C 2 .................(10) 2 1 2 2 From equation (9) and (10) we can write n1 = n2 i.e. the no. of molecules of Gas I = No. of molecules of Gas II This is how Avogadro's law can be derived. 4. Dalton's law of Partial pressure : Partial pressure of a gas is defined as the pressure exerted by it when it occupies the whole volume of the container at that temperature. The kinetic gas equation is PV = 1 mCn 2 3 2 1 m nC 2 = 3 x 2 = 2 K.E (\\ 1 mn C 2 = K.E ) 3 2 or Kinetic energy = 3 PV 2 If a number of gases are under consideration then the total kinetic energy of the gaseous mixture must be equal to the sum of kinetic energy of individual gases. i.e. For a gaseous mixture – (KE)Total = (KE)1 + (KE)2 + (KE)3 + ............. where (KE)1 , (KE)2 , (KE)3 are the kinetic energies of individual gases 1, 2, 3.............. respectively. or, 3 PV = 3 P1V + 3 P2V + 3 P3V + .................. 2 2 2 2 Where P1 , P2 ,P3 ......... are the Partial pressures of gases 1, 2, 3......respectively. or, 3 PV = 3 V (P1 + P2 + P3 + ......... ) 2 2 or, P = P1 + P2 + P3 ...... This is Dalton's law of partial pressure. 5. Graham's law of diffusion : The kinetic gas equation is PV = 1 mnC 2 = 1 MC 2 3 3 where, M = mass of 'n' number of molecules m = mass of individual molecule or, c2 =µµ3MP1d1Vd = 3P = 3P Where d = Density of gas or, c2 M/V d c at constant pressure at constant pressure But root mean square velocity c @ Mean velocity 'C' 1 \\C µ d at constant pressure .....................(11)

GASEOUS STATE 269 The rate of diffusion of gas evidently depends upon the velocity of the molecules. i.e. r µ 'C' ................... (12) From equations (11) and (12), we have rµ 1 which is Graham's law of diffusion. d 6. Combined gas law (Gas equation) : The kinetic gas equation is PV = 1 mn C 2 3 2 1 = 3 x 2 mnC 2 = 2 KT (Q 1 mnC 2 µT) 3 2 or PV = 2 K (a constant) ® Combined Gas law. T 3 For 1 gm molecule of the ideal gas, this constant is denoted by R (Universal gas constant) \\ PV = R T or, PV = RT For 'n' moles of an ideal gas, PV = nRT ® Ideal gas equation. 6.4.6 MOLECULAR VELOCITIES AND THEIR CALCULATION There are three types of molecular velocities. (i) Most Probable Velocity (CP) : The velocity possessed by the maximum fraction of molecules of a gas at a particular temperature is known as most probable velocity (CP). (ii) Average Velocity (Ca) : The arithmetic mean of different velocities of molecules present in a given sample of gas at a particular temperature represents the average velocity (Ca). Suppose there are 'n' number of molecules having velocities V1, V2 , V3 , ..........Vn . The average velocity , Ca = V1,+V2,+V3 +.........+Vn n

270 +2 CHEMISTRY (VOL. - I) (iii) Root mean square Velocity (Cr or C ) : The square root of the mean value of the square of velocities of the molecules in a given sample of gas at a particular temperature represents the root mean square velocity or RMS velocity. C= V12 +V22 +.........+Vn2 n where V1 , V2 ............. Vn are the individual molecular velocities. Calculation of RMS Velocity : The Kinetic gas equation is PV = 1 mn C 2 3 or, C2 = 3PV mn or, C = 3PV mn If we consider one mole of the gas, the number of molecules involved will be N (ie. Avogadro's number) So, for 1 mole of the gas n = N and mN = M = Molecular mass of the gas \\ C= 3PV = = 3RT = 3P = 3P where D ® Density of the gas. M M M/V D From detailed calculation, Ca = 8RT and Cp = 2RT pM M Relationship between various types of velocities : Average velocity (Ca) = 0.9213 x RMS velocity (Cr or C ) Most probable velocity (Cp) = 0.8164 x RMS velocity From the above relation, the ratio of three kinds of velocities Cp : Ca : C = 1 : 1.128 : 1.224 NUMERICAL PROBLEMS Example 1. Calculate the average velocity of oxygen molecule at 270C. Solution : We know C= 3RT M = 3 x 8.314 x 107 x 300 ( Q T = 27 + 273 = 300 K 32 and M = 32 for Oxygen ) = 4.835 x 104 cm/sec. Average Velocity = (0.9213 x 4.835 x 104) cm/sec = 44544.8 cm sec–1.

GASEOUS STATE 271 Example 2. Calculate the average velocity of hydrogen at NTP. Solution : c = 3RT M Average velocity = 3 x 8.314 x 107 x 273 = 18.45 x 104 cm/sec. 2 = 0.9213 x 18.45 x 104 = 169979.85 cm/sec. Example 3. Calculate the RMS velocity of nitrogen at 300c and 75cm pressure. Solution : Given condition at NTP P1 = 75 cm P2 = 76 cm V2 = 22, 400 ml V1 = ? T2 = 273 K T1 = 30 + 273 = 303 K We know that P1 V1 = P2 V2 T1 T2 Þ V1 = P 2 V 2 T1 = 76 x 22,400 x 303 = 25193.02 ml P1 T 2 75 x 273 RMS velocity c = 3PV = 3 x 75 x 13.6 x 981 x 25193.02 M 28 = 51978.64 cm/sec. [ Q P = 75 cm = (75 x 13.6 x 981) dynes / cm2.] Example 4. At what temperature will the RMS velocity of hydrogen be the same as that of oxygen at 300C ? Solution : Let the RMS velocities of hydrogen and oxygen be C1 & C2 respectively C1 = 3RT1 , C2 = 3RT2 M1 M2 When C1 = C2 or 3RT1 = 3RT1 M1 M1 or T1 = T2 M1 M2 M1 = molecular mass of hydrogen = 2 M2 = molecular mass of oxygen = 32 T1 = ? T2 = 30 .+ 273 = 303 K

272 +2 CHEMISTRY (VOL. - I) So, T1 = 303 2 32 or T1 = 303 x 2 = 18.93 K. 32 = 18.93 – 273 = – 254.07 0C. Example 5. The density of H2 at 00C and 760 mm pressure is 0.00009 gms/ml. Find the RMS velocity of H2 molecule. Solution : We know that c= 3P , P = 760 mm = 76 cm (76 x 13.6 x 981) dyne/cm2 D c= 76 x 13.6 x 981 x 3 = 183,100 cm/sec. 0.00009 BEHAVIOUR OF REAL GASES What is an ideal gas ? A perfect or an ideal gas is one which obeys the gas laws i.e. Boyle's law, Charle's law etc for all values of temperature and pressure. In other words. a gas that obeys the equation of state PV = nRT is said to be an ideal gas since both the gas laws are contained in it . The chief characteristics of an ideal gas are as follows : (i) If PV is plotted against 'P' at constant temperature, a straight line is obtained which is parallel to the pressure axis. This shows that the product of P and V at constant temperature for a given mass of gas is constant. (ii) One mole of the gas at NTP occupies 22.4 litres. (iii) The volume of a given mass of gas decreases uniformly with decrease in temperature at constant pressure. At –2730C, the volume becomes zero. DEVIATION OF REAL GASES In actual practice no gas is 100 % ideal or perfect. The word ideal is purely hypothetical. Careful experiments have shown that the real gases obey the ideal gas equation only approximately, particularly when the pressure is low and the temperature is high. The higher the pressure and lower the temperature the more will be the deviation from ideal behaviour. With a view to explaining the deviation of real gases from ideal behaviour, scientists like Regnault, Rayleigh, Amagat had studied the effect of change of pressure on PV of several gases at 00C.

GASEOUS STATE 273 Deviation from gas laws : N2 CO2 (i) Deviation from Boyle's law : H2 He 1 PV ideal gas (in lit-atm) P (in atm) ® Fig 6.8 Plot of PV against P From the above graph it can be seen that for H2 and He there is continuous rise of PV value with increase of pressure. For N2 and CO2 , PV value gradually decreases, passes through a minimum and then continuously rise with increase in pressure. In otherwords for N2 and CO2 at low pressure the PV value decreases from that of ideal gases, but at high pressure there will be marked deviation from ideal behaviour. The horizontal dotted line is for an ideal gas. If the gases behave ideally i.e. if Boyle's law is obeyed then the value of PV would remain constant over all ranges of pressure and the curves for CO2 , N2, etc. would be straight horizontal lines. Thus PV – P plots for serveral gases give us an indication about the deviation of real gases from Boyle's law i.e. from ideal behaviour. (ii) Deviation from Charle's law : According to Charle's law the volume of a given mass of gas increases or decreases by 1 273 of its value at constant pressure and at 00C for each one degree rise or fall in temperature. In other 1 words the coefficient of increase or decrease in volume should be 273 for almost all gases. But in actual practice this is a fact only at low pressure. As the pressure increases there will be marked deviation. The deviation of real gases from Charle's law was successfully explained by scientists like Amagat and others. (iii) Deviation from Avogadro's law : According to Avogadro's law the volume occupied by 1 mole of an ideal gas should be 22.4 litres. But in actual practice it has been found that the volume occupied by 1 mole of different gases is not exactly 22.4 litres but deviate from that. TABLE 6.3 Gram molecular volume Gas at NTP. (in litres) 1. H2 22.427 2. O2 22.393 3. N2 22.401 4. NH3 22.084

274 +2 CHEMISTRY (VOL. - I) Effect of temperature on deviation from ideal behaviour : The following figure depicts the PV-P plots of N2 at various temperature. It is quite evident that with the increase of temperature the dip in the curves gradually diminish and at high 1 PV (in lit-atm) P (in atm) ® Fig 6.9 PV - P Plot of N2 at various temperature temperature the curves almost become straight approaching the ideal behaviour. For N2 at 500 the curve becomes almost horizontal for an appreciable change of pressure. This temperature is called the Boyle Temperature for N2. From the above discussion we may come to a conclusion that the real gases behave ideally only approximately that too under conditions of low pressure and high temperature. It has been found that easily liquefiable and highly soluble gases show larger deviation. So, gases like CO2, SO2 and NH3 show larger deviations than H2, N2, O2 etc. Compressibility factor (Z) and deviation Compressibility factor (Z) is defined as Z= PV = PV = PVm nRT RT ( )PV ideal ( )where Vm is the molar volume ie volume occupied by 1 mole of the gas = V . n Z = 1, for an ideal gas under all conditions of temperature and pressure (Since PV = 1. nRT So, PV = nRT which is equation of state for ‘n’moles of an ideal gas) The deviation of Z from unity is thus a measure of imperfection of the gas under consideration. Thus, compressibility factor successfully explains the deviation of real gases from ideal behaviour.

GASEOUS STATE 275 Effect of change of Pressure on deviation The compressibility factors are determined for a number of gases over a wide range of pressures at a constant temperature i.e 00 C. A graph plotted between compressibility factors and pressures is shown below (Fig. 6.10) 2.0 PV H2 CO He CH ( )PV ideal NH Z = 1.0 CO CH ideal gas NH 0 200 400 600 800 P in atmosphere (Fig : 6.10 ) (i) At extremely low pressure ‘Z’ is very close to unity which indicates that all gases behave alomost ideally. (ii)At very high pressure ‘Z’is more than unity which indicates that the gases are less compress- ible than the ideal gas. This is due to dominance of molecular repulsive forces. (iii) At moderately low pressure the gases are more compressible than the ideal gas. PV is less than PVideal and Z < 1. This is due to dominance of long range attractive forces at low pressure which favour compression. The value of ‘Z’ goes on decreasing with increase in pressure, passes through the minimum and then begins to increase with further increase in pressure. (iv) For H2 and He, Z > 1. These gases are seen to be less compressible than ideal gas at all pressure and at 00C . However, if the temperature is sufficiently low (i.e below – 1650C for H2 and below –2400C for He) then same type of Z – P plots are shown by both these gases as for other gases. On the other hand, if the temperature is sufficiently high then other gases show Z – P plots similar to those given by H2 and He. CAUSES FOR THE DEVIATION A simple and logical explanation for these deviation was given by a Dutch scientist vanderWaals. He pointed out two faulty assumptions of kinetic theory of gases. 1. The actual volume of the molecule is negligible compared to the total volume of the gas. vanderWaals, showed that this assumption is valid so long as the pressure is low and temperature is high. However the assumption does not hold good at high pressure. When pressure

276 +2 CHEMISTRY (VOL. - I) is too high, the volume of the gas decreases to an appreciable amount. But as the molecules are treated as incompressible, the volume of the molecule is not affected. And under such condition the volume of the molecule cannot be neglected compared to the total volume of the gas. The same thing happens for lowering of temperature. 2. The intermolecular forces of attraction are almost negligible in case of gases. This is another faulty assumption of kinetic theory of gases that vanderWaals could point out. According to vanderWaals, the above assumption holds good only at low pressure and high temperature. If the pressure becomes sufficiently high the volume occupied by the given amount of gas becomes low and as a result a definite number of gas molecules will be confined within a small volume. Under such condition the molecules come closer to each other and hence the molecular forces of attraction between them cannot be completely ignored. At low temperature also the velocity of gas molecule decreases as a result of which each molecule exerts appreciable attractive force on the other. Thus vanderWaals gave satisfactory explanation for the deviation of real gases from ideal behaviour. 6.4.7 EQUATION OF STATE FOR REAL GASES To explain the behaviour of real gases a number of equations have been suggested by various scientists. But the best one amongst all is the equation suggested by vanderWaals. vanderWaals' Equation of state : vanderWaals deduced a modified equation of state for real gases by just introducing two correction factors, one for volume and other for pressure into the ideal gas equation in order to rectify the error for neglecting (i) the intermolecular forces of attraction (ii) individual molecular volume. (a) Volume Correction : Consider the motion of the gas molecule 'A' in a closed space CDEF having volume 'V'. FE AB C l cm D ¬ ® Fig : 6.11 Motion of a particle in a closed space Let l cm 2 Length of side CD

GASEOUS STATE 277 The molecule 'A' moves along a straight line parallel to CD and strike the face ED at B. As the molecule is perfectly elastic it will come back to its original position after striking the face. Now the distance traversed by the molecule is not '2l' but (2l–d) where 'd' in the diameter of the molecule. If the motion of all the particles present in the container is considered in a similar way, we will find that the actual volume available for the movement of molecules is less than the original volume. According to kinetic molecular theory the molecules are nothing but point masses. They occupy no volume and move in the free space equal to volume 'V' of the container. But at high pressure the volume occupied by the molecules becomes a considerable fraction of the total volume occupied by the gas and hence can no longer be neglected. Under such condition the molecules are assumed to have finite size. The space occupied by a molecule at a given instant will not be available to other molecules at the same instant for movement. Thus each molecule excludes certain volume for the movement of other molecules. Since molecules are considered to be noncompressible and spherical particles of radius 'r' (say) the closest approach of two molecules may be represented as given below in figure 6.12. 4 p r3 3 r Excluded volume shown 4 r under dotted curve = 3 p (2r)3 2r Fig 6.12 Closest approach of two molecule The upper molecule excludes a volume 8bovefox4l43u43xpmp er4(323porvf)ro338liui..xeme. e4438fptxoimrr 343tehpsteorimt3tshotmoeveouthmlpeepeclenuortlwaomrefovreloameclcuohumlloeeetch.ufeolrer. for movement. The lower molecule excludes a movement. The two molecules mutually exclude So the excluded volume for each molecule will The excluded volume per mole of gas = 4N ( 4 p r 3) = 'b' (vanderWaals' constant) 3 It can be concluded that the actual free space available for movement of the molecules is not 'V' but something less than 'V'. vander Waal suggested a correction factor 'b' for 1 mole of gas where 'b' is called the excluded volume or co-volume.

278 +2 CHEMISTRY (VOL. - I) \\Ideal volume = Actual volume –b = (V–b) per mole of gas Thus in the ideal gas equation we will write (V–b) in place of 'V' for real gases. (b) Pressure correction Another faulty assumption in Kinetic theory of gases is the absence of intermolecular force of attraction. Forces of attraction between the molecules, however weak, are always present. This is much pronounced particularly when the pressure is made high or volume small i.e. when the molecules are brought closer to each other. AB Fig 6.13 Pressure correction Consider a molecule 'A' lying somewhere inside the vessel. The net force of attraction on this molecule is zero. This is because this molecule is uniformly attracted from all sides by the other molecules present in the neighbourhood. As a result the forces of attraction on this molecule mutually neutralise each other. However, the molecule 'B' which is about to strike the wall of the vessel is subjected to an inward pull from the bulk of the molecules. Hence the molecule strikes the wall with a low velocity and will exert a low pressure. The observed pressure is thus less than the ideal pressure. It becomes necessary, therefore, to add a certain quantity to the observed pressure 'P' of the gas inorder to get the ideal pressure. vander Waal introduced a correction factor 'p' for pressure. Thus, the corrected presssure = (P + p) i.e. Ideal Pressure = observed pressure + pressure correction factor = (P + p) Now two things are to be noted : (i) The inward pull which molecule 'B' experiences depends upon the number of molecules per unit volume ie. directly upon density of the gas.

GASEOUS STATE 279 (ii) The number of molecles which are about to strike the wall also depends upon the density of the gas. Thus the correction factor for pressure 'p' which is a direct consequence of the inward pull is proportional to the square of density of the gas. i.e. p µ d2 or, p µ 1 (Q density = mass ) V2 volume or, p = a V2 where, a = constant depending upon the nature of the gas — It is known as vander Waal constant. Hence, Ideal pressure = ( P + a ) V2 Incorporating necessary corrections for volume and pressure in the ideal gas equation we get (P+ a ) (V–b) = RT ............ (1) V2 This is vanderWaals' equation of state for 1 mole of real gas. van derWaals' Equation of state for 'n' moles of real gas : If 'n' moles of gas occupy the volume 'V', the free space available for their movement will be (V–nb), where nb 2 excluded volume. Again, p µ n2d2 or, p µ n2 V2 or, p= an2 V2 Thus, van der Waals' equation of state for 'n' moles of the gas will be (P+ an2 ) (V – nb) = nRT ............ (2) V2 van der Waals' Constant (i) The constant 'a' is related to the intermolecular attractive forces. More the value of 'a', more is the strength of molecular interaction. If for a gas the value of 'a' is more, then the gas can liquify easily. This is due to greater attractive force operating between the molecules. We know that p (pressure correction) = an2 V2

280 +2 CHEMISTRY (VOL. - I) Þ a= p x v2 = Pressure x (volume)2 n2 (no. of moles)2 Thus, the unit of 'a' is atm. lit.2 mole-2 (ii) The constant 'b' is related to size of the molecule. Larger the size of the molecule larger is the value of 'b'. 'b' is incompressible or excluded volume per mole of the gas. Therefore, it has same unit as volume i.e. lit. mole–1 EXPLANATION OF REAL GAS BEHAVIOUR ON THE BASIS OF VAN DER WAALS' EQUATION The deviation of real gases from ideal gas behaviour at various ranges of temperature and pressure can be explained on the basis of van der Waals' equation. (i) At low pressure : The vanderWaals' equation for 1 mole of real gas is (P + a ) (V –b) = RT V2 When pressure is low, volume is sufficiently large. Then the term 'b' can be neglected as compared to 'V' in the above expression. Thus, we have (P + a ) V = RT V2 or, PV + a = RT V or, PV = RT – a ............................. (3) V i.e. PV is less than RT by an amount a . In the fig.1.8 from the graph it is found that actually V at low pressure PV is having a lower value than 'RT'. As pressure increases volume will decrease, a so V will increase. Thus PV value will become less and less with increase of pressure. The dip in the curve for N2 and CO2 can thus be explained. (ii) At high pressure : When the pressure is made high, the volume is small. Under such condition 'b' cannot be a neglected compared to V. However, V2 is neglected compared to pressure P. Hence equation of state can be written as : P(V–b) = RT Þ PV = RT + Pb ................... (4) i.e the value of PV is greater than RT by an amount 'Pb'. Now it can be realised why after reaching a minima the PV value continuously increases with increase in pressure.

GASEOUS STATE 281 (iii) At high temperature : a V2 When the temperature is high, the voume 'V' is large. Thus both the terms and 'b' can be neglected. Under such condition PV = RT i.e. the gases behave ideally. (iv) Exceptional behaviour of hydrogen and helium : The mass of the molecules of such gases have small values. Consequently the forces of attraction between the molecules are negligible. a The correction factor V2 is thus neglected. The equation reduces to the form. PV = RT + Pb This explains why the value of PV will continuously rise with increase of pressure in case of these two gases. 6.4.8 MAXWELL'S LAW OF DISTRIBUTION OF MOLECULAR VELOCITIES 1. Random movement of gas molecules : In a gas assembly the molecules are always in a state of constant rapid motion in all possible directions with different molecular velocities. During their motion they collide amongst themselves and with the walls of the container. As a result, their direction of motion is changed, so also their velocities. Thus the variation of velocity of gas molecules is due to collision amongst themselves. Let us consider two molecules A and B having the same velocity 'c' but moving in opposite directions. When they collide with each other, the velocity of both the molecules will become zero. But when they collide with each other at right angles (Fig.6.14) 'A' will be brought to rest while 'B' will move off with a velocity 2 c in a direction inclined at 450 to the original direction of motion. B 450 B AA (Before collsion) (After collision) Fig 6.14 Collision among molecules As a result of further collisions with other molecules present in the gas, 'A' will acquire momentum and begin to move again whereas 'B' will either lose or gain momentum depending upon the nature of subsequent impacts.

282 +2 CHEMISTRY (VOL. - I) Agas contains a large number of molecules out of which only a few molecules have either low or high velocities but majority of molecules have an average velocity. The zig-zag motion of molecules in all directions with different velocities gives rise to random movement of molecules. (2) Distribution of velocity : From the above discussion it is very much clear that as a result of impacts, all the molecules of a gas cannot have same velocity. It is also interesting to know how between the molecules the velocities are distributed when a steady state is reached. Distribution of velocity thus refers to steady state of the molecules. By steady state we do not mean that the velocity of individual molecule remains unaltered but rather the distribution of velocity remains unchanged. The concept of distribution of molecular velocities between the molecules was first of all suggested by Maxwell. He worked out the same by simply applying probability considerations. Assumptions made by Maxwell : (i) The number of molecules per unit volume at any point is the same. (ii) The molecules are in chaotic motion even in a small volume. Distribution of velocity would equally apply to them. So far as molecular velocity is concerned isotropic (no change in physical properties) behaviour is assumed. (iii) The velocities of individual molecule are changing continuously but at any instant the fraction of total no. of molecules, possessing a given velocity, remains unaltered. This is sometimes referred to as steady state of the gas. Basing on the above assumptions and applying probability considerations Maxwell showed that distribution of molecular velocity is given by the following expression :- ( )dnc 32 e-Mc2 /2RTc2dc .................... n = 4p M (1) 2p RT where dnc = The number of molecules having velocities between c and (c+dc) n = Total number of molecules M=Molecular mass of the gas. T = Temperature in the absolute scale. dnc Fraction of total number of molecules having velocity within the range c and (c+dc) n The above equation may be rewritten as : ( )1dnc 32 e - Mc2 /2RTc2 n n = 4p M .................... (2) 2p RT This is the usual form of Maxwell's law of distribution of velocities. The left hand side of the above expression simply represents the probability of finding the fraction of molecules having velocity 'c'. If molecular mass of any gas is known, it is possible to calculate the fraction of molecules having any particular velocity 'c' at any desired temperature.

GASEOUS STATE 283 Distribution Curve : Most probable velocity. T3 > T2 > T1 Average velocity. - RMS velocity . 1 dnc n dc T1 T2 T3 C2 Fig 6.15 Maxwells' distribution of velocities For a given gas, the theoretical curves can be drawn at various temperatures by plotting 1 dnc n dc against 'c'. Three curves are drawn at various temperature T1, T2 and T3 in the above figure. Nature of distribution curves : An examination of above curves reveals the following : (i) From the nature of the curve, it follows that the total molecules having velocity greater than zero goes on increasing, attains a maximum value and then again starts to fall towards zero for very high velocities. (ii) The fraction of the molecules having very small velocities (c ® 0 ) or very high velocities ( c ® ¥ ) is practically zero (iii) The curve at one particular temperature has a maximum. The velocity corresponding to this maximum represents the speed possessed by majority number of molecules. This velocity is called the most probable velocity. We may represent this velocity by (CP). Thus most probable velocity of a gas is defined as the velocity possessed by maximum number of molecules of gas at a given temperature. (iv) The areas under different curves drawn at different temperatures are the same because the area represents the total number of molecules. Effect of Temperature on distribution of molecular velocities : With variation of temperature, separate curves are obtained for every temperature (see Fig. 6.15) At high temperature there will be wider distribution of velocities. That is the no. of molecules having high speed is increased. The maxima, therefore, shifts to the right and becomes flattened. The value of most probable velocity also increases with increase in temperature. Thus the effect of temperature on the distribution of molecular velocities can be well explained by the graph. The effect can also be well understood from the equation.(2) In this equation there is an exponential term .e-MC2/2RT Here 'e' is raised to a –ve power and T is in the denominator. If we raise the temperature the value of exponential term simply increases. Thus the fraction of molecules in the curve having high velocity increases.

284 +2 CHEMISTRY (VOL. - I) LIQUEFACTION OF GASES Increase of pressure and decrease of temperature are the main causes of liquefaction of a gas. In a gas, the molecules are far apart from each other and are always in a state of constant rapid motion. The molecules have independent existence particularly when the pressure is low and temperature is high. As the temperature decreases, the kinetic energy of gas malecules decreases. Also there is decrease in volume occupied by the gas. At a sufficiently low temperature the slow moving molecules come closer under forces of attraction and ultimately the gas is converted to liquid. Again, as the pressure increases there is decrease in volume. The molecules of the gas come closer to each other. This becomes an additional helpful factor in changing the gas into liquid. For example, SO2 can be liquefied at –80C if the pressure is 1 atm. From the concept of critical temperature it can be concluded that the gas is to be necessar- ily cooled below its critical temperature before it can be liquefied. Gases like NH3, SO2, CO2 etc are gases which have fairly high critical temperature. Hence these gases can be easily liquefied only upon application of pressure. But gases like H2, O2, He, N2 etc have very low value of critical temperature. Hence they can not be liquefied by this simple technique . These gases are to be cooled below their critical temperature before they are compressed to get liquefied. There are two principles which are applied in cooling the gases below their critical tem- perature. 1. Joule-Thomson effect According to Joule and Thomson, when a gas at certain pressure is allowed to expand adiabatically to a region of low pressure through a porous plug its temperature falls and the phenomenon is known as Joule - Thomson effect. In a gas the molecules are held together by cohesive or attractive forces. When a gas ex- pands the molecules fall apart from each other. So work has to be done with a view to overcom- ing the attractive forces operating between the molecules. Since the process is adiabatic, no heat is allowed to enter nor to leave the system. Hence work is done in this case at the expense of kinetic energy of gas molecules, Consequently, kinetic energy decreases and again since average kinetic energy is a direct measure of absolute temperature, the temperature of the gas decreases and cooling effect in caused. It has been found from the experiment that the gases can be cooled by Joule-Thomson effect provided they are below a certain temperature known as Inversion temperature ‘Ti’. Inversion temperature is characteristic of each gas which is related to vanderWaals constants ‘a’ and ‘b’ by the expression. Ti = 2a Rb

GASEOUS STATE 285 2. Adiabatic expansion involving mechanical work During the expansion of gas against a pressure as in case of an engine, it does some exter- nal work. This work is at the expense of its kinetic energy which decreases and consequently there is a fall in temperature. Critical Temperature (Tc) The critical temperature of a gas (Tc) is defined as the temperature above which the gas can not be liquefied however higher the pressure may be. For example, the critical temperature of CO2 is 31.10C. This means that CO2 can not be liquefied by any means above 31.10C. The critical temperatures of O2 and H2 are –1180C and –2400C respectively. CHAPTER (6) AT A GLANCE 1. Expansibility, compressibility and diffusion are three important properties of a gas. 2. Mass, volume, pressure and temperature are the measurable properties of a gas. 3. Boyle's law :- Temperature remaining constant the volume of a given mass of any gas in inversely proportional to pressure. 4. Charle's law :- Pressure remaining constant the volume of a given mass of gas increases or decreases by 1 of its value at 00C for each 10C rise or fall in temperature. 273 5. Absolute Zero of Temperature :- The temperature at which the gas ceases to exist i.e at which the volume of the gas becomes Zero. It is -2730C. 6. Gay - Lussac's law :- At constant volume, the pressure of the given mass of gas is directly proportional to its absolute temperature. 7. STP - or NTP (Standard Temp and Pressure) standard temp = 00C = 273K, standard Press = 1 atm. = 760 mm Hg. 8. Ideal gas equation PV = n RT. 9. Value of R :- (a) 0.0821 lit atm deg–1 mole–1. (b) 82.1 cm3 atm K–1 mole–1. (c) 62.39 lit mm K–1 mole–1. (d) 8.31 x 107 ergs K–1 mole–1. (e) 8.31 JK–1 mole–1. (f) 1.99 Cal K–1 mole–1. (g) 8.31 Nm K–1 mole–1. (h) 8.31 Pa m3 K–1 mole–1. (i) 8.31 K Pa dm3 K–1 mole–1. 10. Dalton's law of Partial Pressure :- The total pressure exerted by a mixture of chemically not reacting gases is equal to sum of partial pressures of individual gases.

286 +2 CHEMISTRY (VOL. - I) 11. Partial Pressure :- Pressure exerted by an individual gas when it is allowed to occupy the container in which the mixture of gas is kept. 12. Partial Pressure = Total pressure x mole fraction. 13. Aqueous Tension = Partial pressure of water vapour. 14. Pdry gas = Pmoist gas – Aqueous Tension. 15. Diffusion :- The property by virtue of which the gases can intermix with each other irre- spective of gravitational force is known as diffusion. 16. Graham's law of diffusion :- Rate of diffusion of the gas is inversely proportional to square root of its density under similar conditions of temperature and pressure. 17. Effusion :- Diffusion through a small pinhole. 18. Average kinetic energy is directly proportional to absolute temperature. 19. Kinetic gas equation is PV = 1 mnc 2 3 20. Most Probable velocity (CP) The velocity possessed by the maximum fraction of mol- ecules of a gas at a particular temperature. 21. Average velocity (Ca) The arithmatic mean of different velocities of molecules present in a given sample of gas at a particular temperature. 22. Root mean square velocity (Cr or c ) :- The square root of the mean value of the square of velocities of the molecules in a given sample of gas at a particular temperature. 23. CP : Ca : c = 1 : 1.128 : 1.224 24. Real gases deviate from ideal gas behaviour at high pressure and low temperature. 25. Two faulty assumptions in kinetic theory of gases are (a) The actual volume of individual molecule is negligible compared to the total volume of the gas. (b) The intermolecular forces of attraction are almost negligible in case of a gas. 26. Equation of State for one mole of real gas as proposed by vander Waal is (P + a ) (V –b) V2 = RT. 27. Maxwells law of distribution of molecular velocities is represented by dnc( )1 32 e-Mc2 /2RT ´ c2 .................... (2) dc n = 4p M 2p RT 28. Compressibility factor (Z) is defined as ( )PV PV ideal For an ideal gas, Z = 1 29. Increases of pressure and decrease of temperature are the main causes of liquefaction of gas. 30. Critical temperature (Tc) is defined as the temperature above which the gas can not be liquefied however high the pressure may be.

GASEOUS STATE 287 QUESTIONS A. Multiple choice type questions with answers (1 mark) 1. For one mole of gas the total Kinetic energy at temperature T is equal to (a) RT (b)3/2 RT, (c) 2/3 RT (d) (Cp - Cv) RT. 2. For a given mass of gas if pressure is reduced to half and the temperature is doubled, the volume will become (a) 4V (b) 2V (c) V (d) 8V 4 3. The value of universal gas constant in litre atmosphere per degree per mole is (a) 8.200 (b) 8.020 (c) 1.987 (d) 0.082 4. Volume of 20 grams of hydrogen gas at NTP is (a) 224 litres (b) 22.4 litres (c) 2.24 litres (d) 112 litres 5. 1cc at NTP of a hydrocarbon vapour is as heavy as 4cc of oxygen. What quantity of hydro- carbon will occupy a volume of 22.4 litres ? (a) 6.4 (b) 32 (c) 128 (d) 64 6. The temperature at which a real gas obeys the ideal gas laws over a wide range of pressure is called the (a) Critical temperature (b) Boyle temperature. (c) Inversion temperature (d) Reduced temperature. 7. A gas at constant temperature 250 K is containted in a closed vessel. If it is heated through 10C, the percentage increase in its pressure is (a) 0.4 % (b) 0.6 % (c) 0.8 % (d) 1.0 % 8. Equal weights of methane and oxygen are mixed in an empty container at 250C. The frac- tion of total pressure exerted by oxygen is (a) 1 (b) 1 (c) 2 (d) 1 ´ 273 3 2 3 3 298 9. Equal weights of methane and hydrogen are mixed in an empty container at 250C. The fraction of the total pressure exerted by hydrogen is (a) 1 (b) 8 (c) 1 (d) 16 . 2 9 9 17 10. Equal weights of ethane and hydrogen are mixed in an empty container at 250C. The frac- tion of the total pressure exerted by hydrogen is (a) 1 : 2 (b) 1 : 1 (c) 1 : 16 (d) 15 : 16 11. In a closed room of 1000 m3 a perfume bottle is opened up. The whole room develops smell. This is due to which property of the gases ? (a) Viscosity (b) Density (c) diffusion (d) None

288 +2 CHEMISTRY (VOL. - I) 12. Helium diffuses twice as fast as another gas B. If the vapour density of He is 2, the molecular weight of B is (a) 4 (b) 8 (c) 16 (d) 24 13. The rate of diffusion of methane at a given temperature is twice that of a gas X. The molecular weight of gas X is (a) 64 (b) 32 (c) 4 (d) 8 14. 50ml of gas A diffuses through a membrane in the same time as taken for the diffusion of 40 ml of another gas B under identical pressure temperature conditions. If the molecular weight of A is 64 that of B would be (a) 100 (b) 250 (c) 200 (d) 80 15. The internal energy of one mole of an ideal gas is given by (a) 3 RT (b) 1 KT (c) 1 RT (d) 3 KT. 2 2 2 2 16. Which is not true in case of an ideal gas ? (a) It cannot be converted into a liquid. (b) There is no interaction between the molecules (c) All molecules of the gas move with same speed. (d) At a given temperature PV is proportional to the amount of the gas. 17. An ideal gas cannot be liquefied because (a) Its critical temperature is always above O0C. (b) Its molecules are relatively smaller in size. (c) It solidifies before becoming a liquid. (d) Forces operative between its molecules are negligible. 18. The vander Waals equation explains the behaviour of (a) Ideal gas (b) Real gases (c) Vapours (d) Non - real gases 19. Real gases approximate to ideal behaviour only (a) If temperature is lower (b) If pressure is high (c) If pressure is low and temperature is high (d) If both temperature and pressure are low. 20. The deviation of behaviour of real gas from ideal gas will depend on (a) Force acting between molecules (b) Space occupied between molecules (c) Mass of molecules only (d) The force acting between molecules, space occupied by molecules and mass of mol- ecules.

GASEOUS STATE 289 21 The pressure at which one mole of a gas at O0C occupies a voume of one litre is : (a) 2.24 atm (b) 4.48 atm (c) 11.2 atm (d) 22.4 atm 22. At what temperature in the celsius scale, volume of certain mass of gas at 270C will be doubled keeping the pressure constant. (a) 540 (b) 3270C (c) 4270C (d) 5270C. 23. Partial pressure of CO2 in a mixture of CO2 and N2 is 1.25 atmosphere while the total pressure of the gas mixture is 5 atmosphere. Mole fraction of N2 in the mixture is : (a) 0.82 (b) 0.75 (c) 0.80 (d) 0.65 (1988 OUAT) 24. A comparision of relative rates of diffusion of H2 and O2 at a given temperature shows that. (a) O2 diffuses four times faster than H2. (b) H2 diffuses two times faster than O2. (c) H2 and O2 diffuse at the same rate. (d) H2 diffuses four times faster than oxygen 25. Two grams of hydrogen diffuse from a container in 10 minutes. How many grams of oxy- gen would diffuse through the same container.' (a) 6 grams (b) 4 grams (c) O.5 grams (d) 8 grams 26. The root mean square velocity is expressed as (a) ( 3 RT)½ (b) ( 3RT )½ (c) ( 2RT )½ (d) ( 3RT ) 2 M M M 27. If 3 litre of oxygen is heated from 270C to 1000C keeping the pressure constant, the new volume will be (a) 3.5 litres (b) 4.5 litres (c) 3.73 litres (d) 5.00 litres 28. In vanderwaals' equation of state for a non ideal gas, the term that accounts for intermolecular forces is (a) (v – b) (b) RT (c) ( p+ a ) (d) (RT)–1 V2 29. A gas is initially at 1 atm pressure. To compress it to 1 th its initial volume, pressure to 4 be applied is : (a) 1 atm (b) 2 atm (c) 4 atm (d) 1 atm 4 30. At what temperature will the average speed of CH4 molecules have the same value as O2 has at 300 K (a) 1200 K (b) 150 K (c) 600 K (d) 300 K B (I). Very short answer type (one mark each) 1. What is the value of R in gas equation in calories ? 2. (a) Give vanderWaals' equation for 1 mole of a real gas ? (b) Write vander Waals equation for n moles of real gas

290 +2 CHEMISTRY (VOL. - I) 3. Define Boyle's law 4. At constant temperature the product of pressure and volume of a given mass of gas is constant. What is this law called ? 5. Real gas will approach the behaviour of ideal gas at _____ 6. The temperature beyond which a gas cannot be liquified even by increase in pressure is called _____ 7. What is the value of 'R' in gas equation when the volume is expressed in litres and pressure in atmospheres. 8. Between SO2 and NH3, which gas will diffuse faster at STP 9. Define Charle's law 10. What is absolute zero of temperature ? 11. Between O2 and NH3, which will diffuse faster at NTP ? 12. Rate of diffusion of SO2(g) < CH4(g). Which is lighter ? 13. What is the relationship between average kinetic energy and temperature ? 14. Name the measurable properties of a gas 15. How is kinetic energy of a gas molecule related to the temperature ? 16. What is the temperature in absolute scale corresponding to –200C 17. What is the relationship between density and temperature of a gas ? 18. \"Rate of diffusion of Carbon dioxide is greater than that of Nitrous oxide.\" Whether the statement is correct or not ? 19. What is the order of ideality of following gases ? CO2, H2, O2 20. The temperature above which a gas can not be liquefied even by increasing the presure in called —————. 21. When pressure of a given mass of gas is tripled its volume becomes one third at constant temperature. What is this law called ? 22. Which of the following has the highest rate of diffusion ? O2,CO2,NH3,N2 23. What is the relationship between root mean square velocity and average velocity ? 24. What is absolute zero of temperatures ? 25. How Kinetic energy is related to absolute temperature ? 26. Under what condition the real gases behave ideally ? 27. Why does CO2 diffuse rapidly as compared to chlorine ? 28. What is the value of 'R' in SI Units ? 29. What is aqueous tension ? 30. Write the vander Waals equation of state for 'n' moles of real gas. 31. What is most probable velocity ? 32. Which symbol represents the excluded volume in vander Waals equation of state for 1 mole of real gas ?

GASEOUS STATE 291 II. Fill in the blanks with answers : (One mark each) 1. The value of PV for 5.6 litres of an ideal gas is ————— RT at NTP 2. The total number of electrons present in 18 ml of water is ——————— 3. The Cp - Cv for an ideal gas is —————— 4. The total energy of one mole of an ideal monoatomic gas at 270C is —————— calo- ries 5. Under similar conditions of temperatures and pressure the rate of diffusion of gases vary — ————— as the square roots of their —————— or ——————. 6. At STP the order of mean square velocity of molecules of H2, N2, O2 & HBr is ———— 7. Eight grams each of oxygen and hydrogen at 270C will have the total kinetic energy in the ratio of —————— 8. The value of the gas constant 'R' in joules per mole per degree is ——————. 9. Between CO2 & NH3, —————— gas will diffuse faster. 10. Temperature remaining constant, the pressure of a gas is —————— proportional to volume. 11. Equal volumes of all gases under similar conditions of —————— and pressure con- tain equal number of ——————. 12. Real gases behave ideally under the conditions of low pressure & ———— temperature. 13. The rate of diffusion of a gas is —————— proportional to the square root of its mo- lecular weight. 14. The temperature at which the real gases obey the ideal gas laws is known as ————. 15. SO3 diffuses —————— than SO2. 16. Average Kinetic energy of gas molecules is a measure of ——————. 17. The vander Waals equation of state for 'n' moles of real gases is (P + –) (V - nb) = nRT. 18. The ratio between root mean square velocity, average velocity and most probable velocity is — : ———— : ———— : —————. 19. The inter molecular forces between the molecules of an ideal gas are ——————. 20. The molecules of a gas are always in a state of constant ——————. C. Short answer type : (Two marks each) 1. State and explain Boyle's law 2. Which of the two graphs will be a straight line at constant temperature, 'P' verses 'V' or 'P' verses 1 ? Explain V 3. Distinguish between ideal gas and real gas. Write an equation for one mole of an ideal gas 4. Why do N2 and C2H4 diffuse at the same rate ? 5. State Graham's law of diffusion

292 +2 CHEMISTRY (VOL. - I) 6. State Dalton's law of partial pressure. 7. Explain absolute zero and absolute scale of temperature 8. Derive an equation for 'n' moles of an ideal gas. 9. At what conditions a real gas will behave as an ideal gas ? Explain in 5 sentences 10. What are the two reasons for relating the gas volumes to the temperature in Kelvin scale rather than Celsius scale ? 11. How does the real gas differ from ideal gas 12. Automobile tyres are inflated to less presure in summer than winter. Give reasons. 13. If certain mass of dry gas at 270C and 760 mm pressure has density 28, what will be its density at 70C and 740 mm pressure ? 14. Under what conditions real gases obey ideal gas equation and why ? 15. Distinguish between diffusion and osmosis. 16. Derive Graham's law of diffusion from kinetic gas equation. 17. Distinguish between Diffusion & Effusion. 18. Liquid ammonia bottle is to be cooled before opening - Explain (I.I.T, 1983) 19. Write three important postulates of Kinetic theory of gas. 20. Deduce Charle's law from Kinetic gas equation. 21. Give units of vander Waals constants. 22. Offer an explanation to the exceptional behaviour of Hydrogen & Helium. 23. What are the two faulty assumptions in Kinetic theory of gases. 24. What is Critical temperature ? Discuss its importance. 25. How is real gas different from ideal gas ? 26. Explain Boyle's law in the light of Kinetic model of gas. 27. A gas collected over water appears to be exerting more pressure in comparison to when it is dry. Explain D. Short answer type questions : (Three marks each) 1. If a given mass of gas, obeying Charle's law and Boyle's law has a volume of 30cc at 270C and 10 atmosphere what will be its volume at NTP ? 2. What will be the volume at 450 K of a gas which occupies 200cc at 300K, the pressure remaining same throughout ? (Ans 300 cc) 3. If the pressure of a given mass of gas becomes half the original pressure and simultaneously the volume has become four times the original volume, what should have happened to the temperature. (Ans-double) 4. The atomic weight of Helium, oxygen and sulphur are 4,16 and 32 respectively. Helium and sulphur dioxide in equal volume are contained in a glass vessel. If the gases leak out through a pinhole, what will be the ratio of rate of escape of helium to that of SO2 ? 5. 300 CC of a gas are heated from 270C to 1270C. What will be the new volume of the gas at constant pressure ? (Ans - 400 CC) 6. The rates of diffusion of an unknown gas and chlorine are in the ratio of 6 : 5. Assuming the density of chlorine to be 36, calculate the molecular weight of the unknown gas ? (Ans - 50)

GASEOUS STATE 293 7. The volume of a gas sample is 100 ml at 1000C. If the pressure is held constant, what will be the temperature of the gas when the sample occupies a volume of 200 ml ? 8. The relative rates of diffusion of two gases A & B are found to be 0.3 and 0.2 respectively. If the density of 'A' is 4, find the relative density of gas B. (Ans - 9) 9. 180 ml of a hydrocarbon diffuse through a porous membrane in 15 minutes while 120 ml of SO2 under identical condition diffuse in 20 minutes. Calculate molecular weight of hydrocarbon. (Ans - 19) 10. What is the root mean square velocity of CH4 at 270C ? At what temperature would ethane molecule have the same velocity as that of methane molecule ? [C = 12, H=1, R=8.314 JK–1 mole–1] 11. A hydrocarbon having the formula Cn H2n+2 diffuses twice as fast as SO2 at the same temperature. Calculate the value of n. (Ans n=1) 12. A sample of pure gas at 270C and 380 torr occupied a volume of 492 cm3. What was the number of moles of gas in this sample ? 13. A desiccator of internal volume of 1 litre and containing nitrogen at one atmospheric pressure is partially evacuated to a final pressure of 7.6 mm of Hg while the temperature remains constant. What is the volume of the gas at this stage ? (Ans-1 litre) 14. A gas cylinder containing cooking gas can withstand a pressure of 14.9 atmosphere. The pressure gauge of the cylinder indicates 12 atmosphere at 270C. Due to a sudden fire in the building, the temperature starts rising. At what temperature will the cylinder explode? (Ans 99.50C) 15. What is the volume of 6 gms of Hydrogen at 1.5 atm and 2730C ? (Ans - 89.6 litres) 16. Calculate the density of NH3 at 300C and 5 atm pressure ? (Ans - 3.42 gms/lit) 17. Calculate the total number of electrons present in 1.6 gms of methane. (1985) (Roorkee) 18. An open vessel at 270C is heated until three fifth of the air in it has been expelled. Assuming that the volume of the vessel remains constant, find the temperature to which the vessel has been heated ? (Ans - 4770C) 19. Oxygen is present in a litre flask at a pressure of 7.6 x 10–10 mm of Hg. Calculate the number of oxygen molecules in the flask at O0C. (Ans - 2.7 x 1010 molecules) 20. When 2 gms of a gas A is introduced into an evacuated flask kept at 250C, the pressure is found to be one atmosphere. If 3 gms of another gas B is added to the same flask, the total pressure becomes 1.5 atm. Assuming ideal gas behaviour, calculate the ratio of molecularweights MA : MB (Ans MA : MB = 1 : 3) 21. 3.7 gms of a gas at 250C occupied the same volume as 0.184 gms of hydrogen at 170C and at the same pressure. What is the molecular weight of the gas ? (Ans. 41. 33) 22. At 270C, hydrogen is leaked through a tiny hole into a vessel for 20 minutes. Another unknown gas at the same temperature and pressure, as that of H2, is leaked through the same hole for 20 minutes. After the effusion of the gases the mixture exerts a pressure of 6 atmosphere. The hydrogen content of the mixture is 0.7 mole. If the volume of the container is 3 litres, what is the molecular weight of the unknown gas ? (Ans - 1088) 23. At room temperature, ammonia gas at one atmospheric pressure and hydrogen chloride gas at P atm pressure are allowed to effuse through identical pin-holes from opposite

294 +2 CHEMISTRY (VOL. - I) ends of a glass table of one metre length and of uniform cross section. NH4Cl is first formed at a distance of 60cm from the end through which HCl gas is sent in. What is the value of 'P' ? (Ans 2.197 atm) 24. A 4 : 1 molar mixture of He and CH4 is contained in a vessel at 20 bar pressure. Due to a hole in the vessel, the gas mixture leaks out. What is the composition of the mixture effusing out initially ? (Ans - He : CH4 = 8 :1) 25 The average velocity at T1 K and the most probable velocity at T2 K of CO2 gas is 9.0 x 104 cm sec–1. Calculate the value of T1 and T2. (Ans T1 = 1682.5K T2 = 2143.4 K) E. Long answer type questions : (Seven marks) 1. State and explain Dalton's law of partial pressure. 500 ml of oxygen at a pressure of 700 mm, 300 ml of hydrogen at a pressure of 750 mm and 700 ml of Nitrogen at a pressure of 600 mm are enclosed in a vessel of 1 litre capacity at 250C. Calculate the total pressure of the gas mixture. 2. State and explain Graham's law of Diffusion. 3. State and explain Dalton's law of partial pressure. A certain quantity of gas occupies 50 ml when collected over water at 150C and 750 mm. pressure. It occupies 49.95 ml in the dry state at NTP. Find the partial pressure of water at 150C. 4. Derive the equation of state for an ideal gas. Why do real gases deviate from ideal gas behaviour ? What is vanderWaal modification ? 5. What are the postulates of Kinetic theory of gases ? How Boyle's law and Charle's law can be explained in the light of Kinetic model of gas ? 6. State and explain Boyle's law and Charle's law ? A gas occupies a volume of 200 ml at 270C and 700 mm pressure. What volume will it occupy at 470C and 600 mm pressure? 7. Define Boyle's law and Charle's law. Derive the combined gas equation from the two laws. Temperature remaining constant 100 ml of oxygen at 100 mm pressure is transferred to a container of 25 ml capacity. What is the pressure of oxygen in the new container ? 8. What are the main postulates of Kinetic theory of gases ? How does it explain the effect of temperature on the volume of a gas. 9. Define Boyle's law and Charle's law. Derive the gas equation from these laws. What will be the volume of 3.2 gms of O2 gas at 00C and 760 mm pressure ? 10. State and explain the equation of state for an ideal gas. Why real gases deviate from the ideal behaviour ? What is vander Waals' modification of the gas equation. 11. Define Boyle's law and Charle's law. Derive the combined gas equation from the two laws. Find out the value of gas constant R. 12. (a) Derive the equation of state for an ideal gas (b) What volume will be occupied by 48 gms of CO2 at 270C and 650 mm pressure ? 13. Write notes on : (a) Dalton's law of Partial pressure (b) Graham's law of Diffusion

GASEOUS STATE 295 14. Describe the kinetic model of gas and derive Boyle's law and Charle's law on this basis. 15. State and explain Graham's law of diffusion. Mention two of its applications. A hydrocarbon with molecular formula CnH2n+2 diffuses twice as fast as SO2 at the same temperature. Find out the molecular formula of the hydrocarbon (At mass of S = 32) 16. How Graham's law of diffusion can be explained in the light of kinetic they of gases ? At STP 0.48g. of oxygen diffuses through the porous partition in 1200 seconds. What volume of carbon dioxide with diffuse in the same time under similar condition ? 17. State and explain Graham's law of diffusion. The rates of diffusion of SO2 and an unknown gas are in the ratio 5:6.03. The density of SO2 is 32. Calculate the molecular mass of the unknown gas ? 18. What are the postulates of kinetic theory of gases ? How are Boyle's law, Charle's law and Dalton's law of partial presures explained in the light of Kinetic model of gas ? 19. Derive the kinetic equation for gases 20. State and explain Graham's law of diffusion. A given volume of oxygen containing 20% by volume of ozone required 175 seconds to diffuse, when an equal volume of oxygen took 167 seconds to diffuse under similar conditions. Find the density of ozone. 21. Derive an equation for 'n' moles of an ideal gas. How can Charles' law be explained in the light of kinetic theory of gases ? Under what condition a real gas shows ideal behaviour ? 22. State and explain Graham's law of Diffusion. Discuss its important application. 23. Derive vander Waals' equation of state and show how it explains the behaviour of real gases. 24. What is an ideal gas ? Why do real gases deviate from ideal behaviour. What are the causes of deviation. 25. (a) State and explain Graham's law of Diffusion (b) 3 moles of a gas are present in a vessel at 270C. What will be the value of 'R', the gas constant in terms of Kinetic energy of the molecules of the gas ? 26. What is numerical value (as obtained from consideration of dimension) of the molar gas constant 'R' ? How do you arrive at this value ? How do you arrive at different units ? (lit-atm, erg, cal, Joules) in which the molar gas constant is expressed ? 27. (a) Does a real gas at a very high pressure occupy more volume or less volume than that occupied by an ideal gas under identical conditions ? (b) Does the temperature have any effect on the above ? (c) Calculate the root mean square speed of a molecule of oxygen at NTP 28. What is vander Waals equation ? Under what conditions it will reduce itself to ideal gas equation ? 29. Write notes on (a) Root mean square velocity. (b) Deviation of real gas from ideal behaviour. (c) Effusion.

296 +2 CHEMISTRY (VOL. - I) 30. State Maxwells' law of distribution of molecular velocities and explain its significance. How distribution of molecular velocities is affected by temperature. 31. Give an account of Kinetic theory of gases. Discuss the behaviour of real gases. 32. How are the gas laws explained on the basis of Kinetic theory of gases ? Calculate the root mean square velocity and average velocity in kms per second of H2 molecule at 300C. 33. State and explain Dalton's law of partial pressure. What are its applications ? ANSWERS (A) Multiple Choice Type Questions. 1. (b) 6. (b) 11. (c) 16. (c) 21. (d) 26. (b) 2. (a) 27. (c) 3. (d) 7. (b) 12. (c) 17. (d) 22. (b) 28. (c) 4.(a) 29. (c) 5. (c) 8. (a) 13. (a) 18. (b) 23. (b) 30. (b) 9. (b) 14. (a) 19. (c) 24. (d) 10. (d) 15. (a) 20. (d) 25. (d) B (II) FILL IN THE BLANKS 1. 0.25 6. H2 > N2 > O2 > HBr 11. temp, molecules 16. Absolute temp. 2. 9 x 1022 17. n2a 7. 1 : 16 12. high V2 3. R 8. 8.314 13. inversely 18. 1 : 1. 120 :1.234 14. Boyle Temp. 4. 900 9. NH3 15. Slower 19. Negligible 5. Inversely, densities, 10 Inversely 20. motion. Molecular wt. qqq

CHAPTER - 7 LIQUID STATE The intermolecular forces of attraction in a liquid is intermediate between the solid and the gas. In crystalline solid the constituent particles occupy fixed position, whereas the particles of the gases are in the state of random motion. But in the liquid state the particles are neither arranged in an orderly manner nor there is a complete disorder as in gases. As a result, the behaviour of the liquid is expected to be in between solids and gases. 7.1 CHARACTERISTIC PROPERTIES OF THE LIQUIDS 1. Volume and shape : Matter in liquid state has a definite volume but no definite geometrical shape. It takes up the shape of the container in which it is contained. The intermolecular forces between the molecules in the liquid state are just sufficient to hold the molecules together, but the forces are less than the forces required to fix the molecules at definite position. 2. Compressibility : Liquids are almost incompressible. It can be compressed to a small extent under extremely high pressure. 3. Fluidity : Liquid possesses the property of flowing. When a solid melts into a liquid, the intermolecular force of attraction decreases and the intemolecular distance increases. Consequently the molecules begin to slide over one another. 4. Diffusion : Gases undergo spontaneous intermixing irrespective of gravity, which is known as diffusion. Liquid also exhibits diffusion. When two miscible liquids like water and alcohol are kept together in a container, they diffuse into one another, but the rate of diffusion is much slower than that of gas. When a few drops of blue ink are added to water taken in a beaker, the coloured ink substances diffuse through water and turn the whole solution blue. The liquid molecules diffuse slowly because of the restricted movement of molecules due to less free space available. 5. Evaporation : The gaseous state of a liquid is known as the vapour. If a liquid is exposed to the atmosphere at room temperature, its volume decreases gradually and after sometime, the liquid disappears. The decrease in volume is due to the spontaneous change of liquid into its vapour. This process in which a liquid changes into the vapour state at the surface is known as evaporation. Evaporation of liquid is due to the fact that the molecules in a liquid possess different kinetic energy, since they move at different

298 +2 CHEMISTRY (VOL. - I) velocities. When the K.E. is sufficient to overcome the attractive forces molecules pass into vapour state provided they are nearer to the surface. When the temperature of a liquid is raised, it evaporates more quickly, since the kinetic energy of the moving molecules of the liquid increases with the rise of temperature. However, when the temperature of a liquid is lowered sufficiently, the liquid changes its state to a solid, that means the liquid freezes to a solid. The lowering of temperature means a decrease in the K.E. of the molecules. When this decrease in energy is enough, the existing weak van der Waals forces between the molecules of the liquid hold the particles together in the fixed position converting it into a solid. 6. Density : The density of liquids is comparatively less than that of the solids but sufficiently greater than that of gases. The weak intermolecular force and relatively poor packing of the molecules account for the less density of a liquid. 7. Structure : In the solid state the lattice points are fixed which is not the case with the liquid state. The weak intermolecular force often acts as a barrier for the formation of a regular crystal lattice. Due to this reason liquids possess a distorted crystal lattice with a tendency to move in different direction. Such an arrangement leads to the semicrystalline state of liquids. 7.2. VISCOSITY Some liquids like kerosene, alcohol, water etc. flow more rapidly, while a few such as oils, honey etc. flow very slowly. The variation in the flow rate of the liquids is due to some internal resistance possessed by all the liquids. This property of the liquids is known as viscosity. The resistance of flow is due to the intermolecular attractive forces in liquids. Viscosity of a liquid is defined as its property by virtue of which it tends to oppose the relative motion between its different layers. Viscosity of liquid is temperature dependent. When the temperature of a liquid increases the viscosity decreases because the kinetic energy of the molecules of the liquid increases which overpowers the attractive forces. High pressure also increases the viscosity of a liquid. The internal structure and the extent of vander Waals force generally controls viscosity. The viscosity of water is high due to greater hydrogen bonding. Viscosity increases with the increase in molecular mass of the same class of compounds. For this reason lower alcohols are mobile than the higher ones. Viscosity increases with the increase in the branching of carbon chain in case of organic liquids. 7.3 SURFACE TENSION Let us consider two molecules A and B of a liquid kept in a beaker. The molecule A is well within the liquid and is attracted equally from all sides by the surrounding molecules. So the resultant force acting on A is zero.

LIQUID STATE 299 B A Fig 7.1 Surface tension The molecule B is situated on the surface of the liquid and is pulled by the molecules below it and those lying on its sides. Since there is no molecule of liquid above B no force exists to balance the downward pull. Hence the molecule experiences a resultant pull downward. All the molecules at the surface of the liquid are bound together from something like a streched membrane tending to compress the molecules below to the smallest possible volume. This force on a liquid is known as surface tension. Surface tension is defined as \"force in dynes acting at right angles to an imaginary line of unit length on the surface of a liquid\". It is expressed in dynes per cm. Examples : (i) The rise or fall of liquids in capillaries, the floating of a shaving blade, the spherical shape acquired by a drop of liquid are the common consequences of surface tension. (ii) The meniscus of a liquid in a glass tube is also due to surface tension. The surface of water is concave because glass - water adhesive force is greater than the intermolecular attraction in water (cohesive force). (iii) The surface of mercury is convex because the intermolecular force in mercury is more than glass – mercury adhesive force. Surface tension decreases with the increase of temperature. Surface tension varies from liquid to liquid according to its nature and structure of molecules. Water Mercury Fig 7.2 Meniscus of liquid in glass tube

300 +2 CHEMISTRY (VOL. - I) 7.4 VAPOUR PRESSURE When a liquid is placed in an open vessel exposed to atmosphere, it gradually disappears. The change of liquid into gas is known as evaporation. At constant temperature, when the evaporation of a liquid takes place in a closed container, the molecule that escape from the surface of the liquid are trapped in the container and occupy the space above the surface of the liquid. The molecules collide with each other and with the walls of the container. A few of them may also hit the surface of the liquid and return to it resulting in condensation. The rate of return of molecules in vapour state (condensation) is proportional to the concentration of molecules in vapour state. Finally a stage reaches, when the rate of evaporation is equal to the rate of condensation. Liquid l Vapour Evaporation Equilibrium Rate Condensation Time Fig. 7.3 Equilibrium curve for evaporation and condensation Vapour pressure of the liquid at any given temperature is the pressure exerted by the vapour of a liquid when it is in equilibrium. with the liquid. Vapour pressure increases with the increase of temperature. B.P. (Boiling point) : The temperature at which the vapour pressure of a liquid becomes equal to the prevailing atmospheric pressure, bubbles begin to form through out the liquid and finally boils. This temperature is called the boiling point of a liquid. Every liquid has a standard or normal boiling point. 800 308 352 373 760 Pressure Diethyl ether1 700 EthWatyleralcohol600 500 400 300 200 100 250 275 300 325 350 375 400 Temperature (K) → Fig 7.4 Plot showing variation of vapour pressure of liquids with temperature

LIQUID STATE 301 Boiling point is defined as the temperature at which the vapour pressure becomes equal to the standard atmospheric pressure (760 mm). At a given temperature, the greater the vapour pressure of a liquid, the lesser would be its boiling temperature. The pressure cookers hasten the cooking process. In pressure cooker higher pressure is built up, consequently the boiling temperature of liquid increases. Molar heat of vapourisation is the quantity of heat supplied to one mole of a liquid at its boiling point so as to change it into vapour state at the same temperature. CHAPTER (7) AT A GLANCE 1. Liquids have definite volume but no definite shape. They are incompressible and they possess the property of flowing. 2. At any given temperature the pressure exerted by the vapour of a liquid when it is in equilibrium with the liquid is known as vapour pressure of the liquid. 3. Vapour pressure increases with increase of temperature. 4. Boiling point is the temperature at which the vapour pressure becomes equal to the atmospheric pressure. 5. Greater the vapour pressure of a liquid, the lesser would be its boiling point. 6. Viscosity of a liquid is its property by virture of which it tends to oppose the relative motion between its different layers. 7. Viscosity decreases with increased in temperature and increases with increase in pressure. 8. Surface tension is the force in dynes acting at right angles to an imaginary line of unit length on the surface of a liquid. 9. Surface tension decreases with increase of temperature.

302 +2 CHEMISTRY (VOL. - I) QUESTIONS A. Multiple choice type questions : (one mark) 1. Compared to Jalandhar, the vapour pressure of water at Simla is, (a) Lower (b) More (c) Equal. 2. Viscosity of a liquid (a) Increases with increase in temperature. (b) Decreases with decrease in temperature. (c) Independent of temperature. (d) Decreases with decrease in pressure 3. With increase in temperature, the vapour pressure of a liquid. (a) Decreases (b) Does not change. (c) Increases 4. Cleaning action of soap is due to. (a) Viscosity of water. (b) Surface tension of water. (c) Polarity of water. (d) high boiling point of water. 5. The unit of viscosity is poise which means (a) Dyne Cm–2 (b) dyne (c) dyne cm–2 sec–1 (d) dyne cm 6. The internal resistance to flow possessed by liquid is called its (a) Surface tension (b) Fuidity (c) (n) (d) Viscosity 7. A liquid in capillary tube rises due to (a) Surface tension (b) Vapour pressure (c) Osmosis (d) Viscosity 8. The vapour pressure of a liquid (a) is directly proportional to the temperature. (b) is inversely proportional to the temperature. (c) increases only up to the boiling point. (d) vary from liquid to liquid.

LIQUID STATE 303 9. At high altitudes the water boils at low temperature because (a) water molecules are bound with strong hydrogen bonds. (b) low atmospheric pressure. (c) high atmospheric pressure. (d) at low altitude the fuel does not give sufficient energy. 10. When temperature increases, the surface tension of a liquid is (a) increased (b) decreased (c) neither increased nor decreased (d) depends on the nature of the liquid 11. The boiling point of water in a pressure cooker is (a) below 1000C. (b) above 1000C. (c) 1000C. (d) depends upon the size of pressure cooker. B. Very short answer type questions : (one mark) 1. At 730 mm pressure, the boiling point of water is .......................................... (more than 1000C, less than 1000C) 2. The boiling point of sea water at 760 mm pressure is ........................................ ( more than 1000C, less than 1000C). 3. The vapour pressure of a liquid is equal to the atmospheric pressure at its .................... (boiling point, freezing point) 4. The viscosity of glycerol is more than that of water due to .......................... (hydrogen bonding, van der Waals forces) 5. Define the term surface tension and viscosity. 6. What is the freezing point of water at 1 atm. pressure in Kelvin scale ? 7. With the increase of temperature, the vapour pressure of a liquid ...................... (a) Decreases, does not change, increases) 8. Between sea water and drinking water which will freeze at a lower temperature ? 9. What do you understand by viscosity of a liquid ? 10. Vapour pressure of a liquid ————— with rise of temperature. 11. How does boiling point of a liquid change with decrease in atmospheric pressure ? 12. What is the effect of pressure on the boiling point of a liquid ? 13. How does viscosity of a liquid vary with temperature ? 14. Between water and ether which has a higher vapour pressure ? 15. What is the SI unit of viscosity ? 16. What is the SI unit of surface tension ? 17. Which of the following has higher vapour pressure at the same temperature ? CH3OH (BP = 64.50C) C2H5OH ( BP = 78.30C)

304 +2 CHEMISTRY (VOL. - I) C. Short answer type questions : (two marks) 1. Write four important characteristics of liquids. 2. Define and explain the term \"vapour pressure\". 3. Describe the effect of temperature on the vapour pressure of a liquid. 4. Define the term 'boiling point'. 5. Explain the term freezing point. 6. Define surface tension. 7. (a) Define the terms (i) viscosity (ii) coefficient of viscosity. (b) What do you understand by viscosity of a liquid ? 8. Explain why ? (a) In summer, earthen pots are used to keep water cool. (b) Liquid drops are spherical. (c) The boiling point of water is more than that of ether. (d) Ammonia is used as a refrigerant. 9. What do you understand by viscosity of a liquid ? (3 marks) 10. Why surface tension of a liquid decreases with increase in temperature ? 11. Between sea water and pure water, which will boil at a higher temperature ? Give reason. 12. Why glycerol is more viscous than water ? 13. What is the effect of temperature on the surface tension of a liquid ? Explain with reason. (3 marks) D. Long answer type questions : (seven marks) 1. Describe the kinetic molecular theory of liquids. 2. Describe the characteristic properties of the liquids. 3. (a) Write a method for measuring the vapour pressure of a liquid. (b) Write a note on boiling point. 4. Explain the term surface tension. What are the factors on which surface tension depends ? 5. Write short notes on : (a) Viscosity (b) Surface tension. (c) Boiling point and freezing point. 6. Describe the characteristic properties of liquids and explain the following ; (i) Evaporation causes cooling. (ii) A drop of liquid assumes spherical shape. (iii) The vapour pressure of a liquid increases with the increase in temperature.

LIQUID STATE 305 (iv) 'Like dissolves like.' (v) \"Liquid state\" lies between solid and gaseous state.\" (vi) Boiling point of a liquid depends on temperature. 7. What is viscosity ? How does it vary with rise of temperature ? Mention some applications of viscosity in eveyday life. A. ANSWERS TO MULTIPLE CHOICE TYPE QUESTIONS 1. (b) 5. (c) 9. (b) 2. (b) 6. (d) 10. (b) 3. (c) 7. (a) 11. (a) 4. (b) 8. (c) qqq

306 +2 CHEMISTRY (VOL. - I) UNIT – VI CHAPTER - 8 THERMODYNAMICS Introduction Thermodynamics is an important branch of physical chemistry that deals with the energy changes accompanying all physical and chemical processes. The word thermodynamics (Thermo- heat, dynamics - motion) refers to the flow of heat or heat motion. It is an exact mathematical science in which inter- relationships between various forms of energies have been fully described. It is not concerned with the total energy of the body rather with the energy changes accompanying a particular process of transformation. There are three fundamental laws of thermodynamics, namely 1st law, 2nd law and 3rd law of thermodynamics which are based on human experience. Importance of Thermodynamics (i) The feasibility of physical and chemical processes is predicted by the laws of thermodynamics. This is its primary objective. (ii) The study of thermodynamics enables us to know whether a chemical reaction is possible under a given set of conditions. (iii) Various laws like Law of chemical equilibrium, gas laws, van’t Hoff law of dilute solutions, Raoult’s law of relative lowering of vapour pressure, Phase rule, Distribution law etc. can be deduced on the basis of the laws of thermodynamics. (iv) The extent to which a given reaction proceeds before attainment of equilibrium can be predicted by thermodynamics. Limitations of Thermodynamics (i) The laws of thermodynamics do not apply to individual atoms or molecules. The laws apply to the matter in bulk only. (ii) The rate at which a given chemical reaction proceeds can not be predicted by thermodynamics. (iii) The systems away from equilibrium can not be explained by thermodynamics. 8.1 SOME BASIC TERMS USED IN CHEMICAL ENERGETICS 1. System and Surrounding : The entire universe may be imagined to be made up of two things : the system and surrounding. Any portion of the universe chosen for thermodynamic consideration is known as the system. The rest of the universe excluded from the system is known as the surrounding. The system can exchange energy with the surrounding.

THERMODYNAMICS 307 The system is separated from the surrounding by well defined real or imaginary boundaries known as thermodynamic walls. This helps us to study the effect of certain variables like pressure, volume, temperature etc upon the contents present in the system. 2. Types of system : ------------ -------V-----L-------a----i-----q-p------------------------- (i) Isolated system : A system which can exchange neither energy nor matter with the surrounding is known as an isolated system. Fig 8.1 e.g. - A liquid in contact with its vapour in an insulated closed vessel. Liquid in contact with Open system : A system which can exchange both energy and matter with the surrounding is known as an open system. vapour e.g. - A liquid in contact with its vapour in an open vessel. Here the liquid -----------------------V-------L--------------------a-i-------------qp----------------------------------------------------------- (ii) absorbs heat from the surrounding and gives out vapour to the surrounding. Fig 8.2 Closed system : A system which can exchange only energy and not matter with the surrounding is known as a closed system. Liquid in contact e.g. - A liquid in contact with its vapour in a closed vessel which is not with vapour in open insulated. vessel --------------------------------V-----L-----------------a-----i----------q----p-------------------------------------------- (iii) Fig 8.3 Liquid in contact with vapour in closed vessel (iv) Homogeneous system : The system which has uniform composition and identical properties through out is known as a homogeneous system. The system consists of only one phase. e.g. - A pure liquid, a solid, two miscible liquids etc. Air is a homogeneous mixture of all gases. (v) Heterogeneous system : The system which has neither uniform composition nor identical properties throughout is said to be a heterogeneous system. The system consists of two or more phases. e.g. - A solid in contact with a liquid, two immiscible liquids, a liquid in contact with its vapour etc. 3. Macroscopic system and Macroscopic properties : When there are a large number of particles (molecules,atoms,ions) the system is referred to as macroscopic system. The properties associated with collective behaviour of particles are known as macroscopic properties e.g. temperature, pressure, volume, density etc. 4. State of a system : A system is said to be in a particular state when it is in equilibrium under a given set of conditions i.e. when all its macroscopic properties have definite values. The system changes from one state to other when there is a change in one or more macroscopic properties.

308 +2 CHEMISTRY (VOL. - I) 5. State Variables or State Functions. The macroscopic properties are the state variables since a change in any macroscopic property brings about a change in state of the system. A state variable that depends on other variables is known as dependent variable whereas the variable which does not depend upon other variables is known as independent variable. For example, in case of an ideal gas P,V or T can be calculated provided any two of these variables are known. These two are independent variables whereas the third one is dependent variable. Again, since the state variables depend only upon the initial and final state of the system and not upon the path followed, these are called state functions. Some common state functions are internal enery (E), entropy (S), enthalpy (H), free energy (G), pressure (P), volume (V) etc. It may be noted that heat (Q) and work (W) are not state functions because they are path dependent. They are rather known as path functions. 6. Properties of a system : The properties of a system may be studied under two heads (i) Extensive property. (ii) Intensive property. (i) Extensive property : The property which depends on the amount of substance present in the system is known as an extensive property. e.g. - mass, volume, energy etc. (ii) Intensive property : The property which is independent of the amount of substance present in the system is known as an intensive property. e.g. - temperature, pressure, density, surface tension, viscosity etc. It is of interest to note that when unit amount of substance is mentioned an extensive property becomes intensive. e.g. - mass and volume both are extensive but mass per unit volume i.e. density and volume per unit mass i.e. specific volume are intensive. 7. Thermodynamic process : A thermodynamic process is the operation which brings about change in the system. The processes are of different types. (i) Adiabatic process : When heat is neither allowed to enter nor to leave the system, it is known as an adiabatic process. (ii) Isothermal process : The process carried out at constant temperature is known as an isothermal process. (iii) Isochoric process : The processcarried outatconstant volumeisknownasan isochoric process. (iv) Isobaric process : The process carried out at constant pressure is known as an isobaric process.

THERMODYNAMICS 309 (v) Cyclic process : A process during which the system can be brought back to the initital (vi) state after undergoing a series of intermediate changes is known as a cyclic process. (vii) Reversible process : When the change is brought about infinitesimally slowly so that the system is always in equilibrium with the surrounding, the process is referred to as a reversible (viii) process . It takes place in both the direction and can never be carried to completion. Irreversible process : If the change is brought about rapidly and the system does not get proper scope to attain equilibrium, the process is referred to as an irreversible process. All the naturally occurring processes are irreversible. These are unidirectional and can be carried to completion. Path : The sequence of various steps involved in the process is known as the path. For the same process there may be more than one path. 8.2 CONCEPT OF INTERNAL ENERGY A definite quantity of any substance, under a given set of conditions is associated with a fixed amount of energy. This amount of energy is different for different substances. The energy stored in the substance by virtue of its chemical nature is called its internal energy. It is denoted by the symbol U. During a chemical reaction there is exchange of energy between the system and the surrounding. So the internal energy of the reactants very much differs from that of the product. If Up ® Internal energy of the product Ur ®Internal energy of the reactant (Up - Ur ) = DU = Change in Internal energy. If the reaction is carried out in such a manner that there is no change in temperature and the system does not perform any work nor has work done upon it then in that case the energy exchange between the system and the surrounding represents the change in internal energy of the system. This is no doubt, a consequence of the law of conservation of energy. In actual practice energy exchange is in the form of heat and work is associated with change in volume against pressure. So change in internal energy of a system (DU) can be measured by carrying out the reaction at constant temperature and constant volume and measuring the heat exchange with the surrounding. Matter, in all its forms, is associated with a certain amount of energy. We can not observe this energy, but we can certainly observe the energy change that appears in the form of heat, light, work etc. during the change of state of matter. Matter consits of molecules, which in turn consists of atoms and within the atoms there are subatomic particles like electrons, protons and neutrons. The origin of energy in matter, therefore, lies in position and motion of molecules, atoms and subatomic particles.

310 +2 CHEMISTRY (VOL. - I) The total energy possessed by the molecules present in the substance is the internal energy (U) and it may be represented as : U = Ur + Uv + Ut + Ue + Un + Upe where, Ur = Rotational energy which arises due to the rotation of molecule as a whole about an axis passing through the centre of gravity. Uv = Vibrational energy which arises due to vibration of molecules to and fro about their mean position. (If vibrational energy exceeds bond energy, the atoms separate out.) Ut = Translational energy which arises due to vibration of molecules from one point to the other. Ue = Electronic energy which arises due to motion and position of elections around the nucleus. Un = Nuclear energy which arises due to binding of the nucleons within the nucleus. Upe = Potential energy which is possessed by the molecules by virtue of their position. Internal Energy 'U' is a state function since its value depends only upon the state of the system and not upon the path followed by the system. If U1 = Internal energy of the system in the initial state. U2 = Internal energy of the system in the final state. Then, DU = Change in Int. energy = (U2 - U1) The absolute value of U can not be determined. However, the value of DU can be determined experimentally using a bomb calorimeter. Factors on which 'U' depends Internal energy U depends upon (i) The quantity of the substance. (ii) The chemical nature of the substance. (iii) Temperature, pressure and volume. Sign of DU and some important conclusions (i) For exothermic reactions UR > UP So DU is -ve. (ii) For endothermic reactions UP > UR So DU is + ve. (iii) In a cyclic process, the change in internal energy, DU is zero since U is a state function.

THERMODYNAMICS 311 (iv) The internal energy (U) of a given system is directly proportional to the absolute temperature. (v) During adiabatic expansion of a gas, work is done at the expense of internal energy. As a result internal energy decreases and cooling effect is caused. (vi) In an isochoric process, volume is kept constant. The quantity of heat supplied to the system is equal to the increase in its internal energy. i.e. Q = DU 8.3 CONCEPT OF ENTHALPY Most of the chemical reactions are carried in open vessels. In an open vessel the reactants are in contact with the atmosphere, therefore the volume of the reactants may change but pressure remains constant i.e. equal to atmospheric pressure. Since atmospheric pressure is constant, the processes carried out in open vessel may be considered as taking place at constant pressure and not at constant volume. There is exchange of energy between the system and the surrounding in every chemical reaction. The amount of heat exchanged between the system and the surrounding at constant pressure and constant temperature must differ from that at constant volume and constant temperature. The reason is obvious. When the process is carried out at constant temperature and constant pressure, its volume changes. If volume increases, the system has to expand against atmospheric pressure and energy is needed for this expansion. Thus heat exchange with the surrounding at constant pressure is less than that at constant volume since a portion of it is used in expansion of the system. On the other hand, in case of contraction, work is done on the system, so the heat exchange at constant pressure becomes greater than that at constant volume. From the above discussion it is clear that energy changes in a reaction at constant pressure and constant temperature is not due to change in internal energy alone but it also includes energy contribution due to expansion or contraction. To explain this, another property called enthalpy or heat content (H) associated with every substance is necessary. 'H' is defined as, H = U + PV Since, U, P and V are the properties of state of the system it follows that 'H' is also a state function. If H1 = Enthalpy of the system in the initial state. H2 = Enthalpy of the system in the final state. DH = (H2 -H1) = (U2 + PV2) - (U1 + PV1) = (U2 - U1) + P (V2 - V1) = DU + PDV. So, DH = DU + PDV - Relationship between Internal energy and Enthalpy.

312 +2 CHEMISTRY (VOL. - I) Sign of DH and some important conclusions : (i) For exothermic reactions HR > HP So, DH = HP -HR = -ve. where, HP = Heat content of products. HR = Heat content of reactants. (ii) For endothermic reactions HP > HR So, DH = HP - HR = +ve. (iii) The quantity of heat supplied to a system at constant pressure i.e Qp is equal to increase in its enthalpy value. QP = DH (iv) DH is more significant than DU since most of the reactions are carried out at constant pressure. (v) When no gaseous components are involved, the change is volume is almost negligible. So, DV = 0 and DH = DU = Q (vi) When gases are involved but the reaction proceeds with no change in volume as in case with H2(g) + I2 (g) ® 2HI(g) 2 vols 2 vols DV = 2 -2 = 0, and DH = DU = Q (vii) When the reaction proceeds with change in gaseous volume, 2SO2(g) + O2(g) ® 2SO3(g) 3 vols 2 vols the no. of moles of reactants and products also change. We know, PV = nRT (For 'n' moles of an ideal gas) If n1 and n2 be the no. of moles in the initial and final states respectively, we have PV1 = n1RT PV2 = n2 RT Þ PV2 - PV1 = n2 RT - n1RT Þ P (V2 - V1) = (n2 - n1) RT Þ PDV = Dn RT So, DH = DU + PDV = DU + DnRT 8.4 FIRST LAW OF THERMODYNAMICS (CONSERVATION OF ENERGY) The law was stated by Meyer and Helmholtz in 1840. Scientists like Joule in England and Colding in Denmark made significant contribution towards the development of the law. The law may be stated in a variety of ways.

THERMODYNAMICS 313 1. This is simply the law of conservation of energy which states that ‘‘Energy can neither be created nor destroyed’’. Only it can be transformed from one form to the other. 2. ‘‘The total energy of an isolated system remains constant although one form of energy is convertible to the other form.’’ 3. ‘‘It is impossible to construct a perpetual motion machine which could produce work without consuming energy.’’ 4. ‘‘Whenever a given quantity of energy of the one kind disappears, an exactly equivalent amount of some other form of energy must appear in the system.’’ 5. ‘‘The change in internal energy in a cyclic process in zero.’’ 6. ‘‘The total mass and energy of an isolated system remains unchanged although these are interconvertible.’’ This is in accordance with Einstein’s mass-energy relationship E = mc2. Mathematical Formulation of the law : Suppose a system absorbs Q amount of heat from the surroundings and undergoes a change from state ‘A’ to state ‘B’. In state ‘A’ - Let U1 = Internal energy of the system V1 = Volume of the gas n1 = Number of moles of the gas P = Pressure of the gas and in state ‘B’– Let U2 = Internal energy of the system V2 - volume of the gas n2 = No. of moles of the gas P = Pressure of the gas. The amount of heat absorbed is used in two ways : (i) It changes the internal energy of the system from U1 to U2 change of internal energy DU = U2 - U1 (ii) It does some work by consuming rest amount of heat energy. Hence, Q = change in internal energy + work done = DU + W Since the pressure is constant, work done is work of expansion only which is P (V2–V1) i.e. PDV Hence, Q = DU + PDV ......................... (1) Equation (1) is the mathematical form of First law of thermodynamics. Again, applying ideal gas equation to both the states ‘A’ & ‘B’,

314 +2 CHEMISTRY (VOL. - I) We have, PV1 = n1RT and PV2 = n2RT So, PV2 – PV1 = (n2 – n1)RT or, P (V2 – V1) = (n2 – n1)RT or, PDV = DnRT ................................... (2) Hence, equation (1) becomes Q = DU + DnRT........... (3) Equation (3) is also another mathematical form of First law of thermodynamics Again, if dq = Infinitesimal amount of heat absorbed by the system. du = Change in internal energy dw = amount of work done then, dq = du + dw ................. (4) Equation (4) is the differential form of First law of thermodynamics. Some special cases Case - I For an isothermal process, DT = 0 Case II So, DU = 0 Equation (1) becomes Q = PDV Case III Heat supplied to the system is used only in doing the external work of Case IV expansion. Case - V For an adiabatic process Q=0 Equation (1) becomes 0 = DU + PDV or, PDV = –DU Thus, work of expansion is due to decrease in internal energy of the system. For a cyclic system, DU = 0 So, Q = PDV Heat absorbed by the system is used only in doing work of expansion. When volume of the gas remains same in both the states ‘A’ and ‘B’, i.e. if V1 = V2 then DV = 0 and equation (1) is reduced to Q = DU i.e. Heat absorbed by the system is used only in raising the internal energy. When the number of moles of reactants in state ‘A’ is same as the number of moles of products in state ‘B’.

THERMODYNAMICS 315 Case - VI i.e. n1 = n2, Dn = (n2 – n1) = 0 Thus, equation (3) is reduced to Q = PDV In case of solid or liquid state, the change in volume is negligible, i.e. DV = 0 So, work done PDV = 0, and the heat supplied to the system is used only in changing the internal energy. Significance of first law (i) It helps in establishing a direct relationship between the heat absorbed by the system and work done. (ii) Work can not be performed by the system without consuming heat energy. (iii) Work and energy can not be produced from nothing. 8.5 HEAT CAPACITY AND SPECIFIC HEAT OF A SYSTEM Heat Capacity : The capacity of a system to absorb heat and preserve the same within is known as heat capacity. As a result of absorption of heat, the kinetic motion of the atoms and molecules present in the system increases. Thus the kinetic energy of the system increases which ultimately results in increasing the temperature of the system. Heat capacity of a system between any two temperatures may be defined as the amount of heat required to raise the temperature of the system from lower to higher temperature divided by the temperature difference. Specific heat : If the mass of the system is one gram, the heat capacity is referred to as the specific heat of the system. Molar Heat Capacity : If the mass of the system is one mole, the heat capacity is referred to as molar heat capacity. It is represented by C. It is defined as the quantity of heat required to raise the temperature of one mole of the substance by 1 K. Thus, C = Q= Q (When the system operates between two temperatures T1 T2 T1 DT and T2) Since molar heat capacity varies with temperature, the true molar heat capacity is represented by the differential equation C= dq .................... (1) dT Molar heat capacity at constant volume (CV) Form lst Law of thermodynamics dq = du + Pdv Substituting the value of dq in equation (1) C= du+ Pdv ............. (2) dT

316 +2 CHEMISTRY (VOL. - I) At constant volume, dv = 0, Thus, CV = du = ( U ( ............. (3) dT TV So, molar heat capacity at constant volume is defined as the rate of change of internal energy with temperature Molar heat capacity at constant pressure (Cp) From equation (2), C= du+ Pdv dT We know that enthalpy of the system H = U + PV Differentiating at constant pressure, dH = dU + PdV Equation (2) becomes CP = ddHT = ( H ( P ................ (4) T Thus, molar heat capacity at constant pressure is defined as the rate of change of enthalpy with temperature. Relationship between CP and CV in gaseous system. Molar heat capacity at constant pressure is always greater than that at constant volume. This is because when the gas is heated at constant volume, the heat absorbed by the gas is used to increase the internal energy only without doing any external work. But heating the gas at constant pressure results in doing the external work and also raising the internal energy of the system. The difference between CP and CV gives the work done by 1 mole of the gas during expansion when heated through 10C at constant pressure. [ [CP – CV = U H [[ T V ....................... (5) TP We know, H = U + PV Differentiating both the sides with respect to temperature at constant pressure, we have [ H [[U P+ P[ V[ P ................................ (6) TP [[ T T[ [ Substituting the value of [ H P from equation (6) in eqn. (5), T We have [CP – CV = U P+P [ V P– [ U [ P .................. (7) T T T Again, we know that U = f (T, V)

THERMODYNAMICS 317 So, dU = [ U V dT + [ U T [[ V T dV Dividing through out by dT and keeping pressure P constant. We have [ U P += [ U V+[ U T[ V T [T V T P .......................... (8) [[ [ [ Substituting the value of [ U P from eqn (8) in eqn. (7), T We have, [CP – CV = [UV+[ U [T[ V [ P +P [ V [ P –[ U [ [ T V [ T T TV [ U [[T[ [V V V [ T T P+P P .......... (9) The term [ U T is zero for an ideal gas. V [So, CP – CV = P V P .................. (10) T Again, for 1 mole of ideal gas, PV = RT Differentiating this equation with respect to temperature at constant pressure, we have P[ V [ P = R ..................... (11) T Thus, equation (10) becomes CP – CV = R Thus, the difference between molar heat capacity at constant pressure (CP) and at constant volume (CV) is equal to the gas constant R (8.314 JK–1 mol–1). Measurement of the Enthalpy change (DH) and change in Internal energy (DU) In a chemical reaction, heat evolved or absorbed in measured by carrying out the reaction in an apparatus called calorimeter. The principle is based on the fact that the heat given out in equal to the heat taken in, If Q = Amount of heat of reaction taken out W = Water equivalent of the calorimeter m = Mass of liquid taken the calorimeter S = Specific heat T1 = Initial tempeature and

318 +2 CHEMISTRY (VOL. - I) T2 = Final temperature Q = (W + m) × S × (T2 – T1) The common types of calorimeter used for the measurement are (i) Water calorimeter (ii) Bomb calorimeter (i) Water calorimeter : This is the simple form of calorimeter normally used in physical laboratories (Fig.8.4) Boiling tube Stirrer It consists of a large vessel in which the Thermometer calorimeter is placed on cork. Packing material such as cotton wool separates the vessel from the calorimeter. There are holes inside the calorimeter Large vassel through which a thermometer, a stirrer and the boiling tube containing reacting substances are Calorimeter fitted. A known amount of water is taken inside the calorimeter. Known amounts of reacting substances are taken in the boiling tube. The heat Water evolved during the reaction will be absorbed by water. The rise in temperature is recorded with the help of the thermometer. The heat evolved is Fig. 8.4 Water calorimeter then calculated using the formula mentioned above. It may be noted that the heat evolved is not lost since the calorimeter is separated from the large vessel with the help of insulating material like cotton wool. (ii) Bomb Calorimeter : This is commonly used to determine the heat of combustion (enthalpy of combustion) of organic substances. The apparatus consists of a sealed combustion chamber called a bomb. A weighed amount of substance is taken in a dish with oxygen at about 20 atm. pressure is placed inside the bomb, which is taken inside the water contained in an insulated copper vessel. The vessel is fitted with Ignition wires Stirrer a stirrer and a sensitive thermometer. The arrangement is shown in Fig. 8.5 Thermometer The temperature of water is noted and the substance is ignited by electric current. After combustion, the rise in temperature of the system Water Heat insulating container is noted. From the heat gained by water and calorimeter, the enthalpy of combustion can be Steel Bomb calculated. Sample in the dish The enthalpy of combustion measured Fig. 8.5 The arrangement of Bomb calorimeter is actually DU, since the reaction takes place at constant volume inside the bomb calorimeter DU = (W+ m)(T2 T1)× S × M kcal. W1

THERMODYNAMICS 319 Where M = Molecular mass of the substance W1 = Mass of the substance taken DH can be calculated using the relationship DH = DU + DnRT 8.6 THERMOCHEMISTRY AND THERMOCHEMICAL REACTIONS Thermochemistry is a branch of physical chemistry which deals with heat changes accompanying chemical reactions. The chemical reactions accompanied with evolution or absorption of heat energy are known as thermochemical reactions. TYPE OF THERMOCHEMICAL REACTIONS : (a) Exothermic reactions (Exoergic reactions) (i) Such reactions proceed with evolution of heat. (ii) The energy of the product is less than the energy of the reactants in HP < HR (iii) DH is -ve. The value of DH is written on the product side with a (+) sign. (iv) If heat content is plotted against extent of reaction, the nature of the graph will be as follows. HR Heat content D H HP ®Extent of reaction Fig 8.6. : Plot of Heat content against Extent of reaction for an exothermic reaction. (v) Examples of such type of reactions are S(s) + O2(g) ® SO2(g) + 297 kJ DH = - 297 kJ 1 H2(g) + 2 O2(g) ® H2O(l) + 286.9 kJ/mole DH = -286.9 kJ/mole C(s) + O2(g) ® CO2(g) + 393.5 kJ DH = - 393.5 kJ (b) Endothermic reactions (Endoergic reactions): (i) Such reactions proceed with absorption of heat. (ii) The energy of the product is more than that of the reactant. i.e. HP > HR

320 +2 CHEMISTRY (VOL. - I) (iii) DH is + ve. The value of DH is written on the product side with a (-) sign. (iv) If heat content is plotted against extent of reaction, the nature of the graph will be as follows. HP Heat content D H HR ®Extent of reaction Fig 8.7 - Plot of heat content against extent of reaction for endothermic reaction. (v) Examples of such reactions are : NH4Cl(S) + aq ® NH4Cl(aq) - 16.11kJ DH = + + 16.11kJ Ice (1 gm) ® Water (1 gm) - 334.7 kJ DH = + 334.7 kJ 8.7 ENTHALPY CHANGES IN CHEMICAL REACTIONS (Heat of Reaction) It is defined as the quantity of heat evolved or absorbed when the number of gram molecules of the substances taking part in the chemical reaction as indicated in the balanced thermochemical reaction have completely reacted giving the products. In otherwords it is the difference between the heat content of the products and that of the reactants. i.e. Heat of reaction = (Heat content of the product) - (Heat content of the reactant) or, DH = åHP - åHR e.g. C(s) + O2(g) ® CO2(g) + 393.5 kJ DH = - 393.5 k.J C(s) + 2S(s) ® CS2(I) - 91.8 kJ DH = 91.8 kJ Factors affecting Enthalpy of reaction : Enthalpy of reaction depends on the following factors. (i) Amount of reactants : More the amount of the reactants more is the heat evolved or absorbed as the case may be. (ii) Physical state of the reactants and products : The change of physical state is accompanied with heat change. So the value of DH depends on the physical state of the substances. (iii) Temperature : The value of DH for a reaction changes with change in temperature.

THERMODYNAMICS 321 Enthalpy change of reaction at constant volume (QV) Consider a chemical reaction taking place at constant temperature and at constant volume. In such a case, work done by the system ‘W’ = 0 According to the 1st law of thermodynamics, Q = DU + W, So, DU = QV .................. (1) Where QV is the heat or enthalpy of reaction at constant volume. If UP = Internal energy of the product UR = Internal energy of the reactant DU = UP – UR = QV Enthalpy change of reaction at constant pressure (QP) Consider a chemical reaction taking place at constant pressure. Let us suppose HR = Enthalpy of reactants HP = Enthalpy of products. Since H = U + PV HR = UR + PV1 HP = UP + PV2 DH = HP – HR = (UP – UR) + P (V2 – V1) = DU + PDV....... (2) If QP be the amount of heat exchanged in the above reaction i.e. the enthalpy change of the reaction at constant pressure, Then, QP = DH = DU + PDV ................ (3) Relationship between QP and QV We known that DH = DU+PDV (Eqn.2) Also, QV = DU and QP = DH (Eqn. 1 & 3) So, we may also write. QP = QV + PDV ................... (4) For ‘n’ moles of ideal gas, PV = nRT Let n1 = no. of moles of the reactants. V1 = volume occupied by the gaseous reactants n2 = no. of moles of the products. V2 = volume occupied by the gaseous products. At constant temperature and pressure, PV1 = n1RT PV2 = n2RT

322 +2 CHEMISTRY (VOL. - I) So, PV2 – PV1 = n2RT – n1RT Or, P(V2 – V1) = (n2 – n1)RT Or, PDV = DnRT .................. (5) Substituting the value of PDV in eqn.(4) QP = QV + DnRT where Dn = Difference between the number of moles of gaseous products and gaseous reactants. Special cases : (i) When the reactants and products are in solid or liquid state, DV becomes almost negligible and under such condition QP = QV (ii) When the number moles of gaseous reactants in same as those of gaseous products, i.e. n1 = n2. Under such condition, Dn = 0 and QP = QV Examples : 1. Dissociation of NH3 2NH3(g) N2(g) + 3H2(g) 2 moles 1 mole 3 moles In this case, n1 = 2, n2 = 1+3 = 4 Dn = 4 – 2 = 2 QP = QV + DnRT = QV + 2RT 2. Combination of Hydrogen and Oxygen 2H2(g) + O2(g) 2H2O(g) 2 moles 1 mole 2 moles In this case, n1 = 2+1 = 3, n2 = 2 Dn = n2 – n1 = 2 – 3 = –1 Qp = QV + DnRT = QV – RT Standard Enthalpy (Heat) of reaction : DH values are usually reported in a standard or reference state. This is because value of DH varies with temperature. The standard state is the most stable state of a substance under l atmospheric pressure and 298 K. So, when a process is carried out at 298 K and l atmospheric pressure the enthapy change DH of a reaction is called the standard enthalpy change and it is represented as DH0.

THERMODYNAMICS 323 8.8 TYPES OF ENTHALPIES OF REACTIONS 1. Enthalpy of formation (DH¦) : It is the quantity of heat evolved or absored (change in enthalpy) when 1 mole of the substance is formed from its constituent elements under given conditions of temperature and pressure. For example - H2(g) + 1 O2(g) ® H2O(l) + 286kJ, DH¦ = 286kJ 2 C(s) + 2H2(g) ® CH4(g) + 74.85 kJ, DH¦ = - 74.85kJ. - 184.8 H2(g) + Cl2(g) ® 2HCl(g) + 184.8 kJ, DH¦ = 2 = - 92.4 kJ. C(s) + 2S(s) ® CS2(l) - 91.8 kJ, DH¦ = + 91.8 kJ Standard enthalpy of formation (DHo¦) When all the substances involved are in their standard states (l atm. pressure and 298 K) the enthalpy change accompanying the formation of 1 mole of the compound from its elements is known as standard enthalpy of formation (DHo¦). The values of standard heat of formation of elementary substances are taken to be zero. Importance of enthapy of formation. The enthalpy of reaction can be calculated by the help of heat of formation values. DHo = åDHo¦ (products) - åDHo¦ (reactants) Example - 1 The enthalpy of reaction for ®C10H8(s) + 1202(g) 10CO2(g) + 4H2O(l) at constant volume is 4345.1 kJ at 250C. Calculate the heat of reaction at constant pressure at 250C Solution : For ®C10H8(s) + 12 O2(g) 10 CO2(g) + 4H2O(l) Given Dn = 10 - 12 = -2 (only in gaseous phase) T = 298 K DU = 4345.1 ´ 103 Joules R = 8.3143 Joules K-1 mole-1 DH = DU + DnRT

324 +2 CHEMISTRY (VOL. - I) = 4345.1 ´ 103 + (-2) (8.3143) (298) = -4350055.3 Joules = - 4350.055 kJ Example - 2 The standard heats of formation for CCl4 (g), H2O(g), CO2(g) and HCl (g) are –25.5, - 57.8, -94.1 and -22.1 kcal respectively. Calculate DH(298) for the reaction in kJ. CCl4(g) + 2H2O(g) ® CO2(g) + 4 HCl(g) Solution : DH0 Reaction = DH0 Product - DH0 Reactants = DDHH00CROea2ct+ion4 ´ -D9H4.01H+Cl4-´D(-H202C.C1)l4-- 2 ´ DH0 H-2O2 ´ ( -57.8) or, = ( - 25.5) or, DH0 Reaction = -41.4 kcals = - 41.4 ´ 4.184 = -173.2 kJ Example - 3 The molar heat of formation of NH4NO3(s) is - 367.54 kJ and those of N2O(g), H2O(1) are 81.46 and -258.8 kJ respectively at 250 C and l atmospheric pressure. Calculate DH and DU of the reaction. NH4NO3(s) ®N2O(g) + 2H2O (l) Solution : DHReaction = Hproduct - Hreactant = (DHN2O + DHH2O ´ 2) - (DHNH4NO3) = (81.46 + 2 ´ -285.8) - (-367.54) = -122.56 kJ. Again, DH = DU + DnRT Here Dn = 1-0 =1 R = 8.314 J T = 298K \\ -122.56 ´ 103 = E + 8.314 ´ 298 or DU = -125037 Joule = -125.037 k Joule. 2. Enthalpy of combustion : It is defined as the enthalpy change (DH) accompanying the complete combustion of one mole of the substance at a given temperature.

THERMODYNAMICS 325 (i) C(s) + O2(g) ® CO2(g) + 393.5 kJ DH = -393.5 kJ (ii) CH4(g) + 2O2(g) ® CO2(g) + 2H2O(l) + 890 kJ DH = -890kJ Since combustion is always accompained with evolution of heat, enthalpy of combustion DH is always -ve. Importance of Enthalpy of Combustion The value of enthalpy of combustion helps us in comparing the efficiencies of fuels. A fuel having higher value of enthalpy of combustion is considered to be a better fuel. Example - 4 When 2 moles of C2H6(g) are completely burnt, 3129 kJ of heat is liberated. Calculate the heat of formation of C2H6(g) DH¦ for CO2(g) and H2O(l) are - 395 and -286 kJ mole-1 respectively. (Roorkee 1988) Solution - We have to find 2C(s) + 3H2(g)® C2H6(g) DH = ? ............................................................... (1) As per question. C(s) + O2(g) ® CO2(g) + 395kJ ................................................................ (2) H2(g)+ 12O2(g) ® H2O (l) + 286 kJ ................................................................(3) C2H6(g) + 7 O2(g) ® 2CO2(g) + 3H2O(l) + 3129 kJ ....................................(4) 2 2 Multiplying equation (2) by 2 and (3) by 3 and then adding. 7 2C(s) + 3H2(g) + 2 O2(g) ® 2CO2(g) + 3H2O(l) + 1648 kJ ........................(5) Subtracting equ. (4) from (5) 2C(S) + 3H2(g) ® C2H6(g) + 83.5 kJ \\ DH¦ of C2H6 = - 83.5 kJ Example - 5 Calculate standard heat of formation of CS2. Given that standard heat of combustion of C, S and CS2 are -393.3, -293.72 and -1108.76 kJ mole-1 respectively (Roorkee 1989) Solution : The required equation is : C + 2S ® CS2, DH = ? As given in the question C + O2 ® CO2 + 393.3 kJ ..................................................................(1)

326 +2 CHEMISTRY (VOL. - I) S + O2 ® SO2 + 293.72 kJ ..................................................................(2) CS2 + 3O2 ® CO2 + 2SO2 +1108.76 kJ .............................................(3) Multiplying equation (2) by 2 and adding equation (1) C + 2S +3O2 ® CO2 + 2SO2 + 980.74 kJ ...........................................(4) Subtracting equation (3) from equation (4) C + 2S ® CS2 - 128.02 kJ \\ DH = 128.02 kJ. 3. Enthalpy of neutralisation : It is defined as the enthalpy change when one gram equivalent of an acid is neutralised by one gram equivalent of a base or vice versa in a fairly dilute solution. e.g. HCl (aq) +NaOH (aq) ® NaCl(aq) + H2O + 57.32 kJ. HCl (aq) + KOH (aq) ® KCl (aq) + H2O + 57.45 kJ The process of neutralisation involves the reaction between H+ ion of the acid with OH - ion of the base leading to the formation of unionised water molecule. Consider the reaction between NaOH and HCl. NaOH + HCl ® NaCl + H2O + 57.32 kJ or, Na+ + OH - + H+ + Cl- ® Na+ + Cl - + H2O + 57.32 kJ or, H+ + OH - ®H2O + 57.32 kJ Thus, enthalpy of neutralisation of a strong acid with a strong base may be defined as the enthalpy of formation of water from H+ and OH - ions. However, when acid or base is weak the enthalpy of neutralisation is lower than 57.32 kJ since a part of the heat evolved by the combination of H+ and OH- ions consumed in causing the ionisation of acid or base. The enthalpy of neutralisation is still lower when both acid and base are weak. For example enthalpy of neautralisation of NH4OH (weak base) by HCl (strong acid) is –51.9 kJ whereas that of CH3COOH by NH4OH is -12.1 kJ (here both are weak) Example - 6 Given that : H+ (aq) + OH- (aq) ® H2O + 57.45 kJ Calculate the heat released when 0.4 moles of HCl is neutralised by 0.4 moles of KOH in aqueous solution. Solution : H+(aq) + OH-(aq) ® H2O (l) 0.4 mole 0.4 mole 0.4 mole

THERMODYNAMICS 327 When 1 mole of water is formed heat relased is 57.1 kJ When 0.4 moles of water is formed heat released in 57.1 ´ 0.4 = 22.84 kJ. Example - 7 Given H+ (aq) + OH - (aq) ® H2O (l) + 57.1 kJ. Calculate the heat released. (i) When 200 ml of 0.2 M HCl solution is mixed with 100 ml of 0.1 M KOH solution. (ii) An aqueous solution containing 0.5 moles of HNO3 is mixed with an aqueous solution containing 0.2 moles of NaOH. Solution : 200 ml of 0.2 M HCl contains 200 ´ 0.2 = 0.04 moles of HCI (i) 1000 (ii) 100 ml of 0.1 M KOH contains 101000´00.1 = 0.01 moles of KOH 0.01 mole of KOH will neutralise 0.01 mole of HCl to from 0.01 mole of water and (0.04 - 0.01) i.e. 0.03 mole of HCl remain unreacted. So, amount of heat released in the formation of 0.01 mole of water = 57.45 ´ 0.01 = 0.574 kJ. (ii) 0.2 moles of NaOH neutralise 0.2 moles of NO3 to from 0.2 moles of H2O and 0.3 moles of HNO3 remain unreacted. So, heat released as a result of formation of 0.2 mole of water is 57.45 ´ 0.2 = 11.49 kJ. 4. Enthalpy of solution : When a solute is dissolved in a solvent heat is either evolved or absorbed. Consequently there is heat change accompanying the system and this heat change is known as enthalpy of solution. It is defined as the amount of heat evolved or absorbed, when 1 mole of the solute is dissolved in a large excess of water so that any further dilution of the solution produces no heat change. e.g. (i) KCl + aq ® KCl(aq) - 18.6 kJ DH = 18.6 kJ (ii) MgSO4 + aq ® MgSO4 (aq) + 84.9 kJ DH = -84.9 kJ 5. Enthalpy of atomisation It is defined as the enthalpy change for the conversion of compounds and elements in to free atoms. This is usually highly endothermic. e.g. H2(g) ® 2H(g), DH = 435 kJ mol–1 In the above case the H atoms are formed by breaking H–H bond in hydrogen molecule. The enthalpy change is known as the enthalpy of atomisation of hydgrogen. It is the enthalpy change in breaking 1 mole of bonds completely to get atoms in the gaseous phase. Here the enthalpy of atomisation is also the bond dissociaton enthalpy.

328 +2 CHEMISTRY (VOL. - I) Other Examples: (i) CH4(g) ® C(g) + 4H(g), DH = 1665 kJ mol–1 (ii) Na(s) ® Na(g), DH = 108.4 kJ mol–1 In this case the enthalpy of atomisation is same as enthalpy of sublimation. 6. Enthalpy of dilution The enthalpy change when some moles of solvent in added to an aqueous solution is the enthalpy of dilution. eg. HCl(g) + 10 H2O ® HCl. 10 H2O, DH1 = – 69.0 kJ mol–1 HCl(g) + 40 H2O ® HCl. 40 H2O, DH2 = –72.8 kJ mol–1 So the enthalpy of dilution of HCl(g) from 10aq to 40aq is (72.8 – 69) = 3.8 kJ mol–1 7. Enthalpy of ionisation We know that enthalpy of neutralisation of strong acid with a strong base is same as the enthalpy of formation of water from H+ and OH– ions. H+(aq) + H–(aq) ® H2O (l), DH = –57.32 kJ mol–1 The standard enthalpies of neutralisation of strong acid by strong base are almost identical as evident from Table 8.1. If however the acid or alkali is weak the enthalpy of neutralisation is less since the reaction involves ionisation of weak acid or weak alkali. For example 1. The enthalpy of neutralisation of CH3COOH by NaOH has been found to be –55.23 kJmol–1. Since the average value for the combination of H+ and OH– ions is taken as –57.32 kJ mol–1, the standard enthalpy of ionisation or dissociation of acetic acid may be taken as (57.32 – 55.23) = 2.09 kJmol–1 2. The enthalpy of neutralisation of HCl by NH4OH is –51.34 kJ mol–1. So the enthalpy of ionisation/dissociation of NH4OH is (57.32–51.34) = 5.98 kJ mol–1. The enthalpies of neutralisation of some weak acids with NaOH are given Table 8.2. Table 8.1 Standard enthalpies of neutralisation of strong acids by strong bases (kJ mol–1) Acid Alkali Enthalpy of neutralisation (kJ mol–1) HCl NaOH HNO3 NaOH – 57.32 HCl KOH – 57.28 HCl LiOH – 57.45 – 57.38

THERMODYNAMICS 329 Table 8.2 Standard enthalpies neutralisation of weak with NaOH Acid Enthalpy of Enthalpy of ionisation neutralisation (kJ mol–1) of dissolution (kJ mol–1) CH3COOH HCOOH – 55.23 + 2.09 – 56.06 + 1.26 HCN – 12.13 + 45.19 H2S – 15.90 + 41.42 8.9 ENTHALPY OF PHASE CHANGES Matter exists in three different states namely solid, liquid and gas. When it changes its state there is heat change. This heat change is known as Enthalpy of phase change. For example solid changes over to liquid at its melting point whereas conversion of liquid to vapour takes place at boiling point of liquid. Each transformation is associated with change in heat content of the system. (a) Enthalpy of fusion or Heat or fusion The enthalpy change accompanying the change of one mole of the substance from solid phase to the liquid phase at ita melting point is known as heat of fusion. Heat of fusion is always positive. e.g. H2O(s) ® H2O(1) - 6 kJ/mole \\ DH = 6 kJ The value of heat of fusion helps us in distinguishing ionic solids from molecular solids. The ionic solids have high value of heat of fusion whereas the molecular solids have low values. This is because of the presence of greater interparticle forces in case of ionic solids. (b) Enthalpy of vaporisation or Heat of vaporisation The enthalpy of vaporisation is defined as the change of enthalpy accompanying the change of one mole of the substance from liquid phase to vapour phase at its boiling point. Heat of vaporisation is also always positive. e.g. H2O(l) ® H2O(g) - 40 kJ/mole. \\ D H = 40 kJ. The more the value of enthalpy of vaporisation the stronger is the intermolecular forces existing between the molecules constituting the liquids. The enthalpy of vaporisation of water is + 40 kJ per mole whereas that of liquids O2 is + 0.67. This is because in case of water the intermolecular forces are hydrogen bonds which are considered to be stronger than the van der Waals forces operating between the oxygen molecules.

330 +2 CHEMISTRY (VOL. - I) (c) Enthalpy of sublimation or Heat of sublimation : The enthalpy change associated with the change of I mole of the substance from the solid phase to the vapour phase at a given temperature below its melting point is known as enthalpy of sublimation. Enthalpy of sublimation of a system is the sum of enthalpy of fusion at its melting point and enthalpy of vaporisation at its boiling point i.e. DHsub = DHfus + DHvap I2(s) ® I2(g) - 62.4 kJ So, DH = 62.4 kJ 8.10 HESS'S LAW OF CONSTANT HEAT SUMMATION In 1840 Hess, a Russian chemist after performing a large number of experiments expressed his experimental findings in the form of a law known as Hess's law. Statement : If a chemical change takes place in two or more different ways whether in one step or in several steps, the amount of heat evolved or absorbed in the total change is the same, no matter, by whichever method the change is brought about. Example : Carbon can be oxidised to CO2 in two different ways. I. II. C(s) + O2(g) ® CO2(g) + 393.5 kJ 1 C(s) + 2 O2(g) ® CO(g) + 110.5 kJ. III. CO(g) + 1 O2(g) ® CO2(g) + 283 kJ. 2 Path - I > 393.5kJ CO2(g) C(S) 110.5 kJ 283 kJ > > CO(g) Path - II Fig 8.8 Demonstration of Hess's law It is evident that 393.5 kJ heat is evolved when the change is brought about by path I. Again path II involves two steps in which the same quantity of heat(110.5 + 283 = 393.5 kJ) is evolved. Proof : Let us start with a substance 'A' which is converted to 'Z' in one step via path -I. The conversion is associated with the evolution of Q1 quantity of heat. (Fig 1.7) So, A ® Z + Q1 (path -I) Again let 'A' be converted to 'Z' through some intermediate steps via path-II with the evolution of Q2 quantity of heat. i.e. A ® B + q1

THERMODYNAMICS 331 B ® C + q2 C ® Z + q3 (path II) Total amount of heat evolved is (q1 + q2 + q3) = Q2 According to Hess's law Q1 = Q2 If Q2 is not equal to Q1, let Q2 > Q1 Path - I Q1 > A C q3 Z q1 > Path - II q2 B Fig 8.9 Proof of Hess's Law Consider the cyclic process AZA. If the system changes its state from A to Z via path II, Q2 amount of heat will be evolved. Again when the system changes to state 'A' via path I, Q1 amount of heat is to be absorbed. If we consider the whole cycle (Q2 - Q1 ) amount of heat will be gained by the system. So repeating the process a number of times, a huge amount of heat can be generated which is against the 1st law of Thermodynamics i.e. against law of conservation of energy. Hence, Q2 can not be greater than Q1, it must be equal to Q1. APPLICATION OF HESS'S LAW : (1) Calculation of Enthalpies of reaction : Enthalpies of many reactions can not be determined experimentally. However, these enthalpies can be calculated by the help of Hess's Law. For example, it is difficult to measure the heat evolved during combustion of 'C' to CO. C(s) + 1 O2(g) ® CO(g), DH = ?? 2 But the combustion of l mole of 'C' to CO2 involves the evolution of 393.5 kJ and that of CO to CO2 involves evolution of 282 kJ of heat energy. Since both these processes can be studied experimentally the heat evolved during the process of conversion of C(s) to CO(g) can be calculated by the use of Hess's Law. If heat evolved is Q then. Q = [Heat evolved during formation of CO2 from C] - [ Heat evolved during formation of CO2 from CO] = - 393.5 - ( -282) = - 111.5 kJ.

332 +2 CHEMISTRY (VOL. - I) (2) Determination of enthalpy changes of slow reaction. There are some reactions which take place so slowly that the enthalpy change accompanying such reactions can not be determined accurately. To determine the enthalpy change of such reactions accurately, Hess's law may be used. For example, consider the transition of rhombic sulphur to monoclinic sulphur. Direct measurement of enthalpy change accompanying the transition is not possible since the process is very slow. But the enthalpies of combustion of rhombic and monoclinic sulphur are found to be - 297.5 and - 300 kJ respectively. (i) S (rhombic) + O2(g) ® SO2(g) + 297.5 kJ (ii) S (monoclinic) + O2(g) ® SO2(g) + 300 kJ. Subtracting (i) from (ii) and rearranging S (monoclinic) ® S (rhombic) + 2.5kJ or, S (rhombic) ® S (monoclinic) - 2.5 kJ. Thus transformation of l gm atom of rhombic to monoclinic sulphur involves absorption of 2.5 kJ of heat. (3) Determination of Enthalpy of formation : It is not possible to determine experimentally the heat of formation of some compounds. The same may be determined by the application of Hess's law. For example, the enthalpy of formation of Benzene can be calculated from the enthalpy of combustion of benzene and enthalpies of formation of water and carbon dioxide. The desired thermochemical equation is 6 C(s) + 3H2(g) ® C6H6(l), DH = ?? We have with us the following data. (i) C6H6(l) + 721O2(g) ® 6CO2(g) + 3H2O(l) + 3267.7 kJ (ii) C(s) + O2(g) ® CO2(g) + 393.5kJ (iii) H2(g) + 1 O2(g) ® H2O(l) + 286.2 kJ 2 Multiplying eqn (ii) by 6 (iii) by 3 we get (iv) 6C(s) + 6O2(g) ® 6CO2(g) + 2361 kJ 3 (v) 3H2(g) + 2 O2(g) ® 3H2O(l) + 858.6 kJ Adding (iv) and (v) and subtracting eqn. (i) we have 6C(s) + 3H2(g) ® C6H6(l) - 48.1 kJ Thus, DH = 48.1kJ.

THERMODYNAMICS 333 8.11 BOND ENERGY Bond energy is a measure of strength of a bond. The more the bond energy the stronger is the bond. Different types of bonds are associated with different bond energies. When a chemical bond is formed between two free atoms in the gaseous state the energy of the system decreases i.e. some energy in the form of heat is given out. So, to break the same bond an equivalent amount of energy must be given to the system. This amount of energy required to break a type of bond in one mole of the compound is usually known as bond energy. Definition : It may be defined in either of the ways given below. (i) It is defined as the average amount of energy required to break one type of bond present in one mole of the compound. (ii) It is referred to as the heat of formation of the bond. So, the enthalpy change associated with conversion of 1 mole of gaseous molecules to constituent gaseous atoms is known as bond energy. In case of a diatomic molecule AB the bond energy is the energy required to break 1 mole of AB into its constituents A and B. But in case of polyatomic molecule, bond energy is the average value. Calculation of Bond energy : Steps involved (i) Determine the enthalpy of formation of compound and also the enthalpy of dissociation of molecules of common elements. (ii) Calculate the total heat of formation by Hess's law. (iii) Consider total no. of bonds and calculate average bond energy. Bond energy of CH4 can be calculated from the following data. (i) C(s) + 2H2(g) ® CH4(g) + 74.9kJ. (ii) 2H 2(g) ® 4H(g) - 870.2 kJ (iii) C(s) ® C(g) - 719.6kJ. Adding equations (ii) and (iii) and subtracting from equation (i). C(g) + 4H(g) ® CH4(g) + 1664.7 kJ. So 1664.7 kJ of heat energy is required to break four moles of C -H bonds in methane. Thus the average bond energy per mole of C - H bond is = 1664.7/4 = 416.2 kJ The value of enthalpies of formation of bonds i.e the bond energies at 250C are given in the following Table 8.3.

334 +2 CHEMISTRY (VOL. - I) Bond Table 8.3 Value of Bond energy at 250C H -H Enthalpy of formation Bond Enthalpy of formation H -F (kJ mole-1) (kJ mole-1) H - Cl 435.1 C -Cl 330.5 H - Br 564.8 C - Br 276.1 O -O 430.9 C -S 259.4 O=O 368.2 C=S 477.0 O -H 138.1 C-N 292.9 C -H 439.7 C=N 615.0 C -O 464.4 C ºN 878.6 C=O 416. 2 N -N 159.0 C -C 351.4 N=N 418.4 C=C 711.3 N ºN 945.6 C=C 347.3 N -H 389.1 C -F 615.0 F -F 154.8 811.7 Cl - Cl 242. 7 439. 3 Br - Br 192.5 Factors affecting the Bond energy : 1. Atomic size - The strength of the bond depends on the size of the atoms. The smaller the size of the atom, greater is the force of attraction operating between the atoms and hence stronger is the bond and more is the bond energy. e.g. Bond energy of H - H is 435.1 kJ mole-1 whereas that of C - C is 347.3 kJ mole-1 2. Extent of overlapping of bonding orbitals : It is a fact that more the extent of overlapping of bonding orbitals more is the energy released and more is the bond energy. e.g. Bond energy of C - C bond is 347.3 kJ mole-1 whereas that of C = C bond is 615.0 kJ mole-1 and that of C º C bond is 811.7 kJ mole-1. This is because the overlapping of bonding orbitals is to a greater extent in case of a triple covalent bond (C º C) than in case of a double covalent bond and single covalent bond (C = C and C - C bond) 3. Electronegativity : The difference in the values of electronegativities between the combining atoms also influence the bond energy. The greater the difference in electronegativies, the stronger is the bond and more is the bond energy. This is beacuse of the greater polarisability of the bond. e.g. bond energy, of HCl. is 430.9 kJ mole-1 whereas that of HBr is 368.2 kJ mole-1. 4. Resonance energy : Resonance energy greatly affects the stability of a molecule. The greater the resonance energy a molecule, the more is the stability and therefore stronger is the bond and more is the bond energy.

THERMODYNAMICS 335 5. Bond length : Bond energy is also known as bond dissociation energy. Bond dissociation energy is less, if the bond length is more and vice versa. e.g. Bond length of C-C is 1.54 A0 whereas that of C= C bond is 1.36A0. Thus bond energy of C -C bond is 347.3 kJ mole-1 whereas that of C = C bond is 615kJ mole-1. Applications of Bond energy : 1. Determination of Enthalpies of reactions : Bond energy helps in the calculation of enthalpies of reactions. Suppose we want to determine the enthalpy of the reaction. H2C = CH2(g) + H2(g) ® H3C - CH3(g), DH = ?? In the above reaction four C - H bonds in C2H4 remain unaffected. A double bond breaks in ethylene and the H - H bond breaks in H2. On the otherhand two new C - H bonds are formed in ethane. So, DH = {DHC=C + DHH-H} - {DHC-C + D2HC-H} Substituting the value of bond energy from the above table. (1.1) we get DH = (615 + 435.1) - (347.3 + 2 ´ 416.2) kJ = - 129.6 kJ. [DH = B.E. of reactants – B.E. of products] 2. Calculation of Enthalpy of formation : Using the bond energy data as mentioned in the above table, the heat of formation of certain compounds can be calculated. e.g. Consider the formation of acetone. HO H 3C(g) + 6H(g) + O(g) ® H __C__C __C __H HH The bonds that break : (i) Three H - H bonds giving six H - atoms. (ii) Half O - O bond giving one 'O' atom. Besides these two, sublimation of three atoms of C(s) giving three atoms of C(g) also takes place. 3C(s) ® 3C(g) The bonds that are formed (i) Two C - C bonds. (ii) Six C - H bonds. (iii) One C = O bond Enthalpy of formation of CH3 COCH3 is : DH¦ = [ 3 (DHH-H) + ½ (DHO-O) + 3 (DHC(s) ® C(g))] -[2(DHC-C) + 6 (DHC-H) + (DHC=O)]

336 +2 CHEMISTRY (VOL. - I) Substituting the values of B. E. from the above table (1.1) we have DH¦ = [(3 ´ 435.1) + (21 ´ 138.1) + (3 x 719.6)] - [(2 ´ 347.3) + (6 ´ 416.2) + 711.3] kJ = (1305.3 + 69.05 + 2157.9) - (694.6 + 2497.2 + 711.3) kJ = 3532.25 - 3903.1 kJ = - 370.85 kJ mole-1 3. Calculation of Resonance energy : If a compound exhibits resonance it is considered to be a stable compound. It is a fact that more the number of resonating structures more is the stability. In case of Benzene there are two Kekule resonating structures. > > For this compound benzene, the enthalpy of formation as calculated from bond energy very much differs from the experimental value. The difference between the two values is a measure of resonance energy of benzene. Let us consider dissociation of Benzene, C6H6(g) ® 6C(g) + 6H(g) During dissociation of Benzene the following bonds break. (i) Three C - C single bonds (ii) Three C = C double bonds. (iii) Six C - H bonds The dissociation energy DHd = 3 (DCC-C) + 3 (DHC=C) + 6(DHC-H) = (3 ´ 347.3) + (3 ´ 615) + ( 6 ´ 416.2) = 5384.1 kJ mol-1 But the experimental value is known to be 5535.1 kJ mol-1 So, the difference (5535.1 - 5384.1) = 151 kJ is attributed to the resonance energy of Benzene. This is beacuse the actual energy required to cause dissociation of Benzene is 151 kJ more than calculated value. Thus, Benzene molecule is more stable than any one of the two Kekule structures. Example 8 Calculate the enthalpy change of the following reaction. CH2 = CH2(g) + H2(g) ® CH3 - CH3(g) mole-1 The bond energy of C - H, C-C, C=C, H-H are 414, 347, 615 and 435 kJ respectively. (Roorkee 1985)

THERMODYNAMICS 337 Solution : DHReaction = [DHC=C + 4(DHC-H) + DHH-H] - [DHC-C + 6DHC-H] Subsitituting the bond energy data in the above expression. DHReaction = [615 + (4 ´ 414) + 435] -[ 347 + (6 ´ 414)] kJ = (615 + 1656 + 435) - (347 + 2484) kJ = -125 kJ Example 9 : The enthalpy change for the following reactions at 250 are given below. 1 H2(g) + 1 O2(g) ® OH(g), DH = -10.06 kcals ................................ (1) 2 2 H2(g) ® 2H(g) DH = 104.18 kcals ........................................................... (2) O2(g) ® 2O(g), DH = 118.32 kcals .......................................................... (3) Calculate the OH bond energy in hydroxyl radical. Solution : We have to arrive at O - H(g) ® O(g) + H(g), DH = ? Dividing eqn (2) by 2 and eqn. (3) by 2 and adding 11 2 H2(g) + 2 O2(g) ® H(g) + O(g) , DH = 111.25 kcals..............................(4) Subtracting eqn (1) from eqn. (4) we have ,1 1 1 1 ® 2 2 2 2 H2(g) + O2(g) - H2(g) - O2(g) H(g) + O(g) - OH(g) DH = 121.31 kcal. Þ OH(g) ® H(g) + O(g), DH = 121.31 kcals. \\ Bond energy of OH radical is 121.31 kcals. Example 10 : The bond dissociation energies of methane and ethane are 360 and 620 kcals mol-1 respectively. Calculate C - C bond energy. Solution : Given, CH4 ® C + 4H, DH = 360 kcals ............................................................................... (1) C2H6 ® C + C + 6H DH = 620 kcals ....................................................................... (2) From (1) the B.E of 4 (C -H) bond = 360 kcals 360 Þ B.E of 1 (C -H) bond = 4 = 90 kcals

338 +2 CHEMISTRY (VOL. - I) From (2) B.E of C2H6 = DHC-C + 6DHC-H Þ 620 = DHC-C + 6 ´ 90 = DHC-C + 540 Þ DHC-C = 620 - 540 = 80 kcals \\ The bond energy of C-C = 80 kcals (334.7 kJ) SECOND LAW OF THERMODYNAMICS 8.12 LIMITATION OF FIRST LAW According to 1st law of Thermodynamics ‘‘Various forms of energies are interconvertible’’. When a given amount of one form of energy disappears, an equivalent amount other form of energy must appear. In otherwords, an exact equivalence between various forms of energies is indicated by 1st law. But sufficient informations are not provided by the law regarding: (i) The feasibility of the process i.e. whether a specified thermodynamic process is possible or not. (ii) The extent to which a given form of energy is consumed. (iii) The condition and direction of change that takes place. (iv) The maximum amount of work that can be obtained from a given quantity of heat. To overcome the insufficiencies of 1st law, there is the need for another law i.e. the second law of thermodynamics. 8.13 SCOPE OF SECOND LAW It helps us (i) To determine the direction in which the energy can be transferred. (ii) To know the equilibrium condition. (iii) To predict whether a thermodynamic process can occur spontaneously or not. (iv) To calculate the maximum fraction of heat that can be converted to work. Thus, spontaneity forms the basis of second law of thermodynamics. 8.14 STATEMENT The law was put forward by Kelvin and Clausius. Scintists like Planck, Carnot, Ostwald etc made significant contribution towards the development of the law. The law may be stated in a variety of ways. (1) It is impossible to convert heat completely to an equivalent amount of work without producing any change in the system or surrounding. (Planck) (2) It is impossible for a self acting machine unaided by an external agency to convey heat from a cold to a hot body. (Clausius)

THERMODYNAMICS 339 (3) All natural and spontaneous processes take place in one direction and are thermodynamically irreversible. (4) It is impossible to construct a machine operating in cycle, which will produce no effect other than absorption of heat from a reservoir and its conversion to an equivalent amount of work. (5) Heat can not of itself flow from a cold to a hot body. 8.15 SPONTANEOUS OR IRREVERSIBLE PROCESS Consider the expansion of a gas. Case-I : When the opposing pressure is infinitely smaller than pressure of the gas the expansion takes place infinitesimally slowly and the process is reversible one. The work obtained in this case is maximum. Case-II : When the opposing pressure is much smaller than the gas presure the expansion of the gas takes place rapidly and the process is irreversible. The work obtained in this case is less. Case-III : When the gas expands in vacuum, there is no opposing force. The work obtained in this case is zero. The natural processes are spontaneous and irreversible. The change proceds with a net decrease in internal energy or enthalpy of the system. Examples : (1) Water flows down the hill spontaneously. The flow can not be reversed without some external aid. (2) In case of a metal bar which is hot at one end and cold at the other, heat flows spontaneously from the hot end to the cold end until the temperature becomes uniform throughout. (3) Expansion of gas is spontaneous from a region of high pressure to a region of low pressure till the pressure is uniform throughout. (4) Flow of heat is spontaneous from a hot reservoir to a cold reservoir when both are connected. (5) Flow of electricity is spontaneous from a point at higher potential to a point at lower potential. (6) Diffusion of solute from a solution of high concentration to a solution of low concentration when these are brought in contact.

340 +2 CHEMISTRY (VOL. - I) 8.16 ENTROPY (S) The concept of entropy was first introduced by Clausius in 1854. It is represented by the symbol 'S'. Like internal energy 'U', 'S' is also a state function since it depends upon the state of the system and independent of the path followed by the system. The actual entropy of the system can not be defined easily, it is rather best represented in terms of entropy change accompanying a particular process. Mathematically, DS = q T where, DS = Change in entropy (Sfinal - Sinitial) q = Amount of heat absorbed by the system. T = Absolute temperature Units of Entropy : Since entropy term is the heat exchanged divided by absolute temperature, entropy is represented in terms of calories per degree i. e. cal deg-1 . This is also known as entropy unit (eu). Again, since entropy is an extensive property i.e it depends upon the amount of the substance present in the system, quantity of the substance is to be mentioned. This quantity is usually one mole. So the unit for entropy change may be written as cal deg -1 mole-1 In SI unit it is expressed as Joules per degree Kelvin (JK-1) Physical significance of Entropy : We know that a process is said to be spontaneous when it has a natural tendency to occur of its own accord. The spontaneous processes are associated with net increase of entropy. If we examine such processes we will find that these are accompanied by increased randomness of distribution. Consider a metal bar which is hot at one end and cold at the other end. There is perfect order in the bar so long as high energy molecules are assembed at one end and low energy molecules at the other end. But as soon as heat starts flowing from hot end to the cold end, there will be uniform distribution of energy throughout the bar. The state of order existing at both the ends is disturbed giving place to disorder. Similar type of change from ordered to disordered state is seen in the following phenomena. (i) Diffusion of solute from a concentrated to a dilute solution. (ii) Flow of electricity from a point at higher potential to a point at lower potential. (iii) Expansion of a gas in vacuum. In case of fusion and vaporisation there is increase in entropy of the system, since both processes are accompained with decrease in order. In solids the molecules, atoms or ions constituting the solid are arranged in a regular order in the crystal lattice. When solid melts, the molecular order is destroyed. Similarly, vaporisation is associated with conversion of liquid to vapour i.e. with more randomness of the molecules. Since these processes are associated with an increase in entropy

THERMODYNAMICS 341 values it is justified to regard entropy as a measure of disorder, chaos or random arrangement of molecules of the system. Important conclusions : (i) The greater the randomness of a system, greater is the entropy. (ii) The desreasing order of entropy in case of three states of matter is gas > liquid > solid. (iii) For all pure crystals, entropy is taken as zero. (iv) In an isolated system, for a process to be spontaneous its DS should be +ve i.e. DS > 0 (v) In case the system is not isolated, DS = DSsyst + DSsurr So, (DSsyst+ DSsurr)> 0 for a spontaneous process, when the system is not isolated. (vi) A thermodynamically irreversible process is always associated with an increase in the entropy of the system and its surrounding taken together whereas in a thermodynamically reversible process, the entropy of the system and the surrounding taken together remains unchanged. 8.17 FREE ENERGY (G) The feasibility or spontaneity of a particular process can be decided by taking into account both enthalpy and entropy change. The important thermodynamic property which accounts for both enthalpy and entropy change of a system is Gibb's Free Energy. It is denoted by the symbol 'G'. It is defined as the amount of energy available from a system that can be put into useful work. Mathematically, G = H - TS where, H ® Enthalpy of the system S ® Entropy of the system T ® Absolute temperature. Since H and S both are state functions, G is also a state function. If G1, H1 and S1 represent the thermodynamic functions for the system in the initial state and G2, H2 and S2 in the final state, the temperature T remaining consant all through, we have -G2 = H2 TS2 -G1 = H1 TS1 \\ DG (change in F.E)= G2 - G1 - - -= (H2 TS2) (H1 TS1) - - -= (H2 H1) T (S2 S1) = DH - T D S i.e. DG = DH - TDS

342 +2 CHEMISTRY (VOL. - I) Standard Free Energy Change (DG0) The free energy change of a system at 250 C and one atmospheric pressure is known as standard free energy change. Mathematically, DG0 = DH0 - T D S0 where, DH0 ® Standard enthalpy change DS0 ® Standard entropy change DG0 ® Standard free energy change T ® 250C i.e. 298 K. Significance of DG : (i) For a system at equilibrium DG = 0 (ii) All spontaneous processes taking place at constant temperature and pressure are associated with net decrease in free energy i.e. for these processes DG is –ve or DG < 0. Greater the –ve value, more is the spontaneity. (iii) A process is not thermodynamically feasible when DG is +ve or DG > 0 i.e. the process is not spontaneous in the forward direction. It may take place in the opposite direction. Effect of temperature on DG : We know, DG = DH - T D S Enthalpy change DH is hardly affected by change of temperature. But the term TDS changes with temperature. (i) For exothermic reaction DH = –ve. If temperature is low TDS will be less and DG becomes –ve. So, exothermic reactions are favoured by lowering the temperature. (ii) For endothermic reactions DH = +ve. If temperature is high, TDS will be more and DG again becomes –ve. So, endothermic reactions are favoured at high temperature. 8.18 GIBB’S FREE ENERGY AND EQUILIBRIUM CONSTANT Consider a general reaction A+B C+D Gibbs free energy for the above reaction at equilibrium, DG = 0 We know that DG = DG0 + RT lnK ..............(1) Where DG0 = Standard free energy change i.e. when all reactants and products are at their standard states of unit activity or concentration. K = Thermodynamic equilibrium constant.

THERMODYNAMICS 343 So at equilibrium, 0 = DG0 + RT lnK .......................(2) Þ DG0 = – RT lnK = –2.303 RT log K ....................... (3) We also know that DG0 = DHO – TDSO .................... (4) So, DG0 = DHO – TDSO = – RT lnK .................... (5) Case I In case of endothermic reaction, DH is large and +ve. In such a case the value of K will be much smaller than 1 and reaction is unlikely to form more product. Case II In case of exothermic reaction, DH is large and –ve. DGO is likely to be large and –ve. K will be much larger than 1, the reaction is likely to form more product. Since DG also depends upon DS, the value of K also gets affected depending on whether DS in +ve or –ve. Equation (5) may be used to estimate DGO from measurement of DHO and DSO. K can then be calculated. Problem 1. Considers the following reaction N2(g) + O2(g) ® 2 NO(g) (SO/JK–1 mol–1) 187.5 200.4 206.3 DHO = 175.8 kJ mol–1 Assuming DHO and DSO to be independent of temperature, calculate the temparature at which the above reaction will be spontaneous. Solution DSO = 2SONO - SON2 - SO O2 = (2 × 206.3) – 187.5 – 200.4 = 24.7 JK–1 mol–1 DGO = DHO – TDSO = 175.8 – (T × 24.7 × 10–3) kJ mol–1 For a reaction to be spontaneous DGO < 0 i.e.– ve & this occurs when TDSO > DHO ÞT> DHO DSO

344 +2 CHEMISTRY (VOL. - I) Þ T > 175.8 ´103 24.7 or T > 7118K So the reaction becomes spontaneous above a temperature of 7118K 2. Can Aluminium oxide be reduced to Aluminium at 298K by metal Na ? Also calculate the value of equilibrium constant. Given DGO (298K) / kJ mol–1 Al2O3(s) = –1498 Na2O(s) = – 360 Element = 0 Solution The reaction is Al2O3(s) + 6Na(s) ® 3Na2O + 2Al(s) DGO = 3GO (Na2O) – DGO (Al2O3) = 3 × (– 360) – (– 1498) = – 1080 + 1498 = + 418 kJ mol–1 Since DGO = +ve, the reaction can not occur. Also DGO = – 2.303 RT log K Þ 418 = 2.303 × 8.314 × 10–3 × 298 log K Þ log K = – 73.33, so K = 4.67 × 10–74 8.19 SPONTANEITY OF A PROCESS We know that a spontaneous reaction is one which proceeds by itself with or without initiation. It is also a fact that all spontaneous chemical reactions proceed in direction in which there is decrese of energy. So all exothermic reactions should be spontaneous. e.g. C(s) + O2(g) ® CO2(g) + 393.5 kJ., DH = - 393.5kJ i.e. on ignition 'C' undergoes complete combustion giving CO2 (case of initiation) On the otherhand, in the reaction, 2NO(g) + O2(g) ® 2NO2(g) + 56.52 kJ DH = - 56.52 kJ Here nitric oxide and oxygen combine readily forming nitrogen dioxide(no initiation is required), However, there are several endothermic reactions which are also spontaneous. e.g. (i) Evaporation of water, H2O(l) ® H2O(g) - 40.58 kJ; DH = 40.58 kJ (ii) Decomposition of Ag2O to oxygen Ag2O(s) ® 2Ag(s) + O2(g) - 30.54 kJ ; DH = 30.54 kJ

THERMODYNAMICS 345 This shows that reactions with –ve as well as +ve DH may be spontaneous. On the otherhand, the spontaneity of a reaction can not be judged only by the sign of DH. There must be some other factor which is involved for deciding the spontaneity of the process. This factor is tendency for maximum randomness. In case of evaporation of water,when water is converted to vapour there is increase in randomness. In case of decomposition of Ag2O, a gas appears in the product side. So, the randomness of the system increases. Thus, tendency to acquire maximum randomness is another factor that determines the spontaneity. The overall tendency for a reaction to be spontaneous is to acquire a state of minimum energy and maximum randomness. This overall tendency is the driving force. As discussed earlier, entropy is a measure of randomness of the system.As the entropy change and enthalpy change, both are contained in the expression for DG (change in free energy)i.e. DG = DH - TDS, we can regard DG as the driving force for deciding the spontaneity of a reaction. The following Table 8.4 summarises how signs of DH and DS for a given reaction determines its spontaneity. Table 8.4 - Signs of DH, DS and DG for considering spontaneity. DH DS DG = DH - TDS Remarks. Reaction is spontaneous at all temperatures. - +- Reaction is nonspontaneous at all temperatures. + -+ Reaction is spontaneous at low temperature - - - (at low temp) Reaction is nonspontaneous at high temperature Reaction is nonspontaneous at low temperature + (at high temp) Reaction is spontaneous at high temperature + + + (at low temp) - (at high temp) This is to be noted that at high temperature the entropy factor predominates whereas at low temperature the energy factor predominates. 8.20 THIRD LAW OF THERMODYNAMICS First law of Thermodynamics leads to concept of energy content of the system whereas second law leads to the concept of entropy of the system. Third law of thermodynamics, however, does not lead to a new concept, rather it places a limitation on the value of entropy of a crystalline solid. Third law is based on the generalisation made by scientist Nernst. According to Nernst heat theorem DG (change in Gibb’s Free energy) and DH (change in enthalpy) of a system become equal to each other at absolute zero of temperature and approach each other asymptotically. This is illustrated in Figure 8.10.

346 +2 CHEMISTRY (VOL. - I) DG or DH 0 Temp (K) ® Fig. 8.10 : Plot of DG or DH against K According to Gibbs - Helmholtz equation DG – DH = Téëê ¶DG ù ........................ (1) ¶T ûú P When T = 0 i.e. at absolute zero of temperature DG = DH. Mathematically the Nernst Heat Theorem can be expressed as .............................. (2) From the concept of entropy é¶DG ù = -DS .............................. (3) êë ¶T ûúP where DS = Entropy change of the reaction From Kirchhoff equation .............................. (4) where DCp = difference in heat capacities of the product and the reactant. From equation 2, 3 and 4, it follows that The Nernst heat theorem holds good only in case of pure solids and is the forerunner of Third law of Thermodynamics. From equation (6), it can be concluded that at absolute zero of temparature the products and reactants in solid state have same value of heat capacites. So DCp tends to approach zero at 0K. From the concept of quantum theory, it is suggested that the heat capacities of solid tend to become zero at 0K. i.e. Lt Cp =0 ............ (7) T®0 From equation (5) it can be concluded that at absolute zero of temparature the products and reaction in solid sate have same value of entropies. So DS tends to approach zero at 0K. On analogy with zero value of heat capacity at absolute zero, the entropy value of all pure solids at 0K approach zero. i.e. Lt S = 0 ............ (8) T®0 This statement has led to following definition of Third law of thermodynamics. At absolute zero of temperature the entropy of every substance may become zero and it does become zero in case of a perfectly crystalline solid.

THERMODYNAMICS 347 In a crystalline solid the molecules, ions or atoms constituting the solid acquire definite positions in the crystal lattice. The arrangement has lowest energy. There is zero disorder and hence associated with zero value of entropy. CHAPTER (8) AT A GLANCE 1. Thermodynamics deals with energy changes accompanying all physical and chemical processes. 2. Internal energy : The total energy stored in the substance by virtue of its chemical nature. 3. First law of thermodynamics (Law of conservation of energy) Energy can neither be created nor destroyed. 4. Heat capacity : Amount of heat required to raise the temperature of the system from lower to higher temperature divided by the temperature difference. Specific heat : When the mass of the system is one gram, heat capacity is called specific heat of the system. Molar heat capacity : When the mass of the system is one mole, the heat capacity is called molar heat capacity. 5. CP – CV = R 6. Relationship between Internal energy and Enthalpy DH = DU + PDV 7. Enthalpy of reaction : Quantity of heat evolved or absorbed when the number of gram molecules of the substances taking part in the chemical reaction have completely reacted. Sign of DH : (a) –ve for Exothermic reaction. (b) +ve for Endothermic reaction. 8. Standard enthalpy of reaction (DH0) : When the process is carried out at 298K and 1 atm. pressure, DH is referred to as standard enthalpy of reaction DH0. 9. DU and DH can be measured in water calorimeter and Bomb calorimeter. 10. Thermochemistry is a branch of physical chemistry which deals with heat changes accompanying chemical reactions. 11. QP = QV + DnRT 12. Standard enthalpy of reaction (DH0) : When the process is carried out at 298K and 1 atm. pressure, DH is referred to as standard enthalpy of reaction DH0. 13. Enthalpy of formation (DH¦) : Quantity of heat evolved or absorbed when 1 mole of the substance is formed from its constituent elements under given conditions of temperature and pressure. 14. Enthalpy of combustion : Enthalpy change accompanying the complete combustion of 1 mole of the substance at a given temperature. 15. Enthalpy of neutralisation : Enthalpy change when one gram equivalent of an acid is neutralised by one gram equivalent of a base in a fairly dilute solution.

348 +2 CHEMISTRY (VOL. - I) 16. Enthalpy of solution : Quantity of heat evolved or absorbed when I mole of the solute is dissolved in a large excess of water so that any further dilution of the solution produces no heat change. 17. Enthalpy of fusion : The enthalpy change accompanying the change of one mole of substance from solid phase to liquid phase at its melting point. 18. Enthalpy of vaporisation : The enthalpy change accompanying the change of one mole of the substance from liquid phase to vapour phase at its boiling point. 19. Enthalpy of sublimation : The enthalpy change associated with the change of 1 mole of the substance from the solid phase to vapour phase at a given temperature below its melting point. 20. Hess's law : If a chemical change takes place in two or more different ways whether in one step or in several steps, the amount of heat evolved or absorbed in total change is the same, no matter, by whichever method the change is brought about. 21. Bond Energy : The average amount of energy required to break one type of bond present in one mole of the compound. 22. Change in entropy : DS= q T 23. Entropy is a measure of disorder or randomness of a system. 24. Free Energy (G) : The amount of energy available from a system that can be put to useful work. 25. DG = DH - T D S 26. (a) DG = –ve : Process is spontaneous. (b) DG = + ve : Process is not thermodynamically feasible. (c) DG = 0, The system is at equilibrium. 27. Enthalpy of atomisation Enthalpy change associated with conversion of compounds and elements into free atoms. This is usually highly endothermic. 28. Enthalpy of dilution The enthalpy change when some moles of solvent is added to an aqueous solution. 29. Enthalpy of ionisation The enthalpy change by which the enthalpy of neutralisation is less than that in case of neutralisation of strong acid with a strong base. This is the case where either the acid is weak or the base is weak. 30. Second law of Thermodynamics It is impossible to convert heat completely to an equivalent amount of work without producing any change in the system or surrounding. 31. Third law of Thermodynamics At absolute zero temperature the entropy of every substance may become zero and it does become zero in case of a perfectly crystalline solid. 32. DG° = DH° – TDS° = – RTlnK.

THERMODYNAMICS 349 QUESTIONS Very short answer type (1 mark each) 1. In which of the following changes there is increase in entropy ? (i) SO2(g) + 1 O 2(g) ® SO3(g) 2 (ii) Ice ® water 2. In which reaction the reactants have less energy than the products ? 3. What is the relationship between enthalpy (DH) and internal energy (E)? 4. In the endothermic reaction DH is ___________. 5. Define enthalpy of combustion. 6. In an exothermic reaction, DH is __________. 7. The enthalpy of combustion is always ____________ 8. What is exothermic reaction ? 9. In an endothermic reaction, DH is always _________. 10. In the reaction CO(g) + 12O2(g) = CO2(g) at constant T and P which one of the following is correct ? (i) DH = DE (ii) DH > DE (iii) DH < DE 11. Define Hess's law. 12. In which type of reaction heat is absorbed ? 13. What is heat of reaction ? 14. The dissociation energy of CH4 is 360 kcals/mole. What is the energy associated with C-H bond ? 15. What is bond energy ? 16. Define heat of combustion. 17. Write the equation relating free energy, entropy and enthalpy. 18. When ice melts to water, its entropy _______. Short Answer type ( 2 marks each) 1. What is understood by exothermic and endothermic reaction ? Give an example of each. 2. Explain the term Bond energy 3. Define Hess's law. 4. What is endothermic reaction. Give example. 5. When a reaction at constant temperature and pressure is at equilibrium the value of DG is __________ .

350 +2 CHEMISTRY (VOL. - I) 6. Define enthalpy of combustion. Give an example. 7. Predict whether DS is +ve or –ve for the following reactions. (i) 2H2(g) + O2(g) ® 2H2O(g) (ii) CaCO3(s) ® CaO(s) + CO2(g) 8. If DH of the reaction CH4(g) + C2H4(g) ® C3H8(g) is - 19.4 kcals, what will be the DH of the reaction C3H8(g) ® CH4(g) + C2H4(g) ? 9. How does the free energy change show the feasibility of a chemical reaction ? 10. What is the free energy of a reaction at 27°C with an enthalpy change 5 Kcals mole-1 and entropy change of 15 eu? 11. Calculate the entropy of vaporisation for a liquid boiling at 300K having enthalpy of vaporisation 27kJ mole–1. 12. The dissociation energy of CH2Cl2 is 354.54kcal mol-1. What is the energy associated with C-H bond if bond energy of C-Cl is 77.99Kcal mol-1. 13. Define enthalpy. What is the SI Unit of enthalpy. Short Answer type (3 marks each) 1. What is enthalpy of ionisation. Discuss with suitable examples. 2. Discuss enthalpy of atomisation with example. 3. What is enthalpy of dilustion. Discuss with example. 4. Derive the expression DH° = – RT lnK Long Answer type (7 marks each) 1. Write a note on Hess's law. 2. Write a note on – Activation energy. 3. Write short notes on (a) Heat of reaction (b) Heat of fusion 4. Define enthalpy of vaporisation. State Hess's law. Mention two conditions under which enthalpy change (DH) may be equal to internal energy change (DE) 5. What are the limitations of 1st law of Thermodynamics? Discuss the scope of 2nd law and write the various forms of statement of 2nd law of thermodynamics. 6. Give a brief account of 3rd law of thermodynamics. ADDITIONAL QUESTIONS Very short Answer type (1 mark each) 1. Define heat of formation 2. The value of DH for combustion of 1 mole of ethanol when both reactants and products are at the same temperature is ________________ 3. What is the heat evolved or absorbed when a compound is formed from the elements called ?

THERMODYNAMICS 351 4. At constant volume q equals to __________ 5. Thermodynamically, adsorption is an _________ phenomenon. 6. Define internal energy. 7. Entropy of a gas is ________ than that of a solid. 8. How DH and DE are related ? 9. Entropy is a measure of ______ of the system. 10. Define heat of fusion. 11. What is bond energy? 12. Predict whether DS is +ve or -ve for the following reactions : i) N2O (g) ® N2O3 (g) + O2 (g) ii) NH3 (g) + HCl (g) ® NH4Cl(s) 13. Which of the following is an intensive property? (mass, energy, temperature) 14. Write the equation relating free energy, entropy and enthalpy. Short answer type (2 marks each) 1. What are the components of internal energy ? 2. Entropy of a pure substance is zero at zero degree Kelvin, Explain why ? 3. State two important conditions which account for the spontaneity of a reaction 4. Explain the term – bond energy. 5. What is heat of neutralisation. 6. Explain law of conservation of energy. 7. What are extensive and intensive properties. Give suitable example. 8. What are the factors on which internal energy 'E' depends ? 9. How DH and DE are related in case of reaction 2SO2(g)+O2(g) ® 2SO3(g) 10. What is standard enthalpy of reaction ? 11. Discuss the importance of Heat of combustion. 12. Heat of neutralisation of a weak acid with a strong base is less than 57.1 kJ -Justify. 13. Explain how resonance energy affect bond energy. 14. What is meant by standard free energy change ? 15. Discuss the effect of temperature on DG. Long answer type (7 marks each) 1. Explain exothermic and endothermic reactions with suitable examples. 2. What do the symbols DH, DS and DG signify ? How are they interrelated ? What is the importance of the relation formed ?

352 +2 CHEMISTRY (VOL. - I) 3. Define Gibb's free energy change. How is the change in free energy related to spontaneity ? For the reaction, M2O(s) ® 2M(s) + 1 O2(g) 2 D H = 30 kJ mole–1 and DS = 0.07 kJ K–1 (at 1 atm) Calculate upto which temperature the reaction would not be spontaneous. 4. What is bond energy ? How it is calculated ? Discuss the various factors that effect bond energy. 5. (a) What is meant by heat of formation ? (b) State Hess's law of constant heat summation by taking suitable examples. 6. Give reasons for the following : (a) Entropy of a system increases with increase in temperature. (b) DG is always –ve for a spontaneous reaction. (c) A strong acid with strong base.Enthalpy of neutralisation of a week acid with a strong base is less than that of a strong acid with strong base. (d) In exothermic reaction DH is –ve (e) DG is zero at equlibrium. 7. What do you understand by (i) Heat of solution (ii) Heat of neutralisation (iii) Heat of combustion (1984 Bihar Intermediate) 8. What is entropy ? What is its unit ? Discuss the physical significance of entropy. Explain why the entropy of ice is less that of water. 9. Define entropy and free energy of a system. Predict the feasibility of a reaction when, (a) both DH and DS increase (b) both DH and DS decrease. (c) DH decreases but DH increases. 10. Write notes on (any three) (a) Enthalpy of formation (b) Enthalpy of combustion (c) Bond energy (d) Enthalpy of neutralisation. 11. State and explain Hess's law of constant heat summation. Discuss its application. Calculate the enthalpy of formation of methane. Given : a) C (graphite) +O2 (g) ® CO2 (g) DHa = -393.5 kJ b) 2H2 (g) + O2 (g) ® 2H2O (l) D Hb = -571.8 kJ c) CH4 (g) + 2O2 (g) ® CO2 (g) + 2H2O (g) D Hc = -890.3 kJ

THERMODYNAMICS 353 NUMERICAL PROBLEMS (UNSOLVED) 1. Calculate the enthalpy of following reaction : CH2 = CH2(g) + H2(g) ® CH3 - CH3(g) (Ans : 154 kJ/mole) The bond energies of C - H, C - C, C = C, and H - H are 416, 356, 598 and 436 kJ/mol respectively. 2. Find out the heat of reaction for the following : N2H4(l) + 2H2O2(I) ® N2(g) + 4H2O at 250C Given at 250C, heat of formation of N2H4(l) = 12 kcal / mole (Ans : - 150 kcals) H2O2(l) = -46 kcal/mole H2O(l) = -57.5 kcal /mole 3. Calculate DH for reaction CH3 - CH = CH2 + Cl2 ® CH3 - CH - CH2 Cl Cl From the following bond energies : EC-C = 347.3, EC=C = 615 EC-Cl = 330.5, ECl-C = 242.7 kJ /mole 4. Standard heat of formation at 298 K for CCl4(g) H2O(g) , CO2(g) and HCl(g) are 25.5, -57.8, -94.1,22.1 kcal/mole respectively. Calculate DH for the reaction CCl4 + 2H2O ® CO2(g) + 4HCl(g) Ans.(-41.9kcal) 5. Free energy change DG for the reaction Ag2 O(s) ® 2Ag(s) + 1 O2(g) 2 is zero DS and DH for this reaction are 0.066 kJ mol-1 K-1 and 31.218 kJ mol-1 respectively. Find out the direction of the reaction below 2000C (Ans : DG < 0, reaction is spontaneous) 6. The molar heats of combustion of C2H2(g), C(graphite) and H2(g) are 310.62 kcals, 94.05 kcal and 68.32 kcal respectively. Calculate the standard heat of formation of C2H2(g). (Ans : 54.20kcals) 7. Determine the enthalpy of the reaction C3H8(g) + H2(g) ® C2H6(g) + CH4(g) at 250C, using the give heat of combustion values under standard conditions. Compound H2(g) CH4(g) C2H6(g) C(graphite) -890.0 -1560.0 -393.5 DH0(kJ/mol) -285.8 The standard heat of formation of C3H8(g) is -103.8 kJ/mol (Ans : -55.7 kJ/mol)

354 +2 CHEMISTRY (VOL. - I) 8. The standard molar heats of formation ethane, carbondioxide and liquid water are -21.1, -94.1, -8.3 kcal respectively. Calculate the standard molar heat of combustion of ethane. (Ans = -372 kcal/mol) 9. Calculate the standard heat of formation of Carbon disulphide(l). Given that the standard heat of combustion of C(s) , S(s) and CS2(l) are -393.3, -293.72 and -1108.76 kJ mole-1 respectively. (Ans : 128.02 kJ mol-1 ) 10. Calculate the difference between heat of reaction at constant pressure and constant valume for the reaction at 250 C in kJ. 2C6H6(l) + 15O2(g) ® 12CO2(g) + 6H2O(l) (Ans : -7.4 kJ) 11. The standard heat of formation of CH4(g), CO2(g) and H2O(g) are -76.2, -394.8 and -241.6 kJ mole-1 respectively. Calculate the amount of heat evolved by burning 1m3 of methane measured under normal conditions. (Ans = 35794.64 kJ) 12. Calculate the resonance energy of N2O from the following data. DHo¦ of N2O = 82kJmol-1 Bond energies of N º N, N=N, O=O and N= O bonds are 946,418, 498 and 607 kJ mol-1 respectively. (Ans = 88 kJ mol-1) 13. When 2 moles of C2H6 are completely burnt 3129 kJ of heat is liberated. Calculate the heat of formation DHf for C2H6. Given DH¦ for CO2 and H2O are -395 and -286 kJ respectively. ( Ans = -83.5 kJ) 14. The molar heat of formation of NH4NO3 is 367.54 kJ and that of N2O(g) and H2 O(l) are 81.46 kJ and -285.75 kJ respectively at 250C and 1 atm. pressure. Calculate DH and DE for the reaction ®NH4NO3(s) N2O(g) + 2H2O(I) (Ans. - 122.56 kJ and -125.037kJ) (1986 Roorkee) 15. The bond dissociation energies of gases H2, Cl2 and HCl are 104,58 and 103 kcal/mole respectively. Calculate the enthalpy of formation of HCl gas. (Ans.-22kcals) (1985 IIT) MULTIPLE CHOICE QUESTIONS 1. A well stoppered thermosflask contains some ice Cubes. This is an example of a (a) Closed system (b) open system (c) isolated system (d) non-thermodynamic system 2. For the reaction N2 + 3H2 = 2NH3, DH = ? (a) DE = 2RT (b) DE - 2RT (c) DE +RT (d) DE + 2RT. 3. For an ideal gas, the relation between the enthalpy change and internal energy at constant temperature is given by (a) DH = DE +PV (b) DH = DE + DnRT (c) DH = DE + PDV (d) DH = DG + TDS

THERMODYNAMICS 355 4. If C(s) + 1OO22(g(g) ) ® CO2(g) , DH = X and CO(g) + ® CO2(g), DH = Y then the heat of formation of CO is (a) X + Y (b) X - Y (c) Y - X (d) XY 5. Enthalpies of elements in their standard states are taken as zero. Hence the enthalpy of formation of a compound (a) should always be negative. (b) should always be positive. (c) will be equal to twice the energy of combustion. (d) may be +ve or -ve 6. Which of the following values of heat of formation indicates that the product is least stable ? (a) -94 kcal, (b) -231.6 kcals (c) +21.4 kcal (d) +64.8 kcal. 7. For an adiabatic process, which of the following is correct (a) P D V = 0, (b) q = + w, (c) DE = q, (d) q = 0 8. Idenify the intensive quantity from the following. (a) Enthalpy and temperature (b) Volume and temperature (c) Enthalpy and volume (d) Temperature and refractive index 9. An exothermic reaction is one which (a) takes place on heating (b) is accompanied by flame (c) is accompanied by absorption of heat (d) is accompanied by evolution of heat. 10. An endothermic reaction is one in which (a) Heat is converted into electricity (b) Heat is absorbed (c) Heat is evolved (d) Heat is converted to mechanical work 11. Which of the following reactions is exothermic ? (a) CaCO3 ® CaO + CO2 (b) Fe + S ® FeS (c) NaOH + HCl ® NaCl + H2O (d) CH4 + 2O2 ® CO2 +2H2O

356 +2 CHEMISTRY (VOL. - I) 12 Which of the following is an endothermic reaction. (a) 2H2 +O2 ® 2H2O (b) N2 +O2 ® 2NO (c) 2NaOH + H2SO4 ® Na2SO4 + 2H2O (d) C2H5OH +3O2 ® 2CO2 + 3H2O 13. In exothermic reactions (a) DE is zero (b) DH is -ve (c) DS is zero (d) DH is +ve. 14. For an endothermic reaction (a) DH is -ve (b) DH is +ve (c) DE is -ve (d) DH = 0 15. Enthalpy for the reaction. C + O2 ® CO2 is (a) + ve (b) -ve (c) zero (d) none. 16. Enthalpy of a compound is equal to its. (a) Heat of combustion. (b) Heat of formation. (c) Heat of reaction. (d) Heat of solution. 17. During isothermal expansion of an ideal gas its. (a) Internal energy increases (b) Enthalpy increases, (c) Enthalpy remains unaffected (d) Enthalpy reduces to zero. 18. Heat of neutralisation of a strong acid and strong base is always. (a) 13.7 kcal/ mole. (b) 9.6 kcal/ mole (c) 6 kcal/ mole, (d) 11.4 kcal / mole. 19. The mutual heat of neutralisation of 40 grams NaOH and 60 grams CH3COOH will be (a) 56.1 kJ (b) Less than 56.1 kJ (c) More than 56.1 kJ (d) 13.7 kJ. 20. In which of the following neutralisation reactions, the heat of neutralisation is the highest. (a) NH4OH and H2SO4 (b) HCl and NaOH (c) CH3COOH and KOH (d) CH3COOH and NH4OH. 21. Hess's law deals with (a) Change in heat of reaction (b) Rate of reaction (c) Equilibrium constant (d) Influence of pressure on volume of a gas.

THERMODYNAMICS 357 22. Hess's law of heat of summation includes (a) Initial reactants only, (b) Initial reactants and final products (c) Final products only (d) Intermediates only. 23. H2(g) + I2(g) ® 2HI(g), DH = -12.40 kcals, Heat of formation of HI will be. (a) 12.4 kcals (b) -12.4 kcals (c) -6.20 kcals (d) 6.20 kcals 24. Combustion of methane (a) is an exothermic process. (b) is an endothermic process. (c) requires catalyst (d) gives H2 25. For which reaction, S will be maximum ? 1 (a) Ca(s) + 2 O2 (g) ® CaO(s) (b) CaCO3(s) ® CaO(s) + CO2(g) (c) C(s) + O2(g) ® CO2(g) (d) N2(g) + O2(g) ® 2NO(g) 26. The total entropy change for a system and its surroundings increases, if the process is (a) reversible (b) irreversible (c) exothermic (d) endothermic 27. When ice melts into water, entropy (a) becomes zero (b) decreases (c) increases (d) remains same. 28. Heat of combustion is always (a) +ve (b) -ve (c) neutral (d) all of the above 29. Energy required to dissociate 4 gms of gaseous hydrogen into free gaseous atoms is 208 kcals. at 250 C. The bond energy of H - H bond will be (a) 104 kcals (b) 10.4 kcals (c) 1040 kcal (d) 104 cals 30. If a refrigerator door is opened then we get (a) room heated (b) room cooled (c) more heat is passed out (d) no effect on room. ANSWER TO MULTIPLE CHOICE QUESTIONS 1.c 6.d 11.b 16.b 21.a 26.b 2.b 7.d 12.b 17.c 22.b 27.c 3.b 8.d 13.b 18.a 23.c 28.b 4. b 9.d 14.b 19.b 24.a 29.a 5.d 10.b 15.b 20.b 25.b 30.a (N.B : E has been replaced by U = Internal energy) qqq

358 +2 CHEMISTRY (VOL. - I) UNIT – VII CHAPTER - 9 EQUILIBRIA 9.1 IRREVERSIBLE AND REVERSIBLE REACTIONS Irreversible reaction : The chemical reactions in which the products formed do not react amongst themselves to produce the original reactants are known as irreversible reactions. Irreversible reactions can otherwise be defined as the chemical reactions, where the reactants are completely changed to the products. Examples : When AgNO3 solution is added to aqueous NaCl solution, a white percipitate is (i) formed due to the formation of AgCl. AgNO3(aq) + NaCl(aq) ® AgCl(s) ¯ + NaCl(aq) (ii) Na2SO4(aq) + BaCl2(aq) ® BaSO4(s) ¯ + 2NaCl(aq) (iii) Zn(s) + dil. H2SO4 ® ZnSO4(aq) + H2(g) - In these reactions, the reactants react completely and the reactions proceed only in forward direction. Backward reactions in these cases are not possible. So, it can be said that, (a) All irreversible reactions can be carried to completion. and (b) They proceed in one direction, i.e. they are unidirectional. Reversible reaction : The chemical reactions in which the products of the reactions can also react with one another under suitable conditions to produce back the original reactants are known as reversible reactions. So, (a) Reversible reactions can never be carried to completion. and (b) They occur in both the directions i.e. forward and backward directions.

EQUILIBRIA 359 Examples : (i) If powdered hot iron is taken in a tube and steam is passed over it, magnetic oxide of iron, Fe3O4 is formed with the liberation of hydrogen gas. 3Fe + 4H2O ® Fe3O4 + 4H2 Again, Fe3O4 can react with H2 to give back iron and water. Fe3O4 + 4H2 ® 3Fe + 4H2O Such reactions, which can take place in both directions under the same conditions are known as reversible reactions. These reactions are represented by the sign of reversibility between the reactants and the products. Thus, 3Fe + 4H2O Fe3O4 + 4H2 Other examples of reversible reactions are, (i) N2(g) + 3H2(g) 2NH3(g) (ii) H2(g) + I2(g) 2HI(g) (iii) PCl5(g) PCl3(g) + Cl2(g) (iv) CH3COOH(l) + C2H5OH(l) CH3COOC2H5(l) +H2O(l) Characteristics of reversible reactions : The main characteristics of reversible reactions are, (i) They never proceed to completion inside a closed vessel. (ii) They can take place in both forward and backward directions. and (iii) The rate of reaction in a particular direction depends upon the molar concentrations of the reacting species. 9.2 EQUILIBRIA IN PHYSICAL PROCESS Phase transformation processes are the most familiar examples. Solid liquid Liquid gas Solid gas Solid liquid equilibrium : Take a perfectly insulated thermos flask; place some ice and water in it at 273K and atm pressure. Both are solid to be in equilibrium with each other. H2O(s) H2O(l). There is no change in mass of ice and since the rate of transfer of molecules from ice into water and of reverse transfer from water into ice are equal at atm. press. and 273K. For any pure substances at atm. pressure the temparature at which the solid and liquid phases are at equilibrium

360 +2 CHEMISTRY (VOL. - I) is called the normal melting point or normal freezing point of the substance. The system is in dynamic equilibrium. Both the opposing process occur simultaneously at the same rate so that the amount of ice and water remains constant. Liquid vapour equilibrium Consider the equilibrium H2O(l) H2O (vap) Here, rate of evaporation = rate of condensation. Liquid water and water vapour are in equilibrium with each other at atmospheric presure and at 100OC in a vessel. For a pure liquid the temperature at which the liquid and its vapour are at equilibrium with each other is called boiling point of the liquid. Solid vapour equilibrium Consider the equilibrium I2(solid) I2 (vapour) NH4Cl (solid) NH4Cl (vapour) Camphor (solid) Camphor (vap) In the first case solid iodine sublimes to give I2 vapour and I2 vapour condenses to give solid I2. CHEMICAL EQUILIBRIA 9.3 STATE OF EQUILIBRIUM It is defined as the state of a system in which its properties do not change with time. The equilibrium with reference to change in physical state is called physical equilibrium. Example : If a small quantity of water is introduced into an evacuated vessel, then evaporation will take place and inside pressure will increase. After some time constant pressure is indicated which proves that the system has attained equilibrium. When equilibrium is attained in a chemical system, it is referred to as chemical equilibrium. Example : When gaseous H2reacts with I2 vapour, HI gas is produced and the reaction is reversible. H2(g) + I2(g) ® 2HI(g) 2HI(g) ® H2(g) + I2(g) If the first reaction is carried out in a closed vessel, it seems to stop after some time. Similarly, the second reaction (decomposition of HI) also is not complete inside a closed vessel and stops after a while. So, it can be said that in reversible reaction, both reactions apparently seem to come to a state of equilibrium after some time. At this stage, the properties of the system remain unchanged with time. This is known as the state of chemical equilibrium or equilibrium state and hence the reaction is represented as,

EQUILIBRIA 361 H2(g) + I2(g) 2HI(g) Types of chemical equilibrium : Chemical equilibrium is of two types, homogeneous and heterogeneous. (i) Homogeneous equilibrium : If all the reactants and products are in the same phase, then the equilibrium is known as the homogeneous equilibrium. A mixture of gases or miscible liquids gives rise to this type of equilibrium. Examples : (a) N2(g) + 3H2(g) 2NH3(g) (b) CH3COOH(l) + C2H5OH(l) CH3COOC2H5(l) + H2O(l) (ii) Heterogeneous equilibrium When the reactants and products are in different phases, then the equilibrium is known as the heterogeneous equilibrium. Examples : (a) CaCO3(s) CaO(s) + CO2(g) (b) 2CuO(s) 2Cu(s) + O2(g) (c) NH4HS(s) NH3(g) + H2S (g) Chemical equilibrium and Rate of reaction : Consider a general reversible reaction, A +B C+D Let the reactants A and B react in a closed vessel. At the beginning the concentrations of A and B are maximum and concentrations of C and D are zero. Since the rate of a reaction depends on the concentrations of the reactants, initially the rate of forward reaction will be high and the rate of backward reaction will be zero. With the progress of the reaction, the rate of forward reaction will decrease and that of backward reaction will increase. Finally, a stage will come, when the rates of both the reactions will be equal. The reaction comes to a stand still position. It appears as if the reaction has stopped, but the reaction has not actually stopped, it proceeds in both the directions with equal speed. This state is known as equilibrium state.

362 +2 CHEMISTRY (VOL. - I) Rate of Forward Reaction reaction Equilibrium point Backward Reaction Time ® Fig 9.1 Attainment of equilibrium state in reversible reaction. So, the state of chemical equilibrium can also be defined as that state at which the rates of the two opposing reactions are same and the concentrations of the reactants and the products do not change. Therefore, it can be called a dynamic chemical equilibrium. Characteristics of Chemical equilibrium : (i) The state of equilibrium is possible only when the reaction is carried out in a closed vessel. (ii) At equilibrium, the properties of the system like concentration, temperature and pressure remain unchanged. (iii) Since equilibrium can shift either in the direction of the reactants or in the direction of products, it is dynamic in nature. (iv) The same state of equilibrium can be attained from either side of a reversible reaction. (v) At equilibrium, both the reactions operate simultaneously and the rate of the forward reaction is the same as that of the backward reaction. (vi) A catalyst added to the system does not affect the state of equilibrium. 9.4 LAW OF MASS ACTION In the year 1864, two Norwegian chemists Guldberg and Waage suggested a qualitative relationship between rates of reaction and the concentrations of the reacting species. This relationship is known as Law of mass action. Statement : The law states that the rate at which a substance reacts is proportional to its molar concentration (active mass) and the rate of a chemical reaction is proportional to the product of the molar concentrations of reacting substances. Let us apply law of mass action to a hypothetical reaction, A + B ® Product (s)

EQUILIBRIA 363 According to the law of mass action, Rate of reaction, r µ [A] [B] or, r = k [A] [B] where, [A] and [B] are the molar concentrations of the reactants A and B respectively, k is a proportionality constant, called rate constant. For any general reaction, mA + n B ® Product (s) where m and n are the no. of moles of A and B respectively, Rate of reaction, r µ[A]m [B]n or, r = k [A]m [B]n Here the number of moles becomes the power to which the concentration terms are to be raised. Application of Law of mass action to reversible reactions : With the limitations of its validity to only one step chemical reactions, the law of mass action can also be applied to reversible chemical reactions. Consider a simple reversible reaction, k1 mA + nB pC + qD. k2 where p and q are the no. of moles of products C and D respectively. Let r1 and r2 be the rates of forward and backward reactions respectively, k1 and k2 are respective rate constants. Then, rate of forward reaction, r1 = k1[A]m [B]n (i)........................... and rate of backward reaction, r2 = k2 [C]p [D]q .(ii)....................... But, at the equilibrium point, the rate of forward reaction will be equal to the rate of backward reaction. This means, r1 = r2 Equating eqn (i) and (ii) we get, k1[A]m [B]n = k2[C]p [D]q [C]p[D]q or, k1 = [A]m[B]n k2 or, KC =[A[C]m]p[[BD]]nq ........................................... (iii) where, KC = k1 / k2 is a constant, known as equilibrium constant. Equation (iii) is known as law of chemical equilibrium.

364 +2 CHEMISTRY (VOL. - I) Equilibrium constant in terms of partial pressure : When all the reactants and products in a reaction are in gaseous form, their concentrations are proportional to their partial pressures. Consider the reaction, mA + nB pC + qD where the reactants and the products are in gaseous state. So, the concentrations of A, B, C and D are proportional to their partial pressures, pA, pB, pC and pD respectively. (pC)p (pD)q ..........................(iv) Applying law of mass action, we get, Kp = (pA)m(pB)n where, Kp is the equilibrium constant in terms of partial pressure. Equilibrium constant (Kx) If the concentrations of the reactants and products at the state of equilibrium are expressed in terms of mole fractions, the equilibrium constant is represented as Kx. Considering the general reaction at equilibrium, mA + nB pC + qD Kp= x p .x q ..................... (v) C y x m .x n A B where, xA, xB, xC and xD are the molefractions of A, B, C and D respectively and m, n, p & q are the no. of moles of A, B, C and D respectively. Relationship between KP, KC and KX (a) Relationship between KP and KX Consider a general reversible reaction mA + nB pC + qD where m,n,p,q be the number of moles of A,B,C and D respectively. Assuming all reactants and the products to be ideal gases, We can write KP = (pC )p× (pD )q (pA )m× (pB )n where pC, pD, pA and pB are the partial pressures of C, D, A and B respectively. In an ideal mixture, each component obeys Dalton’s law of partial pressure according to which partial pressure of individual gas is mole fraction times the total pressure (P) of equilibrium mixture. i.e. pi = xi. P where pi = partial pressure of the ith component in the mixture. xi = Mole fraction of ith component

EQUILIBRIA 365 P = Total pressure So, pC = xC P, pD = xDP, pA = xAP and pB = xBP (( )) (( )) (( )) (( ))Hence Kp = pC p. pD q = xCp p. xDp q = xCp. xDq P[(p+q) – (m+n)] pA m. pB n xAp m. xBp n x m . x n B A or, KP = KX.PDn(g).................................. (1) where Dn(g) = [(p+q) – (m + n)]g i.e. difference between the number of moles of gaseous products and reactants. (b) Relationship between KP and KC For an ideal gaseous mixture piV = niRT (Q PV = nRT) or pi = ni RT= Ci RT V where Ci is the molar concentration of the ith component in the mixture of total volume V. Thus, pA = CART, pB= CBRT, pC = CCRT and pD = CDRT KP = pC ppDq = (CCRT) p (C D RT ) q = CC P .CDq .RT [( p+ q) (m+ n)] p A m .p B n (C A RT) m .(CBRT)n CAm .CBn or, KP = KC RTDn(g)................. (2) From quation (1) and (2) KP = KxPDn(g) = KC(RT)Dn(g) When Dn(g) = 0, i.e. for the reactions which proceed with no change in the number of moles, KP = KX = KC Table : 9.1 Relationship between Kp, Kx & Kc in different equilibria Equilibrium Dn(g) KP = KC (RT) Dn(g) KP = KX (P)Dn(g) KP = KX 1. H2(g) + I2(g) 2HI(g) 0 KC = KC KX = PKP KX = P2KP 2. 2SO2(g) + O2(g) 2SO3(g) –1 KC = KpRT KP = KXP2 Kx = PKP 3. N2(g) + 3H2(g) 2NH3(g) –2 KP = Kp(RT)2 4. NH4HS(s) + NH3(g) + H2S(g) 2 KC = Kp(RT)2 5. NH3(g) + H2O(l) NH4OH(l) –1 KC= KPRT Reaction Quotient and its relationship with equilibrium Consider a general reversible reaction mA+nB pC+qD

366 +2 CHEMISTRY (VOL. - I) At each point in a reaction a ratio of concentration terms can be formulated. It is having the same form as the equilibrium constant expression. This ratio is called reaction quotient and is represented by ‘Q’ [C]p [D]q So, Q = [A]m.[B]n The following cases may arise. (i) If the reaction is in equilibrium, Q = KC (ii) The net reaction to proceed from left to right i.e. for the forward reaction to take place, Q < KC (iii) The net reaction to proceed from right to left i.e. for the backward reaction to take place, Q > KC Equilibrium constant expression for some reactions : While writing the expressions for equilibrium constant, the following points should be kept in mind. (i) Concentration of a pure solid or a liquid is taken as unity. (ii) Concentration of solvent is considered as constant, because it is present in large excess and its concentration does not change appreciably during the reaction. (a) For the reaction, NH3(aq) + H2O(l) NH+(aq) + OH-(aq) Rate of forward reaction, r1 = k1[NH3(aq)] [H2O(l)] and, rate of backward reaction, r2 = k2 [NH+(aq)] [OH- (aq)] We, know at equilibrium, r1 = r2 k2 [NH+4(aq)] Hence, k1 [NH3(aq) ] [H2O] = ] [OH- (aq)] [OH-aq)] [NH+(aq) or, Kc = k1 = k2 [NH3(aq)] [H2O] or, Kc = [NH+(aq)] [OH- (aq)] (Q [H2O] = 1) [NH3(aq)] (b) Reaction between gaseous hydrogen and iodine to form hydrogen iodide : H2(g) + I2(g) 2HI(g) Kc = [HI]2 [H2][I2]

EQUILIBRIA 367 (c) Formation of ammonia : N2(g) + 3H2(g) 2NH3(g) Kc = [NH3]2 [N2][H2]3 (d) Decomposition of BaCO3 : BaCO3(s) BaO(s) + CO2(g) Kc = [BaO] [CO2] [BaCO3] Since, [KBca=C[OC3O(s2)](g=)]1=apnCdO[B2 aO(s)] = 1, Characteristics of equilibrium constant (K) : The main characteristics of the equilibrium constant are as follows. (i) The equilibrium constant has a definite value for every reaction at a given temperature. However, it varies with change in temperature. (ii) KC does not depend upon the individual concentrations of the reacting substances. (iii) The value of equilibrium constant is not affected by adding a catalyst. (iv) The value of equilibrium constant for the forward reaction is reciprocal of that for the backward reaction. Example : H2 + I2 2HI and [HI]2 k1 = [H2] [I2] For forward reaction. For backward reaction k2 = [H2] [I2] [HI]2 Hence, k1 = 1 k2 (v) The value of K tells us the extent to which the forward or backward reaction has taken place. Greater value of Kc and Kp means that the reaction has proceeded to a greater extent in the forward direction. 9.5 LE - CHATELIER'S PRINCIPLE The principle was generalised by Le-Chatelier and Braun in 1884. It helps us to predict the effect of change in temperature, pressure and concentration on a system at equilibrium.

368 +2 CHEMISTRY (VOL. - I) Statement : If a system at equilibrium is subjected to a change in concentration, pressure or temperature, then the equilibrium shifts in that direction of reaction which tends to undo or reduce the effect of change. 1. Effect of change of concentration : If the concentration of any one of the substances (reactants and products) is increased at equilibrium, then the equilibrium shifts in the direction in which the added substance will be consumed. Example : Consider a reaction between gaseous H2 and I2 in a closed vessel at equilibrium. Hence, HK2(cg=) +[IH[2H(2gI])][2I2] 2HI(g) In the above reaction, if the concentration of H2 or I2 increases, then the equilibrium will shift in that direction in which the added H2 and I2 will be consumed forming HI. That means, the equilibrium will shift in the forward direction. Similarly, if the concentrations of HI increases, the reaction will proceed in the backward direction as in that direction the added HI can be consumed by forming H2 and I2. In both the cases the value of Kc remains constant. The effect of concentration can be illustrated by studying the reaction between FeCl3 and NH4SCN. [Fe(CNS)]2+ + NH4Cl FeCl3 + NH4SCN Yellow Colourless Red Colourless At equilibrium, if FeCl3 is added to the system, the equilibrium shifts in the forward direction and more [Fe(CNS)]2+ is formed. So the red colour deepens. But if NH4Cl is added, the equilibrium shifts in the backward direction and more FeCl3 is formed. So the red colour will gradually fade. Conclusions : i. Increase in concentration shifts the Forward direction. equilibrium to of any of the reactants ii. Increase in concentration shifts the Backward direction. of any of the products equilibrium to 2. Effect of change of pressure : If the system in equilibrium consists of gases, then the state of equilibrium is disturbed by the change of pressure. Three types of situations may arise. (a) When the number of moles in the reactant side in the gaseous state is more than that in the product side in an equilibrium reaction, the increase in pressure favours the formation of products. Consider the formation of ammonia molecules,

EQUILIBRIA 369 N2(g) + 3H2(g) 2NH3(g) 1 mole 3 moles 2 moles 4 moles By increasing pressure, the volume of the system will decrease considerably. The total number of moles per unit volume will now be more than before. The change can be counteracted if equilibrium shifts in that direction in which the total number of moles is decreased. This can take place by the combination of H2 and N2 molecules to produce NH3 molecule. Thus, higher the pressure, the greater would be the yield of ammonia. (b) When the number of moles in the reactant side is less than the number of moles on the product side in an equilibrium reaction, then the increase in pressure favours the backward reaction, N2O4(g) 2NO2(g) For such reactions in equilibrium high pressure favours the backward reaction and the equilibrium shifts to left. So, the above reaction is favoured by the decrease in pressure. (c) In an equilibrium reaction in which the number of moles on the reactant side is equal to those on the product side, then the affect in pressure does not change the state of equilibrium. Consider a reaction, H2(g) + I2(g) 2HI(g) The above reaction proceeds in either direction without any change in the number of moles. According to Le-Chatelier's principles, therefore, the pressure would have no effect on this equilibrium. Conclusions : i. Increase in pressure shifts the equilibrium to Less number of gaseous moles. in a direction of ii. Decrease in pressure shifts the equilibrium to large number of gaseous moles in a direction of 3. Effect of change of temperature : To find the effect of temperature, reactions are divided into two categories. (a) Exothermic reversible reaction. and (b) Endothermic reversible reaction. (a) Exothermic reversible reaction : In this type of reactions the heat is liberated in the forward direction, so it can also be said that, the heat is absorbed in the backward direction.

370 +2 CHEMISTRY (VOL. - I) Examples : N2(g) + 3H2(g) 2NH3(g)+ DH (DH = - 92 kJ) 2SO2(g) + O2(g) 2SO3(g) + DH (DH = - 192.5 kJ) In these reactions, DH amount of heat, is liberated in the forward direction. So, when the temperature is increased,the equilibrium shifts in that direction in which this extra heat will be absorbed, that is in the backward direction. Such reactions are, therefore, favoured by decrease in temperature. Thus, the formation of NH3 and SO3 are favoured by a decrease in temperature. (b) Endothermic reversible reactions : In this types of reactions, the heat is absorbed in the forward direction. So, it can be said that, the heat is liberated n the backward direction. Examples : N2(g) + O2(g) + DH 2NO(g) (DH = + 179.1 kJ) In this reaction DH amount of heat, is absorbed in the forward direction. So, when the temperature is increases, the equilibrium shifts in the forward direction, so that extra amount of heat is used up. These reactions are, therefore, favoured by increase in temperature. It must be remembered that when a reaction is exothermic in the forward direction, it is endothermic in the backward direction. Conclusion : (i) Increase in temperature shifts the Endothermic reaction. equilibrium in the direction (ii) Decrease in temperature shifts the Exothermic reaction equilibrium in the direction 4. Effect of catalyst : The addition of a catalyst increases the rates of the opposing reactions to the same extent. This hastens the approach of equilibrium. Thus, a catalyst has no effect on the state of equilibrium. Applications of Le-Chatelier's principle : 1. Manufacture of ammonia (Haber's Process) : Ammonia is manufactured by mixing one volume of nitrogen with three volumes of hydrogen at 5000C under 200-900 atmospheric pressure in presence of catalyst,. The best catalyst known is highly porous finely divided iron containing small amounts of promoters usually molybdenum or oxide of aluminium. N2(g) + 3H2(g) 2NH3(g) DH = -92kJ For this reaction, Kc = [[NN2H]3[]H2 2]3

EQUILIBRIA 371 (a) Efect of change of concentration : By increasing the concentration of N2 or H2 or both, the equilibrium shifts towards right, so as to keep the value of Kc constant. Hence, concentration of NH3 increases by increasing the molar concentration of H2 or N2 or both. (b) Effect of change of pressure : Four volumes of reactants give two volumes of product. Thus formation of ammonia is facilitated by decrease in volume. Hence increase of pressure shifts the equilibrium towards right and favours the formation of ammonia. (c) Effect of change of temperature : This reaction is an exothermic reaction. The increase in temperature will not favour the forward reaction. So, production of NH3 will decrease by increasing temperature. In this case, low temperature favours the formation of ammonia as the equilibrium shifts towards right. But in order to start the reaction, some amount of heat is necessary. Therefore, the reaction mixture is kept at 5000C and is never increased beyond it. This is called optimum temperature. Conclusion : The ammonia can be manufactured industrially by Haber's process at. (i) high pressure, (ii) low temperature and (iii) by increasing the concentration of reactants. 2. Manufacture of sulphuric acid by Contact process : For the manufacture of H2 SO4, SO2 is first oxidised to SO3 in presence of platinised asbestos. The reaction is exothermic. 2SO2 + O2 2SO3, DH = -192.5 kJ Here, 2 volumes of SO2 combine with one volume of oxygen to give two volumes of sulphur trioxide. In this, volume is decreased in the forward reaction. So, high pressure favours the forward reaction. As this is exothermic reaction, low temperature favours the forward reaction. High concentrations of SO2 and O2 also favour the forward reaction and give more yield of SO3. 3. Manufacture of nitric oxide : Nitrogen and oxygen are made to combine at 30000C. This high temperature is achieved by means of an electric spark. N2(g) + O2(g) 2NO(g) DH = +179.1 kJ This reaction is endothermic and takes place by absorption of heat. So, high temperature favours the forward reaction. One volume of nitrogen combines with one volume of oxygen to give two volumes of nitric oxide. Since there is no volume change, pressure has no effect on the equilibrium. 4. Dissociation of nitrogen tetroxide : Nitrogen tetroxide is a colourless gas. This on heating dissociates to nitrogen peroxide,which is a reddish brown gas. N2O4(g) 2NO2(g) DH = + 58.6 kJ

372 +2 CHEMISTRY (VOL. - I) This is an endothermic reaction. Hence increase in temperature favours the forward reaction giving more yield of NO2. One molecule of nitrogen tetroxide gives two molecules of nitrogen peroxide. So in this case volume increases in the forward reaction. So, low pressure will facilitate the forward reaction giving more yield of NO2. If the concentration of N2O4 is increased, the yield of NO2 will increase. Kc = [NO2]2 [N2O4] Thus, according to Le Chatelier's principle, the equilibrium shifts toward the right, in order to keep Kc constant. Thermal dissociation : It is a process in which the molecule of a substance breaks up into simpler molecules or atoms on heating. It is a reversible process. Example : NH4Cl dissociates on heating to give NH3 and HCl NH4Cl NH3 + HCl It is a reversible reaction. On cooling NH3 and HCl combine to give NH4Cl. Other examples are : (i) PCl5 PCl 3 + Cl2 (ii) CaCO3 CaO + CO2 Thermal decomposition : It is a process in which the molecule of a substance on heating breaks up into simpler molecules or atoms. It is not a reversible process. Example : Pb (NO3)2 decomposes into PbO, NO2 and O2 2Pb(NO3)2 D 2PbO + 4NO2 + O2 Distinction between dissociation and decomposition : Dissociation Decomposition i. It is a reversible process i. It is an irreversible process. ii. Reaction never goes to completion. ii. Reaction always goes to completion. iii. Le-Chatelier's principle is applicable iii. Le-Chatelier's principle is not applicable. Application of Law of mass action to different equilibria : 1. Homogenous equilibria : The equilibria in which both the reactants and the products are in the same phase are called homogenous equilibria. (a) Gaseous phase equilibria : This type of equilibria involves the following cases: case (1) When D n = 0.

EQUILIBRIA 373 For example, H2(g) + I2(g) 2HI(g) ab 0 (Initially) ( a-x) (b-x) 2x (at equilibrium) Let 'a' moles of H2(g) react with 'b' moles of I2 to give HI(g) and let 'x' moles of HI(g) is produced at equilibrium. Let the volume of the container be 'v' litre, Hence, [H2] = a-x [I2] = b-x V V and [HI] = 2x V Kc = [HI]2 = 4x2 .................................... (1) [H2][I2] (a - x) (b - x) Again, let 'P' be the pressure of the gas at equilibrium. Total moles at equilibrium = (a -x) + ( b-x) + 2x = a +b. Mole fraction of [H2 ] = a-x a+b Mole fraction of [I2] = b- x a+b and Mole fraction of [HI] = 2x a+b We know, partial pressure of a component = Total pressure ´ its mole fraction. pH2 = P ´ a - x , pI2 = P ´ b - x and p HI = P ´ 2x a + b a + b a+b \\ KP = (pHI)2 pH2 ´ pI2 = 4 x2 ............................. (2) (a - x) ( b-x) From eqn (1) and (2), we get, Kp and Kc are same for the formation of HI. Hence, when number of moles of the reactants and products are same, that is, when D n= 0, Kp = Kc. Case (ii) when Dn ¹ 0, For example, N2(g) + 3H2(g) 2NH3(g) ab o (Initially) (a - x) (b - 3x) 2x (at equilibrium) .Initially let 'a' moles of N2(g) react with 'b' moles of H2(g) to produce NH3(g) Let. x moles of NH3(g) is produced at equilibrium and the volume of the container be 'v' litre.

374 +2 CHEMISTRY (VOL. - I) \\[N2 ] = a - x , [H2 ] = b - 3x and [NH3 ] = 2x v v v [NH3 ]2 (2x / v)2 [N2 ][H2 ]3 a x b 3x 3 vv ( ) ( )\\ Kc= = or, Kc = 4x2. v2 ...................................... (3) (a -x) ( b - 3x)3 Let pressure at equilibrium is P. Total moles at equilibrium = a - x +b - 3x + 2x = a + b - 2x. pN2 =P ´ a-x a + b - 2x pH2 = P ´ b - 3x a + b - 2x and pNH3 = P ´ 2x a + b - 2x ( ( ) )\\ Kp = (pNH3)2 = P ´ a 2x - 2 +b 2x ( pN2) (pH2)3 P ( a- x) P(b- 3x) 3 a + b - 2x´ a + b - 2x or, Kp = 4x2 ( a + b - 2x)2 ....................... (4) P2 ( a-x) ( b -3x)3 From eqn. (3) and (4) we get, Kp ¹ Kc for the formation of NH3. Hence, when the Dn ¹ 0, Kp ¹ Kc Solved numerical problems : 1. The Kc for A2(g) + B2(g) 2AB(g) at 1000C is 50. If one litre flask containing one mole of A2 is connected with a two liter flask containing 2 moles of B2, how many moles of AB will be formed at 1000C (IIT., 1985). Solution : A2(g) + B2(g) 2AB(g) 1.0 2.0 0 (Inital moles) (1 - x) ( 2 - x) 2x (moles at equilibrium) The volume (v) becomes 3 liter on joining two containers. At equilibrium, [A2] = 1- x , [B2] = 2 -x and [AB] = 2x 3 3 3

EQUILIBRIA 375 Hence, according to law of mass action, ( 2x )2 Kc = [AB]2 = 3 x)= 4x2 50 = (1 -x) (2- 2 - 3x + x-2 [A2] [B2] 33 4x2 .................(1) or, 2 - 3x + x2 Solving eqn (i) we get, x = 0.93 and 2.326. Since the value of x should not be equal to or more than 2, the value of x will be 0.93. \\ Number of moles of AB = 2x = 2 ´ 0.93 = 1.86 2. Calculate the equilibrium constant for the reaction, CO2(g) + H2(g) H2O(g) + CO(g) at 1395K, if the equilibrium constant at 1395 K for the for the following are, 2H2O(g) 2H2(g) ++OO22(g(g),), k1 = 2.1 ´ 10 2CO2(g) 2CO(g) k2 = 1.4 ´ 10 Solution For the reaction, 2H2O(g) 2H2(g) + O2(g) k1 = [H2]2 [O2] ......................... (i) [H2O2]2 For the reaction 2kC2 O=2(g[)C[OC]O22[]O222C]O(g)..+....O...2.(.g..)................ (ii) For the reaction CO2(g) + H2(g) H2O(g) + CO(g) .............................. (iii) K= [H2O][CO] [CO2][H2] Dividing eqn (ii) by (i) we get, k2 = [CO]2 [O2] ´ [H2O]2 k1 [CO2]2 [H2]2[O2] or, k2 = [CO]2 [H2O]2 = K2 ............................. (iv) \\ K= k1 [CO2]2 [H2]2 k2 = 1.4 ´ 10-12 = 2.58 k1 2.1 ´ 10-13 3. For the reaction, A + B 2C, 2 moles of A and 3 moles of B are allowed to react. If the equilibrium constant is 4.0 at 4000. C, what will be the moles of C at equilibrium.

376 +2 CHEMISTRY (VOL. - I) Solution : A + B 2C 2 3 0 (Initial moles) ( 2 -x) (3 - x) 2x (moles at equilibrium) Let volume of the container be V litre. (2x/v)2 \\ Kc = [C]2 = [A] [B] 2-x 3-x v ´ v \\ 4.0 = 4x2 (2-x) (3-x) \\ x = 1.2 or, Moles of C at equilibrium = 2x = 2 ´ 1.2 = 2.4 4. For a reaction 2HI H2 + I2 at equilibrium 7.8g, 203.2g and 1638.4g of H2, I2 and HI respectively were found. Calculate Kc. Solution : 2HI H2 + I2 (moles at equlibrium) 1638.4 7.8 203.2 128 2 254 = 12.8 3.9 0.8 Let volume of the container be v litres \\ [H2] = 3.9 , [I2] = 0.8 and v v [HI] = 12.8 v ( )\\ Kc = [H2] [I2] = 3v.9 ´ 0v.8 = 0.019 [HI]2 12.8 2 ( )v 5. 0.5 moles of H2, 0.5 moles of I2 reacts in 10 litre flask at 4480C. The equilibrium constant (Kc) is 50 for H2 + I2 2HI (a) What is the value of Kp ? (b) Calculate moles of I2 at equilibrium. Solution : H2 + I2 2HI 0.5 0.5 0 (Initial moles) (0.5 - x) (0.5 - x) 2x (moles at equlibrium) (a) Since = 0, Kp = Kc = 50

EQUILIBRIA 377 (b) Kc = [HI]2 [H2] [I2] The volume of the flask is 10 litre. [H2]= 0.5 - x, [I2] = 0.5 - x 10 10 or, [HI] = 2x 10 2x 2 \\ Kc = 0.5 10 0.5 x = 4x 2 10 x× 10 (0.5 x)2 or, 4x2 = 50 (0.5 - x)2 or, 2x = 50 or, x = 0.39 0.5 - x Hence, moles of I2 at equilibrium. = 0.50 - 0.39 = 0.11 moles. 6. Kc for CO(g) + H2O(g) CO2(g) + H2(g) at 9860 C is 0.63 A mixture of 1 mole H2O(g) and 3 moles CO(g) is allowed to react to come to an equilibrium. The equilibrium pressure is 2.0 atm. (a) How many moles of H2 are present at equilibrium ? (b) Calculate partial pressure of each gas at equilibrium. Solution : CO(g) + H2O(g) CO2(g) + H2(g) 31 0 0 (Initial moles) (3 - x) (1 - x) x x (moles at equilibrium) Total moles at equilibrium = 3 - x + 1 - x + x + x = 4 (a) Kc = (3 x2 - x) = 0.63 - x)(1 or, x = 0.681 Thus, moles of H2 formed at equilibrium = 0.681. (b) Partial pressure of each gas = Pressure at equilibrium ´ mole fraction of gas.

378 +2 CHEMISTRY (VOL. - I) \\ pH2 = pCO2 = 2.0 ´ 0.681 ( Q mole fraction = total no of moles ) 4 moles at equlibrium = 0.34 atm pCO = 2.0 ´ (3-x) = 2(3 - 0.681) = 1.16 atm 4 4 pH2O = 2(1-x) = 2(1- 0.681) = 0.16 atm 4 4 7. A mixture of SO3, SO2 and O2 gases is maintained at equilibrium in 10 litre flask at a temperature at which Kc for the reaction 2SO2(g) + O2(g) 2SO3(g) is 100. At equilibrium (a) if number of moles of SO3 and SO2 in flask are same, how many moles of O2 are present. (b) if number of moles of SO3 in flask is twice the number of moles of SO2, how many moles of O2 are present ? Solution : 2SO2(g) + O2(g) 2SO3(g) Kc = [SO3]2 [O2] = 100 ..............(i) [SO2]2 (a) [SO3] = [ SO2] Putting these values in eqn (i), we get Kc = 1 = 100 [02] = 1 = 0.01 or, [O2] 100 Hence, moles of O2 = 0.01 volume or, Moles of O2 = 0.01 ´ 10 (\\ v = 10 litre) = 0.1. (b) If, [SO3] = 2[SO2] Putting these values in eqn(i) we get, K = 4 = 100 [O2] or, [O2] =1400 = 0.04 Hence, moles of O2 = 0.04 or, Moles of O2 = 0.04 ´ 10 = 0.4 volume 8. An equilibrium mixture at 300K contains N2O4 and NO2 at 0.28 and 1.1 atmosphere respec tively. If the volume of container is doubled, calculate the new equilibrium pressure of two gases.

EQUILIBRIA 379 Solution : N2O4 2NO2 0.28 1.1 (Partial pressure at equilibrium) If volume of theKcpo=nt(appiNNne2OrO2i4)s2d=ou0(b1.l2.e18d),2its=p4re.3s2suartemw. ill be reduced to half. According to Le-chateliers principle, by decreasing the pressure, the equilibrium will shift towards the side consisting of large number of moles. That means the decomposition of N2O4 will be favoured by decreasing pressure. Let P = Pressure used up for decomposition of N2O4. Thus, N2O4 2NO2 0.28 1.1 2 - P 2 + 2P (new pressure at equilibrium) (( ))=\\ K∴p Kp=121+ 2p 2= 4.32 atm. 0.28 P 2 or, P = 0.045 atm. \\ =pN02.1O44-at0n.0e4w5e=qu0i.l0ib9r5iuamtm. pNO2 at new equilibrium = 0.55 + 2 ´ 0.045 = 0.64atm. 9. For the reaction CO(g) + H2(g) CH3OH(g), H2 gas is introduced into a five litre flask at 3270C, containing 0.2 mole of CO(g) and a catalyst till pressure is 4.92 atomsphere. At this point 0.1 mole of CH3OH(g) is formed. Calculate Kc and KP . Solution : CO(g) + 2H2(g) CH3OH(g) 0.2 a 0 (initial moles) (0.2 - 0.1) (a - 0.2) 0.1 (moles at equlibrium) Total moles at equilibrium = 0.1 + a - 0.2 + 0.1 = a We know, the ideal gas eqn, PV = nRT or, n = PV/RT =0.40.89221´´5600 = 0.499 \\ The moles at equilibrium, a = 0.499 [CH3OH] = 0.1 [H2] = 0.499 - 0.2 = 0.299/5 5 5

380 +2 CHEMISTRY (VOL. - I) and [CO] = 0.1/5. \\ Kc =[C[COH]3[OHH2]]2 = 0.1/5 ´ 0.1/5 = 279.64 litre2 mole-2 (0.299/5)2 We know, Kp = Kc.(RT)Dn In the above reaction, D n = -2 \\ Kp =279.64 ´ (0.0821 ´ 6000) = 0.115 atm-2 10. At a certain temperature, equilibrium constant (Kc) is 16 for the reaction. SO2(g) + NO2(g) SO3(g) + NO(g) If we take one mole each of all the four gases in one litre container, what would be the equilibrium concentration of NO and NO2 gases ? (IIT, 1987) (Initial moles) Solution : SO2(g) + NO2(g) SO3(g) + NO(g) 1 1 11 (1 - x) (1 - x) 1 + x 1+ x (moles at equilibrium) Let 'v' be the volume of the container at equilibrium. \\ Kc = [SO3] [NO] ; Since v = 1 litre, Kc = (1- x) (1+x) [SO2][NO2] (1 - x) (1-x) or, 16 = (1+x)2 or 1+x =4 or, x = 0.6 (1 - x)2 1-x Thus, the equilibrium concentration of NO= 1+x = 1 + 0.6 = 1.6 moles/ litre v 1 The equilibrium concentration of NO2 = 1- x = 1- 0.6 = 0.4 mole /litre. v 1 2. Heterogeneous Equilibria If the reactants and the products are in different phases, the equilibrium is said to be heterogeneous. For example (i) CaCO3 (s) CaO(s) + CO2(g) (ii) NH4HS(s) NH3(g) + H2S(g) (i) Study of equilibria CaCO3(s) CaO(s) + CO2(g) Consider the equilibrium reaction Initial no. of moles CaCO3(s) CaO(s) + CO2(g) 1 00 Equilibrium no. of moles 1–x xx Total no. of moles at equilibrium = x (since no. of moles of solids are not considered)

EQUILIBRIA 381 Molefraction of CO2 = x =1 x Partial pressure = molefraction × total pressure =1×P=P KP = pCO = P (ii) Study of equilibria NH4HS(s) NH3(g) + H2S(g) Consider the equilibrium reaction Initial no. of moles NH4HS(s) NH3(g) + H2S(g) 1 00 Equilibrium no. of moles 1–x xx Total no. of moles at equilibrium = x + x = 2x Molefraction of NH3 = x = 1 2x 2 Partial pressure of NH3 = mole fraction × total pressure = P 2 Mole fraction of H2S = x = 1 2x 2 Partial pressure of H2S = P 2 KP = pNH3× pH2S = P × P = P2 2 2 4 IONIC EQUILIBRIA 9.6 THEORIES OF ACIDS AND BASES Introduction : Various theories have been put forward by different workers in order to explain the nature of acids the bases. These theories are based on the configuration or inner structure of acids and bases. These theories are, (i) Arrhenius theory (ii) Bronsted - Lowry theory (iii) Lewis theory.

382 +2 CHEMISTRY (VOL. - I) (i) Arrhenius theory : The important postulates of Arrhenius theory are, (a) Acids are those substances which yield H+ ions in the aqueous solution.Thus, HCl, H2SO4, are acids, because they contain replaceable HNO3, H3PO4, CH3 COOH, HCN and HOCl H+ ion in the aqueous state. Water HCl(g) Water H+(aq) + Cl-(aq) HNO3(l) H+(aq) + NO3-(aq) CH3COOH(l) Water H+(aq) + CH3COO-(aq) (b) Bases are those substances which provide OH- ions in the aqueous state. Thus,NaOH, (c) NKeOuHtr,aBliNNsaaaH(tOOi4oHOHn)H(o2sf,()INa)nwHwaa4actOtieedrHranOOedtHHca.--pb((araaoqqs)ve)+i+idsNeNbHOaas+4He(+da-(qaoi)qon)nthseinkethyereaaqcuteioonusbsettawteeeannHd a+rieobnassaensd. OH- ions to form undissociated water molecules. (d) During H+(aq) + OH-(aq) ® H2O(l) of an acid, H+ ions proceed to the cathode and electrolysis of an aqueous solution negative ions, to the anode. Thus, HCl(aq) ® H+(aq) + Cl-(aq) At cathode : 2H+(aq) + 2e- ® H2(g) - (reduction) At anode : 2Cl- - 2e- ® Cl2(g) - (oxidation) Limitations of Arrhenius theory : The limitations of Arrherius theory are as follows, (a) This theory explains the acidic and the basic characters of the substances in aqueous solution. It has been found that NaOH exists as Na+ ions OH- ions even in the solid state. This theory fails to explain such facts. (b) According to Arrhenius, an acid releases H+ ions in aqueous medium. H+ ion is proton, which can not exist independently in the aqueous state. It accepts a pair of electrons form water molecule to form hydrated proton or hydronium ion. Hence, H+ ions in acidic solution can not exist independently, but exist in combination with water molecules, hydronium ions. H+(aq) + H2O(l) ® H3O+ > +> H H+ + H__ O __ H .O. HCl H2O H3O+(aq) + Cl-(aq) HH Hydronium ion

EQUILIBRIA 383 Hence, HCI H2O H3O+(aq) + Cl-(aq) (c) There are number of substances which do not provide H+ ions and OH- ions, but still behave as acids and bases. For example, CO2 does not contain any hydrogen atom, but it acts as an acid. It reacts with NaOH to form salt. 2NaOH + CO2 ® Na2CO3 + H2O Base Acid Salt Water Similarly, NH3, although does not contain OH- ions, behaves as a base, beacuse, it reacts with HCl to form a salt. NH3 + HCl ® NH4Cl. (ii) Bronsted - Lowry theory (Proton transfer theory) : According to this theory, (a) Acid is a substance (molecule or ion) which has tendency to donate a proton to any other substance. (b) Base is a substance (molecule or ion) which has a tendency to accept a proton from any other substance. In other words, an acid is a proton donor and base is a proton acceptor. So, this theory is also known as proton transfer theory. since it involves the transfer of proton. Hence, acids are protogenic and bases are protophilic. Examples : Molecular acids : HF, HCl, H2SO4,H2O etc. H3O+, NH+4, HSO4-, HCO-3 etc. Ionic acids : Molecular bases : R-NH2, R2NH, R3N, NH3, H2O etc. OH-, S2-, CO32-, Cl-, NO-3, etc. Ionic bases : It is important to note that, no single substance is an acid or a base. A single substance can not donate a proton unless and until some other substance which accepts the proton is also present. For example, NH3 + HCl ® NH4+ + Cl- Here, HCl donate a proton to NH3 Thus, HCl is an acid and NH3 is a base. Other examples of acid-base reactions are given below. NH3 + H2O NH4+ + OH- CH3COOH + H2O CH3COO- + H3O+ HCl + H2O H3O+ + Cl- H2SO4 + H2O H3O+ + HSO4-

384 +2 CHEMISTRY (VOL. - I) From the above examples, we know that water is an acid with respect to NH3. But, it is a base with respect to CH3COOH or HCl or H2SO4. Such substances (like H2O, HCO3- , HSO4- etc.) which can act both as acids as well as bases are called amphoteric or amphiprotic substances. Conjugate acid-base pairs : When an acid loses a proton, the residual part of it will have a tendency to accept a proton. Thus, the residual part will behave as a base. Such pairs of substances which differ from one another by a proton are known as conjugate acid-base pairs. Consider a general example of an acid, HA. HA H+ + A- Acid Proton Conjugate base Consider the reaction between HCl and H2O H3O+ + Cl- HCl + H2O Acid Base Acid Base Here, HCl donates a proton to water in the forward reaction. Thus, HCl is an acid and H2O is a base. In the backward reaction, H3O+ ion donates its proton to Cl- ion. Thus, H3O+ ion is an acid and Cl- is a base. It is clear that, HCl loses a proton and forms Cl- ion, which is a base. HCl and Cl- ion are called conjugate acid - base pair. Other examples of conjugate acid - base pairs are, HNO3 + H2O H3O+ + NO3-1 Acid-I Base-II Acid-II Base-1 H3O+ + CH3COO- CH3COOH + H2O H2CO3 + H2O H3O+ HCO3- + Relative strength of acids and bases : According to proton transfer theory, (a) the strength of the acid depends upon the tendency to donate proton, and (b) strength of the base depends upon its tendency to accept a proton. For example, consider the reaction between acetic acid and water. CH3 COO- + H3O+ CH3 COOH + H2O weak acid strong base Acetic acid has a poor tendency to donate a proton. Therefore, it is a weak acid. The above CH3COO- ion must equilibrium lies tmooasctclyepttoawparrodtsonth.eHleenftc.eI,tCfHol3loCwOsO, t-h,earecfoonreju, gthataet have a strong tendency base of acetic acid is a strong base. Hence, weak acid H+ + strong conjugate base. Similarly, HCl is a strong acid in water. HCl + H2O H3O+ + Cl- The above equilibrium lies mostly towards the right. It follows, therefore, that Cl- ion must have a little tendency to accept a proton. Hence, Cl- ion is a weak conjugate base. H++ weak conjugate base Strong acid

EQUILIBRIA 385 Thus, we conclude that, every strong acid has a weak conjugate base and vice-versa. Important note : All Arrhenius acids are also Bronsted acids, but all Bronsted bases are not Arrhenius bases. Reason : According toArrhenius theory, base must furnish OH- ion in aqueous state. But, according to Bronsted, a base must accept a proton. So, CO32- ion is a base, as it accepts a proton, according to Bronsted theory. But CO32- is not a base according toArrhenius theory, as it does not furnish OH- ion in aqueous solution. CO32- + H+ ® HCO3- Limitations of Bronsted -Lowry theory : (a) This concept has a greater field of application to aqueous as well as non-aqueous solutions. (b) It fails to explain the acidic nature of oxides like, CO2, SO2 etc. and basic nature of ZnO, CaO etc. (c) It also fails to explain the acidic character of AlCl3, FeCl3 ,BF3 etc. (iii) Lewis Theory : G. N. Lewis (1923) gave new definitions of acids and bases, taking into account their electronic configurations. According to this theory, an acid is a substance (molecule or ion) which can accept a pair of electrons, while a base is any substance (molecule or ion) which can donate a pair of electrons. In short, acid is an electron pair acceptor and base is an electron pair donor. Classification of Lewis bases : (a) All simple anions are Lewis bases. For example, Cl-, CN-, CH3COO-, Cl-, Br-, NO3- , HSO4- (b) Neutral molecules having one or more lone pairs of electrons. For example, NH3, R-NH2, R2NH, R3N, H2O, R- OH, R- O -R etc.(R = any alkyl group) .. ..R - O - H + H+ ® [R - O - H]+ .. ¯ Lewis base Co-ordinated complex Classification of Lewis acids : (a) All simple cations are lewis acids. For example, Cu++, Ag+,Ca++ Fe+++ etc. (b) Molecules, whose central atoms have incomplete octet, also act as lewis acids. For example, BF3, AlCl3, ZnCl2, SO3, FeCl2 etc. (c) Molecules having multiple bonds between atoms of different electronegativities act as Lewis acids. For example CO2, SO2, SO3,etc. (d) Molecules having atoms which can accomodate more electrons in the vacant d-orbitals in the valency shell. For example SiCl4, SiF4, etc.

386 +2 CHEMISTRY (VOL. - I) (e) Element with six electrons in its valence shell also acts as Lewis acid. For example, SO3 ion is oxidised to thiosulphate ion by the acidic nature of sulphur. S + SO3 ƒ S2O3 (Lewis acid )  (Lewis base)  (Co-ordinated complex) Acid-base reactions : According to the orbital concept, (a) any substance having vacant orbital in valence shell can act as an acid. and (b) an atom with a lone-pair of electrons in its valence shell can act as a base. The examples of acid-base reactions are given below. (a) Action of BF3(acid) with NH3(base) to form co-ordinated complex, [F3B NH3] Here, BF3 behaves as an acid as boron atom is electron deficient and NH3 acts as a base, as there is a lone pair of electrons on nitrogen atom ..F.......x...xxFBF.........  H FH (Lewis acid) ....Nxxx. H ƒ F B N H N FH (Co-ordinated complex) (Lewis base) (b) Action of NH3 (base) on H (acid) to from NH4 ion  HH H H+  : N H ƒ H  N HH (Lewis acid) (Lewis base) (Co-ordinated complex) (c) Fe3 ion (Lewis acid) reacts with CN ion (Lewis base) Fe3  6CN ƒ CN C‚N CN 3- ‚‚ ‚‚ (Lewis acid) (Lewis base) Fe CN C‚N CN (Co-ordinated complex)

EQUILIBRIA 387 (d) Action of Ag+ (acid) with NH3 (base) Ag+ + 2NH3 ® [H3N ® Ag ¬ NH3]+ (Lewis acid) (Lewis base) di-ammine silver complex Limitations of Lewis concept : Lewis theory has the following limitations. (a) The relative strength of acids and bases can not be explained as it does not consider ionisation. (b) According to this theory an acid - base reaction is associated with the formation of a dative bond, but no such bond is formed when HCl reacts with NaOH. (c) Acid - base reactions are instantaneous, but dative bond formation is a slow process. (d) The catalytic properties of acids are not explained by Lewis theory. (e) It fails to explain the amphoteric nature of H2O, HCO3- ion etc. 9.7 IONISATION OF WEAK ACIDS AND BASES Ionisation of weak acid Consider the ionisation of weak monobasic acid HA in water. HA + H2O H3O+ + A– Applying law of chemical equilibrium [ ][ ]KC = [HH3AO]+[HA2O] .................. (1) where KC = Equilibrium constant in terms of molar concentration. In dilute solution since water is present in large excess, its concentration is taken as constant. i.e. [H2O] = k (say), Further, H3O+ indicates that H+ ion is hydrated and it may be replaced by H+, Thus equation (1) can be written as [H+ ][A ] KC = [HA] k ....................... (2) or, KC. k = [H+ ][A ] [HA] [ ][ ]or, Ka = H+ A .......................... (3) [HA] where KC.k = a new constant Ka (say)

388 +2 CHEMISTRY (VOL. - I) Ka is known as dissociation constant or ionisation constant of weak acid HA. Ka is constant at a constant temperature. It varies only with temperature. The Ka values of a few weak acids are given below (Table 9.2.) Table 9.2 Ka of some carboxylic acids at 250C Acid Value of Ka at 250C HCOOH 17.7 × 10–5 CH3COOH 1.75 × 10–5 ClCH2COOH 136 × 10–5 Cl2CHCOOH 5530 × 10–5 Cl3C.COOH 23200 × 10–5 CH3CH2CH2COOH 1.52 × 10–5 C6H5COOH 6.3 × 10–5 (p-NO2)C6H4 COOH 36 × 10–5 (p-Cl) C6H4COOH 10.3 × 10–5 (p-OH) C6H4COOH 2.6 × 10–5 Relative strengths of weak acids The ionisation constant of the weak acid may also be represented in terms of degree of ionisation and molar concentration. Consider the ionisation of acetic acid. CH3COOH CH3COO– + H+ Original conc. c 00 Equil. conc. c – ca ca ca where c = molar concentration and a = degree of ionisation. [ ][ ]Ka = C[HC3HC3OCOOOHH]+ = ca.ca = cc(12aa2 )= ca2 ................... (4) c ca 1a For a weak acid ‘a’ is very small and may be neglected in comparison to 1. So, Ka = ca2 or, a2 = K a and a= Ka ......................... (5) c c If two weak acids 1 and 2 have the ionisation constants Ka1 and Ka2 respectively and have same molar concentration ‘c’ from equation (5). a1 = Ka1 a2 Ka2

EQUILIBRIA 389 where a1 = degree of ionisation of acid 1 and a2 = degree of ionisation of acid 2. Further, degree of ionisation is a measure of strength of acid. Here, Strength of Acid 1 = a1 = K a1 Strength of Acid 2 a2 Ka2 Problem 1 : Formic acid and acetic acid have ionisation constant 17.7 × 10–5 and 1.75 × 10–5 respectively at 298K. Calculate the relative strength of two acids. Solution : Strength of Formic Acid = 17.7× 10 5 = 3.18 Strength of Acetic Acid 1.75× 10 5 Hence formic acid is 3.18 times stronger than acetic acid. Equation (5) may be used to calculate the hydrogen ion concentration ([H+]) of the weak acid. [H+] = ca = c Ka = Kac ................ (6) c Problem-2 : A weak monobasic acid has a dissociation constant equal to 1.6 ×10–5 at 250C. Calculate its degree of dissociation at a concentration of 0.1M at the same temperature. What will be the concentration of H+ ions furnished by it ? Solution : a= Ka = 1.6× 10 5 5 c 0.1 = 16× 10 So, the degree of dissociation of 0.1M acid = 0.00016 Again, [H+] = Ka.c= 1.6× 10 5× 0.1 = 0.0531 mol dm–3. Ionisation of Weak base The weak monoacidic base is represented by BOH. According to Arrhenius concept, BOH B+ + OH– [ ]The ionisation constant Kb =B+ [OH ] [BOH ] If c = Initial concentration of the base in moles per litre and a = degree of ionisation BOH B+ + OH– Original conc c oo Equilibrium conc c – ca ca ca

390 +2 CHEMISTRY (VOL. - I) [ ]( )Kb= B+ OH = ca.ca = c2a 2 ca 2 ............. (7) c ca 1a [BOH] c(1 a)= For a weak base ‘a’ is very small compared to 1. So, Kb = ca2 or, a = Kb c Also, [OH–] = ca = c Kb = c.K b ..................... (8) c Ionisation constants of some common weak bass are given in Table 9.3 below. Table 9.3 Ionisation constants of some common weak bases at 250C Bases Kb values at 250C 1.81×10–5 NH4OH 3.83×10–10 C6H5NH2 5.12×10–4 (CH3)2NH 4.38×10–4 CH3NH2 5.60×10–4 C2H5NH2 5.21×10–5 (CH3)3N 1.30×10–9 C5H5N 9.8 IONISATION OF POLYBASIC ACIDS Acids having more than one ionisable proton per molecule are known as polybasic acids. e.g. H2C2O4 (oxalic acid), H2SO4 (Sulphuric acid) H3PO4 (phosphoric acid) For a dibasic acid H2X the ionisation reactions are represented as (i) H2X (aq) H+ (aq) + HX– (aq) (ii) HX– (aq) H+ (aq) + X2– (aq) Equilibrium constant for (i) is [ ][ ]K1 = H+ HX- [H2X] Equilibrium constant for (ii) is [ [][ ] ]K2 = H+ HX2- HX-

EQUILIBRIA 391 where K1 and K2 are called 1st and 2nd ionisation constant of the acid H2X. The values of ionisation constants for some polybasic acids are given in the following table 9.4. Table 9.4 - Ionisation constant of some polybasic acid (298K) Acid K1 K2 K3 Oxalic acid 5.9 × 10–2 6.4 × 10–5 4.0 × 10–7 Ascorbic acid 7.4 × 10–4 1.6 × 10–12 4.2 × 10–13 Sulphuric acid Very large 1.2 × 10–2 Carbonic acid 4.3 × 10–7 5.6 × 10–11 Phosphoric acid 7.5 × 10–3 6.2 × 10–8 The value of K2, K3 are smaller than K1 in case of polybasic acids. The is because it is increasingly difficult to remove positively charged proton from a negative ion due to electrostatic forces. 9.9 FACTORS AFFECTING ACID STRENGTH We know that the extent of dissociation or ionisation of an acid HA depends upon the strength and polarity of HA bond. When strength of H–A bond decreases the energy required to break the bond decreases. So HA becomes a stronger acid. Also more the polarity of bond between H and A, more prominent will be the charge separation, easier will be the bond cleavage, thereby increasing acid strength. If we consider the elements in the same group of periodic table, the H–A bond strength is the important factor in deciding acidity as compared to bond polarity. On going down a group the size of A increases gradually thereby decreasing the H–A bond strength. So the acid strength increases. e.g. Size increases HF < HCl < HBr < HI Acid strength increases In case of Gr 16 elements H2S is a stronger acid as compared to H2O. If we consider elements in the same period of the periodic table, polarity of H–A bond becomes the deciding factor in determining the acid strength. As the electronegativity of A increases, the polarity of HA bond increases and acid strength increases. e.g. Electronagativity of A increases CH4 < NH3 < H2O < HF Acid strength increases

392 +2 CHEMISTRY (VOL. - I) 9.10 IONISATION OF WATER Pure water is a weak electrolyte, since it ionises to a very small extent. It has a low electrical conductivity. Pure water ionises as, H2O(l) H+(aq) +OH-(aq) Applying the law of mass action, we have the equilibrium constant, K= [H+] [OH-] ....................(1) [H2O] Since water dissociates or ionises to a very small extent, its concentration remains almost unchanged and can be taken as constant. Thus, equation (1) can be written as, K ´ [H2O] = [H+] [OH-] or, Kw = [H+] [OH-] ................(2) where, Kw is called the ionic product of water. The value of ionic product (KW) of water at 250C found to be 1.0 x 10-14 In pure water, the concentration of H+ ions and OH- ions are same. Hence, In pure water [H+] = [OH-] So, Kw = [H+] [OH-] = 1.0 ´ 10-14 Hence, in pure water, [H+] = [OH-]= 1.0 ´ 10-7 moles/litre Although Kw is a constant, it is temperature dependent. The numerical value of KW increases with increase in temperature. Table 9.5- Numerical value of Kw at various temperatures. Temperature Kw (in mol2 lit-2) 00C 0.11 ´ 10-14 10OC 0.30 ´ 10-14 250C 1.00 ´ 10-14 400C 3.00 ´ 10-14 1000C 7.5 ´ 10-14 The solution in which the concentration of H+ ions and OH- ions are same is called a neutral solution. But, the solution in which H+ ion concentration is more than 1.0 ´ 10-7 is called acidic solution and where it is less than 1.0 ´ 10-7 is called basic or alkaline solution. Since Kw is a constant at constant temperature, by increasing H+ ion concentration in a solution, the concentration of OH- ion will decrease.

EQUILIBRIA 393 Knowing H+ ion concentration of a solution, its OH- ion concentration can be calculated as follows. [OH-] = Kw [H+] Example : 1 Calculate the H+ion concentration in NaOH solution, 2 gms of which are dissolved in 2 litres solution. solution : Amount of NaOH in 2 litres solution = 2 gms. Amount of NaOH in 1 litres solution = 1 gm. As NaOH is a strong base, [NaOH] = [OH-] \\ [OH-] = 1.0 gm/litre = 1/40 40 \\ H+= Kw = 1.0 ´ 10-14 = 40 ´ 10-14 = 4.0 ´ 10-13M [OH-] 1 /40 9.11 HYDROGEN ION EXPONENT : pH Sorensen introduced a new term to express the hydrogen ion concentration. The term pH has been derived from the French word Puissanced hydrogen, meaning power of hydrogen. According to him, pH of a solution is defined as the negative logarithm of hydrogen ion concentration in moles per litre. Mathematically, pH = -log [H+] ....................(1) We know, Kw = [H+] [OH-] or, [H+] [OH-] = KW = 1.0 ´ 10-14 Taking logarithm, we get, log [H+] + log[OH-] = -14 or, -log [H+] - log[OH-] = 14

394 +2 CHEMISTRY (VOL. - I) or, pH + pOH = 14 .............................(2) In neutral solution [H+] = 1.0 ´ 10-7 Hence, the pH value of neutral solution (pure water) = -log (10-7) = 7.0 In acidic solution : In all acidic solutions H+ ion concentration is more than 1.0 ´10-7 M. H+ ion concentration in an acidic solution may be 10-6M, 10-5M, 10-4M etc. Consider an acidic solution whose H+ ion concentration = 10-6M. Its PH = -log [H+] = -log [10-6 ] = 6.0. Hence, pH value of all acidic solutions will be less than 7.0. In basic solution : In all basic solutions, the H+ ion concentration is less than 1.0 ´ 10-7M Hence their OH- ion concentration will be more than 1.0 ´ 10-7M. In order to keep KW value constant at a particular temperature, the increase in H+ ion concentration decreases the OH- ion concentration in a solution. Consider a solution whose OH- ion concentration is more than 1.0 ´ 10-7M, say 10-5M Hence, its H+ ion concentration will be 10-9M . \\ pH of the solution will be, - log [H+] = - log [10-9] = 9.0 Thus, if [H+] > 10-7M solution in acidic, pH < 7.0 [H+] < 10-7M solution is basic, pH > 7.0 [H+] = 10-7M solution is neutral, pH = 7.0 as H+ ion concentration varies from 1.0M to 10-14M at 250C in aqueous solution,the pH value vary from 0 to 14.0.

EQUILIBRIA 395 The pH scale for neutral, acidic and basic solutions are given below. [H+]100 101 10-2 10-3 10-4 10-5 10-6 10-7 10-8 10-9 10-10 10-11 10-12 10-13 10-14 <IncArceiadsiicng. Neutral ® Alkaline Increasing. pH 0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 11.0 12.0 13.0 14.0 We know that, the ionic product of water increases with increase in temperature. Therefore,H+ ion concentration will increase with increase in temperature. Hence, pH value will decrease with increase in temperature. Thus, pH value of boiling water is 6.56, although it is neutral. Table 9.6 pH and pOH value of aqueous solutions [H+] pH pOH [OH-] 100 0 14 10-14 10-1 1 13 10-13 10-2 2 12 10-12 10-3 3 11 10-11 10-4 4 10 10-10 10-5 59 10-9 10-6 68 10-8 10-7 77 10-7 10-8 86 10-6 10-9 95 10-5 10-10 10 4 10-4 10-11 11 3 10-3 10-12 12 2 10-2 10-13 13 1 10-1 10-14 14 0 10-0 Importance of pH value : i. Soil testing : In agriculture pH of soil is often tested for the application of acidic or basic fertiliser for a particular crop.

396 +2 CHEMISTRY (VOL. - I) ii. Food preservation : A definite value of pH is to be maintained for food preservation. iii. Qualitative and quantitative analysis : pH finds its use in various qualitative and quantita- tive analysis. iv. Biochemical reactions : In human metabolic system, pH value plays an important role. Blood is slightly alkaline, its pH value is 7.4.People suffering from gastritis have lower value of pH, that is pH < 7.0 Example :2 Calculate pH of (i) 10-2 N HCl (ii) 10-2M H2SO4 and (iii) 10-2 N H2SO4 Solution : We know, strong acids on dilution ionises completely. (i) HCl ® H+(aq) + Cl-(aq) 10-2 N 0 0 (before ionisation) 0 10-2 N 10-2 N (after ionisation) \\ [H+] = 10-2 N= 10-2 M \\ pH = -log[H+] = -log [10-2] = 2.0 (ii) H2SO4 ® 2H+(aq) + SO42- (aq) 10-2 M 0 0 (before ionisation) 0 2 ´ 10-2 M 10-2 M (after ionisation) \\ [H+] = 2 ´ 10-2 M ( 1 mole of H2SO4 gives 2 moles of H+ ion) pH = -[logH+] = -log [2 x 10-2] = -log 2 + 2 log 10 = 1.6989 (iii) H2SO4 ® 2H+ (aq) + SO42-(aq) 10--2N 0 00 (before ionisation) (after ionisation) 10--2 N 10--2 N 1 equivalent of H2SO4 gives 1 equivalent of H+ ion. \\ [H+] = 10-2 M \\ pH = -log [H+] = -log [10-2] = 2.0 Example : 3 Calculate the pH value of (i) 0.001N NaOH and (ii) 0.01M Ca(OH)2 Solution : Since NaOH and Ca(OH)2 are strong bases, they ionise completely on dilution. 1N0a-O-3MH(aq) ® Na--(aq) + OH--(aq) (i) 0 0 (before ionisation) 0 10-3 N 10-3N (after ionisation)

EQUILIBRIA 397 \\ [OH–] = 10–3 N = 10–3 M pOH = -log [10–3] = 3.0 pH = 14 - pOH = 14 - 3.0 = 11.0 (ii) 1C0a–(O2NH)2(aq) ® Ca++(aq) + 2OH– (aq) (before ionisation) Example : 4 00 (after ionisation) 0 10–2M 2 ´ 10-2M \\ [OH-] = 2 x10–2 M pOH = - log (2 ´ 10-2) = 1.6989 \\ pH = 14 - 1.6989 = 12.3011 2 gms of NaOH are dissolved in water to make 1 litre solution. What is pH of the solution? Solution : Molarity of NaOH = 2.0 1 = 0.05 M = 5 ´ 10-2 M 40 ´ \\ [OH–] = 5 ´ 10–2M or, pOH = -log [OH-] = - log [5 ´ 10-2] = 1.3010 \\ pH = 14 - pOH = 14 -1.3010 = 12.6989 Example : 5 3.2 gms of hydrogen chloride dissolved in 1 litre of water. What is pH of the solution ? Solution : Molarity of HCl = 3.2 = 0.089 M = 8.9 ´ 10–2 365 \\ [H+] = 8.9 ´ 10–2 M or, pH = -log [H+] = -log [8.9 ´ 10–2] = 1.2138 Example 6 Calculate the pH value of a solution containing 3.65 gms of hydrochloric acid per litre of the solution. Solution : Molecular mass of HCl = 36.5 36.5 gms of HCl per litre = 1.0 M solution 3.65 gms of HCl per litre = 0.1 M solution \\ [H+] = 0.1 M = 10–1 M Hence, pH = -log [H+] = - log[10–1] = 1.0

398 +2 CHEMISTRY (VOL. - I) Exampe 7 How many grams of KOH must be dissolved in one litre of solution to give it a pH value of 12.0. Solution : Strength of KOH = ? pH value of solution = 12.0 We know, pH = - log [H+] or, 12.0 = - log [H+] or, [H+] = 10–12 moles/ litre. At 250C, KW = [H+] [OH -] = 1.0 ´ 10–14 \\ [OH-] = KW = 1.0 ´ 10–4 = 1.0 ´ 10–2 moles / litre. [H+] 1.0 ´ 10–2 Since KOH is a strong base, [KOH] = [OH–] = 1.0 ´ 10–2 M. We know, Strength = Molarity ´ Molecular mass of solute. Molecular mass of KOH = 56 \\ Strength of KOH = 56 ´ 10–2 = 0.56 gms / litre Hence, 0.56 gms of KOH must be dissolved in one litre of the solution to give it a pH value of 12.0. 9. 12 BUFFER SOLUTION Hydrogen ion concentration of aqueous solution changes largely by the addition of very small amount of an acid or a base to it. It is very often necessary to have solutions, whose pH value does not change much on addition of small amount of strong acid or alkali to it. Such solutions are known as buffer solution or simply buffer. For example, take a soluation of NaCl in water. Its pH value is 7.0. When 1 ml of 1N, HCl is added to 1 litre of NaCl solution, the pH of resulting solution decreases from 7.0 to 3.0 But when the same amount of acid or alkali is added to ammonium acetate solution(having pH value 7.0), the pH value of the resulting solution does not undergo an appreciable change. Thus, ammonium acetate solution is a buffer solution. Statement : A buffer solution may be defined as a solution, whose pH value does not change appreciably by the addition of a small amount of either acid or alkali from outside sources. Illustrations : (a) Why aqueous solution of NaCl does not act as buffer ? NaCl on dilution dissociates completely giving Na+ and Cl– ions in aqueous solution. (i) Let HCl(aq) be added to NaCl solution. By the addition of HCl(aq),H+ ion and Cl–

EQUILIBRIA 399 ion concentration in the solution increases. The increase in concentration of Cl– ion in the solution does not affect the pH value. But the increase in concentration of H+ ion, decreases the pH value. (ii) Let NaOH(aq) be added to NaCl solution. Here, by the addition of NaOH(aq), Na+ ion and OH– ion concentration in the solution increases. The increase in concentration of OH-ion decreases the concentration of H+ ion, Hence, pH value increases. Therefore, NaCl solution is not a buffer. (b) Why aqueous solution of CH3COONH4 acts as a buffer ? Ammonium acetate on dilution dissociates completely as follows; CH3COONH4(aq) ® CH3COO-(aq) + NH+4(aq) (i) Let HCl(aq) be added to ammonium acetate solution. CH3COONH4(aq) ® CH3COO–(aq) + NH+4(aq) ® H+(aq) + Cl–(aq) HCl(aq) Here, the H+ ions furnished by acid combine with the acetate ions to form acetic acid molecules. Since acetic acid is a weak electrolyte, it dissociates to a very small extent. Hence, H+ ion concentration inside the solution increases very slightly. hence, its pH value does not change appreciably. The increase in the concentration of Cl– ions does not affect the pH value. (ii) Let NaOH(aq) be added to ammonium acetate solution. CH3COONH4(aq) + NH+4(aq) ® CH3COO–(aq) ® Na+(aq) + OH–(aq) NaOH(aq) Here, the OH- ions furnished by the base combine with NH+4 ions to form ammonium hydroxide molecules. Since ammonium hydroxide is a weak base, it dissociates to a very small extent. Hence, OH- ion concentration inside the solution does not increase appreciably. So, the pH value of the solution does not increase appreciably. Hence, buffer solutions are considered to have reserve acidity and reserve alkalinity. For example, CH3COONH4 has reserve acidity due to the presence of NH+4 ions and reserve alkalinity due to CH3COO- ions. Type of buffers : Buffers are of two types (i) Simple buffer. (ii) Mixed buffer. (i) Simple buffer : Simple buffer solution can be prepared by mixing salt of weak acid and weak base with water. For example : CH3COONH4, NH4CN can act as simple buffer.

400 +2 CHEMISTRY (VOL. - I) (ii) Mixed buffer : These are of two types. (a) Acidic buffer (b) Basic buffer. (a) Acidic buffer : An acidic buffer is an equimolar mixture of a weak acid and its salt of strong base. For example, a mixture containing one mole of CH3COOH and one mole of CH3COONa forms an acidic buffer. Its pH value is 4.75. (b) Basic buffer : A basic buffer is an equimolar mixture of a weak base and its salt strong acid. For example, a mixture of one mole of NH4OH and one mole of NH4Cl forms a basic buffer. Its pH value is about 9.25. Characteristics of Buffer solution : Buffer solutions have the following characteristics. (i) Its pH value is fixed. (ii) Its pH value does not change either keeping it for a long time or on exposure to atmosphere or by dilution to a small extent. (iii) The addition of a little amount of even strong acid or alkali does not change the pH value appreciably. Buffer solution of a mixture of weak acid and its salt. (Acid Buffer) A very common buffer is prepared by mixing equimolar solution of CH3COOH and CH3COONa. CH3COONa is completely dissociated. The mixture, thus contains CH3COO–, Na+ and CH3COOH molecule. When a strong acid is added: The H+ ions will be taken up by CH3COO– ions forming CH3COOH which is weakly ionising. H+ + CH3COO– ® CH3COOH. Thus, pH value remains almost unaffected. When a strong base is added : The OH- ions added are neutralised by CH3COOH present in the mixture. OH- + CH3COOH ® CH3COO– + H2O. Thus, again the pH value remains unaltered. Here the reserve acidity is due to the presence of CH3COOH and reserve alkalinity is due CH3COO– ions. to Buffer solution of a mixture of weak base and its salt. (Basic Buffer) A mixture containing equimolar solution of NH4OH and NH4Cl constitutes another good buffer. The mixture contains NH+4, Cl– ions and undissociated NH4OH.

EQUILIBRIA 401 When a strong acid is added : H+ ions are neutralised by NH4OH. H+ + NH4OH ® H2O + NH + 4 When a strong base is added : OH- ions are neutralised by NH+4 ions forming weakly dissociated NH4OH. OH- + NH + ® NH4OH 4 Here, the reserve acidity is due to NH4+ ions and reserve alkalinity is due to NH4OH. Henderson Equation forAcid and Basic Buffers (Henderson - Hasselbalch equation) I. pH of Acid buffers (Buffers mixture of weak acid & its salt) Let HA be the weak acid and NaA be the salt of weak acid (highly ionised) Since HA is a weak acid, it is slightly ionised. HA H+ + A– Applying law of chemical equalibrium Ka = [H+ ] [A ] ........................ (1) [HA] or, [H+] = Ka[HA] .................... (2) [A ] Again, NaA is completely ionised. NaA Na+ + A– So, in presence of NaA, the+ionisation of HA is further suppressed due to the presence of common ion A–. This is common ion effect. We have thus, [Acid] = [HA]; [Salt] = [A–] Taking logarithm of both sides of equation (2), log [H+] = log Ka + log [HA] – log [A–] .................. (3) or, – log [H+] = –log Ka – log [HA] – log [A–] ............. (4) or, pH = pKa + log [A ] .................................... (5) [HA] [Salt] or, pH = pKa + log [Acid] .................................... (6) This equation (6) is known as Henderson - Hasselbalch equation or simply Henderson equation. This helps in calculation of pH of a buffer solution made by mixing a known quantity of weak acid and its salt.

402 +2 CHEMISTRY (VOL. - I) II. pH of Basic buffers (Buffer mixture of weak base and its salt) Let BOH be the weak base and BA be the salt of weak base (highly ionised). Since BOH is weak base, it is slightly ionised. BOH B+ + OH– and Kb= [B+ ][OH ] ........................ (7) [BOH] or, [OH–] = = K b[BOH] ....................... (8) [B+ ] Again, BA is completely ionised. BA B+ + A– So, in presence of BA the ionisation of BOH is further suppressed due to the presence of the common ion B+ (Common ion effect) We have thus, [Base] = [BOH], [Salt] = [B+] Taking logarithm of both sides of equation (8), we have log [OH–] = log [Kb] + log [BOH] – log [B+] .................... (9) or, –log [OH–] = – log [Kb] – log [BOH] + log [B+] ................ (10) or, pOH = pKb + log [B+ ] ................................ (11) [BOH] [Salt] or, pOH = pKb + log [Base] ............................. (12) Equation (12) is known as Henderson - Hasselbalch equation or simply Henderson equation. It helps in calculation of pH value of the buffer obtained by mixing a known quantity of weak base and its salt. Since pH = 14 – pOH or, pOH = 14 – pH Equation (12) may be represented as [Salt] 14 – pH = pKb + log [Base] [Salt] or, pH = 14 – pKb – log [Base] ................ (13)

EQUILIBRIA 403 Applications of Buffers : Buffer solutions are used in - (a) qualitative analysis of mixtures. (b) quantitative analysis or estimations. (c) lndustrial process such as manufacture of paper, dyes, inks, paints, drugs etc. (d) digestion of food. (e) preservation of foods and fruits. (f) agriculture and dairy products preservation. 9.13 HYDROLYSIS OF SALTS Water dissociates to a small extent producing H+ and OH- ions. H2O(1) H+(aq) + OH-(aq) In pure water, concentration of H+ ions and OH- ions are equal. Hence, pure water is neutral. Salts are strong electrolytes. When dissolved in water, they dissociate almost completely giving +vely charged ions (cations) and - vely charged ions(anions). The anions and cations of the salt react with H+ ions and OH- ions furnished by water to form either acidic or basic or even neutral solution. Statement : Hydrolysis is a phenomenon of the interaction of anions and cations of the salt with H+ and OH- ions furnished by water producing acidic, basic or neutral solution. So, hydrolysis is the reverse of neutralisation. Neutralisation involves the combination of H+ ions and OH- ions to form water, while hydrolysis results in the formation of H+ ions and OH- ions. The salts are classified into the following four types; (a) Salts of strong acids and strong bases. (b) Salts of strong acids and weak bases. (c) Salts of weak acids and strong bases. (d) Salts of weak acids and weak bases. Here, we will discuss the hydrolysis of different types of salts . (a) Salts of strong acids and strong bases : The salts of this type are NaCl, KCl, KNO3 etc. This type of salts do not undergo hydrolysis. The reason is that, the possible products of hydrolysis are strong electrolytes and themselves get fully ionised. Thus, the number of hydrogen ions remains equal to the hydroxyl ions in solution. Such a solution is neutral in character. (b) Salts of strong acids and weak bases : The examples of such salts are ammonium chloride, calcium nitrate etc. Consider the hydrolysis of ammonium chloride in water.

404 +2 CHEMISTRY (VOL. - I) NH4Cl ® NH4+ + Cl– H+ + OH– H2O NH4Cl dissociates fully to form NH4 ions and Cl– ions and water dissociates slightly giving H+ ions and OH- ions. The OH– ions of water associate with the NH+4 ions of the salt to form NH4OH. NH4OH is a weak electrolyte, and it dissociates to a very small extent. As OH– ions are taken up, more of water ionises to maintain the constant value of Kw. Due to this, the concentration of H+ ions in solution becomes more than OH– ion concentration. Therefore, such a solution will be acidic in character. (c) Salts of weak acids and strong bases: The examples of such salts are sodium acetate, potassium carbonate etc. Consider the hydrolysis of sodium acetate in water. CH3COONa ® CHCOO– + Na+ H2O H+ + OH– Sodium acetate dissociates completely to form sodium ions and acetate ions. Water ionises slightly. The H+ions of water attack the CH3COO– ions to form acetic acid, which is a weak acid and dissociates to a very small extent. As H+ ions are taken up, more of water ionises to maintain the constant value of KW. Due to this, the concentration of OH– ions in solution becomes more than the H+ ion concentration. Therefore, such a solution will be basic in character. (d) Salts of weak acids and weak bases : An example of such a salt is ammonium acetate. Consider its hydrolysis. CH3COONH4 ® CH3COO– + NH4+ H+ + OH– H2O Ammonium acetate on dissociation gives CH3COO– ions and NH + ions. Water being a 4 weak electrolyte, dissociates slightly giving H+ ions and OH– ions. The acetate and ammonium ions are attacked by H+ ions and OH– ions of the water to form CH3COOH and NH4OH. Since, both CH3COOH and NH4OH are weak electrolytes, they dissociate to a very small extent, giving rise to equal number of H+ ions and OH-- ions. Such a solution will be neutral in character. 9.14 SOLUBILITY PRODUCT If a small quantity of a sparingly soluble salt is shaken with water, an equilibrium is established between the solid phase and the ions in solution. For example, AB(s) A+(aq) + B–(aq) (undissociated) (dissociated) This is heterogeneous equilibrium in which the ions A+ and B- have variable concentra- tions. Therefore, only these will be taken into consideration. Suppose a small amount of silver

EQUILIBRIA 405 chloride is added to water. Since it is a sparingly soluble salt, a very small amount of it passes into solution. Whatever amount of silver chloride which passes into solution will exist as Ag+ and Cl- ions. The state of equilibrium can be represented as, AgCl(s) Ag+(aq) + Cl-(aq) Applying the law of chemical equilibrium, we have, K = [Ag+] [Cl-] [AgCl(s)] The concentration of the solid (AgCl in this case) is taken as constant. Hence, K [AgCl] = [Ag+] [Cl-] or, Ksp = [Ag+] [Cl-] Ksp is another constant and is called solubility product, which is temperature dependent. Consider the case of lead chloride, which is also a sparingly soluble salt. The equilibrium between the ions and the undissolved salt is written as, PbCl2 Pb2+ + 2Cl- \\ Ksp = [Pb2+] [Cl-]2 In general, if a salt of type Ax By is considered, xAy+ + yBx- Ax By [Ay+]x [Bx-]y Ksp = Hence, the solubility product of a sparingly soluble substance may be defined as the product of the molar concentrations of its ions raised to the power equal to the number of times each ion occurs in the equation representing the dissociation of the substance at a given temperature. Relation between Solubility and Solubility product : Let us suppose that the solubility of a sparingly soluble salt AxByis 'S' mol/L. Then we have, xAy+ + yBx- AxBy = (S moles) (xS moles) (yS moles) Thus, [Ay+] = xS and [Bx-] = yS [Ay+]x [Bx-]y \\ Ksp = [xS]x [yS]y = = xxSxyySy = xx yy Sx + y So, for Ag2 CrO4 2Ag+ + CrO42- x = 2, y = 1 \\ Ksp = 22. 11. S2+1 = 4S3 This relationship between solubility and solubility product can be used for the calculation of solubility product.

406 +2 CHEMISTRY (VOL. - I) Difference between Ionic product and Solubility product : Solubility product is constant for a given electrolyte at a given temperature. But ionic product, which is the product of the concentration of ions, can be varied by changing the concen- trations of the ions. Ionic product of an electrolyte can not exceed its Ksp value irrespective of the nature of the source of the ions. When the ionic product temporarily exceeds the solubility prod- uct due to the increase in concentration of either one or both types of ions, the excess ions atonce combine with oppositely charged ions and the solid salt separates out. Depending upon the values of ionic product, the solutions can be classified into three different categories as follows : (i) If ionic product = Ksp, the solution is just saturated and no precipitation takes place . (ii) If ionic product > Ksp, the solution is super saturated and precipitation takes place. (iii) If ionic product < Ksp' the solution is unsaturated and more of the solute can dissolve. 9.15 COMMON ION EFFECT The phenomenon in which degree of dissociation of a weak electrolyte is suppressed by addition of a substance having an ion common to weak electrolyte. Consider the dissociation of a weak electrolyte, say, acetic acid, CH3COOH CH3COO- + H+ The equilibrium constant of weak acid Ka is given by Now, suppose sKodai=um[Ca[HCce3HtCa3tOCeOOis-Oa]Hd[dH]e+d]to this solution. CH3COONa ® CH3COO- + Na+ By the addition of sodium acetate, the concentration of CH3COO- ions in solution increases. Thus, in order to keep Ka constant, H+ ions concentration must decrease or concentration of undissociated acetic acid must increase. In other words, the dissociation of acetic acid is suppressed by additon of CH3COONa to its solution. Here, CH3COO- ion acts as the common ion. Similarly, the dissociation of NH4OH, which is a weak electrolyte is suppressed by the addition of NH4Cl to it. NH4OH(l) NH4+ + OH- Kb= [NH+4 ] [OH-] [NH4OH] If NH4Cl is added to it, NH4Cl ® NH4+ + Cl- By the addition of NH4Cl, the concentration of NH4 ion increases.

EQUILIBRIA 407 Thus, in order to keep Kb constant, OH- ions concentration must decrease or concentration of undissociated NH4OH must increase. This indicates that, by the addition of NH4Cl to NH4OH solution (having NH + ion in common), the dissociation of weak electrolyte (NH4OH) is further 4 suppressed. Application of the common ion effect and solubility product principle : (a) Recovery of pure NaCl from sea water : The crude salt obtained by the evaporation of sea water contains mainly sodium chloride along with impurities like CaCl2, KBr etc. For obtaining pure NaCl, HCl gas is passed through or conc. HCl is added to it. Hence, all impurities remain in solution and pure NaCl is precipitated out. NaCl ® Na+ + Cl- HCl ® H+ + Cl- The ionic product, [Na+] [Cl-] in the new system on passing HCl, exceeds the solubility product of NaCl. [Na+] [Cl-] > Ksp for NaCl Hence, pure NaCl separates out and then purified by recrystallisation. (b) In qualitative analysis : The qualitative analysis of mixture is based on the principle of solubility product. Fol- lowing are some of the important applications in mixture analysis which involve the principles of solubility product as well as the common ion effect. (a) Precipitation of Group - I radicals (Pb2+, Ag+ and Hg+) The group reagent is dil. HCl. The Group- I radicals are precipitated out by dil HCl because the ionic products of the chlorides of these exceed their corresponding solubility prod- ucts. [Ag+] [Cl-] > Ksp for AgCl No other radicals are precipitated out because the ionic products of their chlorides do not exceed the corresponding solubility products. (b) Precipitation of Group - II radicals by H2S in presence of dil. HCl Hydrochloric acid suppresses the ionisation of weakly dissociated H2S (Group II reagent) H2S 2H+ + S2-. Thus, only the ionic products of the sulphides of Group II radicals exceed their corresponding solubility products and hence only these are precipitated out leaving the sulphides of Group III, IV etc. in solution because their solubility products are high. It is important to note that dil.HNO3 and dil.H2SO4 may also suppress the ionisation of H2S but these are not used because these are oxidising agents and oxidise hydrogen sulphide to sulphur.

408 +2 CHEMISTRY (VOL. - I) (c) Precipitation of Group III A radicals by NH4OH in presence of NH4Cl : Addition of NH4Cl in Group - III A further suppresses the ionisation of NH4OH. As a result of which, the ionic products of only the Group III A radicals exceed their corresponding solubility products. Hence, other group (III B, IV etc.) radicals remain in solution phase while Group III A radicals (Fe3+, Al3+ and Cr3+) are precipitated out as their hydroxides. (d) Precipitation of Group III B radicals by H2S in presence of NH4OH : The Group-IIIB radicals are Co2+ Ni2+ , Mn2+ and Zn2+. The group reagent is H2S in presence of NH4OH. 2H+ + S2- H2S NH+4 + OH- NH4OH OH- ions furnished by NH4OH combine with the H+ ion furnished by H2S to form unionisable water. As a result of which more H2S is ionised to maintain the equilibrium according to Le- Chatelier's principle. Thus, the concentration of S2- ions will be high in the solution and a stage will be reached when the ionic products of Group-III B radicals and sulphide ions exceed the corresponding solubility products and Group III B radicals are precipitated out. (e) Precipitation of Group IV radicals : The group reagent is ammonium carbonate in presence of NH 4Cl. (NH4)2 CO3 2NH4+ + CO32– (weak electrolyte) NH4Cl NH4+ + Cl– (strong electrolyte) Due to common ion (NH4+) effect, ionisation of (NH4)2 CO3 is suppressed. Thus, only the Group - IV radicals precipitate out as carbonates, which have low solubility products.

EQUILIBRIA 409 CHAPTER (9) AT A GLANCE 1. Buffer solution : Buffer solution may be defined as solution, whose pH value does not change appreciably upon the additon of small amounts of either acids or bases or even water from outside source. 2. Common ion effect : It is an effect due to which the ionisation of a weak electrolyte is further suppressed when a salt having common ion is added to it. 3. Equilibrium : A system is said to be in a state of equilibrium if the rate of the forward reaction is equal to that of the backward reaction. 4. Hydrolysis : It is defined as a process in which a salt reacts with water to form an acid and a base. 5. Law of Mass Action : The rate of reaction is proportional to the product of the concen- tration of reactants each raised to the power equal to the number of moles of the reactant species involved in the chemical equation. 6. Le- Chatelier's Principle : If a system at equilibrium is subjected to a change of concentra- tion, pressure or temperature, the equilibrium shifts in the direction that tends to undo the effect of such a change. 7. pH : It is defined as the negative logarithm of hydrogen ion concentration. pH = - log [H+] 8. Solubility Product : The solubility product of a sparingly soluble substance is defined as the product of the molar concentrations of its ions raised to power equal to the number of times each ion occurs in the equation for the dissociation of a substance at a given tempera- ture. [Salt ] 9. Henderson equation : for acid buffers : pH = pKa + log [Acid] [Salt ] : for basic buffers : pOH = pKb + log [Base] [Salt ] and pH = 14 – pKb – log [Base]

410 +2 CHEMISTRY (VOL. - I) QUESTIONS Very short type questions - (1 mark each) 1. What do you call the state of a chemical reaction when the forward and the backward rates are equal ? 2. According to Bronsted theory, what is a base ? 3. What is the pH of 0.01(N) HCl ? 4. Calculate the pH of 0.001(N) HCl 5. What is the pH value of pure water ? 6. Which catalyst is used in contact process for manufacture of sulphuric acid ? 7. What is the basicity of sulphuretted hydrogen ? 8. What is the effect of pressure on the solubility of a solid ? 9. What substance on mixing with ammonium hydroxide forms a buffer solution ? 10. Which catalyst is used in the manufacture of ammonia by Haber's Process. 11. What substance on mixing with weak acid forms buffer solution ? 12. In which reaction, the value of DH will be negative ? 13. The pH of an acidic solution will be more or less than 7.0 ? 14. What is the pH of 0.0001 M NaOH ? 15. Give two examples of Lewis acid. 16. What is the pH of 1N HCl ? 17. What is active mass ? 18. What is the relation between equilibrium constant for forward reaction and equilibrium constant for backward reaction in a reversible process ? 19. What happens to the solubility of calcium acetate if temperature increases. 20. Whether CO2 is a Lewis acid or base? 21. What is the pH of 1M HCl solution? 22. Write the relation between Kp & Kc of the following reaction. PCl5 (g) PCl3 (g) + Cl2 (g) 23. In which condition the rate of forward and backward reactions are equal? 24. Arrange the following solutions in water in increasing pH CaCl2, NaOAc, KCl, NaOH 25. What is the relation between pH & pOH? 26. Write the relation between Kp & Kc of the following reaction. H2(g) + I2(g) 2HI(g)

EQUILIBRIA 411 Short Answer Questions (2 marks each) 1. Under what conditions of temperature and pressure would you get the maximum yield of ammonia in the following reaction ? 3 H2(g)+ N2(g) 2NH3(g), DH = -22.1 kcal. 2. What do you mean by the equilibrium constant of a reaction ? Does it change with 3. (a) temperature? (b) Explain the term buffer solution. 4. Give two examples of buffer solution. For a general reaction mA + nB xC + yD What is the expression for the equilibrium constant ? 5. The solubility product of AgCl is 1.7 ´ 10-10 at 298K. What will be its solubility at that temperature in a 1M aqueous solution of hydrochloric acid ? 6. Arrange the following aqueous solutions in the increasing order of their pH values. (a) 0.1 N sodium chloride (b) 0.1 N hydrochloric acid. (c) 0.1 caustic soda 7. When silver nitrate solution is added to a mixture of chloride and iodide ions in solution, silver iodide precipitates first followed by the silver chloride. Give reasons for that in brief. 8. How do you define an acid in the modern concept ? Give one example. 9. Define solubility product. How the idea of solubility product is used for purification of common salt ? 10. Why it is necessary to add excess of NH4OH before passing H2S for precipitation of Group III B cations ? 11. Give Bronsted and Lowry concept of an acid. 12. Calculate the solubility of AgCl in moles per litre, if the solubility product constant for AgCl is 1.0 ´ 10-10. 13. What is the role of ammonium chloride in the precipitation of Group III A radicals in the qualitative analysis ? 14. Why is sodium carbonate solution alkaline ? 15. 10 ml of N/5 alkali is mixed with 20ml of N/10 acid. What is the pH of the resultant solution ? 16. Why is the pH of a solution of cupric sulphate solution less than 7.0 ? 17. What will be the value of DH, when a reaction at constant temperature and pressure is at equilibrium ? 18. Which substance makes a buffer with acetic acid ? What happens to the pH when a small quantity of HCl is added to such a mixture ?

412 +2 CHEMISTRY (VOL. - I) 19. W hyitisnecessarytoadddiluteH ClbeforepassingH 2S for precipitating Group. II cations ? 20. Name one factor which influences the solubility of a solid in a solvent ? 21. Write the conjugate acid of OH- and NH3. 22. Distinguish between ionic product and solubility product. 23. Why is it necessary to add solid NH4Cl before adding excess of NH4OH for precipita- tions Group IIIA cations ? 24. Explain why the change of pressure in the reaction H2 +I2 2HI will not affect the equilibrium constant ? 25. Write the conjugate base of H2SO4 and HCO3 26. What happens to H+ ion concentratoin when temperature increases and why? 27. Synthetic clothes dry up quicker than cotton clothes. Give reasons. 28. Give two applications of pH. 29. Calculate the pH of a solution containing 0.365g of HCl per litre of the solution. 30. Explain what happens when HCl gas is passed through conc NaCl solution. 31. Give one example each of acid buffer and basic buffer. 32. Write the Lewis structure of H2SO4 molecule. 33. Calculate the pH of 0.001N NaOH solution. 34. How is pure NaCl recovered from sea water? Short Answer Questions (3 marks each) 1. Discuss equilibria is physical processes with reference to solid liquid equilibrium with example. 2. Discuss equilibria in physical processes with reference to liquid vapour equilibrium with example. 3. Discuss solid vapour equilibrium with example. 4. Write a note on ionisation of polybasic acid with example. 5. What are the factors that affect the acid strength. Dscuss with example. Long Questions (7 marks each) 1. State and explain Le-Chatelier's principle. Discuss the effect of change of temperature and pressure on the following system at equilibrium: N2(g) + 3H2(g) 2NH3(g) + 22.4 kcal 2. What is meant by equlibrium reaction ? State Le-Chatelier's principle and discuss its applications to synthesis of ammonia 3. State law of mass action. Explain the law with three examples.

EQUILIBRIA 413 4. What is common ion effect ? Explain giving examples. 5. State Law of mass action. Explain the law with one example. Derive the expression for equilibrium constant of reaction, A +2B 3C +D + E 6. Name the basic radicals of Group III B of qualitative analysis. What is the group reagent ? 7 Write shorts notes on : (a) Buffer solution (b) Le-Chatelier's principle (c) Theories of acids and bases (d) Common ion effect. (e) Solubilty product 8. Define and explain law of mass action. What is an equilibrium constant ? 9. State Le-Chatelier's principle. How is it applied to the synthesis of ammonia ? Give the expression for Kc and Kp for such reaction. ADDITIONAL QUESTIONS 1. Describe the various postulates of the Arrhenius theory of acids and bases. What are the limitations of this theory ? 2. Discuss in detail the phenomenon of hydrolysis of salts. 3. Discuss in detail the Lowry-Bronsted theory of acids and bases. What is meant by a conjugate pair? 4. Write the importance of solubility product in qualitative analysis . 5. What is buffer solution ? Write the types of buffer solution with example. What are the applications of buffer solutions ? 6. The molar concentration of A and B are 0.80 mole litre -1 each. On mixing them, the reaction starts to proceed as A +B C +D and attain equlibrium . At equlibrium molar concentration of C is 0.60 mole litre-1. Calculate Kc of the reaction ? (Ans : Kc = 9.0) 7. A mixture of 3 moles of H2 and 1 mole of N2 react at constant pressure of 100 atmosphere. At equilibrium 0.5 mole of ammonia is formed. Calculate the value of Kp for the equilibrium reaction 3/2 H2 + 1/2 N2 NH3 (Ans : Kp = 0.005 atm) 8. The value of Kp for dissociation of 2HI H2 + I2 is 1.84 ´ 10-2 If the equilibrium concentration of H2 is 0.4789 mole litre-1, Calculate the conc. of HI at equilibrium. (Ans : 3.53 mole litre-1) 9. Ksp of AgCl is 2.8 ´ 10-10 at 250C. Calculate solubility AgCl in pure water (Ans : S = 1.673 x 10-5 mole litre-1)

414 +2 CHEMISTRY (VOL. - I) MULTIPLE CHOICE QUESTIONS 1. Irreversible reaction is one which : (a) proceeds in one direction only. (b) proceeds in both the direction. (c) is instantaneous reaction. (d) is slow reaction. 2. When rate of forward reaction is equal and opposite to the rate of backward reaction,the state is said to be : (a) Reversible state (b) Equilibrium. (c) Chemical equilibrium. (d) None of the above. 3. Which of the following reactions will be favoured by low pressure ? (a) H2 + l2 2HI (b) N2 + 3H2 2NH3 (c) PCl5 PCl3 + Cl2 (d) N2 + O2 2NO 4. Which of the following factor will be useful in manufacture of ammonia by Haber's process ? (a) High pressure. (b) Low pressure. (c) High temperature. (d) Increase in the concentration of ammonia. 5. The reaction in which heat is absorbed is known as : (a) Exothermic. (b) Endothermic. (c) Reversible. (d) None of the above. 6. The rate at which a substance reacts is proportional to its active mass. This statement is : (a) Le-Chatelier's principle. (b) Faraday's Law. (c) Law of multiple proportions. (d) Law of mass action. 7. When chemical equilibrium is reached the (a) Reaction stops. (b) Rate of forward reaction is equal to the rate of backward reaction. (c) Rate of forward reaction is more than that of backward reaction. (d) None of the above.

EQUILIBRIA 415 8. In a reversible reaction if there is no change in total number of molecules, the reaction will be favoured by (a) High pressure. (b) Low pressure (c) High temperature. (d) Higher concentration of a reactant. 9. Which of the following will be favoured by high pressure ? (a) PCl5 PCl3 + Cl2 (b) N2 + O2 2NO (c) N2 + 3H2 2NH3 (d) H2 + I2 2HI 10. Chemical equilibrium is (a) Stationary (b) Dynamic (c) Inertness (d) State of rest. 11.(i) For the reaction, H2 + I2 2HI, the Kp and Kc are related as (a) Kp = Kc (RT)2 (b) Kp = Kc (RT)0 (c) Kp = Kc (RT)-2 (d) Kp = Kc (RT)-1 (ii) In which of the following reactions Kp = Kc ? (a) N2(g) + 3H2(g) 2NH3(g) (b) 2NOCl 2NO(g) + Cl2(g) (c) I2(s) + H2(g) 2HI(g) (d) H2(g) + Cl2(g) 2HCl(g) 12. The partial pressure of PCl3,Cl2 and PCl5 are 0.1,0.2 and 0.008 atmosphere respectively for a reaction, PCl5 PCl3 + Cl2. The value of Kp is (a) 2.5 (b) 5.0 (c) 0.25 (d) 25

416 +2 CHEMISTRY (VOL. - I) 13. For which of the following reactions the value of Kp is greater than Kc ? (a) N2 + O2 2NO (b) 2SO2 + O2 2SO3 (c) 2NO2 N2O4 (d) PCl5 PCl3 + Cl2 14. For the reaction PCl5 PCl3 + Cl2, the forward reaction at constant temperature is favoured by : (a) Introducing an inert gas at constant volume. (b) Introducing chlorine gas at constant volume. (c) Introducing an inert gas at constant pressure. (d) Increasing the volume of the container. 15. Lewis acids are (a) Electron acceptors (b) Proton acceptors (c) Electron donors. (d) Proton donors. 16. The pH of solution containing 0.4 gm NaOH per litre is, (a) 2.0 (b) 12.0 (c) 10.0 (d) 11.0 17. Conjugate base of HCO3- ion is (c) CO32- (a) CO2 (d) HCO3- (b) H2CO3 18. Aqueous solution of FeCl3 is (a) Acidic (b) Basic (c) Amphoteric (d) Neutral 19. When 1.0 ml of dil H2SO4 is added to 100ml of a buffer solution of pH 7.2, the pH (a) becomes 7.0 (b) is less than 7.0 (c) is more than 7.0 (d) does not change

EQUILIBRIA 417 20. What is the pH of 0.01M NaOH assuming complete ionisation ? (a) 0.01 (b) 2.0 (c) 12.0 (d) 14.0 21. The pH of the solution is 3.0. If its pH is changed to 6.0, then the [H+] of the original solution has to be (a) Doubled (b) Halved. (c) Increased 1000 times. (d) Decreased 1000 times. 22. The compound that is not a Lewis acid is (a) BF3 (b) AlCl3 (c) BeCl2 (d) SnCl4 23. The conjugate acid of NH2- is (a) NH3 (b) NH2OH (c) NH4+ (d) N2H4 24. An acidic buffer can be prepared by making solution of (a) HCl and NaCl (b) NaOH and NaCl (c) HCOOH and HCOONa (d) NH4Cl and NH4OH 25. A compound is precipitated when its (a) Ionic product exceeds the solubility product (b) Ionic product is less than its solubility product. (c) Ionic product is equal to the solubility product. (d) None of the above.

418 +2 CHEMISTRY (VOL. - I) 26. A basic buffer can be prepered by mixing (a) CH3COONa and CH3COOH (b) Na2SO4 and H2SO4 (c) NaOH and NaCl (d) NH4Cl and NH4OH 27. Which of the following solutions has the maximum pH value. (a) Solution of caustic soda. (b) Pure water (c) Water satured with CO2 gas (d) Solution of sodium chloride. 28. Hydrolysis is regarded as interaction between (a) H+ and OH- ions. (b) Ions of acid with ions of base (c) Ions of salt with ions of water. (d) Acid and base. 29. Which of the following solutions have pH close to 1.0 ? (a) 100 ml of M/10 HCl + 100 ml of M/10 NaOH (b) 55 ml of M/10 HCl + 45 ml of N/10 NaOH (c) 10 ml of M/10 HCl + 90 ml of M/10 NaOH (d) 75 ml of M/10 HCl + 25 ml of M/5 NaOH 30. The decrease in the ionisation of H2S in the presence of HCl is due to (a) Solubility product. (b) Dilution (c) Common ion effect (d) Saturation ANSWER TO MULTIPLE CHOICE QUESTIONS 1. a 6.d 11.(i) b (ii) d 16. b 21. d 26. d 2. c 7. b 12. a 17. c 22. c. 27. a 3. c 8. d 13. d 18. a 23. a 28. c 4. a 9. c 14. d 19.d 24. c 29. d 5. b 10. b 15. a 20. c 25. a 30. c qqq

UNIT – VIII CHAPTER - 10 REDOX REACTION The transformation of matter into one or more new substances by means of any chemical change is termed as Chemical Reaction. 10.1 REDOX REACTIONS : Chemical reactions such as burning of coal, sulphur, magnesium, rusting of iron etc. fall under a specific category called oxidation reduction reactions or simply redox reaction. Redox reactions form the basis of electrochemical and electrolytic cells. OXIDATION : Classical concept : In a classical sense oxidation is a chemical process involving (a) Addition of oxygen : When hydrogen combines with oxygen water is formed. Carbon burns in air or oxygen forming carbon dioxide. Magnesium burns in oxygen producing magnesium oxide. 2H2 + O2 ® 2H2O C + O2 ® CO2 2Mg + O2 ® 2MgO (b) Increase in the proportion of electronegative element or radical : Iron combines with electronegative element sulphur forming ferrous sulphide. Similarly, cuprous chloride is oxidised by chlorine to give cupric chloride, where the proportion of electronegative radical (Cl–) is increased. Stannous chloride is oxidised by mercuric chloride to form stannic chloride. Fe + S ® FeS 2CuCl + Cl2 ® 2CuCl2 SnCl2 + HgCl2 ® SnCl4 + 2HgCl. (c) Removal of hydrogen – When HCl is heated with MnO2 , Cl2 is liberated. Thus hydrogen is removed from HCl. Similarly, S is liberated from H2S by halogens. 4HCl + MnO2 ® MnCl2 + 2H2O + Cl2 H2S + X2 ® 2HX + S (where X2 = Cl2, Br2 or I2)

420 +2 CHEMISTRY (VOL. - I) (d) Decrease in the proportion of electropositive element or radical – Hydrogen peroxide liberates l2 from KI. Conversion of Cu2O to CuO involves decrease in the proportion of electropositive element Cu. 2KI + H2O2 ® 2KOH + I2 Cu2O ® CuO + Cu. Thus, in a limited sense oxidation is a process of addition of oxygen or increase in the proportion of the electronegative elements or groups in an element or a compound or removal of hydrogen or decrease in the proportion of electropositive elements or groups in a compound. REDUCTION : Classical concept : Reduction is a chemical process just opposite to oxidation involving : (a) Addition of hydrogen : When sulphur is heated in an atmosphere of hydrogen, hydrogen sulphide is formed. Similarly, hydrogen combines with chlorine in presence of sunlight producing hydrogen chloride. In these cases hydrogen is directly added to sulphur and chlorine. H2 + S ® H2S H2 + Cl2 ® 2HCl (b) Increase in the proportion of electropositive element : When CuCl2 is heated with Cu, CuCl is formed where there is increase in the proportion of Cu. Similarly HgCl2 is reduced by SnCl2 to give HgCl. CuCl2 + Cu ® 2CuCl 2HgCl2 + SnCl2 ® 2HgCl + SnCl4. (c) Removal of oxygen : Oxygen is removed from HgO when heated, Similarly, ZnO is reduced to Zn by heating with carbon. 2HgO ® 2Hg + O2 ZnO + C ® Zn + CO (d) Decrease in the proportion of electronegative element : Cupric iodide decomposes to give cuprous iodide where there is decrease in the proportion of the electronegative element. Similarly, ferric chloride is reduced to ferrous chloride by nascent hydrogen. 2CuI2 ® 2CuI + I2 FeCl3 + H ® FeCl2 + HCl. Thus, reduction is the process of addition of hydrogen or increase in the proportion of electropositive element of groups in an element or a compound or removal of oxygen or decrease in the proportion of electronegative elements or groups in a compound.

REDOX REACTION 421 Oxidation and Reduction in the light of Electronic theory : According to electronic theory oxidation is the loss of electrons by an atom or group of atoms participating in a chemical reaction. When electrons are lost, there is increase in the positive charge or decrease in the negative charge of the atom or ion. Fe2+ – e ® Fe3+ Sn2+ – 2e ® Sn4+ Hg+ – e ® Hg2+ Reduction is the gain of electrons by an atom or group of atoms participating in a chemical reaction. When electrons are gained, there is increase in the negative charge or decrease in the positive charge of the atom or ion. Sn4+ + 2e ® Sn2+ Hg2+ + e ® Hg+ Cu2+ + e ® Cu+ Fe3+ + e ® Fe2+ Oxidising Agent : A substance which can oxidise another substance and is itself reduced is termed as an oxidising agent. H2S + Cl2 ® 2HCl + S Cl2 is an oxidising agent which oxidises H2S to S and is itself reduced to HCl. According to the electronic concept, an oxidising agent or oxidant is one which can accept electrons readily. In the above reaction, Cl2 is an oxidant because it accepts electrons readily forming Cl–. Similarly, Cl2 + 2e ® 2Cl– Sn4+, Hg2+, Fe3+ ions are oxidants since they accept electrons as Sn4+ + 2e ® Sn2+ Hg2 + e ® Hg+ Fe3 + e ® Fe2+ Common oxidising agents are H2O2, O3. KMnO4, K2Cr2O7, H2SO4, HNO3, Cl2 etc. Reducing Agent : A substance which can reduce another substance and is itself oxidised is termed as a reducing agent. In the reaction H2S + Cl2 ® 2HCl + S H2S is a reducing agent because it reduces Cl2 to HCl and is itself oxidised to S. According to the electronic concept a reducing agent or reductant is one which can donate electrons readily. In the above reaction S2– donates 2 electron and is converted to elementary S. S2– – 2e ® S0

422 +2 CHEMISTRY (VOL. - I) Similarly, Fe2+, Sn2+ ions are reducing agents since they readily donate electron like Fe2+ – e ® Fe3+ Sn2+ – 2e ® Sn4+ Some common reducing agents are H2, C, CO, SO2, H2S, SnCl2, HgCl, etc. Oxidation and Reduction occur simultaneously : Oxidation involves loss of electrons whereas reduction involves gain of electrons. It is evident that if one substance loses elecrons, another substance at the same time must gain the electrons, because electron can not be produced in a chemical change and the net reaction is electrically neutral. Thus, whenever a substance is oxidised, another substance must be correspondingly reduced. Consider the reaction between Na and Cl2 to form NaCl. 2Na + Cl2 ® 2NaCl In this reaction Na loses electron and behaves as a reducing agent and chlorine acts as an oxidising agent since it gains electrons. 2(Na – e ® Na+) (oxidation) Cl2 + 2e ® 2Cl– (reduction) ———————————————— 2Na + Cl2 ® 2Na+ + 2Cl – or 2NaCl When SnCl2 solution is added to HgCl2 solution, a white precipitate of HgCl is obtained. In this case, mercuric chloride is reduced to mercurous chloride while stannous chloride is oxidised to stannic chloride simultaneously. 2HgCl2 + SnCl2 ® 2HgCl + SnCl4 2Hg2+ + 2e ® 2Hg+ (reduction) Sn2+ – 2e ® Sn4+ (oxidation) ——————————————————––– 2Hg2+ + Sn2+ ® 2Hg+ + Sn4+ (net reaction) The reaction involving simultaneous oxidation and reduction is referred to as redox reaction. A redox reaction comprises of oxidation half and reduction half reaction as shown above. When copper is dipped in AgNO3 solution, the following reaction takes place. Cu + 2AgNO3 ® Cu (NO3)2 + 2Ag This can be written as two half reactions, Cu0(s) – 2e ® Cu2+ (aq) (oxidation half reaction) 2Ag+ (aq) + 2e ® 2Ag0(s) (reduction half reaction) ——————————————————————––– Cu0(s) + 2Ag+ (aq) ® Cu2+(aq) + 2Ag0(s) (net reaction)

REDOX REACTION 423 10.2 COMPETATIVE ELECTRON TRANSFER REACTIONS : When a strip of metallic zinc is placed in aqueous copper nitrate solution and left for an hour, the blue colour of the solution disappears and reddish metallic copper gets deposited on the zinc strip. The reaction between metallic zinc and the aqueous solution of copper nitrate is : Zn (s) + Cu2+ (aq) ® Zn2+ (aq) + Cu(s) When H2S gas is passed through the resulting colourless solution formation of white precipilate of ZnS confirms the presence of Zn2+ ion. In this electron transfer reaction zinc is oxidised, releasing two electrons and copper ion is reduced by gaining electrons from zinc. release of 2e– Zn (s) + Cu2+ (aq) ® Zn2+ (aq) + Cu(s) gain of 2e– In contrast if copper rod is placed in aqueous solution of zinc sulphate for hours together no visible reaction is noticed. On passing H2S gas through the solution we do not get a black pricipitate of copper sulphide. Cu(s) + Zn2+ (aq) ® No reaction Now, if another experiment is performed by placing copper rod in silver nitrate solution, the solution develops a blue colour. when H2S gas is passed through it a black precipitate is obtained, thus revealing the presence of Cu2+ ions in the solution release of 2e– Cu(s) + 2Ag+(aq) ® Cu2+(aq) + 2Ag(s) gain of 2e– A comparision of the above two cases draws our attention towards a new field, i.e. competition between metals to release electrons. The electron releasing tendency of the above metal is in the order : Zn > Cu > Ag. Of course, Electro-chemical series or metal activity series is a list of metals arranged in the order of their electron releasing tendency. 10.3 OXIDATION NUMBER OR OXIDATION STATE : Consider the follwing reaction : 2Mg(s) + O2(g) ® 2Mg2+ + 2O2– (s).......................(1) Mg(s) + Cl2(g) ® Mg2+ + 2Cl– (s) .........................(2) 2H2(g) + O2(g) ® 2H2O(l) ......................................(3) H2(g) + Cl2(g) ® 2HCl(g) .......................................(4)

424 +2 CHEMISTRY (VOL. - I) In the reactions (1) and (2) Mg atoms are oxidised to form Mg2+. Reactions where ions are not formed can also be called oxidation reactions. Reactions (3) and (4) where hydrogen reacts with oxygen forming water and with chlorine forming hydrogen chloride are also oxidation reactions. Here, hydrogen atoms form covalent bonds with oxygen atom and chlorine atom respectively. Oxidation of magnesium differs from oxidation of hydrogen in the sense that in the former case ionic compounds are formed and in the latter covalent compounds are formed. Application of the same term oxidation seems to be unwise. Chemists, therefore, developed a new definition of oxidation to include all these reactions under one category which is based upon a concept of oxidation number. Oxidation number of an element in a compound is the charge positive or negative which would be assigned to the atoms of that element in the compound, if all the bonds were ionic bonds. Though all the bonds are not ionic, for convenience, it is arbitrarily assumed that all the compounds are ionic while computing oxidation number. Oxidation number is different from valency. Valency expresses the combining capacity of an element and does not display positive and negative nature of the atom in the compound. But oxidation number expresses the positive and negative nature of atoms in the compound along with their combining capacity. The value of the oxidation number and valency of an atom in a compound may not be same. For example, the valency of carbon is always 4 but the oxidation numbers of carbon in various compounds of carbon are different. Table 10.1 Oxidation number of carbon Compounds Oxidation number of carbon CH4 – 4 CH3Cl –2 CH2Cl2 0 CHCl3 +2 CCl4 +4 CO2 +4 Unlike valency, oxidation number can be fraction also. For ascribing positive or negative oxidation number to an atom in the compound the electronegative scale is used. The more electronegative element of a pair is arbitrarily given a negative oxidation number and less electronegative is given a positive oxidation number.

REDOX REACTION Table 10.2 Pauling's electronegativity scale 425 Element Electronegativity Element Electronegativity F 2.5 O 4.0 C 2.4 Cl 2.1 N 3.5 Se Br 2.1 I 3.0 Te 2.1 S 2.0 3.0 2.8 H 2.5 P 2.5 As Fluorine has the highest electronegativity (4.0) and therefore, in all its compounds, fluorine is given negative oxidation number. Next in the scale is oxygen and is therefore assigned negative oxidation number in all its compounds except in oxygen - fluorine compounds. For example, in OF2 oxygen being less electronegative than fluorine is assigned positive oxidation number. The following general rules are adopted for the assignment of oxidation numbers. 1. The oxidation number of an element in the free state or elementary state is always zero. Thus, oxidation number of H in H2 is zero, Cl in Cl2 is 0 and so on. They are represented as H0, Cl0, S0, Na0 etc. 2. The oxidation number of hydrogen in its compound is always +1 except in metallic hydrides where hydrogen has the oxidation number – 1 . In NaH, Na has the oxidation number + 1 and hydrogen, – 1. 3. All alkali metals have oxidation number of +1, alkaline earth metals have oxidation number of +2, Aluminium has oxidation number of +3, in their compounds. 4. Oxygen has the oxidation number – 2 in all its compounds except in peroxides where it is –1, in OF2 where it is + 2, in O2F2 where it is + 1 and in superoxide, KO2, RbO2, CsO2 where it is – 0.5. 5. The oxidation number of a monoatomic ion in an ionic compound is equal to its electrical charge. For a polyatomic ion the algebraic sum of the oxidation numbers of all the atoms is equal to the charge on the ion. For example. Cl– has the oxidation number –1. In SO42–, the algebraic sum of the oxidation numbers of sulphur and four oxygen atoms is equal to – 2. 6. In all compounds formed by the combination of a metal and a non-metal, the oxidation number of the metal atom is always positive and that of the non-metallic atom is negative. In KCl, K has always the positive oxidation number (+1) and Cl, the negative oxidation numbers (–1).

426 +2 CHEMISTRY (VOL. - I) 7. In compounds formed by two non-metallic atoms, the atom with lower electronegativity is assigned positive oxidation number and the atom with higher electronegativity is assigned a negative oxidation number. In CH4, H being less electronegative is assigned positive oxidation number (+1) and C, being more electronegative is assigned a negative oxidation number (– 4) 8. Fluorine has an oxidation number of –1 in all its compounds. Chlorine, Bromine and Iodine also have an oxidation number of –1, when they occur as halide ions in their compounds. Cl. Br and I when combine with oxygen i.e. in oxoacids and oxoanions, have positive oxidation numbers. 9. The algebraic sum of the oxidation numbers of all the atoms in a molecule is zero. In KClO4, the oxidation number of K is +1, Cl is +7 and oxygen is –2, so that the algebraic sum of the oxidation number is zero. [+ 1 + 7 + (– 2 x 4) = 0] IMPORTANT NOTE : 1. Metal can have zero oxidation number as in their nitrosyls and carbonyls. 2. The maximum oxidation number of an element can not exceed its group number in the periodic table. Thus, while calculating the O.N of S in Peroxymonosulphuric acid, H2SO5 (Caro's acid) and peroxydisulpharic acid, H2SO8 (Marshall's acid) by the basic rule, it has been found that the O.N. of S is 8 and 7 respectively. This is absurd, since the maximum O.N of S should be 6 because S belongs to Group VI (A) of the periodic table. H2SO5 is a peroxyacid and has one peroxy linkage (–O–O–) i.e. O22- , where the O.N of Oxygen is –1. If the O.N. of S is x, then H2SO5 º +1 x H2 S O23- + O2- 1 Hence, 2 x (+1) + x + (–2 x 3) + (–1) x 2 = 0 or, x – 6 = 0 and x = + 6 Thus, the O.N. of S is +6 which is compatible with the above rule. Again, H2S2O8 another peroxy acid of sulphur containing one peroxy linkage O2-1 Thus, H2S2O8 º +1 x 2- - 1 H S2 O 6 + O 2 2 Hence, (+1) x 2 + (x x 2) + (–2) x 6 + (–1) x 2) = 0 or, 2x = + 12 and x = + 6 Therefore, O.N. of S in this compound is + 6, compatible with the fact that O.N. of S can be + 6 at the maximum.

REDOX REACTION 427 Method for calculating oxidation numbers : For calculating the oxidation number of an atom in the molecule, the following procedure may be adopted. (i) Write down the formula of the molecule or ion with oxidation numbers (positive or negative) written on th top of each atom except for the atom whose oxidation number is to be determined. Let x be the oxidation number of that atom. (ii) Find out the algebraic sum of the oxidation numbers of all the atoms in the molecule (x being written as the oxidation number of the atom in question) (iii) Equate it to zero in case of a neutral molecule or to the charge in case of an ion. (iv) Solve for the value of x to get the oxidation number. Example 1 Calculate the oxidation number of Cr in K2Cr2O7. Solution Let x be the oxidation number of Cr. O.N. of K = + 1 and O has the O.N = – 2. Example 2 +1 x -2 Solution K2 Cr2 O7 (+1 x 2) + 2x + (–2 x 7) = 0 or, 2x = 14 – 2 = + 12 or, x = + 6 Hence, oxidation number of Cr in K2Cr2O7 is + 6 Find out the oxidation number of S in Na2S4O6. Let x be the O.N of S. O.N. of Na = + 1 and of O = – 2. Example 3 +1 x -2 Solution Na 2 S4 O6 . (1 x 2) + 4x + (– 2 x 6) = 0 or, 4x = 12 – 2 = 10 or, x = + 5/2 Hence, the O.N. of S in Na2S4O6 is +5/2. Calculate the oxidation number of chlorine in (a) NaClO2 (b) Cl2O (c) ClO3– Let x be the O.N of Cl. (a) +1 x -2 Na ClO2 1 + x + (–2 x 2) = 0 or, x = + 3 Hence, O.N of Cl in NaClO2 is + 3 (b) x -2 Cl2 O

428 +2 CHEMISTRY (VOL. - I) Example 4 2x – 2 = 0 Solution or, x = + 1 Hence, O.N. of Cl in Cl2O is +1. (c) x -2 Cl O3- x + (– 2 x 3) = – 1 or, x = + 5 Hence, O.N. of Cl in ClO3– is + 5. Calculate the oxidation number of (a) Ni in Ni (CO)4 (b) Fe in [Fe(CN)6]3– (a) Let x be the oxidation number of Ni x0 Ni (CO)4 x + (O x 4) = 0 (Since CO has the oxidation number 0) Hence, oxidation number of Ni is zero (b) Let the oxidation number of Fe be x x -1 [Fe (CN)6 ]3- x + (–1– x 6) = –3 or x = +3 Hence, the oxidation number of Fe is +3. NOTE : 1. The reaction by which SO2 and SO3 dissolve in water to give sulphurous and sulphuric acids respectively are not redox reactions. S+ 6O3(g) +NOHC2OHA®NGHE+I4NSOO4.(Na.q) + H+(aq) ...........(2) +4 +4 SO2 (g) + H2O ® HSO3– (aq) + H+ (aq) .........(1) NO CHANGE IN O.N. +6 HSO4 (aq) + H+ (aq) ...........(2) +6 SO3 (g) + H2O 2. Conversion of K2CrO4 to K2Cr2O7 is not a redox reaction. NO CHANGE IN O.N. +2 +6 -8 +2 +12 -14 K2Cr O 4 K2 Cr 2 O 7

REDOX REACTION 429 In the above examples the oxidation numbers of S and Cr do not change. Hence, they are not redox reactions. 10.4 TYPES OF REDOX REACTIONS : The redox reactions can be classified into different types. They are (1) Synthesis : When a new substance is formed by the direct union of simple substances, the reaction is termed as synthesis. Example : Magnesium burns in oxygen or air producing magnesium oxide. 00 +2–2 2M g + O 2 ® 2MgO Similarly, hydrogen combines with oxygen forming water. 00 +1–2 2H2 + O2 ® 2H2O and nitrogen combines with hydrogen in presence of a catalyst under suitable conditions of temperature and pressure to produce ammonia. 00 –3+1 N2 + 3H2 ® 2NH3 2. Analysis or Decomposition : This type of reaction is just the reverse of synthesis. Here the compound splits up into its constituents or simpler substances. Example : Mercuric oxide when strongly heated decomposes to mercury and oxygen. +2–2 0 0 +1 +5 –2 +1 –1 0 2HgO ® 2Hg + O2 ; 2KClO3(s) 2KCl(s) + 3O2(g) NH4Cl ® NH3 + HCl All decomposition reactions are not a redox reaction. +2 +4 –2 +2 –2 +4 –2 CaCO3 ® CaO + CO2 . 3. Displacement or Substitution : When one element displaces or substitutes another element from its compound the reaction is termed as displacement or substitution reaction. Example : Zinc reacts with dilute sulphuric acid producing hydrogen and zinc sulphate. 0 +2 Zn + H2SO4 ® H2 + ZnSO4 Here, zinc displaces hydrogen from H2SO4. Similarly, iron displaces copper from copper sulphate solution. 0 +2 0 +2 Fe + CuSO4 ® Cu + FeSO4. Some metals are capable of displacing hydrogen from acids. 0 +1 –1 +2 –1 0 Zn + 2HCl ® ZnCl2 + H2 0 +1 –1 +2 –1 0 Mg + 2HCl ® MgCl2 + H2

430 +2 CHEMISTRY (VOL. - I) These reactions are used to prepare dihydrogen in the laboratory. All alkali metals and some alkaline earth metals (Ca, Sr and Ba) which are very good reductants, will displace hydrogen from cold water. 0 +1 –2 +1 –2 +1 0 2Na + 2H2O ® 2NaOH + H2 0 +1 –2 +2 –2 +1 0 Ca + 2H2O ® Ca (OH) + H2 Among halogens fluorine is most reactive and can replace chloride, bromide and iodide ions in solution. In fact, fluorine can displace oxygen of water : +1 –2 0 +1–1 0 2H2O + 2F2 ® 4HF + O2 For this reason the displacement reactions of chlorine, bromine and iodine using fluorine are not carried out in aqueous solutions. 4. Disporportionation reactions : This type of redox reactions involve an element in one oxidation state which is simultaneously oxidised and reduced. Decomposition of hydrogen peroxide belongs to this category. +1 –1 +1 –2 0 2H2O2 ® 2H2O + O2 Here the oxygen atom in hydrogen peroxide molecule is present in –1 oxidation state and decreases to –2 in water molecule whereas in oxygen molecule it increases to 0. Non-metas like phosphorus, sulphur and chlorine undergo disproportionation in the alkaline medium. 0 –3 +1 Ex. P4 + 3NaOH + 3H2O ® PH3 + 3H3PO2 0 –2 +2 S8 + 12 NaOH ® 4Na2S + 2Na2S2O3 + H2O 0 +1 –1 Cl2 + 2 NaOH ® NaClO + NaCl + H2O 10.5 USES OF OXIDATION NUMBER : The concept of oxidation number is used for 1. Writing the formula of the compounds. 2. Classifying the types of reagents or reactants in redox reactions. 3. Balancing equations for redox reactions. 1. Writing formulae of the compounds : Since the algebraic sum of positive oxidation number and negative oxidation number of atoms or groups in a compound is zero, the formula of a compound can easily be written knowing the oxidation number of atoms or groups present in the compound. For example, in calcium chloride the oxidation number of Ca is + 2 and chlorine is – 1, Hence, calcium chloride can be written as

REDOX REACTION 431 CaCl2, so that the algebraic sum of oxidation numbers of Ca and Cl will be zero. Knowing the oxidation number of Al to be + 3 and of sulphate – 2, aluminum sulphate can be written as Al2(SO4)3. 2. Classifying the types of reagents or reactants in redox reactions : Oxidation is a chemical reaction in which there is increase in the oxidation number of an element. When Mg burns in oxygen to form MgO, O.N. of Mg increases from 0 to +2. Therefore this is an oxidation reaction where Mg is oxidised. Conversion of CH4 to CO2 or CCl4 by reaction with O2 and Cl2 respectively increases the oxidation number of C form – 4 to + 4 and hence they are oxidation reactions where C is oxidised. Reduction is a chemical reaction in which there is a decrease in the oxidation number of an element. Consider the same example of the formation of MgO from Mg and oxygen. The oxidation number of oxygen decreases from 0 to – 2. Thus, oxygen is said to have undergone reduction. ¾O¾x¾ida¾tio¾n¾: an¾i¾nc¾rea¾se¾in¾o¾xid¾at¾ion¾n¾um¾b¾er ® ¬¾.....¾.....¾.-4¾-¾3 -¾1¾¾0 +¾1 ¾+2¾+3¾+¾4 ¾.....¾.¾....¾ Reduction : a decrease in oxidation number Reagents which can increase the oxidation number of an element in a compound are called oxidising agents. In the foregoing examples, O2 and Cl2 are oxidising agents, because they increase the oxidation number of Mg and C respectively. The oxidation numbers of the oxidising agents are decreased. Similarly, reagents which can lower the oxidation number of an element in a compound are called reducing agents. SnCl2 reduces HgCl2 to HgCl, Since the oxidation number of Hg decreases from + 2 to + 1. Therefore, SnCl2 is a reducing agent. 2HgCl2 + Sn2+Cl2 ® 2HgCl + Sn4+Cl4 The oxidation number of the reducing agent has increased (+ 2 to + 4). The foregoing facts can be summerised as follows : (i) Oxidation = Loss of electron = Increase in oxidation number. (ii) Oxidising agent = Electron acceptor = Decrease in oxidation number. (iii) Reduction = Gain in electron = Decrease in oxidation number. (iv) Reducing agent = Electron donor = Increase in oxidation number. 3. Balancing Redox reactions : For balancing equations for oxidation - reduction reactions, two methods are generally adopted. They are : (I) Oxidation number method. (II) Ion - electron method.

432 +2 CHEMISTRY (VOL. - I) (I) Oxidation number method : This method is based on the fact that the total increase in the oxidation number must be equal to the total decrease in the oxidation number in the oxidation - reduction reaction. Like other chemical equations, the reactants and the products must be known. The various steps followed for balancing a redox reaction are : Step - 1. Write skeleton equation representing the chemical change. Step - 2. Assign oxidation numbers to all atoms in the reaction and find out which atoms undergo change in oxidation number in the reaction. Step - 3 Find out the increase and decrease in oxidation number for the reaction of 1 mole Step - 4 of reducing and oxidising agents respectively. Introduce suitable co-efficients to make the increase equal to the decrease in oxidation number. Balance the equation for all other atoms and finally balance for hydrogen and oxygen. Complete the equation. The following examples illustrate the procedure outlined. Example 1 : Balance the equation by oxidation number method. CuO + NH3 ® Cu + N2 + H2O (a) Skeleton equation - CuO + NH3 ® Cu + N2 + H2O (b) Assign oxidation number and find out the changes in oxidation number. Decrease in O.N from +2 to 0 2+ 2- 3- 3+ o o 2+ 2- Cu O + N H3 Cu + N2 + H2 O increase in O.N. from –3 to 0 (c) Balance the increase in oxidation number with decrease in oxidation number. For one molecule of CuO, the O.N. of Cu reduces from +2 to O and for one molecule of NH3, the O.N. of N increases from – 3 to 0. Therefore, they must react in the ratio 3 CuO : 2 NH3. Hence, 3 CuO + 2NH3 ® 3Cu + N2 + H2O and (d) Balance the hydrogen and oxygen and complete the equation. 3CuO + 2NH3 ® 3Cu + N2 + H2O

REDOX REACTION 433 Example 2 : Balance the following equation by oxidation number method. FeCl3 + H2S ® FeCl2 + S + HCl (a) Skeleton equation FeCl3 + H2S ® FeCl2 + S + HCl (b) Assign oxidation number and find out the changes in O.N. Decrease in O.N from +3 to +2 +3 -3 +2 -2 +2 -2 o +1 -1 FeCl2 + S + HCl FeCl3 + H2S increase in O.N. from –2 to 0 (c) Balance the increase in oxidation number with the decrease in oxidation number. Oxidation number of Fe decreases from + 3 to + 2 ( F+e3 Cl3 ® F+e2 Cl2) and the oxidation number of S increases from –2 to 0 (H2 S-2 ® S0 ) Therefore, 2 molecules of FeCl3 should react with one molecule of H2S. Hence. 2FeCl3 + H2S ® 2FeCl2 + S + HCl. (d) Balance the number of hydrogen atoms and complete the equation. 2FeCl3 + H2S ® 2FeCl2 + S + 2HCl. Example 3 : Balance the reaction between K2Cr2O7 and KI in presence of dilute H2SO4 by oxidation number method. K2Cr2O7 + KI + H2SO4 ® Cr2(SO4)3 + K2SO4 + I2 + H2O. (a) Skeleton equation K2Cr2O7 + KI + H2SO4 ® Cr2(SO4)3 + K2SO4 + I2 + H2O. (b) Assign oxidation numbers and find out changes in O.N. decrease in O.N from + 12 to + 6 +2 +12 -14 +1 -1 +2 +6 -8 +6 +6 -8 - +2 +6 -8 o +2 -2 K2 Cr2 O7 + KI + H2SO4 Cr2 (SO4)3 + K2SO4 + I2 + H 2O . Increase in O.N. from -1 to 0 (c) Balance the increase in O.N with the decrease in O.N. For each molecule of K2Cr2O7, there is 6 units of reduction of O.N. (K2 C+r122 O7 ® C+r62 (SO4)3) and for each molecule of KI, there is one unit of oxidation. Therefore, for balance, one molecule of K2Cr2O7 must react with 6 molecules of KI. Hence, the equation will be

434 +2 CHEMISTRY (VOL. - I) K2Cr2O7 + 6KI + H2SO4 ® Cr2(SO4)3 + 312 + K2SO4 + H2O. (d) Balance the number of hydrogen and oxygen atoms and complete the equation. Since the reaction is carried out in acidic medium, 14H+ must be added on the reactant- side which means 7 H2SO4 molecules must be required. K2Cr2O7 + 6KI + 7H2SO4 ® Cr2(SO4)3 + 4K2SO4 + 312 + 7H2O. (II) Ion - electron method : In 1927. Jette and La Mer developed this method for balancing redox reactions. The ion - electron method is based on the fact that oxidation involves loss of one or more electrons and reduction, gain of one or more electrons. The method is limited to ionic reactions in aqueous solutions. Ionic equations can be converted to molecular equations and vice versa. The equation is divided into two half reactions, one representing the oxidation and the other, the reduction reaction. Though ions take part in the reaction, this method avoids the spectator ions (these which do not take part in the reaction), but molecules of H2O are added like H+ in acid medium and OH– in basic medium for balancing partial ionic equations. The unit change in oxidation or reduction is a change of one electron, represented by e. Let us consider a common redox reaction of FeCl3 by SnCl2. The equation can be written as FeCl3 + SnCl2 ® FeCl2 + SnCl4 In aqueous solution, ionic representation is Fe3+ + 3Cl– + Sn2+ + 2Cl– ® Fe2+ + 2Cl– + Sn4+ + 4Cl– Chloride ions appear on both sides and therefore they are spectator ions and hence omitted. Only those ion taking part in the reaction are taken into account. Thus, a simplified equation can be written as : Fe3+ + Sn2+ ® Fe2+ + Sn4+ This reaction can be divided into two half reactions. Fe3+ ® Fe2+ .........................(i) representing reduction reaction and Sn2+ ® Sn4+ ........................(ii) representing the oxidation reaction. Equation (i) and (ii) must now be balanced not only with regard to the number of atoms, but also with respect to electrical charges on both sides of the equation. Therefore, Fe3+ + e ® Fe2+ .........................(i a) Sn2+ ® Sn4+ + 2e ........................(ii a)

REDOX REACTION 435 Both the partial equation (i – a) and (ii – a), though balanced with regard to the number of atom, are not balanced electrically, since on adding both the partial equations there is the residual charge on the right hand side. Therefore, the whole equation (i – a) must now be multiplied by 2 and we have, 2Fe3+ + 2e ® 2Fe2+ .........................(i b) Sn2+ ® Sn4+ + 2e ........................(ii b) Adding (i – b) and (ii – b), we obtain 2Fe3+ + Sn2+ + 2e ® 2Fe2+ + Sn4+ +2e and by cancelling the electrons common to both the sides, we have, 2Fe3+ + Sn2+ ® 2Fe2+ + Sn4+ Example 1 : Oxidation of concentrated hydrochloric acid by potassium permanganate. KMnO4 + HCl ® MnCl2 + KCl + H2O + Cl2 Eliminating the spectator ions, the equation can be written as MnO4– + Cl– ® Mn2+ + H2O + Cl2 The partial half reaction for reduction is MnO4– ® Mn2+ To balance this in respect of atoms, we need 8H+. Then, MnO4– + 8H+ ® Mn2+ + 4H2O and to balance it electrically, 5e will be required since O.N of Mn changes from +7 to +2, Hence MnO–4 + 8H+ + 5e ® Mn2+ + 4H2O ....................(I) The partial half reaction for oxidation is Cl– ® Cl2 To balance it in respect of atoms, 2Cl– will be necessary 2Cl– ® Cl2 and to balance it electrically, 2e will be taken out since Cl– ® Cl0 needs removal of one e. Hence, 2Cl– – 2e ® Cl2 ..................................(ii) The above two partial equations are to be adjusted in such a way that the gain of electron is equal to the loss of electron. This is done by multiplying (i) by 2 and (ii) by 5. 2[MnO4– + 8H+ + 5e ® Mn2+ + 4H2O] 5[2Cl– – 2e ® Cl2] —————————————————————————— On adding, 2MnO4– + 16H+ + 10Cl– ® 2Mn2+ + 5Cl2 + 8H2O is obtained as the balanced equation.

436 +2 CHEMISTRY (VOL. - I) Example 2 : Oxidation of ferrous sulphate by potassium dichromate in presence of sulphuric acid K2Cr2O7 + H2SO4 + FeSO4 ® Cr2(SO4)3 + K2SO4 + Fe2(SO4)3 + H2O Eliminating the spectator ions, we find that Cr2O72–, Fe2+ and H+ are taking part in the reaction and in the product side we have Cr3+, Fe3+ and H2O. Balancing the reduction reaction atomically and electrically, we have Cr2O72– + 14H+ + 6e ® 2Cr3+ + 7H2O .................(i) Balancing the oxidation reaction atomically and electrically. we have, Fe2+ – e ® Fe3+ ....................................(ii) One dichromate ion utilises 6e and ferrous ion liberates 1e, hence partial equation (i) and (ii) must be in the ratio 1 : 6. Cr2O72– + 14H+ + 6e ® 2Cr3+ + 7H2O 6[Fe2+ – e ® Fe3+] ————————————————————— Cr2O72– + 6Fe2+ + 14H+ ® 2Cr3+ + 6Fe3+ + 7H2O is the balanced equation. Example 3 : Oxidation of sodium oxalate by potassium permanganate in presence of sulphuric acid. KMnO4 + Na2C2O4 + H2SO4 ® MnSO4 + K2SO4 + Na2SO4 + CO2 + H2O Ions involved in the reaction are MnO4–, C2O42– and H+. Reduction half reaction : MnO4– + 8H+ + 5e ® Mn2+ + 4H2O .......................(I) Oxidation half reaction : C2O42– – 2e ® 2CO2 .............................................. (ii) To make the final equation electrically neutral, multiply (i) by 2 and (ii) by 5. 2[MnO–4 + 8H+ + 5e ® Mn2+ + 4H2O] 5[C2O–4– –2e ® 2CO2] ———————————————————————— 2MnO–4 + 5C2O–4– + 16H+ ® 2Mn2+ + 10CO2 + 8H2O Example 4 Redox reactions involved in iodometric estimation of copper. (a) CuSO4 + KI ® CuI + K2SO4 + I2 (b) I2 + Na2S2O3 ® Na2S4O6 + NaI

REDOX REACTION 437 (a) The ions involved are Cu2+ and I– and the two half reactions are Reduction : Cu2+ ® Cu+ Oxidation : I– ® I2 Balancing both the half reaction atomically and electrically, we have 2Cu2+ + 2e ® 2Cu+ 2I– – 2e ® I2 ——————————— 2Cu2+ + 2I– ® 2Cu+ + I2 (b) The reactants involved are I2 and S2O32–. The two half reactions are Oxidation : S2O32– ® S4O62– Reduction : I2 ® 2I– Balancing both the half reaction atomically and electrically, 2S2O32– – 2e ® S4O62– I2 + 2e ® 2I– —————————————— 2S2O32– + 12 ® S4O62– + 2I– Example 5 : Reaction of hydrogen peroxide with potassium iodide in presence of sulphuric acid. H2O2 + KI + H2SO4 ® K2SO4 + I2 + H2O Ions and molecules involved are I– , H+ and H2O2 Oxidation half reaction : I– – e ® 1 I2 ...................................(i) 2 Reduction half reaction : H2O2 + 2H+ + 2e ® 2H2O ...................(ii) The two partial equation (i) and (ii) must take place in the ratio 2 : 1 to make the equation electrically neutral. Therefore, 2l– – 2e ® I2 H2O2 + 2H+ + 2e ® 2H2O ———————————— H2O2 + 2I– + 2H+ ® I2 + 2H2O. Some of the common half reactions for oxidising and reducing agents are given below. Oxidising agents : MnO4– + 8H+ + 5e ® Mn2+ + 4H2O (i) KMnO4 : In acid medium MnO4– + 2H2O + 3e ® MnO2 + 4OH– In neutral medium MnO4– + e ® MnO42– In alkaline medium

438 +2 CHEMISTRY (VOL. - I) (ii) K2Cr2O7 : In acidic medium Cr2O72– + 14H+ + 6e ® 2Cr3+ + 7H2O. (iii) Halogens X2 : (X = Cl , Br or I). X2 + 2e ® 2X– (iv) H2O2 : H2O2 + 2H+ + 2e ® 2H2O (v) NaOCl : ClO– + H2O + 2e ® Cl– + 2OH– (vi) HNO3 : (a) Concentrated. NO3– + 2H+ + e ® NO2 + H2O (b) Mod. Conc. NO3– + 4H+ + 3e ® NO + 2H2O (c) Dilute. NO3– + 10H+ + 8e ® N2O + 5H2O Reducing agents : (i) H2C2O4 : C2O42– – 2e ® 2CO2 (ii) HI : 2HI ® I2 + 2H+ + 2e H2O2 ® 2H+ + O2 + 2e (iii) H2O2 : 2S2O32– ® S4O62– + 2e (iv) Na2S2O3 : H2S ® 2H+ + S + 2e (v) H2S : SO2 + 2H2O ® SO42– + 4H+ + 2e (vi) SO2 : H2SO3 + H2O ®SO42– + 4H+ + 2e (vii) H2SO3 : Equivalent mass of Oxidants and Reductants The equivalent mass of an oxidising agent or oxidant and a reducing agent or reductant can be calculated from the change in oxidation number or from the change in the number of electrons per ion in any redox reaction. Equivalent mass of an oxidising agent or oxidant = Molecular mass of the oxidising agent Change in oxidation number per mole = Molecular mass of the oxidising agent No. of electrons gained per mole Similarly, Equivalent mass of a reducing agent or reductant = Molecular mass of the reducing agent Change in oxidation number per mole = Molecular mass of the reducing agent No. of electrons lost per mole

REDOX REACTION 439 Example 1 : Calculate the equivalent mass of KMnO4 in acid medium. The partial ionic equation for the reaction can be written as M7+n O–4 + 8H+ + 5e ® Mn2+ + 4H2O . Here; the O.N of Mn changes from +7 to + 2. Thus change in O.N = 7 – 2 = 5. The number of electrons gained by MnO–4 is 5. Therefore, Equivalent mass of KMnO4 = MnO4- = Mol mass of KMnO4 = 158 = 31.6. 5 5 5 Example 2 : Calculate the equivalent mass of K2Cr2O7 in acid medium. The above reaction can be ionically represented as C+r122 O27- + 14H + + 6e ® 2Cr3+ + 7H 2O Here, the change of O.N per mole = 12 – 6 = 6. the number of electron gained by Cr2O27– is 6. Therefore, Eq. mass of K2Cr2O7 = Cr2O627- = Mol mass of K2Cr2O7 = 294 = 49. 6 6 Example 3 : Find out the equivalent mass of FeSO4. 7H2O in acid medium. FeSO4 is oxidised to Fe2(SO4)3 in acid medium by oxidising agents like K2Cr2O7 or KMnO4. This can be represented as Fe2+ – e 2 Fe3+ Here, change of O.N is from + 2 to +3 i.e 1 per mole of Fe 2+. Therefore, Eq. mass of FeSO4. 7H2O. = Fe 2 + = Mol. mass of FeSO4 .7H2O 1 1 = 278 = 278. 1 Example 4 : Find out the equivalent mass of Na2S2O3. The partial ionic equation for the reaction can be written as 2S2O32– – 2e 2 S4O62– Here, the number of electrons lost is 2 per 2 moles of thiosulphate. Therefore, equivalent mass of sodium thiosulphate = 2 Molecular mass of sod. thiosulphate 2 = Mol. mass of sod. thiosulphate 1 Hence, the equivalent mass of sodium thiosulphate is same as its molecular mass.

440 +2 CHEMISTRY (VOL. - I) Example 5 : Find out the equivalent mass of sodium oxalate in acid medium. The partial ionic equation for the reaction can be represented as C2O42– – 2e 2 2CO2 Here, the number of electrons lost is 2 per 2 mole of oxalate. Therefore, equivalent mass of Na2C2O4 is equal to the molecular mass of Na2C2O4 divided by 2. In otherwords, Equivalent mass of Na2C2O4 = Mol. mass of Na 2C 2O 4 2 10.6 APPLICATIONS OF REDOX REACTIONS : Redox reactions have vast applications in chemistry, in industry and also in our daily life. 10.6.1. Quantitavie analysis : Tritration involving Redox Reactions. Redox reactions are very useful in quantitavie analysis. Redox titrations involve the reactions between oxidants and reducing agents and help in the estimation of unknown substances in solution. (a) Potassium permanganate titrations : As discussed earlier KMnO4 acts as an oxidising agent in acidic, alkaline and neutral medium. It is violet in colour in Mn+7 state and changes to colourless compound in Mn+2 state. So it is used as a self-indicator. When it is used against reducing agents like oxalic acid, ferrous sulphate, oxalates etc, just beyond the equivalence point the first lasting tinge of pink colour due to MnO–4 appears, which enables it to be used as a self- indicator. COO– 5 C| OO– (aq) + 2MnO4 (aq) + 16H+ (aq) 2 2Mn2+ (aq) + 10 CO2(g) + 8 H2O Oxalate ion permanganate ion 5 Fe2+ (aq) + MnO4 (aq) + 16 H+(aq) 2 2Mn2+(aq) + 10 CO2(g) + 8 H2O(l) Ferrous ion Permanganate ion (b) Potassium dichromate titrations : In potassium dichromate titrations indicators like diphenyl amine or N–phenyl anthranilic acid are used. At the end point diphenylamine changes colour from bluish green to × purple or to intense blue. In this method strength of reducing agents like FeSO4, Mohr's salt (NH4)2SO4×FeSO4×6H2O is determined. 6 Fe2+ (aq) + Cr2O72– (aq) + 14H+ (aq) 2 2Cr3+ (aq) + 6Fe3+(aq) + 7 H2O (l)

REDOX REACTION 441 (c) Iodometric tritrations : These titrations are carried out in two steps. In the first step, oxidising agents like KMnO4, K2Cr2O7, CuSO4, peroxides etc. are treated with an excess of KI when I2 is liberated quickly and quantitatively. In the second step, the liberated iodine is titrated against standard sodium thiosulphate solution using starch as indicator. At the end point the blue colour disappears. 2 Cu2+ (aq) + 4I– (aq) 2 Cu2I2 (s) + I2 (aq). I2 (aq) + 2 S2O32– (aq) 2 2I– (aq) + S4O62– (aq). 10.6.2 Redox Reactions and Electrode Processes : An electrochemical cell or battery is a device which converts chemical energy to electrical energy. These are widely used in our daily life to run a number of small and big gadgets and equipments. In these cells redox reactions take place. Ex : Galvanic Cell. In this cell Zn rod dipped in ZnSO4 solution and Cu rod dipped in CuSO4 solution serve as anode and cathode respectively. Zn (s) 2 Zn2+ (aq) + 2e– (Anode 2 oxidation) Cu2+(aq) + 2e– 2 Cu (s) (Cathode 2 Reduction) Here, electrons flow from anode to cathode thereby causing flow of electric current in the reverse direction (i.e. from cathode to anode) 10.6.3 Redox Reactions and Extraction of metals: By using a suitable reducing agent, oxides of metals are reduced to metals. For example, ferric oxide Fe2O3 is reduced to iron in the blast furnace using coke as the reducing agent. Fe2O3(s) + 3C(s) 2 2Fe(s) + 3CO(g) Similarly, Al2O3 is reducd to Al by cathodic reduction in an electrolytic cell. Other metals like Li, Na, K, Mg, Ca etc. are also obtained commercially by electrolytic reduction methods. 10.6.4 Redox Reaction and supply of energy : The energy required in our daily life is obtained by the oxidation of fuels. For example, oxidation of fuels like wood, gas, kerosene, petrol etc. produces a large amount of energy required for various purposes in our life. Fuels (wood, gas, kerosene etc) + O2 2 CO2 + H2O + Other products + Energy Human body needs energy for proper functioning. This is obtained by the oxidation of glucose in our body to CO2 and H2O. C6H12O6(aq) + 6 O2(g) 2 6 CO2(g) + 6 H2O(l) + Energy

442 +2 CHEMISTRY (VOL. - I) 10.6.5 Redox Reaction and Photosynthesis : Green plants convert water and carbondioxide into carbohydrate in presence of chlorophyll and the process is defined as photosynthesis. This is represented as 6 CO2(g) + 6 H2O(l) + light energy Chlorophyll C6H12O6(aq) + 6 O2(q) We can see CO2 is reduced to carbohydrates while H2O gets oxidised to O2 and hence it is a redox reaction. The energy is provided by sunlight for this reaction. This reaction is the source of food for plants and animals. It also maintains a constant supply of 21% of O2 by volume in the atmosphere required for conbustion of fuels and breathing of all the living creatures in the world. Other important applications include production of chemicals which are based on redox reactions. Chemicals like caustic soda, chlorine etc. are commercially produced by redox reactions in electrolytic cell. 10.7 THEORETICAL ASPECTS OF ACID-BASE TITRATION : VOLUMETRIC ANALYSIS-TITRATION Determination of Strength and Equivalent Mass of Acid and Base The equivalent mass of an acid or base can be determined by the volumetric method, known as titration.The progressive addition of an acid to a base or vice versa is termed as titration which provides a sensitive method of determining the strength of one when the strength of the other is known. This is carried out in presence of an indicator. An indicator is a substance which indicates the end point or neutralisation point by a colour change. The indicator shows different distinct colours in different media. The following table shows the change of colour of different indicators in different media. Table 10.3 Colour change of some indicators Indicator Colour in Colour in Colour in basic medium neutral medium acid medium (at the end point) Methyl orange Yellow Red Methyl red Yellow Orange Red Litmus Blue Red Red Violet Phenolphthalein Pink Colourless Colourless The choice of indicator will depend upon the nature of the acid and alkali. For example, in the titration of a strong acid with a strong base, any indicator can be used. In the titration of strong acid and weak base (NH4OH), methyl orange and methyl red are suitable. In the titration of weak acid (acetic acid) and strong base, phenolphthalein is suitable. It is important to note that no titration is possible if both the acid and base are weak. The principle underlying titration is that an acid and a base react in the ratio of their gram-equivalents for complete neutralisation. Consider the reaction HCl + NaOH ® NaCl + H2O 36.5 g 40 g

REDOX REACTION 443 From the equation we find that 36.5 g of HCl will completely neutralise 40 g of NaOH (36.5 and 40 are the equivalent mass of HCl and NaOH respectively). Thus, 1000 ml of (N) HCl will completely neutralise 1000 ml. of (N) NaOH, since they contain one gram - equivalent of the substance in the given volume of the solution. In other words, 1000 ml of (N) HCl º 1000 ml of (N) NaOH or, 10 ml of (N) HCl, º 10 ml of (N) NaOH. or 10 ml of (N/10) HCl º 10 ml of (N/10) NaOH. Suppose V1 ml of acid of strength S1 in terms of normality require V2 ml of alkali of strength S2 in terms of normality for complete neutralisation. Then strength of the acid S1 = no. of gram - equivalents of acid = volume in litres no. of gram- equivalents of acid V1 / 1000 (in terms of normality) \\ Number of gram - equivalents of acid = S1 x V1/1000 Similarly, no. of gram - equivalents of alkali = S2 ´ V2/1000 Since acid and alkali react in the ratio of their gram - equivalents for complete neutralisation. S ´ V = S ´ V 1 1 2 2 1000 1000 or, S ´ V = S2 ´ V2 1 1 Therefore, Strength of acid (in terms of normality) ´ Volume of acid (in ml) = Strength of alkali (in terms of normality) ´ Volume of alkali (in ml). In this equation, if three factors are known, the fourth factor can be calculated. Antiparallax card Burette (Acid) Conical Flask Pipette (Alkali) Fig. 10.1 A Volumetric titration. Glazed Tile

444 +2 CHEMISTRY (VOL. - I) The determination of equivalent mass of acids and bases involves the following steps : (i) For the determination of the equivalent mass of the acid, the standard solution of the base is required and vice versa. (ii) A solution of the acid or base is prepared by dissolving an accurately weighed quantity of acid or base in the solvent and making up to a known volume. (iii) The acid is taken in a burette and a known volume of the alkali, in a conical flask with a drop or two of a suitable indicator. Acid is added from the burette till the neutralisation point is reached as indicated by the change of colour of the indicator. (iv) Volume of acid, added from the burette is accurately known. A known volume of alkali is taken in the conical flask. If the strength of the acid is known, the strength of the alkali and vice versa can be calculated using the relation, V1S1 = V2S2 . where V1 = Volume of the acid (in ml) S1 = Normality of the acid V2 = Volume of the alkali (in ml) S2 = Normality of the alkali. (v) The strength in grams per litre is divided by the normality to give the equivalent mass. In general, Grams/litre = Normality x Equivalent mass. Fundamental Principles in Titration : 1. V1 x S1 = V2 x S2 Volume of acid (in ml) x Strength of acid (in terms of normality) = Volume of alkali ( in ml) x Strength of alkali (in terms of normality) 2. Acid and alkali solutions of the same strength in terms of normality neutralise one another in equal volumes. 3. Grams per litre = normality x equivalent mass. Acidimetry is a method by which the strength of an alkali is determined by titrating it against a standard acid. Alkalimetry is a method by which the strength of an acid is determined by titrating against a standard alkali. Problem 1 : 25.2 ml. of dilute sulphuric acid is completely neutralised by 24.0 ml of 0.105 (N) NaOH solution. Find out the mass of sulphuric acid per litre of the solution. (U.U. I.D, 1978)

REDOX REACTION 445 Solution : Suppose V1 = volume of H2SO4 = 25.2 ml S1 = x (N) = Strength of H2SO4 V2 = Volume of NaOH = 24.0 ml S2 = 0.105 (N) = Strength of NaOH. By applying , V1S1 =V2S2 S1 = V2S2 V1 or x (N) = 24ml ´ 0.105 (N) = 0.1. (N) 25.2 ml Hence, normality of the acid is 0.1. Grams per litre of H2SO4 = Normality x Equivalent mass = 0.1 x 49 (since eq mass of H2SO4 = 49) = 4.9. Hence, mass of H2SO4 per litre of the solution = 4.9 g. Problem 2 How many mililiters of 1.1 (N/10) HCl will react with 1 g of marble ? (Ca = 40, C = 12, O = 16) (U.U.I.Sc., 1981 S) Solution : Marble is CaCO3 and its molecular mass = 100 CaCO3 + 2HCl ® CaCl2 + CO2 + H2O 100 2 x 36.5 Thus, 100 g of CaCO3 require (2 x 36.5)g of HCl for complete reaction = 2000 ml of (N) HCl Therefore, 1 g of CaCO3 will require 2000 ml of (N) HCl = 20 ml of (N) HCl 100 Suppose 20 ml of (N) HCl = x ml of 1.1(N/10) HCl = x ml of 0.11 (N) HCl \\x = 20 ml = 181.8 ml. 0.11 Hence, 181.8 ml of 1.1 (N/10) HCl will react with 1 g of marble. Problem 3 : How many mililitres of concentrated sulphuric acid of specific gravity 1.84 containing 98% H2SO4 by mass are required to prepare 200 ml of 0.5 (N) H2SO4 ? (H = 1, O = 16 and S = 32)

446 +2 CHEMISTRY (VOL. - I) Solution : Eq. mass of H2SO4 = Mol.mass = 98 = 49 basicity 2 1000 ml of (N)H2SO4 contain 49 g of H2SO4 200 ml of 0.5(N) H2SO4 will contain 49 x 200 x 0.5 g 1000 = 4.9 g of H2SO4. The given acid contains 98% H2SO4 by mass, which means 98 g of H2SO4 is present in 100 g of H2SO4 solution. \\ 4.9 g of H2SO4 will be present in 100 x 4.9 g = 5 g of acid. 98 Specific gravity = mass = 1.84 volume Hence, volume of acid required = smpa×gsrs = 5 = 2.72 ml. 1.84 Problem 4 20 ml of a solution containing 3 g per litre of a dibasic acid neutralises 10 ml of a solution of Na2CO3 containing 5.3 g per litre. Calculate the molecular mass of the acid. (Na = 23. C = 12, and O = 16) (U.U., ISc., 1981) Solution : Equivalent mass of Na2CO3 = 106 = 53 2 53 g of Na2CO3 per litre gives (N) solution \\ 5.3 g of Na2CO3 per litre will give (N/10) solution. Let the strength of the dibasic acid be S1(N). 20 ml of S1(N) dibasic acid º 10 ml of (N/10) Na2CO3 solution. \\ S1(N) = 10 x N = 0.05(N) 10 x 20 Thus, normality of the acid = 0.05 Gms/litre = Normality x Eq mass, of the dibasic acid. \\ Equivalent mass of the dibasic acid = 3 = 60 0.05 Molecular mass of the dibasic acid = Eq. mass basicity = 60 x 2 = 120 Problem 5 : 6 g of impure sodium carbonate is present in one litre of solution. 10ml of the resulting solution on titration against a (N/10) solution of H2SO4 neutralises 9.5 ml of the latter. Calculate the percentage of purity of Na2CO3. (Equivalent mass of Na2CO3 = 53) (U.U.I. Sc. 1981 S)

REDOX REACTION 447 Solution : Let x = Normality of Na2CO3 solution. 10 ml of x(N) Na2CO3 soln º 9.5 ml of (N/10) H2SO4 Therefore, x(N) = 9.5 (N) = 0.095 (N) 10 x 10 . Normality of Na2CO3 soln . = 0.095. Equivalent mass of Na2CO3 = 53 g / litre of Na2CO3 = Normality x Eq.mass = 0.095 x 53 = 5.035 i.e. 5.035g of pure Na2CO3 is present per litre of solution. 6g impure Na2CO3 contain 5.035 g of pure Na2CO3 100 g of impure Na2CO3 will contain 5.035 ´ 100 = 83.92 g of pure Na2CO3 . 6 Hence, purity of Na2CO3 is 83.92 %. Problem - 6 : 1.575 g of (COOH)2 xH2O crystals are dissolved in water and the volume made up to 250 ml. 8.34 ml of this solution is required for complete neutralisation of 10 ml of (N/12) KOH solution. Calulate the number of water molecules associated with the molecule of the acid. Solution : g/litre of (COOH)2 xH2O = 1.575 ´ 1000 = 6.3 250 If S1 = normality of the acid and S2 = normality of KOH and V1 = volume of the acid and V2 = volume of KOH, Then, V1S1 = V2S2 S1 x 8.34 = 10 x 1/12 or, S1 = 10 = 0.099 =0.1 (approx) 12 ´ 8.34 \\ Eq mass of acid = g./litre = 6.3 = 63.00 normality 0.1 Basicity of the acid = 2 Molecular mass = Eq. mass x basity = 63 x 2 = 126. As per formula, molecular mass of acid (COOH)2 xH2O = (12 + 32 + 1) 2 + 18x = 90 + 18 x 90 + 18x = 126 or, x= 126 - 90 =2 18 Hence, x = no of water molecules in a molecule of acid is 2.

448 +2 CHEMISTRY (VOL. - I) Problem 7 : 3.5 g of a mixture of NaOH and KOH were dissolved and made up to 250 ml. 25 ml of this solution were completely neutralised by 17 ml of (N/2) HCl solution. Calculate the percentage of KOH in the mixture. (Na = 23, K = 39 , Cl = 35.5, O = 16) Solution : Suppose the amount of NaOH in the mixture = x g. \\ Amount of KOH = (3.5 – x) g. 25 ml of alkali solution º 17 ml of (N/2) HCl. or, 250 ml of alkali solution º 170 ml of (N/2) HCl 1000 ml of (N) HCl contain 36.5 g of HCl 170 ml of N HCl will contain 36.5 ´ 170 = 3.1025 g of HCl. 2 1000 ´ 2 Eq mass of NaOH = 40 and that of KOH = 56 40 g of NaOH require 36.5 g of HCl x g of NaOH will require 36.5 ´ x g of HCl 40 Similarly, 56 g of KOH require 36.5 g of HCl (3.5 – x ) g of KOH will require 36.5(3.5 - x) g of HCl 56 Thus, 36.5x + 36.5(3.5 - x) = 3.1025 of HCl. 40 56 or, x = 3.15 g = Amount of NaOH in the mixture \\ Amount of KOH = (3.5 – 3.15)g = 0.35 g. 3.5 g of the mixture contain 0.35 g of KOH 100 g of the mixture will contain 0.35 ´ 100 = 10 g of KOH. 35 Hence , % of KOH = 10. Problem 8. Two acids A and B are titrated separately each time with 25 ml of (N) Na2CO3 solution and require 10 ml and 40 ml respectively for complete neutralisation. What vloume of A and B would you mix to produce one litre of normal acid solution ?

REDOX REACTION 449 Solution : For acid A If , V1 = Volume of acid in ml S1 = Strength of acid in normality V2 = Volume of Na2CO3 solution in ml S2 = strength of Na2CO3 solution in normality V1 S1 = V2S2 10 ml x S1 = 25 ml x (N) or, S1 = 25 (N) = 2.5 (N) = Strength of acid A. 10 Similarly. for acid B, Strength of acid B = 25 (N) = 0.625(N) 40 Suppose, x ml of acid A is mixed with y ml of acid B to give 1000 ml of (N) acid. According to the question, x ml x 2.5(N) + y ml x 0.625(N) º 1000 ml x (N) or, 2.5x + O.625y = 1000 ..........(1) Again, x + y = 1000 ..........(2) (Since the total volume = 1000 ml) From (1) and (2), 0.75 y = 600 or, y = 800 ml. Henc, x = (1000 – 800) ml = 200 ml. Therefore, 200 ml of acid A and 800 ml of acid B are to be mixed to produce 1000 ml of (N) acid solution.

450 +2 CHEMISTRY (VOL. - I) CHAPTER (10) AT A GLANCE 1. Oxidation : Process in which an atom or group of atoms loses one or more electrons. 2. Reduction : Process in which an atom or group of atoms gains one or more electrons. 3. Redox reaction : Oxidation and reduction always occur simultaneously. The overall reaction which involves oxidation and reduction is termed as redox reaction. 4. Oxidation Number : Oxidation number of an element in a compound is the charge positive or negative which would be assigned to the atoms of that element in the compound, if all the bonds were ionic bonds. Oxidation number can be a fraction also. The oxidation number of an element can not exceed its group number in the periodic table. 5. Equivalent mass of oxidising agent : Molecular mass of the oxidising agent Number of electrons gained per mole 6. Equivalent mass of reducing agent : Molecular mass of the reducing agent Number of electrons lost per mole 7. There are four types of redox reactions (a) Combination redox reaction (b) Decomposition redox reactions (c) Displacement redox reaction (d) Disproportionation reactions 8. Applications of redox reactions (a) Quantitative analysis (b) Eelectrode processes (c) Extraction of metals (d) Supply of energy (e) Photosynthesis (f) Production of chemicals 8. Equivalent mass of acid : How many parts by mass of the acid contain one part by mass of replaceable hydrogen. Number of repalceable hydrogen = Basicity. 9. Equivalent mass of the base : How many parts by mass of a base can just be neutralised by one equivalent mass of an acid. Number of moles of monobasic acid neutralising one mole of the base completely = Acidity = Number of replaceable hydroxyl group present in one molecule of the base. 10. Titration : Progressive addition of acid to a base or vice versa. 11. Indicator : Substance which indicates the neutralisation point or end point by a colour change. 12. Principles in Titration : (a) V1 x S1 = V2 x S2 where V1 = Volume of acid in ml. S1 = Strength of acid in terms of normality. V2 = Volume of alkali in ml. S2 = Strength of alkali in terms of normality. (b) Acids and alkali solutions of the same strength in terms of normality neutralise one another in equal volumes. (c) Grams/litre = Normality x Equivalent mass.

REDOX REACTION 451 QUESTIONS PART - A (A) Short Answer type Questions : (1 mark each). 1. Calculate the oxidation number of the underlined elements in the following compound: (i) Na2S2O3 (ii) KMnO4 (iii) K2MnO4 (iv) K2Cr2O7 (v) N2H4 (vi) Na2SO4 (vii) NaH (viii) Cl2O7 (ix) Na2SO3 (xi) CrO3 (xii) K4[Fe(CN)6] (x) CrO2CL2 (xiv) Ni(CO)4 (xv) Na2O2 (xiii) AgNO3 (xvii) K3[Fe (CN)6] (xviii) [Ag (NH3)] Cl. (xvi) OF2 2. Fill in the blanks : (i) Oxidation is due to ––––– of electron whereas reduction is due to ––––– of electrons. (ii) MnO4– + H+ + ––––– ® Mn2+ + 4H2O (iii) Cr2O72– + ––––– + ––––– ® 2Cr3+ + 7H2O (iv) The algebraic sum of oxidation numbers of all the atoms of a neutral molecule must be –––––. (v) Reaction in which oxidation and reduction take place simultaneously is termed as ––––– reaction. (vi) Conversion of KMnO4 to K2MnO4 is a process of –––––. (B) Multiple Choice Questions : (1 mark each). 1. Oxidation involves (i) Loss of electrons (ii) gain of electrons (iii) increase in the valency of negative part (iv) decrease in the valency of positive part. (NCERT, 1971) 2. When iron is added. to copper sulphate, copper is precipitated due to (i) Oxidation of Cu2+. (ii) Reduction of Cu2+. (iii) hydrolysis of CuSO4. (iv) Ionisation of CuSO4. 3. The oxidation number of carbon in CHCl3 is (i) + 2 (ii) + 3 (iii) + 4 (iv) – 3.

452 +2 CHEMISTRY (VOL. - I) 4. The process in which the oxidation number increases is called (i) Oxidation (ii) Reduction (iii) Auto-oxidation (iv) Neutralisation. 5. The conversion of sugar C12H22O11 to CO2 is (i) oxidation (ii) reduction (iii) neither oxidation nor reduction (iv) both oxidation and reduction. 6. The oxidation number of chlorine is + 5 in the compound (i) Cl2O7 (ii) ClO3– (iii) ClO4– (iv) ClO–. 7. In the following half reaction H2O + SO32– + ® SO42– + 2H+ + X, X is (i) 1e (ii) 2e (iii) 3e (iv) 4e. 8. Oxidation number of nitrogen in N3H is (i) +1/3 (ii) +3 (iii) – 1 (iv) – 1/3. 9. The following reaction describes the rusting of iron. 4Fe + 3O2 ® 4Fe3+ + 6O2– Which of the following statements is incorrect ? (i) This is an example of redox reaction. (ii) Metallic iron is reduced to Fe3+ (iii) Fe3+ is an oxidising agent. (iv) Metallic iron is a reducing agent. 10. Iron does not show the oxidation number of (i) 0 (ii) +1 (iii) + 2 (iv) + 3 (v) +8/3. (C) Short answer type : (2 marks each). 1. Balance the following by ion-electron method. (i) Fe3+ + Sn2+ ® Fe2+ + Sn4+ (ii) S2O32– + I2 ® S4O62– + I– (iii) Cr2O72– + H+ + I– ® Cr3+ + I2 + H2O 2. Balance the following equations by oxidation number method. (i) KMnO4 + H2SO4 + H2S ® K2SO4 + MnSO4 + H2O + S (ii) K2CrO4 + HCL ® K2Cr2O7 + KCl + H2O (iii) K2KMnO4 + HCl ® KMnO4 + MnO2 + KCl + H2O (iv) HNO3 + I2 ® HIO3 + NO2 + H2O (v) H2S + HNO3 ® H2O + NO + S (vi) MnO4– + SO2 ® Mn2+ + H2SO4

REDOX REACTION 453 (D) Short answer type : (3 marks each). 1. Consider the reactions : 2S2O32– (aq) + I2(S) ® S4O62– (aq) + 2I– (aq) S2O32– (aq) + 2Br2 (l) + 5H2O (l) ® 2SO42– (aq) + 4Br–(aq) + 10H+(aq) Why does the same reductant, thiosulphate react differently with iodine and bromine? 2. Sulphur dioxide and hydrogen peroxide can act as oxidising as well as reducing agent in their reactions. Why ? 3. Determine the volume of M/8 KMnO4 solution required to react completely with 20 Cm3 of M/4 FeSO4 solution in acidic medium. 4. Which of the following species, do not show disporportionation reaction and why ? CIO–, ClO2–, ClO3– and ClO4– Also write reaction for each of the species that disproportionates. ANSWERS A. (1) (i) +2, (ii) +7, (iii) +6, (iv) +6, (v) –2, (vi) +6, (vii) –1, (viii) +7, (ix) +4, (x) +6, (xii) +2, (xiii) +5, (xiv) 0, (xv) –1, (xvi) +2, (xvii) +3, (xviii) +1 (2) (i) loss, gain (ii) 5e– (iii) 6e– + 14H+ (iv) zero (v) redox (vi) reduction

454 +2 CHEMISTRY (VOL. - I) QUESTIONS PART - A (A) Short Answer type Questions 1. What is the oxidation number of S in Na2S2O3 ? 2. What is the oxidation of Mn in K2MnO4 and KMnO4 ? 3. What is the oxidation number of Cr in K2Cr2O7 and N in N2H4 ? 4. What is the oxidation number of nitrogen in nitrous oxide ? 5. What is the oxidation number of sulphur in SO42– ? 6. What is the oxidation number of hydrogen in sodium hydride ? 7. Balance the following equation : (a) Cr2O72– + H+ + S2– ® 2Cr3+ + S + H2O (b) MnO4– + Cl– + H+ ® Mn2+ + Cl2 + H2O (c) MnO4– + H+ + Fe2+ ® Mn2+ + H2O +Fe3+ (d) Cr2O72– + Cl– + H+ ® Cr3+ + Cl2 + H2O (e) MnO4– + H+ + H2O2 ® Mn2+ + O2 + H2O 8. What is the oxidation number of oxygen in hydrogen peroxide ? 9. Balance the following by ion - electron method : (a) Fe3+ + Sn2+ ® Fe2+ + Sn4+ (b) S2O32– + I2 ® S4O62– + I– (c) Cr2O72– + H+ + I– ® Cr3+ + I2 + H2O 10. (a) Balance the equation and identify the oxidising and reducing agent in the reaction. Cr2O72– + Fe2+ + H+ ® Cr3+ + Fe3+ + H2O. (b) In the following reaction what is oxidised and what is reduced ? 3 CuO + 2NH3 ® 3Cu + N2 + 3H2O 11. Fill in the blanks : (a) Oxidation is due to ———— of electron whereas reduction is due to ———— of electron. (b) MnO4– + H+ + —— ® Mn2+ + 4H2O. (c) Cr2O72– + ................. + .............. ® 2Cr3+ + 7H2O 12. Balance the following equations by oxidation number method. (a) KMnO4 + H2SO4 + H2S ® K2SO4 + MnSO4 + H2O + S (b) K2CrO4 + HCl ® K2Cr2O7 + KCl + H2O (c) K2MnO4 + HCl ® KMnO4 + MnO2 + KCl + H2O. (d) HNO3 + I2 ® HIO3 + NO2 + H2O (e) H2S + HNO3 ® H2O + NO + S (f) MnO4– + SO2 ® Mn2+ + H2SO4.

REDOX REACTION 455 13. Give atleast six oxidation states of nitrogen with one example for each state. 14. Why is decolourisation of KMnO4 by oxalate accompanied by effervecence whereas no effervescence takes place when KMnO4 is decolourised by FeSO4 ? 15. State the reason whether HCl can be used in place of H2SO4 in titration involving KMnO4. 16. Calculate the oxidation number of the underlined elements in the following compounds : (i) Cl2O7, Na2SO3 (ii) K2Cr2O7, CrO2Cl2, Cr O3 (iii) K4[Fe(CN)6], K MnO4, AgNO3. (iv) H N3, K2 Mn O4 17. State whether the following statements are true or false. (i) The oxidation number of S in H2SO4 is + 8. (ii) The equivalent mass of KMnO4 in neutral medium is M/3. (iii) The reaction Ni + 4CO ® Ni(CO)4 is a redox reaction. (iv) Conversion of K2CrO4 to K2Cr2O7 is a redox reaction. (v) Oxidation and reduction take place simultaneously. (vi) Oxidation number of an element in a compound can be a fraction. (Ans : (i) False (ii) True (iii) False (iv) False (v) True (vi) True) 18. Fill in the blanks : (a) The algebraic sum of oxidation numbers of all the atoms of a neutral molecule must be ———— . (CHSE, 1999 I) (b) The oxidation number of oxygen in Na2O2 is ———— . (c) Reduction is ———— of electron. (d) Reaction in which oxidation and reduction take place simultaneously is termed as ———— .reaction. (e) Conversion of KMnO4 to K2MnO4 is a process of ———— . (f) In Cl2O , the oxidation number of oxygen is ———— . (g) In OF2, the oxidation number of oxygen is ———— . (h) Oxidation is ————— of electron. 19. What is the oxidation number of iron in potassium ferricyanide ? 20. What is the oxidation number of S in H2SO4 ? B. Very short question (1 mark) 1. What is the change in oxidation number of marked element in the following ? (i) K2CrO4 ® K2Cr2O7 (ii) Na2S2O3 ® Na2SO4 2. What is the Oxidation number of oxygen in OF2 ?

456 +2 CHEMISTRY (VOL. - I) 3. What is the oxidation number of silver in [Ag (NH3)2] Cl ? 4. What is the oxidation number of nickel in Ni(CO)4 ? C. Short question type (2 marks) 1. Write the ion-electron equation for the following Cu + H2SO4 ® CuSO4 + SO2 + 2H2O 2. Give two examples of oxidation reduction reaction ? 3. Balance the following equation by ion-electron method MnO4– + H+ + C2O42– ® Mn2+ CO2 + H2O 4. Balance the following equation : (i) Cr2O72– + H+ + ........ ® Cr3+ + H2O (ii) Cr2O72– + NO2– ........ ® Cr3+ + NO3– (acidic) ANSWER A. 1. +2 , 2. +6 & +7 , 3. +6 & –2 , 4. +1 , 5. +6 , 6. –1 , 9. 53 , 11.–1 10 (a). Cr2O72– - oxidising agent , Fe2+ – reducing agent (b) NH3 is oxidised (–3 to 0) , CuO is reduced (+2 to 0) 11. (a) loss, gain (b) 5e (c) 6e, 14H+ 13. +5 : N2O5, HNO3 +2 : NO , +4 : N2O4 HNO2 + 12 : N2O, +3 : N2O3, –3 : NH3 , –2 : N2H4 , – 13 : N3H 14. With oxalate, effervescence is due to evolution of CO2 whereas with Fe2+, it is only converted to Fe3+. No gas is evolved. 15. HCl chemically reacts liberating Cl2 gas – 2MnO - + 16H+ + 10Cl– ® 2Mn2+ + 5Cl2 + 4 8H2O. 16. (i) +7, +4 (ii) +6, +6, +6, (iii) +2, +7, +5 (iv) – 13 , +6. 17. (i) False, (ii) True, (iii) False, (iv) False, (v) True, (vi) True 18. (a) Zero, (b) –1, (c) gain, (d) redox, (e) reduction (f) –2, (g) +2 (h) loss. 19. +3. 20. +6 B. 1. (i) + 6 to +6 (ii) +2 to +6 2. + 2 3. +1 4. 0 MULTIPLE CHOICE QUESTION 1. (i) 2. (ii) 3. (i) 4. (i) 5. (i) 6. (ii) 7. (ii) 8. (iv) 9. (ii) 10. (ii)

REDOX REACTION 457 QUESTIONS PART-B A. Short questions type (2 marks) 1. Calculate the normality of the following (a) 0.585 g NaCl / 100 cc soln. (b) 0.49 g H2SO4 / 1000 cc soln. 2. Calculate the normality of the resulting solution obtained by mixing 10cc of N/2 HCl with 30cc of N/10 H2SO4 ? 3. Indicate the type of chemical reaction from the following equations. (a) NaNO3 ® NaNO2 + O2 (b) C + O2 ® CO2 (c) NaCl + AgNO3 ® AgCl + NaNO3 (d) BiCl3 + H2O ® BiOCl + 2HCl (e) NH4CNO ® NH2 CONH2 (f) Cu + 2AgNO3 ® Cu(NO3)2 + 2Ag. 4. How many ml of 1 M H2SO4 would be required to neutralise 100 ml of 1 N sodium hydroxide solution ? 5. A solution containing 10.5 g of an alkali is completely neutralised by 500 ml of 0.5 N acid. What is the equivalent mass of the alkali ? (CHSE, 1990 S) 6. 10ml of 0.1 N NaOH solution is titrated against HCl solution of 0.05 N strength. How much HCl will be required to get the end point ? 7. State whether the following statement, are true or false. (i) Standard solution contains one gram equivalent mass of the solute per litre of the solution. (ii) The reaction of an acid with equivalent quantity of a base gives a solution which is always neutral. (iii) If X ml of an N/2 solution completely react with 2x ml of an unknown solution the normality of the latter is 1. (iv) The molecular mass and equivalent mass of a monobasic acid are equal. (v) The equivalent mass and normality of an acid are X and N respectively. Its solution contains X/N g of it per litre. (vi) Sodium carbonate solution can be titrated with acetic acid using phenolphthalein as indicator. (Ans : (i) False (ii) False (iii) False (iv) True (v) False (vi) True)

458 +2 CHEMISTRY (VOL. - I) B. Long questions : 1. Write notes on : Molality of a solution Oxidation number 2. What are different types of chemical reactions ? Explain giving one example in each case. 3. What is the equivalent mass of a base ? 3 grams of a mixture of NaOH and NaCl were dissolved in water and volume was made 250 ml. 25 ml of this solution required 23 ml of N/10 HCl for neutralisation. Calculate the percentage composition of the mixture. (CHSE, 1990 A, 2001 IER) Ans. 30.66% NaOH, 69.34% NaCl 4. Define 'molality' of a solution and explain it with an example. 2.7 grams of a mixture of NaOH and KCl is dissolved in water and the volume made up to 500 ml. 10 ml of this solution required 9 ml of decinormal HCl for neutralisation. Calculate the percentage composition of the mixture. (CHSE, 1992 S, 2001 R) Ans. 66.67%, NaOh, 33.33% KCl 5. Define : (a) Equivalent mass of an acid (b) Normal solution. 50 ml of H2SO4 of unknown strength was mixed with 25 ml of 1.28 (N/10) NaOH and the mixture was diluted to 100 ml. 10 ml of this diluted solution which was acidic required 8.5 ml of 0.8 (N/10) NaOH for complete neutralisation. Find out the strength of the original H2SO4 solution. Ans. 0.2(N) 6. A sample of organic acid weighing 3.0 g was dissolved in water. Using suitable indicator, it was titrated with 0.5 (N) NaOH. solution. Exactly 40 ml were required for neutralisation. What is the equivalent mass of the acid ? Ans. 150 7. (a) 0.4 g of a metal was dissolved in 500 ml of (N/10) H2SO4. When the reaction was over, the excess required was 150 ml (N/15) alkali for complete neutralisation. Calculate the equivalent mass of the metal. Ans. 10 (b) 0.12gm of Magnesium was dissolved in 500 ml of 0.1 N acid. What volume of N/ 2 NaOH solution will neutralise the resultant solution ? Ans : 80 ml] 8. A 0.50 g sample of impure CaCO3 is dissolved in 50 ml of 0.985 (N) HCl. After the reaction is complete the excess HCl requires 6.0 ml of 0.105 (N) caustic soda for neutralisation. Find the percentage of CaCO3 in the sample. Ans. 42.95%

REDOX REACTION 459 9. Acid is generated in the stomach when the concentration of the acid goes up, a man suffers from acidity. The doctor prescribes him antacid. By taking 3 antacid tablets, the strength decreases from. N/10 to N/200. If the fluid material is 100 ml, calculate the amount of base present per antacid tablet. (Equivalent mass of base in antacid = 100). Ans. 3.179 C. Multiple choice questions : 1. Which of the following in aqueous solution is basic in character ? (i) NaCl (ii) Na2CO3 (iii) H2CO3 (iv) FeSO4. Ans. (ii) 2. The mixture of 100 ml of 0.5 (N) HCl and 500 ml of 0.1 (N) NaOH is (i) acidic (ii) alkaline (iii) neutral Ans. (iii) 3. 0.45 g of the acid of molecular mass 90 was neutralised by 20 ml of 0.5 (N) caustic potash. The basicity of the acid is (i) 1 (ii) 2 (iii) 3 (iv) 4. Ans. : (ii) 4. The volume of 0.05 M H2SO4 needed to completely neutralise 15 ml of 0.1 M NaOH is (i) 15 ml (ii) 75 ml (iii) 30 ml (iv) 7.5 ml Ans (i) qqq

460 +2 CHEMISTRY (VOL. - I) UNIT – IX CHAPTER - 11 HYDROGEN 11.1 POSITION OF HYDROGEN IN THE PERIODIC TABLE Hydrogen is the lightest and smallest element of the periodic table having atomic number one. It contains one proton in its nucleus and only one electron in 1s orbital of K shell. The hydrogen atom therefore has equal tendency to lose this electron to form H+ or gain one electron thereby forming H–. When hydrogen atom loses its one electron it resembles alkali metals (Group 1) in some of its properties and when it gains an electron its other properties resemble those of halogens (Group 17). Hydrogen also differs from both alkali metals and halogens in some other respects. Further, it shows resemblance with carbon (Group 14) in certain respects. Resemblance with Alkali Metals : 1. Electronic configuration : Like alkali metals hydrogen has only one electron in its outermost shell. Hydrogen : 1s1 Lithium : 1s2 2s1 Sodium : 1s2 2s2 2p6 3s1 2. Electropositive character : Hydrogen like alkali metals tends to lose its only electron to form H+ion. H® H+ + e– 1s1 1s0 Na ® Na+ + e– 1s12s22p63s1 1s12s22p6 3. Affinity for non-metals : Hydrogen and alkali metals both have great affinity for non- metals : H2O : Na2O, K2O H2S : Na2S, K2S HCl : NaCl, KCl

HYDROGEN 461 4. Oxidation state : Both hydrogen and alkali metals exhibit an oxidation state of +1 in their compounds : HCl : NaCl, KCl 5. Reducing character. Just like alkali metals hydrogen acts as a reducing agent as shown in the following reactions : Fe3O4 + 4H2 ¾H¾e¾at® 3Fe + 4H2O B2O3 + 6K ¾H¾e¾at® 2B + 3K2O. Resemblance with Halogens : 1. Electronic configuration. Hydrogen atom like halogen atoms requires one electron to attain the electronic configuration of the next inert gas. H : 1s1 He : 1s2 F : 1s22s22p5 Ne : 1s22s22p6 Cl : 1s22s22p6, 3s23p5 Ar : 1s22s22p63s23p6 2. Electronegative character : Like halogen atoms hydrogen atom has tendency to gain one electron to form hydride ion (H–). H + e– ® H– 1s1 1s2 F + e– ® F– 1s22s22p5 1s22s22p6 Cl + e– ® Cl– 1s22s22p63s23p5 1s22s22p6, 3s23p6 3. Ionisation energy :The ionisation energy of hydrogen atom is quite comparable with those of halogens as given below : H : 1312 kJ mole–1 F : 1681 kJ mole–1 Cl : 1255 kJ mole–1 4. Oxidation state :Hydrogen like halogens shows an oxidation state of –1 in some of its compounds : NaH ; NaCl, KBr, KI 5. Combination with metals : With alkali and alkaline earth metals hydrogen forms hydrides similar to that of halogens. NaH CaH2 NaCl CaCl2. 6. Combination with nonmetals. Just like halogens hydrogen forms covalent compounds with non-metals such as carbon, silicon, germanium etc. CH4 SiH4 Methane Silane CCl4 SiCl4 Carbon tetrachloride Silicon tetrachloride.

462 +2 CHEMISTRY (VOL. - I) 7. Atomicity. Like halogens hydrogen exists as a diatomic molecule under ordinary condition. H2 ; F2, Cl2, Br2, I2 Dissimilarity with Alkali Metals and Halogens : Hydrogen also differs from both alkali metals and halogens in the following respects : 1. Hydrogen is less electropositive than alkali metals and less electronegative than halogens. 2. Unlike alkalimetals and halogens, hydrogen has only one proton in its nucleus and only one electron in the extra nuclear part. 3. Oxide of hydrogen e.g. H2O is neutral whereas the oxides of alkali metals and halogens are basic and acidic respectively. H2O Na2O Cl2O7 Neutral Basic Acidic. Similarity with Carbon : Hydrogen also shows resemblance with carbon in the following respects : 1. Half filled outer shell. Outer shells of hydrogen and carbon are half-filled. HC 1s1 1s2 2s2 2px1 2py1 2. Electronegativity. Both the elements have similar values of electronegativity. H = 2.1 and C = 2.5 3. Formation of covalent compounds. Both form covalent compounds such as HCl and CCl4. In the light of the above discussion it is difficult to allot a proper place to hydrogen in the periodic table. Thomson however allotted a special position to hydrogen in his periodic table as shown below. H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar Thomson's Periodic Table (First three lines) Electropositive Character of Hydrogen Hydrogen atom having atomic number one has the simplest electronic configuration 1s1. When the hydrogen atom loses its only electron, H+ ion is formed which shows electropositive character like alkali metals.

HYDROGEN 463 Hydrogen combines with electronegative elements like oxygen, chlorine and sulphur forming compounds such as H2O, HCl and H2S respectively. Similarly, sodium also forms Na2O, NaCl and Na2S with oxygen, chlorine and sulphur respectively. This indicates the electropositive character of hydrogen. When hydrogen halides (e.g. HCl, HBr etc) or oxide (e.g. H2O) are electrolysed hydrogen is liberated at the cathode. This also shows that hydrogen has electropositive character. Electronegative character of hydrogen : Hydrogen atom has also the tendency to gain one electron, thereby attaining the stable configuration of helium (1s2). H– ion then formed exhibits electronegative character like halogens. Hydrogen combines with electropositive elements such as alkali and alkaline earth metals to form hydrides (e.g. NaH, CaH2 etc) thereby exhibiting its electronegative character. Further, when hydrides (e.g.NaH, LiH) are electrolysed, hydrogen is liberated at the anode. This also indicates the electronegative character of hydrogen. 11.2 OCCURRENCE : Hydrogen is the most abundant element in the universe and constitutes about 70% of total mass of it. The jovian planets like Jupiter and Saturn contain mostly hydrogen. About half of the mass of the sun and stars is composed of hydrogen. The extremely high temperature of the sun brings about fusion of hydrogen atoms liberating large amount of energy. 411H ® 4 He + 0 Energy 2 2+1 e + Hydrogen is an essential constituent of water, coal, petroleum, clay and all animal and vegetable matter. Hydrogen gas is much less abundant (0.15% by mass) in earth's atmosphere because it is very light and earth's gravitational pull is not enough to retain it. 11.3 ISOTOPES OF HYDROGEN Isotopes are the different forms of an element having the same atomic number but different mass numbers. Hydrogen has three isotopes having mass numbers 1, 2 and 3. These are called hydrogen or protium (1H1), deuterium (1H2) and tritium (1H3). 1. Hydrogen or Protium. It is represented as 1H1 indicating thereby that this isotope of hydrogen has atomic number 1 and the mass number 1. Protium has one proton and no neutron in the nucleus and one electron in the extra nuclear part. The most abundant isotope of hydrogen is protium. Natural hydrogen contains protium to the extent of 99.984% 2. Deuterium or Heavy hydrogen. It is represented by 1H2 which indicates that this isotope of hydrogen has atomic number 1 but mass number 2. This isotope has one proton and one neutron in the nucleus and one electron in the extra nuclear part. Natural hydrogen contains 0.016% of heavy hydrogen.

464 +2 CHEMISTRY (VOL. - I) 3. Tritium. This isotope of hydrogen has one proton and two neutrons in the nucleus and one electron in the extra nuclear part. It is radioactive and unstable. It constitutes only 10–15 % of total natural hydrogen. Table 11.1 Isotopes of Hydrogen Protium Deuterium Tritium Symbol 1H1 1H2 1H3 Atomic number 1 1 1 Mass number 1 2 3 No. of protons in the nucleus 1 1 1 No of neutrons in the nucleus 0 1 2 No. of electrons 1 1 1 Electronic configuration 1s1 1s1 1s1 Percentage of abundance 99.984 0.016 10–15 Stability Stable Stable Radioactive and hence unstable. Isotopic Effect : All the three isotopes of hydrogen have similar electronic configuration (1s1) and hence they have identical chemical properties. But due to the difference in the masses of isotopes they differ appreciably from one another in the rates of their reactions and equilibrium constants for reversible reactions. They also differ in their physical properties such as melting and boiling points, latent heat of fusion, evaporation, sublimation etc. on account of differences in mass numbers. The difference in properties on account of the mass difference in isotopes of an element is known as isotopic effect. Different Reactive Forms of Hydrogen : 1. Nascent Hydrogen. Hydrogen which is just liberated as a result of chemical reaction is called nascent (newly born) hydrogen. Preparation : Nascent hydrogen can be prepared by the following methods : i. By the action of zinc and dilute sulphuric acid. Zn + H2SO4 ® ZnSO4 + 2H. ii. By the action of water on sodium amalgam. Na + H2O ® NaOH + H. iii. By the action of caustic soda on zinc or aluminium. Zn + 2NaOH ® Na2ZnO2 + 2H.

HYDROGEN 465 iv. Metallic sodium reacts with absolute alcohol to form nascent hydrogen. Na + C2H5OH ® C2H5ONa + H. v. Tin reacts with concentrated hydrochloric acid to form nascent hydrogen. Sn + 2HCl ® SnCl2 + 2H. Reducing properties of nascent hydrogen : (i) When molecular hydrogen (H2) is passed through potassium permanganate solution acidified with dilute sulphuric acid, its pink colour is not discharged. To the same solution if some zinc pieces are added the pink colour gets discharged after some time. This is because nascent hydrogen which is just generated by the reaction of zinc and dilute sulphuric acid reduces acidified potassium permanganate solution. KMnO4 + H2SO4 + H2 ® No reaction (molecular hydrogen) 2KMnO4 + 3H2SO4 + 10H ® K2SO4 + 2MnSO4 + H2O. (nascent hydrogen.) (ii) Ferric chloride solution (yellow) is reduced to ferrous chloride (light green) by adding zinc pieces and dilute hydrochloric acid. This is due to reducing property of nascent hydrogen just produced by the action of dilute hydrochloric acid on zinc. FeCl3 + H ® FeCl2 + HCl (iii) Nascent hydrogen generated from zinc and dilute sulphuric acid can also reduce KClO3 to KCl. KClO3 + 6H ® KCl + 3H2O 2. Atomic Hydrogen Preparation : Atomic hydrogen is best produced by passing molecular hydrogen through tungsten electric arc (2000 – 30000C) at low pressure. The dissociation of molecular hydrogen is an endothermic reaction. H2 ® H + H – 433 kJ This form of hydrogen is more reactive than ordinary and nascent hydrogen. Properties : (i) Stability : The life period of atomic hydrogen is only 0.03 second which can be extended to 10 seconds under special condition. (ii) Recombination : Atomic hydrogen is extremely unstable and reunites readily to form molecular hydrogen with the liberation of large amount of heat. Temperature rises to 4000 – 50000C. Metals like Pt. Pd etc. accelerate this recombination. This is the principle under which the atomic hydrogen torch works. H + H u H2 + 433 kJ.

466 +2 CHEMISTRY (VOL. - I) (iii) Formation of hydrides : It combines with metals (such as Li, Na, K) and nonmetals (such as S, P) to give hydrides. H + Na ® NaH H + Li ® LiH 2H + S ® H2S 3H + P ® PH3. (iv) Reducing property : It reduces oxides, chlorides and sulphides of some metals such as Cu,. Ag, Hg etc. to the corresponding metals. Ag2O + 2H ® 2Ag + H2O HgO + 2H ® Hg + H2O CuS + 2H ® Cu + H2S It also reduces carbon monoxide to formaldehyde and carbondioxide to formic acid. CO + 2H ® HCHO CO2 + 2H ® HCOOH 3. Ortho and Para Hydrogen Hydrogen molecule consists of two hydrogen atoms. Each hydrogen atom in the hydrogen molecule has one proton in its nucleus and one electron revolving around the nucleus. According to Pauli's exclusion principle the spins of the two electrons should be in the opposite directions for the formation of a stable molecule. However, the spins of the protons may be either in the same direction (parallel direction) or in the opposite direction (anti - parallel direction). When the spins of the protons are in the same direction we get ortho hydrogen and when the spins of the protons are in the opposite direction it gives rise to para hydrogen. Pure para hydrogen can be prepared by adsorbing ordinary hydrogen in activated charcoal in quartz vessel kept at a temperature of 20K for 3 to 4 hours. Ortho hydrogen in pure form has not been prepared so far. At normal temperature, ordinary hydrogen is mixture of about 75% of ortho hydrogen and 25% of para hydrogen. As the temperature increases the proportion of ortho form increases and that of para form decreases. At the temperature of liquefaction of air the ratio of ortho and para forms is 1 : 1. Difference betwen Ortho and Para Hydrogen : i. The ortho form has higher internal energy than the para form. ii. The ortho form is more stable than the para form and hence the para form has the tendency to change to ortho form. iii. The two forms differ in physical properties like melting point, boiling point, thermal conductivity, specific heat etc. However, they have similar chemical properties.

HYDROGEN 467 PP PP Ortho Hydrogen Para Hydrogen Fig 11.1 Ortho and Para forms of hydrogen 11.4 DIHYDROGEN : Preparation : There are a number of methods for preparing dihydrogen from metal and metal hydrides. 11.4.1 Laboratory methods of preparation : (1) It is usually prepared by the reaction of granulated zinc with dil. hydrochloric acid. Zn + 2 HCl ® ZnCl2 + H2 - (2) Reaction of zinc with aqueous alkali also produces dihydrogen. Zn + 2 NaOH ® Na2 ZnO2 + H2 - 11.4.2 Commercial methods of preparation : (1) Electrolysis of acidified water : Electrolysis of acidified water using plantinum electrodes gives dihydrogen. Electrolysis 2H2O(l) dil. H2 SO4 2H2(g) + O2(g) (2) Reaction of steam on hydrocarbons : Reaction of steam on hydrocarbons at high temperature in the presence of catalyst yields dihydrogen. 1270 K CnH2n+2 + n H2O Ni n CO(g) + (2n+1) H2(g) 1270 K e.g. CH4 (g) + H2O (g) Ni CO(g) + 3H2(g) The mixture of CO and H2, commonly known as water gas or synthesis gas is used in the synthesis of methanol and also a number of hydrocarbons.

468 +2 CHEMISTRY (VOL. - I) 11.4.3 PROPERTIES Physical Properties : Dihydrogen is a colourless, odourless, tasteless combustible gas. It is lighter than air and insoluble in water. Its other physical properties along with those of Deuterium and Tritium are mentioned on Table 11.2. TABLE 11.2 Atomic and Physical Properties of Hydrogen Property Hydrogen Deuterium Tritium Relative abudance (%) 99.985 0.0156 10–15 Relative atomic mass (g mol–1) 1.008 2.014 3.016 Melting point / K 13.96 18.73 20.62 Boiling point / K 20.39 23.67 25.0 Density / gL–1 0.09 0.18 0.27 Enthalpy of fusion / kJ mol–1 0.117 0.197 - Enthalpy of vaporization / kJ mol–1 0.904 1.226 - Enthalpy of bond dissociation / kJ mol–1 at 198.2K 435.88 443.35 - Internuclear distance / pm 74.14 74.14 - Ionization enthalpy / kJ mol–1 1312 - Electron gain enthalpy / kJ mol–1 –73 - - Covalent radius / pm - - Ionic radius (H–) / pm 37 - - 208 - Chemical Properties : 1. Action with halogens : It reacts with halogens to give hydrogen halides. H2(g) + X2(g) ® 2HX(g) [X = F, Cl, Br, I] Reaction with Fluorine occurs even in dark, while with Iodine, it requires a catalyst. 2. Action with dioxygen : It reacts with dioxygen to form water and the reaction is highly exothermic in nature. Catalyst 2H2(g) + O2(g) or heat 2H2O (l) DHO = –285.09 kJ mol–1 3. Action with dinitrogen : It reacts with dinitrogen to yield ammonia and the process is known as Haber's process. Fe 3H2(g) + N2(g) 673 K, 200 atm 2NH3(g) DHO = –92.6 kJ mol–1

HYDROGEN 469 4. Action with metals : Dihydrogen combines with many metals at high temperature to produce corresponding metal hydrides. H2(g) + 2M(g) ® 2MH(s) where M is an alkali metal. 5. Action with metals oxides : It reduces oxides of metals (less active than iron) into corresponding metals. eg. y H2(g) + MxOy(s) ® xM(s) + y H2O(l) 6. Action with organic compounds : The unsaturated hydrocarbons undergo hydrogenation in presence of suitable catalyst and hydrogen gas and produce the compounds of commercial importance. Vegitable oil + H2 Ni Vanaspati ghee 11.4.4 Uses : 1. A starting material for manufacture of ammonia by Haber's process. and also in the preparation of highly useul chemicals like hydrogen chloride, methanol etc. 2. In the manufacture of Vanaspati ghee. 3. Manufacture of metal hydrides. 4. In the fuel cell for producing electrical energy. 11.5 HYDRIDES : Almost all elements of the periodic table (except noble gases and elements of Groups 7, 8, 9) react with dihydrogen under suitable conditions forming 'Hydrides'. The general formula of these hydrides is MHx, where 'M' stands for metal and 'x' for the number of hydrogen atoms. Hydrides can be classified into following three categories. (1) Ionic hydrides (2) Covalent hydrides and (3) Metallic hydrides 11.5.1 Ionic or Saline or Salt-like hydrides : Dihydrogen reacts with most of the s-block elements having highly electropositve character producing stoichiometric Ionic hydrides. Ex. NaH, CaH2 etc. Properties : 1. The ionic hydrides are cystalline and non-volatile being bad conductors in their solid state. 2. In the molten state they are good conductors of electricity and produce dihydrogen gas at anode. At anode : 2H– (meH) ® H2(g) + 2e– At cathode : Na+ (meH) + e– ® Na (l)

470 +2 CHEMISTRY (VOL. - I) 3. They react violently with water to produce corresponding metal hydroxide with liberation of hydrogen gas. 4. CaH2(s) + 2H2O(l) ® Ca(OH)2(aq) + 2H2(g) Lithium hydride forms useful complex metal hydrides such as lithium aluminium hydride (LiAlH4) and lithium borohydride (LiBH4). These compounds are widely used as reducing agents in organic synthesis. 11.5.2 Covalent or Molecular hydrides : These are the compounds formed by dihydrogen with most of the p-block elements e.g. CH4, NH3. These compounds are covalent in nature. They are further classified into following three categories according to the relative numbers of electrons and bonds in their Lewis structure. (a) Electron-deficient hydrides (b) Electron - precise hydrides (c) Electron - rich hydrides (a) Electron-deficient hydrides : Elements of group 13 from electron-deficient hydrides. They act as Lewis acids i.e. electron acceptors. e.g. BH3, AlH3, etc. (b) Electron-precise hydrides : These compounds have the required number of electrons around the central atom. e.g. CH4. They show regular terahedral geometry. Elements of group 14 form such hydrides. (c) Electron-rich hydrides : In these compounds excess electrons are present as lone pairs around the central atom. So they behave as Lewis bases. Elements of group 15-17 form such compounds. e.g. : NH3 has 1 lone pair, H2O has 2 lone pairs and HF has 3 lone pairs of electrons. 11.5.3 Metallic or Non-stoichiometric or Interstitial Hydrides : The metals present in d-block and f-block (except the metal of group 7, 8 and 9) usually form metallic hydrides. These Hydrides are non-stoichiometric being deficient in hydrogen. e.g. LaH2.87, YbH2.55 etc. They not conduct heat and electricity as efficiently as their parent metals do. Metals like Platinum and Palladium can accomodate a very large volume of hydrogen and therefore used for hydrogen storage in different energy sources. 11.6 WATER Water is the most important natural resource. A major part of all living organisms is made up of water. The estimated world water supply is given in Table 11.3

HYDROGEN 471 Table 11.3 Estimated World Water supply. Source % of total water supply Oceans 97.33 Saline lakes and inland seas 0.008 Polar ice and glaciers 2.04 Ground water 0.61 Rivers 0.0001 Lakes 0.009 Soil moisture 0.0005 Atmospheric water vapour 0.001 11.6.1 Physical properties of Water : 1. It is a colourless odourless and tasteless liquid. 2. H2O possesses high values of freezing point, boiling points, heat of vaporisation and heat of fusion in comparison to hydrides of other group 16 elements such as H2S, H2Se, H2Te etc., because of the presence of extensive hydrogen bonding between its molecules. 3. Water has higher specific heat, thermal conductivity, surface tension, dipole moment and dielectric constant in comparison to other liquids. These properties enable water to play a vital role in biosphere. 4. Water is regarded as a universal solvent as it dissolves most of the inorganic and also covalent compounds. 11.6.2 Chemical properties of Water : 1. Amphoteric Nature : Water has the ability to act as an acid as well as a base. Thus it behaves as an amphoteric substance. According to Br.o.nsted – Lowry theory, it acts as an acid with NH3 and a base with H2S. H2O (l) + NH3 (aq) OH– (aq) + N H + (aq). 4 H2O (l) + H2S (ag) H3O+ (aq) + HS– (aq). 2. Redox Reactions : When water react with highly electropositive metals like Na, K etc. it easily gets reduced to dihydrogen and acts as an oxidising agent. 2H2O (l) + 2Na (s) ® 2 NaOH (aq) + H2 (g). It reduces a number of electronegative elements liberating O2 gas. 2 F2(g) + 2 H2O (l) ® O2(g) + 4H+(aq) + 4F–(aq)

472 +2 CHEMISTRY (VOL. - I) During photosynthesis water gets oxidised to O2. 6 CO2(g) + 12 H2O (l) ® C6H12O6 (aq) + 6 H2O(l) + 6 O2(g) 3. Hydrolysis Reactions : Some ionic and covalent compounds get hydrolysed when dissolved in water. SiCl4 (l) + 2H2O (l) ® SiO2(s) + 4 HCl (aq). 11.6.3 Hard and soft water : Water containing soluble bicarbonate, chloride, sulphate etc. salts of calcium and magnesium is known as Hard water. This type of water does not give lather with soap. Water free from soluble salts of calcium and magnesium is called Soft water. It gives lather with soap easily. Hardness of water is due to the following reaction. 2 C17H35COONa (aq) + Ca2+ ® (C17H35COO)2 Ca¯ + Na+(aq) Hard water is harmful for boilers and also unsuitable for laundry. There are two types of hardness of water : (i) temporary hardness and (ii) permanent hardness. Temporary hardness is due to the presnece of calcium and magnesium bicarbonates. It can be removed by (a) Boiling (b) Adding calculated amount of lime. Permanent hardness is due to the presence of chlorides and sulphates of calcium and magnesium. It can be removed by (a) Treatment with washing soda (b) Adding sodium hexametaphosphate (Na6P6O18) (c) Adding Zeolite (NaAlSiO4) 11.7 HEAVY WATER : Heavy water is deuterium oxide (D2O). An American chemist, Urey established that ordinary water contains 0.016 percent of heavy water. Preparation : 1. Multistage Electrolysis of Ordinary Water : Heavy water is prepared on a commercial scale by multistage (continuous and prolonged) electrolysis of natural water. When ordinary water is electrolysed by passing current through it, lighter water molecules decompose more readily than D2O or HDO molecules. On continuous and prolonged electrolysis the residue gets enriched in heavy water. The decomposition products are burnt and water then formed is transferred to the previous electrolyser. The electrolytic cell consists of a steel tank which serves as the cathode. A perforated cylindrical sheet of nickel acts as the anode. A large number of such cells are used for electrolysis of water in several stages. The residual water in the last electrolyser is almost pure heavy water. 2. Fractional Distillation of Ordinary Water. Separation of heavy water from ordinary water may be effected by fractional distillation method. This method takes the advantage of the difference in the boiling points of ordinary water (1000C) and heavy water (101.420C). The lighter fraction (H2O) distils over first leaving behind a residue richer in heavy water (D2O).

HYDROGEN 473 3. Exchange Reactions : Heavy water can also be prepared by the exchange reaction: H2O + D2 l D2O + H2 This reaction is catalysed by finely divided nickel. Physical Properties : Heavy water is a colourless, odourless and tasteless mobile liquid. It has slightly higher values of density, melting point and boiling point due to its higher molecular mass as compared to that of ordinary water. D2O has a lower dielectric constant than ordinary water and hence ionic compounds are less soluble in D2O than in H2O. Table 11.4 Some physical constants of Heavy Water (D2O) and Ordinary Water (H2O) Property Heavy Water(D2O) Ordinary Water (H2O) 1.017 0.998 Density at 293 K 276.8 K 273 K 374.4 K 373 K Melting point 80.5 82 1.3284 1.3329 Boiling point 12.6 10.09 Dielectric constant 30.5 35.9 Refractive index at 293 K Viscosity at 293 K(milli poise) Solubility NaCl/100 g at 298 K Chemical Properties : Heavy water behaves like ordinary water in most of its chemical properties However, heavy water reacts more slowly than ordinary water in chemical reactions. 1. Action with metals : D2O reacts with alkali and alkaline earth metals liberating heavy hydrogen (D2) 2D2O + 2Na ® 2NaOD + D2 Sodium deuteroxide 2D2O + Ca ® Ca(OD)2 + D2 2. Action with metallic oxides : D2O reacts with basic oxides of metals like Na2O, CaO etc. to form heavy alkalies. D2O + Na2O ® 2NaOD. D2O + CaO ® Ca(OD)2 3. Action with nonmetallic oxides : D2O reacts with acidic oxides of nonmetals such as SO3, P2O5 etc. to form deutero acids D2O + SO3 ® D2SO4 Deutero sulphuric acid. 3D2O + P2O5 ® 2D3PO4 Deutero phosphoric acid.

474 +2 CHEMISTRY (VOL. - I) 4. Action with metallic carbides, nitrides and phosphides : With carbides deutero hydrocarbons are formed. CaC2 + 2D2O ® Ca(OD)2 + C2D2 (deutero acetylene) Al4C3 + 12 D2O ® 4Al (OD)3 + 3CD4 (deutero methane) With metallic nitrides D2O forms deutero ammonia. Mg3N2 + 6D2O ® 3Mg(OD)2 + 2ND3 (deutero ammonia) D2O reacts with metallic phosphides forming deutero phosphine. Ca3P2 + 6D2O ® 3Ca(OD)2 + 2 PD3 (deutero phosphine) 5. Electrolysis : Heavy water containing dissolved P2O5 on electrolysis decomposes to deuterium and oxygen which are liberated at cathode and anode respectively. 2D2O ® 2D2 + O2 6. Formation of deutero hydrates : Compounds such as CuSO4, Na2SO4, MgSO4, etc. when crystallised from heavy water, form crystalline salts with definite number of heavy water molecules called deutero hydrates e.g. CuSO4, 5D2O ; Na2SO4, 10D2O etc. 7. Exchange reaction : Compounds containing hydrogen undergo exchange reactions with heavy water. HCl + D2O l DCl + HOD NaOH + D2O l NaOD + HOD. Summary : Na NaOD Ca Ca(OD)2 Na2O NaOD CaC2 Ca(OD)2 D2O SO3 D2SO4 P2O5 D3PO4 CaC2 Ca (OD)2 + C2D2 Al4C3 Al (OD)3 + CD4 Mg3N2 Mg(OD)2 Ca3P2 Ca(OD)2 Electrolysis D2 + O2 DCl HCl NaOH NaOD + HOD

HYDROGEN 475 Uses of Heavy Water : Heavy water is used (i) As a moderator of neutron in nuclear reactions . (ii) As a source for the preparation of D2. (iii) As a tracer compound for studying reaction mechanism in organic chemistry. 11. 8 HYDROGEN PEROXIDE, H2O2 Hydrogen peroxide occurs in minute quantities in air and water. It is also present in juices of certain plants. A French chemist, Thenard in 1818 prepared hydrogen peroxide by reacting barium peroxide with dilute hydrochloric acid. 11.8.1 Preparation : Hydrogen peroxide can be prepared in the laboratory by the following methods : 1. From Sodium peroxide : Calculated amount of sodium peroxide is reacted with ice - cold 20% solution of sulphuric acid. Na2O2 + H2SO4 ® H2O2 + Na2SO4 Sodium peroxide Sulphuric acid Hydrogen peroxide Sodium sulphate. On cooling the resulting solution to 271K, most of sodium sulphate crystallises out as Na2SO4, 10H2O which is removed. A 30% solution of hydrogen peroxide containing a little amount of sodium sulphate is now obtained. 2. From Barium peroxide : A thin paste of hydrated barium peroxide, BaO2. 8H2O is prepared in ice cold water and then added slowly to an ice-cold solution of 20% sulphuric acid. BaO2 . 8H2O + H2SO4 ® BaSO4 ¯ + H2O2 + 8H2O The white precipitate of barium sulphate is removed by filtration. The resulting solution contains 5% solution of hydrogen peroxide. Hydrogen peroxide prepared by this method contains some Ba2+ ions which catalyse the decomposition of hydrogen peroxide. Hence, the solution can not be stored for a long time. Phosphoric acid may be used in place of sulphuric acid. 3BaO2 + 2H3PO4 ® Ba3(PO4)2 + 3H2O2 . Barium phosphate is then decomposed by dilute sulphuric acid. Ba3(PO4)2 + 3H2SO4 ® 3BaSO4 + 2H3PO4. Insoluble barium sulphate is removed by filtration and phosphoric acid so formed is used again. 3. Merck's Process : Carbon dioxide is passed through a suspension of barium peroxide in ice-cold water when hydrogen peroxide and barium carbonate are formed. BaO2 + H2O + CO2 ® BaCO3 + H2O2 Barium carbonate is filtered and the filtrate is concentrated.