CSIR NET CHEMISTRY (JUNE -2012)
Previous Year Question Paper with Solution.
21. In the reaction (A) and (B),
water behaves as
(a) An acid in both (A) and (B)
(b) An acid in (A) and a base in (B)
(c) A base in (A) an acid in (B)
(d) A base in both (A) and (B)
Ans. b
Sol. In presence of Cl–, water behaves as lewis acid because it accepts electron pair from Cl– and forms [Cl(H2O)n]–
When H2O is present with Mg2+, it behaves as lewis base and forms [Mg(H2O)6]2+
Correct option is (b)
22. The size of the d orbitals in Si, P, S and Cl follows the order.
(a) Cl > S > P > Si
(b) Cl > P > S > Si
(c) P > S > Si > Cl
(d) Si > P > S > Cl
Ans. d
Sol. Size of d-orbitals decrease with decrease in size of element. The decreasing order of size is Si > P > S > Cl
Therefore decreasing order of size of d-orbitals is Si > P > S > Cl
Correct option is (d)
23. The correct structure of basic beryllium nitrate is
(a) 
(b) 
(c) 
(d) 
Ans. a
Sol. Correct option is (a)
24. The total number of lone pairs of electrons in I3– is
(a) zero
(b) three
(c) six
(d) Nine
Ans. d
Sol. 
Number of lone pair of electrons = 9
Correct option is (d)
25. If Mossbauer spectrum of Fe(CO)5 is recorded in the presence of a magnetic field, the original spectrum with two lines changes into the one with
(a) Three lines
(b) Four lines
(c) Five lines
(d) Six lines
Ans. d
Sol. 
Correct option is (d)
26. The spectrophotometric response for the titration of a mixture of Fe3+ and Cu2+ ions against EDTA is given below.
The correct statement is
(a) 
(b) 
(c) 
(d) 
Ans. c
Sol. Correct option is (c)
27. in 'carbon-dating' application of radioisotopes, 14C emits
(a) 
(b) 
(c) 
(d) Positron
Ans. a
Sol. In carbon dating, 
undergoes 
decay
Correct option is (a)
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
Ans. d
Sol. Correct option is (d)
29. The oxidation state of iron in met-hemoglobin is
(a) Three
(b) Two
(c) Four
(d) Zero
Ans. 1
Sol. Correct answer is (a)
30. The reactions of Ni(CO)4 with the ligand (L = PMe3 or P(OMe)3)L yields Ni(CO)3L. The reaction is
(a) Associative
(b) Dissociative
(c) Interchange (Ia)
(d) Interchange (Id)
Ans. b
Sol. Ni(CO)4 [L = PMe3, P(OMe)3]
Ni(CO)4 Total valence electron around
Ni = 10 + 4 × 2 = 18 electron
Substitution reaction at coordinatively saturated tetrahedral complexes with a 18 electron count like Ni(CO)4 or Ni(CO)2(PR3)2 follows a simple first order kinetics.
Since rate does not depend upon the concentration and nature of the ligand suggesting a dissociative mechanism.
Correct option is (b)
31. As a ligand Cl– is
(a) 
(b) 
(c) 
(d) 
Ans. c
Sol. Ligand Cl– is both 
According to LFT, each ligand is always 
some ligands like: NH3 which has only one lone pair of electrons is only 
Some ligands like F–, Cl–, OH– etc which hence two or more lone pair of electrons are 
donors Some ligands like CO, CN–, SH4 bpy, phen etc. how vacant 
orbitals so these are 
Correct option is (c)
32. The correct d-electron configuration showing spin-orbit coupling is
(a) 
(b) 
(c) 
(d) 
Ans. c
Sol. Spin-orbit coupling of d-electrons takes place in which t2g orbitals are unsymmetrically filled.
Correct option is (c)
33. The correct statement for the aggregating nature of alkyl lithium (RLi) reagent is
(a) The carbanion nucleophilicity increases with aggregation.
(b) The observed aggregation arises from its electron deficient nature
(c) Carbanion nucleophilicity does not depend on aggregation
(d) The extent of aggregation is maximum in polar dative solvents
Ans. b
Sol. 
the tendency of organolithium compounds to associate in the solid state as well as in solution is due to the fact that in a single molecule (LiR), the number of valence electrons is too low to use all the available Li valence orbitals for 2e-2c bonding. This electron deficiency is compensated by the formation of multicentre bonds e.g. (2e-4c) bonding.
In this structure methyl group of one (LiCH3)4 unit interacts with the Li atoms of neighbouring (Li)4 tetrahedraon.
Correct option is (b)
34. For the reaction, trans-[IrCl(CO)(PPh3)2] + Cl2
trans-[IrCl3(CO)(Phh3)2], the correct observation
(a) 
(b) 
(c) 
(d) 
Ans. a
Sol. 
Since in product metal Ir is present in the higher oxidation state. So, there is less back bonding to 
orbital of CO.
Correct option is (a)
35. The nucleophilic attack on olefins under mild conditions:
(a) Is always facile
(b) Is more facile than electrophilic attack on olefins
(c) Is facile for electron-rich olefins
(d) Requires activation by coordination to metal
Ans. d
Sol. Nucleophilic attack on olefins under mild conditions
So, it requires activation by coordination to metal.
Correct option is (d)
36. Among the following the strongest oxidizing agent is:
(a) [WO4]2–
(b) [CrO4]2–
(c) [MoO4]2–
(d) [ReO4]–2
Ans. b
Sol. In [CrO4]2–, [MoO4]2–, [WO4]2– and [ReO4]2– all the metals have same oxidation state but Cr6+ is of smallest size. Therefore it has strongest tendency to attract electrons. Therefore strongest oxidizing agent.
Correct option is (b)
37. The least basic among the following is
(a) Al(OH)3
(b) La(OH)3
(c) Ce(OH)3
(d) Lu(OH)3
Ans. a
Sol. Al(OH)3
Amphoteric whereas hydroxides of lanthanoids are all base.
Correct option is (a)
38. For any operator A and its adjoint 
the incorrect statement is:
(a) 
(b) 
(c) 
(d) 
Ans. d
Sol. 
Because for being Hermitian an operator should full till the condition that 
So, it is antihermitian in nature
Correct option is (d)
39. For hydrogen-like atom with a nuclear charge Z, the energy of orbital with principal quantum number 'n' follows the relation
(a) 
(b) 
(c) 
(d) 
Ans. d
Sol. As we know that, the energy of hydrogen like atom with nuclear charge 'z' the energy of orbital with principal quantum number 'n' is 
Correct option is (d)
40. The average value of the radius <r> in the 1 s state of the hydrogen atom is (a0 is Bohr radius)
(a) a0
(b) 1.5 a0
(c) 0.75 a0
(d) 0.5 a0
Ans. b
Sol. The average value of radius 
in the 1s state of hydrogen atom is
Correct option is (b)
41. Among the following, the correct statement is
(a) The number of irreducible representation is equal to classes of symmetry operations.
(b) The number of irreducible representation is equal to the order of the symmetry point group.
(c) The irreducible representations contained in any point group are always of one dimension
(d) A symmetry point group may not contain a totally symmetric irreducible representation.
Ans. a
Sol. The number of irreducible representation is equal to number of classes of symmetry operations in the group. This is first postulate of Great orthogonality theorem (GOT).
Correct option is (a)
42. For a diatomic molecule AB, the energy for the rotational transition from J = 0 to J = 1 state is 3.9 cm–1. The energy for the rotational transition from J = 3 to J = 4 state would be
(a) 3.9 cm–1
(b) 7.8 cm–1
(c) 11.7 cm–1
(d) 15.6 cm–1
Ans. d
Sol. J = 0 to J = 1 appears at 2B = 3.9 cm–1
Value of rotational constant B = 3.9/2 = 1.95
Correct option is (d)
43. For the vibrational Raman spectrum of a homonuclear diatomic molecule, the selection rule under harmonic approximation is
(a) 
(b) 
(c) 
(d) 
Ans. b
Sol. For diatomic simple harmonic oscillator. All the vibrational energy levels are equally spaced and selected rule 
only.
Correct option is (b)
44. With increase in temperature, the Gibbs free energy for the adsorption of a gas on to a solid surface
(a) Becomes more positive from a positive value
(b) Becomes more negative from a positive value
(c) Becomes more positive from a negative value
(d) Becomes more negative from a negative value
Ans. c
Sol. In adsorption of a gas (adsorbate) on a solid surface (adsorbent), adsorbate changes from the more random state to the less random state.
Therefore, adsorption is accompanied by decrease in entropy of the system, i.e. 
is negative. Also, there is decrease of surface energy which appears as heat.
Therefore, adsorption is accompanied by decrease in enthalpy of the system.
For a spontaneous process, the 
should be negative.
Therefore, 
should have a sufficiently high negative value.
As T increases, 
becomes positive and also 
Correct option is (c)
45. The vapour of a pure substance, when cooled under a pressure less than its triple point pressure
(a) Liquiefies
(b) Liquifies first and then solidifies
(c) Solidifies directly
(d) Remains unchanged
Ans. c
Sol. Correct option is (c)
46. The quantities, which are held fixed in a canonical ensemble are
(a) N, T and P
(b) V, T and N
(c) N, V and E
(d) 
Ans. b
Sol. 
Canonical ensemble of replies in thermal contact
Correct option is (b)
47. The correct value of Eº, of a half cell in the following graph of E vs log m(molality) is:
(a) CC´/AC´
(b) AB´
(c) BB´
(d) CC´
Ans. c
Sol. If we take a half-cell
Then E vs log m graph gives intercept Eº. Intercept in the given figure is BB´
Correct option is (c)
48. One of the assumptions made in the conventional activated complex theory is:
(a) Equilibrium is maintained between reactants and the activated complex
(b) Equilibrium is maintained between the reactants and the products.
(c) Equilibrium is maintained between the products and the activated complex
(d) Equilibrium is maintained between the reactants, the activated complex and the products.
Ans. a
Sol. Equilibrium is maintained between reaction and activated complex.
Correct option is (a)
49. For a reaction, the rate constant k at 27ºC was found to be k = 5.4 × 1011 e–50
The activation energy of the reaction is
(a) 50 J mol–1
(b) 415 J mol–1
(c) 15,000 J mol–1
(d) 125,000 J mol–1
Ans. d
Sol. We know that Arrhenius equation is
compare equation (1) and (2), then we get
Correct option is (d)
50. During the addition polymerissation, the reaction proceeds via
(a) Step-growth process
(b) Free-radical chain reaction
(c) Cascade process
(d) Addition reaction
Ans. b
Sol. In addition polymerization reaction proceeds via free radical chain reaction.
Correct option is (b)
51. How many atoms are there in an element packed in a fcc structure
(a) 1
(b) 2
(c) 4
(d) 8
Ans. c
Sol. Atoms that are packed in fcc structure. There are 8 atoms in the
corner of a cube. Since a corner of cube is shared by 8 unit cells.
Hence total atoms contribution to a unit cell = 8 × 1/8 = 1 atom.
Atom in a face is shared by two unit cell and there are 6 faces in the cube. Hence atoms = 6 × 1/2 = 3 atoms.
Total number of atoms in fcc = 3 + 1 = 4 atoms.
Correct option is (c)
52. The structure obtained when all the tetrahedral holes are occupied in a fcc structure is on the type
(a) NaCl
(b) CsCl
(c) CaF2
(d) ZnS
Ans. c
Sol. If the number of close packed spheres be N then the number of tetrahedral voids are 2N.
Since all the tetrahedral holes are occupied. Hence the correct option is CaF2.
Correct option is (c)
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
Ans. b
Sol. 
Correct option is (b)
54. The data obtained from two sets of experiments A and B have the following characteristics
It may be concluded that
(a) A is more precise than B
(b) A is less precise than B
(c) A and B are of the same precision
(d) Relative precision of A and B cannot be assessed
Ans. b
Sol. 
More RSD less precise. Less RSD more precise.
Correct option is (b)
55. The IUPAC name of the compound given below is:
(a) ethyl (R)-2-methyl-4-oxocyclohex-2-enecarboxylate
(b) ethyl (S)-2-methyl-4-oxocyclohex-2-enecarboxylate
(c) (R)-4-ethoxycarbonyl-3-methylcyclo-hex-2-enone
(d) (S)-4-ethoxycarbonyl-3-methylcyclo-hex-2-enone
Ans. b
Sol. Ester is the main functional group
IUPAC name for ester:
Correct option is (b)
56. The major product formed in the following reaction is:
(a) 
(b) 
(c) 
(d) 
Ans. d
Sol. Chemical reaction involved in the above transformation can be illustrated as
Correct option is (d)
57. The number of signals that appear in the broadband decoupled 13C NMR spectrum of phenanthrene and anthracene, respectively are
(a) ten and four
(b) ten and ten
(c) seven and four
(d) seven and seven
Ans. c
Sol. Number of signals appears in the broad decoupled 13C NMR spectrum of phenanthrene and anthracene can be counted as
Here identical carbon has given same numeric as they same signal in 13C NMR.
Correct option is (c)
58. The co-enzyme that is involved in the reduction of a double bond in fatty acid biosynthesis is
(a) NADH
(b) Biotin
(c) Pyridoxal
(d) FADH2
Ans. d
Sol. Correct option is (d)
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
(a) Enantiomers
(b) Diastereomers
(c) Constitutional isomers
(d) Homomers
Ans. b
Sol. The molecule has double bond with diastereotopic faces.
Correct option is (b)
60. The major product formed in the following concerted reaction is
(a) 
(b) 
(c) 
(d) 
Ans. a
Sol. It is an electrocyclic reaction. Both the systems 
are present. In case of 
electrons, conrotatory process will occur, but it leads to a trans junction leading to steric strain in the molecule. So, it must be 
electron system which will undergo photochemical reaction through disrotatory process.
Correct option is (a)
61. The structure of meso-tricarboxylic acid that is formed on potassium permanganate oxidation of abietic acid is:
(a) 
(b) 
(c) 
(d) 
Ans. b
Sol. The molecule plane of symmetry and opposite discriptor at the chiral centre. So, it is an example of meso compound.
Correct answer is (b).
62. The major product formed in the following reaction is:
(a) 
(b) 
(c) 
(d) 
Ans. b
Sol. 
Correct option is (b)
63. The major product formed in the following reaction is:
(a) 
(b) 
(c) 
(d) 
Ans. d
Sol. 
Correct option is (d)
64. Among the following, the synthetic equivalent for acyl anion is:
(a) Nitroethane and base
(b) 
(c) Ethylmagnesium bromide
(d) Acetyl chloride and triethylamine
Ans. a
Sol. 
Correct option is (a)
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) 
(b) 
(c) 
(d) 
Ans. b
Sol. Hydrogenolysis 
cleavage of the bond between the carbon and heteroatom and addition of hydrogen.
The usefulness of benzyloxy carbonyl group in deprotection is the result of the ease by which it can be removed by hydrogenolysis over a palladium catalyst and the intermediate carbonic acid that loses carbondioxide to give the amine product.
Correct option is (b)
66. Among the following, the amino acid which is basic in nature is:
(a) Tyrosine
(b) Asparagine
(c) Leucine
(d) Arginine
Ans. d
Sol. 
More number of NH2 groups, more basic, the amino acid hence more basic is arginine.
Correct option is (d)
67. "Phosphorescence" is represented as
(a) 
(b) 
(c) 
(d) 
Ans. a
Sol. Phosphorescence is in a transfer from triplet state to singlet which can be represented as
Correct option is (a)
68. Anong the following diacids, the one that forms an anhydride fastest on heating with acetic anhydride is:
(a) 
(b) 
(c) 
(d) 
Ans. a
Sol. 
Among I, III and (II, IV), I and III are planar molecule. II and IV are non-planar.
For dehydration two –COOH groups must come close together and it is easy in planar molecule rather than non-planar.
So, I is appropriate for dehydration both –COOH group are cis. III has not cis geometry for both COOH group.
Correct option is (a)
69. The major product formed in the following reaction sequence is:
(a) 
(b) 
(c) 
(d) 
Ans. c
Sol. Chemical reaction involved in the above transformation can be illustrated as
m-CPBA attacks at more electron rich double bond.
Correct option is (c)
70. In the 400 MHz 1H NMR spectrum, an organic compound exhibited a doublet. The two lines of the doublet are at 
2.35 and 2.38 ppm. The coupling constant (J) value is
(a) 3 Hz
(b) 6 Hz
(c) 9 Hz
(d) 12 Hz
Ans. d
Sol. Coupling constant can be calculated as
Correct option is (d)
71. The strength of 
bonding in E-O (E = Si, P, S and Cl) follows the order
(a) Si – O > P – O > S – O > Cl – O
(b) P – O > Si – O > S – O > Cl – O
(c) S – O > Cl – O > P – O > Si – O
(d) Cl – O > S – O > P – O > Si – O
Ans. d
Sol. Smaller the inter nuclear distance between two atoms, stronger will be the 
The increasing or due of inter nuclear distance is Cl – O < S – O < P – O < Si – O
Therefore strength of 
bonding is Cl – O > S – O > P – O > Si – O
Correct option is (d)
72. In the following reactions carried out in liquid NH3.
KNH2 and NH4NO3 act respectively as
(a) Solvo-acid and solvo-base
(b) Solvo-base and solvo-acid
(c) Conjugate acid and conjugate base
(d) Conjugate base and conjugate acid
Ans. a
Sol. Correct option is (a)
73. The pair of lanthanides with the highest third-ionization energy is
(a) Eu, Gd
(b) Eu, Yb
(c) Dy, Yb
(d) Lu, Yb
Ans. b
Sol. 
Correct option is (b)
74. The lanthanide (III) ion having the highest partition coefficient between tri-n-butylphosphate and concentrated HNO3 is
(a) La(III)
(b) Eu(III)
(c) Nd(III)
(d) Lu(III)
Ans. d
Sol. Smaller the size of ln2+ ion, higher will be the partition coefficient.
Correct option is (d)
75. The quantative determination of N2H4 with KIO3 proceeds in a mixture of H2O/CCl4 as follows
(a) Consumption of N2H4
(b) ICl formation
(c) Disappearance of the Yellow color due to Cl2 in CCl4 layer.
(d) Displacement of the Red color due to I2 in CCl4 layer.
Ans. d
Sol. Correct answer is (d).
76. Among the halides, NCl3(A), PCl3(B) and AsCl3(C), those which produce two different acids.
(a) A and B
(b) A and C
(c) B and C
(d) A, B and C
Ans. c
Sol. 
(two different Acids) 
(two different acids)
Correct option is (c)
77. The decreasing order of dipole moment of molecules is
(a) NF3 > NH3 > H2O
(b) NH3 > NF2 > H2O
(c) H2O > NH3 > NF3
(d) H2O > NF3 > NH3
Ans. d
Sol. 
Therefore, order of dipole moment is 
Correct option is (d)
78. The cluster having arachno type structure is:
(a) [Os5(CO)16]
(b) [Os3(CO)12]
(c) [Ir4(CO)12]
(d) [Rh6(CO)16]
Ans. b
Sol. Step-I: Calculate T.V.E. (total valency electron)
= 8 × 3 + 2 × 12 = 48
Step-II: Polyhedral electron count
= T.V.E. – (n × 12) = 48 – 36 = 12 n = number of the metal atom
Step-III: = P.E.C./2 = 12/2 = 6 = (n + 3) = Archano.
Correct option is (b)
79. The carbonyl resonance in 13C NMR spectrum of 
(103Rh, nuclear spin, I = 1/2, 100%) shows a triplet at –65ºC owing to the presence of
(a) Terminal CO
(b) 
(c) 
(d) 
Ans. b
Sol. Triplet of carbonyl carbon indicates the each CO is attached with chemically equivalent two Rh atoms that means 
complex. Actual structure is
80. Low oxidation state complexes are often air-sensitive, but are rarely water sensitive because
(a) Air is reducing in nature while water is inert
(b) Both air and water are oxidizing in nature
(c) Both air and water are not 
(d) Complexes with low oxidation states will easily lose electrons to O2 but will not bind to 
molecule like H2O.
Ans. d
Sol. Correct option is (d)
81. The metal complex that exhibits a triplet as well as doublet in its 31P NMR spectrum is
(a) mer – [IrCl3(PPh3)3]
(b) trans – [IrCl(Co)(PPh3)2]
(c) fac – [IrCl3(PPh3)3]
(d) [Ir(PPh3)4]+
Ans. a
Sol. 
Indicates total three phosphorus atoms in the molecules in two sets –2P and 1P
2P – doublet, 1P – Triplet
That is possible in mer – [IrCl3(PPh3)3] fac – [IrCl3(PPh3)3] will give singlet of 3P.
Trans – [IrCl(CO)(PPh3)2] will give singlet of 2P.
[Ir(PPh3)4]+ will give singlet of 4P.
Correct option is (a)
82. The complex that does not obey 18-electron rule is
(a) 
(b) [W(CO)3(SiMe3)(Cl)(NCMe)2]
(c) [IrCl3(PPh3)2(AsPh2)]–
(d) [Os(N)Br2(PMe3)(NMe2)]–
Ans. d
Sol. 
T.V.E. = 6 + 6 + 2 + 2 + 2 = 18
Total valence electron = 4 + 6 + 2 + 2 + 4 = 18
Total valence electron = 6 + 6 + 4 + 2 = 18
Total valence electron = 8 + 3 + 2 + 2 + 1 + 1 = 17
Correct option is (d)
83. The number of spin-allowed ligand field transitions for octahedral Ni(II) complexes with 3A2g ground state is:
(a) Two
(b) Three
(c) One
(d) Four
Ans. b
Sol. For Ni2+ ion in free state, the thus having same spin multiplicity an 3F and 3P. In octahedral field, 3F splits into 3A2g, 3T2g and 3T1g terms but 3P does not split but it transforms into 3T1g(P). Therefore or orgel diagram is.
Three electronic transitions are 
Correct option is (b)
84. The correct structure of P4S3 is:
(a) 
(b) 
(c) 
(d) 
Ans. a
Sol. Structure of P4S3.
Correct option is (a)
85. The final product of the reaction 
(a) 
(b) 
(c) [Mn(CO)6]
(d) [(MeCO)Mn(CO)5]
Ans. d
Sol. 
Correct option is (d)
86. The reaction that yields Li[AlH4] is:
(a) 
(b) 
(c) 
(d) 
Ans. c
Sol. 
Correct option is (c)
87. The number of microstates for d5 electron configuration is
(a) 21 × 63
(b) 14 × 63
(c) 7 × 62
(d) 28 × 63
Ans. c
Sol. We know that number of microstates 
For d5, r = 5, l = 2
Therefore, n = 2(2l + 1) = 2(2 × 2 + 1) = 10
Number of microstates 
Correct option is (c)
88. The carbon-14 activity of an old wood sample is found to be 14.2 disintegrations min–1 g–1. Calculate age of old wood sample, if for a fresh wood sample carbon-14 activity is 15.3 disintegrations min–1 g–1 (t1/2 carbon-14 is 5730 years), is:
(a) 5,000 years
(b) 4,000 years
(c) 877 years
(d) 617 years
Ans. d
Sol. r0 = 15.3 dpm; r = 14.2 dpm
rate at any time 
Number of atoms
Correct option is (d)
89. The reaction 
[25ºC, 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 favourable but enthalpically unfavorable such that 
(d) Thermodynamically unfavourable 
Ans. b
Sol. Because metal carbonyls are stronger bond then break this bond huge amount of energy are required. But metal-metal bonds are weaker bond. So, reaction is endothermic and enthalpy predominant and entropy unfavourable.
Correct option is (b)
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 natural. 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
(c) 1.50 meq/g
(d) 1.75 meq/g
Ans. d
Sol. Correct answer is (d)
91. The molar extinction coefficient of B (MW = 180) is 4 × 103 lit. mol–1 cm–1. One liter solution of C which contains 0.1358 g pharmaceutical preparation of B, shows an absorbance of 0.411 in a 1 cm quartz cell. The percentage (w/w) of B in the pharmaceutical preparation is:
(a) 10.20
(b) 14.60
(c) 20.40
(d) 29.12
Ans. b
Sol. 
Wt. of sample dissolved in 1 lit solution = 0.1358 g
Absorbance of solution 1 lit = 0.441 at 1 cm.
Absorbance (A) = 
gm quantity of compound = conc. × mol.wt. = 0.11025 × 10–3 × 180 = 0.019845 g.
Correct option is (b)
92. The changes (from A-D given below) which occur when O2 binds to hermerythrin are
(a) One iron atoms is oxidized
(b) Both the iron atoms are oxidized
(c) O2 binds to one iron atom and is also hydrogen bonded.
(d) O2 binds to both the iron atoms and is also hydrogen bonded.
Ans. a
Sol. 
Correct option is (a)
93. In photosynthetic systems the redox metalloproteins involved in electron transfer are cytochrome (cyt, b), cytochrome bf complex (cyt bf) and plastocyanin (PC). The pathway of electron flow is
(a) 
(b) 
(c) 
(d) 
Ans. c
Sol. 
Correct option is (c)
94. The total numbers of fine and hyperfine EPR lines expected for octahedral high-spin Mn(II) complexes are respectively (I = 5/2 for Mn)
(a) 3 and 30
(b) 5 and 33
(c) 5 and 30
(d) 4 and 24
Ans. 3
Sol. The total number of fine and hyperfine lines from high spin Mn(II) complex, I = 5/2.
Number of fine lines = Number of unpaired electron in MN = 5
Number of hyperfine lines = Number of unpaired electron × (2nI + 1)
Note: n = number of Mn atom = 1, I = Nuclear spin of Mn = 5/2 =
Correct option is (c)
95. The Mossbauer spectra of two iron complexes are shown below. They may arise from (i) high spin iron (III), (ii) high-spin iron (II) and (iii) low-spin iron (III)
The correct matches of spectra (A) and (B) with the iron complexes are
(a) A with (i) and B with (ii)
(b) A with (ii) and B with (i)
(c) A with (iii) and B with (ii)
(d) A with (ii) and B with (iii)
Ans. b
Sol. 
No quadrupole splitting, singlet signal as shown in (B).
quadrupole splitting, gives doublet signal as shown in (A)
Axial electronic configuration quadrupole splitting, doublet signal as shown in (A)
Correct option is (b)
96. The probability of finding the particle in a one dimensional box of length 'L' in the region between 
and 
for quantum number n = 1 is:
(a) 
(b) 
(c) 
(d) 
Ans. b
Sol. 
Correct option is (b)
97. A particle in three dimensional cubic box of length L has energy of 
The degeneracy of the state is
(a) 2
(b) 3
(c) 6
(d) 9
Ans. c
Sol. Particle in a 3D cubic box of length 'L' has a energy of 
The degeneracy may be calculated by taking the energy formula for 3D cubic box into consideration
nx, ny, nz combination may (1, 2, 3) (1, 3, 2) (2, 1, 3) (2, 3, 1) (3, 1, 2) (3, 2, 1)
At the energy level 
there is 6 energy state. So, degeneracy = 6.
Correct option is (c)
98. The following are the three statements about perturbation theory
(A) Second order perturbation correction to the ground state energy is Always negative.
(B) Sum of the zeroth order and the first order corrections to the ground state energy is always greater than the exact ground state energy.
(C) Sum of the zeroth order and first order corrections to the ground state energy is less than the exact state energy.
From the following which one is correct?
(a) Only A is true
(b) Both A and B are true
(c) Only C is true
(d) Both B and C are true
Ans. b
Sol. Energy correction for second order energy
So, second order correction to energy is negative.
So, statement (A) and (B) is true.
Correct option is (b)
99. Using Huckel molecular orbital approximation, the two roots of secular equation of ehene are
(a) 
(b) 
(c) 
(d) 
Ans. c
Sol. H2C=CH2
The secular determinant will take the form.
Dividing each element of the above determinant by 
and then setting 
we get.
Which on expanding gives.
x2 – 1 = 0
The two roots of the above equation are
x = –1 and +1
Thus, 
Correct option is (c)
100. For H2 molecule in the excited state 
the part of the triplet state with ms = 0 is proportional to
(a) 
(b) 
(c) 
(d) 
Ans. d
Sol. For molecule H2 in the excited state 
The spin part of triplet state with ms = 0 is proportion to
The three triplet states are 
So, triplet state with ms = 0 is 
Correct option is (d)
101. A square pyramidal, MX4, molecule belongs to C4V point group. The symmetry operations are: E, 2C4, C2, 
The trace for the reducible representation, when symmetry operations of C4V applied to MX4, is:
(a) 5 1 1 1 3
(b) 1 1 1 1 1
(c) 5 1 1 1 1
(d) 4 1 1 1 3
Ans. a
Sol. 
Correct answer is (a).
102. Character table of C2v point group is:
If the initial and final states belong to A1 and B1 irreducible representation respectively, the allowed electronic transition from A1 to B1 is:
(a) z-polarized
(b) y-polarized
(c) x-polarized
(d) x, z-polarized
Ans. c
Sol. Transition A1 to B1 can be evaluated by direct product A1 × B1 and its transformation.
B1 transforms according to x-axis hence transition will be x-axis polarized.
Correct option is (c)
103. Using cuvettes of 0.5 cm path length, a 10–4 M solution of a chromophore shows 50% transmittance at certain wave length. The molar extinction coefficient of the chromophore at this wave length is (log 2 = 0.301)
(a) 1500 M–1 cm–1
(b) 3010 M–1 cm–1
(c) 5000 M–1 cm–1
(d) 6020 M–1 cm–1
Ans. d
Sol. A = log10 T
A = log10 50 = 0.301
Correct option is (d)
104. The set of allowed electronic transitions among the following is
(a) A, B, E
(b) A, C, E
(c) B, C, D
(d) C, D, E
Ans. c
Sol. The transitions 
etc. are allowed only when the spin multiplicity of two states are same.
Correct option is (c)
105. The following data were obtained from the vibrational fine structure in the vibronic spectrum of a diatomic molecule:
where 
is the energy associated with natural frequency of vibration and Xe is the anharmonicity constant. The dissociation energy (De) of the molecule is
(a) 4096 cm–1
(b) 6144 cm–1
(c) 8192 cm–1
(d) 16384 cm–1
Ans. c
Sol. 
Correct option is (c)
106. An ideal gas was subjected to a reversible, adiabatic, expansion and then its initial volume was restored by a reversible, isothermal compression. If 'q' denotes the heat added to the system and 'w' the work done by the system, then
(a) w < 0, q < 0
(b) w > 0, q < 0
(c) w < 0, q > 0
(d) w > 0, q > 0
Ans. b
Sol. q denotes the heat added to the system and 'w' is work done by system.
Then, q < 0, w > 0
Correct option is (b)
107. The gas phase reaction 
is an exothermic process. It an equilibrium mixture of NO2 and N2O4, the decomposition of N2O4 can be induced by
(a) Lowering the temperature
(b) Increasing the pressure
(c) Introducing an inert gas at constant volume
(d) Introducing an inert gas at constant pressure.
Ans. d
Sol. Correct answer is (d).
108. Indicate which one of the following relations is NOT correct.
(a) 
(b) 
(c) 
(d) 
Ans. b
Sol. Applying Euler Theorem Maxwell equation we get Maxwell relation.
Correct option is (b)
109. The energy levels of the harmonic oscillator (neglecting zero point energy) are 
for n = 0, 1, 2, ...... 
Assuming 
the partition function is
(a) e
(b) 
(c) 
(d) 
Ans. d
Sol. 
Correct option is (d)
110. The correct entropy for 6 identical particles with their occupation number {0, 1, 2, 3} in four states is
(a) 
(b) 
(c) 
(d) 
Ans. c
Sol. 
Correct option is (c)
111. The correct Nernst equation for the concentration cell:
without liquid junction would be
(a) 
(b) 
(c) 
(d) 
Ans. c
Sol. 
Correct option is (c)
112. Main assumption(s) involved in the derivation of Debye-Huckel equation is(are) the validity of
(a) Only Poission equation
(b) Poission equation and Boltzmann distribution
(c) Poission equation, Boltzmann distribution and 
(d) Poission equation Boltzmann distribution and 
Ans. d
Sol. 
Than we expand them and put 
condition. Then we apply poission equation to calculate the potential. The equation is 
Correct option is (d)
113. In the base (OH–) hydrolysis of a transition metal complex [ML6]Z+, the slope between log(k / k0) and 
is found to be –2.1. The charge on the complex is
(a) +1
(b) +2
(c) +3
(d) +4
Ans. b
Sol. 
Correct option is (b)
114. The rate law for one of the mechanism of the pyrolysis of CH3CHO at 520ºC and 0.2 bar is 
The overall activation energy E, in terms of the rate law is:
(a) 
(b) 
(c) 
(d) 
Ans. c
Sol. 
Correct option is (c)
115. In the Michaelis-Menten mechanism for enzyme kinetics, the expression obtained is:
The values of k3 (kexp, mol L–1 s–1) and K(Michaelis constant, mol L–1), respectively are
(a) 1.4 × 1012, 104
(b) 1.4 × 108, 104
(c) 1.4 × 108, 10–4
(d) 1.4 × 1012, 10–4
Ans. c
Sol. 
Compare equation (1) and (2)
Putting the value of km in equation (3), we get
Correct option is (c)
116. The most used acid catalyst in oil industry and the relevant process are respectively
(a) Aluminophosphate and reforming
(b) Aluminosilicate and cracking
(c) Aluminosilicate and reforming
(d) Aluminophosphate and cracking
Ans. b
Sol. Correct answer is (b)
117. The wavelength and the spectral region for a single electron transfer across the band gap in a semiconductor (Ex – 1.98 × 10–19) are [h = 6.626 × 10–34 Js, c = 3 × 108 ms–1]
(a) 1000 nm, UV
(b) 1000 nm, IR
(c) 500 nm, visible
(d) 500 nm, FAR IR
Ans. b
Sol. 
= 10.039 × 10–7 m = 1003.9 × 10–9 m = 1003.9 nm
Correct option is (b)
118. The lattice parameter of an element stabilized in a fcc structure is 4.04 Å. The atomic radius of the element is:
(a) 2.86 Å
(b) 1.43 Å
(c) 4.29 Å
(d) 5.72 Å
Ans. b
Sol. Given a = 4.04 Å
Correct option is (b)
119. The number-average molar mass 
and weight-average molar mass 
of a polymer are obtained respectively by
(a) Osmometry and light scattering measurements
(b) Osmometry and viscosity measurements
(c) Light scattering and sedimentation measurements
(d) Viscosity and light scattering measurements
Ans. b
Sol. Correct answer is (b).
120. Two data sets involving the same variables X and Y are given below
X 4.1 4.2 4.3 4.4 4.5 4.6
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
If the slopes and intercepts of the regression lines for the two sets are denoted by (mA, mB) and (CA, CB) respectively, then
(a) mA > mB, CA > CB
(b) mA < mB, CA > CB
(c) mA > mB, CA < CB
(d) mA < mB, CA < CB
Ans. c
Sol. 
For different value of y and x, mA > mB and for value of intercept cA < cB.
Correct option is (c)
121. Compounds A and B exhibit two singlets, each in their 1H NMR spectra. The expected chemical shifts are at 
(a) 6.9 and 2.1 for A; 7.7 and 3.9 for B
(b) 7.7 and 3.9 for A; 6.9 and 2.1 for B
(c) 6.9 and 3.9 for A; 7.7 and 2.1 for B
(d) 7.7 and 2.1 for A; 6.9 and 3.9 for B
Ans. a
Sol. 
ring H's are shielded due to resonance.
Ring becomes electron rich, proton get shielded; appears at 6.9 ppm.
Ring becomes electron defficient, proton get deshielded, appears at 7.7 ppm.
Correct option is (a)
122. In the following reaction sequence, the major products A and B are
(a) 
(b) 
(c) 
(d) 
Ans. b
Sol. 
Correct option is (b)
123. The structure of the tricyclic compound formed in the following two step sequence is
(a) 
(b) 
(c) 
(d) 
Ans. d
Sol. Chemical reaction involved in the above transformation can be illustrated as
In benzyl position (C–H) bond is relatively weaker
Since we have two equation of NBS – Br attacks other –CH3 group also.
Correct option is (d)
124. The two step conversion of 7-dehydrocholesterol to vitamin D3 proceeds through
(a) Photochemical electrocyclic disrotatory ring opening; and thermal antarafacial [1, 7]-H shift.
(b) Photochemical electrocyclic conrotatory ring opening; and thermal antarafacial [1, 7]-H shift.
(c) Thermal electrocyclic conrotatory ring opening; and photochemical superafacial [1,7]-H shift
(d) Thermal electrocyclic disrotatory ring opening; and thermal suprafacial [1,7]-H shift
Ans. b
Sol. Chemical reaction involved in the above transformation can be illustrated as
Since there is a trans ring junction, so it is a photochemical conrotatory process.
Correct option is (b)
125. The intermediate A and the major product B in the following reaction are
(a) 
(b) 
(c) 
(d) 
Ans. d
Sol. Chemical reaction involved in the above transformation can be illustrated as
Correct option is (d)
126. For the following two reactions A and B, the correct statement is:
(a) 
(b) 
(c) 
(d) 
Ans. b
Sol. Chemical reaction involved in the above transformation can be illustrated as
Because Br and H are antiperiplanar. So, E2 elimination will take place.
Correct option is (b)
127. The major compound B formed in the reaction sequence given below exhibited a carbonyl absorption band at 1770 cm–1 in the IR spectrum. The structure A and B are
(a) 
(b) 
(c) 
(d) 
Ans. b
Sol. 
9BBN being a larger group will attack from least hindered side i.e. attack from rear side.
Correct option is (b)
128. Consider the following reaction sequence starting with monoterpene 
Identify the correct statement
(a) A has a disubstituted double bond; B and C are dicarboxylic acids.
(b) A has a trisubstituted double bond; B is a methyl ketone; and C is a dicarboxylic acid.
(c) A has a disubstituted double bond; B is a methyl ketone; and C is a dicarboxylic acid
(d) A has an exocyclic double bond; B and C are monocarboxylic acids.
Ans. b
Sol. Chemical reaction involved in the above transformation can be illustrated as
Correct answer is (b).
129. The major product formed when (3R, 4S)-3, 4-dimethylhexa-1, 5-diene is heated at 240º 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
Ans. c
Sol. (3R, 4S)-3, 4-dimethyl hexa-1, 5-diene
Now we have to find stereochemistry of methyl groups it is E or Z. For this we have to draw cyclic chair transition state.
Correct option is (c)
130. Structure of the starting material A in the following photochemical Norrish reaction, is
(a) 
(b) 
(c) 
(d) 
Ans. c
Sol. Chemical reaction involved in the above transformation can be illustrated as
Correct option is (c)
131. Considering the following reaction, among A-C the correct statements are
(A) The carbonyl group has enantiotopic faces;
(B) The hydride attack is re-facial
(C) It is a diastereoselective reduction
(a) A and B only
(b) A and C only
(c) B and C only
(d) A, B and C
Ans. c
Sol. Chemical reaction involved in the above transformation can be illustrated as
(i) Since there is already chiral centre present in reactant. So, it has diastereotopic faces option(a) is wrong.
Since H– is attacking from back side of Si-face. Or we can say attacking from Re-face. Option (b) is right.
(iii) Since reactant has diastereotopic faces. So definitely it is a diastereoselective reduction. Option (c) is also right
Correct option is (c)
132. The major product formed in the following reaction sequence is
(a) 
(b) 
(c) 
(d) 
Ans. d
Sol. Chemical reaction involved in the above transformation can be illustrated as
BF3 is a lewis acid. So it will accept lone pair of electron from oxygen atom.
Bond breaking can be of two ways leading to two different cation but cation at benzylic position is more stable.
Correct option is (d)
133. The major product formed in the following reaction sequence is
(a) 
(b) 
(c) 
(d) 
Ans. a
Sol. Chemical reaction involved in the above transformation can be illustrated as
DIBAL 
is selective reducing agent for lactones.
Correct option is (a)
134. Match the following
(a) A-iii, B-ii, C-iv, D-i
(b) A-iii, B-iv, C-i, D-ii
(c) A-i, B-ii, C-iv, D-ii
(d) A-iii, B-i, C-iii, D-iv
Ans. a
Sol. (i) Acetic acid = 175 ppm
(ii) Acetonitrile = 115 ppm
(iii) Acetone = 205 ppm
(iv) CCl4 = 95 ppm
Correct option is (a)
135. The major products A and B in the following reaction sequence are
(a) 
(b) 
(c) 
(d) 
Ans. a
Sol. Chemical reaction involved in the above transformation can be illustrated as
Correct option is (a)
136. The major product formed in the following reaction is
(a) 
(b) 
(c) 
(d) 
Ans. b
Sol. Chemical reaction involved in the above transformation can be illustrated as
This is an example of favorskii type rearrangement.
Correct option is (b)
137. The reagents A and B in the following reactions are
(a) A = CH2I2, Zn – Cu; B = Me3S+I–, NaH
(b) A = CH2I2, Zn – Cu; B = Me3S+ (O) I–, NaH
(c) A = Me3S+I–, NaH; B = Me3S+ (O) I–, NaH
(d) A = Me3S+ (O) I–, NaH, B = CH2I2, Zn – Cu
Ans. b
Sol. Chemical reaction involved in the above transformation can be illustrated as
Stabilized ylide will give cyclopropane and Simmon's Smith reagent will attack at allylic double bond and form cyclopropane.
Correct option is (b)
138. The major products A and B formed in the following reaction sequence are
(a) 
(b) 
(c) 
(d) 
Ans. c
Sol. Chemical reaction involved in the above transformation can be illustrated as
Mechanism:
Triple bond is reduced by LiAlH4 when oxygen containing group is attached to triple bond. Since there is co-ordination between oxygen and metal atom. After co-ordination with oxygen atom hydride ion is transfered leading to formation of trans double bond.
Sharpless asymmetric epoxidation occur at the allylic alcohol not on the simple double bond.
Correct option is (c)
139. The major products A and B formed in the following reaction sequence are
(a) 
(b) 
(c) 
(d) 
Ans. b
Sol. Chemical reaction involved in the above transformation can be illustrated as
Correct option is (b)
140. The correct reagent combination/reaction sequence are effecting the following conversion is
(a) (1) Me3SiCH2OMe, BuLi; (2) H3O+; (3) NaBH4, MeOH
(b) (1) Ph3P+CH2MeCl, BuLi; (2) H3O+; (3) NaBH4, MeOH
(c) (1) NH2NHTs; (2) NaOEt; (3) ClCOOEt
(d) (1) NH2NHTs; (2) 2 eq. BuLi; (3) HCHO
Ans. d
Sol. Chemical reaction involved in the above transformation can be illustrated as
Correct option is (d)
141. The major product formed in the following reaction is:
(a) 
(b) 
(c) 
(d) 
Ans. d
Sol. Chemical reaction involved in the above transformation can be illustrated as
1, 4-addition of Me ion.
Correct option is (d)
142. The correct sequence of reagents for effecting the following conversion is:
(a) (1) (CH2OH)2, PTSA, 
(3) H3O+; (4) KOH
(b) (1) (CH2OH)2, PTSA, 
(2) Ph3P–CH2; (3) H3O+; (4) KOH
(c) 
(Tebbe's reagent); (2) H3O+; (3) KOH
(d) Ph3P=CH2; (2) H3O+; (3) KOH
Ans. a
Sol. 
Since we have two functional groups, so first we have to protect one carbonyl group so last two options can not be the right one.
Ketonic group is more reactive than ester group. So, it will be protected first.
In this hydrolysis process two reactions will occur. One is deprotection of carbonyl group and an other reaction formation of ketone.
Correct option is (a)
143. The major products A and B formed in the following reaction sequence are
(a) 
(b) 
(c) 
(d) 
Ans. d
Sol. Chemical reaction involved in the above transformation can be illustrated as
Note: (3) position is blocked. So, electrophilic substitution will be occur at (2) position.
144. The reagent A required, and the major product B formed in the following reaction sequence are
(a) 
(b) 
(c) 
(d) 
Ans. c
Sol. 
Correct option is (c)
145. Among the choices, the correct statements for A formed in the following reaction.
(a) A is a single enantiomer
(b) A is a racemic mixture
(c) A is a mixture of two diastereomers
(d) A is a mixture of two epimers.
Ans. b
Sol. Chemical reaction involved in the above transformation can be illustrated as
Correct option is (b)