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Chemical Sciences CSIR UGC-NET/JRF Exam. Solved Paper

June 2012 Chemical Science Useful Fundamental Constants Chromium Cr 24 52.00 Cobalt Co 27 58.93 m Mass of electron 9.11 × 10–31 Kg Copper Cu 29 63.54 Curium Cm 96 (247) h Planck’s constant 6.63 × 10–34 Jsec Dysprosium Dy 66 162.50 Einsteinium Es 99 (254) e Charge of electron 1.6 × 10–19 C Erbium Er 68 167.26 Europium Eu 63 151.96 k Boltzmann constant 1.38 × 10–23 J/k Fermium Fm 100 (253) Fluorine F 9 19.00 c Velocity of Light 3.0 × 108 m/sec Francium Fr 87 (223) Gadolinium Gd 64 157.25 IcV 1.6 × 10–14 J Gallium Ga 31 69.72 Germanium Ge 32 72.59 amu 1.67 × 10–27 kg Gold Au 79 196.97 Hafnium Hf 72 178.49 G 6.67 × 10–11 Nm2 kg–2 Helium He 2 4.003 Holmium Ho 67 164.93 Ry Rydberg constant 1.097 × 107 m–1 Hydrogen H 1 1.0080 Indium In 49 114.82 NA Avogadro number 6.023 × 1023 mole–1 Iodine I 53 126.90 Iridium Ir 77 192.2 ε0 8.854 × 10–12 Fm–1 8.314 JK–1 mole–1 Iron Fe 26 55.85 μ0 4π × 10–7 Hm–2 Krypton Kr 36 83.80 R Molar Gas constant Lanthanum La 57 138.91 Lawrencium Lr 103 (257) List of the Atomic Weights of the Elements Lead Pb 82 207.19 Lithium Li 3 6.939 Element Symbol Atomic Atomic Lutetium Lu 71 174.97 Number Weight Magnesium Mg 12 24.312 Actinium Ac Manganese Mn 25 54.94 Aluminium Al 89 (227) Mendelevium Md 101 (256) Americium Am 13 26.98 Mercury Hg 80 200.59 Antimony Sb 95 Molybdenum Mo 42 95.94 Argon Ar 51 (243) Neodymium Nd 60 144.24 Arsenic As 18 121.75 Neon Ne 10 20.183 Astatine At 33 Barium Ba 85 39.948 Neptunium Np 93 (237) Berkelium Bk 56 74.92 Beryllium Be 97 (210) Nickel Ni 28 58.71 Bismuth Bi 4 137.34 Boron B 83 (249) Niobium Nb 41 92.91 Bromine Br 5 9.012 Cadmium Cd 35 208.98 Nitrogen N 7 14.007 Calcium Ca 48 10.81 Californium Cf 20 79.909 Nobelium No 102 (253) Carbon C 98 112.40 Cerium Ce 6 40.08 Osmium Os 76 190.2 Cesium Cs 58 (251) Chlorine Cl 55 12.011 17 140.12 132.91 35.453

4J | Chemical Sciences 2012 Oxygen O8 15.9994 (B) High altitude and low air temperature (C) Low altitude and low air temperature Palladium Pd 46 106.4 (D) High altitude and high air temperature Phosphorus P 15 30.974 2. How many squares are there in this figure ? Platinum Pt 78 195.09 Plutonium Pu 94 (242) Polonium Po 84 (210) Potassium K 19 39.102 Praseodymium Pr 59 140.91 Promethium Pm 61 (147) Protactinium Pa 91 (231) Radium Ra 88 (226) Radon Rn 86 (222) Rhenium Re 75 186.23 Rhodium Rh 45 102.91 Rubidium Rb 37 85.47 (A) 9 (B) 14 (C) 15 (D) 17 Ruthenium Ru 44 101.1 Samarium Sm 62 150.35 3. A mountain road has 3 sections of different slopes as shown. What is the average slope m Scandium Sc 21 44.95 of the entire climb ? Selenium Se 34 78.96 Silicon Si 14 28.09 Silver Ag 47 107.870 Sodium Na 11 22.9898 Strontium Sr 38 37.62 Sulfur S 16 32.064 Tantalum Ta 73 180.95 Technetium Tc 43 (99) Tellurium Te 52 127.60 Terbium Tb 65 158.92 Thallium Tl 81 204.37 ( ) ( )(A) 1 (B) 1 1 3 <m< 2 Thorium Th 90 232.04 Thulium Tm 69 168.93 (D) ⎛⎝⎜⎯√13⎠⎟⎞ < m < 1 Tin Sn 50 118.69 (C) 1 < m < ⎯√3 Titanium Tl 22 47.90 Tungsten W 74 183.85 Uranium U 92 238.03 4. Which of the following graphs shows the concentration of a sugar solution as a function Vanadium V 23 50.94 of the cumulative amount of sugar added in the process of preparing a saturated solution Xenon Xe 54 131.30 (the temperature remaining constant) ? Ytterbium Yb 70 173.04 Yttrium Y 35 88.91 Zinc Zn 30 65.37 Zirconium Zr 40 91.22 * Based on mass of C12 at 12.00… . The ratio of these weights of those on the order chemical scale (in which (A) (B) oxygen of natural isotopic composition was assigned a mass of 16.0000…) is 1.000050. (Values in parentheses represent the most stable known isotopes). Part A 1. In still air, fragrance of a burning incense (C) (D) stick will be smelt by an observer quickest when the experiment is carried out at— (A) Low altitude and high air temperature

Chemical Sciences 2012 | 5J 5. There are sand-piles which are geometrically (A) Chewing gum is a pain killer similar but of different heights. The ratio of (B) Chewing equilibrates pressure on both the masses of the sand comprising two randomly chosen piles will be equal to the sides of the ear drum ratio of the— (C) Chewing gum closes the ear drum (D) Chewing distracts the person (A) Pile heights 10. The reason why a lunar eclipse does not occur at every full moon is— (B) Squares of the pile heights (A) The position of the sun is not favourable (C) Cubes of the pile heights at all full moons (B) The orbital planes of the moon and that (D) Cube-roots of the pile heights of the earth are inclined to each other by 6. There are two identical vessels of volume V a small angle each, one empty, and the other containing a (C) The shape of the earth is not a perfect block of wood of weight w. The vessels are sphere then filled with water up to the brim. The two (D) The moon reflects only from one hemi- arrangements are shown as A and B in the sphere figure. If the density of water is ρ and g is the 11. A boy throws a stone vertically upwards with acceleration due to gravity, then— a certain initial velocity. Which of the following graphs depicts the velocity as a (A) A and B have equal weights function of time, if the acceleration due to gravity is assumed to be uniform and (B) A is heavier than B by an amount w constant ? (C) A is heavier than B by an amount (A) Vρg – w (B) (D) B is heavier than A by an amount Vρg – w (C) 7. If the father has blood group O and the (D) mother has blood group AB, what are the possible blood groups of their children ? 12. A rigid uniform bar of a certain mass has two bobs of the same size, but with different (A) O, AB, A (B) A, B densities ρ and 2ρ suspended identically from its ends. (C) A, O (D) B, AB 8. Nuclei of 32P and 32S, accelerated through the same potential difference enter a uniform, transverse magnetic field (Z = 15 for P and Z = 16 for S). As they emerge from the magnetic field— (A) Both nuclei emerge undeflected (B) 32P is deflected less than 32S (C) 32P is deflected more than 32S (D) Both are equally deflected 9. A person chewing a bubble gum did not experience ear pain in a jet plane while landing whereas another person not chewing a gum had ear pain. The reason could be—

6J | Chemical Sciences 2012 When the bar is level on a fulcrum as shown 16. The 14C dating method is not usually used for dating organic substances older than–60,000 in the figure, d and d′ are related by— years, because— (A) 2d = d′ (B) d > 2d′ (A) Such objects rarely contain carbon (B) Such objects accumulated 14C after their (C) d = 2d′ (D) d < 2d′ formation (C) In those times there was no production of 14C (D) Most of the 14C in the sample would have decayed 13. There are two points A and A′ on the equator 17. A seismograph receives a S-wave 60 s after it at longitudes 0° and 90°E, and two other points B and B′ on the same longitudes, receives the P-wave. If the velocities of P-and respectively, but at latitude 60°S. The S-waves are 7 km/s and 6 km/s respectively, distances (along the latitudes) between the points A, A′ and B, B′ are related by— then the distance of the seismic focus from the seismograph is— (A) AA′ = BB′ (A) 2520 km (B) 42 km (C) 7070 km (D) 72 km (B) AA′ = 2BB′ 18. The decay of a radioactive isotope P produces a stable daughter isotope D. The ratio of the ( )(C) AA′ = √⎯ 3 BB′ number of atoms of D to the number of atoms of P after 2 half lives would be— ( )(D) AA′ = ⎯√ 2 BB′ 1 3 (A) 4 (B) 4 14. (C) 3 (D) 2 Water is flowing through a tube as shown. 19. The scatter plots represents the values The cross-sectional areas at A and C are measured by two similar instruments. Point A equal, and greater than the cross-sectional are in the figure represents the true value. Which at B. If the flow is steady, then the pressure of the following is a correct description of the on the walls at B is— quality of these measurements ? (A) Less than that at A and that at C Fig. 1 Fig. 2 (B) More than that at A and that at C (A) Fig. 1 : good accuracy, good precision (C) Same as that at A and that at C Fig. 2 : good accuracy, good precision (D) More than that at A but less than that at C 15. Match the two lists— (B) Fig. 1 : poor accuracy, poor precision Raw Material Product Fig. 2 : good accuracy, poor precision (a) Limestone 1. Porcelain (C) Fig. 1 : poor accuracy, good precision (b) Gypsum 2. Glass Fig. 2 : poor accuracy, poor precision (c) Silica sand 3. Plaster of Paris (D) Fig. 1 : poor accuracy, poor precision (d) Clay 4. Cement Fig. 2 : poor accuracy, good precision (a) (b) (c) (d) 20. Even though the concentration of CO2 is the (A) 1 2 3 4 same at sea level and at high altitude, the (B) 4 3 2 1 photosynthetic rate is higher in a plant grown (C) 1 3 4 2 at sea level than in a plant (of the same (D) 4 1 3 2 species) grown at high altitude. The reason for this is—

Chemical Sciences 2012 | 7J (A) Light intensity is more at sea level 24. The total number of lone pairs of electrons in (B) Temperature is lower at higher altitude I–3 is— (C) Atmospheric pressure is higher at sea (A) Zero (B) Three level (D) Relative humidity is higher at sea level (C) Six (D) Nine Part B 25. If Mössbauer spectrum of Fe(CO)5 is recorded in the presence of a magnetic field, the 21. In the reactions (A) and (B), original spectrum with two lines changes into nH2O + Cl– → [Cl (H2O)n]– …(A) the one with— 6H2O + Mg2+ → [Mg (H2O)6]2+ …(B) (A) Three lines (B) Four lines water behaves as— (A) An acid in both (A) and (B) (C) Five lines (D) Six lines (B) An acid in (A) and a base in (B) (C) A base in (A) and an acid in (B) 26. The spectrophotometric response for the (D) A base in both (A) and (B) titration of a mixture of Fe3+ and Cu2+ ions 22. The size of the d orbitals in Si, P, S and Cl follows the order— against EDTA is given below : (A) Cl > S > P > Si (B) Cl > P > S > Si The correct statement is— (C) P > S > Si > Cl (D) Si > P > S > Cl (A) Volume ab ≡ [Fe3+] and volume c d ≡ 23. The correct structure of basic beryllium [Cu2+] nitrate is— (B) Volume ab ≡ [Cu2+] and volume cd ≡ (A) [Fe3+] (B) (C) Volume ab ≡ [Fe3+] and volume c d ≡ excess EDTA (C) (D) Volume ab ≡ [Cu2+] and volume cd ≡ (D) excess EDTA 27. In ‘carbon-dating’ application of radioiso- topes, 14C emits— (A) β-particle (B) α-particle (C) γ-radiation (D) Positron 28. The actual base pairs present in the double helical structure of DNA containing adenine (A), thymine (T), cytosine (C) and guanine (G), are— (A) AG and CT (B) AC and GT (C) AG and AC (D) AT and GC 29. The oxidation state of iron in met-hemoglobin is— (A) Three (B) Two (C) Four (D) Zero

8J | Chemical Sciences 2012 30. The reactions of Ni(CO)4 with the ligand L [L 37. The least basic among the following is— = PMe3 or P(OMe)3] yields Ni(CO)3 L. The reaction is— (A) Al(OH)3 (B) La(OH)3 (A) Associative (B) Dissociative (C) Ce(OH)3 (D) Lu(OH)3 (C) Interchange (Ia) (D) Interchange (Id) 38. For any operator A and its adjoint A+, the incorrect statement is— 31. As a ligand Cl– is— (A) Only a σ-donor (A) AA+ is hermitian (B) Only a π-donor (B) AA+ + A+A is hermitian (C) Both a σ-donor and a π-donor (D) a σ-donor and a π-acceptor (C) A + A+ is hermitian (D) A – A+ is hermitian 39. For hydrogen-like atom with a nuclear charge 32. The correct d-electron configuration showing Z, the energy of orbital with principal spin-orbit coupling is— quantum number ‘n’ follows the relation— (A) t2 6 eg2 (B) t2 6 eg0 (A) En ∝ n2Z2 (B) En ∝– Z2 g g n (C) t2 4 eg0 (D) t2 3 eg2 Z Z2 g g n n2 (C) En ∝ – (D) En ∝– 33. The correct statement for the aggregating nature of alkyl lithium (RLi) reagent is— 40. The average value of the radius <r> in the 1s state of the hydrogen atom is (ao is Bohr (A) The carbanion nucleophilicity increases radius)— with aggregation (A) a0 (B) 1.5 a0 (B) The observed aggregation arises from its electron deficient nature (C) 0.75 a0 (D) 0.5 a0 (C) Carbanion nucleophilicity does not 41. Among the following, the correct statement depend on aggregation is— (D) The extent of aggregation is maximum in (A) The number of irreducible representa- polar dative solvents tions is equal to classes of symmetry operations 34. For the reaction, trans-[IrCl(CO) (PPh3)2] + Cl2 → trans-[IrCl3(CO) (PPh3)2], the correct (B) The number of irreducible representa- observation is— tions is equal to the order of the symmetry point group (A) VCO (product) > VCO (reactant) (B) VCO (product) < VCO (reactant) (C) The irreducible representations contained in any point group are always of one (C) VCO (product) = VCO (reactant) dimension (D) VCO (product) = VCO (free CO) (D) A symmetry point group may not contain a totally symmetric irreducible represen- 35. The nucleophilic attack on olefins under mild tation conditions— 42. For a diatomic molecule AB, the energy for (A) Is always facile the rotational transition from J = 0 to J = 1 (B) Is more facile than electrophilic attack on olefins state is 3.9 cm–1. The energy for the rotational (C) Is facile for electron-rich olefins transition from J = 3 to J = 4 state would be— (D) Requires activation by coordination to (A) 3.9 cm–1 (B) 7.8 cm–1 metal (C) 11.7 cm–1 (D) 15.6 cm–1 36. Among the following, the strongest oxidizing 43. For the vibrational Raman spectrum of a agent is— homonuclear diatomic molecule, the selection rule under harmonic approximation is— (A) [WO4]2– (B) [CrO4]2– (A) Δv = 0 only (B) Δv = ± 1 only (C) [MoO4]2– (D) [ReO4]–1 (C) Δv = ± 2 only (D) Δv = 0, ±1

Chemical Sciences 2012 | 9J 44. With increase in temperature, the Gibbs free 49. For a reaction, the rate constant k at 27°C was energy for the adsorption of a gas on to a found to be : solid surface— k = 5·4 × 1011 e– 50 (A) Becomes more positive from a positive value The activation energy of the reaction is— (B) Becomes more negative from a positive (A) 50 J mol–1 value (B) 415 J mol–l (C) Becomes more positive from a negative (C) 15,000 J mol–1 value (D) 125,000 J mol–1 (D) Becomes more negative from a negative value 50. During the addition polymerisation, the reac- tion proceeds via— 45. The vapour of a pure substance, when cooled (A) Step-growth process under a pressure less than its triple-point (B) Free-radical chain reaction pressure— (C) Cascade process (A) Liquefies (D) Addition reaction (B) Liquefies first and then solidifies (C) Solidifies directly 51. How many atoms are there in an element (D) Remains unchanged packed in a fcc structure ? 46. The quantities, which are held fixed in a (A) 1 (B) 2 canonical ensemble are— (A) N, T and P (C) 4 (D) 8 (B) V, T and N (C) N, V and E 52. The structure obtained when all the tetrahe- (D) μ, V and P dral holes are occupied in a fcc structure, is of the type— 47. The correct value of E0 of a half cell in the following graph of E vs log m (molality) is— (A) NaCl (B) CsCl (C) CaF2 (D) ZnS 53. Dispersion of a solid in a liquid, a liquid in a gas and a liquid in a liquid are respectively known as— (A) Aerosol, emulsion, sol (B) Sol, aerosol, emulsion (C) Emulsion, sol, aerosol (D) aerosol, sol, emulsion (A) CC′/AC′ (B) AB′ 54. The data obtained from two sets of experi- (C) BB′ (D) CC′ ments A and B have the following charac- teristics : 48. One of the assumptions made in the conven- tional activated complex theory is— Experiment AB (A) Equilibrium is maintained between the Mean 50 units 100 units reactants and the activated complex Standard deviation 2 units 2 units (B) Equilibrium is maintained between the reactants and the products It may be concluded that— (C) Equilibrium is maintained between the (A) A is more precise than B products and the activated complex (B) A is less precise than B (D) Equilibrium is maintained between the reactants, the activated complex and the (C) A and B are of the same precision products (D) Relative precision of A and B cannot be assessed

10J | Chemical Sciences 2012 55. The IUPAC name of the compound given (A) Enantiomers below is— (B) Diastercomers (C) Constitutional isomers (D) Homomers 60. The major product formed in the following concerted reaction is— (A) ethyl (R)-2-methyl-4-oxocyclohex-2- (A) enecarboxylate (B) (C) (B) ethyl (S)-2-methyl-4-oxocyclohex-2- enecarboxylate (C) (R)-4-ethoxycarbonyl-3-methylcyclohex- 2-enone (D) (S)-4-ethoxycarbonyl-3-methylcyclohex- 2-enone 56. The major product formed in the following reaction is— (A) (D) (B) 61. The structure of meso-tricarboxylic acid that is formed on potassium permanganate oxida- tion of abietic acid is— (C) (A) (D) 57. The number of signals that appear in the (B) broad-band decoupled 13C NMR spectrum of (C) phenanthrene and anthracene, respectively, (D) are— (A) ten and four (B) ten and ten (C) seven and four (D) seven and seven 58. The co-enzyme that is involved in the reduc- tion of a double bond in fatty acid biosynthesis is— (A) NADH (B) Biotin (C) Pyridoxal (D) FADH2 59. Epoxidation of (R)-cyclohex-2-enol with peracetic acid yields a 95 : 5 mixture of compounds A and B. Compounds A and B are—

Chemical Sciences 2012 | 11J 62. The major product formed in the following reaction is— (D) (A) (B) 66. Among the following, the amino acid which is basic in nature is— (C) (D) 63. The major product formed in the following (A) Tyrosine (B) Asparagine reaction is— (C) Leucine (D) Arginine 67. “Phosphorescence” is represented as— (A) T1 → S0 + hv (B) T1 → S0 + Δ (C) S1 → S0 + hv (D) S1 → T1 + Δ 68. Among the following diacids, the one that forms an anhydride fastest on heating with acetic anhydride is— (A) (A) (B) (B) (C) (D) (C) 64. Among the following, the synthetic equiva- (D) lent for acyl anion is— 69. The major product formed in the following (A) Nitroethane and base reaction sequence is— (B) α-chloroacrylonitrile (C) Ethylmagnesium bromide (A) (B) (D) Acetyl chloride and triethylamine 65. Among the following, the compound that undergoes deprotection easily on treatment with hydrogen in the presence of 10% Pd/C to generate RNH2 is— (A) (C) (D) 70. In the 400 MHz 1H NMR spectrum, an (B) organic compound exhibited a doublet. The two lines of the doublet are at δ 2.35 and 2.38 ppm. The coupling constant (J) value is— (C) (A) 3 Hz (B) 6 Hz (C) 9 Hz (D) 12 Hz

12J | Chemical Sciences 2012 Part C 77. The decreasing order of dipole moment of molecules is— 71. The strength of pn– d2 bonding in E—O (E = (A) NF3 > NH3 > H2O Si, P, S and Cl) follows the order— (B) NH3 > NF3 > H2O (A) Si—O > P—O > S—O > Cl—O (C) H2O > NH3 > NF3 (B) P—O > Si—O > S—O > Cl—O (D) H2O > NF3 > NH3 (C) S—O > Cl—O > P—O > Si—O (D) Cl—O > S—O > P—O > Si—O 78. The cluster having arachno type structure is— 72. In the following reactions carried out in liquid (A) [Os5(CO)16] (B) [Os3(CO)12] NH3, Zn(NH2)2 + 2KNH2 → K2 [Zn(NH2)4] (C) [Ir4(CO)12] (D) [Rh6(CO)16] K2 [Zn(NH2)4] + 2NH4NO3 → Zn(NH2)2 + 2KNO3 + 4NH3 79. The carbonyl resonance in 13C NMR spec- KNH2 and NH4NO3 act respectively as— trum of (η5–C5H5) Rh(CO)]3 (103Rh, nuclear (A) Solvo-acid and solvo-base spin, I = 1/2, 100%) shows a triplet at – 65 °C (B) Solvo-base and solvo-acid owing to the presence of— (C) Conjugate acid and conjugate base (A) Terminal CO (B) μ2– CO (C) μ3– CO (D) η5– C5H5 (D) Conjugate base and conjugate acid 80. Low oxidation state complexes are often air- 73. The pair of lanthanides with the highest third- sensitive, but are rarely water sensitive ionization energy is— because— (A) Eu, Gd (B) Eu, Yb (A) Air is reducing in nature while water is inert (C) Dy, Yb (D) Lu, Yb (B) Both air and water are oxidizing in nature (C) Both air and water are not π-acceptors 74. The lanthanide (III) ion having the highest partition coefficient between tri-n-butylphos- (D) Complexes with low oxidation states will phate and concentrated HNO3 is— easily lose electrons to O2 but will not (A) La (III) (B) Eu (III) bind to a π-donor molecule like H2O (C) Nd (III) (D) Lu (III) 81. The metal complex that exhibits a triplet as well as a doublet in its 31P NMR spectrum 75. The quantitative determination of N2H4 with KIO3 proceeds in a mixture of H2O/CCl4 as is— follows : (A) mer–[IrCl3(PPh3)3] N2H4 + KIO3 + 2HCl ⎯→ (B) trans–[IrCl(CO)(PPh3)2] N2 + KCl + ICl + 3H2O (C) fac–[IrCl3(PPh3)3] (D) [Ir(PPh3)4]+ The end point for the titrimetric reaction is— 82. The complex that does not obey 18 electron (A) Consumption of N2H4 rule is— (B) ICl formation (A) [(η5–C5H5)RuCl(CO)(PPh3)] (C) Disappearance of the yellow colour due (B) [W(CO)3(SiMe3)(Cl)(NCMe)2] to Cl2 in CCl4 layer (C) [IrCl3(PPh3)2(AsPh2)]– (D) [Os (N)Br2(PMe3)(NMe2)]– (D) Disappearance of the red colour due to I2 in CCl4 layer 76. Among the halides, NCl3(A), PCl3(B) and 83. The number of spin-allowed ligand field AsCl3 (C), those which produce two different acids upon hydrolysis are— transitions for octahedral Ni (II) complexes with 3A2g ground state is— (A) A and B (B) A and C (A) Two (B) Three (C) B and C (D) A, B and C (C) One (D) Four

Chemical Sciences 2012 | 13J 84. The correct structure of P4S3 is— (A) 5,000 years (B) 4,000 years (A) (C) 877 years (D) 617 years (B) (C) 89. The reaction 3[Rh4(CO)12] → 2 [Rh6(CO)16] + 4CO [25 °C, 500 atm CO] is— (A) Exothermic as more metal-metal bonds are formed (B) Endothermic as stronger metal carbonyl bonds are cleaved while weaker metal- metal bonds are formed (C) Is entropically favorable but enthalpi- cally unfavorable such that ΔG = 0 (D) Thermodynamically unfavorable (ΔG > 0) 90. A column is packed with 0.5 g of a strongly acidic ion exchange resin in H+ form. A 1.0 M NaCl solution is passed through the column until the eluant coming out becomes neutral. The collected eluant is completely neutralized by 17 ml of 0.5 M NaOH. The ion exchange capacity of the resin is— (A) 1.00 meq/g (B) 1.25 meq/g (D) (C) 1.50 meq/g (D) 1.75 meq/g 85. The final product of the reaction [Mn(CO)6]+ 91. The molar extraction coefficient of B (MW = + MeLi → is— 180) is 4 × 103 lit mol–1 cm–1. One litre (A) [Mn(CO)6]– Me solution of C1 which contains 0.1358 g (B) [Mn(CO)5 Me] pharmaceutical preparation of B, shows an absorbance of 0.441 in a 1 cm quartz cell. The (C) [Mn(CO)6] (D) [(MeCO) Mn(CO)5] percentage (w/w) of B in the pharmaceutical preparation is— (A) 10.20 (B) 14.60 (C) 20.40 (D) 29.12 86. The reaction that yields Li [AlH4] is— 92. The changes (from A–D given below) which (A) HCl (excess) + AlCl3 + Li → occur when O2 binds to hemerythrin are— (B) H2 + Al + Li → (C) LiH (excess) + AlCl3 → 1. One iron atom is oxidized (D) LiH (excess) + Al → 2. Both the iron atoms are oxidized 3. O2 binds to one iron atom and is also hydrogen bonded 87. The number of microstates for d5 electron 4. O2 binds to both the iron atoms and is configuration is— also hydrogen bonded (A) 21 × 63 (B) 14 × 63 (A) 2 and 3 (B) 2 and 4 (C) 7 × 62 (D) 28 × 63 (C) 1 and 4 (D) 1 and 3 88. The carbon-14 activity of an old wood sample 93. In photosynthetic systems the redox metallo- is found to be 14.2 disintegrations min–1 g–1. proteins involved in electron transfer are cytochrome (cyt b), cytochrome bf complex Calculate age of old wood sample, if for a (cyt bf) and plastocyanin (PC). The pathway of electron flow is— fresh wood sample carbon–14 activity is 15.3 disintegrations min–1 g–1 (t1/2 carbon–14 is (A) PC → cyt b → cyt bf 5730 years), is—

14J | Chemical Sciences 2012 (B) cyt bf → cyt b → PC 2. Sum of the zeroth order and first order (C) cyt b → cyt bf → PC corrections to the ground state energy is (D) PC → cyt bf → cyt b ALWAYS greater than the exact ground state energy. 94. The total numbers of fine and hyperfine EPR 3. Sum of the zeroth order and first order lines expected for octahedral high-spin Mn corrections to the ground state energy is less than the exact ground state energy. (II) complexes are respectively (I = 5/2 for From the following which one is correct ? Mn)— (A) Only 1 is true (A) 3 and 30 (B) 5 and 33 (B) Both 1 and 2 are true (C) 5 and 30 (D) 4 and 24 (C) Only 3 is true 95. The Mossbauer spectra of two iron complexes are shown below. They may arise from (i) (D) Both 2 and 3 are true high-spin iron (III), (ii) high-spin iron (II) and (iii) low-spin iron (III) 99. Using Huckel molecular orbital approxi- mation, the two roots of secular equation of ethene are— (A) α + ⎯√⎯2β, α – ⎯√⎯2β (B) α + β, α (C) α + β, α – β (D) α + 2β, α – 2β The correct matches of spectra (A) and (B) 100. For H2 molecule in the excited state σ1g σ1z, with the iron complexes are— the spin part of the triplet state with ms = 0 is proportional to— (A) A with (i) and B with (ii) (A) α(1) β(2) (B) A with (ii) and B with (i) (C) A with (iii) and B with (ii) (B) [α(1) β(2) – β(1) α(2)] (D) A with (ii) and B with (iii) (C) α(1) α(2) 96. The probability of finding the particle in a one (D) [α(1) β(2) + β(1) α(2)] dimensional box of length L in the region 101. A square pyramidal, MX4, molecule belongs to C4V point group. The symmetry opera- between L and 3L for quantum number n = 1 tions are : E, 2C4, C2, 2σY, and 2σΔ. The 4 4 trace for the reducible representation, when is— symmetry operations of C4, applied to MX4, is— (A) 1 (B) 1 + 1 2 2 π (C) 1 – 1 (D) 2 (A) 5 1 1 1 3 (B) 1 1 1 1 1 2 π 3 (C) 5 1 1 1 1 (D) 4 1 1 1 3 97. A particle in three dimensional cubic box of length L has energy of 14h2 . The degeneracy 102. Character table of C2V, point group is 8mL2 C2 E C2 σV σV of the state is— A1 1 1 1 1 z A2 1 1 –1 –1 – (A) 2 (B) 3 B1 1 –1 1 –1 x B2 1 –1 –1 1 y (C) 6 (D) 9 If the initial and final states belong to A1 and 98. The following are the three statements about B1 irreducible representations respectively, perturbation theory— 1. Second order perturbation correction to the ground state energy is ALWAYS negative.

Chemical Sciences 2012 | 15J the allowed electronic transition from A1 to 108. Indicate which one of the following relations B1 is— is not correct— (A) z-polarized (B) y-polarized ( ) ( )(A) –∂T = ∂ρ (C) x-polarized (D) x, z polarized ∂V ∂S V S 103. Using cuvette of 0.5 cm path length, a 10–4 ( ) ( )(B) –∂T = ∂V M solution of a chromophore shows 50% ∂ρ ∂S P transmittance at certain wavelength. The S molar extinction coefficient of the chromo- phore at this wavelength is (log 2 = 0.301)— ( ) ( )(C) –∂S ∂ρ ∂V =– ∂T V (A) 1500 M–1 cm–1 (B) 3010 M–1 cm–1 T (C) 5000 M–1 cm–1 (D) 6020 M–1 cm–1 ( ) ( )(D) –∂S = ∂V ∂ρ ∂T P T 104. The set of allowed electronic transitions 109. The energy levels of the harmonic oscillator among the following is— (neglecting zero point energy) are εn = nhv for n = 0, 1, 2, …, ∞. Assuming hv = kaT, 1. 4∑ → 2Π 2. 3∑ → 3Π the partition function is— 3. 1Δ → 1Δ 4. 2Π → 2Π 5. 2∑ → 3Δ (A) e (B) 1 e (A) 1, 2, 5 (B) 1, 3, 5 (C) 2, 3, 4 (D) 3, 4, 5 (C) 1– 1 (D) 1 e 105. The following data were obtained from the 1 – 1 e vibrational fine structure in the vibronic spectrum of a diatomic molecule : 110. The correct entropy for 6 identical particles with their occupation number |0, 1, 2, 3| in ωe = 512 cm–1, ωexe ≡ 8 cm–1 four states is— where ωe is the energy associated with the (A) ka ln 6 (B) ka ln 12 natural frequency of vibration and xe is the (C) kb ln 60 (D) kb ln 720 anharmonicity constant. The dissociation energy (De) of the molecule is— 111. The correct Nernst equation for the (A) 4096 cm–1 (B) 6144 cm–1 concentration cell : (C) 8192 cm–1 (D) 16384 cm–1 68 106. An ideal gas was subjected to a reversible, Pt |H2 (P) | HCl (a±)1 | AgCl(S) | Ag Ag | adiabatic expansion and then its initial volume was restored by a reversible, isother- AgCl(S) | HCl (a±)2 | II2 (P) | Pt without mal compression. If ‘q’ denotes the heat liquid junction would be— added to the system and ‘w’ the work done by the system, then— (A) E = 2RT Ln (a±)1 F (a±)2 (A) w < 0, q < 0 (B) w > 0, q < 0 (B) E = RT Ln (a±)2 (C) w < 0, q > 0 (D) w > 0, q > 0 F (a±)1 107. The gas phase reaction 2NO2(g) → N2O4(g) (C) E = 2RT Ln (a±)2 is an exothermic process. In an equilibrium F (a±)1 mixture of NO2 and N2O4, the decomposi- tion of N2O4 can be induced by— (D) E = RT Ln (a±)2 (A) Lowering the temperature 2F (a±)1 (B) Increasing the pressure 112. Main assumption(s) involved in the deriva- tion of Debye-Huckel equation is(are) the (C) Introducing an inert gas at constant validity of— volume (A) Only Poisson equation (D) Introducing an inert gas at constant pressure

16J | Chemical Sciences 2012 (B) Poisson equation and Boltzmann distri- [h = 6.626 × 10–34 Js, c = 3 × 108 ms–1] bution (A) 1000 nm, UV (B) 1000 nm, IR (C) 500 nm, visible (D) 500 nm, FAR IR (C) Poisson equation, Boltzmann distribu- tion and |±Zeφ| >> kBT 118. The lattice parameter of an element stabi- lized in a fcc structure is 4.04 Å. The atomic (D) Poisson equation, Boltzmann distribu- radius of the element is— tion and |± Zeφ| << kBT 113. In the base (OH–) hydrolysis of a transition (A) 2.86 Å (B) 1.43 Å metal complex [ML6]Z+, the slope between (C) 4.29 Å (D) 5.72 Å log(k/k0) and √⎯ 1 is found to be –2.1. The — charge on the complex is— 119. The number-average molar mass (Mn) and (A) +1 (B) +2 — (C) +3 (D) +4 weight-average molar mass (Mw) of a poly- mer are obtained respectively by— 114. The rate law for one of the mechanisms of (A) Osmometry and light scattering mea- the pyrolysis of CH3CHO at 520 °C and 0.2 surements bar is Rate = – k2 ⎛⎝⎜2kk1d⎞⎠⎟1/2 [CH3CHO]3/2 (B) Osmometry and viscosity measure- ments The overall activation energy Ea in terms of (C) Light scattering and sedimentation mea- the rate law is— surements (A) Ea(2) + Ea(1) + 2Ea(4) (D) Viscosity and light scattering measure- ments (B) Ea(2) + 1 Ea(1) – Ea(4) 120. Two data sets involving the same variables 2 X and Y are given below : X 4.1 4.2 4.3 4.4 4.5 4.6 (C) Ea(2) + 1 Ea (1) – 1 Ea(4) 2 2 Y (Set A) 10.2 10.6 10.9 11.5 11.8 12.2 Y (Set B) 10.2 10.6 11.1 11.3 11.8 12.2 (D) Ea(2) – 1 Ea (1) + 1 Ea(4) 2 2 If the slopes and intercepts of the regression lines for the two sets are denoted by (mA, 115. In the Michaelis-Menten mechanism for en- mB) and (cA, cB), respectively, then— zyme kinetics, t he expression obtained is (A) mA > mB, cA > cB (B) mA < mB, cA > cB V = 1.4 × 1012 – 104 V (C) mA > mB, cA < cB [E]0 [S] [E]0 (D) mA < mB, cA < cB The values of k 3 (k cat, mol L–1s–1) and K (Michaelis constant, mol L–1), respectively, are— 121. Compounds A and B exhibit two singlets, each in their 1H NMR spectra. The expected (A) 1.4 × 1012, 104 (B) 1.4 × 108, 104 chemical shifts are at δ— (C) 1.4 × 108, 10–4 (D) 1.4 × 1012, 10–4 116. The most used acid catalyst in oil industry (A) 6.9 and 2.1 for A; 7.7 and 3.9 for B and the relevant process are respectively— (B) 7.7 and 3.9 for A; 6.9 and 2.1 for B (A) Aluminophosphate and reforming (C) 6.9 and 3.9 for A; 7·7 and 2.1 for B (B) Aluminosilicate and cracking (D) 7.7 and 2.1 for A; 6.9 and 3.9 for B (C) Aluminosilicate and reforming (D) Aluminophosphate and cracking 117. The wavelength and the spectral region for a single electron transfer across the band gap in a semiconductor (Eg = 1·98 × 10–19 J) are—

Chemical Sciences 2012 | 17J 122. In the following reaction sequence, the (A) Photochemical electrocyclic disrotatory major products A and B are— ring opening; and thermal antarafacial [1, 7]-H shift (A) (B) (B) Photochemical electrocyclic conrota- tory ring opening; and thermal antara- facial [1, 7]-H shift (C) Thermal electrocyclic conrotatory ring opening; and photochemical suprafacial [1, 7]-H shift (D) Thermal electrocyclic disrotatory ring opening; and thermal suprafacial [1, 7]- H shift 125. The intermediate A and the major product B in the following reaction are— (C) (A) (D) (B) (C) 123. The structure of the tricyclic compound (D) formed in the following two steps sequence is— 126. For the following two reactions A and B, the correct statement is— (A) (B) (C) (D) (A) (B) 124. The two steps conversion of 7-dehydro- cholesterol to vitamin D3 proceeds through—

18J | Chemical Sciences 2012 130. Structure of the starting material A in the (C) following photochemical Norrish reaction, is— (D) 127. The major compound B formed in the (A) (B) reaction sequence given below exhibited a carbonyl absorption band at 1770 cm–1 in the IR spectrum. The structures A and B are— (C) (D) (A) 131. Considering the following reaction, among a–c, the correct statements are— (B) (C) 1. The carbonyl group has enantiotopic faces; (D) 2. The hydride attack is re-facial; 128. Consider the following reaction sequence 3. It is a diastereoselective reduction starting with monoterpene α-pinene. Identify (A) 1 and 2 only (B) 1 and 3 only the correct statement— (C) 2 and 3 only (D) 1, 2 and 3 132. The major product formed in the following reaction sequence is— (A) A has a disubstituted double bond; B (A) (B) and C are dicarboxylic acids (C) (D) (B) A has a trisubstituted double bond; B is 133. The major product formed in the following a methyl ketone; and C is a dicar- boxylic acid reaction sequence is— (C) A has a disubstituted double bond; B is a methyl ketone; and C is a dicaroxylic acid (D) A has an exocyclic double bond; B and C are monocarboxylic acids 129. The major product formed when (3R, 4S)-3, 4-dimethylhexa-1, 5-diene is heated at 240 °C is— (A) (2Z,6Z)-octa-2,6-diene (B) (2E,6E)-octa-2,6-diene (C) (2E,6Z)-octa-2,6-diene (D) (3Z,5E)-octa-3,5-diene

Chemical Sciences 2012 | 19J (A) (C) (B) (D) (C) 136. The major product formed in the following reaction is— (D) 134. Match the following— (A) (B) Compound (C) (D) (a) Acetic acid (b) Acetonitrile 137. The reagents A and B in the following (c) Acetone reactions are— (d) Carbon tetrachloride (A) A = CH2I2,Zn–Cu; B = Me4S+I, NaH 13C NMR chemical shift (δ ppm) (B) A = CH2I2,Zn–Cu; B = Me3S+(O)I–, 1. 95 2. 115 NaH (C) A = Me3S+I–,NaH; B = Me3S+(O)I–, 3. 175 4. 205 NaH (a) (b) (c) (d) (D) A = Me3S+(O)I–,NaH; B = CH2I2, (A) 3 2 4 1 Zn–Cu 138. The major products A and B formed in the (B) 3 4 1 2 following reaction sequence are— (C) 1 2 4 3 (D) 3 1 2 4 135. The major products A and B in the following reaction sequence are— (A) (B) (A)

20J | Chemical Sciences 2012 141. The major product formed in the following (B) reaction is— (C) (A) (B) (D) 139. The major products A and B formed in the following reaction sequence are— (C) (A) (D) (B) 142. The correct sequence of reagents for effec- ting the following conversion is— (C) (D) (A) (a) (CH2OH)2, PTSA, Δ; 140. The correct reagent combination/reaction (b) CO2Ti Cl_ AlMe2 (Tebbe’s sequence for effecting the following conver- sion is— reagent); (c) H3O+; (d) KOH (B) (a) (CH2OH)2, PTSA, Δ; (b) Ph3P–CH2; (c) H3O+; (d) KOH (C) (a) CO2Ti Cl_ AlMe2 (Tebbe’s reagent); (A) (a) Me3SiCH2OMe, BuLi; (b) H3O+; (b) H3O+; (c) KOH (c) NaBH4, MeOH (D) (a) Ph3P–CH2; (b) H3O+; (c) KOH (B) (a) Ph3P+CH2OMeCl–, BuLi; (b) H3O+; 143. The major products A and B formed in the (c) NaBH4, MeOH following reaction sequence are— (C) (a) NH2NHTs; (b) NaOEt; (c) ClCOOEt (D) (a) NH2NHTs; (b) 2eq. nBuLi; (c) HCHO

Chemical Sciences 2012 | 21J (A) Answers with Explanations (B) 1. (D) (C) 2. (C) Total number of squares are 15. (D) 3. (D) 4. (B) 5. (C) 6. (A) 7. (B) 144. The reagent A required and the major 8. (B) product B formed in the following reaction sequence are— 9. (B) Chewing equilibrates the pressure on both side of the ear drum. (A) A = CH2Br2 and KOtBu 10. (A) 11. (A) 12. (D) 13. (B) 14. (A) (B) A = CH2Br2 and KOtBu 15. (B) In the most general sense of the word, a (C) A = CHBr3 and KOtBu cement is a binder, a substance that sets and hardens independently, and can bind other (D) A = CHBr3 and KOtBu materials together. The word “cement” traces 145. Among the choices, the correct statement for to the Romans, who used the term opus caementicium to describe masonary resem- A formed in the following reaction— bling modern concrete that was made from crushed rock with burnt lime as binder. (A) A is a single enantiomer (B) A is a racemic mixture 16. (D) Radiocarbon dating (usually referred to as (C) A is a mixture of two diastereomers simply carbon dating) is a radiometric dating (D) A is a mixture of two epimers method that uses the naturally occurring radioisotope carbon-14 (14C) to estimate the age of carbon-bearing materials up to about 58,000 to 60,000 years. Raw i.e., uncali- brated, radiocarbon ages are usually reported in radiocarbon years “Before Present” (BP), with “present” defined as AD 1950. Such raw ages can be calibrated to give calendar dates. One of the most frequent uses of radiocarbon dating is to estimate the age of organic remains from archaeological sites. When plants fix atmospheric carbon dioxide (CO2) into organic material during photosynthesis they incorporate a quantity of 14C that approximately matches the level of this isotope in the atmosphere (a small difference occurs because of isotope fractionation, but this is corrected after laboratory analysis. 17. (A) 18. (C) 19. (C) 20. (C) 21. (B) 22. (D) 23. (A) 24. (D) 25. (D) 26. (C) 27. (A) Carbon 14, unlike carbon 12 and carbon 13, is radioactive; meaning that over time the carbon 14 atoms will decay. When the isotope carbon 14 decays it gives off a beta particle and in doing so becomes nitrogen 14. The amount of carbon 12 and carbon 13, however, remains constant. 28. (D) The base pairing-rules for DNA are that, only the Nitrogen Bases of DNA which are;

22J | Chemical Sciences 2012 Adenine “A”-which only pairs with-Thymine triple point involving two different fluid “T”, and Cytosine “C”-which only pairs with- Guanine “G” can only pair to one another phases. In general, for a system with p within that sequence. possible phases, there are triple points. 29. (A) 30. (B) ( )p = 1 p (p – 1) (p – 2) 3 6 31. (C) In coordination chemistry, a ligand is an 46. (B) 47. (C) ion or molecule that binds to a central metal atom to form a coordination complex. The 48. (A) The key assumption of Activated bonding between metal and ligand generally Complex Theory (ACT), that the AC is in involves formal donation of one or more of thermodynamic equilibrium with the reac- the ligand’s electron deficient pairs. The tants, needs to be reconsidered. This is nature of metal-ligand bonding can range because the formation of the AC is slower from covalent to ionic. Furthermore, the than its collapse to product. However, this can metal-ligand bond order can range from one be remedied by assuming that the AC is to three. Ligands are viewed as Lewis bases, formed in a rapid pre-equilibrium as a although rare cases are known involving thermally activated species, which collapses Lewis acidic “ligands”. to products in a slow step involving the diffusion of another AC molecule (or solvent 32. (C) 33. (B) 34. (A) 35. (D) in the case of a unimolecular reaction). 36. (B) Chromate salts contain the chromate 49. (D) anion, CrO42–. Dichromate salts contain the dichromate anion, Cr2O72–. They are oxya- 50. (B) Additional Polymerization : A chemical reaction in which simple molecules (mono- nions of chromium in the oxidation state +6. mers) are added to each other to form long- chain molecules (polymers) without by They are moderately strong oxidizing agents. products. The molecules of the monomer join together to form a polymeric product in which 37. (A) For hydrogen and other nuclei stripped to the molecular formula of the repeating unit is one electron, the energy depends only upon identical with that of the monomer. The the principal quantum number n. molecular weight of the polymer so formed is thus the total of the molecular weights of all En = –13.6 Z2 eV of the combined monomer units. n2 This fits the hydrogen spectrum unless you 51. (C) First, consider the atoms at the 8 corners. take a high resolution look at fine structure where the electron spin and orbital quantum Each corner atom is occupied by 8 unit cells, numbers are involved. At even higher resolu- tions, there is a tiny dependence upon the which means 1/8 of the atom is actually in orbital quantum number in the Lamb shift. one unit cell. There are 8 of these corner ( )atoms, so 8 1 = 1. 8 38. (D) 39. (D) 40. (B) 41. (A) 42. (D) Next, consider the atoms at the 6 faces of the 43. (B) 44. (C) unit cell. Each face atom is occupied by 2 unit cells, which means 1/2 of the atom is in one 45. (C) In thermodynamics, the triple point of a unit cell. There are 6 of these face atoms, so 6 substance is the temperature and pressure at (1/2) = 3. which the three phases (gas, liquid and solid) of that substance coexist in thermodynamic Add up the two values… 1 + 3 = 4 equilibrium. For example, the triple point of mercury occurs at a temperature of So, there are 4 atoms contained in a face- –38.8344°C and a pressure of 0.2 mPa. centered. In addition to the triple point between solid, 52. (C) Liquids, solids and gases all may be liquid and gas, there can be triple points mixed to form colloidal dispersions. involving more than one solid phase, for substances with multiple polymorphs. Aerosols : solid or liquid particles in a gas. Helium-4 is a special case that presents a Examples : Smoke is a solid in a gas. Fog is a liquid in a gas.

Chemical Sciences 2012 | 23J Sols : solid particles in a liquid. Example : 58. (D) Milk of Magnesia is a sol with solid magne- 59. (B) Due to generation of two chiral centers by sium hydroxide in water. epoxidation it will generate diastereomers. A Emulsions : liquid particles in liquid. diastereomer is simply any stereoisomer that Example : Mayonnaise is oil in water. is not an enantiomer. Technically, cis-trans isomers are diastereomers. However, typically Gels : liquids in solid. Examples : gelatin is the term is reserved for stereoisomers that protein in water. Quicksand is sand in water. differ at some but not all stereocenters. 53. (B) 54. (B) Diastereomers formed by inverting some but not all stereocenters 55. (B) As ester is first priority group compared to ketone and alkene, it will go in first priority 60. (A) 61. (B) and after applying Cahn-Prelog Rule (A 62. (B) Due to the cis bromonium addition and molecule may contain any number of stereo centers and any number of double bonds, and after that SN2 reaction of OMe - gives the each gives rise to two possible configurations) product two due to oxygen lone pair. option 2 is correct. 63. (D) 64. (A) 65. (B) 66. (D) 67. (A) 68. (A) 56. (D) Due to the generation of allelic and 69. (C) Due to Birch reduction first intermediate benzylic carbonium that is resonance stabi- is the one which prefers the formation of lized so it will attack on the bromo allyl in product 3 as para directing strength of methyl very facile condition that’s why product 4 is group is more than that of methoxy. formed as major product. 70. (D) 71. (D) 72. (A) 73. (B) 74. (D) 75. (D) 76. (C) 77. (C) 78. (B) 79. (B) 57. (C) In the phenathrene there are 7 types of 80. (D) 81. (A) 82. (D) 83. (B) 84. (A) magnetically equivalent while in anthracene 85. (D) 86. (C) 87. (C) 88. (D) 89. (B) there are 4 types of magnetically equivalent 90. (D) 91. (A) 92. (A) carbons. Phenanthrene is a polycyclic aromatic hydrocarbon composed of three fused benzene rings. The name phenanthrene is a composite of phenyl and anthracene. In its pure form, it is found in cigarette smoke and is a known irritant, photo-sensitizing skin to light. Phenanthrene appears as a white powder having blue fluorescence. 93. (C)

24J | Chemical Sciences 2012 94. (C) 95. (B) 96. (B) 97. (C) 98. (B) 133. (A) In case of DIBAL reaction opening of lactam took place because here is only 1 99. (C) 100. (D) 101. (A) 102. (C) 103. (D) equation DIBAL is used so further reduction of aldehyde is not possible second reaction 104. (C) 105. (C) 106. (B) 107. (D) 108. (B) is Grignard on aldehyde group which give secondary allelic alcohol which on oxidation 109. (D) 110. (C) with PCC gives ketone. 111. (B) The two (ultimately equivalent) equa- 134. (A) 135. (A) 136. (B) tions for these two cases (half-cell, full cell) are as follows : 137. (B) The alpha beta unsaturated ketones on reaction with Ch2I2 Zn–Cl give cyclopropyl Ered = Eθred – RT ln aRed (half-cell reduction on alpha beta unsaturated bond on the other zF aOx hand alpha beta unsat alcohols on reaction with tetramethyl sulphonium iodide in pre- potential) sence of hydride gives cyclopropane at alpha beta unsat positions. Ecell = Eθcell – RT ln Q (total cell potential) zF 138. (C) Reaction with LAH gives cis ally alcohol which on sharpless epoxidation 112. (D) 113. (B) 114. (C) 115. (C) 116. (B) gives up epoxide. 117. (B) 118. (B) 119. (A) 120. (C) 121. (A) 139. (B) The hydroboration of alkynes gives trans hydroborene due to steric hindrance of 122. (B) 123. (D) 124. (B) 125. (D) 126. (B) bulkier cyclopropyl and hydroborating group which on further substitution gives same 127. (B) side conjugated alkene. 128. (B) A has a trisubstituted double bond which 140. (D) A simple stark examine reaction after is endocyclic and after KMnO4 oxidation it that anionic attack on formaldehyde gives gives B a methyl ketone and after the the alcohol. reaction with NaOH and Br2 gives dicarboxylic acid C. 141. (D) 142. (A) 129. (C) 143. (D) Abstraction of H from indolic N attacks on oxirane giving product 4A which after 130. (C) This is the Norrish type reaction in addition with benzaldehyde gives product which abstraction of gama hydrogen takes 4B. place by the formation of double bond that gives product 3. 144. (C) 145. (B) 131. (C) The hydride ion attacks from the lower face of the plane that is called re facial and it reduces the ketone that results in another chiral centre so it will be diastereoselective reduction. 132. (D)

Chemical Sciences CSIR UGC-NET/JRF Exam. Solved Paper

December 2012 Chemical Science Time : 3 Hours] [Maximum Marks : 200 Direction List of the Atomic Weights of the Elements 1. This Test Booklet contains one hundred and forty five (20 Part ‘A’ + 50 Part ‘B’ + 75 Part Element Symbol Atomic Atomic ‘C’) Multiple Choice Questions (MCQs). You Number Weight are required to answer a maximum of 15, 35 Actinium Ac and 25 questions from part ‘A’, ‘B’ and ‘C’ Aluminium Al 89 (227) respectively. If more than required number of Americium Am 13 26.98 questions are answered, only first 15, 35 and Antimony Sb 95 25 questions in Parts ‘A’, ‘B’ and ‘C’ Argon Ar 51 (243) respectively, will be taken up for evaluation. Arsenic As 18 121.75 Astatine At 33 2. Each question in Part ‘A’ and ‘B’ carries 2 Barium Ba 85 39.948 marks and Part ‘C’ questions carry 4 marks Berkelium Bk 56 74.92 each respectively. There will be negative Beryllium Be 97 (210) marking @ 25% for each wrong answer. Bismuth Bi 4 137.34 Boron B 83 (249) 3. Below each question in Part ‘A’, ‘B’ and ‘C’ Bromine Br 5 9.012 four alternatives or responses are given. Only Cadmium Cd 35 208.98 one of these alternatives is the ‘correct’ option Calcium Ca 48 10.81 to the question. You have to find, for each Californium Cf 20 79.909 question, the correct or the best answer. Carbon C 98 112.40 Cerium Ce 6 40.08 Useful Fundamental Constants Cesium Cs 58 (251) Chlorine Cl 55 12.011 m Mass of electron 9.11 × 10–31 kg Chromium Cr 17 140.12 Cobalt Co 24 132.91 h Planck’s constant 6.63 × 10–34 J-sec Copper Cu 27 35.453 Curium Cm 29 52.00 e Charge of electron 1.6 × 10–19 C Dysprosium Dy 96 58.93 Einsteinium Es 66 63.54 K Boltzmann constant 1.38 × 10–23 J/K Erbium Er 99 (247) Europium Eu 68 162.50 c Velocity of Light 3.0 × 108 m/sec Fermium Fm 63 (254) Fluorine F 100 167.26 IeV 1.6 × 10–19 J Francium Fr 151.96 Gadolinium Gd 9 (253) amu 1.67 × 10–27 kg Gallium Ga Germanium Ge 87 19.00 G 6.67 × 10–11 Nm2 kg–2 Gold Au Hafnium Hf 64 (223) Ry Rydberg constant 1.097 × 107 m–1 31 157.25 NA Avogadro number 6.023 × 1023 mole–1 ε0 8.854 × 10–12 Fm–1 8.314 JK–1 mole–1 32 69.72 μ0 4π × 10–7 Hm–1 R Molar Gas constant 79 72.59 72 196.97 178.49

4 | CSIR–Chemical Sciences (Dec.–12) Helium He 2 4.003 Tellurium Te 52 127.60 Holmium Ho 67 164.93 Hydrogen H 1 1.0080 Terbium Tb 65 158.92 Indium In 49 114.82 Iodine I 53 126.90 Thallium Tl 81 204.37 Iridium Ir 77 192.2 Iron Fe 26 55.85 Thorium Th 90 232.04 Krypton Kr 36 83.80 Lanthanum La 57 138.91 Thulium Tm 69 168.93 Lawrencium Lr 103 (257) Lead Pb 82 207.19 Tin Sn 50 118.69 Lithium Li 3 6.939 Lutetium Lu 71 174.97 Titanium Ti 22 47.90 Magnesium Mg 12 24.312 Manganese Mn 25 54.94 Tungsten W 74 183.85 Mendelevium Md 101 (256) Mercury Hg 80 200.59 Uranium U 92 238.03 Molybdenum Mo 42 95.94 Neodymium Nd 60 144.24 Vanadium V 23 50.94 Neon Ne 10 20.183 Xenon Xe 54 131.30 Neptunium Np 93 (237) Ytterbium Yb 70 173.04 Nickel Ni 28 58.71 Yttrium Y 39 88.91 Niobium Nb 41 92.91 Zinc Zn 30 65.37 Nitrogen N 7 14.007 Zirconium Zr 40 91.22 Nobelium No 102 (253) * Based on mass of C12 at 12.00… . The ratio of these Osmium Os 76 190.2 weights of those on the order chemical scale (in which Oxygen O 8 15.9994 oxygen of natural isotopic composition was assigned a Palladium Pd 46 106.4 mass of 16.0000…) is 1.000050. (Values in parentheses Phosphorus P 15 30.974 represent the most stable known isotopes). Platinum Pt 78 195.09 Plutonium Pu 94 (242) Part–A Polonium Po 84 (210) Potassium K 19 39.102 1. Which of the following numbers is the Praseodymium Pr 59 140.91 Promethium Pm 61 (147) largest ? Protactinium Pa 91 (231) 234, 243,324, 342, 423, 432. Radium Ra 88 (226) (A) 234 (B) 342 Radon Rn 86 (222) (C) 432 (D) 423 Rhenium Re 75 186.23 Rhodium Rh 45 102.91 2. The cube ABCDEFGH in the figure has each Rubidium Rb 37 85.47 Ruthenium Ru 44 101.1 edge equal to a. The area of the triangle with Samarium Sm 62 150.35 Scandium Sc 21 44.95 vertices at A, C and F is— Selenium Se 34 78.96 Silicon Si 14 28.09 (A) ⎯√ 3 a2 (B) √⎯ 3 a2 Silver Ag 47 107.870 Sodium Na 11 22.9898 4 2 Strontium Sr 38 87.62 Sulphur S 16 32.064 (C) ⎯√ 3 a2 (D) 2√⎯ 3 a2 Tantalum Ta 73 180.95 Technetium Tc 43 (99) 3. What is the number of distinct arrangements of the letters of the word UGC-CSIR so that U and I cannot come together ? (A) 2520 (B) 720 (C) 1520 (D) 1800 4. Suppose the sum of the seven positive numbers is 21. What is the minimum possible value of the average of the squares of these numbers ? (A) 63 (B) 21 (C) 9 (D) 7

CSIR–Chemical Sciences (Dec.–12) | 5 5. Let A = 113 + 213 + 313 + ··· + 10013, 10. Amar, Akbar and Anthony are three friends, 100 one of whom is a doctor, another is an engineer and the third is a professor. Amar is B = 113 + 313 + 513 + ··· + 9913, not an engineer. Akbar is the shortest. The 50 tallest person is a doctor. The engineer’s height is the geometric mean of the heights of C = 213 + 413 + 613 + ··· + 10013, the other two. Then which of the following is 50 true ? Which of the following is true ? (A) Amar is a doctor and he is the tallest (A) B < C < A (B) A < B < C (B) Akbar is a professor and he is tallest (C) B < A < C (D) C < A < B (C) Anthony is an engineer and he is shortest 6. A circle of radius 5 units in the XY plane has (D) Anthony is a doctor and he is the tallest its centre in the first quadrant, touches the x-axis and has a chord of length 6 units on the 11. If 100 cats catch 100 mice in 100 minutes, y-axis. The coordinate of its centre are— then how long will it take for 7 cats to catch 7 mice ? (A) (4, 6) (B) (3, 5) (A) 100/7 minutes (B) 100 minutes (C) (5, 4) (D) (4, 5) (C) 49/100 minutes (D) 7 minutes 7. A wire of length 6 metres is used to make a tetrahedron of each edge 1 metre, using only 12. What does this digram demonstrate ? one strand of wire for each edge. The minimun number of times the wire has to be cut is— (A) 1 + 2 + 3 + ··· + n = n · (n + 1) 2 (A) 2 (B) 3 (C) 1 (D) 0 (B) 12 + 22 + 32 + ··· + n2 = n· (n + 1)·(2n + 1) 2 8. If the sum of the next two terms of the series (C) 1 + 3 + ··· + (2n + = –1) = n2 below is x, what is the value of log2x ? (D) 22 + 42 + ··· + (2n)2 = 2· n(n + 1)·(2n + 1) 2, – 4, 8, – 16, 32, – 64, 128, ………… 3 (A) 128 (B) 10 13. Suppose there are socks of N different colours (C) 256 (D) 8 in a box. If you take out one sock at a time, 9. what is the maximum number of socks that you have to take out before a matching pair is found ? Assume that N is an even number. (A) N (B) N + 1 (C) N – 1 (D) N/2 A conical vessel with semi-vertical angle 30° 14. At what time after 4 O’ clock, the hour and and height 10·5 cm has a thin lid. A sphere the minute hands will lie opposite to each kept inside it touches the lid. The radius of other ? the sphere (in cm) is— (A) 4 – 50′ – 31′′ (B) 4 – 52′ – 51′′ (C) 4 – 53′ – 23′′ (D) 4 – 54′ – 33′′ (A) 3·5 (B) 5 15. Which of the following curves just touches the x-axis ? (C) 6·5 (D) 7 (A) y = x2 – x + 1

6 | CSIR–Chemical Sciences (Dec.–12) (B) y = x2 – 2x + 2 Part–B (C) y = x2 – 10x + 25 (D) y = x2 – 7x + 12 21. For an odd nucleon in ‘g’ nuclear orbital and parallel to I, spin and parity are— 16. (A) 9/2 and (+) (B) 7/2 and (+) (C) 9/2 and (–) (D) 7/2 and (–) 22. For the deposition of Pb by electroplating, the best suited compound among the following is— (A) PbCl2 (B) PbSO4 If AB is paralleled to CD and AO = 2OD, (C) Pb(Et)4 (D) Pb(BF4)2 then the area of triangle OAB is bigger than the area of triangle OCD by a factor of— 23. Appropriate reasons for the deviation form the Beer’s law among the following are— (A) 2 (B) 3 1. Monochromaticity of light 2. Very high concentration of analyte (C) 4 (D) 8 3. Association of analyte 4. Dissociation of analyte 17. (A) 1, 2 and 4 (B) 2, 3 and 4 (C) 1, 3 and 4 (D) 1, 2 and 3 24. Which one of the following shows the highest solubility in hot concentrated aqueous NaOH? A semi-circular arch of radius R has a vertical (A) La(OH)3 (B) Nd(OH)3 pole put on the ground together with one of its legs. An ant on the top of the arch finds the (C) Sm(OH)3 (D) Lu(OH)3 angular height of the tip of the pole to be 45°. The height of the pole is— 25. In the vibrational spectrum of CO2, the number of fundamental vibrational modes (A) √⎯ 2R (B) ⎯√ 3R common in both infrared and Raman are— (C) ⎯√ 4R (D) √⎯ 5R (A) three (B) two (C) one (D) zero 18. Suppose we make N identical smaller spheres 26. The light pink colour of [Co(H2O)6]2+ and the from a big sphere. The total surface area of deep blue colour of [CoCl4]2– are due to— the smaller spheres is X times the total surface area of the big sphere, where X is— (A) MLCT transition in the first and d-d transition in the second (A) ⎯√ N (B) 1 (D) N3 (B) LMCT transitions in both (C) N1/3 (C) d-d transitions in both 19. What is the next number in the sequence 24, 30, 33, 39, 51, ········ ? (D) d-d transition in the first and MLCT transition in the second (A) 57 (B) 69 27. In Mo2(S2)6]2– cluster the number of bridging S22– and coordination number of Mo (C) 54 (D) 81 respectively, are— 20. Four lines are drawn on a plane with no two parallel and no three concurrent. Lines are (A) 2 and 8 (B) 2 and 6 drawn joining the points of intersection of the previous four lines. The number of new lines (C) 1 and 8 (D) 1 and 6 obtained this way is— 28. 1H NMR spectrum of HD would show— (A) 3 (B) 5 (A) a singlet (B) a doublet (C) 12 (D) 2

CSIR–Chemical Sciences (Dec.–12) | 7 (C) a triplet with intensity ratio : 1 : 2 :1 37. Degradation of penicillin G— (D) a triplet with intensity ratio 1 : 1 : 1 29. The number of possible isomers of [Ru(PPh3)2(acac)2] (acac = acetylacetonate) is— (A) 2 (B) 3 gives penicillamine that can utilize nitrogen, oxygen or sulfur atoms as donors to bind with (C) 4 (D) 5 lead(II), mercury(II) or copper(II). The structure of penicillamine is— 30. The total number of Cu – O bonds present in the crystalline copper (II) acetate monohy- (A) drate is— (B) (A) 10 (B) 6 (C) (C) 8 (D) 4 31. The electronegativity difference is the highest for the pair— (A) Li, Cl (B) K, F (C) Na, Cl (D) Li, F 32. Which ones among CO32–, SO3, XeO3 and (D) NO3– have planar structure ? (A) CO32–, SO3 and XeO3 38. The molecule that has an S6 symmetry (B) SO3, XeO3 and NO3– element is— (C) CO32–, XeO3 and NO3– (D) CO32–, SO3 and NO3– 33. The substitution of η5 – Cp group with nitric (A) B2H6 (B) CH4 oxide is the easiest for— (C) PH5 (D) SF6 (A) η5 – Cp2Fe (B) η5 – Cp2CoCl 39. The electric dipole allowed transition in a d2 (C) η5 – Cp2Ni (D) η5 – Cp2Co atomic system is— 34. The molecule (A) 3F → 1D (B) 3F → 1P (C) 3F → 3D (D) 3F → 3P obeys 18 e rule. The two ‘M’ satisfying the 40. When a hydrogen atom is placed in an electric field along the y-axis, the orbital that mixes condition are— most with the ground state 1s orbital is— (A) Cr, Re+ (B) Mo, V (A) 2s (B) 2px (C) 2py (D) 2pz (C) V, Re+ (D) Cr, V 35. The correct set of the biologically essential 41. For water, ΔHvap ~~ 41 kJ mol–1. The molar elements is— entropy of vaporization at 1 atm pressure is (A) Fe, Mo, Cu, Zn (B) Fe, Cu, Co, Ru approximately— (C) Cu, Mn, Zn, Ag (D) Fe, Ru, Zn, Mg (A) 410 J K–1 mol–1 (B) 110 J K–1 mol–1 36. The number of lines exhibited by a high (C) 41 J K–1 mol–1 (D) 11 J K–1 mol–1 resolution EPR spectrum of the species, 42. If A and B are non-commuting hermitian [Cu(ethylenediamine2]2+ is [Nuclear spin (I) operators, all eigenvalues of the operator given of Cu = 3/2 and that of N = 1]— by the commutator [A, B] are— (A) 12 (B) 15 (A) complex (B) real (C) 20 (D) 36 (C) imaginary (D) zero

8 | CSIR–Chemical Sciences (Dec.–12) 43. The value of the commutator [x , px2] is given (C) hydrochloric acid in water (D) benzoic acid in benzene by— (A) 2 i (B) 2 i h– 50. A dilute silver nitrate solution is added to a (C) 2 i –h x (D) 2 i –h px slight excess of sodium iodide solution. A 44. The correlation coeffcient between two solution of Agl is formed whose surface arbitrary variable x and y is zero, if— (A) <xy> = <yx > (B) <x2> = <x>2 adsorbs— (B) NO3– (C) <y2> = <y >2 (D) <xy > = <yx > <y> (A) I – (C) Na+ (D) Ag+ 45. A Carnot engine takes up 90 J of heat from 51. The absorption spectrum of O2 shows a vibra- the source kept at 300 K. The correct tional structure that becomes continuum at statement among the following is— 56875 cm–1. At the continuum, it dissociates (A) it transfers 60 J of heat to the sink at 200 K into one ground state atom (Og) and one (B) it transfers 50 J of heat to the sink at excited state atom (Oe). The energy difference 200 K between Oe and Og is 15125 cm–1. The dissociation energy (in cm–1) of ground state (C) it transfers 60 J of heat to the sink at 250 K of O2 is— (D) it transfers 50 J of heat to the sink at (A) 56875 (B) 15125 250 K 15125 56875 (C) 72000 (D) 41750 52. The angle between the two planes represented 46. The relative population in two states with by the Miller indices (110) and (111) in a energies E1 and E2 satisfying Boltzman distribution is given by n1/n2 = (3/2) exp. simple cubic lattice is— [–(E1 – E2 )/KBT]. The relative degeneracy g2/g1 is— (A) 30° (B) 45° (C) 60° (D) 90° (A) 2 (B) 2/3 53. The correct representation of the variation of molar conductivity (y-axis) with surfactant (C) 3/2 (D) 3 concentration (x-axis) is [CMC = Critical Micelle Concentration]— 47. The Daniell cell is— (A) Pt1 (s)⎜Zn(s) ⎜Zn2+ (aq)⎜⎜Cu2+ (aq)⎜Cu(s)⎜ (A) (B) Pt11 (s) (B) Pt1 (s)⎜Zn(s) ⎜Zn2+ (aq) ⎜⎜Ag+ (aq)⎜Ag(s)⎜ Pt11 (s) (C) Pt1 (s)⎜Fe(s)⎜Fe2+ (aq)⎜⎜Cu2+ (aq)⎜Cu(s)⎜ Pt11 (s) (D) Pt1 (s)⎜H2(g)⎜H2SO4 (aq)⎜⎜Cu2+ (aq) ⎜Cu(s)⎜Pt11 (s) 48. If the concept of half-life is generalized to (C) (D) quarter-life of a first order chemical reaction, it will be equal to— 54. The major product formed in the following reaction is— (A) ln 2 / k (B) ln 4 / k hν (C) 4 / k (D) 1 / 4k 49. Kohlrausch’s law is applicable to a dilute solution of— (A) potassium chloride in hexane (B) acetic acid in water

CSIR–Chemical Sciences (Dec.–12) | 9 59. The correct statement(s) A-D are given for (A) the following reaction. The correct one(s) is (are)— (B) (C) 1. aromatic ipso substitution reaction 2. aromatic nucleophilic substitution (D) 3. aromatic electrophilic substitution 4. aromatic free radical substitution 55. If the pKa value for p-methoxybenzoic acid is 4·46 and that of benzoic acid is 4·19, the σpara (A) 1 and 2 only (B) 1 and 3 only for methoxy group is— (C) 3 and 4 only (D) 3 only (A) 8·65 (B) 4·32 60. The following photochemical transformation proceeds through (C) 0·27 (D) –0·27 56. The biosynthetic precursor of cadinene is— Cadinene (A) Norrish type I reaction (B) Norrish type II reaction (A) shikimic acid (B) mevalonic acid (C) Barton reaction (C) arachidonic acid (D) prephenic acid (D) Paterno-Buchi reaction 57. The correct order of acidity of the compounds 61. A tripeptide gives the following products on A-C is— Edman degradation. The tripeptide is— (B) Phe-Gly-Ala (A) Phe-Ala-Gly (D) Gly-Ala-Phe (C) Ala-Gly-Phe A BC 62. In the 1H NMR spectrum recorded at 293 K, (A) A > B > C (B) B > C > A an organic compound (C3H7NO), exhibited (C) C > A > B (D) B > A > C signals at δ 7·8 (1 H, s), 2·8 (3 H, s) and 2·6 (3 H, s). The compound is— 58. The mechanism involved in the following conversion is— (A) (B) (C) (D) 63. In the IR spectrum of p-nitrophenyl acetate, the carbonyl absorption band appears at— (A) E2-elimination (B) E1-elimination (A) 1670 cm–1 (B) 1700 cm–1 (C) syn-elimination (D) E1cb-elimination (C) 1730 cm–1 (D) 1760 cm–1

10 | CSIR–Chemical Sciences (Dec.–12) 64. The absolute configuration at the two chiral 69. The most appropriate reagent to effect the centres of (–)-camphor is— following chemoselective conversion is— (A) 1R, 4R (B) 1R, 4s (A) HCl, EtOH, reflux (C) 1s, 4R (D) 1s, 4s (B) Bu4NF 65. The major product formed in the following (C) K2CO3 MeOH reaction is— (D) CF3COOH, EtOH, rt (A) (B) 70. Among the following, an example of a ‘Green (C) (D) Synthesis’ is— (A) synthesis of malachite green (B) Friedel-Craft’s acylation of anisole with Ac2O/anhydrous AlCl3 (C) Jones’ oxidation of benzyl alcohol to benzoic acid (D) Diels-Alder reaction of furan and maleic acid in water 66. The first person to separate a racemic mixture Part–C into individual enantiomers is— (A) J. H. Van’t Hoff (B) L. Pasteur 71. The recoil energy of a Mössbauer nuclide of (C) H. E. Fischer (D) F. Wöhler mass 139 amu is 2·5 Mev. The energy emitted by the nucleus in keV is— 67. Consider the following statements for [18]- (A) 12·5 (B) 15·0 annulene : (C) 20·5 (D) 25·0 1. it is aromatic 72. Complexes of general formula, fac- [Mo(CO)3(phosphine)3] have the C-O 2. the inner protons resonate at δ 9·28 in its stretching bands as given below : 1H NMR spectrum Phosphines : PF3 (A); PCl3 (B); P(Cl)Ph2 (C); PMe3 (D) 3. there are six protons in the shielded zone. v(CO), cm–1 : 2090 (i); 2040 (ii); 1977 (iii); 1945 (iv) The correct statements are— The correct combination of the phosphine and (A) 1, 2, 3 (B) 1 and 2 only the streching frequency is— (C) 2 and 3 only (D) 1 and 3 only (A) (A – i) (B – ii) (C – iii) (D – iv) 68. In the compound given below, the relation (B) (A – ii) (B – i) (C – iv) (D – iii) between HA, HB, and between Br1, Br2 is— (C) (A – iv) (B – iii) (C – ii) (D – i) (A) HA, HB are enantiotropic; and Br1, Br2 are diastereotopic (D) (A – iii) (B – iv) (C – i) (D – ii) (B) HA, HB are diastereotopic; and Br1, Br2 73. On subjecting 9·5 ml solution of Pb2+ of X M are enantiotropic to polorographic measurements, Id was found (C) HA, HB are diastereotopic; and Br1, Br2 to be 1 μA. When 0·5 ml of 0·04 M Pb2+ was are homotopic added before the measurement, the Id was found to be 1·25 μA. The molarity X is— (D) HA, HB are enantiotropic; and Br1, Br2 are homotopic (A) 0·0035 (B) 0·0400 (C) 0·0067 (D) 0·0080

CSIR–Chemical Sciences (Dec.–12) | 11 74. Match each item from the List-I (compound Consider the following statements about its in solvent) with that from the List-II (its be- room temperature spectral data : haviour) and select the correct combination using the codes given below— 1. 1H NMR has singles at 5·48 and 3·18 ppm. List-I 2. 1H NMR has multiplet at 5·48 and singlet (a) CH3COOH in pyridine at 3·18 ppm. (b) CH3COOH in H2SO4 (c) HClO4 in H2SO4 3. IR has CO stretching bands at 1950 and (d) SbF5 in HF 1860 cm–1. List-II 4. IR has only one CO stretching band at 1900 cm–1. The correct pair of statements is— 1. Strong acid 2. Weak acid (A) 1 and 3 (B) 2 and 3 3. Strong base 4. Weak base (C) 1 and 4 (D) 2 and 4 Codes : (b) (c) (d) 79. In the cluster [Co3(CH)(CO)9] obeying 18e (a) 23 4 rule, the number of metal-metal bonds and the 13 4 bridging ligands respectively, are— (A) 1 42 1 (B) 2 23 1 (A) 3 and 1 CH (B) 0 and 3 CO (C) 3 (D) 4 (C) 3 and 1 CO (D) 6 and 1 CH 75. Structure of a carborane with formula, 80. Consider the ions Eu(III), Gd(III), Sm(III) C2B4H8 is formally derived from— and Lu(III). The observed and calculated magnetic moment values are closest for the (A) Closo-borane (B) Nido-borane pair— (C) Arachno-borane (D) Conjuncto-borane (A) Gd (III), Lu (III) 76. Boric acid is a weak acid in aqueous solution. (B) Eu (III), Lu (III) But its acidity increases significantly in the presence of ethylene glycol, because— (C) Sm (III), Gd (III) (A) ethylene glycol releases additional H+ (D) Sm (III), Eu (III) (B) B(OH)4– is consumed in forming a 81. Silicates with continuous 3D framework compound with ethylene glycol are— (C) ethylene glycol neutralizes H+ released (A) Neso-silicates (B) Soro-silicates by boric acid (C) Phyllo-silicates (D) Tecto-silicates (D) Boric acid dissociates better in the mixed-solvent 82. The correct spinel structure of Co3O4 is— (A) (Co2+)t(2Co3+)oO4 77. Coordination number of ‘C’ in Be2C3 whose (B) (Co3+)t(Co2+Co3+)oO4 structure is correlated with that of CaF2, is— (C) (Co2+Co3+)t(Co3+)oO4 (D) (2Co3+)t(Co2+)oO4 (A) 2 (B) 4 83. In the solid state, the CuCl53– ion has two (C) 6 (D) 8 types of bonds. There are— (A) three long and two short 78. For the molecule given below— (B) two long and three short (C) one long and four short (D) four long and one short 84. In metalloenzymes, the metal centers are covalently linked through the side chains of the amino acid residues. The correct set of

12 | CSIR–Chemical Sciences (Dec.–12) amino-acids which are involved in the 88. Successive addition of NaCl, H3PO4, KSCN primary coordination spheres of metalloen- and NaF to a solution of Fe(NO3)3·9H2O zymes is— gives yellow, colourless, red and again (A) Ala, Leu, His (B) Glu, His, Cys (C) Leu, Glu. Cys (D) Ala, His, Glu colourless solutions due to the respective 85. Consider the catalyst in Column-I and reaction formation of— in Column-II : (A) [Fe(H2O)5Cl]2 + , [Fe(H2O)5(PO4)], Column-I [Fe(H2O)5(SCN)]2+, [Fe(H2O)5F]2+ (a) [(R)-BINAP]Ru2– (b) [Rh(CO)2I2]– (B) [Fe(H2O)4Cl(OH)]1+, [Fe(H2O)5(PO4)], (c) Pd(PPh3)4 [Fe(H2O)5(SCN)]2+, [Fe(H2O)5F]2+ (C) [Fe(H2O)5Cl]2 + , [Fe(H2 O)6]3+, [Fe(H2O)5(SCN)]2+, [Fe(H2O)5F]2+ (D) [Fe(H2O)5Cl]2 + , [Fe(H2O)5(PO4)], [Fe(H2O)5(SCN)]2+, Fe(H2O)4(SCN)F]1+ (d) 89. Which one of following will not undergo oxidative addition by methyl iodide ? Column-II (A) [Rh(CO)2I2]– 1. Hydroformylation (B) [Ir(PPh3)2(CO)Cl] 2. Asymmetric hydrogenation (C) [η5 –CpRh(CO)2] 3. Asymmetric hydrogen transfer (D) [η5 –Cp2Ti(Me)(Cl)] 4. Heck coupling 90. In hydroformylation reaction using The best match of a catalyst of Column-I with [Rh(PPh3)3(CO)(H)] as the catalyst, addition the reaction under Column-II is— of excess PPh3 would— (A) increase the rate of reaction Codes : (B) decrease the rate of reaction (C) not influence the rate of reaction (a) (b) (c) (d) (D) stop the reaction (A) 2 14 3 (B) 1 23 4 91. Find out the number of lines in the 31P NMR (C) 3 14 2 signal for— (D) 4 32 1 86. A solution of 2·0 g of brass was analysed for Cu electrogravimetrically using Pt-gauze as electrode. The weight of Pt-gauze changed from 14·5 g to 16·0 g. The percentage weight (A) 3 (B) 6 (C) 18 (D) 90 of Cu in brass is— (A) 50 (B) 55 92. The rate of exchange of OH2 present in the (C) 60 (D) 75 coordination sphere by 18OH2 of, 87. The platinum complex of NH3 and Cl– ligands (i) [Cu(OH2)6]2+; (ii) [Mn(OH2)6]2+; is an anti-tumour agent. The correct isomeric formula of the complex and its precursor are— (iii) [Fe(OH2)6]2+; (iv) [Ni(OH2)6]2+, follows (A) cis–Pt(NH3)2Cl2 and PtCl42– an order— (B) trans–Pt(NH3)2Cl2 and PtCl42– (C) cis–Pt(NH3)2Cl2 and Pt(NH3)42+ (A) (i) > (ii) > (iii) > (iv) (D) trans–Pt(NH3)2Cl2 and Pt(NH3)42+ (B) (i) > (iv) > (iii) > (ii) (C) (ii) > (iii) > (iv) > (i) (D) (iii) > (i) > (iv) > (ii)

CSIR–Chemical Sciences (Dec.–12) | 13 93. Based on the behaviour of the metalloen- 96. A metal crystallizes in f.c.c. structure with a zymes, consider the following statements : unit cell side of 500 pm. If the density of the 1. In the enzymes, the zinc activates O2 to crystal is 1·33 g/cc, the molar mass of the form peroxide species. metal is close to— 2. In the enzymes, the zinc activates H2O (A) 23 (B) 24 and provides a zinc bound hydroxide. (C) 25 (D) 26 3. In the oxidases, the iron activates O2 to 97. The activation energy for the bimolecular break the bonding between the two reaction A + BC → AB + C is E0 in the gas phase. If the reaction is carried out in a oxygens. confined volume of λ3, the activation energy is expected to— 4. Zinc ion acts as a nucleophile and attacks (A) remain unchanged at the peptide carbonyl. (B) increase with decreasing λ The set of correct statements is— (C) decrease with decreasing λ (A) 1 and 2 (B) 2 and 3 (D) oscillate with decreasing λ (C) 3 and 4 (D) 1 and 4 94. Fe2+–porphyrins fail to exhibit reversible oxygen transport and cannot differentiate CO from O2. However, the haemoglobin is free 98. In a many-electron atom, the total orbital both these pitfalls. Among the following : angular momentum (L) and spin (S) are good 1. Fe2+–porphyrins undergo μ–oxodimer quantum numbers instead of the individual formation and the same is prevented in electron orbital (l1, l2) and spin (s1, s2) angular case of the haemoglobin. momenta in the presence of— 2. Fe–CO bond strength is much low in (A) inter-electron repulsion case of haemoglobin when compared to (B) spin-orbit interaction the Fe2+–porphyrins. (C) hyperfine couping (D) external magnetic field 3. While Fe–CO is linear, Fe-O2 is bent and is recognized by haemoglobin. 4. The interlinked four monomeric units in 99. The packing fraction of a simple cubic lattice the haemoglobin are responsible to is close to— overcome the pitfalls. (A) 0·94 (B) 0·76 The correct set of statements is— (C) 0·52 (D) 0·45 (A) 1 and 2 (B) 1 and 3 100. The number of IR active vibrational modes of (C) 3 and 4 (D) 2 and 4 pyridine is— 95. Reactions A and B are, termed as respecti- C2v E2 C2 σv σ′v vely— 1. A1 1 1 1 1 z A2 1 1 –1 –1 Rz B1 1 –1 1 –1 x, Ry B2 1 –1 –1 1 y, Rx 2. (A) 12 (B) 20 (C) 24 (D) 33 (A) Insertion, Metathesis 101. One of the excited states of Ti has the (B) Metathesis, Insertion electronic configuration [Ar]4s2 3d1 4p1. The (C) Oxidative addition, Metathesis number of microstates with zero total spin (S) (D) Oxidative addition, Insertion for this configuration is— (A) 9 (B) 15 (C) 27 (D) 60

14 | CSIR–Chemical Sciences (Dec.–12) 102. For the reaction A2 2A in a closed (A) A has fcc lattice while B has bcc lattice container, the relation between the degree of (B) A has bcc lattice while B has fcc lattice dissociation (α) and the equilibrium constant (C) A and B both have fcc lattice Kp at a fixed temperature is given by— (D) A and B both have bcc lattice (A) α = [Kp / (Kp + 4p)] 108. The decomposition of NH3 on Mo surface (B) α = [Kp / (Kp + 4p)]1/2 follows Langmuir-Hinshelwood mechanism. The decomposition was carried out at low (C) α = [Kp + 4p) / Kp] pressures. The initial pressure of NH3 was (D) α = [(Kp + 4p) / Kp]1/2 10–2 torr. The pressure of NH3 was reduced to 103. The fugacity of a gas depends on pressure and 10–4 torr in 10 minutes. The rate constant of — decomposition of NH3 is— (A) 9·9 × 10–4 torr min–1 the compressibility factor Z (= pV /RT) (B) 0·4606 min–1 (C) 9·9 × 10–3 torr min–1 — (D) 0·693 min–1 through the relation [V is the molar volume]— [ ]∫ƒ = p . expp Z– 1dp o p 109. A polymer sample has the following com- For most gases at temperature T and up to position. moderate pressure, this equation shows that— (A) ƒ < p, if T → 0 (B) ƒ < p, if T → ∞ Number of molecules Molecular weight (C) ƒ > p, if T → 0 (D) ƒ = p, if T → 0 10 1000 104. The internal pressure (∂U/∂V)T of a real gas is 50 2000 40 4000 — The polydispersity Index (P.D.I.) of the related to be compressibility factor Z = pV/RT — by [V is the molar volume]— (A) (∂U/∂V)T = RT (∂Z/∂V)T polymer is— (B) (∂U/∂V)T = RT / — Z) (A) 85000 (B) 85 27 81 (V (C) (∂U/∂V)T = (RT2 / — (∂Z/∂T)V 850 729 729 850 V) (D) (∂U/∂V)T = — /RT2) (∂Z/∂T)V (C) (D) (V 105. Suppose, the ground stationary state of a 110. The equilibrium constant for an electro- harmonic oscillator with force constant k is chemical reaction 2Fe3+ + Sn2+ 2Fe2+ + given by ψo = exp [–Ax2] Sn4+ is [E0 (Fe3+ / Fe2+) = 0·75V, E 0 (Sn4+ / Sn2+) = 0·15V, (2·303RT / F) = 0·06V]— Then, A should depend on k as— (A) A ∝ k –1/2 (B) A ∝ k (A) 1010 (B) 1020 (C) A ∝ k1/2 (D) A ∝ k1/3 (C) 1030 (D) 1040 106. Combining two real wave functions φ1 and φ2, 111. A bacterial colony grows most commonly by the following functions are constructed : cell division. The change in the population A = φ1 + φ2, B = φ1 + iφ2, C = φ1 – iφ2, D = due to cell division in an actively growing i(φ1 + φ2). The correct statement will then colony is d N = λgN d t. The population of bacterial colony at time t is [N0 = N(t = 0)]— be— (A) A and B represent the same state (A) N0 λgt (B) N0 exp[–λgt] (B) A and C represent the same state (C) N0 exp[λgt] (D) N0 (λgt)2 (C) A and D represent the same state 112. The Arrhenius parameters for the thermal (D) B and D represent the same state decomposition of NOCl, 2NOCl(g) → 2NO(g) 107. Crystal A diffracts from (111) and (200) + Cl2(g), are A = 1013 M–1 s–1, Ea = = 105 kJ planes but not from (110) plane, while the mol–1 and RT = 2·5 kJ mol–1. The enthalpy (in Crystal B diffracts from (110) and (200) planes but not from the (111) plane. From the kJ mol–1) of the activated complex will be— (A) 110 (B) 105 above, we may conclude that— (C) 102·5 (D) 100

CSIR–Chemical Sciences (Dec.–12) | 15 113. The rotational partition function of H2 is— 119. The trial wave function of a system is (A) ∑ (2J + 1)e–βhcBJ(J + 1) expanded as ψt = c1φ1 + c2φ2. The matrix J = 0‚1‚2‚ … elements of the Hamiltonian are <φ1⏐H⏐φ1〉 = 0; (φ1⏐H⏐φ2〉 = 2·0 = <φ2|H|φ1〉 and (B) ∑ (2J + 1)e–βhcBJ(J + 1) <φ2⏐H⏐φ2〉 = 3·0. The approximate ground- J = 1‚3‚5‚ … (C) ∑ (2J + 1)e–βhcBJ(J + 1) state energy of the system from the linear variational principle is— J = 0‚2‚4‚ … (A) –1·0 (B) –2·0 [(D) 1 ∑ (2J + 1)e–βhcBJ(J + 1) 4 (C) +4·0 (D) +5·0 J = 0‚2‚4‚ … 120. One molecular orbital of a polar molecule AB ]+ 3 ∑ (2J + 1)e–βhcBJ(J + 1) has the form CAψA + CBψB , where ψA and J = 1‚3‚5‚ … ψB are normalized atomic orbitals of centred 114. The potential in Debye-Hückel theory is on A and B, respectively. The electron in this proportional to— orbital is found on atom B with a probability (A) 1/kr (B) exp[–kr] of 90%. Neglecting the overlap between ψA (C) exp[–kr]/r (D) kr and ψB, a possible set of CA and CB is— 115. The vibrational frequency and anharmonicity (A) CA = 0·95, CB = 0·32 constant of an alkali halide are 300 cm–1 and (B) CA = 0·10, CB = 0·90 0·0025, respectively. The positions (in cm–1) (C) CA = –0·95, CB = 0·32 (D) CA = 0·32, CB = 0·95 of its fundamental mode and first overtone are respectively— (A) 300, 600 (B) 298·5, 595·5 121. 4-Hydroxybenzoic acid exhibited signals at (C) 301·5, 604·5 (D) 290, 580 δ 171, 162, 133, 122 and 116 ppm in its 116. The adsorption of a gas is described by the broadband decoupled 13C NMR spectrum. The Langmuir isotherm with the equilibrium cons- correct assignment of the signals is— tant K = 0·9 kPa–1 at 25°C. The pressure (in (A) δ 171 (C–4), 162 (COOH), 133 (C–3 kPa) at which the fractional surface coverage & 5), 122 (C-1) and 116 (C-2 & 6) is 0·95, is— (B) δ 171 (COOH), 162 (C-4), 133 (C–2 (A) 1/11·1 (B) 21·1 & 6), 122 (C-1) and 116 (C-3 & 5) (C) 11·1 (D) 42·2 (C) δ 171 (C–4), 162 (COOH), 133 (C–2 117. The energy of a harmonic oscillator in its & 6), 122 (C-1) and 116 (C-3 & 5) ground state is 1 hω. According to the virial (D) δ 171 (COOH), 162 (C–4), 133 (C–3 2 & 5), 122 (C-1) and 116 (C-2 & 6) theorem, the average kinetic (T) and potential 122. An organic compound (C9H10O3) exhibited (V) energies of the above are— the following spectral data : (A) T = 1 hω; V = 1 hω IR : 3400, 1680 cm–1; 4 4 1H NMR : δ 7·8 (1 H, d, J = 8 Hz), 7·0 (1 H, (B) T = 1 hω; V = 3 hω d, J = 8 Hz), 6·5 (1 H, s), 5·8 (1 H, s, D2O 8 8 exchangeable), 3·9 (3 H, s), 2·3 (3 H, s). (C) T = hω; V = – 1 hω The compound is— 2 (D) T = 3 hω; V = 1 hω 8 8 (A) (B) 118. The energy of a hydrogen atom in a state is –hcRH (RH = Rydberg constant). The degen- 25 eracy of the state will be— (C) (D) (A) 5 (B) 10 (C) 25 (D) 50

16 | CSIR–Chemical Sciences (Dec.–12) 123. The [α ]D of a 90% optically pure (C) 2-arylpropanoic acid solution is +135°. On (D) treatment with a base at RT for one hour, [α]D changed to +120°. The optical purity is reduced to 40% after 3 hours. If so, the optical purity of the solution after 1 hour, and its [α ]D after 3 hours, respectively, would 126. The major product formed in the following be— reaction is— (A) 80%; and 60° (B) 70%; and 40° (C) 80%; and 90° (D) 70%; and 60° 124. In the following pericyclic reaction, the (A) structure of the allene formed and its configu- ration are— (B) (optically pure) (C) (A) R (D) 127. The following conversion involves— (B) S (C) R (A) a 1, 3-dipolar species as reactive inter- (D) S mediate and a cycloaddition. (B) a carbenium ion as reactive intermediate, and a cycloaddition (C) a 1, 3-dipolar species as reactive inter- mediate, and an aza Wittig reaction (D) a carbanion as reactive intermediate, and an aza Cope rearrangement 128. The following transformation involves— 125. In the following sequence of pericyclic reactions X and Y are— X Y (A) an iminium ion, [3, 3]-sigmatropic shift (A) and Mannich reaction (B) (B) a nitrenium ion, [3, 3]-sigmatropic shift and Michael reaction (C) an iminium ion, [1, 3]-sigmatropic shift and Mannich reaction (D) a nitrenium ion, [1, 3]-sigmatropic shift and Michael reaction 129. With respect to the following biogenetic conversion of chorismic acid (A) to 4-

CSIR–Chemical Sciences (Dec.–12) | 17 hydroxyphenylpyruvic acid (C), the correct (D) statement is— 132. In the following reaction A and B are— (A) X is Claisen rearrangement; Y is oxida- (A) (B) tive decarboxylation (A) (B) X is Fries rearrangement; Y is oxidative (B) decarboxylation (C) X is Fries rearrangement; Y is dehydra- tion (D) X is Claisen rearrangement; Y is dehy- dration 130. Match the following— (C) List-I (a) β-amyrin (b) squalene (D) (c) morphine (d) ephedrine List-II 133. Match the following biochemical transforma- 1. alkaloid, secondary alcohol tions with the coenzymes involved— 2. alkaloid, phenol 3. triterpene, secondary alcohol List-I 4. acyclic triterpene, polyene (a) α-ketoglutarate to glutamic acid (b) uridine to thymidine (c) pyruvic acid to acetyl coenzyme A Codes : List-II (a) (b) (c) (d) 1. tetrahydrofolate (A) 3 42 1 2. NADH (B) 2 13 4 3. thiamine pyrophosphate (C) 3 24 1 4. pyridoxamine (D) 1 42 3 Codes : 131. In the following reaction, the structure of B, (a) (b) (c) and the mode of addition are— (A) 4 1 3 (B) 1 2 4 (C) 2 1 3 (D) 4 2 3 (A) 134. The structure of the major product B formed in the following reaction sequence is— (B) (C)

18 | CSIR–Chemical Sciences (Dec.–12) (A) (B) (A) (B) (C) (D) (C) (D) 135. Given the energy of each gauche butane 138. 12·0 g of acetophenone on reaction with interaction is 0·9 k-cal/mol, ΔG value of the 76·2 g of iodine in the presence of aq. NaOH following reaction is— gave solid A in 75% yield. Approximate amount of A obtained in the reaction and its structure are— (A) 80 g, CI4 (B) 40 g, CI4 (C) 60 g, CHI3 (D) 30 g, CHI3 (A) 0·9 k-cal/mol (B) 1·8 k-cal/mol 139. Consider the following reaction mechanism. (C) 2·7 k-cal/mol (D) 3·6 k-cal/mol 136. In the following reaction, the reagent I and | the major product II are— l I I The steps A, B and C, respectively, are— (A) II (A) oxidative addition; transmetallation; re- (B) II ductive elimination (C) (B) oxidative addition; carbopalladation; β- hydride elimination (C) carbopalladation; transmetallation; re- ductive elimination (D) metal halogen exchange; transmetallation; metal extrusion 140. The major product formed in the following reaction sequence is— (D) (A) 137. The major product formed in the following (B) reaction sequence is— (C) (D)

CSIR–Chemical Sciences (Dec.–12) | 19 141. The major product B formed in the following 144. The major product formed in the following reaction sequence is— reaction is— (A) (B) (A) (C) (D) (B) 145. In the following enantioselective reaction, the major product formed is— (C) (D) (A) (B) 142. The major product B formed in the following reaction sequence is— (C) (D) Answers with Explanations (A) (B) Part–A (C) (D) 1. (A) 234 = 281, 243 = 264, 324 = 316, 342 = 316, 143. The osazone A could be obtained from— 423 = 48 = 216, 432 = 49 = 218. Therefore, 234 is the largest. A (A) glucose and mannose 2. (B) D C A a 2 B (B) mannose and galactose (C) gulose and fructose AB (D) galactose and fructose H G a2 a2 EF F Equilateral triangle is formed with diagonal. ∴ Area ACF = √⎯ 3 (Side)2 4 = ⎯√ 3 (a√⎯ 2)2 4 = ⎯√23a2

20 | CSIR–Chemical Sciences (Dec.–12) 3. (D) Total number of distinct arrangements OB2 = OC2 + BC2 52 = OC2 + 32 7 OC = ± 4 = = 2520 ⇒ OC = 4 2 –ve not taken in first quadrant So, centre is (4, 5). Number of arrangements when U and I come 7. (C) The minimum single cut is required together because wire ABCDB = 5 m 6 A = × 2 = 720 2 Thus, the no. of arrangements when U and I cannot come together = 2520 – 720 = 1800 D 4. (C) Let numbers are a, b, c, d, e, f, g BC and Wire AD = 1m Given, a + b + c + d + e + f + g = 21 8. (D) 2, – 4, 8, –16, 32, – 64, 128, – 256, 512 Q Arithematic Progression ≥ Geometrical – 256 + 512 = x Progression x = 256 A.P. = G.P. log2 x = log 256 = log2 28 (When all the numbers are equal) log2 x = 8 log2 2 = 1 Now, a+b+c+d+e+f+g = 3 9. (A) Let r is radius, then 7 OR = 10·5 – r If a = b = c = d = e = f = g = 3 r ⇒ a2 + b2 + c2 + d2 + e2 + f2 + g2 = 9 O 7 P 5. (C) 50B = 113 + 313 + 513 + … 9913 …(1) 50C = 213 + 413 + 613 + … 10013 …(2) Each term of (2) is greater to term of (1) Then, 50C > 50B ⇒ C > B Now, 100A = 113 + 213 + 313 + 413 30° + … 10013 R 100A = 50B + 50C In Δ OPR 2A = B + C OP r 1 OR 10·5 2 Therefore, 2A > 2B and 2A < 2C sin 30° = ⇒ – r = Then, A > B and A < C r 10·5 – r Therefore, B<A<C ⇒ 0·5 = 6. (D) In right angled triangle BOC 10·5 r y ⇒ 2 = – 1 II A Ist quadrant ⇒ 3 = 10·5 35 r CO ⇒ r = 3·5 3 35 x 10. (A) Amar Doctor (tallest) B (Shortest) Akbar Engineer e2 = dp Anthony Professor III IV Therefore, Amar is a doctor and he is the tallest.

CSIR–Chemical Sciences (Dec.–12) | 21 11. (B) If 100 cats catch 100 mice in 100 minutes, ⇒ R3 = Nr3 then 7 cates will catch 7 mice also in ⇒ 100 minutes. Also, ( )r 3 = 1 ⇒ R N ⇒ 12. (A) Each half diagram shows increase in one 4πR2·X = N(4πr2) ⇒ unit, thus 1 + 2 + 3 + … +n = n(n + 1). R2X = Nr2 2 ( )X = r 2 13. (B) Since, socks is in pair, so atleast N + 1 R number of socks would be taken out. N 14. (D) 11 12 ( )= 1 2/3 N ·N 10 X = N1–2/3 = N1/3 93 19. (A) 24 30 33 39 51 57 4 2 4 6 3 0 3 3 3 6 3 9 12 5 1 6 5 Thus, the next number is 57. 6 20. (A) No parallel and No concurrent lines. 4 – 54′ – 33′′ 3 15. (C) When curve touches the x-axis, y = 0 for 42 real x. ∴ 0 = x 2 – 10x + 25 P QR 1 ⇒ (x – 5)2 = 0 UT S ⇒ x=5 No other option have real value of x. New lines are PU, QS and RT. 16. (C) AB // CD Part–B CD 21. (A) Like atomic and molecular orbital, there are also nuclear orbital with same rules. We O know that Unitary operator have two parity, AB P = ±1 When even serial number (azimuthal quantum Let OD = r No.) is found for orbital, it is positive (P) and vice-versa. We know that spin for orbitals are Thus, ∠ COD = ∠ AOB Orbitals Serial No. Parity Spin (Azimuthal and ∠ DCO = ∠ ABO Quantum No.) s 0 +1 Therefore, Δ OCD and Δ OBA are symmetric 2 triangle p 1 –3 2 ( )Thus, d 2 +5 Area of triangle Δ OAB = AO 2 2 Area of triangle Δ OCD OD f 3 –7 2 ( )= g 4 +9 2OD 2 2 OD h 5 – 11 2 =4 17. (*) 18. (C) Let r is the radius of small sphere and R is the radius of big sphere. Thus, 4 πR3 = N 4 πr3 3 3 Volume of sphere = 4 π (radius)3 3

22 | CSIR–Chemical Sciences (Dec.–12) 22. (D) Electroplating is a process that uses Therefore, there are no fundamental vibra- electrical current to reduce dissolved metal tional modes common in both IR and Raman cations so that they form a coherent metal spectra. coating on an electrode. 26. (C) [CO(H2O)6]2+, d7 system Also, PbCl2 and PbSO4 are ionic and insolu- eg Transition from + g to eg ble in cold water Pb(Et)4 is toxic. ee 2g Cathode Anode Octahedral environment Pb [COCl4]2–, d7 system 2 Transition from e to + 2 Pb 2 BF4 e So, both complexes show d-d transition. Fig : Electroplating of a Metal with Lead 27. (A) [MO2(S2)6]–2 in a Pd(BF4)2 bath S22– is 2e– donor It is generally agreed that electro deposited SS metals are crystalline and external appearance depends mainly on the rate at which the SS S crystal grow and on the rate of the formation S MO S MO S of fresh nuclei. S S S S 23. (B) Beer’s law is, Briolging S2–2 ligand are 2 and coordination number are 8. ( )logI0 I = ecl= A 28. (D) For D, I = 1 Multiplicity = 2nI + 1 Since, absorbance is directly proportional to n = No. of D atom concentration, very high concentration of I = spin of D atom analyte deviates the Beer’s law. Further Asso- ciation or dissociation of analyte increases or Therefore, M = 2 × 1 × 1 + 1 decreases the concentration of analyte, so it =3 also cause deviation. and Intensity ratio will be non-pascal. 24. (D) In Lanthanoids in (+3) oxidation state So, 1H NMR spectrum of HD gives a triplet shows decrease in size when we go down with intensity ratio 1 : 1 : 1. from La to Lu. In hot concentrated aqueous NaOH, OH have 29. (B) [Ru(PPh3)2(acac)2] (acac = acetylaceto- small size and strongly binds with Lu+3 and nate) forms [Lu(OH)6]–3. That is the reason PPh3 acac O Lu(OH)3 shows the highest solubility in hot OO O PPh3 acac O Ru O acac Ru concentrated aqueous NaOH. O PPh3 PPh3 25. (D) The structure of CO2 is acac O minor O C O Symmetric Trans Cis-d Since, symmetric stretching gives no change Ph3P O acac in dipole moment so it is IR-inactive mode. Ph3P O Since, CO2 is centrosymmetric molecule, Ru therefore all the modes that is inactive in IR, O are active in Raman. O acac Cis-l 3 possible isomer are shown above.

CSIR–Chemical Sciences (Dec.–12) | 23 30. (A) This complex exists or dimer and structure 34. (A) (OC)5M C OCH3 is Ph O OH2 This is an example of metal carbene complex. O Since, the molecule obeys 18e– rule and CH3C CCH3 carbene is 2e– donor. CH3C Cu CCH3 O O Therefore, we need a metal with 6e–. O δ bond This is only Cr and Re+ pair. O O 35. (A) Fe is found in haemoglobin myoglobin Cu and cytrochromes. Mo is for nitrogen fixation. O OH2 Cu for heamocyanin and Zn for metallo- enzymes. There are 10 Cu-O bond presents in this complex. 36. (D) H2 N 31. (B) Among the given elements K has least NH2 2 3 electronegativity and F has most electro- CH2 CH2 of Cu = 2 negativity. Therefore, the electronegativity Cu I difference is greater in K and F. , CH2 I of N = 1 CH2 NH2 N H2 32. (D) O O Therefore, C S ( )(2n1I1 + 1) (2n2I2 + 1) = 2 × 1 × 3 + 1 2 OO OO sp2-hybridization sp2-hybridization (2 × 4 × 1 + 1) Trigonal planar Trigonal planar = 36 O n1 = no. of Cu atom, I1 = I of Cu Xe N n2 = no. of N atom, I2 = I of N. 37. (A) Degradation of penicillin G OOO OO donor atom sp3-hybridization sp2-hybridization Trigonal planar Trigonal pyramidal Therefore, CO32–, SO3 and NO3– have planar donor atom structure. donor atom 33. (C) Cl Penicillamine can utilize N, O, S atoms as Fe CO CO donors to bind with Pb(II), Hg(II) and Cu(II). 38. (D) B2H6 → D2h point group and S2 axis 18°-e– system 18e– system 19e– system CH4 → Td point group and S4 axis stable stable unstable PH5 → D3h point group and S3 axis SF6 → Oh point group and S6 axis But Cp is not replaceable with NO in this case. Therefore, (symmetry element) 39. (C) For electronic transition selection rule is, Ni NO Ni 18 e system stable ΔS = O, Δl = ± 1 and Δ J = 0, ± 1 20 e system For d2, 3F → 3D is allowed transition. unstable N Because ΔS = 0, Δl = –1. 40. (C) When a hydrogen atom is placed in an O electric field along the y-axis, electron density will be oriented along y-axis and therefore 2py orbital mixes most with the ground state 1s orbital.

24 | CSIR–Chemical Sciences (Dec.–12) 41. (B) For water, ΔHvap ≈ 41kJ mol–1 ∴ g1 = 3 For vaporization, Tb = 273 + 100 g2 2 = 373 K ⇒ g2 = 2 Therefore, molar entropy of vaporization, g1 3 ΔSvap = ΔHvap 47. (A) In the Daniell cell, Copper and Zinc elec- Tb trodes are immersed in a solution of Copper 41 × 1000J mol–1 (II) sulphate and Zinc sulphate respectively. 373K = Oxidation → Zn(s) → Zn+2(aq) + 2e– at anode ≈ 109·9 J K–1 mol–1 Reduction → Cu+2(aq) + 2e– → Cu(s) 42. (C) For commuting operators, at cathode [A, B] = 0 Overall cell rh is For non-commuting hermitian operators, there Zn(s) + Cu+2(aq) → Zn+2(aq) + Cu(s) is always a real value multiplie with i is found. 48. (B) We know that for first order reaction, Therefore, it is always imaginary. A+ = A0e–k1+ According to question, 43. (D) [x, px2] = ih·2px A0 = A0e–K1 + 1/4 4 = 2ihpx 1 1 It is non-commuting operator. ⇒ 4 = eK1 + 1/4 44. (D) The correlation coefficient between two ⇒ 4 = eK1 + 1/4 arbitrary variables x and y is given by ⇒ ln 4 = K1+ 1/4 r = n[Σxy – Σx Σy] ⇒ t1/4 = ln 4 √⎯⎯[n⎯Σ⎯x⎯2⎯–⎯(⎯Σ⎯x)⎯2]⎯[⎯n⎯Σ⎯y⎯2 ⎯–⎯(Σ⎯y⎯)2] K1 =0 49. (C) According to Kohlrausch’s law at infinite ⇒ Σxy = ΣxΣy dilution, when dissociation is complete, each or, <xy > = <x> <y > ion makes a definite contribution towards 45. (A) T2 = 300 K molar conductance of the electrolyte irrespec- q2 = 90 J tive of the nature of the other ion with which it is associated and the molar conductance at (hot) infinite dilution for any electrolyte is given by T2 the sum of the contribution of the two ions. Source q2 In given option, only HCl in H2O completely Sink work done by dissociated and contributes equal ion. T1 Carnot engine 50. (A) A dilute silver nitrate solution is added to (cold) the engine a slight excess of sodium iodide solution, a q1 negatively charged solution of silver iodide is formed. We know q2 – q1 = T2 – T1 Na ∴ q2 T2 Na I I I Na ( )q2 – q1 = 90 × 300 – 200 Na I I Na 300 = 30 I AgI I Na Na q1 = 60 J 46. (B) Boltzman Distribution law is I I ni gi Na I I I Na nJ gJ Na = e–(Ei – EJ)/KBT ( )and given thatn1 = 3 e–(E1 – E2)/KBT Fig. : Electrical property of colloid based on n2 2 concept of electrical double layer AgNO3 + NaI → AgI + NaNO3

CSIR–Chemical Sciences (Dec.–12) | 25 The negatively charged AgI attracts Na+ and factant with part of the charge of the micelle repels NO3–. neutralized by associated counter ions. Therefore, Na+ try to form a rigid layer with log CMC = –a log ci + b I–. 54. (A) 51. (D) We know that, Me Me Dissociation energy, Ph De = Emax – E0 Ag2O Walf O C Ph E0 is ground state energy MeOH or Rmeeanrtrange- It is clear from question that hν N2 Me Emax = 56875 cm–1 and E0 = 15125 MeOH ∴ De = 56875 – 15125 = 41750 cm–1 Me OMe Me Me Ph OH Ph C OHTPrarnostofenrPh CO Me O OMe HOMe Ph OMe 55. (D) We know that, log KP-OMe = σ KCOOH 52. (*) The angle between two planes having σ = log KP-OMe – log KCOOH miller indices (h1k1l1) and (h2k2l2) is given as = –log KCOOH – (–log KP-OMe) cos θ = h1h2 + k1k2 + l1l2 = PKa(COOH) – PKa(P-OMe) √⎯⎯h⎯12⎯+⎯⎯k1⎯2⎯+⎯l12⎯√⎯h⎯22⎯+⎯⎯k2⎯2⎯+⎯l22 = 4·19 – 4·46 | X | + | X | + OX σ = – 0·27 ⇒ cos θ = √⎯⎯1⎯2 ⎯+⎯1⎯2⎯+⎯02 ⎯√⎯1⎯2 ⎯+⎯1⎯2⎯+⎯12 Electron releasing group have –ve σ value. = 2 = ⎯√ 2 56. (B) Biosynthetic precursor of Cadinene ⎯√ 2·⎯√3 ⎯√ 3 H HO O θ = cos–1 (0·816) is mevalonic acid, HO OH. ⇒ θ = 35·31 O O O , All options are incorrect. 57. (C) H 53. (B) The molar conductance of an ionic sur- O OO factant of the type Na+R– in water is plotted A B against the square root of the normality of the more acidic solution. The curve obtained, instead of being with reactive aromatic ring the smoothly decreasing curve characteristic methylene gp makes proton less of ionic electrolytes of this type, has a sharp acidic due to e– break in it at low concentrations. This sharp HO O withdrawing nature break in the curve accompanied by reduction in the conductance of the solution, indicating OO a sharp increase in the mass per unit charge of the material in soluton, is interpreted as H evidence for the formation of miceller at that point from the unassociated molecular of sur- HO O O O C acidic like H2SO4 aromatic in nature Therefore, C > A > B

26 | CSIR–Chemical Sciences (Dec.–12) 58. (D) δ 7·8(1H, s) HN Ph Piperidine δ 2·8(3H, s) HO COOEt N H δ 2·6(3H, s) O From the above observation the structure of compound is Ph OO N CH3(2.8) Chemically HN H NMe2 H CH3(2.6) non-equivalent O COOEt (7.8) O O Carbanion (Stable) 63. (D) O2N OC CH3 ↓ E1CB A H2N Ph CO2 Normally, ester groups show absorption at COOEt 1750 – 1770 in phenyl acetate, is shower absorption at 1765 but in compound (A) due 59. (B) to conjugation with ring and O-atom absor- ption decreases and it appears at 1761 cm–1. F O DMSO K2CO3 N MeO2S NO 64. (D) (–)-Camphor 1O MeO2S H 4 This reaction is aromatic ipso substitution and aromatic nucleophilic substitution. 2 60. (B) O Ph O CH3 24 Ph N Me hν Ph N Ph 1 O3 1 3 Norrish type-II reaction view 4 γ-H abstraction OH CH2 OH 1s Ph by O Ph N Ph 2 N Ph 2 O3 1, 4-biradical H4 1 1 61. (A) Edman’s reagent is NCS 3 4 view 4 Phenyisothio cyanate 4s HN Ph O Absolute configuration is 1s, 4s. S NO H2N CH C OH 65. (A) This reaction is Vielsmeir-Haac Reaction. Ph CH2 Ph Cl Me Cl Cl Phe(Phenyl alamine) Cl H C N Me O Me Cl O P Cl P Me O O O Cl O N PO H H2N N OH H2N OH + H2N OH C H Cl O O Ala(alanine) Gly (Glycine) H H C NMe2 Therefore, the sequence is Phe-Ala-Gly, Since Cl H C NMe2 N Edman degradation occur from N-T-A H C NMe2 H Cl N Cl From N-terminal to Acid terminal. H aromatization reactive 62. (A) C3H7NO species CHO H C NMe2 No. of C-atoms + 2 – No. of H-atoms H3O N DBE = + No. of N-atoms N H 2 H = 6 + 2 – 7 + 1 = 1 2

CSIR–Chemical Sciences (Dec.–12) | 27 66. (B) The first person to separate a racemic 69. (C) mixture into individual enantiomers is L. Pasteur. Boc OAc Boc N N OTBDMS H Pasteur separated the left and right crystal H OH shapes from each other to form two piles of OTBDMS crystals; in solution one form rotated light to the left, the other to the right, while an equal The hydrolysis of ester is faster in alkaline mixture of the two forms cancelled each medium therefore the correct one is K2CO3, other’s effect and does not rotate the polarized MeOH. light. 70. (D) Diels-Alder reaction of furan and maleic acid in water is an example of ‘Green Synthesis’. OO O OH H2O COOH Furan OH O COOH Maleic acid Endo product Mirror Part–C image 67. (D) 71. (D) Recoil energy, H H Er = E2 H H 2Mc2 H HHH H or, Er = 536E2 eV H HHH H M E= 2·5 × 106 × 139 × 931·5 536 = 24·57 keV HH 72. (A) CO is π-acid ligend and have tendency to back-bonding with metal. HH [18]-Annulene CO * aromatic, planar * shows aromatic substitution H3P CO * 12 H outer, deshielded zone δ = 9·28 Mo * 6 H inner, shielded zone δ = –3·0 PH3 CO * 37 k-cal/mol resonance energy * diatropic PH3 68. (B) HA HB Electron donating phosphine group increase Me Me e– density on metal and back bonding with M and CO becomes strong. As a result of this C = O bond becomes weak. Br1 Br2 Therefore, stretching frequency, υC = 0 decreases. Molecule has a plane of symmetry bisecting HA and HB reflecting Br1/Br2 and CH3/CH3. Phosphine V(CO) (cm–1) (a) Br1 and Br2 are reflected by plane, so they are enantiotropic to each other. PF3 (A) 2090 (i) (b) HA and HB are bisecting by plane. Hence, they are diastereotopic to each other. PCl3 (B) 2040 (ii) P(Cl)Ph2 (C) 1977 (iii) PMe3 (D) 1945 (iv)

28 | CSIR–Chemical Sciences (Dec.–12) 73. (C) Id = K(C*—C) Thus, Be+2 have 4 coordination number and C3–2 have 8 coordination number. where (C*—C) is concentration gradient and Be 2 K is constant Id α concentration (mol ml–1) C3 2 ⎧⎪ C1 = X ⎨⎪⎩ C2 9·5 = 0·04 0·5 Id1 = C1 78. (A) H Id2 C2 HH ⇒ X = 0·0067 H Mo H Singlets CH3 N 74. (C) (A) CH3COOH in pyridine at 3.18 ppm CH3 CO CO Strong base ⊕ → CH3COO– + C5H5NH solvocation All 5-protons r are equal and resonate at 5·48 ppm. (B) CH3COOH in H2SO4 Also, both CO are is different environment, Strong base therefore gives two signal at 1950 and 1860 ⊕ cm–1. → CH3COOH2 + HSO4– solvo anion (C) HClO4 in H2SO4 79. (A) —CH is 3e– donor ligand. → ClO4– + H3S+O4 (weak acid) No. of M—M bond (D) SbF5 in HF → SbF6– + H2F+ strong acid = 18 × 3 – (9 × 3 + 9 × 2 + 3) 2 2 mole 75. (B) C2B4H8 = (BH) × 4 + (CH) × 2 + 2 = 54 – 48 = 3 2 = 2×4+3×2+2 = 16e– or 8e– pair Therefore, structure of [CO3(CH)(CO)9] or (m + 2)e– pair (Nido) CO CO CO —BH contributes 2e–s to framework. —CH contributes 3e–s to framework. Co 76. (B) Boric acid is a weak acid in aq. solution. OC CO OC Co Co CO H OC C CO O B(OH)4 OH B O with HO—CH2— H HO H Thus, No. of M—M bond = 3 CH2—OH and Bridging ligand = 1 80. (A) Since, Gd+3 and Lu+3 have 4 f 7 and 4f 14 B(OH)4– consumed in forming a compound electronic configuration. Therefore, no orbital with ethylene glycol. contribution observed and magnetic moment value is closest w.r.t. observed and calculated. HO CH2 81. (D) Neso-silicates = ortho silicates HO B OH HO CH2 HO B OCH2 SiO4–2 unit HO OH H2O HO OCH2 Soro-silicates = Si2O7–2 unit HO CH2 HO CH2 CH2O B OCH2 Tecto-silicates = SiO2 unit and 3D H2O CH2O OCH2 framework 77. (D) Be2C3 structure is correlated with CaF2. 82. (A) The structure of Co3O4 is normal spinel. Therefore, Be+2 in tetrahedral voids. In normal spinel the Co+2 ions occupy tetra- Since, molecules have FCC arrangement. hedral voids and Co+3 ions occupy octahedral voider.

CSIR–Chemical Sciences (Dec.–12) | 29 Therefore structure is, 88. (A) In solution Fe(NO3)3·9H2O exists as [Fe(H2O)6]+3·(NO3–)3·3H2O (Co+2)t (2Co+3)OO4 [Fe(H2O)6] 3 For Inverse spinel structures it is NaCl H3PO4 KSCN NaF (Co+3)t (Co+2 Co+3)O O4 83. (A) In the solid state, the CuCl53– ion have [Fe(H2O)5Cl] 2 [Fe(H2O)5(PO4)] [Fe(H2O)5(SCN)] 2 [Fe(H2O)5F]2 trigonal bipyramidal geometry and d-orbitals Yellow Colourless Red Colourless are 89. (D) For oxidative addition reaction, metal dz1 > dxy2 = dx2– y2 > dxz2 = dyz2 should have e– rich and have tendency to charge its oxidation state. The two ligands on the z-axis and three are in between xy plane and directly in path of Therefore, [η5 – Cp2Ti(Me)Cl] do not have orbitals lobe and feels. There are four e–s in tendency to show oxidative addition reaction. dxz2 and dyz2 and repel each other, due to this bond length becomes long. Thus, three long 90. (B) Ph3P CO and two short bond is found. Rh PPh3 84. (B) In metalloenzymes, the metal centers are covalently linked through the side chains of Ph3P H the amino acid residues. Therefore Glu, His Cys are involved in the primary coordination In hydroformylation reaction, 16e– species is spheres of metalloenzymes. reactive catalysts. Due to excess PPh3, it becomes 20 e–s species and lost it catalytic 85. (A) [R-BINAP]Ru2– is used for Asymmetric reactivity. hydrogenation. So, the rate of reaction decreases in the pre- [Rh(CO)2I2]– for Hydroformylation sence of excess PPh3. [Pd(PPh3)4] for Heck coupling. 91. (D) Species I 1H 1 15N Ts N Ru Cl for Asymmetric hydrogen transfer 19F 2 aH 15 Ha NH2 1 N 2 bH P F 1 F 86. (D) The weight of pt-gauze as electrode is 2 Ha N15Ha changed as = 16·0 – 14·5 31P 1 2 = 1·5g No. of lines Thus, weight of Cu is 2·0 g brass is 1·5 g. = (2n1I1 + 1) (2n2I2 + 1) (2n3I3 + 1) (2n4I4 + 1) Therefore, percentage weight of Cu in brass = 1·5 × 100 2N 2F 4H 1H 2·0 = (2 × 2 × 1 + 1) (2 × 2 × 1 + 1) (2 × 4 × 1 + 1) = 75% 2 2 2 87. (A) Cl Cl 2 Cl NH3 (2 × 1 × 1 + 1) 2 Pt NH3 Pt =3×3×5×2 Cl Cl Cl Cl = 90 Cl– have trans effect > NH3 92. (A) Since in [Cu(H2O)6]+2, John-Teller distor- tion observed, so rate of exchange is fast. Cl NH3 NH3 Pt Rest of others are 3d-series dipositive metal cations. Thus, rate of exchange with 18H2O Cl NH3 decreases with decrease in size and increase Cis-platin in atomic number. (anti-tumour agent)

30 | CSIR–Chemical Sciences (Dec.–12) 93. (B) Since, Zn is a lewis acid, so 97. (C) Bimolecular reaction Zn 2 OH2 Zn H H Zn(OH) A + BC → AB + C Activates H2O OH Zinc bound There will be greater possibilities of collisions hydroxide between molecule when volume is smaller and therefore activation energy decreases In oxidases an enzyme that catalyse the with decreasing λ. reduction of O2 to H2O or H2 O2, the iron 98. (A) There is a strong electron-electron i.e., activates O2 to break the bonding between the inter-electron repulsion in many-electron two oxygen's. atom. Due to this the orbital angular momen- tum (l1, l2) and sign angular momentum (s1, 94. (B) Fe+2-porphyrins fail to exhibit reversible s2) are couple together to for total orbital angular momentum (L) and (S). oxygen transport and cannot differentiate CO 99. (C) Packing fraction from O2. However, the haemoglobin is free from both these pitfalls. O (No. of particles per unit cell) Fe 3 Fe 3 μ-oxodimer = × (Volume of 1 particle) (hematin) Total volume of unit cell Unable to bind O2 For simple cubic lattice * Fe+2-porphyrins undergo μ-oxodimer formation and the same is prevented in case 4 3 of haemoglobin. 1 × πr3 *1 Fe-CO is linear, Fe—O2 is bent recognized = a3 by haemoglobin. 4 22 r3 95. (A) SnCl2 behaves as a carbene and it insert = 3 × 7 × (2r) 3 into the CO—CO bond. = 0·52 Insertion : (OC)4CO CO(CO)4 SnCl2 100. (C) H Insertion HH Cl HNH (OC)4CO Sn CO(CO)4 Order of group = 4 (C2v = 2 × 2 = 4) Cl C2v E C2 σV(xz) σV′(yz) √ Metathesis : A1 1 1 1 1z A2 1 1 –1 –1 Rz Re(CO)5 Me B1 1 –1 1 –1 x, Ry B2 1 –1 –1 1 y, Rx Me Cl 2NaRe(CO)5 (OC)5Re Sn Re(CO)5 11 3 3 Sn No. of unshifted 11 atoms Me Cl Re(CO)5 Me 96. (C) We know that, Character per 3 –1 1 1 atoms Density p = nM Na3 Reducible 33 –3 3 11 where n is no. of particles per unit cell Representation M = Molar Mass No. of A1 = 1 [33 × 1 × 1 + (–3) × 1 × 1 + 3 4 N = Avogadro’s number a = edge length of cube × 1 × 1 + 11 × 1 × 1] = 11 ∴ M = pNa3 No. of A2 = 1 [33 × 1 × 1 + (–3) × 1 × 1 + 3 n 4 (1·33g cm–3) (6·023 × 1023 mole–1) × (–1) × 1 + 11 × (–1) × +1] = 4 = (500 × 100–10 cm)3 No. of B1 = 1 [33 × 1 × 1 + (–3) × (–1) × 1 4 4 M = 25 + 3 × 1 × 1 + 11 × (–1) × 1] = 7

CSIR–Chemical Sciences (Dec.–12) | 31 1 — 4 pV No. of B2 = [33 × 1 × 1 + (–3) (–1) × 1 + 3 103. (A) Z= RT × (–1) × 1 + 11 × 1 × 1] = 11 p Z–1 P p·e0 ∫ Therefore, ( )f= dp R.R., τred = 11A1 + 4A2 + 7B1 + 11B2 For Van der Waal gas τtrans = B1 + B2 + A1 f = pePb/RT (Right side of box) ⇒ If T → ∞, f = p τrot = B2 + B1 + A2 If T → 0, then f < p. — ∴ τvib = 10A1 + 3A2 + 5B1 + 9B2 pV 104. (C) Z = RT Since, A2 is not IR-active. ZRT Therefore, total IR active vibration modes are ⇒ p = — 10 + 5 + 9 = 24 V 101. (B) 22Ti – [Ar] 4s2 3d1 4p1 ( ) [( ) ]⇒∂P = R ∂Z ∂T — ∂T T+Z Excited state V V V 4l + 2 ( )i.p. = ∂U No. of microstates = ∂V T 4l + 2 – q q where l is azimuthal quantum no. of valence ( )=T ∂P –p shell and q is no. of e–s in valence shell ∂T V 4×1+2 [( ) ]=TR ∂Z –p = — ∂T ·T + Z V 4+2–2 2 V 6×5× 4 ( )=RT2 ∂Z + ZRT – p = — ∂T — V V V 4 ×2×1 ⎧⎨Q ZRT = p — = 15 ⎩V 102. (B) A2 2A ( ) ( )∂U=RT2 ∂Z ∂V — ∂T V at t = 0 10 T V at equilibrium 1 – α 2α 105. (C) Harmonic Oscillator wave function is Since, PA = XAP ⇒ XA = 2α = 2α given as : α+ 1+α 1 – 2α ψn(x) = NnHn e–αx2/2 PA2 = XA2P and given wave function is ⇒ XA2 = 1–α = 1 – α ψ0 = e–Ax2 – α + 2α 1 + α 1 Thus, A ∝ α 2 (( ))Thus, Kp = 2α 2 PA2 = 1+α 4α2 P Since, α = ⎜⎛⎝hkμ2 ⎞⎟⎠1/2 PA2 P2 1 – α2 1–α = 1+α P (1 – α2)Kp = 4α2p Therefore, A ∝ k1/2 Kp – Kpα2 – 4α2p = 0 106. (C) A = φ1 + φ2 ⇒ α = ⎢⎡⎣⎢(KpK+p 4p)⎦⎥⎥⎤1/2 D = i(φ1 + φ2) Kp = α2(Kp + 4p) Combing two real wave functions φ1 and φ2, constructed functions φ1 + φ2 and i(φ1 + φ2) represent the same state because if the two function multiplied by any constant, no

32 | CSIR–Chemical Sciences (Dec.–12) effect on the state and on the energy because 111. (C) dN = λgNdt multiply only in eigen value, not in energy Integrating it with appropriate limits and no effect on wave function. ∫ ∫N dN t 107. (A) For BCC, h + k + l should be even FCC, = λg N0 N dt h, k, l all either even or odd 0 where 0 is even number. [ln N]NN0 = λgt N = N0eλgt Thus, A has fCC lattic while B has bcc lattice. 108. (B) At low pressure, Langmuir-Hinshelwood 112. (D) 2NOCl(g) → 2NO(g) + Cl2(g) ΔH = Ea + (Δng* – 1)RT mechanism follows first order kinetics. Δng* = –1 (Bimolecular reaction) ∴ K1 = 2·303 log10 P0 ∴ ΔH = 105 kJ mol–1 T Pf + (–1 – 1) × 2·5 kJ mol–1 = 2·303 log10 10–3 = 100 kJ mol–1 10 10–4 = 2·303 × 2 113. (D) Rotational partition function of H2 10 = 0·4606 min–1 qr = 1 [Σ (2J + 1) e–βhc BJ(J + 1) 4 109. — = n1M1 + n2M2 + n3M3 J = 0‚ 2‚ 4 (C) Mn n1 + n2 + n3 + 3 Σ (2J + 1) e–βhc BJ(J + 1)] 10 × 1000 + 50 × 2000 + 40 J = 1‚ 3‚ 5 = × 4000 114. (C) The potential in Debye Hückel theory is 10 + 50 + 40 ψr = Zie0 e–kr = 2700 qr — = n1M12 + n2M22 + n3M32 ψr ∝ e–kr Mw n1M1 + n2M2 + n3M3 r = 85000 115. (B) Fundamental vibration : 27 ΔE~0 → 1 = ~υe(1 – 2xe) Therefore, = 300(1 – 2 × 0·0025) — Mw P.D.I. = — = 298·50 cm–1 Mn First overtones = 85000 · 1 = 850 ~ΔE0 → 2 = 2v~e (1 – 3xe) 27 2700 729 = 2 × 300(1 – 3 × 0·0025) 110. (B) 2 × (Fe+3 + e– → Fe+2) reduction at = 595·50 cm–1 cathode 116. (B) According to Langmuir Isotherm, frac- Sn+2 → Sn+4 + 2e– Oxidation tional surface coverage is given as : at anode 2Fe+3 + Sn+2 → 2Fe+2 + Sn+4 1 = 1 + 1 θ KeqP E0Cell = Ecathode – Eanode 1 1 = 0·75V – 0·15V ⇒ θ – 1 = Keq P = 0·60V Thus, E0Cell = 2·303RT log Keq ⇒ P = θ nF (1 – θ)Keq log Keq = 0·60V × 2 = 20 = 0·95 0·06V 0·05 × 0·9 Keq = 1020 = 21·1

CSIR–Chemical Sciences (Dec.–12) | 33 117. (A) According to virial theorem for simple 121. (B) δ 171 at COOH harmonic oscillator, the kinetic energy and δ 162 at C-4 potential energy is equal and half of the total δ 133 at C-2 and C-6 δ 122 at C-1 energy. δ 116 at C-3 and C-5 E = K+U QU=K hw OH 2 = 2K 5 hw 5 4 4-hydroxy benzoic acid K = 4 =U 3 3 2 118. (C) Energy of a hydrogen atom in a state is COOH 1 ( )–hcRH h2 = –hCRH Q E = –13·6 · Z2 OH 25 n2 4 Degeneracy = n2 = 25 53 119. (A) ψt = c1φ1 + c2φ2 62 1 COOH ⎪⎪⎪⎪ H11 – ES11 H12 – ES12 ⎪⎪⎪⎪ = 0 122. (D) C9H10O3 H21 – ES21 H22 – ES22 DBE = 18 + 2 – 10 2 Q S11 = 1, S22 = 1, S12 = S21 = 0 = 5 double bond or ring H11 = <φ1 | H^ | φ1> = 0, H12 = H21 = <φ1 | H^ | φ2 = 2·0 = <φ2 |H^ | φ1>, H22 = <φ2 | H^ | φ2> IR : 3400, 1680 cm–1 = 3·0 1H NMR : ∴ ⎪⎪⎪⎪ 0 – E 2 – 0 ⎪⎪⎪⎪ = –3E + E2 – 4 = 0 }δ 7·8 (1H‚ d‚ J = 8Hz) aromatic protons 2 – 0 3 – E 7·0 (1H‚ d‚ J = 8 Hz) with m-position E2 – 3E – 4 = 0 6·5 (1H, S) aromatic proton with O- position to —OH to —OCH3 ⇒ E2 – 4E + E – 4 = 0 5·8 (1H, S, D2O exchangeable) —OH ⇒ (E + 1) (E – 4) = 0 3·9 (3H, S) with electronegative atom like Since, ground state energy is lowest. oxygen. Thus, E = –1·0 2·3 (3H, S) adjacent to carbonyl group. 120. (D) Polar molecule AB → A+ B– has H H molecular orbital CAψA + CBψB O HO CH3 CA2 + CB2 = 1 (normalization) (6.5H) OCH3 Since, CB2 = 90% 123. (A) Since, specific rotation [α]D of a 90% optically pure 2-arylpropanoic acid is +135°. = 90 = 0·9 100 So, 100% optically pure isomer will show ⇒ CB = 0·95 CA2 = 10% 135 × 100 = 150° [a]D. 90 = 10 = 0·1 After one hour [α]D reduced to 120° 100 120 ⇒ CA = 0·32 Thus, Optical purity = 150 × 100 ⇒ CA = 0·32, CB = 0·95 = 80%

34 | CSIR–Chemical Sciences (Dec.–12) After 3 hours optical purity is reduced to 128. (A) 40%. O OH So, Specific rotation = 150 × 40 = 60° CH OH Me CH N Me 100 N N MeHN 124. (A) OH Me AcO 1 4Me 2 View H Me H CH C CH2 C Me PhNEt2 Allene N N 3 Ac2O 3Ph Δ 21 Claisen [3, 3] N 31 N OH 2 rearrangement OH [3, 3] 3 Me Me 2 Mannich Intermediate Iminium ion 1O 3 21 O Me 2 3 4 CH N Me N N Ph Me R-configuration H N CH Thermal condition, so retention in configu- Me OH ration w.r.t. Me and Ph. COOH 125. (C) 129. (A) HOOC COOH 3 Δ 2 3 Claisen O 1 Ph Conrotation H Ph 2 Rearrang- OH 8-membered ring COOH O OH 1 COOH ment Oxidation [3, 3] COOH 8π-system formation H A Disrotation Δ Six membered ring B 6π-system formation H COOH Ph COOH O HOOC COOH H Decarboxylation O 126. (C) CO2 Y OH OH O K H, THF C H2 O Claisen rearrangement followed by oxidative decarboxylation. Keto-enol 1 130. (A) O 3 2 1O 3 Tautomerism 32 Δ [3, 3]-sigmatropic 2 OH H3O H 1 β-amyrin O HO 127. (A) Triterpene and Secondary alcohol. Ph NO2 Et3N Ph NO2 Acyclic triterpene Cl polyene N Acid-base N H rh Cl Squalene COOEt Δ Ph N N(EBOta3sN2e)PChl N NO2 Ph NO2 OH CH3 Cl H Cl N Phenol O Alkaloid H NHCH3 Cl Cycloaddition 1, 3-dipolar OH N CH3 H species OH Ph N H Ephedine NO2 NO2 Ph Ph N H3O Morphine Alkaloid, Secondary alcohol

CSIR–Chemical Sciences (Dec.–12) | 35 131. (C) 3 OLi OH O 136. (A) N H Cu(acac)2 :C H H N C COOEt N2 COOEt Ph 2 O B Ph Me Si face attack 1 Carbene Re face COOEt Re face of carbonyl and Si-face of enolate, syn product will be formed. MeO : CHCOOEt Chelotropic reaction MeO 132. (A) N i-PrMgCl N MgCl hydrogenolysis O -i-PrH A O MeOH↓H2, 10% Pd/C COOOEt Grignard reagent O N Mg 2Cl N O MeO N OMe O O least hindered bond will be break Me 137. (C) 133. (A) α -Ketoglutarate P⎯yr⎯ido⎯xa⎯mi→ne Glutamic Me O H acid O Me LiAlH4 Me OH Uridine ⎯Tet⎯rah⎯yd⎯ro⎯fol→ate Thymidine Pyruvic acid ⎯T⎯hi⎯am⎯ine→ Acetyl coenzyme A Et2O Ac2O, Py Pyrophosphate 20 (For structure see Lehninger text book) O Me Me2CuLi Flipping Me O C CH3 Me Et2O Me O C CH3 equatorial O 1, 3-a, e attack Me O trans CHO C OH Me 138. (D) NaOH + I2 → NaOI + HI H OH Br2 H OH 134. (A) H OH PhCOCH3 + 3NaOI → PhCOCI3 + 3NaOH OH H2O H OH PhCOCI3 + NaOH → PhCOONa + CHI3 H Mild H OH CH2OH oxidising COCH3 M.w.t. = 120 M.w.t. = 394 agent CH2OH 12g CHI3 CO2 H2O2 Fe2(SO4)3 CHO 120g acetophenone gives 394g iodoform. 12g acetophenone willl give H OH 394 × 12 = 39·4g H OH 120 Since, yield is 75% CH2OH So, 12g acetophenone will give H CH3 = 39·4 × 75 H CH3 100 135. (B) CH3 H2,Pd = 29·5 g CHI3 CH3 250 C 139. (A) No. 1, 3 interaction Two, H/Me Pd(O)L2 PhI Ph PdL2 ligand transfer from (A) I one metal to another Oxidative SnBu3 addition 1, 3 interaction Since, 1 gauche butane interaction energy = Oxidation state increased by two units. 0·9 k-cal/mol. (B) Transmetallation So, ΔG for given reaction (C) = 2 × 0·9 k-cal/mol Pd(O)L2 Reductive Ph PdL2 elimination = 1·8 k-cal/mol Ph

36 | CSIR–Chemical Sciences (Dec.–12) 140. (B) This is Reimer-Tiemann Reaction NNHPh NNHPh NNHPh O CHCl3 OH NaOH OH HO H HO H CH OH H OH NH2NHPh H OH O H OH H CH2OH H2O2 NaOH O O H OH OH CH2OH OO Ozazone OH C H O O HO OH O HO H HO H OH OH H C O H OO H OH O H OH C OH H HCOO CH2OH D-mannose OH OH 141. (A) O 144. (B) Li, liq.NH3 H /H3O Ph C C D O BD N tBuOH N N H H Syn addition P O O B Ph D H , H2O H3O H3O OO H O N NH2 H P I Ph P B(OH)2 D Ph D Ph Pd(OAc)2 Ph MeOH KOH cis product PPh3, Et3N Suzuki H coupling O OH H 142. (A) O More acidic O OO O O H H2C N2 CH3 N2 MeOH KOH O Et2O O O OH O O Me 145. (*) Doubtful question ? MeMgCl N2 CH3 N2 O OCH3Et2O OCH3 O 1, 2-addition C OEt n-BuLi COOEt COOEt H3O OH2 Me Me Me N OMe HN SiMe3 H2O H2O OMe N O CH3 OO CH3 CH3OH N OMe N SiMe3 N OMe N N SiMe3 143. (A) H O NNHPh H OH H OH COOEt H3O COOEt H HO H NH2NHPh HO OH H2, Raney Ni H OH H OH NH2 H OH H CH2OH COOEt CH2OH D-glucose

Chemical Sciences CSIR UGC-NET/JRF Exam. Solved Paper

June 2013 Chemical Sciences PART A them, 19.9% of the women and 8.8% of the men were willing to pay for the facilities— 1. During an evening party, when Ms. Black, 1. What is the ratio of the men to women Ms. Brown and Ms. White met, Ms. Brown remarked, “It is interesting that our dresses customers who wanted child care are white, black or brown, but for each of us facilities ? the name does not match the colour of the 2. If the survey had been conducted during dress !”. Ms. White replied, “But your white the weekend instead, how will the result dress does not suit you !”. Pick the correct change ? answer. With the above data— (A) Only 1 can be answered (A) Ms. White’s dress was brown (B) Only 2 can be answered (C) Both 1 and 2 can be answered (B) Ms. Black’s dress was white (D) Neither 1 nor 2 can be answered (C) Ms. White’s dress was black 6. The map given below shows contour lines which connect points of equal ground surface (D) Ms. Black’s dress was black elevation in a region. Inverted ‘V’ shaped portions of contour lines represent a valley 2. Of all the triangles that can be inscribed in a along which a river flows. What is the semicircle of radius R with the diameter as downstream direction of the river ? one side, the biggest one has the area— (A) R2 (B) R2⎯√2 (C) R2⎯√3 (D) 2R2 N 200 3. A square pyramid is to be made using a wire 180 such that only one strand of wire is used for 160 each edge. What is the minimum number of times that the wire has to be cut in order to make the pyramid ? Scale = 1 : 5000 (A) 3 (B) 7 (A) North (B) South (C) East (D) West (C) 2 (D) 1 4. Identify the next figure in the sequence— 7. During a summer vacation, of 20 friends from a hostel, each wrote a letter to each of all others. The total number of letters written was— (A) 20 (B) 400 (A) (B) (C) 200 (D) 380 (C) (D) 8. D C 5. In a customer survey conducted during North Monday to Friday, of the customers who A B East asked for child care facilities in super markets, 23% were men and the rest, women. Among

4 | CSIR-Chemical Sciences (June-2013) A person has to cross a square field by going (A) 56 (B) 70 from A to C. The person is only allowed to (C) 86 (D) 100 move towards the east or towards the north or use a combination of these movements. The 14. Define a ⊗ b = lcm (a, b) + gcd (a, b) and total distance travelled by the person— a ⊕ b = ab + ba. What is the value of (1 ⊕ 2) ⊗ (3 ⊕ 4) ? Here lcm = least common (A) depends on the length of each step multiple and gcd = greatest common (B) depends on the total number of steps divisor— (C) is different for different paths (A) 145 (B) 286 (D) is the same for all paths (C) 436 (D) 572 9. A crow is flying along a horizontal circle of 15. There is an equilateral triangle in the XY radius R at a height R above the horizontal plane with its centre at the origin. The ground. Each of a number of men on the distance of its sides from the origin is 3.5 cm. ground found that the angular height of the The area of its circumcircle in cm2 is— crow was a fixed angle θ (< 45°) when it was (A) 38.5 (B) 49 closest to him. Then all these men must be on (C) 63.65 (D) 154 a circle on the ground with a radius— (A) R + R sin θ (B) R + R cos θ 16. What is the value of 1 2+ 1 3+ 1 4+ … × × × (C) R + R tan θ (D) R + R cot θ 1 2 3 10. How many pairs of positive integers have gcd to ∞ ? 20 and lcm 600 ? (A) 2 (B) 1 3 (gcd = greatest common divisor; lcm = least common multiple) (C) 2 (D) ∞ (A) 4 (B) 0 17. A sphere of iron of radius R/2 fixed to one end of a string was lowered into water in a (C) 1 (D) 7 cylindrical container of base radius R to keep exactly half the sphere dipped. The rise in the 11. Two integers are picked at random from the level of water in the container will be— first 15 positive integers without replacement. What is the probability that the sum of the two numbers is 20 ? (A) 3 (B) 1 4 21 (C) 1 (D) 1 105 20 12. A daily sheet calendar of the year 2013 contains sheets of 10 × 10 cm size. All the (A) R /3 (B) R /4 (C) R /8 (D) R /12 sheets of the calendar are spread over the floor of a room of 5 m × 7.3 m size. What percentage of the floor will be covered by 18. Choose the largest number— these sheets ? (A) 2500 (B) 3400 (A) 0.1 (B) 1 (C) 4300 (D) 5200 (C) 10 (D) 100 19. A crystal grows by stacking of unit cells of 10 × 20 × 5 nm size as shown in the diagram 13. How many rectangles (which are not squares) given below. How many unit cells will make are there in the following figure ? a crystal of 1 cm3 volume ? 5 nm 20 nm 10 nm Unit cell (not to scale)

CSIR-Chemical Sciences (June-2013) | 5 24. The oxidation state of molybdenum in [(η7– tropylium) Mo (CO3]+ is— (A) +2 (B) +1 20 nm (C) 0 (D) –1 5 nm 25. The reaction of [PtCl4]2– with two equivalents 10 nm of NH3 produces— Crystal (not to scale) (A) cis-[Pt(NH3)2Cl2] (B) trans-[Pt(NH3)2Cl2] (A) 106 (B) 109 (C) Both cis-[Pt(NH3)2Cl2] and (C) 1012 (D) 1018 trans-[Pt(NH3)2Cl2] (D) cis-[Pt(NH3)2Cl4]2– 20. A solid cylinder of basal area A was held dipped in water in a cylindrical vessel of basal 26. The electronic transition responsible for the area 2A vertically such that a length h of the cylinder is immersed. The lower tip of the colour of the transition metal ions is— cylinder is at a height h from the base of the vessel. What will be the height of water in the (A) dπ → dσ (B) dπ → dσ* vessel when the cylinder is taken out ? (C) dπ → dπ* (D) dσ → dπ* 27. The number of metal-metal bonds in [W2(OPh)6] is— h (A) 1 (B) 2 h (C) 3 (D) 4 (A) 2h (B) 3 h 28. The Mulliken symbols for the spectroscopic 2 states arising from the free-ion term F are— 4 5 (A) T2g + Eg (B) T1g + T2g + T1u 3 4 (C) h (D) h (C) T1g + T2g + A2g (D) A1g + T2g + T1g PART B 29. Which of the following is used as propellant for whipping creams ? (A) N2O (B) NO 21. Which of the following pairs has the highest (C) N2O3 (D) N2O5 difference in their first ionization energy ? 30. Flame proof fabrics contain— (A) Xe, Cs (B) Kr, Rb (A) H2NC(O)NH2·Na2SO4 (B) H2NC(S)NH2·Na2SO4 (C) Ar, K (D) Ne, Na (C) H2NC(O)NH2·H3PO4 (D) H2NC(S)NH2·H3PO4 22. The ligand in uranocene is— (A) C8H82– (B) C5H52– (C) C6H6 (D) C4C42– 31. Among the compounds A–D, those which hydrolyse easily are— 23. In metal–olefin interaction, the extent of increase in metal → olefin π-back-donation (A) NCl3 (B) NF3 would— (C) BiCl3 (D) PCl3 (A) Lead to a decrease in C = C bond length 32. The coordination geometry of copper(II) in (B) Change the formal oxidation state of the the type I copper protein plastocyanin is— metal (A) Square planar (B) Tetrahedral (C) Change the hybridisation of the olefin (C) Octahedral carbon from sp2 to sp3 (D) Distorted tetrahedral (D) Increase with the presence of electron donating substituents on the olefin

6 | CSIR-Chemical Sciences (June-2013) 33. The metal ions present in the active site of respectively. What is the probability that the nitrogenase enzyme co-factor are— system energy will be observed to be E1 ? (A) Fe, Mo (B) Fe, W 3 3 16 4 (C) Fe, Cu (D) Fe, Ni (A) (B) 34. The reaction [(CO)5 Mn(Me)] + CO → (C) 1 1 [(CO)5 Mn{(C(O)Me}] is an example for— 4 (D) 4 (A) Oxidation addition 40. What is the atomic term symbol for helium atom with electronic configuration 1s2 ? (B) Electrophilic substitution (C) Nucleophilic substitution (A) 2S1/2 (B) 1P0 (D) Migratory insertion (C) 1S0 (D) 1S1 35. The number of EPR signals observed for octahedral Ni (II) complexes is— 41. A molecule contains the following symmetry operations : E, 2C6, 2C3, C2, 3σd, 3σv. The (A) One (B) Two number of classes and order of the symmetry point group is— (C) Three (D) Zero 36. For neutron activation analysis of an element, (A) 3, 12 (B) 5, 12 the favourable characteristics of both the target and the product are from the (C) 6, 12 (D) 6, 6 following— 42. A triatomic molecule of the type AB2 shows 1. High neutron cross-section area of target two IR absorption lines and one IR-Raman line. The structure of the molecule is— 2. Long half-life of the product 3. Low neutron cross-section area of target (A) B–B–A (B) B–A–B 4. Low half-life time of the product BA The correct characteristics from the above (C) (D) are— (A) 1 and 2 (B) 2 and 3 B AB B (C) 3 and 4 (D) 1 and 4 37. The concentrations of a species A undergoing 43. In NMR spectroscopy, the product of the nuclear ‘g’ factor (gN), the nuclear magneton the reaction A → P is 1.0, 0.5, 0.33, 0.25 mol (βN) and the magnetic field strength (B0) dm–3 at t = 0, 1, 2 and 3 seconds, respectively. gives the— The order of the reaction is— (A) Energy of transition from α to β state (A) Two (B) One (B) Chemical shift (C) Spin-spin coupling constant (C) Zero (D) There (D) Magnetogyric ratio 38. The difference in energy levels of n = 2 and 44. An aqueous mixed solution of NaCl and HCl n = 1 of a particle in a one dimensional box is 6 units of energy . In the same units, what is is exactly neutralized by an aqueous NaOH the difference in energy levels of n = 3 and n = 2 for the above system ? solution. The number of components in the (A) 4 (B) 5 final mixture is— (C) 9 (D) 10 (A) 1 (B) 2 39. The wave function ψ of a certain system is (C) 3 (D) 4 the linear combination 45. The lowest pressure at which the liquid phase ψ= 1 ψ1 + 3 ψ2 of a pure substance can exist is known as— 4 4 (A) Critical point pressure (B) Super-incumbent pressure where ψ1 and ψ2 are energy eigen functions (C) Triple-point pressure with eigen values (non-degenerate) E1 and E2, (D) Saturation vapour pressure

CSIR-Chemical Sciences (June-2013) | 7 46. A chemical reaction involving non-linear 54. The correlation coefficient of two parameters molecule + non-linear molecule non-linear is found to be –0.99. It may be concluded that activated complex. The number of vibrational the two parameters are— degrees of freedom in the activated complex, containing N atoms, is— (A) Strongly correlated (A) 3N – 5 (B) 3N – 6 (B) Almost uncorrelated (C) 3N – 7 (D) 3N – 8 (C) Connected by a cause-effect relationship 47. Calculate the total number of microstates for (D) Not connected by a cause-effect relation- 6 identical particle with their occupation ship numbers {1, 2, 3} in three states is— 55. The IUPAC name for the compound given (A) 6 (B) 12 below is— (C) 60 (D) 720 OH PH 48. If the concentration (c) is increased to 4 times its original value (c), the change in molar (A) (2R, 3Z)-7-phenylhept-3-en-2-ol conductivity for strong electrolytes is (where (B) (2S,3Z)-7-phenylhept-3-en-2-ol b is Kohlrausch constant)— (C) (2R,3E)-7-phenylhept-3-en-2-ol (D) (2S,3E)-7-phenylhept-3-en-2-ol (A) 0 (B) b√⎯ c (C) 2b√⎯ c (D) 4b√⎯ c 56. Among the following esters, the one that undergoes acid hydrolysis fastest is— 49. In atom recombination reactions— (A) Ea = 0, ΔS# = +ve, ΔH# = +ve OCOCH3 (B) Ea = 0, ΔS# = –ve, ΔH# = –ve OCOCH3 (B) (C) Ea = +ve, ΔS# = –ve, ΔH# = –ve (A) OH (D) Ea = +ve, ΔS# = +ve, ΔH# = +ve OCOCH3 50. In the Lindemann mechanism of unimolecular (C) (D) OCOCH3 reactions, the observed order at low concen- CH3 tration is— OH (A) 0.5 (B) 1 (C) 1.5 (D) 2 57. Reaction of cyclohexyl benzyl ether with hydrogen in the presence of 10% Pd/C 51. The aggregation of surfactant molecules is yields— known as— (A) cyclohexanol and toluene (B) cyclohexanol and benzyl alcohol (A) Micelles (B) Clusters (C) cyclohexane and benzyl alcohol (D) cyclohexane and toluene (C) Gel (D) Colloid 58. Among the following dibromocyclohexanes, 52. The coordinates for the atoms in a body- the one that reacts fastest with sodium iodide centred cubic unit cell are— to give cyclohexene is— (A) (0, 0, 0) and (1/2, 0, 0) Br Br (B) (B) (0, 0, 0) and (1/2, 1/2, 1/2) (A) Br (C) (0, 0, 0) and (0, 1/2, 0) Br (D) (0, 0, 0) and (0, 0, 1/2) Br Br 53. The interplanar distance (A° ) for (100) plane (C) (D) Br Br in a cubic structure with the lattice parameter of 4 A° is— (A) 1 (B) 2 (C) 4 (D) 8

8 | CSIR-Chemical Sciences (June-2013) 59. Match the following drugs with their 63. In the compound given below, the hydrogens medicinal activity— marked A and B are— (a) 5-fluorouracil 1. Anti-bacterial O (b) Amoxicillin 2. Cholesterol lowering 3. Anticancer 4. Anti-inflammatory Ph N Ph (a) (b) HB HA COOH (A) 1 2 (A) Homotopic (B) Isotopic (C) Enantiotopic (D) Diastereotopic (B) 4 3 (C) 3 4 64. In the IR spectrum, the absorption band due to carbonyl group in phenyl acetate appears (D) 3 1 at— 60. The major product formed in the following (A) 1800 cm–1 (B) 1760 cm–1 reaction sequence is— Me 1. CH3CO3H (C) 1710 cm–1 (D) 1660 cm–1 O 2. LiAIH4 65. The reactive intermediate involved in the Me OH Me OH following reaction is— NaNH2 (A) OH (B) OH Br N-Me N H Me Me Me OH (A) A carbocation (B) A carbanion (C) OH OH (C) A free radical (D) An aryne (D) 66. Number of isoprene units present in lupeol OH is— 61. The biosynthetic precursor for the steroids H is— H (A) Secologanin (B) Shikimic acid H (C) Mevalonic acid (D) α-ketoglutaric acid HO H Lupeol 62. The major product formed in the following (A) Two (B) Four reaction sequence is— (C) Six (D) Eight Br 67. The heterocyclic ring present in the amino acid histidine is— 1⎯. n⎯Bu⎯3S⎯nC⎯l‚ N⎯aB⎯H⎯4‚ ⎯Al→BN 2. H2SO4; CrO3 (A) Pyridine (B) Tetrahydropyrrole O OEt (C) Indole (D) Imidazole (A) (B) 68. The gauche conformation (ϕ = 60°) of n-butane posseses— OO OO (A) Plane of symmetry; and is achiral (C) O O (D) O O (B) C2-axis of symmetry; and is chiral (C) Centre of symmetry; and is achiral (D) Plane of symmetry; and is chiral

CSIR-Chemical Sciences (June-2013) | 9 69. The following photochemical conversion 74. Among the following, those can act as proceeds through— Mossbauer nuclei are— O H Me O 1. 129I 2. 57Co hν 3. 57Fe 4. 121Sb (A) 1, 2, 3 and 4 H (B) 2, 3 and 4 only (A) Barton reaction (B) Paterno-Buchi reaction (C) 1, 2 and 4 only (C) Norrish type I reaction (D) Norrish type II reaction (D) 1, 3 and 4 only 70. Among the following dienes, the one that undergoes a degenerate Cope rearrangement 75. Which of the pairs will generally result in is— tetrahedral coordination complexes, when ligands are Cl– or OH–— H 1. Be(II), Ba(II) 2. Ba(II), Co(II) (A) (B) 3. Co(II), Zn(II) 4. Be(II), Zn(II) H (A) 1 and 2 (B) 2 and 3 (C) (D) (C) 3 and 4 (D) 1 and 4 PART C 76. Silica gel contains [CoCl4]2– as an indicator. When activated, silica gel becomes dark blue 71. A radioisotope 41Ar initially decays at the rate while upon absorption of moisture, its colour of 34,500 disintegrations/minute, but decay changes to pale pink. This is because— rate falls to 21,500 disintegrations/minute after 75 minutes. The t1/2 for 41Ar is— (A) Co(II) changes its coordination from tetrahedral to octahedral (B) Co(II) changes its oxidation state to Co(III) (C) Tetrahedral crystal field splitting is NOT equal to octahedral crystal field splitting (D) Co(II) forms kinetically labile while Co(III) forms kinetically inert complexes (A) 90 minutes (B) 110 minutes 77. For the metalloprotein hemerythrin, the statement that is not true is— (C) 180 minutes (D) 220 minutes (A) There are two iron centers per active site 72. The orders of reactivity of ligands, NMe3, PMe3 and CO with complexes MeTiCl3 and (B) Both iron centres are hexacoordinated in (CO)5 MO (thf) are— the active state (A) CO > PMe3 > NMe3 and CO > NMe3 > (C) One iron is hexacoordinated while the PMe3 other is pentacoordinated in the active state (B) PMe3 > CO > NMe3 and NMe3 > CO > PMe3 (D) It is found in marine invertebrates (C) NMe3 > PMe3 > CO and CO > PMe3 > 78. For a tetragonally distorted Cr(III) complex, NMe3 zero-field splitting results in the following (D) NMe3 > CO > PMe3 and PMe3 > NMe3 > CO number of Kramers doublets— (A) 1 (B) 2 (C) 3 (D) 4 73. The number of lone-pairs are identical in the 79. Intense band at 15000 cm–1 in the UV-visible pairs— spectrum of [Bu4N]2 Re2Cl8 is due to the transition— (B) XeO4, ICl4– (A) XeF4, ClF3 (A) π → π* (B) δ → δ* (C) XeO2F2, ICl4– (D) XeO4, ClF3 (C) δ → π* (D) π → δ*

10 | CSIR-Chemical Sciences (June-2013) 80. Electron change in reduction of Ce(SO4)2, 86. The species with highest magnetic moment KMnO4, HNO2 and I2 with hydrazine in (spin only value) is— acidic medium, respectively is— (A) VCl64 – (B) (η5–C5H5)2Cr (A) 1e, 1e, 2e and 4e (C) [Co(NO2)6]3– (D) [Ni(EDTA)]2– (B) 1e, 3e, 2e and 4e (C) 2e, 3e, 1e and 4e (D) 2e, 4e, 1e and 3e 81. The compound that will behave as an acid in 87. The number of metal-metal bonds in Ir4(CO)12 is— H2SO4 is— (A) 4 (B) 6 (A) CH3COOH (B) HNO3 (C) 10 (D) 12 (C) HClO4 (D) H2O 82. Among the oxides of nitrogen, N2O3, N2O4 88. Three bands in the electronic spectrum of and N2O5, the compound(s) having N–N bond [Cr(NH3)6] 3 + are due to the following is/are— transitions : (A) N2O4 and N2O5 1. 4A2g → 4T1g 2. 4A2g → 4T2g (B) N2O3 and N2O5 3. 4A2g → 2Eg (C) N2O3 and N2O4 (D) N2O5 only Identify the correct statement about them— (A) Intensity of 1 is lowest 83. The treatment of PhBr with n-BuLi yields— (B) Intensity of 3 is lowest (A) 2n-BuPh + Br2 + Li2 (C) Intensities of 1, 2 and 3 are similar (B) PhPh + octane + 2LiBr (D) Intensities of 2 and 3 are similar (C) n-BuPh + LiBr 89. Identify the pairs in which the covalent radii of elements are almost similar— (D) PhLi + n-BuBr 84. Though cyclobutadiene (C4H4) is highly 1. Nb, Ta 2. Mo, W unstable and readily polymerizes in its free state, its transition metal complexes could be 3. La, Lu 4. Sc, Y isolated because— (A) 1 and 2 only (A) It engages in long-range interaction with transition metals (B) 1 and 3 only (B) It gains stability due to formation of (C) 2 and 3 only C4H42– on binding to transition metals (D) 1, 2 and 3 only (C) Its polymerization ability reduces in presence of transition metal 90. Consider the following lanthanide(III) ions : (D) It becomes stable in presence of 1. Nd(III) 2. Gd(III) transition metals due to formation of C4H42+ 3. Dy(III) The magnetic moment closest to the spin only value is(are) for— (A) 2 only (B) 1 and 2 only 85. Identify the order representing increasing (C) 1 and 3 only (D) 2 and 3 only π-acidity of the following ligands— C2F4, NEt3, CO and C2H4 91. The Δt of the following complexes— (A) CO < C2F4 < C2H4 < NEt3 (B) C2F4 < C2H4 < NEt3 < CO 1. [CoCl4]2– (C) C2H4 < NEt3 < CO < C2F4 (D) NEt3 < C2H4 < C2F4 < CO 2. [CoBr4]2– and 3. [Co(NCS)4]2– follows the order (A) 3 > 1 > 2 (B) 1 > 2 > 3 (C) 2 > 1 > 3 (D) 3 > 2 > 1

CSIR-Chemical Sciences (June-2013) | 11 92. In complexometric titration (B) AB + AB A2B2 (fast) S(substrate) + T(titrant) → P(product) slow A2B2 + B2 ⎯→ 2AB2 The end point is estimated spectrophoto- (C) AB + B2 ⎯slo→w AB3 metrically. If S and P have ε = 0, the shape of AB3 + AB ⎯fas→t 2AB2 the titration curve would look like— (D) AB + B2 AB3 (fast) AB3 + AB ⎯slo→w 2AB2 (A) A (B) A 97. Observe the following statements— T T 1. In the H2–O2 reaction, explosion occurs when the rate of chain branching exceeds (C) A (D) A that of chain termination. 2. The order of the reaction, nA → products, TT is 2.5. For this reaction, t1/2 ∝ [A]0–3/2 93. Identify the chiral complexes from the 3. Unimolecular gas phase reactions are following— second order at low pressure but become first order at high pressure. 1. [Cr(EDTA)]– 2. [Ru(bipy)3]3+ Which of the following is correct ? 3. [PtCl(diene)]+ (A) 1, 2 and 3 are correct (A) 1 only (B) 1 and 2 only (B) Only 2 is correct (C) 1 and 3 only (D) 2 and 3 only (C) Only 3 is correct 94. Distribution ratio of ‘A’ between CHCl3 and water is 9.0. It is extracted with several, 5 mL (D) 1 and 2 are correct aliquot of CHCl3. The number of aliquots needed to extract 99.9% of ‘A’ from its 5 mL 98. For the particle-in-a-box problem in (0, L), an aqueous solution are— approximate wave function is giv–en as x (L/2 – x) (L – x). The average energy E for such a (A) 2 (B) 3 (C) 4 (D) 5 state will obey— 95. The correct equilibrium order for the inter- (A) h2 < – h2 (B) – h2 conversion of different forms of SiO2 is— 8mL2 E 2mL2 E> 2mL2 (A) Tridymite quartz cristobalite h2 – h2 – h2 4mL2 E 2mL2 E 8mL2 liquid SiO2 (C) < < (D) 0 < < (B) Quartz tridymite cristobalite 99. For two variables x and y, the following data set is given : liquid SiO2 (C) Quartz cristobalite tridymite xy liquid SiO2 tridymite quartz –1 1 (D) Cristobalite 02 liquid SiO2 13 96. The rate equation for the reaction, 2AB + B2 The correct statement for the covariance A → 2 AB2, is given by and correlation coefficient B of x and y is— rate = k [AB] [B2] (A) A = 2/3, B=1 A possible mechanism consistent with this (B) A = –2/3, B =1 rate law is— (A) 2AB + B2 ⎯slo→w 2AB2 (C) A = –2/3, B = –1 (D) A = 0, B=0

12 | CSIR-Chemical Sciences (June-2013) 100. The hydrogenic orbital with the form of the 107. Identify the Hückel determinant for cyclo- butadiene— radial function α–E β 0 0 r2 (α1 – r) (α2 – r) exp [–βr], where α1α2 ⎪⎪ ⎪⎪β α – E β 0 and β are constants, may be identified as a— ⎪ ⎪(A) 0 β α – E β (A) 3d orbital (B) 4f orbital ⎪ ⎪0 0 β α – E (C) 5d orbital (D) 5f orbital α–E β 0 β 101. The operator [x, [x, p2]] is identical with— ⎪⎪ ⎪⎪β α – E β 0 ⎪ ⎪(B) 0 β α – E β (A) [px, [x, p]] (B) [xp, [x, p]] ⎪ ⎪β β 0 α – E (C) –[p, [x2, p]] (D) [x, [x2, p]] α–E β 0 β 102. For the particle-in-a-box problem in (0, L), the value of (x3) in the n → ∞ limit would ⎪⎪ ⎪⎪β α – E β 0 ⎪ ⎪(C) 0 β α – E β be— ⎪ ⎪β 0 β α – E (A) L3/6 (B) L3/3 α–E β 0 β (C) L3/4 (D) L4/4 ⎪⎪ ⎪⎪β α – E β 0 103. Identity the Mulliken notation for the ⎪ ⎪(D) 0 β α – E β following irreducible representation ⎪ ⎪0 0 β α – E E En nC2 i σh 1 1 –1 –1 –1 (A) A'1u (B) A\"2u 108. On mixing 120 mL of 0.05 M CH3COOH (C) B'2u (D) A'2u and 40 mL of 0.05 M of NaOH, the pH of 104. Identify the point group symmetry of the the solution is— following molecule (all C–C bond lengths are equal)— ( pKa = –log Ka) X (A) pKa + 0.69 (B) pKa + 0.301 (C) pKa (D) pKa – 0.69 X 109. A system consists of gaseous H2, O2, H2O X and CO2 where the amount of CO2 is speci- fied and the equilibrium constant for the reaction 2H2(g) + O2 (g) 2H2O(g) is known. The number of degrees of freedom X of the system is— (A) C2v (B) S4 (A) 2 (B) 3 (C) D2d (D) D4d (C) 4 (D) 5 105. The ground state term symbol for Nb 110. “Colloids are thermodynamically unstable (atomic number 41) is 6D. The electronic with reference to bulk but kinetically configuration corresponding to this term stable”. Identify the correct pair— symbol is— Statements Reasons (A) [Kr] 4d3 5s2 (B) [Kr] 4d4 5s1 (a) Thermodynamically (c) Interfacial surface (C) [Kr] 4d5 5s0 (D) [Kr] 4d3 5s15p1 tension 106. In the presence of an external magnetic (b) Kinetically stable (d) Electrical double layer field (normal Zeeman effect), the transition 1D2 → 1P1 splits into— (A) (a) ↔ (d) and (b) ↔ (c) (B) (a) ↔ (c) and (b) ↔ (d) (A) 9 lines (B) 8 lines (C) (a) ↔ (c) and (b) ↔ (c) (D) (a) ↔ (d) and (b) ↔ (d) (C) 7 lines (D) 6 lines

CSIR-Chemical Sciences (June-2013) | 13 111. An AX system gave 4 lines at 4.72, 4.6, 1.12 117. A plane of spacing d shows first order Bragg diffraction at angle θ. A plane of spacing and 1.0 ppm away from TMS using an nmr 2d— spectrometer operating at 100 MHz. What are the values of JAX (in Hz) and δAX (in (A) Shows Bragg diffraction at 2θ ppm), respectively ? (B) Shows Bragg diffraction at θ/2 (A) 12 and 3.6 (B) 6 and 3.6 (C) Shows Bragg diffraction at sin–1 (C) 12 and 2.86 (D) 6 and 2.86 ( )sin θ 112. The equilibrium population ratio (nj/ni) of a 2 doubly-degenerate energy level (Ej) lying at (D) Shows Bragg diffraction at sin–1 energy 2 units higher than a lower ( )sin 2θ nondegenerate energy level (Ei), assuming 2 kBT = 1 unit, will be— (A) 2e–2 (B) 2e2 118. In the formation of H2 molecule from 2H atoms placed at positions A and B, and (C) e2 (D) e–2 separated by a distance rAB, a part of the spatial wave function is— 113. Which of the following statements is true for ϕA(1) ϕA(2) + ϕB(1) ϕB(2) a cyclic process ? (A) This is a covalent term and is important (A) ƒOdq = 0 as rAB → ∞ (B) Oƒdw = 0 (B) This is an ionic term and is important as rAB → ∞ (C) Heat can be completely converted into work (C) This is a covalent term and is important as rAB → 0 (D) Work can be completely converted into heat (D) This is an ionic term and is important as rAB → 0 114. Identify, from the following, the correct ionic strengths for (a) a 0.01 molal solution 119. A 0.1 M solution of compound A shows of NaCl and (b) a 0.01 molal solution of 50% transmittance when a cell of 1 cm Na2SO4— width is used at λ 1 nm. Another 0.1 M (A) (a) 0.010 mol kg–1 (b) 0.010 mol kg–1 solution of compound B gives the optical density value of 0.1761 using 1 cm cell at λ1 (B) (a) 0.010 mol kg–1 (b) 0.030 mol kg–1 nm. What will be the transmittance of a solution that is simultaneously 0.1 M in A (C) (a) 0.010 mol kg–1 (b) 0.025 mol kg–1 and 0.1 M in B using the same cell and at the same wavelength ? (D) (a) 0.010 mol kg–1 (b) 0.015 mol kg–1 115. A system has 100 degenerate energy levels and 100 bosons are kept in it. Find the entropy of the system at equilibrium— (A) 10–2 kB (B) 102 kB (C) 460.6 kB (D) 4.606 kB 116. Which is correct Nernst equation for redox (log 20 = 1.301; log 30 = 1.4771; log 50 = reaction O + ne R ? 1.699) (A) E = E0 – RT ln [O] (A) 33.3% (B) 50% nF [R] (C) 66.7% (D) 70% nF (B) [O] = e RT (E – E0) 120. Using standard equation for intrinsic [R] viscosity [η ] = K—Mva, for a solution of (C) [O] = e – nF (E – E0) polymer and any information from the graph [R] RT — (D) [O] = e RT (E – E0) identify viscosity-average molar mass ( Mv) [R] nF [given that a = 0.5, K = 5 × 10–5 Lg–1].

14 | CSIR-Chemical Sciences (June-2013) 0.055 (A) A is the major product and it is Cram product η − η0 1 η0 c (B) A is the major product and it is anti- Cram product 0.050 c 10 0 (C) B is the major product and it is a Cram product (A) 103 g/mol (B) 104 g/mol (C) 105 g/mol (D) 106 g/mol (D) B is the major product and it is anti- Cram product 124. Which one of the following statements is true for the following transformation ? 121. Among the following, the correct statement O O O for the following reaction is— O O O + 1. MeMgBr, Et2O + AB 2. H3PO4 AB (A) Suitable reagent is m-CPBA and B is (A) A is the major product and it will have the major product five signals in the proton decoupled 13C NMR spectrum (B) Suitable reagent is m-CPBA and A is the major product (B) A is the minor product and it will have eight signals in the proton decoupled (C) Suitable reagent is aq. H2O2/NaOH and 13C NMR spectrum B is the major product (C) B is the major product and it will have (D) Suitable reagent is aq. H2O2/NaOH and five signals in the proton decoupled A is the major product 13C NMR spectrum 125. The compound formed in the following (D) B is the minor product and it will have reaction sequence is— five signals in the proton decoupled 13C NMR spectrum 1. Li, liq.NH3, t-BuOH 2. 10% aq. H2SO4 122. For the following three step conversion of A MeO to B, the appropriate sequence of reactions is— HO3S OH OO (A) OH CHO H2N AB HO3S (B) HO (A) MnO2; (CH2OH)2/p-TSA; PCC (C) (B) PCC; MnO2; (CH2OH)2/p-TSA (C) PCC; (CH2OH)2/p-TSA; Jones’ reagent (D) (D) Jones’ reagent; (CH2OH)2/p-TSA; O MnO2 126. Among the following compounds, the one which has highest dipole moment is— 123. Which one of the following statements is true for the following transformation ? (A) (B) CHO OH OH (C) (D) Ph MgBr, Et2O + Ph Me Ph Me Me B A

CSIR-Chemical Sciences (June-2013) | 15 127. In the UV-Vis spectrum, a diterpenoid 130. An organic compound exhibited the follo- exhibited a λmax at 275 nm. The compound, wing 1H NMR spectral data : among the choices given below, is— δ 7.80 (2, H, d, J = 8 Hz), 6.80 (2 H, d, J = 8 (A) Hz), 4.10 (2 H, q, J = 7.2 Hz); 2.4 (3 H, s), 1.25 (3 H, t, J = 7.2Hz). The compound, H among the choices given below, is— HOOC O (B) (A) H EtO HOOC O (B) MeO O (C) (C) HO H HOOC O (D) OMe (D) Et H 131. α-Pinene on reaction with dilute alkaline HOOC KMnO4 produces a diol, which on further oxidation with chromium trioxide gives 128. The major product formed in the following product A, which undergoes a positive reaction is— haloform test. The compound A is— + 1. Δ NO2 2. TiCl3, HCl O (B) COOH (A) COOH O (A) (B) O O O NH2 NH2 (C) COOH (D) (C) (D) 129. In the broadband decoupled 13C NMR 132. The major product formed in the reaction of spectrum, the number of signals appearing guanosine with one equivalent of methyl for the two pyrenediols A and B, iodide is— respectively, are— O Me OH OH (A) HN NI H2N N N R HO OH O N A B Me N (A) Eight and eight (B) Eight and sixteen N R (C) Five and ten (D) Five and eight (B) H2N N

16 | CSIR-Chemical Sciences (June-2013) O N (B) Phe-Leu + toluene + carbon dioxide HN NI (C) Phe-Leu + benzyl alcohol + carbon (C) R dioxide H2N N (D) Gly-Leu + benzyl alcohol + carbon Me dioxide O N 135. Among the following, the most suitable (D) HN reagent for carrying out resolution of N racemic 3-methylcyclohexanone is— Me-HN N R HO 133. The major product formed in the following (A) reaction is— OH CN O OH + Δ COOMe O (B) MeOOC OH O HH Me OH (A) (C) Ph HH O O OH O (D) HH HO (B) 136. In the following reaction sequence, structures HH of the major products X and Y are— O CN O + (COOEt)2 ⎯Na⎯OE→t X ⎯Zn⎯/A⎯cO→H Y H NO2 (C) HO (A) X is O HO COOEt NC NO2 (D) (B) X is Y is COOEt N OH O H COOEt 134. Reaction of the dipeptide, given below, with NO2 OH hydrogen in the presence of 10% palladium NO over carbon, produces a mixture of— Y is H Ph HO OH ON N O HO Ph (A) Gly-Leu + toluene + carbon dioxide

CSIR-Chemical Sciences (June-2013) | 17 (C) X is O O amount of p-TSA furnished the major O product A. The structure of A is— O OEt Y is N HO NO2 H (A) O O O Ph O (D) X is COOEt HO OO OEt Y is NH (B) Ph NO2 137. Consider the following reaction sequence : OH OH OH (C) O Sn/HCl O OH HNO3 O2N (90%) H2N (70%) Ph O OH (D) Ph O OH Ac2O (90%) AcHN OH OH Ac2O OH 140. The major product formed in the following Sn/HCl NH2 (90%) NHAc reaction is— (25%) NO2 (90%) H OTs The overall yield for the formation of p- ⎯K⎯H/T⎯H→F hydroxyacetanilide and o-hydroxyaceta- nilides from phenol, respectively, are appro- ximately— (A) 57 and 20% (B) 57 and 68% OH (C) 83 and 68% (D) 83 and 20% H H (A) O 138. The most stable conformations of 1, 2–diflu- (B) oroethane and dl-2,3-butanediol are— OH F OH Me H HF (A) and H H Me H (C) (D) H OH O O F OH 141. The major product formed in the following HH Me OH reaction is— (B) and HH H Me OH ⎯1. ⎯Na⎯H‚⎯CS⎯2‚⎯M→el FH 2. 200°C F OH HH H OH (C) and H H Me H F Me F Me (A) (B) HF H OH (D) and HH H OH H Me 139. Reaction of (S)-1,2,4-butanetriol with (C) (D) benzaldehyde in the presence of a catalytic

18 | CSIR-Chemical Sciences (June-2013) 142. The major product formed in the following H reaction is— H N2 OO (A) Con-rotatory in photochemical; and dis-rotatory in thermal conditions 1. hv‚ aq. THF 2. Δ (B) Con-rotatory in thermal; and dis- rotatory in photochemical conditions O O N (C) Con-rotatory in thermal; and con- (B) rotatory in photochemical conditions (A) N O (D) Dis-rotatory in photochemical; and dis- rotatory in thermal conditions O (D) O (C) OH OH 143. The major product formed in the following Answers with Hints reaction is— Part–A O hv‚ Et2O 1. (C) Ms. Black Ms. Brown Ms. White ↓↓↓ brown white black OH Each bears different colour dress according to (A) (B) O their name. 2. (A) OH 3. (D) E (D) (C) O DC H square pyramid 144. Predict the condition A and the structure of the major product B in the following AB sequence— Only one cut is required as ABCEDA is A 1. O3; Me2S B formed without cut and then after one cut CD (A) A is hv B is is made. 2. Ph3P = CHCOOEt 4. (C) For every next figure, dark cut is rotating (excess) clockwise with one part adjacent to next and small circles are increasing 2 and 1 adjacent COOEt to dark area and next after adjacent to dark COOEt area respectively. (B) A is hv B is COOEt 5. (A) COOEt 6. (B) N (C) A is Δ B is COOEt 200 COOEt 180 160 (D) A is Δ B is COOEt Since inverted ‘V’ shaped portions of contour COOEt lines represent a valley along which a river flows and that direction is north according to 145. The most appropriate mode of cyclisation in given figure. Thus, the downstream direction the following transformation is— of the river is south.

CSIR-Chemical Sciences (June-2013) | 19 7. (D) 24. (C) [(n7–tropylium) Mo(CO)3]+ 8. (D) D C North MO East OC CO CO AB Since, there is square and allowed only to In zero oxidation state of Mo, complex have North and East direction move the total distance travelled by the person is same for all 18 e– (stable) paths. 7 + 6 + 6 – 1 = 18 tropylium is a 7e – donor ligand in neutral 9. (D) 10. (A) 11. (B) 12. (C) 13. (B) method. 14. (C) 15. (D) 16. (B) 17. (D) 18. (B) 19. (D) 20. (B) 25. (A) Cl Cl −2 Cl NH3 Cl NH3 Pt +NH3 Pt +NH3 Pt Part–B Cl Cl Cl Cl Cl NH3 Cis-platin 21. (D) As we go down the group in periodic table for sodium family, the number of electrons (anticancer drug) increases and their attraction towards nucleus increases and removal of e – become difficult. Cl– have more trans directing ability than So Na loses e – s very easily. NH3. 26. (A) 27. (C) 28. (C) 29. (A) 30. (C) In the case of inert gas family, Ne has its 31. (C) Among given compounds NF3 is not complete octet, strongly connected with hydrolyse easily because, the more electro- nucleus e– attraction and do not give e–s to negative F atom strongly attracts the e–s lose easily. present in N atom and they are no longer available to get hydrolysed. Thus Na, Ne pair have highest difference in their first ionization energy in given option. 22. (A) The structure of uranocene is N 102 F U+4 (C8H8– 2)2. F F .5° 32. (D) 33. (A) 34. (D) O U C (CO)4Mn CH3 (CO)4Mn C CH3 O Most stable actenocene. O 23. (C) In Metal-Olefin interaction, the extent of (Co insertion) increase in metal → Olefin π-back donation would strengthen the Metal-Olefin bond O HH C C CH3 C M (CO)4Mn CH3 (CO)4Mn C (methyl migration) HH = vacant site which produces a increase in C = C bond So overall migratory insertion. length and somewhat single bond character which results a change in hybridisation from 35. (A) sp2 to sp3. 36. (A) Neutron activation analysis is a sensitive multi-element analytical technique used for both qualitative and quantitative analysis of major, minor, trace and rare elements. It requires high neutron cross-section area of target and long half-life of the product.

20 | CSIR-Chemical Sciences (June-2013) 37. (C) A→P 42. (B) In triatonic centrosymmetric molecule, the ⇒ vibrational mode which are active in IR, are rate = – d [A] inactive in Raman and vice-versa. Thus AB2 dt molecule showing two IR absorption lines and one IR–Raman line have following structure— [A]t = [A]0 – n Kt ←→ B —A—B Conc. asymmetric stretching and bending → IR active symmetric stretching → IR-Raman active Time 43. (A) In NMR spectroscopy energy of transition from α to β state is given as : From given observation, we find this graph and it indicates that order of reaction is zero. Vα → β = gNβNB0 38. (D) For a particle in a one dimensional 44. (B) NaCl + HCl ⎯aq.⎯N⎯aO→H 2 NaCl + H2O box energy is given as (aq) Neutrali- 2 components n2h2 zation 8 ma2 En = There are no HCl and NaOH in final mixture. where a is length of box, m is mass of 45. (C) particle. Thus, solid liquid E1 = h2 P3 vapour 8 ma2 and E2 = 4h2 T3 8 ma2 Temperature So, E2 – E1 = 3h2 The lowest pressure at which the liquid phase 8 ma2 of a pure substance can exist is known as ‘Triple-point pressure’. = 6 units of energy 46. (C) 47. (C) Number of microstates Therefore, E3 – E2 = (9 – 4) 8 h2 ma2 = 5×2 39. (C) = 10 units of energy n 40. (C) 2He : 1S2 1S = n1 n2 n3 …… Electronic term symbol = 2S + 1LJ where n is total identical particles and n1, n2, n3 are occupation number. S = n = 0 Thus, 6 2 = ⇒ 2S + 1 = 1 123 L = 0 (S) 6×5×4× 3 J=0 = Thus, term symbol for He = 1S0 1×2×1× 3 41. (C) Symmetry operation : E, 2C6, 2C3, C2, 3σd, 48. = 60 3σv symmerty operations with different sym- 50. (B) 49. (B) (D) Lindeman Mechanism : metry element is equal to number of classes = 6 A + A K1 A* + A K– 1 order of the symmetry point group A* ⎯K→2 D = 1+2+2+1+3+3 = 12 Rate of formation of product = K2 [A*]

CSIR-Chemical Sciences (June-2013) | 21 Applying SSA on [A*] 56. (B) 57. (A) d [A*] = K1[A]2 – K – 1 [A*] [A] – K2[A]* H2 dt O CH2 10% Pd/c =0 Cyclohexyl benzyl ether [A*] = K1 [A]2 K – 1 [A] + K2 OH + H3C K2K1 [A]2 Rate = K – 1 [A] + K2 Cyclohexanal Toluene If K2 >> K – 1 [A] at low concentration Br Br Br Na+I− Rate = K1 [A]2 58. (C) Br Second order kinetics. 51. (A) no available position to form cyclohexene 52. (B) When coordinates are changed with one Br Br I−Na+ axis, the unit cell is simple cubic, when changes with two axis, it is face-centred cubic. −IBr Br Thus, when coordinates are changed with Br three axis, the unit cell is body centred cubic. Trans-1,2, dibromo cyclohexane ( )(0, 0, 0) to 21‚ 0‚ 0 – simple cubic ( )(0, 0, 0) to 12‚ 21‚ 0 – f.c.c. Br Cyclohexene ( )(0, 0, 0) to 21‚ 21‚ 1 – b.c.c. 59. (D) Me 2 60. (B) OO Me 53. (C) For cubic system, interplanar distance is Me O CH3CO3H OH given as esterification CH2 OH d2hkl = a2 h2 + k2 + l2 Thus, d1200 = 12 + 42 + 02 = 16 LiAlH4 02 LiAlH4 reduces as ⇒ d100 = 4 Å O OH 54. (A) 55. (D) OH H 123 45 67 Ph Me OH CHOH −CHO For, 2-position O OH CH2 1 4H 4 OH OH OH 32 32 61. (C) Steroids are type of organic compound view 25 1 that contains a characteristic arrangement of four cycloalkane ring that are joined to each For, 3–position other as— 2 H 1 3E H CD AB 2 cholestane, a polytypical steroid skeleton Thus, IUPAC Name is originated as (2S, 3E)–7–Phenylhept-3-en-2-ol

22 | CSIR-Chemical Sciences (June-2013) Mevalonic Acid— OH O N Me H HO 5 4 3 2 1 OH Six-membered ring (3R)–3, 5–Dihydroxy-3-methyl pentanoic acid is stable used as biosynthetic precursor for steroids. 62. (B) Br nBu3Sn Cl NaBH4 N N H Me H Me AIBN O OEt C Cl O OEt nBu3SnBr 66. (C) 1 H 2 CrO3 OO Isoprene 5 H3 HO 4 O OEt H2SO4 6H C N N C hυ C + N2 CN H CN CN Lupeol (AIBN) Six isoprene units in Lupeol is identified. 63. (C) O O 67. (D) N Histidine (His) +X N CH2 CH COOH −HA Ph N Ph Imidazole ring NH2 Ph N Ph HB X (A) 68. (B) Gauche conformation of n-butane with HA COOH 60° dihedral angle (φ) HB COOH +X O CH3 −HB H CH3 Ph N Ph φ = 60° HA (B) HH X COOH H Structures (A) and (B) have non-superimpo- Staggered sable mirror images, Thus, HA and HB are enantiotopic C2–axis bisects molecule with one methyl at each side (front & back) and is chiral, no O plane of symmetry, no centre of symmetry. 64. (B) O C CH3 69. (D) Ester group have carbonyl stretching H frequency at 1765 to 1770. But due to phenyl ring, —O—atom involves in conjugation and O H Me O CH2 OH CH2 – CO frequency decreases to 1760 cm– 1. hυ NT-II 65. (D) H γ-hydrogen ring opening abstraction Br N+aNH2− −Br H N Me −NH3 Br N Me O OH H H Proceeds through aryne intermediate O regeneration of C group (aryne) Benzyne intermediate

CSIR-Chemical Sciences (June-2013) | 23 70. (A) H H 74. (D) Mössbauer spectra are shown by nuclei which have value of I > O and also have H H different value of I in excited and ground state of nucleus. 1,5-diene Same product Among given molecule, only 57Co nuclei A molecular rearrangement in which the are mössbauer inactive because it decays as principal product is indistinguishable (in the absence of isotopic labelling) from the prin- 5277Co ⎯1e→– 26Fe57 cipal reactant, is known as ‘degenerate rearr- which is M.B. active, 129I and 121Sb is also angement’. active. Part–C 75. (C) 71. (B) Initial decay, 76. (A) [CoCl4]– 2 ⎯m⎯ois⎯tur→e [Co(H2O)6]+ 2 dark blue e(xHc2eOss) pale pink NO = 34500 disintegrations/minute tetrahedral octahedral After 75 minutes, + 2 O.S. + 2 O.S. Nt = 21500 disintegrations/minute We know that, disintegration constant, Tetrahedral complexes are darker in colour. λ = 2·303 log ⎝⎜⎛NNOt ⎟⎞⎠ 77. (B) 78. (B) t 79. (B) [Bu4N]2 Re2Cl8 → 2 NBu4+ + Re2Cl8– 2 ( )= Re : 5d5 6s2 2·303 log 34500 Re+++ : 5d4 75 21500 σ2π4δ2 → σ2π4δ1δ*1 = 0·00626 15000 cm– 1 (Royal blue) 0·693 0·693 And, also t1/2 = λ = 0·00626 σ∗ π∗ = 110·7 minute δ∗ δ∗ 72. (C) Since MeTiCl3 have no e –s for back δ → δ* transition donation with π-accepting ligands like PMe3 π and CO, therefore, It reacts in order σ NMe3 > PMe3 > CO 80. (A) (more π-accepting tendency) 81. (C) (base) CH3COOH + H2SO4 Now, in (CO)5Mo(thf), Mo have sufficient e–s and vacant site to react with π-acceptor ligand ⊕ Thus, the order is → CH3COOH2 + HSO4– (base) HNO3 + H2SO4 → H2NO3⊕ + HSO4– CO > PMe3 > NMe3 (acid) HClO4 + H2SO4 → ClO4– + H3⊕SO4– 73. (A) In given option, only I is correct, if we behaves as acid consider only attached molecules lone-pairs, (base) H2O + H2SO4 → H3O⊕ + HSO4– 82. (C) FI O OO O O O F Xe F C NN NN NN FFF F 12 lone-pairs OO OO O O 12 lone-pairs N2O3 N2O4 N2O5 O Cl O N2O3 and N2O4 have N–N bond. O Xe O Cl I Cl F Xe F 83. (D) Br Li O Cl O + n-BuLi + n-BuBr 8 lone-pairs 12 lone-pairs 10 lone-pairs

24 | CSIR-Chemical Sciences (June-2013) 84. (B) highly unstable and readily poly- Thus, merises in its free state when binds with M–M bonds = 4 × 18 – (4 × 9 + 2 × 12) 2 transition metal complexes, it gains stability due to formation of C4H42–, like = 72 – 60 = 6 2 Co 18e – species 4T1g(P) 4e – donor 88. (B) Energy 4P 4T1g(F) 4F (stable) 4T2g 4A2g 85. (D) Among the given ligand CO is most π- acid ligand. Between C2H4 and C2F4, C2F4 has Strength of ligand more tendency to behave as π-acid ligand due to more electronegative F atoms which holds [Cr (NH3)6]+ 3 e– density towards F atoms and alkene behaves as a better π-accepting position from Cr+ 3 : +2 +1 0 −1 −2 metal. NEt3 has very little behaviour towards π-back bonding. Thus, the order is 2S + 1 = 4 NEt3 < C2H4 < C2F4 < CO 86. (A) eg e1g eg L = 3 (F) +2g a1g +2g 4F e2g [Ni (EDTA)]– 2 }4A2g ⎯→ 4T2g strong bands ||| μ = 2·9 BM ⎯→ 4T1g VCl6–2 (n5 – C5H5)2 Cr O μ = 3·9 BM μ = 2·9 BM Thus, 4A2g → 2Eg intensity is lowest OO 89. (A) Due to lanthanide contraction, the covalent radii of Nb(1·34 Å), Ta(1·34Å) and Mo(1·29), W(1·30) are almost similar. ( )Chemical Reactivity = f NO Charge Ni Radius NO No increase in covalent radii due to L.C. eg O O 90. (B) μS = n (n + 2) +2g O [Co (NO2)6]– 3 μ=0 ( )7 7 7 × 11 2 2 2 Thus, VCl6– 2 has highest magnetic moment. Gd (III) : μS = 2 + 2 = 87. (B) Metal-Metal bonds n × 18 –Total no. of valence e– sin = 77 = 4·38 complex 2 = 2 n = total no. of metals Nd (III) : μS = 3 × 7 = 21 2 2 2 = 2·29 Ir4(CO)12, CO is 2e– donor ligand Ir have 9 e–s in valence shell 91. (A) As strong field ligand have high crystal field stabilisation energy, according to spec- OC CO CO trochemical series of ligand, the given ligand in order Ir Br – < Cl– < NCS– OC CO Thus, Δt of given complexes follow the order : OC Ir Ir CO [Co(NCS)4]– 2 > [CoCl4]– 2 > [CoBr4]– 2 OC CO 92. (C) 93. (B) (A) [Cr (EDTA)]– is a chiral molecule, Ir having d-l pair (enanntiones) see structure of OC CO CO question 86 (iv) 18 e– species (Stable)

CSIR-Chemical Sciences (June-2013) | 25 (B) N N [ ]A0 – 3/2 3 ([A]–0 3/2 + 3 nK t1/2 N N NN 2 2 2 = N R4 R4 N + [A0]– 3/2 = n Kt1/2 NN t1/2 ∝ [A]–0 3/2 NN (iii) According to Lindeman theory, Unimole- Enantiomers (d–l pair) cular reactions (gas phase) are second order at low pressure and become first order at high where N N = bypyridine pressure. 98. (B) 99. (A) 100. (C) NN Radial function = r2 (α1 – r)(α2 – r) e– βr Indicater that r2 = r l Optically active ⇒ l = 2 means d-orbital (C) [Pt Cl (diene)]+ has square planar struc- Since it have two nodes, r = α1 and r = α2 tures so it is a chiral. Thus, orbital (hydrogenic) is identified as 5d 94. (C) 95. (B) because radial node 96. (C) 2 AB + B2 → 2 AB2 rate is given as : = n–l–1 rate = K [AB] [B2] = 5–2–1 According to this rate law, mechanism is =2 AB + B2 ⎯sl⎯ow→ AB3 101. (C) [x, (x, p2)] = [x, ih– . 2p] K1 = x . 2p i–h – 2i–h px = 0 AB3 + AB ⎯f⎯ast→ 2 AB2 – [ p [x2, p]] = – [ p, – ih– . 2x] K2 = – [– ih– . 2x . p – (– p . i–h rate = K1 [AB] [B2], . 2x)] slowest step is rate determining step. = + 2i–h xp – 2 ih– xp = 0 97. (A) (i) As reaction proceed, pressure of system increases and radical become closer, so rate 102. (C) Wave function, of propagation become fast. At a certain point 2 radicals value become infinite and this respon- L sin sible for explosion due to increase in pressure. ( )Ψ(x) = nπx L (ii) nA → P (products) Thus, < x3> rate = K [A]5/2 ∫L d [A] nK [A]5/2 = Ψ(x) x3 . Ψ(x)* dx dt 0 = L sin2 0 t ∫ ( )= 2 nπx . x3 dx) L L ∫ ∫+ dt At d [A] = – nK L3 Ao [A]5/2 0 4 = Integrating it gives, 103. (B) E Cn nC2 i σh 1 1 –1 –1 –1 ∫+ A A5/2 + 1 = – n Kt Ao – 5/2 + 1 – 2 [A– 3/2]AAto = – n Kt If below principle axis is + ve sign, there is A 3 notation in irreducible representation. 2 [[At]– 3/2 – [A0]– 3/2] = n Kt If below C2 : – ve, then 2 3 If below σh : – ve then double dash ” [At]– 3/2 = 3 ([A]–0 3/2 + n Kt) If below i : – ve then u 2 Thus, overall irreducible representation = A2\"4

26 | CSIR-Chemical Sciences (June-2013) 104. (C) 41Nb : [Kr] 4d4 5s1 JAX = (difference between any two adja- 105. (B) cent peaks) × (instrument frequency) 4d 5s = (4·72 – 4·60) ppm × 100 MHz +2 −1 0 −1 −2 0 = 0·12 ppm × 100 MHz 2S + 1 = 6 = 12 Hz L = +2×1=2D δAX = 4·72 + 4·60 + 1·12 + 1·00 4 Thus, ground state term symbol = 6D 106. (A) = 11·44 = 2·86 ppm 4 107. (C) 1 2 3 4 2134⎪⎪⎪⎪⎪ x 1 0 1 ⎪⎪⎪⎪⎪ 12 112. (A) nj = g e– (Ej – Ei)/KBT 1 x 1 0 43 ni 0 1 x 1 1 0 1 x Since g = 2 (doubly-degenerate) and Ej – Ei = 2 units α–E KBT = 1 unit β ... x = Thus, nj = 2e– 2 ni ⇒ E = α – βx 113. (D) ∴ Huckel determinant 114. (B) Ionic strength is given as α–E i ⎪ ⎪β ⎪⎪⎪ ⎪⎪⎪= 1 0 1 I= 1 ∑ mi zi2 2 1 α–E 1 0 i=0 β α–E mi is molality of ion and zi is charge 0 1 β 1 Thus, for NaCl ⎪⎪ ⎪⎪1 α–E Z+ = 1, β 0 1 Z– = 1, ⎪⎪⎪⎪⎪ α – E β 0 β ⎪⎪⎪⎪⎪ m+ = 0·01 mol = m– β α–E β 0 0 α–E β I = 1 (0·01 × 12 + 0·01 × 12) β β β – 2 = 0 0·02 = 2 = 0·010 mol kg– 1 α E For Na2SO4 108. (C) Z+ = 1, Z– = 2 120 mL of 0·05 M CH3COOH m+ = 2 × 0·01 = 0·02, 40 mL of 0·05 M NaOH m– = 0·01 molal moles of H+ = 0·05 × 120 = 6·0 mM I = 1 (0·02 × 12 + 0·01 × 22) 2 – moles of OH = 0·05 × 40 = 2·0 mM remaining H+ = 4 mM = 0·02 + 0·04 2 [H+]net = 4 × 10– 3 3= 4 = 1 160 × 10– 160 40 = 0·030 mol kg– 1 = 0·025 mole l– 1 115. (B) 116. (B) pH = – log (2·5 × 10– 2) O + ne– R = 2 + log 2·5 = 2·39 E = E° – RT ln [R] nF [O] 109. (B) 110. (B) 111. (C) AX system gives 4 lines at 4·72, 4·6, (According to Nernst equation) 1·12 and 1·0 ppm 100 MHz instrument is given. Thus, E – E° = – RT ln [R] nF [O]

CSIR-Chemical Sciences (June-2013) | 27 nF (E – E°) = – ln [R] OH H+OH RT [O] ⇒ [O] = – nF Ph H [R] Me Ph e RT (E – E°) 117. (C) We know Bragg equation that Major 2d sin θ = n λ Cram product (front side attack) where n = order of diffraction 124. (D) 125. (D) Given that 2d sin θ = λ …(1) 126. (B) Among all the options, only option (B) have aromaticity character after charge A plane of spacing 2d shows distribution or keeps in that position and this leads to highest dipole moment in its. 2 . 2d sin φ = λ From (1), 4d sin φ = 2d sin θ sin φ = sin θ 2 ( )⇒ sin θ φ = sin– 1 2 aromatic compound 118. (D) 119. (A) 120. (D) (C) The broadband decoupled 13C NMR spectrum. 121. (D) O O OH OH Me H3PO4 CH2 MeMgBr 4 432 OH 5 67 HH 4 Et2O 1 8 3 52 3 97 25 2 10 6 3 8 13C NMR 43 1 HO 1 2 3 4 4 1 2 3 4 5 OH Signal Major CH2 122. (A) +5 2 Five signals Ten signals OH 5 4 3 Minor 127. (C) 128. (A) 129. (C) 130. (A) 1H NMR spectral data is given as 5 13C NMR Signal O δ 7·80 (2H, d, J = 8Hz) aromatic m-hydrogens w.r.t. OH MnO2 OH > C = 0 group Selectively (CH2OH)2/P-TSA 6·80 (2H, d, J = 8 Hz) oxidises benzylic protecting group for Ketone aromatic o–hyrogens w.r.t. alcohol into Ketone > C = 0 group O O OO 4·10 (2H, q, J = 7·2 Hz) 123. (A) CHO OH PCC – CH2 environment adjacent to electronegative group Oxidises − OH into − CHO 1·25 (3H, t, J = 7·2 Hz) HO J value indicates that coupling Me Back +2 − with – CH2 group 2·4 (3H, S) adjacent to > C = 0 group or CHO MgBr Ph H Me Ph H2O benzylic position Group ‘L’ must trans/anti Less hindered 7.80 d HO O to – C = 0 group side-major H 2.4 H OH OH q CH3 H Ph t EtO Ph Me CH2 O S CH3 Minor 4.10 Hd 1.25 H 6.80 131. (C) 132. (A)

28 | CSIR-Chemical Sciences (June-2013) 133. (C) CN O 139. (B) CN 140. (D) OTs Δ K H (base) O H OTs THF O CN O −H2 O H OH Δ HO 134. (B) HH O O H Ph O N OH 141. (A) 142. (B) N 143. (D) H OO O Ph O 144. (A) hυ H εH Dipeptide Et2O ε γ α O NT-II clearage δ β H2 Ph CH2 H + CO2 + H2N OH abstraction due to ring cyclisation 10% Pd/c Toluene Carbon N OH reduction OH dioxide H Ph O H1 6 543 2 Phe-Leu H 135. (B) 136. (A) ring cyclisation 137. (A) Let overall phenol is x (6-membered) is important in this case Thus, overall formation of p-hydroxyace- tanilide = x × 70 × 90 × 90 × 100 100 100 100 = x × 567 × 100 = 56·7% hυ disrotation 1000 cyclization And, overall formation of o–hydroxyaceta- nilide O3, Me2S = x × 25 × 90 × 90 × 100 EtOOC reductive 100 100 100 EtOOC oxidation = 20·25% Ph3P = CHCOOEt O 138. (D) F excess witting HF reaction O HH 145. (D) H H H hυ 1, 2–difluoroethane most stable conformation disrotation due to φ = 60° (staggered) H H Me Me H OH HO H Δ disrotation cis-product Stereochemistry is important H in pericyclic reactions H OH HO H Me Me θ = 60° dl–2, 3–butanediol H (gauche/staggered) cis-product

CSIR UGC-NET/JRF Exam., December 2013 Solved Paper Chemical Sciences PART–A Which of the following statements is correct ? 1. A cylinder of radius 1 cm and height 1 cm is (A) The speed is never zero broken into three pieces. Which of the (B) The acceleration is zero once on the path following must be true ? (C) The distance covered initially increases and then decreases (A) At least one piece has volume equal to (D) The car comes back to its initial position 1 cm3 once (B) At least two pieces have equal volumes (C) At least one piece has volume less than 1 cm3 6. If a + b + c + d + e = 10 (all positive numbers), then the maximum value of a × b (D) At least one piece has volume greater × c × d × e is— than 1 cm3 2. For real number x and y, x2 + (y – 4)2 = 0. (A) 12 (B) 32 Then the value of x + y is— (C) 48 (D) 72 (A) 0 (B) 2 7. How many nine-digit positive integers are there, the sum of squares of whose digits is (C) ⎯√ 2 (D) 4 2? 3. Every time a ball falls to ground, it bounces back to half the height it fell from. A ball is (A) 8 (B) 9 dropped from a height of 1024 cm. The (C) 10 (D) 11 maximum height from the ground to which it can rise after the tenth bounce is— 8. A circle of radius 7 units lying in the fourth quadrant touches the x-axis at (10, 0). The (A) 102·4 cm (B) 1·24 cm centre of the circle has coordinates— (C) 1 cm (D) 2 cm 4. A farmer gives 7 full, 7 half-full and 7 empty (A) (7, 7) (B) (– 10, 7) bottles of honey to his three sons and asks (C) (10, – 7) (D) (7, – 7) them to share these among themselves such 9. One of the four—A, B, C and D committed a crime. A said, “I did it.” B said, “I didn’t.” C that each of them gets the same amount of said, “B did it.” D said, “A did it.” Who is lying ? honey and the same number of bottles. In how many ways can this be done ? (bottles cannot be distinguished otherwise, they are sealed and cannot be broken)— (A) A (B) B (A) 0 (B) 1 (C) C (D) D (C) 2 (D) 3 1 1, × 2×3 5. A car is moving along a straight track. Its 10. What is the arithmetic mean of 1 2 , speed is changing with time as shown. 1 , 1 , … , 1 3 × 4 4 × 5 100 × 101 ? Speed (A) 0·01 (B) 1 101 49 1 50 + 50 1 51 × × O Time (C) 0·00111… (D) 2

2 | CSIR-Chemical Sciences (D-13) 11. A circle circumscribes identical, closepacked 17. Consider the sequence of ordered sets of circles of unit diameter as shown. What is the total area of the shaded portion ? natural numbers— {1}, {2, 3}, {4, 5, 6}, … What is the last number in the 10th set ? (A) 10 (B) 19 (C) 55 (D) 67 18. A student buys a book from an online shop at (A) 2 (B) 2π 20% discount. His friend buys another copy (C) 1/2 (D) π/2 of the same book in a book fair for 192 paying 20% less than his friend. What is the 12. There are 2 hills, A and B, in a region. If hill full price of the book ? A is located N30°E of hill B, what will be the (A) 275 (B) 300 direction of hill B when observed from hill (C) 320 (D) 392 A ? (N 30°E means 30° from north towards 19. 366 players participate in a knock-out east). tournament. In each round all competing (A) S 30°W (B) S 60°W players pair together and play a match, the (C) S 30°E (D) S 60°E winner of each match moving to the next 13. What is the next number in the following round. If at the end of a round there is an odd sequence ? number of winners, the unpaired one moves 39, 42, 46, 50, …… to the next round without playing a match. (A) 52 (B) 53 What is the total number of matches played ? (C) 54 (D) 55 (A) 366 (B) 282 14. What is the perimeter of the given figure, (C) 365 (D) 418 where adjacent sides are at right angles to each other ? 20. What does the diagram establish ? 5 cm 4 cm 1cm Note—The diagram is a circle inside a square. (A) 20 cm (A) π > 3 (B) 18 cm (C) 21 cm (B) π ≥ 2√⎯ 2 (D) Cannot be determined (C) π < 4 15. Three fishermen caught fishes and went to (D) π is closer to 3 than to 4 sleep. One of them woke up, took away one fish and 1/3rd of the remainder as his share, PART–B without others’ knowledge. Later, the three of them divided the remainder equally. How 21. The boiling point of a solution of non-volatile many fishes were caught ? solid is higher than that of the pure solvent. It always indicates that— (A) 58 (B) 19 (A) the enthalpy of the solution is higher than that of the pure solvent (C) 76 (D) 88 (B) the entropy of the solution is higher than that of the pure solvent 16. (25 ÷ 5 + 3 – 2 × 4) + (16 × 4 – 3) = (C) the Gibbs free energy of the solution is higher than that of the pure solvent (A) 61 (B) 22 (D) the internal energy of the solution is higher than that of the pure solvent (C) 41/24 (D) 16

CSIR-Chemical Sciences (D-13) | 3 22. When Frenkel defects are created in an (B) band gap of gold changes with size of the otherwise perfect ionic crystal, the density of nanoparticle the ionic crystal— (C) gold in nanocrystalline form undergoes (A) increases transmutation to other elements (B) decreases (D) colloidal suspensions diffract light (C) remains same 29. A reactor contains a mixutre of N2, H2 and NH3 in equilibrium (KP = 3·75 atm– 2). If (D) oscillates with the number of defects sufficient he is introduced into the reactor to 23. The correct thermodynamic relation among double the total pressure, the value of KP at the following is— the new equilibrium would be— ( ) ( )(A)∂U (B) ∂H =–P ∂V =–P ∂V (A) 0·94 atm– 2 (B) 3·75 atm– 2 S S (C) 7·50 atm– 2 (D) 15·00 atm– 2 ( ) ( )(C)∂G ∂A ∂V =–P (D) ∂V =–S 30. Electrolysis of an aqueous solution of 1·0 M NaOH results in— S S (A) Na at the cathode and O2 at the anode 24. The molecule in which the bond order (B) H2 at the cathode and O2 at the anode increases upon addition of an electron is— (C) Na and H2 at the cathode and O2 at the (A) O2 (B) B2 anode (C) P2 (D) N2 (D) O2 at the cathode and H2 at the anode 25. The volume of a gas adsorbed on a solid surface is 10·0 mL, 11·0 mL, 11·2 mL, 14·5 31. In a potentiometric titration, the end point is mL and 22·5 mL at 1·0, 2·0, 3·0, 4·0 and 5·0 obtained by observing— atm. pressure, respectively. These data are (A) change in colour best represented by— (B) jump in potential (C) increase in current (A) Gibbs’s isotherm (D) increase in turbidity (B) Langmuir isotherm (C) Freundlich isotherm (D) BET isotherm 32. The orbital with two radial and two angular nodes is— 26. A compound of M and X atoms has a cubic unit cell. M atoms are at the corners and body (A) 3p (B) 5d centre position and X atoms are at face centre positions of the cube. The molecular formula (C) 5f (D) 8d of the compound is— 33. The energy of 2s and 2p orbitals is the same (A) MX (B) MX2 for— (C) M3X2 (D) M2X3 (A) Li (B) Li2+ (C) Be2+ (D) H– 27. The angle at which the first order Bragg 34. If a homonuclear diatomic molecule is reflection is observed from (110) plane in a simple cubic unit cell of side 3·238 Å, when oriented along the Z-axis, the molecular chromium Kα radiation of wavelength 2·29 Å is used, is— orbital formed by linear combination of Px orbitals of the two atoms is— (A) σ (B) σ* (A) 30° (B) 45° (C) π (D) δ (C) 60° (D) 90° 35. In the mechanism of reaction, H2 + Br2 → 2HBr, the first step is— 28. Michael Faraday observed that the colour of (A) dissociation of H2 into H-radicals colloidal suspensions of gold nanoparticles (B) dissociation of Br2 into Br-radicals changes with the size of the nanoparticles. (C) reaction of H-radical with Br2 This is because— (D) reaction of Br-radical with H2 (A) gold forms complex with the solvent

4 | CSIR-Chemical Sciences (D-13) 36. The cell voltage of Daniel cell 42. Amongst the following, the compound which has the lowest energy barrier for the cis-trans [Zn]ZnSO4(aq)|| CuSO4(aq)[Cu] is 1·07 V. If isomerisation is— reduction potential of Cu2+ [Cu is 0·34 V, the (A) (B) reduction potential of Zn2+ |Zn is— (A) 1·41 V (B) – 1·41 V (C) 0·73 V (D) – 0·73 V 37. According to Arrhenius equation (k = rate (C) (D) constant and T = temperature)— (A) In k decreases linearly with 1/T 43. For estrone, among the statements A-C, the (B) In k decreases linearly with T correct ones are— (C) In k increases linearly with 1/T (D) In k increases linearly with T 1. it is a steroidal hormone 38. The IUPAC name of the compound given 2. it has two hydroxyl groups below is— 3. it has one ketone and one hydroxyl HO groups OH (A) 1, 2 and 3 (B) 1 and 2 Cl (C) 1 and 3 (D) 3 and 3 (A) (2E, 4E)-3-chlorohexa-2, 4-diene-1, 6- diol 44. An organic compound having the molecular (B) (2Z, 4E)-3-chlorohexa-2, 4-diene-1, 6- formula C10H14 exhibited two singlets in the diol 1H NMR spectrum, and three signals in the 13C NMR spectrum. The compound is— (C) (2Z, 4Z)-4-chlorohexa-2, 4-diene-1, 6- diol (D) (2E, 4Z)-4-chlorohexa-2, 4-diene-1, 6- diol 39. A suitable organocatalyst for enantioselective (A) (B) synthesis of Wieland-Miescher ketone (A) is— OO O (C) (D) OO 45. The major product formed in the following A (optically active) reaction is— (A) (–)-proline (B) (+)-menthone Cl (C) guanidine (D) (+)-BINOL 40. The major product formed in the reaction of N NaOMe styrene with an excess of lithium in liquid ammonia and t-butylalcohol is— MeO MeOH Cl (A) (B) MeO Cl OMe (A) N N (C) (D) Cl (B) Cl MeO MeO 41. In the IR spectrum of p-nitrophenyl acetate, Cl MeO the carbonyl absorption band appears at— N (C) N (D) Cl (A) 1660 cm– 1 (B) 1700 cm– 1 MeO MeO (C) 1730 cm– 1 (D) 1770 cm– 1 OMe

CSIR-Chemical Sciences (D-13) | 5 46. L-DOPA is used for the treatment of— (A) diastereotopic in I and enantiotopic in II (A) tuberculosis (B) diastereotopic in II and enantiotopic in I (B) Parkinson’s disease (C) diastereotopic in both I and II (C) diabetes (D) enantiotopic in both I and II (D) cancer 51. Deuterium kinetic isotope effect for the 47. The following reaction proceeds through a— following reaction was found to be 4·0. Based Δ on this information, mechanism of the reaction is— O CHO H(D) (A) 1, 3-sigmatropic rearrangement +− KOH (B) 2, 3-sigmatropic rearrangement MeOH (C) 3, 3-sigmatropic rearrangement NMe3 (D) 3, 5-sigmatropic rearrangement (A) E1 (B) E2 (C) E1CB (D) Free radical 48. The major product formed in the following 52. The number of nodes present in the highest reaction is— occupied molecular orbital of 1, 3, 5- hexatriene in its ground state is— OMe OO 1. heat + 2. H2SO4 heat (A) one (B) two NH2 OMe (C) three (D) four 53. The major product formed in the following reaction is— OMe OMe O O O MeO (A) N (B) N Zn(BH4)2 OMeH OMeH OH OH OMe (A) MeO (B) MeO OMe (C) (D) NH2 O OH N HO OMe (C) (D) HO OMe 49. The constituent amino acids present in the 54. For acylation with acetic anhydride/ following dipeptide, respectively, are— triethylamine, and oxidation with chromium trioxide of the trans- and cis-alcohols A and COOH NH2 B, the correct statement is— H N OH OH H2N O COOH (A) (R)-aspartic acid and (S)-lysine AB (B) (S)-aspartic acid and (R)-lysine (C) (R)-glutamic acid and (S)-arginine (A) A undergoes acylation as well as (D) (S)-glutamic acid and (S)-arginine oxidation faster than B 50. The two benzylic hydrogens HA and HB in the (B) B undergoes acylation as well as compounds I and II, are— oxidation faster than A HA HA (C) A undergoes acylation faster than B, N HB N HB whereas B undergoes oxidation faster than A Ph Ph (I) (II) (D) B undergoes acylation faster than A, whereas A undergoes oxidation faster than B

6 | CSIR-Chemical Sciences (D-13) 55. Patients suffering from Wilson’s disease 63. In a cluster, H3CoRu3(CO)12, total number of have— electrons considered to be involved in its formation is— (A) low level of Cu-Zn superoxide dismutase (A) 57 (B) 60 (B) high level of Cu-Zn superoxide dismutase (C) 63 (D) 72 (C) low level of copper-storage protein, ceruloplasmin 64. Among the following the correct acid strength trend is represented by— (D) high level of copper-storage protein, ceruloplasmin (A) [Al(H2O)6] 3 + < [Fe(H2O)6] 3 + < [Fe(H2O)6]2+ 56. High dose of dietary supplement ZnSO4 for the cure of Zn deficiency— (B) [Fe(H2O)6] 3 + < [Al(H2O)6] 3 + < [Fe(H2O)6]2+ (A) reduces myoglobin (C) [Fe(H2O)6] 2 + < [Fe(H2O)6] 3 + < (B) increases iron level in blood [Al(H2O)6]3+ (C) increases copper level in brain (D) [Fe(H2O)6] 2 + < [Al(H2O)6] 3 + < [Fe(H2O)6]3+ (D) reduces copper, iron and calcium levels in body 65. An octahedral metal ion M2+ has magnetic moment of 4·0 B.M. The correct combination 57. The bond order of the metal-metal bond in the of metal ion and d-electron configuration is dimeric complex [Re2Cl4(PMe2Ph)4]+ is— given by— (A) 4·0 (B) 3·5 (A) Co2+, t2g5 eg2 (C) 3·0 (D) 2·5 (B) Cr2+, t2g4 eg2 58. Among the molten alkali metals, the example (C) Mn2+, t2g3 eg1 of an immiscible pair (in all proportions) is— (D) Fe2+, t2g4 eg2 (A) K and Na (B) K and Cs (C) Li and Cs (D) Rb and Cs 59. Among the following, an example of a 66. The reaction of FeCl3·6H2O with SOCl2 hypervalent species is— yields— (A) BF3·OEt2 (B) SF4 (A) FeCl2(s), SO2(g) and HCl(g) (C) [PF6]– (D) Sb2S3 (B) FeCl3(s), SO2(g) and HCl(I) (C) FeCl2(s), SO3(g) and HCl(g) 60. Commonly used scintillator for measuring (D) FeCl3(s), SO2(g) and HCl(g) γ-radiation is— (A) Nal(Al) (B) NaI(Tl) (C) CsI(Tl) (D) CsI(Al) 67. Treatment of ClF3 with SbF5 leads to the formation of a/an— 61. A sample of Aluminium ore (having no other metal) is dissolved in 50 mL of 0·05 (A) polymeric material MEDTA. For the titration of unreacted EDTA, 4 mL of 0·05 M. MgSO4 is required. (B) covalent cluster The percentage of Al in the sample is— (C) ionic compound (A) 27 (B) 31 (D) Lewis acid-base adduct (C) 35 (D) 40 68. According to VSEPR theory, the geometry (with lone pair) around the central iodine in 62. Which of the following is not suitable as I3+ and I3– ions respectively are— catalyst for hydroformylation ? (A) HCo(CO)4 (A) tetrahedral and tetrahedral (B) HCo(CO)3PBu3 (C) HRh(CO)(PPh3)3 (B) trigonal bipyramidal and trigonal (D) H2Rh(PPh3)2Cl bipyramidal (C) tetrahedral and trigonal bipyramidal (D) tetrahedral and octahedral

CSIR-Chemical Sciences (D-13) | 7 69. Two tautomeric forms of phosphorus acid F are— FF OH O OH F Al F (A) P OH and P (C) FF F FF OH H OH Al F Al OH O H FF FF (B) P OH and P F Al F H H OH FF F OH O O FFF F (C) P OH and H P (D) Al F Al F Al F Al OH H H F F F O OH OO FFF F (D) P and H P 74. Oxidised form of enzyme catalase (structure A), prepared by the reaction of [Fe(P)]+ (P = HO OH HO OH porphyrin) with H2O2, has green color 70. The reason for the chemical inertness of because— gaseous nitrogen at room temperature is best given by its— NN Fe (A) high bonding energy only NON (B) electronic configuration A (substituents on ring are removed (C) HOMO-LUMO gap only for clarity) (D) high bond energy and HOMO-LUMO (A) oxidation state of iron changed from FeIII gap to FeIV PART–C (B) porphyrin ring is oxidized by one 71. The reactive position of nicotinamide adenine electron dinucleotide (NAD) in biological redox reactions is— (C) π - π* transition appears in the visible (A) 2-position of the pyridine ring region (B) 6-position of the pyridine ring (C) 4-position of the pyridine ring (D) FeIV is coordinated with anionic (D) 5-position of the pyridine ring tyrosinate ligand in axial position 72. The electrophile Ph3C+ reacts with [(η5 75. Substitution of L with other ligands will be – C5H5)Fe(CO)2 (CDMe2)]+ to give a product easiest for the species— A. The product A is formed because— (A) Fe is oxidised (A) MLn Me Me (B) alkyl is substituted with Ph3C (B) Me (C) Fe-Ph bond is formed (D) alkyl is converted to alkene Me MLn Me 73. The solid state structure of aluminum fluoride (C) MLn (D) is— MLn FF F 76. The ground state terms of Sm3+ and Eu3+, (A) Al Al respectively, are— FF F (B) Al F Al F Al F Al (A) 7F0 and 6H5/2 (B) 6H5/2 and 7F0 F F F (C) 2F5/2 and 5I4 (D) 7F6 and 2F7/2

8 | CSIR-Chemical Sciences (D-13) 77. The orbital interactions shown below (C) Mn is in + 3 oxidation state and 1 Cr is in represent + 2 and the other is in + 3 state sp3 orbitals (D) Cr3+ will have a LFSE in the octahedral site whereas the Mn2+ ion will not E 83. Reaction of Ph2PCH2CH2PPh2 with s orbitals [RHCl(CO)2]2 in a 2 : 1 molar ratio gives a crystalline solid A. The IR spectrum of (A) CH3—Al interactions in Al2(CH3)6 complex A shows vCO at 1985 cm– 1. The 31P{1H} NMR spectrum of A consists of two (B) B—H interactions in B2H6 doublets of doublets of equal intensities (103Rh is 100% abundant and I = 1/2). The (C) CH3—Li interactions in Li4(CH3)4 structure of complex A is— (D) CH3CH2—Mg interactions in PPh2 Cl EtMgBr·(OEt2)2 (A) Rh 78. Compounds K2Ba[Cu(NO2)6] (A) and P CO Cs2Ba[Cu(NO2)6] (B) exhibit tetragonal Ph2 elongation and tetragonal compression, respectively. The unpaired electron in A and Ph2 Ph2 P CO P B are found respectively, in orbitals— (B) Rh2 Cl PP (A) dz2 and dx2.y2 (B) dx2. 2 and dz2 Ph2 Ph2 y (C) dz2 and dz2 (D) dx2.y2 and dx2.y2 79. Among the following, the correct statement Ph2 Cl is— P CO (A) CH is isolobal to Co(CO)3 (B) CH2 is isolobal to Ni(CO)2 (C) Rh PPh2 (C) CH is isolobal to Fe(CO)4 PP (D) CH2 is isolobal to Mn(CO)4 Ph2 Ph2 80. In Mossbauer experiment, a source emitting Ph2P O PPh2 at 14·4 Kev (3·48 × 1018 Hz) had to be moved C Cl towards absorber at 2·2 mm s– 1 for resonance. (D) Cl Rh C Rh The shift in the frequency between the source Ph2P O PPh2 and the absorber is— 84. In a polarographic estimation, the limiting currents (μA) were 0·15, 4·65, 9·15 and 27·15 (A) 15·0 MHz (B) 20·0 MHz when concentrations (mM) of Pb(II) were 0, 0·5, 1·0 and 3·0, respectively. An unknown (C) 25·5 MHz (D) 30·0 MHz solution of Pb(II) gives a limiting current of 13·65 μA. Concentration of Pb(II) in the 81. In the atomic absorption spectroscopic estimation of Fe(III) using O2/H2 flame, the unknown is— absorbance decreases with the addition of— (A) 1·355 mM (B) 1·408 mM (A) CO32– (B) SO42– (C) 1·468 mM (D) 1·500 mM (C) EDTA (D) Cl– 85. The gases SO2 and SO3 were reacted separa- tely with ClF gas under ambient conditions. 82. MnCr2O4 is likely to have a normal spinel The major products expected from the two structure because— reactions respectively, are— (A) Mn2+ will have a LFSE in the octahedral (A) SOF2 and ClOSO2F site whereas the Cr3+ will not (B) SOF2 and SO2F2 (C) SO2ClF and SO2F2 (B) Mn is in + 2 oxidation state and both the Cr are in + 3 oxidation state (D) SO2ClF and ClOSO2F

CSIR-Chemical Sciences (D-13) | 9 86. In a specific reaction, 91. Among the following the correct combination hexachlorocyclotriphosphazene, N3P3Cl6 was of complex and its colour is— reacted with a metal fluoride to obtain mixed Complex Colour halo derivatives namely N3 P3Cl5F(1), (A) [Co(CN)4]2– Red N3P3Cl4F2(2), N3P3Cl3F3(3), N3P3Cl2F4(4), (B) [CoCl4]2– Orange N3P3ClF5(5). (C) [Co(NCS)4]2– Blue Compositions among these which can give (D) [CoF4]2– Yellow isomeric products are— (A) 1, 2 and 3 92. On reducting Fe3(CO)12 with an excess of (B) 2, 3 and 4 sodium, a carbonylate ion is formed. The ion (C) 3, 4 and 5 is isoelectronic with— (D) 5, 1 and 2 (A) [Mn(CO)5]– (B) [Ni(CO)4] 87. Xenon forms several fluorides and (C) [Mn(CO)5]+ (D) [V(CO)6]– oxofluorides which exhibit acidic behavior. The correct sequence of descending Lewis 93. The correct statement for ozone is— acidity among the given species is represented (A) It absorbs radiations in wavelength region 290–320 nm by— (B) It is mostly destroyed by NO radical in (A) XeF6 > XeOF4 > XeF4 > XeO2F2 atmosphere (B) XeOF4 > XeO2F2 > XeOF4 > XeF6 (C) XeF4 > XeO2F2 > XeOF4 > XeF6 (C) It is non-toxic even at 100 ppm level (D) XeF4 > XeF6 > XeOF4 > XeO2F2 (D) Its concentration near poles is high due 88. Among the following clusters, to its paramagnetic nature A = [(H)Co6(CO)15]–, B = [(H)2Os6(CO)18], C = [(H)2Os5(CO)16] 94. The most appropriate structure for the H is encapsulated in— complex [Pt2(NH3)2(NCS)2(PPh3)2] is— (A) A only (B) B only (A) H3N Pt NCS Pt PPh3 H3N SCN PPh3 (B) Ph3P Pt NCS Pt NH3 H3N SCN PPh3 (C) B and C only (C) H3N Pt NCS Pt PPh3 (D) A and B only Ph3P SCN NH3 89. Number of isomeric derivatives possible for (D) H3N Pt NCS Pt NH3 the neutral closo-carborane, C2B10H12 is— Ph3P NCS PPh3 (A) three (B) two 95. For higher boranes 3c-2e ‘BBB’ bond may be (C) four (D) six a part of their structures. In B5H9, the number of such electron deficient bond(s) present 90. The correct statement regarding the terminal/bridging CO groups in solid is/are— CO4(CO)12 and Ir4(CO)12 is— (A) four (B) two (A) both have equal number of bridging CO groups (C) zero (D) one (B) number of bridging CO groups in 96. An organic compound (C8H10O2), which does CO4(CO)12 is 4 not change the color of ferric chloride solution, exhibited the following 1H NMR (C) the number of terminal CO groups in spectral data : δ 7·3 (1 H, t, J = 8 Hz), 7·0 (1 Co4(CO)12 is 8 H, d, J = 8 Hz), 6·95 (1 H, s), 6·9 (1 H, d, J = (D) the number of bridging CO groups in 8 Hz), 5·3 (1 H, brs, D2O exchangeable), 4·6 Ir4(CO)12 is zero (2 H, s), 3·9 (3 H, s). Structure of the compound is—

10 | CSIR-Chemical Sciences (D-13) MeO 100. The major product formed in the sulfuric (A) acid mediated rearrangement of the sesquiterpene santonin A is— HO MeO OH H H2SO4 (B) O MeO OH HO (C) OMe O HO H (D) 97. In the following reaction, the intermediate (A) H and the major product A are— O HO O N CHCl3/aq. NaOH A H n-Bu4N+ Br− HO (B) H HO O (A) CHCl and N Cl (C) HO H (B) CHCl and Cl HO N O (C) CCl2 and N Cl (D) H (D) CCl2 and Cl HO H O N O 98. Methyl 4-oxopentanoate exhibited signals at δ 101. In the following transformation, the reagent 208, 172, 51, 37, 32 and 27 ppm in its 13C A and the major product B, respectively, NMR spectrum. The signals due to the are— methoxy, Cl, C4 and C5 carbons are— A + Ac2O/Δ B (A) OMe - 32; Cl - 208; C4 - 172; C5 - 51 NN (B) OMe - 51; Cl - 208; C4 - 172; C5 - 32 O− (C) OMe - 32; Cl - 172; C4 - 208; C5 - 51 SO3H OAc (A) and (D) OMe - 51; Cl - 172; C4 - 208; C5 - 32 Cl N 99. The major product formed in the following SO3H reaction sequence is— (B) and OAc 1. BH3, THF; H2O2/NaOH Cl N 2. Ph3P, DEAD; 4-nitrobenzoic acid CO3H 3. K2CO3, MeOH (DEAD = diethyl azodicarboxylate) (C) and OAc OH OH Cl N (A) (B) OH OH CO3H (C) (D) (D) and OAc Cl N

CSIR-Chemical Sciences (D-13) | 11 102. The peptide A on reaction with I-fluoro-2, 4- 106. The major product formed in the following dinitrobenzene followed by exhaustive reaction sequence is— hydrolysis gave phenylalanine, alanine, serine and N-(2,4-dinitrophenyl)glycine. On BnO OH 1. KH, THF the other hand, peptide A after two cycles of SiMe3 2. m-CPBA Edman degradation gave Phe-Ser as the product. The structure of the peptide A is— (A) BnO OH OH OH H (A) Phe-Ser-Ala-Gly H (B) BnO O O (B) Phe-Ser-Gly-Ala HH (C) Gly-Ala-Phe-Ser (D) Ala-Gly-Phe-Ser 103. The compound B (labeled) is precursor for HH biosynthesis of the natural product A. The labeled carbons in the product A are— (C) BnO O O H (D) BnO H COOH OH OH Me14COOH OH 107. The major product formed in the following AB reaction sequence is— (A) C1, C3, C5 and Me HO O OH 1. PhCHO (1 equiv.), p-TSA (B) C2, C4, C6 and Me (C) C2, C4, C6 and COOH HO OH 2. NalO4 (D) C1, C3, C5 and COOH OH 3. NaBH4 (A) O OH 104. The major product formed in the following Ph O OH reaction sequence is— OH Ph 1. SOCl2 OH 2. NEt3 COOH HO Ph 3. CH2 = CH-OEt OO Ph Ph (B) (A) Ph (B) Ph Ph EtO O O HO OH OEt OO O (C) Ph Ph (D) Ph Ph HO OO Ph (C) OEt Ph (D) O OEt HO OH 105. The major product formed in the following 108. The major product formed in the following reaction sequence is— reaction sequence is— 1. TsNHNH2, EtOH 1. CH2I2‚ Zn-Cu‚ CH2Cl2 2. 2 eq. nBuLi, THF cis-but-2-ene-1,4-diol ⎯⎯2. P⎯CC⎯‚ C⎯H2C⎯l2→ O 3. DMF HO HO (A) CHO (B) (A) O (B) O H H CHO H COOH H CHO COOH CHO O (C) (D) H H (C) (D) NMe2 NMe2 O

12 | CSIR-Chemical Sciences (D-13) 109. The conditions A-B, required for the (C) HO OH HO OH following pericyclic reactions are— A Me B H Me & Me 90% 10% Me Me Me (D) HO OH HO OH H (A) A - Δ; B - Δ (B) A - hv; B - Δ & (C) A - hv; B - hv (D) A - Δ; B - hv 95% 5% 110. The major product formed in the following 112. The major product formed in the following reaction sequence is— reaction sequence is— Br H nBu3SnH,AlBN 1. N-bromosuccinimide, CH2Cl2 toluene, reflux OH 2. potassium t-butoxide COOEt OH (A) (B) O H COOEt (C) (D) (A) OH O OH 113. The number of π electrons participating and the pericyclic mode in the following reaction H COOEt are— (B) H OH heat H COOEt (C) OH H H COOEt (A) 4 and conrotatory (D) (B) 4 and disrotatory (C) 6 and conrotatory OH (D) 6 and disrotatory 111. Stereoselective reduction of the dione A 114. The major product formed in the following with a chiral reducing agent provides the reaction sequence is— corresponding diol B in 100% diastereoselectivity and 90% ce favouring O R,R-configuration. The composition of the product is— + 1. hv OAc 2. KOH, MeOH OO HO OH AB O O H (A) HO OH HO OH (A) (B) O & H OH H 90% 10% O (D) (B) HO OH HO OH OH H OMeH (C) & 5% H 95%

CSIR-Chemical Sciences (D-13) | 13 115. The major product formed in the following 118. The major product formed in the following photochemical reaction is— reaction sequence is— O hv O 1. H2O2, NaOH Ph 2. TsNHNH4 3. potassium t-butoxide 4. H2, Pd/CaCO3, quinoline (A) Ph (B) Ph (A) (B) OO O O H (C) Ph (D) Ph (C) (D) OO COOH 116. The most suitable reagent combination of A- H C, required in the following conversions are— OH OA O 119. The major product formed in the following B reaction sequence is— SePh C O 1. H2O2 O SO2 2. Heat (A) O (B) S (A) A = Li/liq. NH3 ; B = NaBH4, O2 O CeCl3.7H2O; C = H2, (PH3P)3RhCl H H (B) A = Li/liq. NH3 ; B = NaBH4, (C) (D) CeCl3.7H2O; C = H2, 10% Pd/C HO HO (C) A = NaBH4, CeCl3.7H2O; B = Li/liq. NH3; C = H2, (Ph3P)3RhCl 120. The major product B formed in the (D) A = NaBH4, CeCl3.7H2O; B = Li/liq. following reaction sequence, and overall NH3; C = H2, 10% Pd/C yield of its formation are— (S)-glutamic acid ⎯a⎯nil⎯ine→ A ⎯L⎯iA⎯IH4→ B 117. The major product formed in the following 180°C 60% 80% reaction sequence is— Ph 1. Hg(OAc)2, H2SO4 (A) NHPh and 48% O 2. BnNH2, heat N H (A) Ph H (B) NHPh and 70% N Ph N H O (C) N (B) Ph H NHPh and 48% O N Ph H (C) N Ph (D) NHPh and 70% Ph N H 121. Metallic gold crystallizes in fcc structure (D) N Ph with unit cell dimension of 4·00 Å. The Ph atomic radius of gold is— (A) 0·866 Å (B) 1·414 Å (C) 1·732 Å (D) 2·000 Å

14 | CSIR-Chemical Sciences (D-13) 122. A first order gaseous reaction is 25% 127. The molecule that has the smallest diffusion complete in 30 minutes at 227°C and in 10 coefficient in water is— minutes at 237°C. The activation energy of the reaction is closest to (R = 2 cal K– 1 (A) Glucose (B) Fructose mol–1)— (C) Ribose (D) Sucrose (A) 27 kcal mol– 1 (B) 110 kcal mol– 1 128. In the reaction between NO and H2 the (C) 55 kcal mol– 1 (D) 5·5 kcal mol– 1 following data are obtained— 123. The most probable value of r for an electron Experiment I : PH2 = Constant in 1s orbital of hydrogen atom is— PNO (mm of Hg) 359 300 152 (A) a0/2 (B) a0 – dPNO 1·50 1·03 0·25 dt (C) √⎯ 2 a0 (D) 3a0/2 Experiment II : PNO = Constant 124. The quantum mechanical virial theorem for a general potential V(x, y, z) is given by PH2 (mm of Hg) 289 205 147 x ∂V + y ∂V + z ∂V = 2(T) where T is the – dPH2 1·60 1·10 0·79 ∂x ∂y ∂z dt kinetic energy operator and < > indicates The orders with respect to H2 and NO are— expectation value. This leads to the following (A) 1 with respect to NO and 2 with respect to H2 relation between the expectation value of (B) 2 with respect to NO and 1 with respect kinetic energy and potential energy for a to H2 quantum mechanical harmonic oscillator (C) 1 with respect to NO and 3 with respect to H2 problem with potential V =21 kxX2 + 1 kyY2 + 2 1 kzZ2— (D) 2 with respect to NO and 2 with respect 2 to H2 (A) 〈T〉 = 〈V〉 (B) 〈T〉 = – 1 〈V〉 2 129. The angular momentum operator L^ y is— (C) 〈T〉 = 1 〈V〉 (D) 〈T〉 = – 〈V〉 2 h (A) – i (y∂/∂z – z∂/∂y) 125. Consider a particle in a one dimensional box (B) h (z∂/∂x – x∂/∂z) of length ‘a’ with the following potential— i V(x) = ∞ x<0 V(x) = ∞ x>a (C) – ih ∂ 2m ∂x V(x) = 0 0 ≤ x ≤ a/2 V(x) = V1 a/2 ≤ x ≤ a (D) h (z∂/∂x – y∂/∂y) i Starting with the standard particle in a box hamiltonian as the zeroth order Hamiltonian 130. Both NaCl and KCl crystallize with the fcc structure. However, the X-ray powder and the potential of V1 from ‘a/2’ to ‘a’ as a diffraction pattern of NaCl corresponds to perturbation, the first-order energy correction the fcc structure whereas, that of KCl corresponds to simple cubic structure. This to the ground state is— is because— (A) V1 (B) V1/4 (A) K+ and Cl– are isoelctronic (C) – V1 (D) V1/2 (B) Na+ and Cl– are isoelctronic 126. The energy for a single electron excitation in cyclopropenium cation in Huckel theory (C) K+ and Cl– are disordered in the crystal is— lattice (A) β (B) 2β (D) KCl has anti-site defects (C) 3β (D) 4β

CSIR-Chemical Sciences (D-13) | 15 131. Consider a two-dimensional harmonic 136. The chemical potential (μi) of the ith component is defined as— oscillator with potential energy V(x, y) = 1 2 ( ) ( )(A) μi = ∂U (B) μi = ∂H kxX2 + 1 kyY2. If Ψax (x) and Ψny (y) are the ∂ni T,P ∂ni T,P 2 ( ) ( )(C) μi = eigensolutions and Enx and Eny are the ∂A (D) μi = ∂G eigenvalues of harmonic oscillator problem ∂ni T,P ∂ni T,P in x and y direction with potential 1 kxX2 and 137. The transition that is allowed by x-polarized 2 light in trans-butadiene is— 1 kyY2, respectively, the wave functions and (The character table for C2h is given below) 2 C2h E C2 i σh eigenvalues of the above two-dimensional Ag 1 1 1 1 Rx, x2, y2, z2, xy harmonic oscillator problem are— (A) Ψnx,ny = Ψnx (x) + Ψny (y) Bg 1 – 1 1 – 1 Rx, Ry, xz, yz Au 1 1 –1 –1 z Enx,ny = Enx + Eny Bu 1 – 1 (A) 1Au → 1Au – 1 1 x, y (B) Ψnx,ny = Ψnx (x) Ψny (y) (C) 1Bu → 1Bg (B) 1Au → 1Bg (D) 3Bg → 1Ag Enx,ny = Enx Eny (C) Ψnx,ny = Ψnx (x) Ψny (y) Enx,ny = Enx + Eny (D) Ψnx,ny = Ψnx (x) + Ψny (y) 138. The masses recorded when a substance is weighted 4 times are 15·8, 15·4, 15·6 and Enx,ny = Enx Eny 16·0 mg. The variance (square of the standard deviation) is closest to— 132. An element exists in two crystallographic modifications with FCC and BCC structures. (A) 0·02 (B) 0·05 The ratio of the densities of the FCC and (C) 0·10 (D) 0·20 BCC modifications in terms of the volumes of their unit cells (VFCC and VBCC) is— 139. The heat capacity of 10 mol of an ideal gas at a certain temperature is 300 JK– 1 at (A) VBCC : VFCC (B) 2VBCC : VFCC constant pressure. The heat capacity of the (C) VBCC : 2VFCC (D) VBCC : ⎯√ 2 VFCC same gas at the same temperature and at 133. For an electronic configuration of two non- constant volume would be— equivalent π electrons [π1, π1], which of the (A) 383 JK– 1 (B) 217 JK– 1 following terms is not possible ? (C) 134 JK– 1 (D) 466 JK– 1 (A) 1Σ (B) 3Σ 140. Work (w) involved in isothermal reversible expansion from Vi to Vf of n moles of an (C) 3Δ (D) 3Φ ideal gas is— 134. Given γ (1H) ~– 2·7 × 108 T– 1S– 1. The (A) w = – nRT In (Vf /Vi) resonance frequency of a proton in magnetic (B) w = nRT (Vf /Vi) field of 12·6 T is close to (π = 3·14)— (C) w = – nRT (Vf/Vi) (A) 60 MHz (B) 110 MHz (D) w = nRT log(Vf/Vi) (C) 540 MHz (D) 780 MHz 135. The atomic masses of fluorine and hydrogen 141. The Maxwell’s relationship derived from the are 19·0 and 1·0 amu, respectively (1 amu = equation dG = VdP – SdT is— 1·67 × 10– 27 kg). The bond length of HF is 2·0 Å. The moment of inertia of HF is— ∂V( ) ( )(A)= ∂S (A) 3·2 × 10– 47 kg m2 ∂T ∂P T (B) 6·4 × 10– 47 kg m2 P (C) 9·6 × 10– 47 kg m2 ∂P( ) ( )(B)= ∂T (D) 4·8 × 10– 47 kg m2 ∂V ∂S P T

16 | CSIR-Chemical Sciences (D-13) ( ) ( )(C) ∂V ∂S 5. (B) We know that ∂T =– ∂P T acceleration = Change in velocity P Change in time ( ) ( )(D) ∂P ∂T ∂V =– ∂S P ∴ the graph represents, acceleration is 0 once T 142. The number of ways in which four on path. molecules can be distributed in two different energy levels is— 6. (B) 7. (A) 8. (C) It is clear from figure, the centre of circle (A) 6 (B) 3 is (10, –7) (C) 16 (D) 8 Y 143. Consider the cell— (−, +) (+, +) Zn | Zn2+ (a = 0·01) || Fe2+ (a = 0·001), Fe3+ X′ (10, 0) (a = 0·01) | Pt (−, −) X Ecell = 1·71 V at 25°C for the above cell. Y′ 7 The equilibrium constant for the reaction : (10,−7) (+, −) Zn + 2Fe3+ Zn2+ + 2Fe2+ at 25°C would 9. (C) A said, “I did it.” B said, “I didn’t.” be close to— C said, “B did it.” D said, “A did it.” (A) 1027 (B) 1054 Thus, it is clear from statements that C is (C) 1081 (D) 1040 lying. 10. (B) 11. (D) 144. The molecule with the smallest rotation 12. (A) Hill A is located N30° E of hill B. partition function at any temperature among the following is— WA (A) CH3-C ≡ C-H (B) H-C ≡ C-H 30° (C) H-C ≡ C-D (D) D-C ≡ C-D 145. The limiting molar conductivities of NaCl, SN NaI and RbI are 12·7, 10·8 and 9·1 mS m2 mol– 1, respectively. The limiting molar 30° conductivity of RbCl would be— BE (A) 32·6 mS m2 mol– 1 Thus, according to question, direction of hill (B) 7·2 mS m2 mol– 1 B from A = S 30° W. (C) 14·4 mS m2 mol– 1 13. (D) 14. (A) (D) 11·0 mS m2 mol– 1 15. (B) Of all the options given in question only Answers with Hints 19 is correct because (19 – 1) × 1 = 6, 7 fish is 3 1. (D) 2. (D) 3. (C) A ball is dropped from a height of 1024 taken by one fisherman. cm. ∴ 19 – 7 = 12 is divided equally among all the three with 4 fishes. 16. (A) (25 ÷ 5 + 3 – 2 × 4) + (16 × 4 – 3) h = (5 + 3 – 2 × 4) + (16 × 4 – 3) h/2 = (5 + 3 – 8) + (64 – 3) = (8 – 8) + (64 – 3) ∴ The maximum height from the ground to = 0 + 61 = 61 which it can rise after the tenth bounce 17. (C) {1}, {2, 3}, {4, 5, 6} 2 × 2 × 2 × 2 × 1024 × 2 × 2 × 2 × 2= 1 cm in every set 1 number increases proceeded to 2×2 words higher natural number. 4. (C) Thus, last number in 10th set is 55.

CSIR-Chemical Sciences (D-13) | 17 18. (B) 19. (C) 20. (C) 21. (B) 27. (B) According to Bragg’s Law, 22. (C) Since displacement of cation takes place, nλ = 2d sin θ so no effect on density. ∴ sin θ = nλ Cation in 2d Cation Vacancy interstitial site A+ B− A+ A+ (first order diffraction, n = 1) A+ sin θ = 2·29 Å × √⎯ 2 B− B− A+ B− 2 × 3·238 Å = 1 = sin 45° 2 A+ B− A+ B− A+ ⎛⎝⎜·.· a 2⎟⎠⎞ d = ⎯√⎯h⎯2 ⎯+⎯k⎯2⎯+⎯l B− A+ B− A+ B− θ = 45° 23. (A) From this rule, given in figure we know dU = TdS – PdV 28. (B) S ΔM P 29. (B) Mixture of N2, H2 & NH3, ΔV ΔG Kp = 3·75 atm– 2 V ΔA (−) T He is introduced into the reactor to double the (−) total pressure. The addition of inert gas Differentiating w.r.t. V at constant S N2 + 3H2 → 2NH3 ( )∴ ∂U = –P Kp ∝ (Ptotal) ∂V S ∴ new Kp, 24. (B) We know that for B2. M.O. diagram is ∴ Kp′ = (Ptotal) 3πg* 3πg* = Ptotal ∴ K′p = 3·75 = 3·75 atm– 2 (1σg2) (1σ4*2) (2σg2) (2σ4*2) 3σ4* 30. (B) Consider ionization as O–H( 3σg NaOH Na+(aq) + aq) 11 water also dissociates into ions O–H 3π4 3π4 H2O(l) H+ (aq) + (aq) Addition of an e– increases the value of bond Since reduction potential of H+ is higher than order due to extra e– will go into bonding- Na+ orbital. at cathode 1 H2 (g) 25. (D) H+ (aq) + e– → 2 26. (D) M atoms are at the corners at anode, 2H2O(l) → O2(g) + 4H+ (aq) + 4e– Net reaction, 1 = 8 × 8 NaOH (aq) + H2O(l) =1 → Na+ (aq) + O–H (aq) + 1 H2(g) + 1 O2(g) 2 2 and at body centre position = 1 31. (B) In potentiometric titration, EMF measure- ∴ M=2 ment is done. The end point is observed after X atoms are at face centre positions sharp jump in potential. = 1 × 6 Sharp 2 E Jump =3 ∴ X=3 V Thus, formula of compound = M2X3

18 | CSIR-Chemical Sciences (D-13) 32. (B) We know that 39. (A) angular node = value of l N H and radial node O O CO2H N CO2H O (−)-proline n–l–1 = 2 DMF solvent n–l = 3 O −H2O O* Condensation to ∴ 5d orbital have 2 radial & 2 angular nodes. form enamine Carbonyls are diastereotopic 33. (B) 34. (C) enamine elects to attack * 35. (B) H2 + Br2 → 2HBr carbonyl First step – initiation Br2 ⎯K→1 2B°r O O O CO2H Optically active dissociation of Br2 into B°r radicals N OH }H2 + Br• ⎯K→2 HBr + H° propagation H° + Br2 ⎯K3→ HBr + B°r 40. (C) excess Li excess Li in liq.NH3 in liq.NH3 HBr + B°r ⎯K→4 H° + Br2 – inhibition t-butyl alcohol t-butyl alcohol (not important) alkyl gp acts e – releasing group, facilitate the 2B°r ⎯K→5 Br2 – termination formation of ortho & meta hydrogenated product. 36. (D) Daniel cell Zn|ZnSO4(aq)||CuSO4(aq)|Cu7 E = 1·07 V O half cell reactions are 41. (D) O2N OC CH3 Zn(s) → Zn+2 + 2e– at anode P-nitro phenyl acetate Cu+ 2 + 2e– → Cu(s) at cathode ·.· E = Ec – Ea esters show band at 1760 – 1780 cm– 1 1·07 = 0·34 – Ezn+ 2/Zn 42. (C) 43. (C) ⇒ Ezn+ 2/Zn = – 0·73 V 20 + 12 – 14 37. (A) According to Arrhenius equation, 44. (A) C10H14, DBE = 2 = 4 K = Ae– Ea/RT lnK −Ea/R 1 two singlets in 1H NMR three signals in 13C NMR lnK 1 CH3 CH3 2 H 2 3 21 H 32 CH3 1/T 1CH3 lnK = ln A – Ea 45. (C) 46. (B) RT [3,3] sigmatropic hint – y = mx + c 47. (C) 3 Δ 2 1 O Formation of 3,3-bond lnK decreases linearly with T . 1 2 & breaking of 1,1-bond 3 38. (B) HH 2H 2 2H 1 H HO 2 4 6 C O CHO 24 1 3 5 OH Cl H 1 2Z Cl 1 1 4E H 2 (2Z, 4E)-3-chlorohexa-2, 4-diene-1, 6-dial 48. (C)

CSIR-Chemical Sciences (D-13) | 19 49. (A) 59. (C) F C2H5 B O COOH SF4 H NH2 hydrolysis sp3d N H2N O COOH FF C2H5 Coordination bond 43 COOH H4 3 NH2 F OH FF H + H2N P 1 1 H2N 2 2 COOH O FF F (R) - aspartic acid (S) - lysine P : 1s2 2s2 2p6 3s2 3p3 50. (B) N HA N HA 5 valence e–s Ph HB Ph HB This compound have hypervalent covalent (I) (II) bond. enantiotopic diastereotopic 60. (B) 61. (B) 62. (D) 2RCH = CH2 + 2CO + H2 ⎯C⎯O2⎯(CO⎯)8→ 51. (B) ‘C—H’ or ‘C—D’ bond cleavage in T.S.—Primary Kinetic isotope effect RCH2CH2CHO + RCH2 CH3 H(D) +− H (D) HC = O K OH + I − fast + I− NH3 In hydroformylation reaction, we need one or NMe3 MeOH two mole of CO from catalyst to give product rds-E2 by migratory insertion. Thus, the given option H2RH(PPh3)2Cl is not suitable as catalyst for 52. (B) 1 24 6 1,3,5-hexatriene hydroformylation. 35 63. (B) H3CORu3(CO)12 Total No. of e–s :— 3 + 9 + (8 × 3) + (12 × 2) 2 node = 60 (valence e–s) 1 node 64. (D) For aqua ion complex, acidity is more when (1) small size & (ii) greater charge 0 node 65. (A) Co+2, Octahedral, eg 53. (A) 54. (C) 55. (C) 56. (D) d7, system t2g 57. (B) dimeric complex [Re2Cl4(PMe2Ph)4]+ M.M. –~ 3·9 B.M. 66. (D) Oxidation state 67. (C) ClF3 + SbF5 → SbF6– + ClF2+ 2x – 4 = + 1 base acid ionic product x = 2·5 68. (C) ∴ Bond order = 1 I 2 I (B.M.O’s e–s – ABMO’s e–s) II = 1 (8 – 1) 2 = 3·5 I I Tetrahedral (I3+) Trigonal bipyromidal (I3−) 58. (C)

20 | CSIR-Chemical Sciences (D-13) 69. (A) Phosphorus Acid (ortho) For Sm+ 3, we have ground state term symbol, O OH 2S + 1LJ = 6H5/2 Eu+ 3, f 6 system PP reducing H OH OH OH 6 agent OH 2 S = = 3 two tautomeric forms 70. (D) 2S + 1 = 7, L = 3 (F), J = 0 HO ∴ 7F0 71. (C) 5 4 3 C NH2 77. (C) O 6 1 2 78. (B) K2Ba[Cu(NO2)6] OPO (A) O N O NN – tetragonal elongation (Zout) OPO O NN unpaired e– will go in dx2 – y2 orbital. O Thus, orbital order – Nicotinamide adenine dinucleotide (NAD) dx2 – y2 > dz2 > dxy > dxz = dyz 4-position of pyridine ring is most favourable Cs2Ba[Cu(NO2)6] position for nucleophilic attack in this case (B) 72. (D) – tetragonal compression (Zin) Fe Ph3C CPh3 dz2 > dx2 – y2 > dxz = dyz > dxy Fe unpaired e– will go in dz2 orbital. OC CO CDMe2 OC CO CDMe2 79. (A) In C—H, the three valency of C is remain 73. (C) In solid state, Alluminium flouride have and in Co(CO)3, 15e– system also have lack three-dimensional network structure unlike of 3e–s for 18 electronic system. AlCl3 (dimeric) So, CH & Co(CO)3 are isolobal. 80. (C) We know that, Doppler shift in frequency, F Δυ = υ υ FF c F Al F ∴ Δυ = 3·48 × 1018 Hz × 2·2 mms– 1 3 × 1011 mms– 1 FF F FF Al F Al = 2·55 × 107 Hz FF FF F Al F = 25·5 MHz FF 81. (B) F 74. (B) 75. (C) 82. (D) MnCr2O4 – normal spinel structure 76. (B) Sm+3, f5 system (Mn+2)Td (Cr2+3)0 O4 because Cr+ 3 will have a LFSE in octahedral site whereas Mn+ 2 ion will +3 +2 +1 0 −1 −2 −3 not. S = 5/2 83. (A) 2S + 1 = 6 Ph2PCH2CH2PPh2 + [RhCl(CO)2]2 O L = +3+2+1+0–1 = 5(H) 2 : 1 molar ratio ∴ J = L–S −2CO CO C Cl (less than half-filled orbital) 2 H2C PPh2 Cl Rh Rh = 5/2 Rh Cl C Cl H2C P CO O Ph2 υCO = 1985 (terminal Co)

CSIR-Chemical Sciences (D-13) | 21 CH2 couple with P to form doublet and this structure is NCS is coordinated with both side doublet with another CH2 gives doublets of N & S with 2e– donation doublets. H3N Pt NCS Pt PPh3 84. (D) 85. (D) 86. (B) 87. (A) 88. (A) PPh3P SCN NH3 89. (A) Closo-carborane – C2B10H12 95. (D) For higher borones 3C–2e ‘BBB’ bond may be a part of their structures. For B5 H9 (i) 1, 2–ortho isomer 1 Styx code : 4120 (ii) 1, 7–meta (iii) 1, 12–para 56 HH 42 H B 3 B H 8 H BH 97 10 11 BB 12 H HH 90. (D) Co4(CO)12 96. (B) C8H10O2, CO DBE = 16 + 2 – 10 = 4 (one ring) OC CO 2 Co δ7·3 (1H, + J = 8Hz) 7·0 (1H, d J = 8Hz) O 6·95 (1H, S) C 6·9 (1H, d, J = 8Hz) OC Co CO 5·3 (1H, brs, D2O exchangeable) Co CO C 4·6 (2H, S) OC O O C Co 3·9 (3H, S) OC 3 double bond CO 6.95,s M–M bond = 72 – (24 + 36) = 6 3.9, s 2 H 4.6, s CH3O H2 5.3, brs OH Terminal CO = 9 bridging CO = 3 H H 6.9, d H Ir4(CO)12 7.0, d 7.3, t CO MeO OC CO OH correct structure Ir 97. (D) CHCl3 aq.NaOH CCl3 _ Cl : CCl2 _ H2O OC CO OC Ir Ir CO Cl OC CO CCl2 Cl Ir N −Cl − N H OC CO CO H M–M bond = 72 – (24 + 36) = 6 Cl 2 + Terminal CO = 12 N bridging CO = 0 H 91. (C) 92. (B) 93. (A) −H 94. (C) Pt complexes are 16e– species generally, Cl therefore there is not Pt-Pt bond formation & N

22 | CSIR-Chemical Sciences (D-13) 98. (D) 51 2 37 O OH OH 27 3 OMe 208 BnO H m-CPBA BnO 4 532 O 1 H 172 O Methy-4-Oxopentanoate 13C NMR spectrum data suggest OH OMe – 51, C1 = 172, C4 = 208, C5 = 32 BnO SiMe3 99. (A) 1. BH3.THF OH 107. (A) O H2O2/NaOH 2. Ph3P, DEAD 108. (A) OH OH 4-nitrobenzoic acid HH 3. K2CO3, MeOH CH2 OH OH OH OH OH CH2Cl2 PCC CH2I2, Zn-Cu CH2Cl2 H H 100. (B) CH2 O H H H OO OH HO O HH CHO H2SO4 HO O (A) O hυ HH Me disrotation CH2 H Me Me HO H H 109. (B) O HO Δ 101. (D) O HO Me disrotation CO3H Cl O H Me AC2O/Δ OAC Me H NN N H O 110. (D) Br COOEt nBu3SnH O 102. (C) AIBN 103. (C) C COOH 1. SOCl2 Ph O Ph C Cl H toluene Ph 104. (B) reflux Ph H COOEt 105. (A) 106. (B) OH BnO Ph O 1. NEt3 Ph HH COOEt 3. CH2=CH OET OEt O H H OH OH H SiMe3 BnO SiMe3 COOEt O OH OH − HH KH, THF

CSIR-Chemical Sciences (D-13) | 23 111. (D) 124. (A) Its first option, OO OH OH 〈T〉 = (V〉, all explanations are true. Chiral 125. (D) 126. (C) 127. (D) reducing 128. (B) From given data, it is clear that agent R, R-configuration 100% diastereoselectivity rate ∝ [NO]2 90% ee means 95% R,R–configured product. and rate ∝ [H2] 112. (B) Thus, orders are 2 w.r.t. NO and 1 w.r.t. H2. HH 129. (B) The angular momentum operator ^Ly is 113. (D) heat ( )^Ly = –ih– ∂ ∂ 6 π ed_issspoatrattiicoinpating ∂x ∂z z – x H ( )= i·i ∂ ∂ 114. (A) 115. (B) – i –h z ∂x – x ∂z 116. (A) O ( )= –h z ∂ – x ∂ (i2 = – 1) i ∂x ∂z OH > c=o bond A 130. (A) NaCl & KCl crystallize with fcc structure reduction B Li/Liq.NH3 double bond NaBH4 reduction (1,4) CeCl3.7H2O NaCl — fcc H2,RhCl (PPh3)3 C KCl — simple cubic less hindered double bond reduction because K+ and Cl– are iso electronic and form simple cubic structure simply. O O 131. (C) V(x,y) = 1 kx X2 + 1 ky Y2 2 2 117. (D) 118. (B) 119. (D) aniline LiAlH4 Thus, ψnx,ny = ψnx (x).ψny (y) 80% and Enx, Eny = Enx + Eny CH2CH2COOH NH2 180°C 132. (B) We know 120. (A) 60% ρ = nM COOH H NV N NHPh ∴ ρFCC = 4M H 48% NVFCC 121. (B) Unit cell dimension, a = 4·00Å 2M NVBCC atomic radius in fcc structure is given by and ρBCC = r= a = 4 ∴ ρFCC = 4M · NVBCC ρBCC NVFCC 2M 2⎯√ 2 2⎯√ 2 = 1·414 Å 2VBCC VFCC 122. (C) = 123. (B) The most probable value of r for e– is 1s 133. (D) π1 π1 orbital of Hydrogen σ4* πg* πg* rmp = P(r) = R(r)2·r2 For this, we have 11 ⇒ 4ddr (e – 2r·r2) = 0 Λ = 0(Σ) π4 π4 2Σ + 1 = 1 or 3 σg (·.· R1s = 2·e– r) = 1π rmp = 1 = a0 = 2Δ expectation value, <r> = 3a0 There is no term symbol like 3φ because it 2 has Λ = 3 which is not possible in this case.

24 | CSIR-Chemical Sciences (D-13) 134. (C) We know that angular frequency, 140. (A) Isothermal reversible expansion work, ω = γ B0 w = – nRT ln ⎜⎝⎛VVif⎞⎟⎠ where γ is gynomagnetic ratio & B0 = magnetic field. Thus, ω = 2·7 × 108 T– 1S – 1 × 12·6T – sign for expansion, = 34·02 × 108 Hz If compression then + (positive) sign. ∴ resonance frequency, 141. (C) Maxwell’s relationships are υ = ω = 34·02 × 108 Hz ( ) ( )(i)∂S ∂P 2π 2 × 3·14 ∂V ∂T V = = 5·14 × 108 Hz T = 514 MHz ( ) ( )(ii)∂S ∂V ∂P =– ∂T P 135. (B) Moment of Inertia, T J = μr2 ( ) ( )(iii)∂V ∂T ∂S ∂P S I = m1m2 × r2 = m1 + m2 P 19 ( ) ( )(iv)∂P ∂T = 20 × 2 × 2 × 10– 20 m2 ∂S V=– ∂V S × 1·67 × 10– 27 kg 142. (C) = 6·6 × 10– 47 kgm2 143. (B) Zn/Zn+ 2 (a= 0·01) || Fe+ 2 (a = 0·001), 136. (D) The chemical potential is given as Fe+ 3 (a = 0·01)/Pt Ecell = 1·71V S ΔH P Zn + 2Fe+ 3 Zn+ 2 + 2Fe+ 2 ΔV ΔG Zn → Zn+ 2 + 2e– V ΔA (−) T 2Fe+ 3 ⎯2e→– 2Fe+ 2 (−) ( )μi = ∂G ∴ log keg = n × Ecell ∂ni T,P 0·0591 at constant temp. & pressure. 2 × 1·7100 ~~ 0·0591 137. (B) The transition that is x-polarised light in = 54 trans-butadiene is for keg = 1054 1Au → 1Bu Au ⎜⎝⎜⎛yxz BBAuuu⎟⎠⎞⎟ Bu = 1 1 11 144. (B) All explanation are correct, but option is 1 1 11 (B) H—C ≡ C—H – smallest partition 1 –1 1 –1 function, less molecular mass. → x-polarised (all + ve) 145. (D) Limiting molar conductivities of reactant 138. (B) = λ∞ of product 139. (B) n = 10 mol, CP = 300 JK– 1, CV = ? ∴ NaI + RbCl → NaCl + RbI mSm2 mol– 1 ∴ CP – CV = nR 10·8 12·7 9·1 CV = CP – nR ∴ λRbCl = 12·7 + 9·1 – 10·8 = 300 – 10 × 8·314 JK– 1 = 11·0 mSm2 mol– 1 = 217 JK– 1

Chemical Sciences CSIR-UGC NET/JRF Exam. Solved Paper

June 2014 Chemical Sciences PART A dispersion (D) = (maximum marks – minimum marks), and 1. The following diagram shows 2 perpendi- relative dispersion (RD) = dispersion · Then, cularly inter-grown prismatic crystals (twins) mean of identical shape and size. What is the volume of the object shown (units are (A) RD of group A = RD of group B arbitrary) ? (B) RD of group A > RD of group B (C) RD of group A < RD of group B (D) D of group A < D of group B 5. In 450 g of pure coffee powder 50 g of chicory 10 is added. A person buys 100 g of this mixture and adds 5 g of chicory to that. What would 2 be the rounded-off percentage of chicory in 2 this final mixture ? (A) 60 (B) 65 (A) 10 (B) 5 (C) 72 (D) 80 (C) 14 (D) 15 2. Suppose in a box there are 20 red, 30 black, 40 6. The time gap between the two instants, one before and one after 12·00 noon, when the blue and 50 white balls. What is the minimum angle between the hour hand and the minute number of balls to be drawn, without hand is 66°, is— replacement, so that you are certain about getting 4 red, 5 black, 6 blue and 7 white (A) 12 min (B) 16 min balls ? (C) 18 min (D) 24 min (A) 140 (B) 97 7. Suppose (C) 104 (D) 124 x Δ y = (x – y)2 3. In the growing years of a child, the height x o y = (x + y)2 increases as the square root of the age while x * y = (x × y)– 1 the weight increases in direct proportion to the age. The ratio of the weight to the square x·y = x×y of the height in this phase of growth— +, – and × have their usual meanings. What is (A) is constant the value of (B) reduces with age [(197 o 315) – (197 Δ 315)} · (197 * 315) ? (C) increases with age (A) 118 (B) 512 (D) is constant only if the weight and height (C) 2 (D) 4 at birth are both zero 8. If A × B = 24, B × C = 32, C × D = 48 then 4. Students in group A obtained the following A × D— marks : 40, 80, 70, 50, 60, 90, 30. Students in (A) cannot be found (B) is a perfect square group B obtained 40, 80, 35, 70, 85, 45, 50, (C) is a perfect cube (D) is odd 75, 60 marks. Define

4 | CSIR-UGC Chemical Sciences (J-14) 9. If all horses are donkeys, some donkeys are 15. Consider a right-angled triangle ABC where monkeys and some monkeys are men, then AB = AC = 3. A rectangle APOQ is drawn which statement must be true ? inside it, as shown, such that the height of the rectangle is twice its width. The rectangle is (A) All donkeys are men moved horizontally by a distance 0·2 as shown schematically in the diagram (not to scale). (B) Some donkeys may be men C (C) Some horses are men Q OS T (D) All horses are also monkeys 10. A rectangular area of sides 9 and 6 units is to be covered by square tiles of sides 1, 2 and 5 units. The minimum number of tiles needed for this is— (A) 3 (B) 11 (C) 12 (D) 15 AP B 11. Suppose n is a positive integer. Then (n2 + n) What is the value of the ratio Area of ΔABC ? (2n + 1)— Area of ΔOST (A) may not be divisible by 2 (A) 625 (B) 400 (B) is always divisible by 2 but may not be divisible by 3 (C) 225 (D) 125 (C) is always divisible by 3 but may not be 16. 80 gsm paper is cut into sheets of 200 mm × divisible by 6 300 mm size and assembled in packets of 500 (D) is always divisible by 6 sheets. What will be the weight of a packet ? (gsm = g/m2) 12. There is a train of length 500 m, in which a (A) 1·2 kg (B) 2·4 kg man is standing at the rear end. At the instant (C) 3·6 kg (D) 4·8 kg the rear end crosses a stationary observer on a 17. Find the missing letter platform, the man starts walking from the rear ABCD to the front and the front to the rear of the train F I LO at a constant speed of 3 km/hr. The speed of the train is 80 km/hr. The distance of the man KPUZ from the observer at the end of 30 minutes PWD ? is— (A) P (B) K (A) 41·5 km (B) 40·5 km (C) J (D) L (C) 40·0 km (D) 41·0 km 18. A merchant buys equal numbers of shirts and trousers and pays Rs. 38,000. If the cost of 3 13. Three identical flat equilateral-triangular shirts is Rs. 800 and that of a trouser is Rs. plates of side 5 cm each are placed together 1,000, then how many shirts were bought ? such that they form a trapezium. The length of the longer of the two parallel sides of this (A) 60 (B) 30 trapezium is— (C) 15 (D) 10 ⎯√(A) 5 3 cm (B) 5⎯√ 2 cm 19. Consider the set of numbers {171, 172, …, 4 (D) 10⎯√ 3 cm 17300}. How many of these numbers end with (C) 10 cm the digit 3 ? 14. An archer climbs to the top of a 10 m high (A) 60 (B) 75 building and aims at a bird atop a tree 17 m (C) 100 (D) 150 away. The line of sight from the archer to the 20. Find the missing number in the triangle. bird makes an angle of 45° to the horizontal. 718 What is the height of the tree ? (A) 17 m (B) 27 m 90 13 ? (C) 37 m (D) 47 m 3 56 42 6

(A) 16 (B) 96 CSIR-UGC Chemical Sciences (J-14) | 5 (C) 50 (D) 80 Br PART B 21. The correct order of basicity for the following Br anions is— O− (A) molecule is chiral and possesses a chiral plane O− O− NO2 (B) Molecule is chiral and possesses a chiral axis I II NO2 NO2 III (C) Molecule is achiral as it possesses a plane of symmetry (A) II > III > I (B) I > II > III (C) II > I > III (D) III > II > I (D) Molecule is achiral as it possesses a centre of symmetry 22. The major product formed in the reaction of 25. Consider the following statements about cis- 2, 5-hexanedione with P2O5 is— and trans-decalins— 1. cis-isomer is more stable than trans- isomer (A) 2. trans-isomer is more stable than cis-isomer O O 3. trans-isomer undergoes ring-flip 4. cis-isomer undergoes ring-flip The correct statements among the above are— (B) (A) 2 and 4 (B) 1 and 3 (C) 1 and 4 (D) 2 and 3 O 26. In bis (dimethylglyoximato) nickel(II), the number of Ni—N, Ni—O and intramolecular O hydrogen bond(s), respectively, are— (C) (A) 4, 0 and 2 (B) 2, 2 and 2 O (C) 2, 2 and 0 (D) 4, 0 and 1 O 27. Among the following species, (1) Ni(II) as (D) dimethylglyoximate, (2) Al(III) as oxinate, (3) Ag(I) as chloride, those that precipitate with the urea hydrolysis method are— (A) 1, 2 and 3 (B) 1 and 2 O (C) 1 and 3 (D) 2 and 3 23. The absolute configuration of the two 28. If an enzyme fixes N2 in plants by evolving stereogenic (chiral) centres in the following H2, the number of electrons and protons molecule is— associated with that, respectively are— O (A) 6 and 6 (B) 8 and 8 (C) 6 and 8 (D) 8 and 6 (A) 5R, 6R (B) 5R, 6S 29. The particles postulated to always accompany (C) 5S, 6R (D) 5S, 6S the positron emission among— (1) neutrino, (2) anti-neutrino, (3) electron, are— 24. The correct statement about the following (A) 1, 2 and 3 (B) 1 and 2 molecule is— (C) 1 and 3 (D) 2 and 3

6 | CSIR-UGC Chemical Sciences (J-14) 30. Toxicity of cadmium and mercury in the body (A) η5 – C5H5 (B) η1 – C5H5 is being reversed by proteins, mainly using (C) η5 – C5H5 and CO (D) η1 – C5H5 and CO the amino acid residue— 37. [CoL6]3+ is red in colour whereas [CoL′6]3+ is (A) Glycine (B) Leucine green. L and L′ respectively corresponds to— (A) NH3 and H2O (C) Lysine (D) Cysteine (B) NH3 and 1, 10-phenanthroline (C) H2O and 1, 10-phenanthroline 31. NiBr2 reacts with (Et)(PH)2P at – 78°C in CS2 (D) H2O and NH3 to give red compound, ‘A’, which upon standing at room temperature turns green to 38. The oxidation state of Ni and the number of give compound, ‘B’, of the same formula. metal–metal bonds in [Ni2(CO)6]2+ that are The measured magnetic moments of ‘A’ and consistent with the 18 electrons rule are— ‘B’ are 0·0 and 3·2 BM, respectively. The (A) Ni(– II), 1 bond (B) Ni(IV), 2 bonds geometries of ‘A’ and ‘B’ are— (C) Ni(–I), 1 bond (D) Ni(IV), 3 bonds (A) Square planar and Tetrahedral (B) Tetrahedral and Square planar (C) Square planar and Octahedral (D) Tetrahedral and Octahedral 32. The correct non-linear and iso-structural pair is— 39. Structures of SbPh5 and PPh5 respectively are— (A) SCl2 and I3 (B) SCl2 and I3+ (A) trigonal bipyramidal, square pyramidal (C) SCl2 and ClF2– (D) I3+ and ClF2– (B) square pyramidal, trigonal bipyramidal 33. Ozone present in upper atmosphere protects (C) trigonal bipyramidal, trigonal bipyramidal people on the earth— (A) due to its diamagnetic nature (D) square pyramidal, square pyramidal (B) due to its blue colour (C) due to absorption of radiation of 40. The typical electronic configurations of the wavelength at 255 nm (D) by destroying chlorofluoro carbons transition metal centre for oxidative addition are— (A) d 0 and d 8 (B) d 5 and d 8 34. If L is a neutral monodentate ligand, the (C) d 8 and d 10 (D) d 5 and d 10 species, [AgL4}2+, [AgL6}2+ and [AgL4]3+ respectively are— 41. Gelatin added during the polarographic measurement carried out using dropping (A) paramagnetic, paramagnetic and mercury electrode— diamagnetic (A) reduces streaming motion of Hg drop (B) paramagnetic, diamagnetic and paramagnetic (B) decreases viscosity of the solution (C) diamagnetic, paramagnetic and (C) eliminates migrating current diamagnetic (D) prevents oxidation of Hg (D) paramagnetic, diamagnetic and diamagnetic 42. The pKa values of the following salt of aspartic acid are indicated below. The 35. Chromite ore on fusion with sodium predominant species that would exist at pH = carbonate gives— 5 is— (A) Na2CrO4 and Fe2O3 (B) Na2Cr2O7 and Fe2O3 (pKa = 9.9) H3N COOH (pKa = 2.0) (C) Cr2(CO3)3 and Fe(OH)3 (D) Na2CrO4 and Fe2(CO3)3 COOH (pKa = 3.9) 36. The ligand(s) that is (are) fluxional in [(η5 H3N COO – C5H5) (η1–C5H5)Fe(CO)2] in the tempera- (A) COOH ture range 221 – 298 K, is (are)—

CSIR-UGC Chemical Sciences (J-14) | 7 H3N COO 45. The major product formed in the following (B) COO reaction is— H Me Ph AcOH OTs H2N COO (C) COO Me H H Me H3N COOH Ph (D) COO OAc (A) Me H H Me 43. The major product formed in the following Ph photochemical reaction is— OAc t-Bu hv (B) H Me t-Bu Me Me t-Bu (C) Ph H t-Bu t-Bu Me H (D) Ph Me (A) 46. The correct order for the rates of electrophilic t-Bu aromatic substitution of the following com- pounds is— H Cl t-Bu (B) N N III t-Bu I Me II t-Bu (A) I > II > III (B) II > I > III t-Bu t-Bu (C) III > II > I (D) I > III > II (C) 47. The commutator of the kinetic energy t-Bu H operator, T^ x and the momentum operator, ^px t-Bu for the one-dimensional case is— (D) t-Bu (A) i –h (B) i –h d t-Bu dx 44. The pair of solvents in which PCl5 does NOT (C) 0 (D) i h–x ionize is— (A) CH3CN, CH3NO2 48. The major product formed in the reaction of (B) CH3CN, CCl4 trans-1-bromo-3-methylcyclobutane with (C) C6H6, CCl4 sodium iodide in DMF is— (D) CH3CN, C6H6 Me Me (A) (B) Me Me (C) (D) ll

8 | CSIR-UGC Chemical Sciences (J-14) 49. When Si is doped with a Group V element— 55. Examine the following first order consecutive reactions. The rate constant (in s– 1 units) for (A) donor levels are created close to the valence band each step is given above the arrow mark— (B) donor levels are created close to the 105 108 conduction band 1. P ⎯→ Q ⎯→ R (C) acceptor levels are created close to the 105 103 valence band 2. P ⎯→ Q ⎯→ R (D) acceptor levels are created close to the 107 107 conduction band 3. P ⎯→ Q ⎯→ R 50. The symmetry point group of propyne is— 102 106 4. P ⎯→ Q ⎯→ R (A) C3 (B) C3v Steady-state approximation can be applied (C) D3 (D) D3d to— (A) 1 only (B) 3 only 51. For a first order reaction A → products, the (C) 2 and 3 only (D) 1 and 4 only plot of ln ⎛⎝⎜[[AA]]t0⎞⎟⎠ vs. time, where [A]0 and [A]t 56. The figure below represents the path followed by a gas during expansion from A → B. The refer to concentrations at time t = 0 and t work done is (L atm.) respectively, is— P(atm) 5 A 4 (A) a straight line with a positive slope passing through origin 3 2 (B) a straight line with a negative slope B passing through origin 1 (C) an exponential curve asymptotic to the 12345 time axis V(L) (D) a curve asymptotic to the ln ⎝⎛⎜[[AA]]t0⎠⎟⎞ axis (A) 0 (B) 9 (C) 5 (D) 4 52. In radical chain polymerization the quantity given by ‘the rate of monomer depletion, 57. An aqueous solution of an optically pure divided by the rate of propagating radical formation’ is called— compound of concentration 100 mg in 1 mL of (A) kinetic chain length (B) propagation efficiency water and measured in a quartz tube of 5 cm length was found to be – 3°. The specific (C) propagation rate constant rotation is— (D) polymerization time (A) – 30° (B) – 60° (C) – 6° (D) + 6° 53. Number of rotational symmetry axes for 58. Two phases (α and β ) of a species are in triclinic crystal system is— equilibrium. The correct relations observed among the variables, T, p and μ are— (A) 4 (B) 3 (A) Tα = Tβ, pα ≠ pβ, μα = μβ (C) 1 (D) 0 (B) Tα ≠ Tβ, pα = pβ, μα = μβ (C) Tα = Tβ, pα = pβ, μα = μβ 54. Generally, hydrophobic colloids are (D) Tα = Tβ, pα = pβ, μα ≠ μβ flocculated efficiently by ions of opposite type and high charge number. This is 59. The number of configurations in the most consistent with the— probable state, according to Boltzmann (A) peptization principle formula is— (B) Krafft theory (A) e S/kB (B) e – S/kB (C) Hardy-Schulze rule (C) e – E/kBT (D) e – ΔG/kBT (D) Langmuir adsorption mechanism

CSIR-UGC Chemical Sciences (J-14) | 9 60. The correct match of the ′HNMR chemical (A) 1 and 3 (B) Only 3 shifts (δ) of the following species/compounds (C) 2 and 4 (D) 2 and 3 is— 64. Pericyclic reaction involved in one of the steps of the following reaction sequence is— −+ I II III + O i. Heat OH S ii. (EtO)3P (A) I : 5·4; II : 7·2; III : 9·2 (B) I : 9·2; II : 7·2; III : 5·4 H (C) I : 9·2; II : 5·4; III : 7·2 (D) I : 7·2; II : 9·2; III : 5·4 (A) [1, 3] sigmatropic shift (B) [3, 3] sigmatropic shift 61. The major products formed in the following (C) [1, 5] sigmatropic shift reaction are— (D) [2, 3] sigmatropic shift O i. Ph3C Na+ 65. Atorvastatin (structure given below) is a— OEt ii. H3O+ OH HOOC O OH N F (A) OH + Ph3C–CH2CH3 Me HO CPh3 Me (B) CPh3 + EtOH PhHN Ph O (A) cholesterol lowering drug OO (B) blood sugar lowering drug (C) OEt + EtOH (C) anti-plasmodial drug (D) anti-HIV drug O 66. The maximum bond order obtained from the (D) OH + CH2=CH2 molecular orbitals of a transition metal dimer, formed as linear combinations of d-orbitals 62. In a Diels-Alder reaction, the most reactive alone, is— diene amongst the following is— (A) (4E)-1, 4-hexadiene (A) 3 (B) 4 (B) (4Z)-1, 4-hexadiene (C) (2E, 4E)-2, 4-hexadiene (C) 5 (D) 6 (D) (2Z, 4Z)-2, 4-hexadiene 67. The term symbol that is NOT allowed for the 63. Consider the statements about the following np2 configuration is— structures X and Y. (A) 1D (B) 3P (C) 1S (D) 3D 68. If the ionization energy of H atom is x, the ionization energy of Li2+ is— Ph CN Ph (A) 2x (B) 3x NH (C) 9x (D) 27x XY 69. If temperature is doubled and the mass of the 1. X and Y are resonance structures gaseous molecule is halved, the rms speed of 2. X and Y are tautomers the molecule will change by a factor of— 3. Y is more basic than X 4. X is more basic than Y (A) 1 (B) 2 The correct statement(s) among the above is (are)— (C) 1 (D) 1 2 4

10 | CSIR-UGC Chemical Sciences (J-14) 70. In the graph below, the correct option, (B) The ‘phen’ is a π-donor ligand that according to Kohlrausch law is— enhances the rate of electron transfer A (C) The ‘phen’ forms charge transfer complex with iron and facilitates the ΛB electron transfer C (D) The ‘phen’ forms kinetically labile D complex with iron and facilitates the electron transfer C 73. The compound [Re2(Me2PPh)4Cl4] (M) having a configuration of σ2π4,δ2δ*2 can be (A) A is a weak electrolyte and B is a strong oxidized to M+ and M2+. The formal metal electrolyte – metal bond order in M, M+ and M2+ respectively are— (B) A is a strong electrolyte and B is a weak electrolyte (A) 3·0, 3·5 and 4·0 (B) 3·5, 4·0 and 3·0 (C) C is a strong electrolyte and D is a weak (C) 4·0, 3·5 and 3·0 (D) 3·0, 4·0 and 3·5 electrolyte 74. In low chloride ion concentration, the (D) C is a weak electrolyte and D is a strong anticancer drug cis-platin hydrolyses to give a electrolyte diaquo complex and this binds to DNA via adjacent guanine PART C O 71. Reduction of [Ru(NH3)5 (isonicotinamide)]3+ N NH with [Cr(H2O)6]2+ occurs by inner sphere mechanism and rate of the reaction is N N NH2 determined by dissociation of the successor H complex. It is due to the— (guanine) (A) inert ruthenium bridged to inert chromium centre The co-ordinating atom of guanine to Pt(II) (B) inert ruthenium bridged to labile is— chromium centre (A) N1 (B) N3 (C) labile ruthenium bridged to inert chromium centre (C) N7 (D) N9 (D) labile ruthenium bridged to labile 75. The 19F-NMR spectrum of CIF3 shows— chromium centre (A) doublet and triplet for a T-shaped structure 72. Consider the second order rate constants for (B) singlet for a trigonal planar structure the following outer-sphere electron transfer (C) singlet for a trigonal pyramidal structure reactions (D) doublet and singlet for a T-shaped structure [Fe(H2O)6]3+/[Fe(H2O)6]2+ 4·0 M– 1 sec– 1 [Fe(phen)3]3+/[Fe(Phen)3]2+ 3·0 × 107 M– 1 76. The low temperature (–98°C) 19F NMR sec– 1 spectrum of SF4 shows doublet of triplets. It is consistent with the point group symmetry— (phen = 1, 10-phenanthroline) (A) C3v (B) C4v The enhanced rate constant for the second (C) Td (D) C2v reaction is due to the fact that— 77. Amongst organolithium (A), Grignard (B) (A) The ‘phen’ is a π-acceptor ligand that and organoaluminium (C) compounds, those allows mixing of electron donor and react with SiCl4 to give compound containing acceptor orbitals that enhances the rate of Si — C bond are— electron transfer (A) A and B (B) B and C (C) A and C (D) A, B and C

CSIR-UGC Chemical Sciences (J-14) | 11 78. In its electronic spectrum, [V(H2O)6]3+ 83. The use of dynamic inert atmosphere in exhibits two absorption bands, one at 17,800 thermogravimetric analysis (TGA)— (υ1) and the second at 25,700 (υ2) cm– 1. The (A) decreases decomposition temperature correct assignment of these bands, respec- (B) decreases weight loss tively is— (C) reduces rate of decomposition (A) υ 1 = 3T1g(F) → 3T2g(F) υ 2 = 3T1g(F) → 3T1g(P) (D) increases weight loss (B) υ 1 = 3T1g(F) → 3T1g(P) υ 2 = 3T1g(F) → 3T2g(P) 84. The correct statements for hollow cathode (C) υ 1 = 3A2g → 3T1g(F) lamp (HCL) from the following are— υ 2 = 3A2g → 3T2g(F) (D) υ 1 = 3A2g → 3T2g(F) 1. HCL is suitable for atomic absorption spectroscopy (AAS) υ 2 = 3A2g → 3T1g(F) 2. lines emitted from HCL are very narrow 79. Reactions of elemental As with hot and conc. HNO3 and H2SO4, respectively, give— 3. the hardening of lamp makes it unsuitable for AAS (A) As4O6 and As2(SO4)3 4. transition elements used in the lamps (B) As(NO3)5 and As2(SO4)3 have short life (C) As4O6 and H3AsO4 (A) 1, 2 and 3 (B) 2, 3 and 4 (D) H3AsO4 and As4O6 (C) 3, 4 and 1 (D) 4, 1 and 2 80. The total valence electron count and the 85. Identify the correct statement about structure type adopted by the complex [Fe5(CO)15C] respectively, are— [Ni(H2O)6]2+ and [Cu(H2O)6]2+ (A) 74 and nido (B) 60 and closo (A) all Ni–O and Cu–O bond lengths of individual species are equal (C) 84 and arachno (D) 62 and nido (B) Ni–O (equatorial) and Cu–O (equatorial) 81. 1H NMR spectrum of [η5–C5H5)Rh(C2H4)2] bond lengths are shorter than Ni–O at – 20°C shows a typical AA ‘XX’ pattern in (axial) and Cu–O (axial) ones respectively the olefinic region. On increasing the temperature to ~70°C, the separate lines (C) all Ni–O bond lengths are equal whereas collapse into a single line which is due to— Cu–O(equatorial) bonds are shorter than Cu–O(axial) bonds (A) free rotation of the ethylene ligand about the metal-olefin bond (D) all Cu–O bond lengths are equal whereas Ni–O(equatorial) bonds are shorter than (B) intramolecular exchange between the Ni–O(axial) bonds ethylene ligands 86. Reaction of nitrosyl tetrafluoroborate to (C) intermolecular exchange between the Vaska’s complex gives complex A with ethylene ligands ∠M–N–O = 124°. The complex A and its N–O stretching frequency are, respectively— (D) change in hapticity of the cyclopenta- dienyl ligand (A) [IrCl(CO)(NO)(PPh3)2]BF4, 1620 cm– 1 (B) [IrCl(CO)(NO)2(PPh3)](BF4)2, 1720 cm– 1 (C) [IrCl(CO)(NO)2(PPh3)](BF4)2, 1520 cm– 1 (D) [IrCl(CO)(NO)(PPh3)2], 1820 cm– 1 82. The nuclides among the following, capable of 87. The correct order of decreasing electro- undergoing fission by thermal neutrons, are— negativity of the following atoms is— (A) As > Al > Ca > S 1. 233U 2. 235U (B) S > As > Al > Ca (C) Al > Ca > S > As 3. 239Pu 4. 232Th (D) S > Ca > As > Al (A) 1, 2 and 4 (B) 1, 3 and 4 (C) 2, 3 and 4 (D) 1, 2 and 3

12 | CSIR-UGC Chemical Sciences (J-14) 88. A 1 : 2 mixture of Me2NCH2CH2CH2PPh2 (A) Sm is a better one electron reductant than and KSCN with K2[PdCl4] gives a square Eu planar complex A. Identify the correct pairs of donor atoms trans to each other in complex (B) Sm is a better one electron oxidant than A from the following combinations— Eu 1. P, N 2. N, S (C) facile oxidation state is +2 for both the elements 3. P, S 4. N, N (D) both of these display similar redox (A) 1 and 2 (B) 1 and 4 behaviour (C) 2 and 3 (D) 3 and 4 93. The cooperative binding of O2 in hemoglobin is due to— 89. For a low energy nuclear reaction, 24Mg(d, α)22Na, the correct statements from the (A) a decrease in size of iron followed by changes in the protein conformation following are— (B) an increase in size of iron followed by 1. kinetic energy of d particle is not fully changes in the protein conformation available for exciting 24Mg (C) a decrease in size of iron that is NOT 2. total number of protons and neutrons is accompanied by the protein conforma- tional changes conserved (D) an increase in size of iron that is NOT 3. Q value of nuclear reaction is much accompanied by the protein conforma- tional changes higher in magnitude relative to heat of 94. Amongst the following which is not isolobal chemical reaction pairs— 4. threshold energy is ≤ Q value (A) Mn(CO)5, CH3 (B) Fe(CO)4, O (C) Co(CO)3, R2Si (D) Mn(CO)5, RS (A) 1, 2 and 3 (B) 1, 2 and 4 95. The correct order of the size of S, S2–, S2+ and (C) 2, 3 and 4 (D) 1, 3 and 4 S4+ species is— (A) S > S2+ > S4+ > S2– 90. At pH 7, the zinc(II) ion in carbonic (B) S2+ > S4+ > S2– > S anhydrase reacts with CO2 to give— (C) S2– > S > S2+ > S4+ (D) S4+ > S2– > S > S2+ H OH O 96. The major product formed in the following (A) Zn O O (B) Zn C reaction is— C O Br H O n-Bu3SnH OH H O AlBN (C) Zn O O O benzene (D) Zn C reflux C O H 91. Molybdoenzymes can both oxidize as well as reduce the substrates, because— HH (A) Mo(VI) is more stable than Mo(IV) (A) (B) (B) Mo(IV) can transfer oxygen atom to the HH HH substrate and Mo(VI) can abstract oxygen atom from the substrate Br SnBu3 (C) conversion of Mo(VI) to Mo(IV) is not favoured (D) Mo(VI) can transfer oxygen atom to the substrate and Mo(IV) can abstract oxygen atom from the substrate 92. A comparison of the valence electron con- (C) (D) H figuration of the elements, Sm and Eu sug- H HH gests that—

CSIR-UGC Chemical Sciences (J-14) | 13 97. The correct combination of reagents to effect 100. The correct 13C NMR chemical (δ) shift the following conversion is— values of carbons labeled a-e in the following O COOH ester are— ed Oa HH Me c b O Me HH (A) a : 19; b : 143; c : 167; d : 125; e : 52 (A) (i) Ph3P+CH2OMeCl–, BuLi; (ii) H3O+; (B) a : 52; b : 143; c : 167; d : 125; e : 19 (iii) Jones’ reagent (C) a : 52, b : 167; c : 143; d : 125; e : 19 (B) (i) H2N–NHTs; (ii) BuLi (2 equiv.); (iii) DMF (D) a : 52, b : 167; c : 125; d : 143; e : 19 (C) (i) H2N–NHTs; (ii) BuLi (2 equiv.); 101. The products A and B in the following (iii) CO2 reaction sequence are— (D) (i) ClCH3CO2Et, LDA; (ii) BF3·OEt2; (iii) DMSO, (COCl)2, Et3N – 78°C to rt OO NH2 OH MeO Cl B Et3N A 98. The major product formed in the following OO O reaction is— (A) A : O OMe B : N H Bu3N+Br− anhyd. CF3COOH OO OO CH2Cl2 (B) A : O Cl B : ON OO H Br O H (A) (C) A : O OMe B : MeO N (D) A : H H OO O (B) O Cl O Br B: H O (C) 102. The biosynthesis of isopentenyl pyrophos- CF3COO phate from acetyl CoA involves— 1. Four molecules of acetyl CoA (D) 2. Three molecules of ATP O 3. Two molecules of NADPH 99. Consider the following reaction. 4. Two molecules of lipoic acid OO Ph The correct options among these are— + Ph-N3 CF3COOH N (A) 1, 2 and 4 (B) 1 and 2 (C) 2 and 3 (D) 1, 3 and 4 The appropriate intermediate involved in this 103. Amongst the following, the major products reaction is— formed in the following photochemical reac- tion are— O H HO Ph (A) N (B) N O hυ Ph H Ph Ph HO O ON HO N N N (1) O (D) (C) O (2) (3) (4)

14 | CSIR-UGC Chemical Sciences (J-14) (A) 1 and 3 (B) 2 and 3 3B (C) 1 and 4 (D) 1 and 2 H X 104. The products A and B in the following reaction sequence are— O BnO OMe H2N-NH2 A Heat O B and H N (A) H Zn O BnO BnO O Zn and B: NH (B) H (A) A : H2N MeO O H HO O O O O (C) Zn and H N O H (B) A: N B: MeO OO BnO O O O O (D) and H N O H Zn (C) A : N HO B: OO BnO O O OMe 107. Using Boltzmann distribution, the probability of an oscillator occupying the (D) A : B: first three levels (n = 0, 1 and 2) is found to H2N MeO HN be p0 = 0·633, p1 = 0·233 and p2 = 0·086. 105. Anthranilic acid, on treatment with iso-amyl The probability of finding an oscillator in nitrite furnishes a product which displays a energy levels n ≥ 3 is— strong peak at 76 (m/e) in its mass spectrum. The structure of the product is— (A) 0·032 (B) 0·048 NO (C) 0·952 (D) 1·000 (A) (B) 108. The major products A and B in the COOH following reaction sequence are— i. PhNCO OH NO2 Et3N A H2, Raney Ni B MeOH, H2O (C) (D) ii. AcOH COOH COOH 106. The organoborane X, when reacted with (A) A : N O OH Et2Zn followed by p-iodotoluene in the O B: presence of catalytic amount of Pd(PPh3)4 furnishes a trisubstituted alkene. The inter- (B) A : N O OH O B: mediate and the product of the reaction, respectively, are—

CSIR-UGC Chemical Sciences (J-14) | 15 N OO Ph O B: (C) A : H3COOC N COOCH3 (D) A : N O OH (A) O B: Ph Ph Ph 109. The correct combination of reagents required H3COOC N COOCH3 to effect the following conversion is— (B) Ph Ph COOCH3 O Ph COOCH3 OH H3COOC N Ph (A) (i) Na, xylene, Me3SiCl, heat; (ii) H3O+ (C) Ph (B) (i) Na, xylene, heat; (ii) H2O2, NaOH H3COOC (C) (i) NaOEt, EtOH; (ii) Na, xylene, heat Ph (D) (i) TiCl3, Zn-Cu, Me3SiCl, heat; (ii) H3COOC N Ph H3O+ (D) H3COOC Ph 110. An organic compound gives following 112. The correct combination of reagents for spectral data : effecting the following sequence of reactions is— IR : 2210, 1724 cm– 1; 1H NMR : δ 1·4 (t, J = 7·1 Hz, 3H), 4·4 (q, J = 7·1 Hz, 2H), 7·7 (d, J AB O = 8·0 Hz, 2H), 8·2 (d, J = 7·0 Hz, 2H); 13C O O NMR : δ 16, 62, 118, 119, 125, 126, 127, 168. The compound is— O O (A) A = O3/O2; B = K+ –OOC–N=N–COO– K+, AcOH (A) O NC (B) A = O2, Rose Bengal, hυ; B = K+ –OOC–N=N–COO–K+, AcOH O (B) O NH2 (C) A = O2, Rose Bengal, hυ; B = H2, Pd/C (D) A = O2, Rose Bengal, Δ; B = H2, Pd/C O 113. The correct combination of reagents required to effect the following conversion is— (C) O I NC O NN OO (D) CN (A) I2, HNO3 O (B) s-BuLi, – 78°C followed by KI (C) NaOEt followed by ICH2CH2I 111. The major product formed in the following (D) s-BuLi, – 78°C followed by ICH2CH2I reaction is— 114. Consider a particle confined in a cubic box. Ph The degeneracy of the level, that has an :N + Ph Ph Δ energy twice that of the lowest level, is— (A) 3 (B) 1 H3COOC COOCH3 (C) 2 (D) 4

16 | CSIR-UGC Chemical Sciences (J-14) 115. Only two products are obtained in the (A) A : CHO ON following reaction sequence. The structures (B) A : B: of the products from the list I-IV are— (C) A : OO CHO O i. NaNH2 B: ii. BrCH2CH2Br O CHO B : O O O OO (D) A : B: N CHO O I II III IV (A) I and II (B) II and IV 118. The spatial part of the wave function of the (C) I and III atom in its ground state is 1s(1)1s (2). The (D) III and IV spin part would be— 116. The major product A formed in the following (A) α(1)α(2) reaction is— (B) β(1)β(2) MeOOC Heat COOMe P (C) 1 [α(1)β(2) + β(1)α(2)] O ⎯√ 2 COOMe (D) 1 [α(1)β(2) – β(1)α(2)] √⎯ 2 MeOOC H 119. The number of phases, components and (A) degrees of freedom, when Ar is added to an equilibrium mixture of NO, O2 and NO2 in H gas phase are, respectively, COOMe (A) 1, 3, 5 (B) 1, 4, 5 MeOOC (C) 1, 3, 4 (D) 1, 4, 4 (B) O 120. The major product formed in the following reaction is— COOMe MeOOC TsCl (C) O pyridine H H OH COOMe (A) TsO (B) (D) OTs (C) COOMe OTs (D) OTs OH 117. The products A and B in the following reaction sequence are— SeO2 A aq. NaCN, MnO2 B dioxane i-PrOH, Me2NH reflux

CSIR-UGC Chemical Sciences (J-14) | 17 121. A particle in a one dimensional harmonic (B) Cl oscillator in x-direction is perturbed by a HX = HCl and P = Br Br potential λx (λ is number). The first-order correction to the energy of the ground state— (C) HX = HBr and P = Br (A) is zero (B) is negative (C) is positive (D) may be negative or positive but NOT zero 122. The products A and B in the following (D) Br sequence of reactions are— HX = HBr and P = Br OH OH 124. Match the following natural products in column A with their structural features in HO CHO MeOH/H+ A PhCH(OMe)2 (1 equiv) B column B. reflux, 24h H+ OH OH Column A Column B OH Ph (a) Colchicine 1. Tetrahydrooxepine HO O O O OMe (b) Strychnine 2. Phenanthrene HO OMe O OH (A) A = B = HO (c) Quinine 3. Tropolone OH (d) Ephedrine 4. Phenylethylamine OH O 5. Quionoline O O (B) A = HO Ph HO O 6. Benzofuran HO HO OMe B= HO OMe Identify the correct match from the following— OH HO (a) (b) (c) (d) O (C) A = HO O OMe Ph O O (A) 3 1 5 4 HO OH B= HO OMe H (B) 6 1 2 5 OH (C) 1 4 6 4 A = HO O OMe B = HO O OMe (D) 3 1 5 6 HO HO O (D) 125. A particle in a one-dimensional box O OH Ph H (potential zero between 0 to a and infinite 123. The mass spectrum of the product A, formed outside) has the ground state energy E0 = in the following reaction, exhibits M, M + 2, M + 4 peaks in the ratio of about 1 : 2 : 1. 0·125h 2 The reagent HX and the product P are— ma2 · The expectation value of the above OH Hamiltonian with ψ(x) = x (x – a) yields an energy E1. Using a linear combination of two even functions x (x – a) and x 2 (x – a)2, HX A we obtain variational minimum to the Br ground state energy as E2. Which of the F following relations holds for E0, E1 and E2 ? (A) E0 < E1 < E2 (B) E0 < E2 < E1 (A) (C) E1 < E0 < E2 HX = HF and P = Br (D) E2 < E0 < E1

18 | CSIR-UGC Chemical Sciences (J-14) 126. The dissociation constant of a weak acid HX Using the approximate data given above, the at a given temperature is 2·5 × 10– 5. The pH of 0·01 M NaX at this temperature is— frequency factor (A) for a reaction of the type : atom + diatomic molecule → (A) 7·3 (B) 7·7 nonlinear transition state → product, according to the conventional transition state (C) 8·3 (D) 8·7 theory is— 127. The ground state energy of hydrogen atom is – 13·598 eV. The expectation values of (A) 2 × 103 (B) 6 × 107 kinetic energy, 〈T〉 and potential energy, 〈V〉, in units of eV, are— (C) 2 × 1012 (D) 6 × 1013 (A) 〈T〉 = 13·598, 〈V〉 = – 27·196 132. The interplanar spacing of (110) planes in a cubic unit cell with lattice parameter a = (B) 〈T〉 = – 27·196, 〈V〉 = 13·598 4,242 Å is— (C) 〈T〉 = – 6·799, 〈V〉 = – 6·799 (A) 3 Å (B) 6 Å (D) 〈T〉 = 6·799, 〈V〉 = – 20·397 (C) 7·35 Å (D) 2·45 Å 128. If ψ = 0·8 ϕ A + 0·4 ϕB is a normalized 133. A compound AxBy has a cubic structure with A atoms occupying all the corners of the molecular orbital of a diatomic molecule AB, cube as well as all the face center positions. constructed from ϕA and ϕB which are also The B atoms occupy four tetrahedral voids. normalized, the overlap between ϕA and ϕB The values of x and y respectively, are— is— (A) 0·11 (B) 0·31 (A) 4, 4 (B) 4, 8 (C) 0·51 (D) 0·71 129. At a given temperature consider (C) 8, 4 (D) 4, 2 Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g); 134. The number of lines in the ESR spectrum of K1 = 0·05 CD3 is (the spin of D is 1)— 2CO2(g) 2CO(g) + O2(g); K2 = 2 × 10– 12 (A) 1 (B) 3 The equilibrium constant for the reaction (C) 4 (D) 7 2Fe2O3(s) 4Fe(s) + 3O2 is— 135. The C = O bond length is 120 pm in CO2. The moment of inertia of CO2 would be (A) 1 × 10– 13 (B) 2 × 10– 38 close to (masses of C and O are 1·9 × 10– 27 kg and 2·5 × 10– 27 kg, respectively)— (C) 4 × 10– 15 (D) 2 × 10– 24 (A) 1·8 × 10– 45 kg m2 130. In a bomb calorimeter, the combustion of (B) 3·6 × 10– 45 kg m2 0·5 of compound A (molar mass = 50 g mol– 1) increased the temperature by 4 K. If (C) 5·4 × 10– 45 kg m2 the heat capacity of the calorimeter along (D) 7·2 × 10– 45 kg m2 with that of the material is 2·5 kJ K– 1, the molar internal energy of combustion, in kJ, 136. The fluorescence lifetime of a molecule in a is— solution is 5 × 10– 9 s. The sum of all the noradiative rate constants (Σknr)for the decay (A) 1000 (B) – 1000 of excited state is 1·2 × 108 s– 1. The fluorescence quantum yield of the molecule (C) 20 (D) – 20 is— 131. The translational, rotational and vibrational (A) 0·1 (B) 0·2 partition functions for a molecule are— (C) 0·4 (D) 0·6 f translation –~ 1010 m– 1, f rotation ~– 10, f vibration ~– 1, (kBT/h) ~– 1013 at room temperature, NA ~– 6 × 1023

CSIR-UGC Chemical Sciences (J-14) | 19 137. Solutions of three electrolytes have the same 142. The E ⊗ E direct product in D3 point group ionic strength and different dielectric contains the irreducible representations constants as 4, 25 and 81. The corresponding relative magnitude of Debye-Huckel D3 E 2C3 3C2 screening lengths of the three solutions are— A1 1 1 1 (A) 4, 25 and 81 A2 1 1 – 1 (B) 2, 5 and 9 E2 2 – 1 0 (C) 1/2, 1/5 and 1/9 (A) A1 + A2 + E (B) 2A1 + E (D) 1, 1 and 1 (C) 2A2 + E (D) 2A1 + 2A2 138. Simple H·u·ckel molecular orbital theory— 143. The result of the product C2(x) C2(y) is— (A) considers electron-electron repulsion (A) E (B) σxy explicitly (C) C2(z) (D) i (B) distinguishes cis-butadiene and trans- butadiene 144. Given (C) distinguishes cis-butadiene and 1. Fe(OH)2 (s) + 2e – → Fe(s) + 2OH– (aq); cyclobutadiene E° = – 0·877 V (D) has different coulomb integrals for non- 2. Al3+ (aq) + 3e– → Al(s); E° = – 1·66V equivalent carbons 3. AgBr(aq) + e– → Ag(s) + Br– (aq); E° = 0·071 V The overall reaction for the cells in the direction of spontaneous change would be— 139. For the nondissociative Langmuir type (A) Cell with A & B : Fe reduced adsorption of a gas on a solid surface at a particular temperature, the fraction of Cell with A & C : Fe reduced surface coverage is 0·6 at 30 bar. The Langmuir isotherm constant (in bar– 1 units) (B) Cell with A & B : Fe reduced at this temperature is— Cell with A & C : Fe oxidized (C) Cell with A & B : Fe oxidized Cell with A & C : Fe oxidized (A) 0·05 (B) 0·20 (D) Cell with A & B : Fe oxidized (C) 2·0 (D) 5·0 Cell with A & C : Fe reduced 145. The reagent X used and the major product Y 140. For a set of 10 observed data points, the formed in the following reaction sequence mean is 8 and the variance is 0·04. The ‘standard deviation’ and the ‘coefficient of are— variation’ of the data set are, respectively— O BrCN (A) 0·005, 0·1% (B) 0·02, 0·2% A B (C) 0·20, 2·5% (D) 0·32, 1·0% N N Me Me 141. In the Lineweaver-Burk plot of (initial (A) A : LiAH4 B : Br CN rate)– 1 vs. (initial substrate concentration)– 1 N for an enzyme catalyzed reaction following (B) A : LiAH4 B :NC Br (C) A : NaBH4 B : Br N Michaelis-Menten mechanism, the y- intercept is 5000 M– 1 s. If the initial enzyme CN concentration is 1 × 10– 9 M, the turnover N number is— O (A) 2·5 × 103 (B) 1·0 × 104 (D) A : H2/Pd-C B : N (C) 2·5 × 104 (D) 2·0 × 105 Me

20 | CSIR-UGC Chemical Sciences (J-14) Answers with Hints 10. (C) area of rectangle = 54 units 1. (C) 2. (D) perimeter of square tiles = 4, 8, 20 3. (D) According to question, Thus, minimum no. of tiles needed = 12. 11. (D) (n 2 + n) (2n + 1), n is a + ve integer h ∝ ⎯√⎯a⎯ge n = 1, (1 + 1) (2 + 1) = 6 ∴ h 2 ∝ age n = 2, 6 × 5 = 30 and w ∝ age n = 3, 12 × 7 = 84 Thus, w = 1 (constant) A number is divisible by 6, when it is also h2 divisible by 2 and 3. if weight and height at birth are both zero. Thus, it is always divisible by 6. 4. (B) Mean of A 12. (B) 40 + 80 + 70 + 50 + 60 + 90 + 30 13. (C) Length of longer of two parallel sides of 7 = = 60 trapezium = 5 × 2 = 10 cm DC Dispersion of A = 90 – 30 = 60 relative dispersion (RD) of A 5 = 60 = 1 AB 60 Trapezium 5 Now, Mean of B 5 55 5 = 40 + 80 + 35 + 70 + 85 + 45 + 50 + 75 + 60 9 = 60 dispersion of B = 85 – 35 = 50 55 RD of B = 50 = 0·83 14. (B) Figure clearly indicates that we have to 60 find out BC + BD = h + BD Thus, RD of A > RD of B. C bird 5. (C) 6. (D) tree h 7. (D) {(197 + 315)2 – (197 – 315)2} 45° · (197 × 315)– 1 Archer A 17 m B = + 4 × 197 × 315 × 197 1 315 = 4 10 m building 10 m × 8. (B) A × B × C × D = 24 × 48 ED B × C 32 ⇒ A×B = 3 × 48 Thus, tan 45° = BC 4 AB = 3 × 12 1 = h ⇒ h = 17 17 = 36 = (6)2 Thus, height of tree ⇒ Perfect square. h + BD = 17 + 10 9. (B) horses = donkeys …(1) = 27 m some donkeys = monkeys …(2) 15. (C) Area of ΔABC some monkeys = men …(3) = 1 × AB × AC 2 go through options, eq. (1), (2) and (3) gives = 1 × 3 × 3 = 9 Some donkeys may be men. 2 2

CSIR-UGC Chemical Sciences (J-14) | 21 In rectangle, h = 2b In case of o and p – substitution – NO2 group acts as electron withdrawing group with Thus, OS = b mesomeric effect and it is more dominant 2 with p-substitution. Hence, the order of basicity is— = 0·2 + 0·2 = 0·2 2 II > III > I and ST = 0·2 22. (A) Area of ΔOST = 1 × OS × ST O 2 P2O5 = 1 × 0·2 × 0·2 O 2 as acid anhydride furan Thus, value of ratio O & dehydrating agent 1 2,5-hexanedione 2 intramolecular aldol × 3 × 3 OH O = 1 2 × 0·2 × 0·2 = 225 Cyclopentanone 16. (B) A →+ 1 B →+ 1 C →+ 1 D 23. (C) 4 17. (B) F →+ 3 I →+ 3 L →+ 3 O 32 K +→5 P +→5 U →+ 5 Z 3 21 O P →+ 7 W +→7 D +→7 K For 5-position view 1 4 56 For 6-position 23 H 1 4 800 12 3 18. (B) Price of one shirt = 3 Price of one trousers = 1‚000 24. (A) CH2 CH2 Thus, go through options— Thus, price of 30 shirts Br = 800 × 30 = 8,000 CH2 CH2 3 and that of 30 trousers Br = 1‚000 × 30 = 30,000 Due to chiral plane molecule is chiral. Total = 30,000 + 8,000 = 38,000 25. (A) (CSIR gave option (D), but it is wrong) Thus, right option is 30. H ring-flipping H a a (I) H e He 19. (B) 20. (D) 21. (A) O cis-decalin O H N (II) cflailplpedinagsn,oLtopcoksinsigblgero-up, O H Mesomerism is not possible in case of m- trans-decalin substitution, hence basicity increases due to availability of electrons. Two factors are responsible for less stability of (I)

22 | CSIR-UGC Chemical Science (J-14) (i) non-bonding interaction in concave area. 32. (B) S 201 Pm (ii) 1,3-diaxial interaction with atoms or Cl 103° Cl I 201 Pm groups. II 26. (A) Structure of bis (dimethylglyoximato) non-linear and iso-structural pair nickel (II) OH O Ni − N bond = 4 33. (C) Ozone absorbs UV-B. UV-B (255 nm) Ni − O bond = 0 would otherwise damage the DNA of all CN NC H − bonds = 2 DNA-based life on/near Earth’s Surface. It (intramolecular) acts as a greenhouse sunlight can come in but Ni it can’t go out. Ozone protects earth by keeping it warm. Ozone is a protectant in CN NC upper atmosphere but if found in lower atmosphere where we live, it is poisonous. OH O 27. (B) 28. (B) 34. (A) [AgL4]2+ [AgL6]2+ [AgL4]3+ Ag(II) Ag(II) Ag(III) 29. (D) Positron (positive electron) result from 4d 4s 4p Same diamagnetic the transformation of a proton to a neutron. octahedral (no unpaired e−) 11P → 10n + 0 e + –υ unpaired e− 1 paramagnetic proton neutron positron anti-neutrino ⎯fu⎯sio→n Na2CO3 anti-neutrino balance the spiner. When the 35. (A) FeCr2O4 Na2[CrO4] + Fe2O3 positron is ejected from the nucleur it very Chromite ore sodium quickly collides with an electron in the chromate surroundings. Sometimes positron emission also gives positron and neutrino with Spinel— FeII(CrIII)2O4 ↓ characteristic energy spectrum. 2Cr ←2⎯Al⎯ Cr2O3 ←2⎯C⎯ Na2Cr2O7 pure – Al2O3 –Na2CO3 30. (D) The thiol group has a high affinity for chromium –CO heavy metals, so that proteins containing cysteine, such as metallo thionein, will bind 36. (B) η5 (Fluxional) metals such as Hg & Cd, Pb tightly and η1 (221-298 K) removes toxicity of such metals. Fe OC CO O 37. (A) [Co(NH3)6]+ 3 – d 6 system, Red colour HS OH L = NH3 (thiol gp) NH2 cysteine [Co(H2O)6]+ 3 – d 6 system, green colour 31. (A) L′ = H2O 38. (C) [Ni2(CO)6]– 2 2(Et)(Ph)2P [NiBr2.P(Ph)2(Et)] NiBr2 −78°C M.M. = 0.0 A (Red) M–M bond = 36 – (12 + 20 + 2) = 1 2 in CS2 Square planar (dsp2) room temperature −2 tetrahedral [NiBr2.P(Ph)2(Et)] OC CO Ni (– I) state OC Ni Ni CO (sp3) green, B OC CO sp3 18 e– system M.M. = 3.2 BM (two unpaired e− + orbital 39. (B) SbPh5 contribution) Shows square pyramidal structure due to prior availability of d-orbital

CSIR-UGC Chemical Sciences (J-14) | 23 Ph 43. (A) 1, 2, 4-tri-t-butyl benzene gives the 1,2,5- Ph Ph tri-t-butyl Dewar structure. The driving force for the reaction is probably the relief of steric Sb compression of the t-butyl groups. While the ring closure could be regarded as an allowed Ph Ph disrotatory closure of a 4N-system, the orbital symmetry rules for aliphatic systems cannot {s p} d be arbitrarily extended to aromatic systems. PPh5 = sp3d t-Bu t-Bu t-Bu t-Bu Ph Ph t-Bu H t-Bu hυ P Ph t-Bu t-Bu H t-Bu 1,2,5-tri-t-butyl T.B.P. Dewar structure Ph Ph 44. (C) Eliminate options, all given options have 40. (C) For an oxidative addition reaction to polar solvents except option (C). So the right occur : answer is C. Benzene and carbon tetra (i) non-bonding e – density on the metal chloride are non-polar solvents and do not (ii) Two vacant coordination site ionize PCl5. (iii) A metal with stable O.S. separated by 2 units. 45. (A) H Me H Me Thus, d 8 and d10 is correct option. Ph Slow Ph 41. (A) Gelatin is added during the polaroghaphic OTs measurement carried out using droping RDS mercury electrode to— Me H Me H (i) reduce streaming motion of falling Hg ACOH fast drop (ii) increase Id H Me (iii) increase E1/2 + OTS Ph (iv) eliminate residual current OAc (Dissociation) Me H (association) Due to planar structure of carbocation, attack at both side is possible. 46. (D) 42. (B) (pKa =H93N.9) COOH (pKa = 2.0) H E H EH E COOH (pKa = 3.9) E E aspartic acid N N NN At pH = 5 (acidic character), unstable electron unstable e– deficient But we know that deficient cation cation lower pKa = strong acid E HE E higher pKa = weak acid Therefore, the species exist at pH = 5 is— N at 3-position N N Me Me CH3 benzene ring H3N COO intact COO Thus, I > III > II ⇒ rate of electrophilic aromatic substitution.

24 | CSIR-UGC Chemical Sciences (J-14) Cl 51. (B) A → products, First order reaction We know that is deactivating towards further EASR. [A]f = [A]0 e– nKt 47. (C) [T^ xP^ x] = T^ xP^ x – P^ xT^ x = 0 ∴ ln ⎝⎜⎛[[AA]]0f ⎞⎠⎟ = – nKt (same axis operator in this case do not commute) ∴ T^ x = – –h2 d2 ( (ln [A]f t 2μ dx 2 [A]0 −nK ^Px ih– d = – dx 48. (C) t Me in DMF SN2 I Me Straight line with –ve slope passing through origin. Br Na+−I cis 52. (A) The kinetic chain length can be defined as the average number of monomer moleculers trans-1-bromo-3-methyl small loosely large Me consumed by each effective free radical cyclobutane held generated by initiator. more nucleophilic rate of propagation rate of initiation I υ = 49. (B) Group V element is added to the Si. The = rate of propagation fifth electron on Group V element is not rate of termination bonded and excited into conduction band where they act as charge carriers (extrinsic rate of initiation = rate of monomer depletion. conduction) 53. (D) Triclinic crystal system-most unsymme- Conduction band trical exists in primitive form (a ≠ b ≠ c, α ≠ β ≠ γ ≠ 90°) Energy γc donor band α Valence band β 50. (B) One C3 axis (passing through the C* and ab — C ≡ C — H group and also through the centre of the triangle formed by joining the Examples : K2Cr2O7, CuSO4·5H2O, H3BO3 three hydrogen atoms), σn absent, 3σv present So, no rotational symmetry axis is found. (each passing through C*, —C ≡ C—H group and one of the H atoms, and bisecting the line 54. (C) Hardy-Schulze Rule— joining the other two H atoms). H (1) Coagulation is brought about by ions having opposite charge to that of the sol. C (2) The efficacy of an ion to cause C coagulation depends upon its valency (high C C3V charge). HH H Complete list of symmetry elements — E, C13 (3) The minimum concentration of an C23, 3σv point group— C3v electrolyte required to cause coagulation or flocculation of a sol is called its flocculation value (millimoles per litre) efficacy varies directly as the square of the valency of the ion. 55. (D) In the state of steady approximation, reactions are investigated under such condi- tions that the slowest rate-determining step

CSIR-UGC Chemical Sciences (J-14) | 25 does not exist, one assumes the steady state 61. (C) Ph3C Na O approximation (s.s.a.) for the transient, i.e., O −Ph3CH OEt short-lived intermediate. H OEt In case of consecutive rn, K1 ≠ K2, so (3) is O wrong and in (2), K1 > K2 which is not OEt possible due to long lifetime of Q. So (1) and (4) shows s.s.a. EtOH H3O EtO + OO OEt 56. (D) Expansion against a particular constant external pressure, Pext ∝ P Then, Work done 62. (C) (2E, 4E) – 2, 4-hexadiene ∫V2 H H most reactive diene W = – Pext dv [expansion (–)] H S-cis conformation V1 H W = – Pext (V2 – V1) (2z, 4z) – 2, 4-hexadiene = – 1 (5 – 1) = – 4 H H it also give Diels-Alder –ve sign only indicates expansion. but least reactive as 57. (B) Specific rotation is given by H compared to previous [α]λT = α H Cl Other two options are not working well for α = – 3°, C = 100 mg in 1 mL = 0·1 g in 1 mL D–A reaction. l = 5 cm = 0·5 dm (decimeter) 63. (D) CH2 C N CH C NH ∴ [α]λT = –3 0·1 × 0·5 = – 60° 58. (C) Two phases (α and β) of a species are in equilibrium. Then following conditions— (i) Thermal Equilibrium ⇒ Tα = Tβ XY (ii) Mechanical equilibrium Clearly, X and Y are toutomers because ⇒ Pα = Pβ sharing of H takes place with hetero atom. (iii) Chemical Equilibrium Y is more basic due to availability of electrons ⇒ μα = μβ on N-atom and forming ability of NH2 with 59. (A) The Boltzmann formula for the entropy, external source. S = KB ln W 64. (D) where W is number configurations in the Δ S most probable state of the system O [2,3] O Ph S ∴ ln W = S KB Ph (EtO)3P ⇒ W = eS/KB 60. (A) 1HNMR − + OH EtO−H O P (EtO)3 + δ 5.4 7.2 9.2 I II III S Ph In Benzene, Hydrogens are deshielded by the Side Reaction : P (OEt)3 OEt PhSEt large anisotropic field generated by the e–s in + the ring’s II system. I, III also show ring S EtOH (EtO)3P = O current. Ph

26 | CSIR-UGC Chemical Sciences (J-14) 65. (A) 66. (C) 71. (B) [CSIR gave wrong answer option (A)] 67. (D) np2 configuration [Ru(NH3)5 (isonicotinamide)]3+ Inert np 2S + 1 = 3 3P [Cr(H2O)6]2+ inner + Labile ⎯⎯⎯⎯⎯⎯→ sphere mechanism +1 0 –1 L = 1(P) 2S + 1 = 1 1D isonicotinamide can act as a bridge for electron transfer. L=2 72. (A) [Fe(H2O)6]3+ 2S + 1 = 1 1S For outer electron transfer reaction : L=0 (i) electron transfer from t 2g takes place. 3D is not possible for np2 configuration. (ii) spin state should be same. 68. (C) According to Bohr, ionization energy of H and H like atom is given by (iii) not much difference in bond length. En = – 2π2 Z2 me4 (iv) unsaturated π-acceptor ligand (mixing of (4πε0)2 n2h2 e– donar and acceptor orbitals). Thus, En ∝ Z2 (Z = atomic number) The presence of all the above factors in For Li +2, Z = 3 [Fe(Phen)3]+ 3 makes it’s reaction faster than [Fe(H2O)6]3+. And for H, En = x 73. (A) [Re2(Me2PPh)4Cl4] Therefore, En for Li+ 2 = 9x Configuration : σ2 π4 δ2 δ*2 69. (B) Root Mean Square Speed of the molecule δ* is δ 3RT 1/2 π M ( )Crms = σ ( )Crms ∝ T 1/2 Re2[Me2PPh)4Cl4] (M+) (M+2) M (M) 3.0 3.5 4.0 T = 2T1, M = 1 M1 M—M bond order = BMO – ABMO ∴ 2 2 Crms = ⎜⎜⎛⎝122⎞⎠⎟⎟1/2 = (4)1/2 = 2 = 8 – 2 = 3 for M 2 74. (C) Cis-isomer is active at low concentration 70. (C) Strong electrolyte— cis–PT (NH3)2Cl2 + H2O cis-[Pt(NH3)2Cl (H2O)]+ Strong + Cl Electrolyte Pt binds at 7-position of guanine. C Weak D When a self-complementary aligomer (a C portion of a DNA chain) reacts with the cis isomer, two adjacent guaniners are bound and linear increase with dilution, curve can be Watson-Crick base pairing is disrupted [See also James E. Huheey–Page No.-700]. extrapolated and the value of Λ ∞ can be 75. (A) F m determined. Weak electrolyte— Cl F T-Shape Variation is non-linear, initially very slow, as C approaches zero increase become very fast F but never approaches limiting value (Y-axis).

CSIR-UGC Chemical Sciences (J-14) | 27 19F NMR spectrum Observed spectrum A υ1 Two axial F 2 × 2 × 1 + 1 = 3 (triplet) υυ32 2 equatorial F. 2 × 1 × 1 + 1 = 2 (doublet) υ 2 υ1 = 3T1g (F) → 3T2g (F) υ 2 = 3T1g (F) → 3T1g (P) 76. (D) low temp. 19F NMR spectrum of SF4— 79. (D) doublet of triplets 80. (A) [Fe5(CO)15C] Total valence electron count F = 8 × 5 + 15 × 2 + 4 C2V F = 74 S structure type = 40 + 30 + 4 F = 74 – 60 = 14 = 7e– pair F = (m + 2) = Nido 81. (A) [(η5 – C5H5) Rh (C2H4)2] equatorial F = 2 × 2 × 1 + 1 = 3 at – 20°C shows 1H NMR 2 typical AA1 XX1 pattern at ~ 70°C – single line axial F = 2 × 2 × 1 + 1 = 3 2 One C2 axis is present (passes through S and HH bisects the line joining the two F atoms each. No. σh.2σV present (molecular plane and the plane bisecting the line joining the two F atoms passing through S). Complete list of symmetry elements = E, C2, H C Rh C H 2σV C HC + +2 − SiCl4 Cl Si R MgBrCl HH H Cl 77. (D) RMgBr Cl due to free rotation of the ethylene ligand like this, all compounds given in question about the metal-olefin bond. form Si—C bond due to R species which attacks at Si centre. 82. (D) Very heavy nuclei have a lower binding energy per nucleon than nuclei with an 78. (A) [V(H2O)6]3+, d2 system intermediate mass. U233 and U235 are used as fission by thermal neutrons. 3A2g(F) 3p υ2 υ3 3T1g(P) 29328U + 01n → 239 U β 239 Np 3F 3T2g(F) 92 ⎯⎯⎯⎯→ 93 υ1 ty2 = 23·5 min 3T1g(F) β 29349Pu ⎯⎯⎯⎯→ +2 +1 0 −1 −2 ty2 = 2·3 days All isotopes of Pu are fissile and used as a 2S + 1 = 3 3F, 3P nuclear fuel. L = 3(F) 83. (A) 84. (A) [3T1g (F) → 3A2g (F) – not observed low intensity and high energy portion of 85. (C) [Ni(H2O)6]2+ is d 8 system and have spectrum] regular octahedral geometry (all Ni–O bond lengths are equal). Whereas [Cu(H2O)6]2+ is

28 | CSIR-UGC Chemical Sciences (J-14) d 9 system, shows large John-Teller 91. (D) distortion with Zout (tetragonally elongated). 92. (B) Sm+2 : [Xe] 4f 6 Therefore, Cu–O (equatorial) bonds are shorter than Cu–O (axial) bonds. Eu+2 : [Xe] 4f 7 86. (A) Ir(Ph3P)2(CO)Cl + NO+ BF–4 → Thus, Sm is better oxidant (oxidising agent) [Ir(Ph3P)2 (CO) (NO)Cl]+BF4– statement 1, 3 and 4 are wrong. Square pyramidal Sm+ 2 + e – → Sm+ 1 : [Xe] 4f 7 bent nitrosyl ligand, ∠M–N–O = 124° at apical position 1 e– donor, less e– density on 93. (A) (i) Binding of dioxygen molecule metal, high Bond Order, so N—O stretching frequency, υ NO ~– 1620 cm– 1. PFeII + O2 PFeII O2 ↔ PFeIIIO2– 87. (B) Across the period from left to right, (ii) Forming μ-peroxo complex with second electro-negativity increases. heme Down the group, electronegativity decreases PFeIIO2 + FeII → PFeIII —O—O—FeIII P combining both factors, we have the order (iii) PFeIII—O—O—FeIII P → 2PFeIII—O· ↔ S > As > Al > Ca 2PFeIV = O 88. (A) Cl Cl −2 (iv) 2FeIV = O + PFeII → PFeIII—O—FeIII P (decrease in size, Fe+ 2 → Fe+ 3) Me2NCH2CH2CH2PPh2 Followed by changes in protein conformation takes place. K2 Pd 1 : 2 ratio 94. (C) Isolobal pairs – electronically equivalent Cl Cl − groups KSCN Mn(CO)5 = 7 + 10 = 17 less than 1 for CH3 = 6 + 3 = 9 inertness SCN Me2 N CH2 Pd CH2 NCS CH2 Fe(CO)4 = 8 + 8 = 16 less than 2 for PPh2 O=8 inert gas configuration On warming, S–bonded isomer is converted to the N- Mn(CO)5 = 17 Co(CO)3 = 9 + 6 = 15 different bonding isomer, which is presumably slightly RS = 17 R2Si = 4 + 2 = 6 more stable. So, P opposite to N and S opposite to N. 95. (C) anion > atom > cation – order of size 89. (A) 1224Mg + (d‚ α) 2121Na + 2He4 Therefore, S2– > S > S2+ > S4+ 1H2 ⎯⎯→ Due to increased ionic radii in anions. Denterium 96. (A) (i) Q >> heat of chemical reaction. Br H H (ii) total no. of P and n is conserved. (iii) no enough energy in d-particle. n-Bu3 SnH 90. (A) At PH = 7 AIBN H Benzene reflux N H CO2 Zn O C O H N Zn O O H H HO HO N H H H interconversion of CO2 and O cis at hydrogen carbonate ion. ring joint Zn O CO H HH H Final HO H product

CSIR-UGC Chemical Sciences (J-14) | 29 Side Rh HO a C N NC C + N2 100. (D) d b CH3 H3Ce CN c O CN CN initiator H AIBN Bu3Sn H C CN Bu3Sn + C CN > C = O group shows maximum value around 170 alkene carbon – 120 – 150 H (almost) 97. (C) Me(a) adjacent to electronegative –O atom (more value). O N−NHTs In case of c and d there is α, β-unsaturated H H2 N−NHTs H carbonyl case and (d) carbon have more value due to more deshielding effect. −H2O HH Thus, a : 52; b : 167; c : 125; d : 143; e : 19 O N=N Ts 2 BuLi 101. (A) CO Li H N−N Ts Li Li H −N2 H H H −Ts H H2O COOH 98. (B) Bu4N+Br− NBu4 102. (C) anti-addition Br retention of anhyd. CF3COOH O OH configuration CH2Cl2 2ATP HO2C H 3X SCoA 2NADPH OH acetyl CoA mevalonic acid ATP Br O OH H elimination 99. (D) OPP HO OPP O HH H Ph H N isopentenyl O N N Ph N pyrophosphate N=N=N N Ph N Intermediate there is no role of lipoic acid. CF3COOH ATP – 3 molecules −N2 NADPH – 2 molecules O acetyl CoA – 3 molecules. Ph H Ph {PP = pyrophosphate group transferred from N O ATP} N −H

30 | CSIR-UGC Chemical Sciences (J-14) 103. (D) 108. (A) O hv O H γ OH O O OCN αβ N N Ph O hυ O H OH HH NO OH O Ph HO touto- (A) N O merisation O NO N (B) (A) H2 Raney Ni O 104. (A) Et3N HN O −CO2 H Ph O −NH2−PH BnO OMe NH O OH OH MeOH, H2O NO O AcOH O H2N NH2 BnO OMe 109. (A) BnO O NH2 NH2 O ONa ONa BnO OCH3 Na OCH3 radical OCH3 NH (B) heat O OCH3 Xylene OCH3 dimeriza- OCH3 tion −HOMe H2N OMe O ONa ONa ketyl radical unstable initial (A) anions product 105. (A) O O O O O OH iso amyl C O 1st e− O 2nd e− transfer nitrite transfer NN O NH2 OO −N2 Me3SiCl −CO2 1,2-diketone anthronilic acid OSiMe3 O H3O O OSiMe3 OH enediolate dimerization m/e = 76 106. (D) 110. (C) IR : 2210 cm– 1 (presence of ≡ bond C ≡ C, C ≡ N) organozinc 13CNMR : 8 types of C–atom. Et2Zn H Zn + 2B δ 1·4 (t‚ J = 7·1 Hz‚ 3H) intermediate H I δ 4·4 (q‚ J = 7·1 Hz‚ 2H) —CH2—CH3 with O H 3B (X) organoborane δ 7·7 (d‚ J = 7·0 Hz‚ 2H) δ 8·2 (d‚ J = 7·0 Hz‚ 2H) aromatic protons Pd(PPh3)4 CH3 Thus, we have structure : CH3 H O product O 107. (B) NC

CSIR-UGC Chemical Sciences (J-14) | 31 111. (B) Br CH2CH2Br Ph 115. (A) H NaNH2 NΔ H3COOC Ph OO N COOCH3 H3COOC COOCH3 Br Ph Ph Ph NaNH2 H CH2 CH2 N O H3COOC COOCH3 Br −HBr Ph Ph (I) O Br O 112. (B) A O CH2 CH2 O2, hv O Br CH2CH2Br O rose bengal O O (II) Rose Bengal is used in synthetic chemistry 116. (A) to generate singlet oxygen from triplet oxygen, which reacts with an alkene. H COOMe COOMe A : O2, Rose Bengal, hυ heat O H OO O COOMe (4 + 2)CA H COOMe (4 + 2)CA O KO C N N C OK O MeOOC COOMe O B, AcOH O H (H2, Pd/C – not useful here) endo 113. (D) H Li O product N N S-BuLi Stable due to secondary interaction between O −78°C O double bond and lone pair of electrons on −S-BuH oxygen atom. ICH2CH2I 117. (D) I O Se O H dioxane N reflux C (A) α−oxidation O O N Strong base like LDA and Buhi can abstract C i−PrOH aq. NaCN H attached with benzene ring. MnO2 Me2 NH (Oxidizing 114. (A) Energy of a particle confined in cubic agent) box is O E3D = (nx2 + ny2 + nz2)8hm2a2 118. (D) The four normalized two electrons spin eigen functions with correct exchange Let K = h2 properties are— 8ma 2 Symmetric : ⎪⎧⎪⎨⎩⎪⎪[βαα((11(1)))βαβ((2(22))) + β(1) α(2)]/√⎯ 2 Thus, the degeneracy of the level, E = σK, twice that of the lowest level = 3 (2, 1, 1) (1, 1, 2) (1, 2, 1) ⎯⎯⎯ E = 6K Antisymmetric : [α(1) β(2) – β(1) α(2)/⎯√2 (1, 1, 1) ⎯⎯⎯⎯ E = 3K

32 | CSIR-UGC Chemical Sciences (J-14) Thus, ψ(0) = 1s(1) 1s(2) · 1 122. (B) The reaction is simply known as acetal formation of glucose which is driven by ⎯√ 2 anomeric effect. [α(1) β(2) – β(1) α(2)] OH OH OH HO CHO O So, spin part = 1 [α(1) β(2) – β(1) α(2)] HO MeOH/H+ OH OH HO √⎯ 2 reflux, 24h Spatial part = 1s(1) 1s(2) OH OH 119. (C) All species are in gaseous phase, Hence Ph O OH bonding P=1 O O PhCH(OMe)2 HO O interaction number of components, C = N — E HO H HO C−O σ* where N = 4 OH OMe OH OMe axial anomer E = equations = 1, viz, anomeric effect 2NO + O2 → 2NO2 123. (C) If M, M + 2 and M + 4 in the ratio of 1 : 2 : 1, this confirms two Br group in A as So, C = 4 – 1 = 3 (a + b)2 = a2 + b2 + 2ab degrees of freedom, = 1:2:1 F = C–P+2 Thus, = 3–1+2=4 OH Br Thus, P = 1‚ C = 3‚ F = 4 HBr 120. (B) 1 (HX) 23 Br −H2O Br NGP 7 4 124. (A) Quinine belongs to the quinoline group TsCl 65 of alkaloids and is known as a cinchona −HCl alkaloid. OH OTs Natural product Group OTs Ephedrine — Phenyl ethly amine OTs Colchicine — Tropolone Strychnine — Tetrahydrooxepine 121. (A) Wave function for ground state (one-D 125. (B) E0 = 0·125h 2 ma2 harmonic oscillator) ψ(x) = x (x – a), E1 ( )ψ(0) =α 1/4 π linear combination of x (x – a) and x2(x – a)2, e– αx2/2 even function E2 ∴ First order correction to the energy Variational principle tells that ground state energy is always less than other higher level ∫ΔE = ψ(0)* H^ (1) ψ(0) dt energies. ∫ ( )= ∞ α 1/4 So, E0 < (E1, E2) –∞ π e– αx2/2.λx . Now, E2 is obtained from even functions (two), it have high value than E1 ( )α 1/4 π e– αx2/2 dx 1/2 ∞ x . e– αx2 dx Thus, E0 < E2 < E1 ( ) ∫= λ α π –∞ 1 Kw) 1 Ka 2 2 ·.· the integrand here is overall an odd 126. (C) pH = 14 – (p + p function, so the integral vanishes. + (log C) × 1 ΔE = 0 2

CSIR-UGC Chemical Sciences (J-14) | 33 p Ka = – log Ka K12 = [Fe]4 [CO2]6 , = – log (2·5 × 10– 5) [Fe2O3]2 [CO]6 = 4·60 K23 = [CO]6 [O2]3 [CO2]6 p Kw = 14 log C = log (0·01) = – 2 [Fe]4 [O2]3 [Fe2O3]2 ∴ pH = 14 – 1 (14) + 1 (4·6) Now, K12·K23 = 2 4 Thus, Keq = K12·K23 1 + (– 2) × 2 = (0·05)2·(2 × 10– 12)3 = 14 – 7 + 2·3 – 1 = 1 × 8 × 10– 36 400 = 8·3 = 2 × 10– 38 127. (A) We know that for H-atom, energy, K.E. and P.E. holds following relation 130. (B) In bomb calorimeter, the heat of combustion, qV at constant volume is 〈T〉 = – 1 〈V〉 = – 〈E〉 internal energy and given by 2 M Given that 〈E〉 = – 13·598 qV = C × θ × m ·.· 〈T〉 = – 〈E〉 K– 1 × × 50 g mol– 1 0·5 = 2·5 kJ 4K ∴ 〈T〉 = 13·598 = 1000 kJ mol– 1 and 〈V〉 = 2 〈E〉 Since qV always has –ve sign. = 2 × (– 13·598) Thus, molar internal energy = – 1000 k J 〈V〉 = – 27·196 131. (B) atom + diatomic molecule → 128. (B) ψ = 0·8 ϕA + 0·4 ϕB non-linear transition state – normalized M.O. ↓ ϕA and ϕB are also normalized product An overlap integral is a direct measure of the Thus, ( )A = RT frot extent of the overlap of the orbitals centered h ftrans·frot on two different nuclei. ( )= kBT ·NA f h ftrans·frot We know that, 132. (A) (h K 1) = (1 1 0) ψ = aϕA + bϕB a = 0·8, b = 0·4 ∴ dhkl = a Thus, overlap integral, ⎯√⎯h⎯2 ⎯+⎯k⎯2⎯+⎯l2 ∫SAB = aϕA·bϕB dt = 4·242 Å = 3Å 1·414 = ab = 0·8 × 0·4 133. (A) AxBy = 0·32 A atoms – all corners and face centre 129. (B) K1 = [F[Fee2O]23[]C[COO2]]33, = 1 × 8 + 1 × 6 8 2 K2 = [CO]2 [O2] = 1+3=4 [CO2]2 B atoms – 4 tetrahedral voids Keq = [F[Fee]42O[O3]22]3, Thus, x = 4, y = 4 A4B4

34 | CSIR-UGC Chemical Sciences (J-14) 134. (D) C• D3, For D, I = 1 (iii) for same and different orbitals – overlap Thus, no. of lines in ESR spectrum integral. = 2nI + 1 (iv) distinguishes cis-butadiene and cyclobutadiene. = 2×3×1+1=7 139. (A) Nondissociative Langmuir adsorption 135. (B) We know, 1 = 1 P + 1 I = μr2 θ Keq 2r = 2 × 120 × 10– 12 m Thus, 1 – 1 = 1 2r ⇒ θ Keq P ←⎯⎯⎯→ O=C=O ( )1 = P 1 – 1 θ μ = m1m2 Keq m1 + m2 ( )1 10 – 0·6 Keq 2·5 × 10– 27 × 2·5 × 10– 27 = 30 bar 1 (2·5 + 2·5) × 10– 27 = ( )= 2 3 Thus, I = 1·25 × 10– 27 × (2·4 × 10– 10)2 30 × bar = 7·2 × 10– 45 kg m2 Thus, Keq = 3 2= 1 30 × 20 Since, C atom is on axis of rotation = do not contribute to I. = 0·05 bar– 1 136. (C) φf = Kf 140. (C) Variance = σ2 Kf + KIC + KISC σ is standard deviation fluorescence lifetime, σ = ⎯√⎯v⎯ar⎯ia⎯n⎯ce τ0 = φf = ⎯√⎯0⎯·04 = 0·20 Kf coefficient of variation 1 τ = Kf + KIC + KISC = σ × 100 — 10 x = 12 × 108 s– 1 (x– is mean of data) = 8·3 × 10– 9 s 0·20 Thus, φf = τ = 0·83 × 10– 8 s = 8 × 100 τ0 5 × 10– 9 s = 2·5% 137. (B) We know that Debye-Huckel Screening 141. (D) [E]0 = 1 × 10– 9 M length K– 1 = ⎛⎝⎜⎜ε kBT 1 e02⎞⎟⎟⎠1/2 1 = Km · 1 + 1 4π nj0 zi2 r rmax [s] 0 rmax Σ i same ionic strength 1 r ε = dielectric constants k – 1 ∝ (ε)1/2 5000 M−1s Thus, it should be 2, 5 and 9. 1/[s]0 138. (C) Simple Huckel molecular orbital theory 1 = 5000 M– 1s is— rmax (i) Energy of similar orbital coulomb rmax = 1 × 10– 3 Ms– 1 integral. 5 (ii) adjacent atoms – Resonance integral

CSIR-UGC Chemical Sciences (J-14) | 35 1 × 10– 3 Ms– 1 143. (C) There are two fold axes at right angles to 5 one another, there must necessarily be a K2 = third at right angles to both. 1 × 10– 9 M = 1 × 106 S– 1 [x1, y1, z1] C2(x) [x1, − y1, − z1] C2(y) 5 C2(x) [− x1, − y1, z1] [x1, − y1, z1] = 2 × 105 S– 1 TON = K2 = rmax Thus, C2(x).C2(y) = C2(z) [E]0 144. (B) AB C 142. (A) D3 E 2C3 3C2 ΔG° = 1·654F 4·98F – 0·071F 11 1 –1 Cell with A and B ⇒ since less +ve value A1 1 –1 0 of A, 10 A2 1 Thus, Fe will reduce 10 Cell with A and C ⇒ Fe oxidized E2 2 11 1 –2 O E⊗E 4 40 145. (A) LiAlH4 Br C N Let us check with options N reducer N Br C = O bond Me (i) A1 + A2 + E 4 Me (ii) 2A1 + E 4 CN Br N (iii) 2A2 + E 4 Br N CN (iv) 2A1 + 2A2 4 Me

Chemical Sciences CSIR-UGC NET/JRF Exam. (December 2014) Solved Paper

December 2014 Chemical Sciences PART–A 4. If n is a positive integer, then 1. Average yield of a product in different years n(n + 1) (n + 2) (n + 3) (n + 4) (n + 5) (n + 6) is shown in the histogram. If the vertical bars indicate variability during the year, then is divisible by— during which year was the per cent variability over the average of that year the least ? (A) 3 but not 7 (B) 3 and 7 (C) 7 but not 3 (D) neither 3 nor 7 5. The area (in m2) of a triangular park of dimensions 50 m, 120m and 130 m is— (A) 3000 (B) 3250 (C) 5550 (D) 7800 (A) 2000 (B) 2001 6. Lunch-dinner pattern of a person for m days is (C) 2002 (D) 2003 given below. He has a choice of a VEG or a NON-VEG meal for his lunch/dinner— 2. A rectangle of length d and breadth d/2 is revolved once completely around its length (1) If he takes a NON-VEG lunch he will and once around its breadth. The ratio of have only VEG for dinner. volumes swept in the two cases is— (2) He takes NON-VEG dinner for exactly 9 days. (3) He takes VEG lunch for exactly 15 days. (4) He takes a total of 14 NON-VEG meals. What is m ? (A) 1 : 1 (B) 1 : 2 (A) 18 (B) 24 (C) 20 (D) 38 (C) 1 : 3 (D) 1 : 4 7. A bank offers a scheme wherein deposits 3. A long ribbon is wound around a spool up to made for 1600 days are doubled in value. The a radius R. Holding the tip of the ribbon, a interest being compounded daily. The interest boy runs away from the spool with a constant accrued on a deposit of 1000 over the first speed maintaining the unwound portion of the 400 days would be — ribbon horizontal. In 4 minutes, the radius of (A) 250 (B) 183 the wound portion becomes R . In what (C) 148 (D) 190 √⎯ 2 8. What is the next number of the following R further time, it will become 2 ? sequence ? 2, 3, 4, 7, 6, 11, 8, 15, 10 … (A) 12 (B) 13 (C) 17 (D) 19 (A) √⎯ 2 min (B) 2 min 9. Two locomotives are running towards each (C) 2√⎯ 2 min (D) 4 min other with speeds of 60 and 40 km/h. An object keeps on flying to and fro from the front tip of one locomotive to the front tip of the other with a speed of 70 km/h. After 30

CSIR-UGC Chemical Sciences (D-14) (II) | 3 minutes, the two locomotives collide and the 15. A person sells two objects at 1035 each. On object is crushed. What distance did the the first object he suffers a loss of 10% while object cover before being crushed ? on the second he gains 15%. What is his net loss/gain percentage ? (A) 50 km (B) 45 km (C) 35 km (D) 10 km (A) 5% gain (B) < 1% gain 10. Weights (in kg) of 13 persons are given (C) < 1% loss (D) no loss, no gain below— 16. Continue the sequence 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 2, 5, 10, 17, 28, 41, –, –, – 94. (A) 58, 77, 100 (B) 64, 81, 100 Two new persons having weights 100 kg and (C) 43, 47, 53 (D) 55, 89, 113 79 kg join the group. The average weight of the group increases by— 17. A ladder rests against a wall as shown. The top and the bottom ends of the ladder are (A) 0 kg (B) 1 kg marked A and B. The base B slips. The central point C of the ladder falls along— (C) 1·6 kg (D) 1·8 kg 11. A code consists of at most two identical A letters followed by at most four identical digits. The code must have at least one letter C and one digit. How many distinct codes can be generated using letters A to Z and digits 1 to 9 ? (A) 936 (B) 1148 (C) 1872 (D) 2574 B 12. Two solid iron spheres are heated to 100°C (A) a parabola (B) the arc of a circle and then allowed to cool. One has the size of a football; the other has the size of a pea. (C) a straight line (D) a hyperbola Which sphere will attain the room tempera- ture (constant) first ? 18. 20% of students of a particular course get jobs within one year of passing. 20% of the (A) The bigger sphere remaining students get jobs by the end of second year of passing. If 16 students are still (B) The smaller sphere jobless, how many students had passed the course ? (C) Both spheres will take the same time (A) 32 (B) 64 (D) It will depend on the room temperature 13. Find the missing letter— (C) 25 (D) 100 A ?QE 19. Binomial theorem in algebra gives (1 + x)n = a0 + a1x + a2x2 + …… + anxn, where a0, a1, CMS C ……, an are constants depending on n. What is the sum a0 + a1 + a2 + …… + an ? EKUA G I WY (A) 2n (B) n (A) L (B) Q (C) n2 (D) n2 + n (C) N (D) O 14. The least significant bit of an 8-bit binary 20. A sphere is made up of very thin concentric number is zero. A binary number whose value shells of increasing radii (leaving no gaps). is 8 times the previous number has— The mass of an arbitrarily chosen shell is— (A) 12 bits ending with three zeros (A) equal to the mass of the preceding shell (B) 11 bits ending with four zeros (B) proportional to its volume (C) 11 bits ending with three zeros (C) proportional to its radius (D) 12 bits ending with four zeroes (D) proportional to its surface area

4 | CSIR-UGC Chemical Sciences (D-14) (II) PART–B 29. The rate of the reaction hv 21. The correct order of the retention of cations on a sulfonated cation exchange resin column Ni(CO)4 + PPh3 ⎯→ [Ni(CO)3 (PPh3)] + CO is— depends on— (A) Ag+ > K+ > Na+ > Li+ (A) concentration of both the reactants (B) K+ > Na+ > Ag+ > Li+ (B) concentration of Ni(CO)4 only (C) Li+ > Na+ > K+ > Ag+ (C) concentration of PPh3 only (D) Li+ > Na+ > Ag+ > K+ (D) the steric bulk of PPh3 22. Among F–, Na+, O2– and Mg2+ ions, those 30. The hapticities ‘x ’ and ‘y ’ of the arene having the highest and the lowest ionic radii respectively are— moieties in the diamagnetic complex [(ηx - (A) O2 – and Na+ (B) F– and Mg2 + C6H6) Ru(ηy - C6H6)] respectively are— (C) O2 – and Mg2 + (D) Mg2 + and O2 – (A) 6 and 6 (B) 4 and 4 (C) 4 and 6 (D) 6 and 2 23. In a polarographic measurement, (aqueous 31. The correct order of the isomeric shift in Mossbauer spectra (57Fe source) of iron KCl solution used as supporting electrolyte) compounds is— an applied potential more than + 0·4 V, (A) Fe(II) > Fe(III) > Fe(IV) results mainly in the formation of— (B) Fe(III) > Fe(II) > Fe(IV) (A) HgI (B) HgII (C) Fe(IV) > Fe(III) > Fe(II) (C) Cl2 (D) O2 (D) Fe(IV) > Fe(II) > Fe(III) 24. The reaction between SbF5 and two equiva- 32. The δ-bond is formed via the overlap of— lents of HF leads to the formation of— (A) dx2 – y2 and dx2 – y2 orbitals (B) dxz and dxz orbitals (A) H2SbF3 + 2F2 (B) HSbF2 + 3F2 (C) dxy and dxy orbitals (C) SbF3 + H2 + 2F2 (D) [SbF6]– [H2F]+ (D) dyz and dyz orbitals 25. The extent of π-electron conjugation in mac- 33. Reductive elimination step in hydrogenation rocyclic rings of (a) heme, (b) coenzyme B12 of alkenes by Wilkinson catalyst results in and (c) chlorophyll follows the order— (neglecting solvent in coordination sphere of Rh) (A) (a) > (c) > (b) (B) (a) > (b) > (c) (C) (c) > (a) > (b) (D) (b) ≈ (a) > (c) 26. The point group symmetries for trans- (A) T-shaped [Rh(PPh3)2Cl] [Cr(en)2F2]+ and [TiCl6]3–, respectively, are— (B) Trigonal-planar [Rh(PPh3)2Cl]2 + (C) T-shaped [Rh(H)(PPh3)Cl]+ (A) D4d and D3d (B) D3d and D4d (D) Trigonal-planar [Rh(H)(PPh3)2] (C) D4h and D3h (D) D3h and D4h 27. The S and L values for 15N atom respectively, are— (A) 1 and 1 (B) 1 and 0 34. In the following reaction 2 2 [PtCl4]2 – + NO2 ⎯→ A N⎯H→3 B compound B is— (C) 1 and 0 (D) 3 and 0 (A) trans–[PtCl2(NO2)(NH3)]– 2 (B) cis–[PtCl2(NO2)(NH3)]– (C) trans–[PtCl2(NH3)2] 28. The product of the reaction of propene, CO (D) cis–[PtCl2(NO2)2]2– and H2 in the presence of CO2(CO)8 as a catalyst is— (A) butanoic acid (B) butanal (C) 2-butanone (D) methylpropanoate

CSIR-UGC Chemical Sciences (D-14) (II) | 5 35. Co4(CO)12 adopts the— (A) 2, 3 and 3, 3 (B) 3, 3 and 2, 3 (A) closo—structure (C) 3, 3 and 2, 2 (D) 2, 4 and 3, 2 (B) nido—structure (C) arachno—structure 41. In the most stable conformation of neomen- (D) hypho—structure thol, stereochemical orientation of the three substituents on the cyclohexane ring are— 36. The major product formed in the following reaction is— (A) OH : equatorial; i-Pr : equatorial and Me : equatorial (A) (B) (B) OH : axial; i-Pr : equatorial and Me : equatorial (C) OH : equatorial; i-Pr : equatorial and Me : axial (D) OH : equatorial; i-Pr : axial and Me : equatorial 42. The reaction of 1-bromo-2-fluorobenzene with furan in the presence of one equivalent of Mg gives— OCH2OCH3 (A) (B) HOOC (C) (D) (C) (D) H3C 37. The compound that is antiaromatic is— I II III IV 43. Identify correct statements for mercury as an environment pollutant : (A) I (B) II (C) III (D) IV 1. Carbanionic biomethylation converts it to MeHg+. 38. The decreasing order of basicity of the following compounds is— 2. Thiol group of cysteine has strong affinity for mercury. H N NN 3. Mercury containing industrial catalyst release caused Minamata disaster. The correct answer is— N N HN N N (A) 1 and 2 (B) 1 and 3 (C) 2 and 3 (D) 1, 2 and 3 I II III IV 44. The increasing order of pK a value of the circled hydrogens in the following compounds (A) III > II > III > IV (B) IV > I > II > III is— (C) III > II > I > IV (D) IV > III > II > I 39. The configurations of carbon atoms C3 and C4 in D-ribose, respectively, are— (A) R and S (B) S and R (C) R and R (D) S and S 40. The number of histidine amino acid nitrogen I II III atoms coordinated to bimetallic active site of oxyhemocyanin and oxyhemerythrin, respec- (A) I < II < III (B) I < III < II tively, are— (C) II < I < III (D) II < III < I

6 | CSIR-UGC Chemical Sciences (D-14) (II) 45. The absolute configurations of the chiral 49. The major product of the following reaction centres of starting ketone in the following is— reaction is— Major product (A) (B) (A) 3R, 6S (B) 3S, 6S Br (C) 3R, 6R (D) 3S, 6R (C) (D) 46. The product for the following sequence of S reactions is— Br 50. The major product of the following reaction is— (A) (B) (A) (B) (C) (D) (C) (D) 47. A compound with molecular formula C4H6O2 51. For the cell reaction, shows band at 1770 cm– 1 in IR spectrum and peaks at 178, 68, 28, and 22 ppm in 13C NMR Sn(s) + Sn4 + (aq) 2Sn2 + (aq), spectrum. The correct structure of the com- pound is— separate electrode reactions could be written (A) (B) with the respective standard electrode poten- tial data at 25°C as— (C) (D) Sn4 + (aq) + 2e → Sn2 + (aq), E0 = + 0·15 V 48. The cyclic product(s) of the following photo- chemical reaction is(are)— Sn2 + (aq) + 2e → Sn(s), E0 = + 0·14 V ⎯⎯⎯h⎯ν⎯⎯→ Vapour phase When RT/F is given as 25·7 mV, logarithm of the equilibrium constant (ln K) is obtained (A) only cis-1, 2-dimethylcyclopentane (B) only trans-1, 2-dimethylcyclopentane as— (C) a mixture of cis- and trans-1, 2-dimethy- (A) 22·6 (B) 226 lcyclopentanes (D) only 2, 6-dimethylcyclohexanol (C) 2·26 (D) 2·26 × 10– 1 52. For a particle of mass m confined in a box of length L, assume Δx = L. Assume further that Δ p (min) = (p 2 )1/2. Use the uncertainty principle to obtain an estimate of the energy of the particle. The value will be— (A) h2 (B) h2 (8mL2) (8mL2) (C) h2 (D) h2 (32mL2) (2mL2)

CSIR-UGC Chemical Sciences (D-14) (II) | 7 53. The mass of metastable ion produced due to 59. decomposition of F1+ in the following mass fragmentation sequence is— Diethyl phthalate ⎯→ F1+ ⎯→ F2+ + CO M+‚ 222 (177) (A) 141·2 (B) 125·4 (C) 45·0 (D) 210·2 54. If the component of the orbital angular mom- Identify the speed distribution functions of entum along the molecular axis of a heteronu- Ne, Ar, and Kr with the curves in the figure clear diatomic molecule is nonzero, the rota- above. tional-vibrational spectrum will show— (A) Ne-A, Ar-B, Kr-C (A) P and R branches only (B) Ne-B, Ar-C, Kr-A (B) P and Q branches only (C) Ne-C, Ar-B, Kr-A (C) Q and R branches only (D) Ne-C, Ar-A, Kr-B (D) all the P, Q and R branches 55. The ratio of the relative intensities of the 60. For a process in a closed system, temperature carbon signals in the first order 13C NMR is equal to— spectrum of CD3CI is— (A) 1 : 4 : 6 : 4 : 1 ( )(A) ∂H ( )(B) – ∂A ∂P S ∂V T (B) 1 : 3 : 3 : 1 ( )(C) ∂G ( )(D) ∂H (C) 1 : 6 : 15 : 20 : 15 : 6 : 1 ∂P T ∂S P (D) 1 : 3 : 6 : 7 : 6 : 3 : 1 56. Bond lengths of homonuclear diatomic 61. A reaction A + B + C → D follows the molecules can be determined with the help of mechanism— both— A + B AB (A) rotational and vibrational spectroscopy AB + C → D (B) rotational and rotational Raman spectro- scopy in which first step remains essentially in equilibrium. If ΔH is the enthalpy change for (C) rotational Raman and electronic spectro- the first reaction and E0 is the activation scopy energy for the second reaction, the activation energy of the overall reaction will be given (D) vibrational and electronic spectroscopy by— 57. The biosynthetic precursor of abietic acid (A) E0 (B) E0 – ΔH is— (C) E0 + ΔH (D) E0 + 2ΔH (A) shikimic acid (B) mevalonic acid (C) chorismic acid (D) cinnamic acid 58. The amino acid constituents of artificial 62. Hydrogen is adsorbed on many metal surfaces sweetener given below are— by dissociation (S represents a surface site)— HH || || H2 + —S—S— —S —S — (A) D-Glutamic acid and L-phenylglycine If the pressure of H2 (p) is small, the fraction (B) L-Glutamic acid and L-phenylalanine of the surface covered by hydrogen is propor- (C) L-Aspartic acid and L-phenylalanine (D) L-Aspartic acid and L-tyrosine tional to— (A) p (B) p2 (C) p1/2 (D) p3/2

8 | CSIR-UGC Chemical Sciences (D-14) (II) 63. The exact differential df of a state function average molar mass and (Mw) for the weight average molar mass. The variance of Mn will f (x, y), among the following, is— then be— (A) xdy (B) dx – x dy (A) 39 (B) 3 y (C) ydx – xdy (D) 1 dx – x dy (C) 1 (D) 87 y y2 70. Wavelength (λ in nm) of the Lyman series for ∂ 64. The angular momentum operator Lz = – ih ∂φ an one-electron ion is in the range 24 ≤ λ ≤ 30. The ionization energy of the ion will be has eigenfunctions of the form exp [iAφ]. The ( )closest to1J= 1019 eV — 1·6 condition that a full rotation leaves such an eigenfuction unchanged is satisfied for all the (A) 32 eV (B) 42 eV values of A— (C) 52 eV (D) 62 eV (A) 0, ± 1 , ± 2 , ± 1, ± 4 , … 3 3 3 (B) 0, ± 1, ± 2, ± 3, … PART–C (C) 0, ± 1 , ± 1, ± 3 , … 71. 1H NMR spectrum of free benzene shows a 2 2 peak at ~ 7·2 ppm. The expected chemical shift (in ppm) of C6H6 ligand in 1H NMR (D) 0, 1 , 3 , 5 , … spectrum of [(η6–C6H6)Cr(CO)3] and the 2 2 2 reason for it, if any, is/are— 65. X-ray diffraction does not give any structural (A) 4·5; disruption of ring current information for— (B) 9·0; inductive effect (A) metallic solids (B) ionic solids (C) 7·2 (C) molecular solids (D) amorphous solids (D) 2·5; combination of inductive effect and 66. For an enzyme-substrate reaction, a plot disruption of ring current between 1 and 1 yields a slope of 40s. If the v [s] 0 enzyme concentration is 2·5 μM, then the 72. Match the metalloproteins in column A with their function in column B— catalytic efficiency of the enzyme is— (A) 40 Lmol– 1 s– 1 (B) 10– 4 Lmol– 1 s– 1 Column A Column B (C) 107 Lmol– 1 s– 1 (D) 104 Lmol– 1 s– 1 (a) Oxyhemocyanin 1. hydrolysis of C- terminal peptide 67. 10 ml of 0·02 M NaOH is added to 10 ml of bond 0·02 M acetic acid (pKa = 4·75). The pH of the solution will be closest to— (b) Carbonic anhydrase 2. methylation (A) 7·0 (B) 8·4 (c) Cytochrome P450 3. conversion of CO2 to H2CO3 (C) 5·6 (D) 9·6 (d) Carboxypeptidase A 4. oxidation of 68. For a polydispersed macromolecular colloid, osmometry gives— alkene (A) weight–average molecular weight 5. oxygen storage (B) number–average molecular weight 6. oxygen transport (C) both weight–average and number aver- Codes : age molecular weight (a) (b) (c) (d) (D) viscosity–average molecular weight (A) 6 3 4 1 69. A sample experiment revealed that PVC (B) 5 3 1 6 formed in the medium has (Mn) = 13, and (C) 6 2 3 1 (Mw) = 16, where (Mn) stands for the number (D) 5 4 3 1

CSIR-UGC Chemical Sciences (D-14) (II) | 9 73. The geometric cross-section (in barn) of a (C) Sigmoidal and pH independent; hyperbo- nucleus A = 125, r0 = 1·4 × 10–15 m approxi- lic and pH dependent. mately is— (D) Hyperbolic and pH dependent; sigmoidal and pH independent. (A) 1·05 (B) 1·54 (C) 2·05 (D) 2·54 79. The compound that undergoes oxidative addi- 74. Na[(η5–C5H5) Fe(CO2)] reacts with Br2 to tion reaction in presence of H2 is— give A. Reaction of A with LiAlH4 results in (A) [Mn(CO)5]– B. The proton NMR spectrum of B consists of (B) [(η5–C5H5) Mo(CO)3]– two singlets of relative intensity 5 : 1. (C) [IrCl(CO) (PPh3)2] Compounds A and B, respectively, are— (D) [η5–C5H5)2 ReH] (A) (η5–C5H5) Fe(CO)2Br and (η5–C5H5) 80. The spin-only magnetic moment and the Fe(CO)2H (η4–C5H5) spectroscopic ground state term symbol of manganese centre in [MnF6]3 – ion respecti- (B) (η4–C5H5) Fe(CO)2Br2 and Fe(CO)2HBr vely, are— (C) (η5–C5H5) Fe(CO)2Br and (η4–C5H5) (A) 4·9 BM and 5D (B) 4·9 BM and 4F Fe(CO)2(H)2 (C) 3·9 BM and 3D (D) 4·9 BM and 3F (D) (η5–C5H5)Fe(CO)2Br and (η5–C5H5) 81. The main products of the reaction of equimo- lar quantities of XeF6 with NaNO3– are— Fe(CO)2HBr 75. Which of the following will result in devia- (A) XeOF4, NaF and NO2F tion from Beer’s law ? (B) XeO2F2, NaF, NOF and F2 1. Change in refractive index or medium. (C) XeOF4, NaNO2 and F2 2. Dissociation of analyte on dilution. (D) XeF4, NaNO2 and F2O 3. Polychromatic light. 82. A borane (X) is reacted with ammonia to give 4. Path length of cuvette. a salt of borohydride (Y). The 11B NMR (A) 1, 2 and 3 (B) 2, 3 and 4 spectrum of Y consists of a triplet and a (C) 1, 3 and 4 (D) 1, 2 and 4 quintet. The borane X is— 76. The number of stereoisomers of trans— (A) B2H6 (B) B3H9 [CoCl2(triethylenetetramine)]Br is— (C) B4H8 (D) B5H9 (A) One (B) Two 83. Base hydrolysis of [CoCl(NH3)5]2 + is an overall second order reaction, whereas that of (C) Three (D) Four [Co(CN)6]3 – is of first order. The rates depend in both cases solely on the concentrations of 77. The gas commonly used in generating plasma in Inductively Coupled Plasma-Atomic Emis- the cobalt complex. This may be due to : sion Spectroscopy (ICP—AES) is— 1. Presence of ionizable proton in (A) argon (B) carbon dioxide [CoCl(NH3)5]2 + but not in [Co(CN)6]3 – (C) nitrous oxide (D) hydrogen 2. SN1CB mechanism in the case of 78. Under physiological condition, oxygen is bin- [CoCl(NH3)5]2 + only ding to deoxyhemoglobin and deoxymyoglo- 3. SN1CB mechanism in the case of bin, the binding curve and its pH dependence, respectively, are— [Co(CN)6]3 – only 4. SN1CB mechanism in both the complexes (A) Sigmoidal and pH dependent; hyperbolic and pH independent. Correct explanation(s) is/are— (B) Hyperbolic and pH independent; sigmoi- (A) 1 and 2 (B) 1 and 3 dal and pH dependent. (C) 2 only (D) 1 and 4

10 | CSIR-UGC Chemical Sciences (D-14) (II) 84. Hindered β—diketonates like dpmH (dpmH = (B) PPh3, [Ph3PI]Me, Ph2P(nBu)3 dipivaloylmethane) are used for the separation (C) PPh3, [Ph3PMe]I, Ph3P==CH2 of lanthanides because complexes formed (D) [PPh4]Cl, [Ph3P==CH2]I, Ph3P(nBu)]Li with dpmH can be separated by— (A) Gel permeation chromatography 89. The reaction between diphenyldichlorosilane (B) Gas chromatography and water in 1 : 2 molar ratio gives product A (C) Gel filtration chromatography which on heating above 100°C yields a cyclic (D) Ion exchange chromatography or polymeric product B. The products A and B respectively, are— 85. The final product in the reaction of [Cp*2ThH] with CO in an equimolar ratio (A) is— (A) (B) (B) (C) and (Ph2SiO)n (n = 3, 4, or ∞) (C) (D) and (Ph2SiO)n (n = 3, 4, or ∞) 86. An aqueous solution of [Mn(H2O)6]2+ comp- (D) lex is pale pink in color. The probable reasons for it are— 90. The spin-only (μS) and spin plus orbital 1. presence of 6A1g ground state. (μS + L) magnetic moments of [CrCl6]3 – are— (A) 3·87 BM and 5·20 BM 2. disallowed transition by spin selection (B) 2·84 BM and 5·20 BM rule. (C) 3·87 BM and 6·34 BM (D) 2·84 BM and 6·34 BM 3. presence of 2T2g ground state. 91. The three dimensional structure of compound 4. charge transfer transition. [Co(Co(NH3)4 (OH)2)3]Br6 has— (A) twelve Co—O and twelve Co—N bonds The correct answer is— (B) ten Co—O and ten Co—N bonds (C) fourteen Co—O and ten Co—N bonds (A) 1 and 2 (B) 1 and 3 (D) twelve Co—O and ten Co—N bonds (C) 2 and 3 (D) 3 and 4 92. 12-Crown-4 binds with the alkali metal ions 87. According to Wade’s rule, anion C2B9H12 in the following order : adopts— Li+ >> Na+ > K+ > Cs+. It is due to the— (A) closo—structure (A) right size of cation (B) change in entropy being positive (B) nido—structure (C) conformational flexibility of crown ether (D) hydrophobicity of crown ether (C) arachno—structure (D) hypho—structure 88. The reaction of phosphorus trichloride with phenyllithium in 1 : 3 molar ratio yields product ‘A’, which on further treatment with methyl iodide produces ‘B’. The reaction of B with nBuLi gives product ‘C’. The products A, B and C, respectively, are— (A) [PPh4]Cl, [Ph2P==CH2]I, Ph2P(nBu)

93. Complexes HM(CO)5 and [(η5–C5H5) CSIR-UGC Chemical Sciences (D-14) (II) | 11 M′(CO)3]2 obey the 18-electron rule. Identify (C) M and M′ and their 1H NMR chemical shifts relative to TMS— (D) (A) M = Mn, – 7·5; M′ = Cr, 4·10 (B) M = Cr, – 4·10; M′ = Mn, – 7·5 (C) M = V, – 7·5; M′ = Cr, 4·10 (D) M = Mn, – 10·22; M′ = Fe, 2·80 94. Gel permeation chromatography can be used to separate which of the following ? 1. Lanthanides 2. Alkaline earths 3. Fatty acids 4. Low molecular weight peptides The correct answer is— (A) 1 and 2 (B) 2 and 3 (C) 3 and 4 (D) 1 and 4 95. The correct schematic molecular energy diagram for SF6 molecule is— (A) 96. The major product formed in the following reaction is— (B) (A) (B) (C) (D) 97. The product B in the following reaction sequ- ence is— ⎯B⎯r2(⎯1 e⎯qu⎯iv→) A e⎯xc⎯es⎯s M⎯e⎯2N→H B hexane‚ rt.

12 | CSIR-UGC Chemical Sciences (D-14) (II) (A) 99. The major product of the following reaction is— (B) ⎯1⎯. H⎯3⎯O+⎯‚ h→ν 2. NaOH (C) (A) (B) (D) 98. The major product formed in the following (C) (D) transformation is— 100. The major product of the following reaction is— (A) ⎯C⎯F3⎯CO⎯2H→ (A) (B) (B) (C) (C) (D) (D) OCOCF3 O MeO

CSIR-UGC Chemical Sciences (D-14) (II) | 13 101. The products A and B in the following reac- 103. The products A and B in the following reac- tion sequence are— tion sequence are— O LDA A ⎯⎯Pd⎯C⎯l2‚⎯O⎯2 → B 1. NaOEt ⎯⎯→ ⎯⎯⎯→ Br CuCl DMF.H2O 2. H3O+ (A) 1. 2. (A) 1. 2. (B) 1. 2. (B) 1. 2. (C) 1. 2. (C) 1. 2. (D) 1. 2. (D) 1. 2. 102. The major product for the following reaction is— 104. The products A and B in the following reac- tion sequence are— (A) 1. 2. (A) (B) (B) 1. 2. (C) (D) (C) 1. 2.

14 | CSIR-UGC Chemical Sciences (D-14) (II) (D) 1. 2. 107. The correct combinations of the reactions and the reagents are— Reactions (A) (B) 105. The major product of the following reac- tion— (C) Reagents (P) PPh3 and EtO2CN== NCO2Et (Q) POCl3.EtyN (R) H2SO4 (A) (B) (A) A—P, B—Q, C—R (B) A—Q, B—R, C—P (C) A—P, B—R, C—Q (D) A—Q, B—P, C—R (C) (D) 108. The products A and B in the following reaction sequence are— 106. The major product formed in the following (A) 1. 2. reaction sequence is— (B) 1. 2. (A) (C) 1. 2. (B) (D) 1. 2. (C) 109. The major product of the following reaction is— (D)

CSIR-UGC Chemical Sciences (D-14) (II) | 15 (A) (A) X = PhSO2H, BF3OEt2 and Y = CH2==CHCOOEt, BF3 OEt2 (B) (B) X = 1. PhSH, PTSA; 2. m-CPBA and Y = CH2== CHCOOEt, BF3 OEt2 (C) (C) X = PhSO3H, BF3 OEt2 and Y = LDA, CH2 == CHCOOEt (D) X = 1. PhSH, PTSA; 2. m-CPBA and Y = LDA, CH2== CHCOOEt 113. The reactive intermediate and the product (D) formed in the following reaction are— 110. The major product of the following reaction ⎯(⎯n-B⎯u⎯3S⎯n)2→ is— BF3OEt2‚ hν (A) Free radical and 4-iodomethyloxepan-2 one (B) Free radical and 5-iodooxacan-2-one (A) (B) (C) Carbene and 3-oxabicyclo [5.1.0] octane 2-one (D) Carbene and (E)-5-iodopent-3-en-1-yl acetate (C) (D) 114. An organic compound having molecular for- mula C10H12O2 exhibits the following spectral 111. The following reaction gives a product data— (racemic) which exhibits the following NMR IR : 3400 (br), 1600 cm– 1. data— 1H NMR : δ 1·85 (3H, d, J = 6Hz), 3·8 (3H, s), 1H NMR : δ 2·67 (2H, s), 5·60 (2H, s) ppm; 5·0 (1H, s, D2O exchangeable), 6·0 (1H, dq, J 13C NMR : δ 170·3, 129·0, 105·0, 25·4 ppm. = 18, 6 Hz), 6·28 (1H, d, J = 18 Hz), 6·75 The structure of the product (racemic) is— (1H, d, J = 8 Hz), 6·8 (1H, s), 6·90 (1H, d, J = 8 Hz) ppm; 13C NMR : δ 146·5, 144·0, 131·0, 130·5, 123·0, 119·0, 114·0, 108·0, 55·0, 18·0 ppm. The structure of the compound is— (A) (B) (A) (B) (C) (D) (C) (D) 112. In the following reaction sequence, the rea- gents X and Y are, respectively— 115. The major product formed in the following reaction sequence is—

16 | CSIR-UGC Chemical Sciences (D-14) (II) (A) (A) (B) (B) (C) (C) (D) 116. The correct combination of the following reac- (D) tions and their ρ values is— Entry Reaction Entry ρ value A ArNH2 + PhCOCl in P + 2·01 119. The major products A and B formed in the following reaction sequence are— benzene B ArO + EtI in EtOH Q – 0·99 C ArCO2Et + aq NaOH R – 2·69 in EtOH S + 0·78 (A) A—P; B—R; C—P (A) 1. 2. (B) A—R; B—Q; C—P (C) A—R; B—P; C—Q (D) A—Q; B—R; C—S 117. In the following reaction sequence, the struc- (B) 1. 2. ture of the product is— (A) (B) (C) 1. 2. (D) 1. 2. (C) (D) 120. The major product of the following reaction is— 118. In the following reaction sequence, the struc- tures of A and B are, respectively— (A) (B)

CSIR-UGC Chemical Sciences (D-14) (II) | 17 (C) (D) (A) 0 (B) 1 (C) L (D) 2 2 L –∞ ∫[Hint : f (x) δ(x – a)dx = f (a)] +∞ 121. Species A undergoes a unimolecular reaction 124. For a gaseous reaction, 2NO(g) + Cl2(g) → as follows— Non-linear T.S. → 2NOCl, the pre- exponential factor in the rate constant is A+A k1 A* + A proportional to— (B) T– 1/2 (A) T1/2 k– 1 (C) T– 5/2 (D) T– 7/2 k2 125. The probability of finding the harmonic A* ⎯→ P oscillator in the energy level n = 1 is (neglect For this reaction, the first order rate constant zero point energy and assume hν = kBT). at high pressure is k∞. The first order rate (A) e (B) e2 constant becomes k∞ when pressure of A is (C) 1 – e– 2 (D) e– 2 (e – 1) 2 126. At high pressure, the fugacity coefficient of a [A]1/2. real gas is greater than one, because— (A) attractive term overweighs the repulsive term (B) repulsive term overweighs the attractive term (C) repulsive term is equal to the attractive term (D) the system is independent of both the attractive and repulsive terms The value of k1 will be— 127. If the bond length of a heteronuclear diatomic molecule is greater in the upper vibrational (A) k∞ (B) k∞[A]1/2 state, the gap between the successive [A]1/2 absorption lines of P-branch— (C) k∞ – [A]1/2 (D) [A]1/2 (A) increases non-linearly k∞ (B) decreases non-linearly (C) increases linearly 122. The low and high temperature limits of vibra- (D) decreases linearly tional partition function are (θ = hν/k). 128. A quantum particle with fixed initial energy (A) e– θ/T and T e– θ/T (B) e– θ/2T and T e– θ/2T E0 < V is allowed to strike the following four θ θ barriers separately. The transmission probabi- lity is maximum in— (C) e– θ/2T and T e– θ/T (D) e– θ/2T and θ e– θ/2T θ T (A) (B) 123. A particle in a 1-dimentional box of length L is perturbed by a delta function potential, ( )δ x – L , in the middle of the box. The first (C) (D) 2 order energy correction to the ground state will be—

18 | CSIR-UGC Chemical Sciences (D-14) (II) 129. EPR spectrum of a free radical containing (A) D0(H2) = D0(H2+) + I(H) – I(H2) nuclei with nonzero nuclear spin is obtained if (B) D0(H2) = D0(H2+) – I(H) + I(H2) the following selection rules are observed— (C) D0(H2+) = D0(H2) + I(H) + I(H2) (A) Δms = 0, Δm1 = 0 (D) D0(H2+) = D0(H2) – I(H) – I(H2) (B) Δms = ± 1, Δm1 = 0 (C) Δms = ± 1, Δm1 = ± 1 135. Fuel cells provide clean electrical energy to a (D) Δms = 0, Δm1 = ± 1 variety of applications including automobiles 130. Given the following two relations, x1dμ1 + x2dμ2 = 0 and stationary power sources. Normally –– hydrogen combines with oxygen to give and x1dV1 + x2dV2 = 0, …(1) electrical energy and water. If we use butane …(2) instead of hydrogen at 1·0 bar and 298 K, the following reaction occurs— for a binary liquid mixture at constant C4H10(g) + 13 O2 (g) → 4CO2(g) + 5H 2O(l) temperature and pressure, the true statement 2 is that, If the change in the Gibbs free energy of this (A) both the relations are correct. reaction is— (B) relation A is correct, but B is not. 2746·06 kJ mol– 1, involving 26 electrons, its open circuit voltage is— (C) relation B is correct, but A is not. (A) 1·55 V (B) 1·09 V (D) both the relations are incorrect except for very dilute solution. 131. The operators S± are defined by— (C) 3·15 V (D) 2·06 V S± = Sx ± iSy, 136. A solid consisting of only X atoms has a where Sx and Sy are components of the spin close-packed structure with X-X distance of 160 pm. Assuming it to be a closed-packed angular momentum operator. The commutator structure of hard spheres with radius equal to [Sz, S+] is— half of the X-X bond length, the number of (A) hS+ (B) hS– atoms in 1 cm3 would be— (C) –hS+ (D) –hS– (A) 6·023 × 1027 (B) 3·45 × 1023 (C) 6·02 × 1021 (D) 3·8 × 1021 132. The configuration [Ne] 2p1 3p1 has a 3D term. 137. The character table of C2v point group is Its levels are— (A) 3D3/2, 3D1/2 given below. In cis-butadiene molecule the (B) 3D5/2, 3D3/2, 3D1/2 vibrational modes belonging to A2 irreducible representation are IR inactive. The remaining (C) 3D3, 3D2, 3D1 IR active modes are— (D) 3D3, 3D2, 3D1, 3D0 C2v E C2 σv σ′y 133. The fraction of groups condensed at time t in A1 1 1 1 1 z, x2, y2, z any stepwise condensation polymerization (overall second order) reaction is— A2 1 1 – 1 – 1 Rz, xy (A) 1 + kt[A]0 (B) 1+ 1 B1 1 – 1 1 – 1 x, Ry, xz kt[A]0 (C) kt[A]0 (D) 1 + kt[A]0 B2 1 – 1 – 1 1 y, Rx, yz 1 + kt[A]0 kt[A]0 (A) 7A1 + 5B1 + 8B2 134. If D0(A) and I(A) refer respectively to the (B) 9A1 + 4B1 + 7B2 dissociation energy and ionization potential of (C) 7A1 + 3B1 + 7B2 A (where A is either H, H2, or H2+ species), (D) 9A1 + 3B1 + 8B2 the correct relation among the following is—

CSIR-UGC Chemical Sciences (D-14) (II) | 19 138. The product σxy.S z (S4z is the four fold impro- (C) 1saα (1) 1sbα (1) 4 per axis of rotation around the z axis, and σxy 1saα (2) 1sbα (2) is the reflection in the xy plane) is— (D) 1saα (1) 1sbβ (1) 1saα (1) 1sbβ (2) (A) Cz4 (B) Cz4.i (C) C4y (D) C2z 143. The number of microstates that are possible, 139. Among the following figures, when two particles are distributed in four states such that the resulting wave functions are antisymmetric with respect to exchange of the particles, is— (A) 16 (B) 12 (C) 8 (D) 6 the variations of mass adsorbed with pressure 144. When T → ∞, value of the single-particle for a monolayer and a multilayer are repre- partition function will be (given : degeneracy sented by— of level j = gj). (A) A and C respectively (A) 1 (B) g0 (B) A and B respectively (C) ∑j gj (C) C and A respectively (D) 1 (D) B and A respectively ∑j gj 140. According to Huckel theory, the π electron 145. The rate constant for a reaction charge on the central carbon atom in propenyl cation (CH2CHCH2)+ is (in units of electronic A1 + + Bn + ⎯→ P charge). is measured in two different aqueous solu- tions of ionic strengths 0·01 M and 0·04 M. If (A) 1 (B) 1 log k0·04 = 0·3, the charge n on B is closest 2 k0·01 ⎯√ 2 to— (C) 1 (D) 2 (A) 1 (B) 2 141. For some one-electron system with l = 0 and (C) 3 (D) 6 m = 0, the functions N0 e– σ and N1(2 – σ)e– σ/2 Answers with Hints refer respectively to the ground (E0) and first excited (E1) energy levels. If a variational wave functio–n N2(3 – σ)e– σ yields and ave- 1. (B) In 2001, rage energy E, it will satisfy— (250 + 75) + (250 – 75) = 250 2 – – (A) E ≥ 0 (B) 0 ≤ E ≤ E0 and 75 × 100 = 30% 250 (C) – ≥ E1 (D) E0 ≤ – ≤ E1 E E (least per cent variability) 142. A Slater determinant corresponding to the Other years it is 33%, 37·5% and 50% ionic part of the ground state valence bond 2. (B) When a rectangle is revolved around any wave function of H2 molecule is (1saα, 1saβ, axis, its create cylinder with height according 1sbα, 1sbβ are atomic spin orbitals of hydro- gen atoms a and b of the hydrogen molecule) to that axis. (A) 1saα (1) 1saβ (1) Thus, Volume V1 = lwh 1saα (2) 1saβ (2) = length × width × height = d × d × d = d3 2 2 4 (B) 1saα (1) 1sbβ (1) 1saα (2) 1sbβ (2) and V2 = d × d × d = d3 2 2

20 | CSIR-UGC Chemical Sciences (D-14) (II) Thus, ratio V1 = d3 · 2 = 1 10. (B) Average V2 4 d3 2 70 + 72 + 74 + 76 + 78 + 80 + 82 ⇒ V1 : V2 : : 1 : 2 = + 84 + 86 + 88 + 90 + 92 + 94 13 3. (B) Its very common that in 4 minutes the = 82 kg R 1R·41. wound portion becomes = Then in Two new persons average weight ⎯√ 2 = 100 + 79 = 89·5 kg further app. 2 minutes, it will become = R2. 2 ( )1·41 Thus, total average = 1066 + 179 = 83 kg 2 15 = 0·70 4. (B) A number is divisible by 3 if the sum of Thus, increase in average = 1 kg its digits is divisible by 3. 11. (C) 1872 codes can be generated using letters A number is divisible by 7 if after doubling A to Z and digits 1 to 9. last digit and subtracting it from remaining leading truncated number is divisible by 7. 12. (B) Thus, given combination is divisible by 3 and 7. 5. (A) B 130 Cooling is much faster in smaller sphere, so 50 attain room temperature first. 13. (D) From below, we find the gap of 1 letter in English alphabet, so, I J K L M N O A 120 C 14. (B) Area of Δ ABC = 1 × base × height 15. (B) 1035 × 10 = 103·5 ⇒ 1138·50 2 100 = 1 × 120 × 50 1035 × 15 = 155·25 ⇒ + 879·75 2 100 2018·25 = 3000 m2 Thus, % gain = 2018·25 × 100 ≈ < 1% 2070 6. (C) According to given statements, it is clear that the value of m is 20 then he takes 14 non- 16. (A) 2 + 3 = 5 + 5 = 10 + 7 = 17 + 11 = 28 + 13 = 41 + 17 = 58 + 19 = 77 + 23 = 100 veg meals. 7. (D) Don’t use formula, just think, how ? addition of prime numbers in increasing order. 17. (B) Amount doubled in 1600 days (4·4 years) Then in 1·1 years (400 days) interest of ` 1000 A should be around 200. Thus right option is C arc of a circle 190. B 8. (D) Series is increasing and decreasing and option is 19. 9. (C) Since locomotives are running towards each other and collide and object has crushed. Then only speed of object which matters. Thus, distance = speed × time = 70 km/h × 1 hour 18. (C) 25 × 20 = 5 students get job within 1 2 100 = 35 km year of passing

CSIR-UGC Chemical Sciences (D-14) (II) | 21 20 × 20 = 4 students get job by end of 2nd 26. (*) F TiCl6 3– 100 D4h : C4 + 4C2 + 4σv year of passing remaining students = 16 en Cr en + σn So, correct option is 25, (we have to check options) 19. (A) (1 + x)n = a0 + a1x + a2x2 + …… anxn F Then, a0 + a1 + a2 + a3 + …… an = 2n D2h : C2 + 2C2 + 2σv + σn None of the given alternative are correct. 20. (D) 27. (D) 14N or 15N arbitrarily no. of protons same chosen shell S = n = 3 Surface area of sphere = 4πr2 2 2 r is radius, mass depends upon surface area L = +1×1+0×1+–1×1=0 Thus, the mass is proportional to its surface area. Thus, S = 3 ‚ L = 0 21. (A) Cation exchange resin–exchanges cations 2 and depends on bigger size which is attached till last time. Thus, the correct order of 28. (B) Hydroformylation (Oxo synthesis) retention of cations is (distribution constant, k) Ag+ > k+ > Na+ > Li+ CH3 —CH== CH2 ⎯C⎯O⎯+ ⎯H2→ CH3CH2CH2CHO 22. (C) All have 10 e–s but ionic radius is greater CO2(CO)8 (normal) butanol in case of highest – ve charge which gives increased radius (ionic). Thus, O2 – has highest + (CH3)2CHCHO and Mg+ 2 has lowest ionic radii. isobutyraldehyde 23. (A) In a polarographic measurement, (aq. KCl sol. used as supporting electrolyte) an applied hν potential more than + 4V, results the 29. (B) Ni(CO)4 + PPh3 ⎯→ [Ni(CO)3PPh3] formation of Hg+. DME worked at potential + 0·4V to – 0·2V + CO 24. (D) SbF5 + 2HF ⎯→ [SbF6]– [H2F]+ 18 e–s complex. strongest acid in the system, is 1016 times stronger than 100% H2SO4. For substitution, it must follow dissociative 25. (A) Heme – porphyrin ring (11 π) mechanism (SN1). Hence, it depends only on Coenzyme B12 — Corrin ring Ni(CO)4 concentration. Chlorophyll — Chlorin ring (10 π) 30. (C) [(ηx – C6H6)Ru (ηy – C6H6)] Thus, π – e– extent of conjugation is Heme > Chlorophyll > Coenzyme B12 – diamagnetic (a) > (c) > (b) Ru – 8 e– in valence shell, want 10 more e– for 18e– system, so hepticity will be 4 and 6. x = 4, y = 6. 31. (A) Isomeric shift is given by— δ = k(Re2 – Rg2) {[ψs2(0)]b – [ψs2(0)]a} oxidised ferric ions (Fe+3) e– density difference have lower isomeric on nucleus (a = effect than Fe+ 2 source, b = sample) because s – e– density at nucleus of Fe+3 is greater due to a weaker screening effect by delectrons. So, δ : Fe(II) > Fe(III) > Fe(IV)

22 | CSIR-UGC Chemical Sciences (D-14) (II) 32. (C) dxy and dxy orbitals (Re2Cl8)2– 39. (C) R and R Two nodal planes which D-ribose contain the internuclear axis 40. (B) and go through both axis. Oxyhemocyanin = delta bond formation 2d x y or 2dx2–y2 orbitals interacting. 33. (A) after reductive climination step Cl (I) PPh3 Rh 16 VE PPh3 S S = Solvent ⇒ neglecting solvent, [PPh3—R|h—PPh3] 34. (A) [P + Cl4]– 2 + NO2– → Cl − Cl NO2 Cl NO2 Pt NH3 Pt Cl Cl NH3 Cl Trans Oxyhemerythrin trans effect of NO2– > Cl 41. (B) ≡ 35. (B) CO4(CO)12 OH : axial, i-Pr : equato- rial, Me : equatorial Total Valence electron = 4 × 9 + 12 × 2 = 60 Putting n = 4 in 14n + 4 Wade’s Rule 14n + 2 ⇒ Closo We have 14n + 4 = 60 ⇒ Nido structure 14n + 6 ⇒ arachno 36. (C) 42. (C) Br MgBr Mg FF arene intermediate O Diels-Alder reaction 37. (A) O (I) (II) (IV) (III) 43. (D) Biomethylation is a process whereby living organisms produce a direct linkage of a 4nπ system, aromatic non-aromatic methyl group to a metal or metalloid, thus Huckel rule forming metal-carbon bonds found in soil extensively. antiaromatic Thus, all statements are common and correct. 38. (C) Availability of lone pairs on N-atom deci- 44. (C) pka = – log ka, greater ka leads to lower des basicity in these cases. Clearly, pyridine value of pka. is least basic due to N- is part of aromatic ring Electron withdrawing group ⇒ greater ka ⇒ (sp2 system). lower pka. Greater e– density leads to greater basic Electron releasing group ⇒ lowers acidity (ka) ⇒ greater pka. character. Thus, III > II > I > IV

CSIR-UGC Chemical Sciences (D-14) (II) | 23 Thus, Birch reduction e– withdrawing gp, reduction 2-cyano guanidine at ipso and 4-position and alkynes into II < I < III 45. (A) Crams Rule, smaller nucleophile, trans alkenes with trans-geometry aldol product 51. (A) Sn(s) → Sn+ 2 (aq) + 2e–, E1° = + 0·14V E2° = + 0·15V Sn+ 4 (aq) + 2e– → Sn+ 2 (aq) Sn(s) + Sn+ 4 (aq) → 2Sn+ 2 (aq) – nFE3° = – nFE1° – nFE2° 4E3° = 2E1° + 2E2° 46. (C) E3° = 0·29 = 0·145 2 Thus ln k = nFE3° = 4 × 0·145 RT 25·7 × 10– 3V Ketone Cyclohexane = 22·56 ≈ 22·6 carboxaldehyde 52. (B) We know that Br—CH2—O—Me ⎯P(⎯O⎯Et)→3 h DME Δp Δx ≥ 2 α-haloether h wittig reagents ⇒ Δp = 2L 47. (C) C4H6O2 Energy of the particle IR : 1770 cm– 1 implier γ-lactone (Δp)2 ⎝⎜⎛2hL⎠⎟⎞2 1 h2 hence, = 2m = 2m = 8mL2 13C NMR : 178, 68, 28, 22 53. (B) 48. (C) hν ⎯⎯→ vapour phase 2,6-dimethyl cyclohexanone α-cleavage free rotation metastable ion peak, ⎯⎯⎯⎯⎯⎯→ Mixture of cis & trans- of C—C bond 1‚ 2-dimethyl cyclopentanes m* = m22 m1 49. (C) ⎯Br⎯2(e⎯xc⎯es→s) (149)2 Benzothiophene CHCl3‚ r+ = 177 –~ 125·429 54. (D) The selection rule have two consequen- ces— Bromination at 2,3-position (1) Both vibrational and rotational quantum numbers must change. The transition— 50. (A) ΔV = ± 1, ΔJ = 0 (Q-branch) is forbidden. (2) The energy change of rotation can be either subtracted from or added to the energy change of vibration, giving the p- & R-bran- ches of the spectrum respectively.

24 | CSIR-UGC Chemical Sciences (D-14) (II) 55. (D) CD3Cl; n = 3, I = 1 ( )⇒ ∂M = – 1 2nI + 1 = 7 non-pascal triangle ∂x y2 y so, intensity ratio : : 1 : 3 : 6 : 7 : 6 : 3 : 1 56. (C) Since homonuclear diatomic molecules ( ) ( )Q ∂N ∂y x = ∂M are both rotational & IR inactive (absence of ∂x y dipole moment). So rotational Raman and electronic spectroscopy can be used for ∴ equation is exact. determination of bond-lengths. 64. (B) LZ = – ih ∂ 57. (B) ∂φ The eigen function equation is LZ Ylm = – ih ∂ Ylm ∂φ Ylm = f(0).eiAφ f is arbitrary function φ must be periodic for Y is to be unchanged with period 2π. Thus, value of A should be A = 0, ± 1, ± 2, … 58. (C) 65. (D) X-Ray Diffraction method is used for crystal structure identification and study. Amorphous solids don’t show XRD. 66. (D) The Lineweaver burk equation is given L-aspartic acid L-phenylalanine by— 2kT 1 = 1 + km · 1 59. (C) ν mp = M , thus, greater the mass, V Vmax Vmax [s] 0 lesser will be the νmp. Given km = 40s, [E] = 2·5 × 10– 6 M Thus, Ne – C, Ar – B, Kr – A. Vmax 60. (D) We know that We know that at maximum velocity, dH = TdS + VdP [ES] = [E] ( )Thus, ∂H ∴ Vmax = k2[E] ∂S P = T (dP = 0) k2 Vmax km Vmax × 61. (C) The difference between the activation Thus, Efficiency = = [E]x 40s energy for forward & backward reaction is ΔH. Since given that E0 is activation energy = 104 Lmol– 1 s– 1 for second reaction. The activation energy of overall reaction will be E0 + ΔH. 67. (B) CH3COOH CH3COO– + H+ 62. (C) Fractional coverage (θ) for diatomic I 0·01 00 C –x +x +x molecule on metal surface, E + 0·01 – x +x +x θ = (KP)1/2 [CH3COO–] = 10 × 0·2 = 0·01 1 + (KP)1/2 10 + 10 since P is small, 1 > (KP)1/2. pka = 4·75 θ = K1/2P1/2 ⇒ ka = 1·8 × 10– 8 Thus, ⇒ θ ∝ P1/2 kb = 5·6 × 10– 10 63. (D) N = 1 = x2 Y 0·01 ( )⇒ ∂N = – 1 x = – = 2·36 × 10– 6 ∂y y2 [OH] x M = – x pOH = 5·62 y2 Hence, pH = 14 – pOH = 8·38 –~ 8·4.

CSIR-UGC Chemical Sciences (D-14) (II) | 25 68. (B) f00 polydispersed macromolecular colloid; 74. (A) Na[(η5-C5H5) Fe(CO)2] ⎯B⎯r2→ — — 17e– system Mw > Mn osmometry gives number average — molecular weight, M n. ⎯L⎯iA⎯lH⎯4→ [(η5-C5H5) Fe(CO)2 Br] 18e– system 69. (A) Mn \" = 13, Mw \" = 16, substitution of Br with H. [(η5-C5H5) Fe(CO)2H] The variance, σ2 for molecular weight deter- 1H NMR appears for 5H of Cp and 1H in mination σ2 = M—n2 ⎛⎜⎜⎜⎝—M—M nw – 1⎠⎟⎟⎟⎞ ration 5 : 1. 75. (A) Path length of curvetle is not responsible for deviation from ideal behaviour. All the ( )= (13)2 16 remaining statement, A, B and C will result in 13 – 1 deviation form. 76. (C) Three isomers shown as— = 169 × 3 = 39 13 70. (C) I.E. = hc = 6·626 × 10– 34 × 3 × 108 × 1019 λ 24 × 10– 9 × 1·6 = 52 eV (a) (b) Ionization energy is the energy required to remove an electron from outermost orbital of gaseous system. 71. (A) All the given options are incorrect except (1). There is ring current disruption after (c) mirror plane formation of complex in [η6 – (C6H6) 77. (A) Also known as inductively coupled Cr(CO)3]. The shift in δ is also attributed to plasma optical emission spectroscopy (ICP- this reason from ~ 7·2 ppm to 4·4 (almost). OES) is an analytical technique used for the detection of trace metals. Argon gas is typically used to create the plasma. 78. (A) Myoglobin is pH independent 72. (A) Oxyhemocyanin—oxygen transport speci- fically in invertebrates. Carbonic anhydrase—In Red Blood Cells, CO2 to HCO3. Cytochrome P450—Oxydation of alkene, monooxygenose, green catalyst. Carboxypeptidase A—Hydrolysis of C- Lungs (gills) Hb + 4O2 → Hb(O2)4 terminal peptide bond. Tissues Hb(O2)4 + 4 Mb → 4 Mb(O2) + Hb pH dependence shown by hemoglobin is 73. (B) Geometric cross-section = πr2 known as Bohr effect (binds one H+ for every dioxygen molecules). and r = r0(A)1/3 79. (C) Iridium complex is 16 e– system and Thus, r = 1·4 × 10– 15 m × (125)1/3 favourable for oxidative addition reaction in presence of H2, commonly known as Vaska’s = 0·7 × 10– 14 m complex. Rest are 18 e– stable complex compound. Hence, GC-S = 3·14 × (0·7 × 10– 14)2 m2 ~– 1·54 × 10– 28 m2 (1 barn = 10– 28 m2) –~ 1·54 barn

26 | CSIR-UGC Chemical Sciences (D-14) (II) 80. (A) Mn+ 3 : 87. (B) C2B9H12, According to Wade’s rule 25 + 1 = 5, L = 2(D), μ = ⎯√⎯n⎯(n⎯+⎯⎯2) = (CH) × 2 + (BH) × 9 + 1 + 1 no. of unpaired electrons = 4 = 3 × 2 + 2 × 9 + 1 + 1 = 26 or 13 e– pair So, μS.O. = 4·9 BM. = (m + 2) e– pair ⇒ Nido structure. 81. (A) XeF6 + NaNO3 → XeOF4 + NaF + NO2F 6Na Nitrodes react with XeF6 to form XeOF4. This 88. (C) PCl3 + 3PhCl ⎯⎯→ PPh3 + 6NaCl safe & convenient route involves reaction of (A) ↓ MeI XeF6 with a stoichiometric deficiency of NaNO3. BuLi [Ph3PMe]+ I– Ph3P== CH2 ⎯⎯⎯⎯⎯→ (B) ylides (C) – BuH‚ – Li⊕ 82. (A) H H H 2NH3 89. (C) Ph2SiCl2 ⎯2⎯H⎯2O→ Ph2Si(OH)2 BB – 2HCl Unsymmetric HHH Cleavage (A) Δ 100°C – H2O Ph [(BH2) (NH3)2] [BH4] 2H → triplet Polymeric | structure 4H → quinlet Si—O h | Other product is Borohydride salt [BH2(NH3)2] [BH4] which gives triplet and quinlet in 11B. Ph (B) NMR spectrum due to 2H and <| H respecti- 90. (A) Cr+ 3 : vely. – OH μS.O. = n(n + 2) = 83. (A) [CoCl(NH3)5]2 + ⎯⎯⎯→ 3(3 + 2) = 3·87 B.M. SN1CB – H2O and μS + L = n(n + 2) + L(L + 1) = 15 + 12 = 27 = 5·20 BM [CoCl(NH2) (NH3)4]+ 91. (A) – Cl ↓ Slow [CO(OH) (NH3)5]2 + ←H⎯2O⎯ fast [Co(NH2) (NH3)4]2 + In [Co(CN)6]3 –, no ionizable proton. Thus, statements A & B are correct. 84. (B) By gas chromatography. 85. (D) O Th—C—H Inorganic optically active complex— (η5 CP2) TH—H + CO → (η5 (CP2) 12Co—O and 12 Co—N bonds CO insertion occurs in case of Thorium hydride A driving force for this reaction is the 92. (A) The size of the cavity is responsible for strong interaction of oxygen of inserted CO binding with alkali metals. So, crown ether with thorium atom. binds different metals. Thus, right size of cation is important factor. 86. (A) Mn+ 2 : 12-crown-4 2s + 1 = 6, L = 0 (s) Spin selection rule, Δs = 0 93. (A) HMn(CO)5 and [(η5-C5H5) Cr(CO)3]2 6s ⎯→ 6A1g ground state δ : – 7·5 4·10 Violated and disallowed transition charge transfer impossible and 2T2g is not ground state. Hence, A & B are correct.

CSIR-UGC Chemical Sciences (D-14) (II) | 27 94. (C) Type of size exclusion chromatography 99. (A) (SEC), that separates analytes on the basis of size. Analysis of polymers is done by this technique. Thus, from given options, fatty acids low molecular weight peptides can be separated by this method. 95. (A) SF6 : 6 S-F bonds, 12 e–s involved hypervalent 4-orbitals (1a1, 1t1) : bonding 4-orbitals (2t1, 2a1) : anti-bonding 2-orbitals (1e) : non-bonding 96. (C) Hoffmann—Loffler—Fretyag reaction ⇒ Free radical mechanism 100. (B) ⎯→ ⎯De⎯m⎯eth⎯yla⎯tio→n 97. (C) Br Br Br Br 101. (A) NMe2 Br anti addition Br NMe2 Me2NH NMe2 Me2N(HSN2) inversion Br NMe2 98. (A) Diastereoselective proplem-formation of cis enolode thus cis-aldol product. 102. (A) small ligand (Bu), good leaving gp (OTf) and hindered amine (i-Pr2NEt) facilitates the reaction.

28 | CSIR-UGC Chemical Sciences (D-14) (II) 103. (A) 2. TMSCl ⎯⎯⎯⎯⎯⎯⎯⎯→ LDA‚ THF – 78°C CH CH3 C OTMS 3. Δ O ⎯⎯⎯⎯⎯⎯⎯⎯⎯→ Claisen rearrangement OMOM O 104. (A) O ⎯⎯4.⎯H⎯3O⎯+ → 105. (A) TMSO C 5. CH2N2 O OMOM methylation Me MeOOC O OMOM Me 107. (D) A is dehydration reaction using POCl3, Et3N. B is esterification using …… PPh3 and EtO2CN==NCO2Et. C is butylation using H2SO4. 108. (A) ⎯C⎯O⎯2(C⎯O⎯)8‚⎯C⎯H2⎯Cl→2 Me3N+—O– Pauson-Khand Reaction (PKR) [2 + 2 + 1] CAR THF EtOH ⎯⎯⎯⎯⎯⎯⎯⎯→ (A) SmI2 reducing agent ←3.⎯ΘH⎯2O⎯ 109. (C) (B) OH 4. H⊕ ⎯⎯→ Variation of Stork enamine reaction. 106. (D) 1. EtCOCl ⎯⎯⎯⎯→ Et3N

110. (A) CSIR-UGC Chemical Sciences (D-14) (II) | 29 ⇒ δ 6·28 & J = 18 Hz indicates trans-geometry of double bond. 115. (A) 111. (C) δ 2·67 (2H, S) indicates13C NMR shows 4 signals. Thus \"== entity structure may be (2) or (3). 112. (D) 116. (B) δ = reaction constant O || (A) ·· + Cl—C—Ph benzene ArN H2 ⎯⎯⎯⎯→ rds moderate negativeδ value (– 2·69) R 113. (B) O 114. (A) C10H12O2 C Ph Ar H2N positive charge near ring EtOH (B) ArO– + EtI ⎯⎯⎯⎯→ ArOEt rds no electron change small δ value (– 0·99) Q. O || EtOH (C) Ar—C—OEt + aq. NaOH ⎯⎯⎯⎯→ rds O || Ar—C—OH Moderate positive δ value (+ 2·01) P/e–s flow into transition state.

30 | CSIR-UGC Chemical Sciences (D-14) (II) 117. (B) N N 118. (A) heat N 119. (C) 120. (D) O O ⎯⎯⎯→ N – CO2 Ts Ts oxazolidone azomethine ylide Me cis-product HN H ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→ N intramolecular dipolar CAR Ts 121. (A) We know that for unimolecular reaction first order rate constant kuni = k1k2 [A] k– 1 [A] + k2 When k– 1[A] = k2, then kuni = k∞ and [A] = [A]1/2 2 kuni = k1[A] 2 ⇒ k00 = k1[A]1/2 2 2 ⇒ k1 = k∞ [A]1/2 122. (B) Vibrational partition function is given by— qvib = e– θ/2T , 1– e– θ/T where θ = hν k (1) At low T, θ/T >> 1, so e– θ/T is negligible Thus, qvib = e– θ/2T CH3—NH (2) At high T, θ/T << 1 | Then, e– θ/T = 1 – θ/T + …… ⎯⎯⎯⎯C⎯H2⎯CO⎯O⎯H→ Hence, qvib = e– θ/2T iminium ion 1 – (1 – θ/T) ⇒ qvib = T e– θ/2T θ 123. (D) First order energy correction is— n E(n1) = ψ(n0) | H | ψ(n0)* \" ( )ψ(n0) = 2 nπx L sin L

CSIR-UGC Chemical Sciences (D-14) (II) | 31 L sin2 –L ∫ ( )Thus, E(n1) = 2 nπx 127. (A) Rotational-vibrational spectrum L L Δ~E = 2mB + ν~e (1 – 2x~e) ( )for ground state n = 1 L δ x – 2 dx In upper vibrational state B1 < B0 L B = h cm– 1 (μr2) sin2 –L ∫ ( )E(n1) = 2 πx 8π2 C L L r = bond length ( )δ x – L dx Thus, the gap between the successive 2 absorption lines of p-branch increases non- ∞ linearly. 128. (B) Transmission probability is inversely ∫Q δ(x – a)dx = 1 –∞ Thus, E(n1) = 2 proportional to barrier width and height. L 129. (B) Selection rule of EPR spectrum for a free kBT qt.s. – ΔE0 radical containing nuclei with non-zero n qNO qCl2 RT 124. (D) Rate constant k(T) = e nuclear spin Δms = ± 1, Since qtrans = ⎜⎛⎝2mπh2kBT⎠⎞⎟3/2 ΔmI = 0 130. (A) Relation (A) is true and it is Gibbs- Duhem equation. k(T) ∝ T.T3/2.T. ∝ 1 We know that T3/2.T.T.T3/2.T.T. T7/2 i k(T) = T– 7/2 ∑ Nidui = – SdT + Vdp 125. (D) ( )En = n + 1 hν 2 i=1 Probability, P(E) = e– E/kBT At constant T and P, relation (B) is also true. ∑ e– E/kBT 131. (A) [Sz, S+] = [Sz, Sx] + i[Sz, Sy] = ihSy + i(– ihSx) = h(Sx + iSy) = hS+ i 132. (C) [Ne] = e– e– 3/2 3/2 = (1 e– 1 1) 2P 3P 1/2 + e– + e– Term = 3D e– 1.e (e – 1) e – 1 = e+1 × (e– 1) = e2 – 1 S = n = 1, 2 e2 >> 1 L = 2(D) e–1 = e2 J = (L – S) to (L + S) = 1, 2, 3 ⇒ p(E) = e– 2 (e – 1) Thus, 3D3, 3D2, 3D1 126. (B) Fugacity coefficient is given by— 133. (C) Stepwise condensation polymerization, P (Z – 1) overall rate law is given as 0 P ∫ln r = dP d[A] = – k[A]2 dt PV Q Z = RT ∴ [A] = [A]0 1 + kt [A]0 At high pressure, z >> 1 Fraction of groups condensed at time t. Thus r>1 [A]0 – [A] k + [A]0 Thus, repulsive term overweighs the attractive f = [A]0 = 1 + k + [A]0 term.

32 | CSIR-UGC Chemical Sciences (D-14) (II) 134. (B) B represents BET curve for multilayer adsorption C also represents multilayer adsorption for benzene on Fe2O3 at 50°C. 140. (C) Wave function for central carbon in propenyl cation ψ2 = 1 φ1 – 1 φ3 Thus, D0(H2) = D0(H2+) – I(H) + I(H2) √⎯ 2 √⎯ 2 π-electron charge r (internuclear distance). qr = ∑ ηi ci2r 135. (B) ΔG = 2746·06 kJ mol– 1 So, q2 = 2 × ⎝⎜⎛√⎯12⎞⎟⎠2 + 0 + 0 We know, ΔG = nFE ΔG 2746·06 × 103 J mol– 1 nF 26 × 96500 So, E = = =1 = 1·0944 V –~ 1·09V 141. (C) 136. (B) ccp means fcc structure 142. (A) ⎪⎪⎪⎪11SSaa α (1) 1Sa β (1) ⎪⎪⎪⎪ α (2) 1Sa β (2) r = 160 pm Q a√⎯ 2 = 4r = 1Sa α(1) 1Sa β(2) – 1Sa β(1) 1Sa α(2) α and β are spin up and down a = 4r According to Pauli’s principle, 2 electrons in ‘a’ atomic orbital with spin up and down. √⎯ 2 143. (D) No. of microstates no. of atoms in fcc = 4 So, no. of atoms in 1 cm3 (one atom) 4 = ( )= 2 4 4 √⎯ 2 2 4–2 a3 = 4×3× 2 (4r) 3 = 4 × (⎯√2)3 = =6 2 ×2×1 4 × 4 × 4 × (160)3 × (10– 10)3 144. (C) Partition function = 3·45 × 1023 ∑f = gj e– εi /kT 137. (D) 1 E C2 σV σV1 j no. of 10 0 0 10 When T → ∞, e– εi /kT → 1 unshifted Thus, f = ∑ gj atoms j Character 3 – 1 1 1 order of 145. (C) We know that per atom 30 0 0 10 group, h = 4 A1 = 10, A2 = 5, B1 = 5, B2 = 10 log k = log k0 + 1·018 | ZAZB| ⎯√I ltrans = B1 + B2 + A1, lrot = B2 + B1 + A2 log k0·04 = 1·018 (1·x) [(0·04)1/2 log k0·01 Thus, IR active model = 9A1 + 3B1 + 8B2 – (0·01)1/2] σxy.S4z = C4z (rotation axis) 138. (A) 0·3 = 1·018 × x[0·2 – 0·1] 139. (B) A represents Langmuir adsorption x = 0·30 = 2·77 ≈ 3 isotherm for monolayer adsorption. 0·1080

Chemical Sciences CSIR-UGC-NET/JRF Exam. (June 2015) Solved Paper

June 2015 Chemical Sciences PART A 06. Find the height of a box of base area 24 cm × 48 cm, in which the longest stick that can 01. Each of the following pairs of words hides a be kept is 56 cm long— number, based on which you can arrange them in ascending order. Pick the correct answer : (A) 8 cm (B) 32 cm (C) 37·5 cm (D) 16 cm I. Cloth reel 07. The product of the perimeter of a triangle, the radius of its in-circle, and a number gives the J. Silent wonder area of the triangle. The number is— K. Good tone L. Bronze rod (A) 1/4 (B) 1/3 (A) L, K, J, I (B) I, J, K, L (C) 1/2 (D) 1 (C) K, L, J, I (D) K, J, I, L 08. An infinite row of boxes is arranged. Each box has half the volume of the previous box. 02. Which of the following values is same as If the largest box has a volume of 20 cc, what 2222 ? is the total volume of all the boxes ? (A) 26 (B) 28 (A) Infinite (B) 400 cc (C) 216 (D) 2222 (C) 40 cc (D) 80 cc 03. A 12 m × 4 m rectangular roof is resting on 09. Find the missing element based on the given pattern— four 4 m tall thin poles. Sunlight falls on the 1. 2. 3. roof at an angle of 45° from the east, creating a shadow on the ground. What will be the area of the shadow ? 1. 2. 3. ? (A) (B) (A) 24 m2 (B) 36 m2 (C) 48 m2 (D) 60 m2 (C) (D) 04. If 2a 10. By reading the accompanying graph, deter- ×b2 mine the INCORRECT statement out of the following— c6 84 liquid 8d6 Pressure Here a, b, c and d are digits. gas solid Then a + b = (A) 4 (B) 9 (C) 11 (D) 16 Temperature 05. The maximum number of points formed by (A) Melting point increases with pressure intersection of all pairs of diagonals of convex (B) Melting point decreases with pressure octagon is— (C) Boiling point increases with pressure (D) Solid, liquid and gas can co-exist at the (A) 70 (B) 400 same pressure and temperature (C) 120 (D) 190

Chemical Sciences (CSIR) J-15 | 3 11. If you change only one observation from a set 16. An ant can lift another ant of its size whereas of 10 observations, which of the following an elephant cannot lift another elephant of its will definitely change ? size, because— (A) Mean (B) Median (A) ant muscle fibres are stronger than elephant muscle fibres (C) Mode (D) Standard deviation (B) ant has proportionately thicker legs than 12. A man starts his journey at 0100 Hrs. local elephant time to reach another country at 0900 Hrs local time on the same date. He starts a return (C) strength scales as the square of the size journey on the same night at 2100 Hrs. local while weight scales as cube of the size time to his original place, taking the same time to travel back. If the time zone of his (D) ants work cooperatively, whereas country of visit lags by 10 hours, the duration elephants work as individuals for which the man was away from his place is— 17. Consider a series of letters placed in the following way : U…G…C…C…S…I…R (A) 48 hours (B) 20 hours Each letter moves one step to its right and the extreme right letter takes the first position, (C) 25 hours (D) 36 hours completing one operation. After which of the following numbers of operations do the Cs 13. Let r be a positive number satisfying not sit side by side ? r (1/1234) + r (–1/1234) = 2 (A) 3 (B) 10 Then r 4321 + r – 4321 = ? (C) 19 (D) 25 (A) 2 (B) 2 (4321/1234) 18. An inclined plane rests against a horizontal cylinder of radius R. If the plane makes an (C) 2 3087 (D) 2 1234 angle of 30° with the ground, the point of contact of the plane with the cylinder is at a 14. A float is drifting in a river, 10 m downstream height of— of a boat that can be rowed at a speed of 10 m/minute in still water. If the boat is rowed (A) 1·500 R (B) 1·866 R downstream, the time taken to catch up with the float— (C) 1·414 R (D) 1·000 R (A) will be 1 minute 19. What is the maximum number of parallel, non-overlapping cricket pitches (length 24 m, (B) will be more than 1 min width 3 m) that can be laid in a field of diameter 140 m, if the boundary is required to (C) will be less than 1 min be at least 60 m from the centre of any pitch ? (D) can be determined only if the speed of (A) 6 (B) 7 the river is known (C) 12 (D) 4 15. ABC is a right angled triangle inscribed in a semicircle. Smaller semicircles are drawn on 20. In a fast moving car with open windows, the sides BC and AC. If the area of the triangle is driver feels a continuous incoming breeze. a, what is the total area of the shaded lumes ? The pressure inside the car, however, does not keep increasing because— B (A) air coming in from the front window C A goes out from the rear (A) a (B) πa (B) air comes in as well as goes out through (C) a/π (D) a/2π every window but the driver only feels the incoming one (C) no air actually comes in and the feeling of breeze is an illusion (D) cool air reduces the temperature therefore the pressure does not increase

4 | Chemical Sciences (CSIR) J-15 PART B 26. The species having he strongest gas phase proton affinity among the following— 21. The biological functions of carbonic anhy- (A) N3– (B) NF3 drase and carboxypeptidase A, respectively, are— (C) NH3 (D) N(CH3)3 (A) interconversion of CO2 and carbonates 27. Consider the following statements regarding and hydrolysis of peptide bond the diffusion current at dropping mercury (B) gene regulation and interconversion of CO2 and carbonates electrode. (C) gene regulation and hydrolysis of peptide 1. It does not depend on mercury flow rate. bond 2. It depends on drop time. (D) interconversion of CO2 and carbonates and gene regulation 3. It depends on temperature. Correct statement(s) is/are— (A) 1 only (B) 2 only (C) 1 and 2 (D) 2 and 3 22. Fe–Nporphyrin bond distances in the deoxy- and 28. Q value for the reaction 13N(n, p)13C is 3·236 oxy-haemoglobin, respectively are— MeV. The threshold energy (in MeV) for the (A) ~ 2·1 and 2·0 Å (B) ~2·0 and 2·0 Å reaction 13C(p, n)13N is— (C) ~2·2 and 2·3 Å (D) ~2·3 and 2·5 Å (A) – 3·236 (B) – 3·485 23. The binding modes of NO in 18 (C) 3·485 (D) 3·845 electron compounds [Co(CO)3(NO)] and [Ni(η5 – Cp)(NO)], respectively are— 29. The 119Sn NMR chemical shift (approximately in ppm) corresponding to (η5-Cp)2Sn (relative (A) linear and bent to Me4Sn) is— (B) bent and linear (A) – 4 (B) + 137 (C) linear and linear (C) + 346 (D) – 2200 (D) bent and bent 30. All forms of phosphorus upon melting, exist as— 24. The role of copper salt as co-catalyst in (A) P Wacker process is— (A) oxidation of Pd(0) by Cu(II) nP P (B) oxidation of Pd(0) by Cu(I) P (C) oxidation of Pd(II) by Cu(I) (D) oxidation of Pd(II) by Cu(II) (B) P P 25. For typical Fischer and Schrock carbenes, P PP P consider the following statements— P Pn 1. Oxidation state of metal is low in Fischer (C) n (P ≡ P) carbene and high in Schrock carbene. (D) P P P P P P P 2. Auxiliary ligands are π -acceptor in Fischer carbene and non-π-acceptor in PPP P P P P PPP PPP Schrock carbene. P P PPP 3. Substituents on carbene carbon are non- 31. For the oxidation state(s) of sulphur atoms in π-donor in Fischer carbene and π-donor in Schrock carbene. S2O, consider the following— 4. Carbene carbon is electrophilic in 1. – 2 and + 4 Fischer carbene and nucleophilic in Schrock carbene. 2. 0 and + 2 3. + 4 and 0 The correct statements are— The correct answer(s) is/are— (A) 1, 2 and 3 (B) 1, 2 and 4 (A) 1 and 2 (B) 1 and 3 (C) 2, 3 and 4 (D) 1, 3 and 4 (C) 2 and 3 (D) 3 only

Chemical Sciences (CSIR) J-15 | 5 32. The correct set of pseudohalide anions is— (B) Equilibrium will shift towards left in (A) CN–, ClO4–, BF4–, PF6– case of both A and B (B) N3–, NO3–, HSO4–, AsF6– (C) SCN–, PO43–, H2PO4–, N3– (C) Equilibrium will shift towards right in A (D) CN–, N3–, SCN–, NCN2– and left in case of B (D) Equilibrium will shift towards right in B and left in case of A 33. In transition metal phoshine (M—PR3) com- 38. The compound that exhibits sharp bands at plexes, the back-bonding involves donation 3300 and 2150 cm– 1 in the IR spectrum is— of electrons from— (A) 1-butyne (B) 2-butyne (A) M(t2g) → PR3(σ*) (B) M(t2g) → PR3(π*) (C) butyronitrile (D) butylamine (C) M(eg) → P(d) (D) PR3(π) → M (t2g) 39. The 1H NMR spectrum of a dilute solution of a mixture of acetone and dichloromethane in 34. The refluxing of RhCl3.3H2O with an excess CDCl3 exhibits two singlets of 1 : 1 intensity. of PPh3 in ethanol gives a complex A. Molar ratio of acetone to dichloromethane in Complex A and the valence electron count on the solution is— rhodium are, respectively— (A) 3 : 1 (B) 1 : 3 (A) [RhCl(PPh3)3], 16 (B) [RhCl(PPh3)5], 16 (C) 1 : 1 (D) 1 : 2 (C) [RhCl(PPh3)3], 18 (D) [RhCl(PPh3)5], 18 40. Intense band generally observed for a carbonyl group in the IR specturm is due to— 35. The β-hydrogen elimination will be facile (A) The force constant of CO bond is large in— (B) The force constant of CO bond is small (A) M (B) M (C) There is no change in dipole moment for HH CO bond stretching (D) The dipole moment change due to CO bond stretching is large (C) M (D) M H 41. The compound that gives precipitate on warming with aqueous AgNO3 is— H 36. The reaction [Co(CN)5H2O]2– + X– → Br (B) Br [Co(CN)5X]2– + H2O follows a/an— (A) (A) Interchange dissociative (Id) mechanism (B) Dissociative (D) mechanism Br Br (C) (D) (C) Associative (A) mechanism (D) Interchange Associative (Ia) mechanism 37. Correct statement on the effect of addition of N aq. HCl on the equilibrium is— OH O 42. Following reaction goes through— HO … Eq. A COOAg Br O O Br2 O + CN− O− (A) Free radical intermediate … Eq. B (B) carbanion intermediate (C) carbocation intermediate CN (D) carbene intermediate (A) Equilibrium will shift towards right in case of both A and B

6 | Chemical Sciences (CSIR) J-15 43. The most stable conformation for the Me Me Me Me Me following compound is— Me Me Me Me Me Me I II III IV (A) I > II > III > IV (B) I > III > II > IV (C) IV > I > III > II (D) IV > II > I > III Me Me 47. Among the following, the correct statement (s) Me Me about ribose is (are)— (A) H (B) Me 1. On reduction with NaBH4 it gives optically inactive product. Me H 2. On reaction with methanolic HCl it gives Me a furanoside. Me Me 3. On reaction with Br2–CaCO3–water it (C) H (D) Me gives optically inactive product. Me H 4. It gives positive Tollen’s test. Me (A) 1, 2 and 4 (B) 1, 2 and 3 (C) 2 and 3 (D) 4 only 44. The major product formed in the following 48. Biogenetic precursors for the natural product reaction is— umbelliferone among the following are— O Me NaBH4, CeCl3 HO O O umbelliferone (A) OH CHO MeOH, H2O Me OH 1. L-tryptophan 2. cinnamic acid 3. L-methionine 4. L-phenylalanine (A) 1 and 2 (B) 2 and 4 Me (C) 2 and 3 (D) 3 and 4 (B) 49. Number of signals in the 13C{H} NMR OH O spectrum of (R)–4-methylpentan-2-ol are— (C) OH (A) 3 (B) 4 Me (C) 5 (D) 6 OH 50. The major product formed in the following reaction is— (D) OH Me Me Me CHO EtO2C NaBH4 H 0°C 45. The correct relation between the following compounds is— MeOH/THF O HH H Cl (A) Me Me (B) Me Me Cl Me Me OH H HO HO HO H H H OH (A) enantiomers OH H (B) diastereomers (C) homomers (identical) (C) Me Me (D) Me Me (D) constitutional isomers EtO2C EtO2C 46. The correct order of heat of hydrogenation for HH H OH the following compounds is— OH H

Chemical Sciences (CSIR) J-15 | 7 51. The major product formed in the following 55. A particle is in a one-dimensional box with a reaction is— potential V0 inside the box and infinite outside. An energy state corresponding to heat n = 0 (n : quantum number) is not allowed because— Me (A) the total energy becomes zero Me H (A) (B) (B) the average momentum becomes zero H Me (C) the wave function becomes zero every where HH (D) the potential V0 ≠ 0 (C) (D) 56. An eigenstate of energy satisfies HΨn = En Ψn. In the presence of an extra constant potential H Me H Me V0— (A) both En and Ψn will change 52. The major product formed in the following reaction is— (B) both En and the average kinetic energy will change Me O H2N-NH2.HCl O (C) only En will change, but not Ψn Et3N, CH3CN, rt (D) only Ψn will change, but not En Me 57. The intensity of a light beam decreases by Me O 50% when it passes through a sample of 1·0 (A) Me cm path length. The percentage of trans- O (B) mission of the light passing through the same sample, but of 3·0 cm path length, would be— (A) 50·0 (B) 25·0 Me Me Me (C) 16·67 (D) 12·5 Me HO Me HO 58. The electric-dipole allowed transition among (C) (D) the following is— (A) 3S → 3D (B) 3S → 3P (C) 3S → 1D (D) 3S → 1F Me Me 59. The product C2xσxy (C2x is the two-fold rotation axis around the x-axis and σxy is the xy mirror 53. The magnitude of the stability constants for plane) is— K+ ion complexes of the following supra- molecular hosts follows the order— (A) σx z (B) σy z (C) C2y (D) Cz2 HH NN S 60. The simplest ground-state VB wave function O OO OO O of a diatomic molecule like HCl is written as Ψ = ΨH (1s, 1) ΨCl (3pz, 2) + B where B O OO OO O stands for— N O H S (A) ΨH (3pz, 2) ΨCl (1s, 1) (B) ΨH (1s, 2) ΨCl (3pz, 1) A BC (C) ΨCl (1s, 2) ΨCl (3pz, 1) (D) ΨCl (1s, 2) ΨH (3pz, 1) (A) B > A > C (B) C > A > B (C) A > B > C (D) C > B > A 54. Antitubercular drug(s) among the following is (are)— 1. Salbutamol 2. Ethambutanol 61. Heat capacity of a species is independent of 3. Isoniazid 4. Diazepam temperature if it is— (A) 1 and 2 (B) 2 and 3 (A) tetratomic (B) triatomic (C) 3 and 4 (D) 4 alone (C) diatomic (D) monatomic

8 | Chemical Sciences (CSIR) J-15 62. In a chemical reaction : PCl5(g) PCl3(g) + (C) ΔG° – RT ln aZn(s) Cl2 (g ), xenon gas is added at constant aCu2+ volume. The equilibrium— (D) ΔG° + RT ln aZn2+ (A) will shift towards the reactant aCu2+ (B) will shift towards the products 68. The lowest energy-state of an atom with electronic configuration ns1 np1 has the term (C) will not change the amount of reactant and products symbol— (D) will increase both reactant and products (A) 3p1 (B) 1p1 63. The temperature-dependence of a reaction is (C) 3p2 (D) 3p0 give by k = AT2 exp (– E0/RT). 69. Energy of interaction of colloidal particles as a function of distance of separation can be The activation energy (Ea) of the reaction is identified as (1) van der Waals, (2) double give by— layer, (3) van der Waals and double layer. The correct order of interactions in the figure (A) E0 + 1 RT (B) E0 corresponding to curves (a), (b) and (c) 2 respectively, is— (C) E0 + 2RT (D) 2E0 + RT 64. For a reaction, 2A + B → 3Z, if the rate of consumption of A is 2 × 10– 4 mol dm– 3 s– 1, E (b) (a) the rate of formation of Z (in mol dm– 3 s– 1) will be— r (A) 3 × 10– 4 (B) 2 × 10– 4 (c) (C) 4 × 10– 4 (D) 4 × 10– 4 3 65. Dominant contribution to the escaping (A) 1, 2, 3 (B) 2, 3, 1 tendency of a charged particle with uniform (C) 3, 1, 2 (D) 1, 3, 2 concentration in a phase, depends on— (A) chemical potential of that phase 70. The packing factor (PF) and number of (B) electric potential of the phase atomic sites per unit cell (N) of an FCC (C) thermal energy of that phase crystal system are— (D) gravitational potential of that phase (A) PF = 0·52 and N = 3 66. The intrinsic viscosity depends on the molar mass as [η] = KMa. (B) PF = 0·74 and N = 3 The empirical constants K and a are dependent on— (C) PF = 0·52 and N = 4 (A) solvent only (D) PF = 0·74 and N = 4 (B) polymer only PART C (C) polymer-solvent pair 71. Differential pulse polarography (DPP) is (D) polymer-polymer interaction more sensitive than D.C. Polarography (DCP). Consider following reasons for it— 67. The correct ΔG for the cell reaction involving 1. Non-faradic current is less in DPP in steps comparison to DCP. Zn(s) → Zn2+ (aq) + 2e– 2. Non-faradic current is more in DPP in comparison to DCP. Cu2+ (aq) + 2e– → Cu(s) is— 3. Polarogram of DPP is of different shape (A) ΔG° – RT ln aZn2+ than that of DCP. aCu2+ Correct reasons(s) is/are— (B) ΔG° + RT ln aZn2+ (A) 1 and 3 (B) 2 and 3 aCu(s) (C) 2 only (D) 1 only

Chemical Sciences (CSIR) J-15 | 9 72. Considering the following parameters with 78. Considering the intert pair effect on lead, the reference to the fluorescence of a solution : most probable structure of PbR2 [R = 2, 6– C6H3 (2, 6—Pr2C6H3)2] is— 1. Molar absorptivity of fluorescent mole- cule. RR 2. Intensity of light source used for R Pb R excitation. (A) Pb Pb R Pb 3. Dissolved oxygen (B) R The correct answer for the enhancement of RR fluorescence with the increase in these R R parameters is/are— R Pb (D) R Pb R R Pb R (A) 1 and 2 (B) 2 and 3 (C) Pb R (C) 1 and 3 (D) 3 only 73. The geometric cross section of 125Sn (in barn) 79. The reaction of SbCl3 with 3 equivalents of EtMgBr yields compound X. Two equivalents is nearly— of SbI3 react with one equivalent of X to give Y. In the solid state, Y has a 1D-polymeric (A) 1·33 (B) 1·53 structure in which each Sb is in a square pyramidal environment. Compounds X and Y (C) 1·73 (D) 1·93 respectively, are— 74. Match column A (coupling reactions) with (A) SbEt3 and [Sb(Et)I2]n column B (reagents)— (B) Sb(Et2)Cl and [Sb(Et2)Cl]n Column A Column B (C) SbEt3 and [SbEt2Br2]n a. Suzuki coupling i. CH2=CHCO2CH3 ii. RB(OH)2 (D) Sb(Et)Br2 and [SbEt(I) (Br)]n b. Heck coupling iii. PhCO(CH2)3ZnI 80. Match the complexes given in column I with c. Sonogashira iv. HC≡CR the electronic transitions (mainly responsible coupling for their colours) listed in column II d. Negeshi coupling v. SnR4 I II The correct match is— a. Fe(II)-protopor- 1. π → π* (a) (b) (c) (d) phyrin IX (A) ii i iv iii b. [Mn(H2O)6]Cl2 2. spin allowed d → d c. [Co(H2O)6]Cl2 3. spin forbidden d → d (B) i v iii iv 4. M → L charge transfer (C) iv iii ii i (D) ii iii iv v The correct answer is— (a) (b) (c) 75. The oxoacid of phosphorus having P atoms in (A) 1 3 2 + 4, + 3 and + 4 oxidation states respectively, (B) 4 2 3 (C) 1 3 4 is— (D) 1 2 3 (A) H5P3O10 (B) H5P3O7 (C) H5P3O8 (D) H5P3O9 76. The geometries of [Br3]+ and [I5]+ respec- 81. The following statements are given regarding tively, are— the agostic interaction C—H…Ir observed in (A) trigonal and tetrahedral [Ir(Ph3P)3Cl]. 1. Upfield shift of C—H proton in 1H NMR (B) tetrahedral and trigonal bipyramidal spectrum. (C) tetrahedral and tetrahedral 2. Increased acid character of C—H. (D) linear and trigonal pyramidal 3. vC–H in IR spectrum shifts to higher 77. According to Wade’s theory the anion wavenumber. [B12H12]2– adopts— The correct answer is/are— (A) closo - structure (B) arachno - structure (A) 1 and 3 (B) 2 and 3 (C) hypo - structure (D) nido - structure (C) 1 and 2 (D) 3 only

10 | Chemical Sciences (CSIR) J-15 82. Amongst the following : 87. Match the action of H2O2 in aqueous medium given in column A with the oxidation/ reduc- 1. [Mn(η5 – Cp) (CO)3], tion listed in column B : 2. [Os(η5–Cp)2], 3. [Ru(η5–Cp)2] and A : action of H2O2 B : type of reaction 4. [Fe(η5–Cp)2], a. Oxidation in acid 1. [Fe(CN)6]3– → the compounds with most shielded and b. Oxidation in base [Fe(CN)6]4 – deshielded Cp protons respectively, are— c. Reduction in acid 2. [Fe(CN)6]4 – → (A) 4 and 1 (B) 4 and 2 d. Reduction in base [Fe(CN)6]3– (C) 3 and 1 (D) 3 and 2 3. MnO4– → Mn2+ 4. Mn2+ → Mn4+ 83. Total number of vertices in metal clusters [Ru6(C)(CO)17], [Os5(C)(CO)15] and The correct answer is— (d) [Ru5(C)(CO)16] are 6, 5 and 5 respectively. (a) (b) (c) 4 The predicted structures of these complexes, 1 respectively, are— (A) 1 2 3 1 (B) 2 4 3 2 (A) closo, nido and nido (C) 3 4 2 (D) 4 1 3 (B) closo, nido and arachno 88. The reduced form of a metal ion M in a (C) arachno, closo and nido complex is NMR active. On oxidation, the (D) arachno, nido and closo complex gives an EPR signal with g|| ≈ 2·2 and g⊥ ≈ 2·0. M·o·ssbauer spectroscopy cannot 84. Among the complexes, characterise the metal complex. The M is— 1. K4[Cr(CN)6] (A) Zn (B) Sn 2. K4[Fe(CN)6] 3. K3[Co(CN)6] (C) Cu (D) Fe 4. K4[Mn(CN)6] Jahn-Teller distortion is expected in— 89. The least probable product from A on reductive elimination is— (A) 1, 2 and 3 (B) 2, 3 and 4 Ph Ph (C) 1 and 4 (D) 2 and 3 P CH3 M 85. The reductive elimination of Ar—R (coupled product) from A is facile when— P CH3 Ph Ph Ph Ph A P Ar (A) H3C CH3 (B) CH4 Pd A PR Ph Ph CH3 (D) CH3 (A) R = CH3 (B) R = CH2Ph (C) H3C H3C CH3 (C) R = CH2COPh (D) R = CH2CF3 90. Water plays different roles in the following reactions— 86. The total number of metal ions and the number of coordinated imidazole units of 1. 2H2O + Ca → Ca2+ + 2OH– + H2 histidine in the active site of oxy-hemocyanin, respectively, are— 2. nH2O + Cl → [Cl(H2O)n]– 3. 6H2O + Mg2+ → [Mg(H2O)6]2+ (A) 2Cu2+ and 6 (B) 2Fe2+ and 5 4. 2H2O + 2F2 → 4HF + O2 (C) 2Cu+ and 6 (D) Fe2+ and 3

Chemical Sciences (CSIR) J-15 | 11 The correct role of water in each reaction is— 1. CO2 combines with Ni(PR3)2 (1, 5-cyclo- (A) (1) oxidant, (2) acid, (3) base and octadiene) (4) reductant 2. Insertion of CO2 occurs (B) (1) oxidant, (2) base, (3) acid and 3. Insertion of Et Et takes place (4) reductant The correct answer is— (C) (1) acid, (2) oxidant, (3) reductant and (A) 1 and 2 (B) 2 and 3 (4) base (D) (1) base, (2) reductant, (3) oxidant and (C) 3 and 1 (D) 1, 2 and 3 (4) base 95. Consider the following statements for 91. With respect to σ and π bonding in Pt—||| in (NH4)2 [Ce(NO3)6] (Z) the structure given below, which of the following represent the correct bonding ? 1. Coordination number of Ce is 12 Ph 2. Z is paramagnetic Ph3P C 3. Z is an oxidising agent Pt 1.32 A 4. Reaction of Ph3PO with Z gives a Ph3P C complex having coordination number 10 for Ce. Ph (A) M(σ) → L(σ) and M(π) → L(π*) The correct statements are— (B) L(σ) → M(π) and L(π) → M(π) (C) L(π) → M(π) and L(σ) → M(π) (A) 1, 2 and 3 (B) 2, 1 and 4 (D) L(π) → M(σ) and M(π) → L(π*) (C) 2, 3 and 4 (D) 1, 3 and 4 92. The complex [Fe(phen)2(NCS)2] (phen = 1, 96. The major product formed in the following reaction sequence is— 10-phenanthroline) shows spin cross-over HO2C 1. (i) SOCl2, behaviour. CFSE and μeff at 250 and 150 K O (ii) NaN3, MeOH respectively, are— 2. t-BuOK (A) 0·4 Δ •, 4·90 BM and 2·4 Δ •, 0·00 BM 3. H3O+ (B) 2·4 Δ •, 2·90 BM and 0·4 Δ •, 1·77 BM (C) 2·4 Δ •, 0·00 BM and 0·4 Δ •, 4·90 BM (A) O (D) 1·2 Δ •, 4·90 BM and 2·4 Δ •, 0·00 BM HO2C NH 93. Consider the following statements with respect H to uranium 1. UO22+ disproportionates more easily than O UO22+ NH 2. U3O8 is its most stable oxide of U (B) H HO2C 3. Coordination number of U in [UO2 (NO3)2 (H2O)2].4H2O is six 4. UO22+ is linear The correct set of statements is— HO2C H (C) N (A) 1, 2 and 4 (B) 1, 3 and 4 O (C) 2, 3 and 4 (D) 1, 2 and 3 94. Et H 2 Et Et Et (D) HO2C H Et + CO2(R3P)2Ni(1,5-cyclooctadiene) OO N Et O For the above conversion, which of the H following statements are correct ?

12 | Chemical Sciences (CSIR) J-15 97. The major products A and B in the following CO2H CO2H reaction sequence are— R (B) A = B= N A R = OH R = Me B H CO2Et N H NaNH2 NaNH2 NH3 (I) NH3 (I) CO2Et CO2H OH Br (C) A = B= N Me Me H CO2H N H (A) A = B= + NH2 NH2 CO2Et NH2 (1:1) OH (D) A = B= Me N H CO2H N H (B) A = B= NH2 NH2 100. The major products formed in the following OH OH Me Me reaction are— (C) A = + B= + 0.5 equiv. PhC(Me)2OOH 1.0 equiv. Ti(OiPr)4 NH2 NH2 OH OMe 1.2 equiv. (−)-DIPT NH2 (1:1) NH2 (1:1) CH2Cl2, −20°C OH Me O (D) (A) A = B= A= B= OH OMe OH OMe NH2 NH2 O 98. The major product formed in the following (B) A = B= reaction is— OH OMe OH OMe AcO p-TsNH-NH2 AcOH O B= O NaBH3CN (C) A = OMe OH OMe OH AcO (B) O B= (A) (D) A = OH OMe OMe OH AcO AcO 101. The correct statement about the following (C) (D) reaction is— O 99. The major products A and B in the following NH2 Br2 reaction sequence are— NaOH OO NF H2N + CO2Et aq KOH aq KOH B (A) The product is 2-fluoropyridin-3-amine EtO2C rt A reflux and reaction involves nitrene inter- mediate CO2H CO2H (B) The product is 2-fluoropyridin-3-amine (A) A = CO2Et B= and reaction involves radical inter- N mediate H N CO2H H

Chemical Sciences (CSIR) J-15 | 13 (C) The product is 2-hydroxynicotinamide 104. The major products A and B formed in the and reaction involves benzyne-like intermediate following reactions are— (D) The product is 2-hydroxynicotinamide O 1. PdCl2, CuCl and reaction involves addition-elimina- tion mechanism i. Li, NH3 (l) A O2, DMF-H2O B ii. allyl bromide 2. ethanolic KOH 102. The major product formed in the following O reaction is— (A) A = B= O (B) A = OAc Pd(OAc)2 (C) PPh3, Et3N A= H N Ph CH3CN O O H B= OH Ph (A) N Me H Ph OO N Me (B) B = OH H O H Ph (D) A = B= O N (C) H H 105. An organic compound shows following HO spectral data : (D) N Ph IR (cm– 1) : 1680 Ac 1H NMR (CDCl3) : δ 7·66 (m, 1H), 7·60 (m, 1H), 7·10 (m, 1H), 2·25 (s, 3H) 103. The major products A and B formed in the 13C NMR (CDCl3) : δ 190, 144, 134, 132, following reactions are— 128, 28 m/z (EI) : 126 (M+, 100%), 128 (M+ + 2, 4·9%) The structure of the compound is— Me Me KH A Br Ph3P OAc (A) THF n-BuLi B (B) O O HO Me 18-crown-6 rt 0 °C S (A) A = Me Me Me (C) (D) S (B) A = Me CHO B = Me O (C) A = Me Me CO2Me O (D) A = Me Me Me 106. The correct set of reagents to effect the Me Me CHO B = Me Me following transformation is— Me Me O O Me CO2Me Me CHO B = Me Me (A) (i) (a) NaOMe, Mel; (b) NaCl, wet DMSO, 160°C; Me (ii) (a) LDA, – 78°C, TMSCl; (b) t- Me BuCl, TiCl4, 50°C CHO B = Me Me

14 | Chemical Sciences (CSIR) J-15 (B) (i) (a) NaOMe, MeI; (b) aq. NaOH then (C) A : LiHMDS, AcOEt B = OH HCl, heat OH (ii) (a) Et3N, TMSCl, rt; (b) t-BuCl, NN TiCl4, 50°C OH (C) (i) LDA, t-BuCl; (ii) LDA, MeI; (iii) OH aq. NaOH then HCl, heat (D) A : n-BuLi, AcOEt B = N N (D) (i) (a) NaCl, wet DMSO, 160°C; (b) NaH, t-BuCl 109. The major product of the following reaction sequence is— (ii) (a) morpholine, H+; (b) MeI then H3O+ Br NHAc OH 107. The correct structures of the intermediates CO2Me [A] and [B] in the following reaction are— Pd(OAc)2 Pd(OAc)2 POCl3 [A] PH NH2 N PPh3, Et3N PPh3, Et3N [B] Ts N O N N Ph (A) CO2Me H H OH (A) A = Cl B = Cl Cl AcHN N P OP(O)Cl2 N H Cl H O N Ts (B) A = B = Cl (B) CO2Me NO N Cl AcHN P(O)Cl2 P(O)Cl2 (C) A = Cl B = N Cl N N Ts OP(O)Cl2 (C) CO2Me H (D) A = B= Cl Cl NHAc NO N P N P(O)Cl2 H Cl O Ts 108. The correct reagent combination A and the (D) OH CO2Me major product B in the following reaction sequence are— NHAc O OO N EtO2C A H2N-NH2 B Ts EtO2C (A) A : LiHMDS, AcCl B = 110. The major product formed in the following reaction is— N N O H EtO2C Cp Me O Cp Ti Cl AI Me (B) A : n-BuLi, AcCl B = HO Pyridine, toluene, −40°C N N H EtO2C PhMe2Si COOEt

Chemical Sciences (CSIR) J-15 | 15 (A) O 112. The major product formed in the following PhMe2Si H reaction is— O EtOOC O hv, acetone (B) O H PhMe2Si H Me EtOOC O Me (A) (B) OO (C) O O PhMe2Si H Me (C) (D) EtOOC O O O 113. The hydrocarbon among the following H having conformationally locked chair-boat- chair form is— (D) HH HH PhMe2Si (A) (B) EtOOC O HH HH HH HH 111. The major products A and B in the following synthetic sequence are— (C) (D) O i. PhMgBr HH HH Me CuI A NaOEt B 114. The major product formed in the following ii. H3O+ Br2 reaction sequence is— OO O 1. (Boc)2O, pyridine HO 2. TBSCl, Imidazole (A) A = Me B = Ch2Br NH2 3. LiAlH(Ot-Bu)3 Ph Ph EtOH, −78°C OH O O (A) TBSO (B) A = Me B = Ch2Br NHBoc (C) A = Ph Ph OH (D) A = O Br O (B) BocO Me B = Me Ph Ph NHTBS O Br O OH Me B = Me (C) TBSO Ph Ph NHBoc OH (D) BocO NHTBS

16 | Chemical Sciences (CSIR) J-15 115. The major product in the following reaction 117. IN the following reaction, the ratio of A:B:C sequence is— is (* indicates labelled carbon) O N2 Me Br Br Br hv, OTIPS NBS ++ AIBN ABC Me CCl4 vycor filter heat ClCH2CH2Cl, 80°C (A) Me Me (A) 1 : 1 : 1 (B) 1 : 2 : 1 HO H (C) 2 : 1 : 1 (D) 3 : 2 : 1 OTIPS 118. Structure of the major product in the following synthetic sequence is— H CO2Me (B) Me Me O OTIPS 1. CuI 2. SeO2 N2 (C) Me Me (A) HO H HO OTIPS Me CO2Me H (D) Me Me HO Me H CO2Me HO OTIPS (B) H HOMe H (C) H CO2Me CO2Me 116. Structures of A and B in the following HO H synthetic sequence are— (D) Me O 1. Ph3P=CHCO2Me i. LiAlH4 H ii. H3O+ AcO N CHO 2. heat A B H 119. Major product formed in the following (A) A = N CO2Me B = N synthetic sequence on the monoterpene pulegone is— O OAc O 1. Br2 HH 2. NaOEt, EtOH O 3. KOH, EtOH (B) A = N CO2Me B = N CO2OH O COOH (A) CO2H (B) OH O H (C) A= N B = Me N H AcO H MeO2C HOH2C HH (D) A = N CO2Me B = N CH2OH (C) (D) O HO O O OEt

Chemical Sciences (CSIR) J-15 | 17 120. Optically pure isomers A and B were heated 125. The spectroscopic technique, by which the with NaN3 in DMF. The correct statement ground state dissociation energies of from the following is— diatomic molecules can be estimated is— NMe2 NMe2 (A) microwave spectroscopy Br Br (B) infrared spectroscopy A B (C) UV-visible absorption spectroscopy NMe2 NMe2 (D) X-ray spectroscopy C N3 D N3 126. The term symbol for the first excited state of Be with the electronic configuration 1s2 2s1 (A) A gives optically pure D and B gives 3s1 is— optically pure C (A) 3S1 (B) 3S0 (B) A gives racemic mixture of C and B gives optically pure C (C) 1S0 (D) 2S1/2 (C) A gives optically pure C and B gives 127. Which of the following statements is racemic C INCORRECT ? (D) A gives optically pure D and B gives (A) A Slater determinant is an antisym- racemic D metrized wavefunction 121. A molecular orbital of a diatomic molecule (B) Electronic wavefunction should be changes sign when it is rotated by 180° represented by Slater determinants around the molecular axis. This orbital is— (C) A Slater determinant always corres- (A) σ (B) π ponds to a particular spin state (C) δ (D) ϕ (D) A Slater determinant obeys the Pauli exclusion principle 122. IR active normal modes of methane belong 128. Compare the difference of energies of the to the irreducible representation— first excited and ground states of a particle Td E 8C3 3C2 6S4 6σd confined in (i) a 1-d box (Δ1), (ii) a 2-d A1 1 1 1 1 1 x2 + y2 + z2 square box (Δ2) and (iii) a 3-d cubic box (Δ3). Assume the length of each of the boxes A2 1 1 1 –1 –1 is the same. The correct relation between the E 2 –1 energy differences Δ1, Δ2 and Δ 3 for the 2 0 0 2z2 – x2 – y2, three cases is— x2 – y2 (A) Δ1 > Δ2 > Δ3 (B) Δ1 = Δ2 = Δ3 T1 3 0 – 1 1 – 1 Rx, Ry, Rz (C) Δ3 > Δ2 > Δ1 (D) Δ3 > Δ1 > Δ2 T2 3 0 – 1 – 1 1 x, y, z, xy, yz, zx (A) E + A1 (B) E + A2 129. The correct statement about both the average (C) T1 (D) T2 value of position (〈x〉) and momentum (〈p〉) of a 1-d harmonic oscillator wavefunction 123. The symmetric rotor among the following is— is— (A) 〈x〉 ≠ 0 and 〈p〉 ≠ 0 (A) CH4 (B) CH3Cl (B) 〈x〉 = 0 but 〈p〉 ≠ 0 (C) CH2Cl2 (D) CCl4 (C) 〈x〉 = 0 and 〈p〉 = 0 124. The nuclear g-factors of 1H and 14N are 5·6 (D) 〈x〉 ≠ 0 but 〈p〉 = 0 and 0·40 respectively. If the magnetic field in an NMR spectrometer is set such that the 130. The value of the commutator [x, [x, px]] is— proton resonates at 700 MHz, the 14N nucleus would resonate at— (A) i hx (B) – i h (A) 1750 MHz (B) 700 MHz (C) i h (D) 0 (C) 125 MHz (D) 50 MHz

18 | Chemical Sciences (CSIR) J-15 131. The equilibrium constants for the reactions 136. For a reaction on a surface HH CH4(g) + 2H2O(g) CO2(g) + 4H2(g) and H2 + S S SS CO(g) + H2O(g) CO2(g) + H2(g) are K1 and K2, respectively. The equilibrium HH H constant for the reaction CH4(g) + H2O(g) CO(g) + 3H2(g) is— SS slow S S +H (A) K1·K2 (B) K1 – K2 (C) K1/K2 (D) K2 – K1 At low pressure of H2, the rate is propor- tional to— 132. Consider the progress of a system along the path shown in the figure. ΔS (B → C) for one (A) [H2] (B) 1/[H2] (C) [H2]1/2 (D) 1/[H2]1/2 mole of an ideal gas is then given by— A(T1, V1) 137. The temperature-dependence of an electro- chemical cell potential is— p Adiabatic (A) ΔG (B) ΔH nFT nF process C(T3, V1) B(T2, V2) (C) ΔS (D) ΔS nF nFT V (A) R ln T1 (B) R ln T3 138. The single-particle partition function (f) for a T3 T1 certain system has the form f = AVeBT. The average energy per particle will then be (k is (C) R ln V2 (D) R ln V1 the Boltzman constant)— V1 V2 133. A thermodynamic equation that relates the (A) BkT (B) Bk T2 chemical potential to the composition of a mixture is known as— (C) kT/B (D) k T/B2 (A) Gibbs-Helmholtz equation (B) Gibbs-Duhem equation 139. The indistinguishability correction in the (C) Joule-Thomson equation Boltzmann formulation is incorporated in (D) Debye-H·u·ckel equation the following way : (N = total number of particles; f = single-particle partition func- 134. According to transition state theory, the tion) temperature-dependence of pre-exponential (A) Replace f by f/ N! (B) Replace f N by fN/N! factor (A) for a reaction between a linear and (C) Replace f by f/ ln (N!) a non-linear molecule, that forms products (D) Replace f N by fN/ln(N!) through a non-linear transition state, is given by— (A) T (B) T2 140. In a photochemical reaction, radicals are formed according to the equation— (C) T– 2 (D) T– 1·5 135. For a given ionic strength, (I) rate of reac- k1 tion is given by C4H10 + hv 2C2H5 log k = –4 × 0·51 (I)1/2. Which of the k–1 k0 C2H5 + C2H5 ⎯k2→ C2H6 + C2H4 following reactions follows the above equation ? (A) S2O82– + I– If I is the intensity of light absorbed, the rate (B) Co(NH3)5Br2+ + OH– of the overall reaction is proportional to— (C) CH3COOC2H5 + OH– (A) I (B) I1/2 (D) H+ + Br + H2O2 (C) I [C4H10] (D) I1/2 [C4H10]1/2

Chemical Sciences (CSIR) J-15 | 19 141. Conductometric titration of a strong acid Answers with Hints with a strong alkali (MOH) shows linear fall of conductance up to neutralization point 1. (A) I. Cloth reel → three because of— J. Silent wonder → two (A) formation of water K. Good tone → one (B) increase in alkali concentration (C) faster moving H+ being replaced by L. Bronze rod → zero slower moving M+ Ascending order L, K, J, I. (D) neutralization of acid 2. (C) 2222 = 242 = 216 142. Find the probability of the link in polymers 12m × 4m where average values of links are (1) 10, (2) 50 and (3) 100— 3. (C) (A) (1) 0·99, (2) 0·98, (3) 0·90 4m 4m 45° (B) (1) 0·98, (2) 0·90, (3) 0·99 4m (C) (1) 0·90, (2) 0·98, (3) 0·99 Area = 12 × 4 = 48 m2 (D) (1) 0·90, (2) 0·99, (3) 0·98 4. (C) 20 a 143. The stability of a lyophobic colloid is the consequence of— 10b 200b 10ab = 840 (A) van der Waals attraction among the 2 40 2a = c6 solute-solvent adducts and 2 × a gives 6 at unit place (B) Brownian motion of the colloidal particles 2×3 = 6 or 2 × 8 = 16 (C) insolubility of colloidal particles in solvent So, a can be 3 or 8. (D) electrostatic repulsion among double- According to question, layered colloidal particles 840 + c 6 = 8d6 144. In a conductometric experiment for estima- tion of acid dissociation constant of acetic which gives c + 4 = d acid, the following values were obtained in four sets of measurements : as digit at hundred place is 8, so c will be 1·71 × 10– 5, 1·77 × 10– 5, 1·79 × 10– 5 and 1·73 × 10– 5. between 0 and 5. The standard deviation of the data would be in the range of— So, from table we have (A) 0·010 × 10– 5 – 0·019 × 10– 5 (B) 0·020 × 10– 5 – 0·029 × 10– 5 40 + 2a = c6 (C) 0·030 × 10– 5 – 0·039 × 10– 5 (D) 0·040 × 10– 5 – 0·049 × 10– 5 c should be 4 or 5 145. Silver crystallizes in face-centered cubic 40 + 2 × 3 = 46 …(I) structure. The 2nd order diffraction angle of a beam of X-ray (λ = 1Å) of (111) plane of and 40 + 2 × 5 = 50 → (not possible) the crystal is 30°. Therefore, the unit cell length of the crystal would be— So, a = 3, c = 4 (A) a = 3·151 Å (B) a = 3·273 Å (C) a = 3·034 Å (D) a = 3·464 Å Now check from option We obtain a + b = 11. 5. (A) We use combination nC4 point of inter- section. For octagon n = 8. 8C4 = 8! 4! × 4! = 8 × 7× 6 ×5 × 4 = 70 4! × 4! 6. (D) base area = 24cm × 48cm Longest stick can be kept inside the body diagonally. So, (24)2 + (48)2 + h2 = 562 after solving we get, h = 16 cm.

20 | Chemical Sciences (CSIR) J-15 7. (C) So, duration of Man = Elapse time including time lag + Return time + staying time = 18 hrs + 18 hrs + 12 hrs Area of triangle = N. = 48 hours. Radius of incircle × P of triangle. 13. (A) Let 1 = x 1234 N=? rx + r– x = 2 We know incircle touches the triangle at mid point. rx + 1 = 2 rx 1 Area of triangle = 2 base × height r2x + 1 = 2 rx 1 = 2 × radius × perimeter r2x – 2rx + 1 = 0 1 (rx – 1)2 = 0 2 N = or rx = 1 and r – x = 1 8. (C) So, r4321 + r – 4321 = 2 Let volume be x 14. (A) Time taken = Distance Speed in still water x2, x x Series : x, 4 , … , 2n – = 10 m 10m/min. x = 20 cc = 1 min. 1 Common ratio = 2 15. (A) 16. (C) Sum of the series (as series is in G.P.) 17. (D) U … G … C … C … S … I … R x 20cc No. of letters = 7 – = 1 r = 1 Position of first C = 3 2 1 – Second C = 4 = 40 cc Permutation 3 × 7 × 4 = 25 9. (B) 1. 2. 3. 18. (B) A 90° CR 1. 2. 3. B Ratation of 180° clockwise. R 10. (A) Melting point decreases with pressure, 30° boiling point increases and also solid, liquid and gas can co-exist at the same pressure and D temp. Height = AB + CD 11. (A) Mean = Sum of observation AB = R cos 30° No. of observation Height = R + R cos 30° If sum of observation change then mean will = R + ⎯√ 3 R also change. 2 12. (A) Journey start 19. (B) Length = 24 m Width = 3 m 0100 Hrs local time ⎯ 0900 Hrs. Boundary is required to be at least 60 m So, elapsed time = (9 – 1) hrs + lag time first pitch is at 60 m from boundary = 0 = 8 hrs + 10 = 18 hrs. Diameter of field = 140 m Elapse time for staying Next pitch at 63, 66, 69, 72, 75, 78, 81 = (21 – 9) 60 + 81 = 141 = 12 hrs So, leave 81

Chemical Sciences (CSIR) J-15 | 21 So, total possible pitches at 60, 63, 66, 69, 70, 25. (B) [reference → Inorganic chemistry + Atkins] 75, 78 m position In Fischer Carbene Oxidation State of metal Total = 7 is low and ligand is π acceptor and nature of Fischer carbene is electrophilic 20. (B) carbonic OH 21. (A) CO2 + H2O Anhydrase H2CO3 C (Fischer Carbene) Hydration occur in blood at high pH value. Ex. : (CO)5 Cr CH3 Dehydration occur in lung at low pH value. Carboxypeptidase A is pancreatic exopep- While in Schrock carbene metal are in high positive oxidation state, ligand are non-π tidase that hydrolise peptide bond of C termi- acceptor and nature of Schrock carbene is nal. nucleophilic. 22. (A) Deoxy. Hb. Oxy. Hb. Ex. : Grubb Catalyst Fe+2 HS Fe+2 LS t2g 4 eg2 t2g 6eg0 Pcy3 Cl H Fe is above the Fe is fitted well Ru C plane which tends inside the plane Cl H to increase the thus reducing Fe—N distance Fe—N bond dis- Pcy3 tance. 26. (A) [Inorganic Chemistry – By Greenwood] 23. (C) When NO is linear it donate 3e – Proton affinity of an anion or a neutral atom or molecule is a measure of its gas phase Since both complex follow 18e – rule. basicity. Higher the proton affinity stronger the base and weaker the conjugate acid in gas So [Co(CO)3NO] ⇒ Total valence electron phase. Co = 9 (d7s2) So N3– is strongest base. CO = 3 × 2 = 6 27. (D) [Ref. → Analytical Chemistry – S.M. NO = 3 Kopkar] [Ni(η5 – Cp)NO] = 18 e– ilkovic equation id = 607 nD1/2 m2/3 t1/6 c Total Valence Electron From this eq. id ∝ t1/6 Ni = 10 (d8s2) i.e., depends upon drop time (t = drop time) η5Cp = 5 and its value also depend upon temperature NO = 3 the value of id (diffusion Current) increases at a rate of 1 – 2% per °C. TV E = 18 e – 24. (A) Wacker process— CH2 CH2 + O2 [PdCl4]2− O H 28. (C) The threshold energy is the minimum CuCl2 CH3 projectile energy necessary to satisfy mass energy and momentum conservation in a i.e., formation of carbonyl compound by Nuclear reaction to form product in their Reaction of Alkene and O2 in presence of ground state. Pd+2. T.E. = – Q ⎝⎛⎜mpm+cmc⎠⎞⎟ Mechanism— CH2 CH2 + [PdCl4]2− + H2O ( )= – (– 3·23) O 1 + 13 13 CH3 H + Pd° + 2HCl = 3·485 Regeneration of Wacker catalyst‚ 13N(n, p)13C Q = 3·236 So 13C(p, n)13N = – 3·236 Pd° + II + 2Cl– → [PdCl4]2– + 2CuCl 2Cu Cl2

22 | Chemical Sciences (CSIR) J-15 29. (D) ⇒ – 2200 ppm Reference—Principles of Structure and Reference → Tin Chemistry—Fundamental Reactivity. INORGANIC CHEMISTRY by Frontiers and Application By – Marcel James E. Huheey IV edition Ch-13, page 552. Gielen. 37. (A) Explanation—After addition of HCl in 30. (A) At high temperature or at melting form equation (A) Primary OH group protonated phosphorous exist as symmetrical P4 type and H2O removed early. i.e., structure HO OH H H2O OH i.e., n P PP OO P O −2 +4 O− −H2O O 31. (A) S2O exist as S S In equation (B), Carbonyl group protonated 32. (D) Pseudohalogen are polyatomic analogous and Electrophilic charactor of C = O group of Halogen whose chemistry resemble with increases so, CN ion attacks easily, true Halogen common example. These anions δ(−) H O CN contain nitrogen atoms. O + CN CN–, N3–, SCN–, NCN2–, OCN, Co(CO)4 etc. 33. (A) M (t2g) → PR3 (σ*) H Phosphines (PR3) primarily function as Lewis base, interacting with metal as σ donar Therefore, equilibrium will shift towards right Ligand. PR3 can accept electron density from in case of both A and B. metal into P—C (σ*) Antibonding orbital having π symmetry. 38. (A) Explanation— 34. (A) 3300 cm–1 peak is for ≡ C—H stretch. 2150 cm–1 is for C ≡ C—H stretch. Ethanol RhCl3·3H2O + PPh3 ⎯⎯→ [RhCl(PPh3)3] Thus, CH3—C ≡ C—H RhCl(PPh3)3 is Wilkinson catalyst and 39. (B) Explanation—Acetone have 6H and common method for preparation of Dichloromethane have 2H, when concentra- tion is same then intensity ratio is 3 : 1. in RhCl(PPh3)3 is Refluxing of RhCl3·3H2O 1H–NM R spectra. with PPh3. Moreover, on counting the total valence electrons, But intensity ratio obtained is 1 : 1. So, concentration of Acetone and Dichloro- [Rh(PPh3)2Cl] = 9 + 3 × 2 + 1 = 16e– methane must be in ratio of 1 : 3. {Cl atom cannot be bridging ∴ its contri- 40. (D) Explanation—The intensity of IR-spec- bution is 1}. Wilkinson catalyst is 16e– trum bands is dependent mainly on the mag- nitude of dipole. species. 35. (A) In M⎯——⎯H, the unit is anti-periplanar and thus β-H elimination not possible. α More the polar charactor of a bond, the β greater the intensity of IR-band. So, In case of CO the intense band observed in IR-spectrum M due to large dipole moment change of CO due to bond stretching. H For β hydride elimination reaction β hydrogen 41. (C) Explanation—Bromo-Heptatriene give should be closer to metal in other 3 option all Bromide ion easily and becomes aromatic, so AgNO3 give precipitate of AgBr. hydrogen atoms are far so elimination not possible (sp 2 carbon) 36. (A) The reaction follows SN, CB mechanism Br +AgNO3 + AgBr in which M—Y bond is fully broken before M—X bond begins to form. Thus, Id mechanism is the most evidential −Br Tropyllium cation mechanism. Aromatic (Most stable)

Chemical Sciences (CSIR) J-15 | 23 42. (A) It is Hunsdiecker reaction. This reaction Me Me Me Me Me Me follows a free radical mechanism. Me COOAg COOBr COO Me (I) (II) (III) (IV) Br2 −Br −AgBr order, I > III > II > IV −CO2 because heat of hydrogenation of alkene COO COOBr ∝ Stability 1 Alkene + of Br i.e., more the stable Alkene less will be heat of hydrogenation. (Product) (Free Radical) Hence order of stability of Alkene → 43. (A) Explanation—In general, 1, 2 Hetero I < III < II < IV substitution, ee is the most stable form but due to repulsion between the Methyl groups, 47. (A) Explanation—The correct statement(s) ae form here is most stable. for ribose is— H Me (A) On reduction with NaBH4 it give Me optically inactive product H Me axial Equitorial CHO CH2−OH (ae) (CHOH)3 NaBH4 (CHOH)3 CH2OH CH2OH (Optical inactive) 44. (D) O OH (B) On reaction with methanolic HCl it gives Me a furanoside. CHO NaBH4, CeCl3 MeOH, H2O Me CHO H CO NaBH4 give chemoselective reduction of ketone in presence of CHO group, when H OH HCl, MeOH MeO O CH2OH CeCl3 is mixed with NaBH4. H OH HH H OH 45. (C) Explanation—Both structures are HH identical; because both are super-imposable to CH2OH each other, when one structure is rotated out OH OH of plane as shown below :— (furanoside) (C) On reaction with Br2-CaCO3–water its gives following compounds— H H H Cl CHO CHO H OH H OH Cl Me , Me H H OH Br2−CaCO3 H OH HO OH H OH water CH2OH (I) (II) CH2OH 180 Rotn H H which is optical active. out of plane Cl Me (D) It gives positive follen’s test. OH (III) CHO COOH H OH H OH I and III are Homomers. So, both structures H OH Ag2O H OH are Homomers. H OH Tollen,s H OH 46. (B) The correct order of heat of hydrogenation CH2OH Reagent CH2OH of the following compounds are— So, statement A, B and D is correct.

24 | Chemical Sciences (CSIR) J-15 48. (B) Cinnamic acid is biogenatic precursors for 53. (A) Explanation—Magnitude of stability the natural product umbelliferone constant for K+ ion complexes of the supra molecular hosts is directly proportional to HO O O ionic interaction and size of cavity. 49. (C) (R)-4-methylpentane-2-ol give 5; 13C- So, more the ionic interaction of hetero atom NMR signal. with K+ ion more will be stability. So, stability order is B > A > C. 5 54. (B) Explanation—Antitubercular drug(s) CH3 4 3 2 CH3 1 among the given compounds are— OH CH3 (A) Salbutamol—It is used for relief of bronchospasm in conditions such as ‘asthma’ 1 signal → 2 Methyl group and ‘chronic abstructive’. 1 signal → CH OH 1 signal → CH2 HO HN 1 signal → CH (attached to OH) HO 1 signal → CH3 (B) Ethambutanol 50. (D) NaBH4 is chemoselective reducing agent OH for carbonyl group in presence of ester. In H presence of steric hindrence it give hydride N ion from less hindered side. So, product (D) is major product N H HO Me Me Me Me (C) Isoniazid O NH EtOOC NH2 EtOOC OH H H N NaBH4 Less O 0°C H (D) Diazepam—It is used treatment of ‘anxiety’, alcohol withdrawal syndrome, hindered MeOH/THF muscle spams, trouble sleeping and restless legs syndrome. H more hindered due bridging CH3 O Me N 51. (A) Me Heat [ene reaction] CH2 Cl N CH2 H 52. (C) O Me O H2N−NH2.HCl 55. (C) We know that [Walf-Kishner reduction] ( )Ψ = n πx 2 sin a if n = 0 a then Ψ = 0 (i.e., wavefunction become zero) Me Me Me and it is not possible. In I-D box for ground O O HO state n = 1. n = 0 exists only for Simple H Harmonic Oscillator. [Product] Et3N 56. (C) When we add potential in the Hamaltonian the energy eigen value will shift but the eigen function will not change.

Chemical Sciences (CSIR) J-15 | 25 57. (D) A1 = E.C. L1 because upon addition of inert gas at constant A2 E.C. L2 volume, the total pressure will increase. But ( )⇒ I0 the concentration of the product and reactant It log L1 (i.e., ratio of their moles to the volume of L2 A2 = container) will not change. log 100 63. (C) k = ATn exp. – E0 50 RT 1 ⇒ = 3 is the modified form of Arrhenius Eqn. A2 k = AT2 exp. – E0 (given) RT log 2 = 1 A2 3 Comparing eqn. (i) and (ii) we get A2 = 3 log 2 = 3 × 0·3010 Ea = 2RT + E0 64. (A) 2A + B → 3Z A2 = 0·9030 1 D[A] D[B] 1 D[Z] We know that A = log 1 – 2 Dt = Dt = 3 Dt …(i) T D[A] 1 = Antilog of A2 – Dt = 2 × 10– 4 (given) T from eqn. (i) 1 ⇒ T = Antilog 0·903 – 1 D[A] = 1 D[Z] 2 Dt 3 Dt 1 T = 8 D[Z] = 3 × 1 × 2 × 10– 4 Dt 2 18% 1 T = = 8 × 100 = 12·5 = 3 × 10– 4 mol dm– 3 s– 1 58. (B) For allowed transition 65. (B) The escaping tendency of a charged particle from a phase can be affected by the ΔS = 0 charge state of the phase and the variable used ΔL = ± 1 to describe the differences is escaping tendency of electrostatic potential φ and the 3S → 3P electrostatic potential at a point (a) is defined ΔS = 3 – 3 = 0 as the work per unit charge required to bring a ΔL = (1 – 0) = ± 1 positive test change reversible from infinity to the point (For P, L = 1 for S, L = 0) …(1) φ(a) = δω∞ → a 59. (A) ⎪⎪⎪⎪⎪zyx⎪⎪⎪⎪⎪ ⎯σ⎯xy→ ⎪⎪⎪⎪⎪–zyx⎪⎪⎪⎪⎪ dθ Reference—‘Principle of thermodynamics’ C2x.σxy = ⎪⎪⎪⎪⎪zxy⎪⎪⎪⎪⎪ C⎯2→(x) ⎪⎪⎪⎪⎪––xzy⎪⎪⎪⎪⎪ ⎯σ→xy ⎪⎪⎪⎪⎪–yxz⎪⎪⎪⎪⎪ by ‘Myron Kaufam’. 66. (C) η = KMa (given) It is Mark-Hauwink equation where K, a is From eqn. (i) and (ii) constant for polymer solvent pair. C2x.σxy = σx z. 67. (D) We know that, 60. (B) HCl is a covalent molecule. Thus, both the ΔG = ΔG° + RT ln Q electrons cannot reside only in Cl atom. Thus, option (C) is wrong. In option (A) and (D) for Q = Reaction quotient, H-atom, 3p orbital is not present (H → 1s; Cl = 3s23p5). Q = [Product] [Reactant] ΔG = ΔG° + RT ln [Zn+2] [Cu] [Zn] [Cu+2] 61. (D) For monoatomic gas it only posses translational degree of freedom so variation of Since activity of Cu and Zn is unity heat capacity with temperature is ignored. So, ΔG = ΔG° + RT ln aZn+2 aCu+2 62. (C) At constant volume addition of inert gas does not cause any effect on equilibrium where a = activity.

26 | Chemical Sciences (CSIR) J-15 68. (D) Term symbol = 2S + 1LJ 125Sn ⇒ A = 125 S = 1, L = 0 + 1 = 1 = P r = 1·2 × 10– 15 m 2S + 1 = 3 (A)1/3 = (125)1/3 = 5 J = |L + S| ……… |L – S| r = 1·2 × 10– 15 × 5 = |1 + 1| ……… |1 – 1| = 6 × 10– 15 m Geometrical cross section = π r2 = 2, 1, 0 1 barn = 10– 28 m2 As this is half filled orbital ∴ Lower value of J will be considered. Cross section Area ∴ 3P0. 69. (B) Double layer interaction is determined by = 22 × (6 × 10– 15 m)2 ≈ 1·53 Gany–Chapmann potential (Ψ0 e– kr) 7 a = represents double layer b = represents vander Wall’s and double layer. 74. (A) c = Van der Wall’s. 70. (D) In FCC, Packing fraction is 74% and No. Suzuki coupling—Palladium catalyzed cross of atoms per unit cell is ‘4’. coupling between organoboranic acid and halides. Heck coupling—Palladium catalyzed C—C coupling between aryl halides or vinyl halides and activated alkenes in the presence of a base. Sonogashira coupling—This coupling terminal alkynes with aryl or vinyl halides is performed with a palladium catalyst, Cu(I) cocatalyst and amine base. No. of atom per unit cell Negeshi coupling—This reaction is the organic reaction of an organohalide with an 1 1 = 8 × 8 + 6 × 2 = 4 organic-zinc compound Pd and Ni catalyst. Packing fraction (φ) 75. (C) H5P3O8 = Volume occupied by the particle × 100 O−2 O−2 O−2 Volume of unit cell +4 +3 P +4 4 × 4 π r3 −1 P 3 P −1 −1 OH = × 100 = 74% HO −1 −1OH OH OH a3 71. (D) Differential pulse polarography (DPP) is 76. (B) Hybridisation = sp3 more sensitive then (DCP) from analytical [Br3]+ Geometry = Td point of view because Non-foradic current is Shape = Bent shap less in DPP then DCP. 72. (A) Fluorescence is directly proportional to [Br3]+ ≈ Br2 Br+≈ the amount of absorbed radiation where F = S KP∈bc. Br S ⇒ SBr2 Br The fluorescence signal can be increased if the radiat power of the incident beam is [I5]+ Hybridisation = sp3d increased, therefore always use more intense Geometry = TBP sources. 77. (A) Wade’s theory, Dissolved oxygen largely limits fluorescence F = 3b + 4c + h + x – 2n since it promoters intersystem crossing because it is paramagnetic. b = No. of boron atoms c = No. of carbon atoms 73. (B) We know that, According to Fermi model h = No. of hydrogen atoms r = r0 (A)1/3 x = Amount of negative ions A = mass number. n = No. of vertices = b + c

Chemical Sciences (CSIR) J-15 | 27 F = 3 × 12 + 4 × 0 + 12 + 2 – 2(12 + 0) [Os5(C)(CO)15] = 26 = 1 [74 – 12 × 5] F = 2n + 2 for Closo. 2 26 = 2 × 12 + 2 = 1 [14] = 7 (M + 2) ⇒ nido = 24 + 2 = 26. 2 Hence, closo. [Ru5(C)(CO)16] 78. (A) R = 1 [76 – 60] R 2 Pb Pb = 1 [16] = 8 (M + 3) Aracho. 2 R R 84. (C) J. T. distortion occurs where there is Pb show the inert pair effect. unsymmetrical filling of electrons in t 2g or eg orbital. Due to inert pair effect Pb(II) is more stable forming banana bond in which structure is K4[Cr(CN)6] only correct one. 79. (A) SbCl3 + 3EtMgBr → SbEt3 + 3MgBrCl Cr = + 2 O.S. low spin complex SbEt3 + 2SbI3 → [Sb(Et)I2]n Cr(II) = d4 t 2 4 eg0 (unsym) g Polymeric structure K4[Fe(CN)6] low spin complex Fe(II) = d 6 t 2 6 eg0 (symmetrical) g 80. (A) Fe–porphyrin complex – colour show is due to π – π* transition. K3[Co(CN)6] low spin complex Co(III) ⇒ d 6 [Mn(H2O)6]Cl2 : t 2 6 eg0 (symmetrical) Mn+ 2 ⇒ d5 configuration which is both spin g and Laporte forbidden K4[Mn(CN)6] low spin complex Mn(II) ⇒ d5 t2g5 eg0 (unsym) [Co(H2O)6]Cl2 : Hence only A and D is having unsymmetrical Co(II) ⇒ d7 configuration which is spin filling, hence undergo distortion. allowed but laporte forbidden. 85. (A) The reductive elimination of (A) gives 81. (C) Agostic interaction can be detected by the toluene presence of a 1H NMR peak. A Ar R CH3 Agostic interaction most commonly refers to So, R = CH3 option A is correct. a C—H bond on a ligand that undergoes an 86. (A) Structure of oxy-haemocyanin. interaction with the metal complex. 82. (A) Cp proton of complex (A) is highly His N deshielded because of CO ligand. CO is very N N N His good π-acceptor ligand. His ++ N His His O ++ Complex (D) will be highly shielded because N Cu N Cu N Fe is small in size. It is more closed to N O nucleus. N N N 83. (B) (M + 1) = closo, (M + 2) = nido, (M + 3) His = Aracho, (M + 4) = Hypo = (M = no. of metal atom) Weadge Rule = 1 [TVE – 12 × No. of metal] Blue colour O.S. Cu = II 2 B.O. of O2 = I Ru6(C) (CO)17 87. (B) In strong acidic condition = 1 [8 × 6 + 4 + 17 × 2 – 12 × 6] 2 H+ KMnO4 ⎯⎯→ Mn+2 = 1 [86 – 72] 2 Base Mn+2 ⎯⎯→ Mn+ 4 = 1 [14] = 7 So, option (B) is correct., Acidified Mn+2 is 2 not oxidised by H2O2 but Alkaline (M ± 1) = closo Mn(OH)2 + H2O2 → Mn2O3

28 | Chemical Sciences (CSIR) J-15 88. (C) Oxidation of Zn to Zn+2 will be diamag- 5 H.S. netic hence EPR inactive. 4 Fe and Sn – Inactive. 3 Sharp change 175 EPR spectroscopy generally used for the μ2 species have one or more unpaired electron. 1 Cu(II) are EPR active because non-paired electron. 0 L.S. 175 T CFSE Δ = (– 0·4p + 0·6q) 89. (C) CH3 where p = No. of e–s in t2g orbital CH3 ⇒ d6H.S. q = No. of e–s in eg orbital Δ = (– 0·4 × 4 + 0·6 × 2) Alkene does not undergo reductive elemina- tion reaction easily as compared to Alkane. = – 1·6 + 1·2 = – 0·4 Δ 90. (A) Reaction (i) ⇒ d6 L.S. Δ = – 0·4 × 6 = – 2·4 Δ. Water acting as an oxidant oxidizing Ca to (–) ve sign is used for CFSE. Ca2+ getting itself reduced to H2. Reaction (ii) Water act as Lewis acid accept- 93. (A) UO2+ disproportionates more easily than ing elecron from Cl – which act as a ligand. UO2+2 and U3O8 is most stable oxide of U. Reaction (iii) Water act as Lewis base donat- But coordination No. of U in ing electron to Mg2+ which act as a. [UO2 (NO3)2·(H2O)2]4H2O is eight. Reaction (iv) Water acting as an reductant reducing F to F –. Structure of UO22(+) is linear (Reffrac. = J.D. Lee) 91. (D) Alkyne can act as 2 or 4 electron doner. L(π) → M(σ) : C 94. (B) CO2 does not combines with [Ni(PR3)2 (1,5 – cyclooctadiene]. Only insertion of CO2 M occurs and Et − − Et is inserted to complex C (Empty metal (Filled acetylene and formed compound. d-orbital) π-orbitals) 95. (D) (NH4)2[Ce(NO3)6] complex NO3 is bidented ligand and structure is Icosahedran σ-Type donation i.e., coordination number of Ce is 12. and L(π) → M(σ) : O 2− C M ON C N OO O O N O O (Metal filled (Empty acetylene O Ce O O O d-orbital) π* orbitals) π-type donation N OO OO N ON 92. (A) Fe(phen)2 (NCS)2 O.S. of Fe = + 2 O Fe(II) = d 6 In Ce+ 4 more stable High spin – low spin equilibria or spin cross Ce = 4f 1 5d 1 6s 2 over Ce (IV) = 4f 0 5d 1 6s 0 d6 H.S. d6 L.S. i.e., Z is diamagnetic Ce + 4 is a strong one electron oxidizing agent. No. of unpaird electron No. of unpaired e– = 0 Ce + 4 used in redox process. μ = 4·9 BM μ = 0·0 BM

Chemical Sciences (CSIR) J-15 | 29 96. (C) Cl O N3 N OC N OC N O C O O t-BuOK O OH O C NaN3 O C SOCl2 O −N2 O C O O O CN [Nitrene] [Isocynate] C NH OC N O H OH O O O CN H2O H O COOH NH O O O O C NH C NH C NH O O O C C O HH H HO OH H 97. (D) O– OH 99. (C) O OH H2N OO NaNH2 NH2 + EtO2C CO2Et CO2Et HN H NH3(l) O NH2 H O CO2Et Benzyne O Br NH2 (OH at para position shows (e– releasing) Me Me Me HN CO2Et CO2Et HN H OH Aq. KOH NaNH2 NH2 Benzyne NH2 CO2Et HO CO2Et H NH3(l) H Br NH2 Aq. KOH 98. (C) O O HO CO2Et HN CO2Et OH CO2Et N CO2Et H H AcO AcO HN p−TsNH−NH2/AcOH N N Ts CO2Et CO2Et H O CO2OH CO2Et CO2Et NaBH3CN aq.KOH aq.KOH reflux AcO AcO N CO2OH N CO2Et N H [A] H H [B] NN N NH Ts 0.5 equiv PhC[Me]2OOH H H 100. (A) 1.0 equiv. Ti[OiPr]4 OH 1.2 equiv. (−) DIPT O OMe CH2Cl2, −20°C AcO OH OMe and OH OMe [A] [B] (Product)

30 | Chemical Sciences (CSIR) J-15 The asymetric synthesis of chiral secondary allylic alcohol or their corresponding epoxide. 101. (A) O O O O NCO NH2 OH N Br−Br N Br OH N Br NF NF N FH N FH NF Isocynate OH Nitrene NH2 H OH NH C O NCO FO OH NF N NF [Product] 102. (B) OAc Pd[OAc]2/PPh3; Ph Ph HH H 103. (B) N Ph Et3N, CH3CN H N Et3N N Me H HO H Me Me Me Me Me Me CHO KH/THF Me K+O Me OHC Me [A] 18−Crown−6 rt Me BrPh3P Me Me Me Me [B] 104. (D) O O O IR 1680 C (Carbonyl group) Li, NH3(l) Allyl bromide O 1H–NMR → δ 7·66 (M, 1H), 7·60 (M, 1H) 7·10 (M, 1H), 2·50 (S, 3H) H3C CH2 CH CH2 Br 13C–NMR → δ·190, 144, 134, 132, 128, 28 Ethanolic KOH M/Z → 126 [M+, 100%], 128 [M+ + 2, 4·9%] O O H3C PdCl2, CuCl 106. (A) O2, DMF.H2O H3C [A] O CO2Me O H OH OO NaOMe H2C H H CH3 OMe– O H3C H CH3 O– [B] CO2Me O Me CO2Me 105. (D) {7.10; (m1 1H)}H H{7.66 (m, 1H)} CH3–I H CH3 NaCl, wet {7.60, (m1 1H)} S (2.50; S1 3H) DMSO, 160°C O

Chemical Sciences (CSIR) J-15 | 31 OSiMe3 Cl O H 110. (C) Me t−BuCl LDA Me LDA O – 78°C O Cp Me TMSCl H Ti Al tBu-Cl TiCl4 (LDA removes e–s from least Cp Cl Me hindered side) Tabbe,s Reagent O O Me PhMe2Si COOEt [Product] O 107. (C) Cl Cl H2C (4+2) Cyclo Addition POCl3 OP(O)Cl2 N OP(O)Cl2 Reaction N PhMe2Si H NO H H O O H Cl [A] COOEt Ph NH2 −H+ NN Ph N Cl N Cl H [B] H PhMe2Si 108. (A) EtOOC O O O 111. (A) O LiHMDS, AcCl O Me PhMgBr/CuI CH3 Cl O EtO2C CH3 O EtO2C CH2 O CH3 Cl Me Ph H3O O N NH2 H2N−NH2 OO HO EtO2C CH3 EtO2C CH3 O CH3 O CH3 Br Br NaOEt Me Ph O Ph CH2−Br CO2Et N CO2Et N HO N N N EtO2C N Ph (Product) HH HH 112. (A) Br I NHAc N Me O Me O O EtO2C N 109. (D) CO2Me hv, Acetone NHAc N Pd |OAc|2, Br H Me PPh3, Et3N Me CO2Me Ts O HH N O Ts O OH NHAc OH Me Me CO2Me Pd (OAc)2 HH PPh3, Et3N N (Product) Ts

32 | Chemical Sciences (CSIR) J-15 113. (D) 116. (B) H H HH O 3 AcO N 12 Ph3P CHCO2Me 6 12 3 CHO Stabilised H 6 54 54 H O HH AcO N H H (Chair-Boat-Chair CH CH CO2Me Conformation) E-form (trans) 114. (C) O (Boc)2O, Pyridine O N CH CO2Me H CO2Me NHBoc O CH2 N HO HO NH2 OAc [A] O O TBSO TBSCl, Imidazole H NHBoc N CH2OH LiAlH4/H3O+ OH [B] LiAlH, (Ot−Bu)3 EtOH, −78°C TBSO 117. (C) NIBS gives bromination at allylic posi- tion. In this reaction the ratio of A : B : C = step - (2) N 2 : 1 : 1. Step 2 : Chelation-controlled reduction of the ketone produces the anti-alcohol diastereo selectivity. Br 115. (D) NIBS, AIBN ++ CCl4, heat O Br O O C [A] [C] N2 hυ Br Me Me Carbene Ketene + O OTIPS Br Me O [A] [B] OTIPS C Ratio of A : B : C = 2 : 1 : 1. Me Ketene [2 + 2] Cycloaddition 118. (A) CO2Me Reaction CuI N2 Me Me CO2Me OTIPS O Me Me H O OTIPS H [Carbene] (Unstable) +H −H Me Me Me H CO2Me HO OTIPS SeO2 Me HO H CO2Me

Chemical Sciences (CSIR) J-15 | 33 119. (B) nA1 = 1 [15 × 1 × 1 + 0 × 1 × 8 + 24 Br2 NaOEt, EtOH (–1) × 1 × 3 + (– 1) × (+1) O O + 6 + 3 × 1 × 6] = 1 Br nA2 = 1 [15 × 1 × 1 + 0 × 1 × 8 + 24 COOH (– 1) × 1 × 3 + (– 1) × 1 × 6 KOH/EtOH OH O + 3 × (– 1) × 6] = 0 Br (Product) O nE = 1 [15 × 1 × 1 + 0 × (– 1) × 8 24 Br 120. (B) A gives Racemic mixture of ‘C’ and ‘B’ + ( 1) × 2 × 3 + (– 1) × (0) × 6 gives optically pure ‘C’. + 3 × 0 × 6] = 1 NMe2 NMe2 nT1 = 1 [15 × 1 × 1 + 0 × 0 × 8 + 24 NMe ( – 1) × (– 1) × 3 + (– 1) × 1 × 6 + 3 × (– 1) × 6] = 1 Br Br nT2 = 1 [15 × 1 × 1 + 0 × 0 × 8 + (A) MN e2 24 MN e2 (– 1) × (– 1) × 3 + (– 1) × (– 1) × 6 + 3 × 1 × 6] = 3 So reducible representation. N3 A1 + E + T1 + 3T2 NMe2 NMe2 Since IR activity is count by x , y , z co- ordinate so T2 mode is IR active by using N3 N3 character table. 123. (B) Explanation CH4, CCl4 → Spherical mols. (All Td mols are spherical). Racemic Mixture H NMe2 NMe2 NMe2 NMe2 CH2Cl2 ⇒ | has C2V point Br Cl C Cl N3 | Br H N3 (C) N3 group and thus is assymmetric top rotor. 121. (B) In diatomic molecule only σ and π-bond Remaining CHCl3 is symmetric top rotor. is possible Source–Molecular spectroscopy by Banwell. σ(gerade) spherical ⎯1⎯80°→ No change occur. 124. (D) We know that π(ungerade) ⎯1⎯80→° change in sign occur. υ = γβ0 2π and g ∝ υ 122. (D) or gH = υH gN υN C = 5·6 × 700 = 50 MHz 0·4 x order = 24 E 8C3 3C2 6S4 6σd 125. (B) The dissociation energy g a diatomic molecule is calculated from vibration No. of unshifted atom 5 2 1 1 3 spectroscopy (IR) Contribution per atom 3 0 – 1 – 1 1 — ωe (1 – x e2) De 4 xe Reducible representation 15 0 – 1 – 1 3 =

34 | Chemical Sciences (CSIR) J-15 126. (A) Term Symbol = 2S + 1LJ Substracting equation (1) and (2), we get For orbital l = 0 0 Eqn. (3) and According to equilibrium 2s 3s K3 = K1 K2 1 1 L = 0 → S, S = 2 + 2 = 1 T2 V2 T1 V1 132. (D) ΔS = Cv ln + R ln 2S + 1 = 2 × 1 × 1 = 3 In this case, for process B → C, J = L+S T2 T3 T1 T2 = 0+1=1 = Term symbol = 3S1 and V2 = V1 127. (C) A slater determinent always corresponds V1 V2 to a particular spin state. T3 V1 * Slater determinate changes by changing ∴ ΔS = Cv ln T2 + R ln V2 sign upon exchange of 2 electron. ∫ ∫ΔS = T3 dV T3 PdV 128. (B) Energy (Δ1) = η x2h2 for 1 - D Box T2 T + T2 T 8m l 2 If Cv ln ⎝⎜⎛TT32⎟⎠⎞ = 0 x Similarly, Δ2 = h2 ⎡⎢⎣⎢Lηxx22 + Lηyy22⎤⎥⎦⎥ 8m V1 h2 ⎢⎡⎢⎣ηLxx22 ηy2 Lηzz22⎤⎦⎥⎥ then, (ΔS)B → C = R ln V2 8m Ly2 Δ3 = + + 133. (B) Gibbs-Duhem equation gives relationship between chemical potential and component So, 1D BOX 2D BOX 3D BOX of a thermodynamic system Δ = nx2h2 h2 ⎣⎡⎢⎢Lnxx22 + nLyy22⎥⎥⎤⎦ h2 ⎢⎡⎣⎢nLxx22 + ny2 + Lnzz22⎥⎦⎥⎤ I 8m lx2 8m 8m Ly2 Σ ni dμi = – SdT + VdP i=1 Ground n = 1 nx = 1, ny = 1 nx = 1, ny = 1, 134. (D) According to Transition state theory state nx = 2, ny = 1 nz = 1 qTranslational ∝ T3/2 Ist excited n = 2 nx = 2, ny = 1, qV/b ∝ T° state nz = 1 qrotational ∝ T (linear) Energy difference qrotational ∝ T3/2 (Non-linear) 3h2 3h 2 3h2 Linear + Non-linear Non-linear T.S. 8m L2 8mL2 8mL2 = Pre exponential factor So, Δ1 = Δ2 = Δ3 A = kb T × Q# 129. (C) 〈x〉 = 0 h Qa.Qb and 〈p〉 = 0 or A = T1·T3/2 × T3/2·T0 T1·T3/2·T3/2 T3/2 Reference ⇒ Physical chemistry by ‘Alkins’. = T – 1·5 130. (D) x.(x, Px) = x.i h = 0 135. (B) Explanation : Primary salt effect Since = [x . Px] = i h log k = 2A | Z+ Z– |⎯√I k0 When position vector commulate with where A = 0·51 constant term, it is always zero. log k = 2 × 0·51 Z+Z– ⎯√I 131. (C) k0 CH4 + 2H2O CO2 + 4H2; K1 …(1) = – 4 × 0·51⎯√I CO + H2O CO2 + H2; K 2 …(2) Z+Z– = – 2 For option (B); + 2 × – 1 = – 2 CH4 + H2O CO + 3H2; K3 = ? …(3)

Chemical Sciences (CSIR) J-15 | 35 136. (C) From Longmuir Chemisorption isotherm 140. (A) ∂P = k [C2H5]2 …(1) ∂t dissociation isotherm θ = k·[P]1/n Applying steady state Approximation on 1 + k [P]1/n C2H5 For Diatomic molecule n = 2 ∂[C2H5] θ = k · [H2]1/2 ∂t = 2k1Ia – 2k–1[C2H5]2 – 2k2[C2H5]2 1 + k [H2]1/2 2k1Ia = 2k–1[C2H5]2 + 2k2[C2H5]2 So, at low pressure 2k1Ia θ = k [H2]1/2 [C2H5]2 = 2(k–1 + k2) or θ ∝ [H2]1/2 Substituting above value in eq. (1) 137. (C) We know that, ∂P kk 1Ia ∂P ∂t k–1 + k ∂t ( )ΔS = nf ∂∈ = or ∝ I ∂T P 2 ( )or ∂∈ = ΔS 141. (C) Basic principle of conductometric ∂T nf titration; when faster moving ion is replace P by slower moving ion the conductance decreases linearly. Temperature dependence of Electrochemical cell. 142. — = 1 (C) Since N 1–P 138. (B) F = Partition function = AVe BT P = Probability of polymer link — where, A.V. = Constant. N = Average size — (a) 10 = 1 ⇒ P = 0·9 Average Energy ( E ) – = – ∂ mf …(1) 1 P ∂β 1 ⇒ But β = 1 (b) 50 = 1 – P P = 0·98 k BT 1 ⇒ ∂β = – 1 (c) 100 = 1 – P P = 0·99 ∂T kB T2 143. (D) Electrostatic repulsion among double- ∂β = – ∂T layered colloidal particle kB T2 Reference : Physical chemistry by Puri, then — = kBT2·∂∂T mf Sharma and Pathania. E = kBT2· ∂ · m (Av eBT) 144. (C) The mean ∂T [1·71 × 10– 5 + 1·77 × 10– 5 + 1·79 × 10– 5 = kBT2 ∂ [mA + mU + BT] x– = + 1·73 × 10– 5 ∂T 4 — = k BT2 ∂ [BT] [1·71 × 1·77 + 1·79 + 1·73] × 10– 5 E ∂T 4 = — = k BT2 B 7·0 × 10– 5 × 10– 5 E 4 = = 1·75 139. (B) For indistinguisable particles, the Standard Deviation (s) canonical partition function over-counts the number of microstates by the total number (1·71 – 1·75)2 × 10– 10 + (1·77 – 1·75) × 10– 10 of possible swaps among the particle = + (1·79 – 1·75)2 × 10– 10 + (1·73 – 1·75)2 × 10– 10 identities ‘N’ correcting this (4 – 1) Q = qN N!

36 | Chemical Sciences (CSIR) J-15 (0·04)2 + (0·03)2 + (0·04)2 d2 = a2 = + (0·02)2 × 10– 10 12 + 12 + 12 3 d2 = a2 3 = ⎢⎣⎢⎢⎡ ·0016 + 0·0009 + 00··00000146⎤⎥⎥⎦⎥ 3 + × 10– 5 d= a √⎯ 3 = 0·0045 × 10– 5 Putting the value of ‘d’ in equation (1) 3 (s) = √⎯⎯0⎯·0⎯0⎯15 × 10– 5 2 × 1 = 2 × a sin 30° On solving, we get standard deviation data ⎯√ 3 of Range = 0·030 × 10– 5 – 0·039 × 10– 5 2 = 2 × a · 1 145. (D) From Bragg’s Equation 2 √⎯ 3 n λ = 2d sin θ …(1) and 1 = h2 + k2 + l2 a = 2√⎯ 3 d2 a2 a2 a2 = 2 × 1·732 a2 or d2 = h2 + k2 + l2 a = 3·464 Å ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯