CSIR NET CHEMISTRY (June-2017)
Previous Year Question Paper with Solution.
21. Which one of the following pairs has two magic numbers for closed nuclear shells?
(a) 8, 10
(b) 10, 20
(c) 50, 82
(d) 82, 130
Ans. c
Sol. 2, 8, 20, 28, 50, 82 and 126 are magic number
Correct option is (c)
22. Identify the correct statement(s) for phosphorimetric measurement from the following:
A. It is done after a time delay when fluorescence, if present becomes negligible
B. Immobilization of analytic increases phosphorescence
C. Phosphorescence decreases in the presence of heavy atoms
(a) A only
(b) A and B
(c) A and C
(d) B and C
Ans. b
Sol. Phosphorescence has been observed form a wide variety of compounds and is differentiated from fluorescence by long lived emission of light after extinction of the excitation source i.e. it done after fluorescence, if parsists.
Phosphorescence increases when heavy atoms like iodine silver and lead etc. are added.
Phosphorescence increases, when analyte used become immobile.
Hence both (A) and (B) are correct.
Correct option is (b)
23. Choose the isoelectronic pair among the following:
(a) A and B
(b) B and C
(c) C and D
(d) A and D
Ans. b
Sol. The species having same number of electrons are called isoelectronic.
Correct option is (b)
24. An organometallic fragment that is isolobal to 
is
(a) 
(b) 
(c) 
(d) 
Ans. c
Sol. Two fragments are isolobal if the number, symmetry properties, approximate energy and shape of their frontier molecular orbitals and the number of electrons in them are similar not identical but similar.
Correct option is (c)
25. The calculated and observed magnetic moments (in B.M.) of aqua complex of a lanthanide ion are 0 and ~3.5, respectively. The lanthanide ion is
(a) Pm3+
(b) Pr3+
(c) Eu3+
(d) Sm3+
Ans. c
Sol. 
Correct option is (c)
26. The compound that gives a basic solution in HF is:
(a) AsF5
(b) PF5
(c) BF3
(d) BrF3
Ans. d
Sol. Species that produce [HF2]– will behave as base in HF
BrF3 acts as base as it donate F– in HF.
Hence, correct option is (d)
27. Based on VSEPR theory, the predicted shapes of 
and BrF5 respectively, are
(a) pentagonal planar and square pyramidal
(b) square pyramidal and trigonal bipyramidal
(c) trigonal bipyramidal and square pyramidal
(d) square pyramidal and pentagonal planar
Ans. a
Sol. 
Correct option is (a)
28. Both potassium and sulfuric acid form intercalation compounds with graphite. The graphite layers are
(a) reduced in both the cases
(b) oxidized in both the cases
(c) oxidized in the case of potassium and reduced in the case of sulphuric acid
(d) reduced in the case of potassium and oxidized in the case of sulphuric acid
Ans. d
Sol. 
H2SO4 + graphite = [graphite]+ HSO4–
Correct option is (d)
29. The resonance Raman stretching frequencies (in cm–1) of the bound O2 species in oxy – hemerthyrin and oxy – hemoglobin, respectively, are
(a) ~850 and 1100
(b) ~750 and 850
(c) ~850 and 850
(d) ~1100 and 850
Ans. a
Sol. In oxyhemerythrin vO-O = 850 cm–1
In oxyhemoglobin vO-O = 1100 cm–1
Correct option is (a)
30. CdS, HgS and BiL3, are coloured due to
(a) L 
M charge transfer transitions
(b) d 
d electronic transitions
(c) M 
L charge transfer transitions
(d) combination of L 
M charge transfer and d 
d electronic transitions
Ans. a
Sol. In CdS, HgS and BiI3 colour is due to LMCT as Cd2+ and Hg2+ has d10 configuration. Hence, no d-d transition. Also, no MLCT as ligand is not 
Hence, correct option is (a).
31. The relative rates of water exchange for the hydrated complexes of (1) Ni2+, (2) V2+ and (3) Cr3+ ions follows the trend
(a) (1) > (2) > (3)
(b) (1) < (2) < (3)
(c) (1) > (2) > (3)
(d) (1) < (2) < (3)
Ans. a
Sol. Change in CFSE on going for Oh to five coordinate intermediate for ions is as follows.
V2+(d3) = –2.00
Cr3+(d3) = –2.00
Ni2+(d8) = –2.00
Correct option is (a)
32. Consider the following sulfur donor atom bearing bidentate ligand where X and name of ligands are given in following columns:
Correct match of entries given in two columns is
(a) A – II, B – III, C – I, D – IV
(b) A – III, B – II, C – IV, D – I
(c) A – I, B – II, C – III, D – IV
(d) A – IV, B – I, C – II, D – III
Ans. a
Sol. NR2 
Dithiocarbamate
OR 
Xanthate
O– 
Dithiocarbonate
SR 
Thioxanthate
Correct option is (a)
33. In vitro reaction of an excess of O2 with free heme B in aqueous medium the end product is
(a) hematin
(b) [ –Fe (III) – protoporphyrin – IX]
(c) heme B(O2)
(d) oxoferrylprotoporphyrin – IX cation radical
Ans. a
Sol. 
Correct option is (a)
34. 13C NMR spectrum of DMSO – d6 gives a signal at 
as a
(a) singlet
(b) triplet
(c) quintet
(d) septet
Ans. d
Sol. 
13C NMR n = 3; I = 1 (for D)
Multiplicity = 2nI + 1 = 2 × 3 × 1 + 1 = 6 + 1 = 7 (septet)
Correct option is (d)
35. Following reaction is an example of
(a) Ramberg – Backlund reaction
(b) [2, 3] – sigmatropic shift
(c) [3, 3] – sigmatropic shift
(d) Pummerer rearrangement
Ans. b
Sol. 
Correct option is (b)
36. Among the following, the synthetic equivalent of acetyl anion is
(a) 
(b) CH3CN
(c) 
(d) CH3CH2NO2
Ans. d
Sol. 
Hence, nitroethane anion is synthetic equivalent of acetyl anion.
Correct option is (d)
37. Following reaction is an example of
(a) [3 + 2] cycloaddition
(b) [4 + 2] cycloaddition
(c) [6 + 2] cycloaddition
(d) [8 + 2] cycloaddition
Ans. d
Sol. 
Correct option is (d)
38. The major product of the following reaction is
(a) 
(b) 
(c) 
(d) 
Ans. d
Sol. 
Correct option is (d)
39. The most stable conformation of 2 – fluoroethanol is
(a) 
(b) 
(c) 
(d) 
Ans. b
Sol. 
Most stable due to intramolecular hydrogen bonding.
Correct option is (b)
40. The IUPAC name of the following compound is
(a) 9 – borabicyclo [3.3.1] nonane
(b) l – borabicyclo [3. 3. 1] nonane
(c) 9 – borabicyclo [3. 3. 0] octane
(d) l – borabicyclo [3. 3. 0] octane
Ans. a
Sol. 
Correct option is (a)
41. The correct match of natural products in Column – I with their biosynthetic precursors in Column – II is
(a) A – IV, B–I
(b) A – IV, B – II
(c) A – III, B – I
(d) A – III, B – II
Ans. a
Sol. 
It is a pyridine and piperidine alkaloid found in the tree tobacco plant. Its principal industrial use is an insecticide.
Lysine serves as the precursor of piperidine ring of anabasine.
Gernyl pyrophosphate is the biosynthetic precursor of (+)-menthol.
Correct option is (a)
42. The correct order to pKa values for the following species is
(a) 
(b) 
(c) 
(d) 
Ans. b
Sol. Basicity order for their conjugate base is
iPr2 NH > PhNH2 > Ph2NH
Hence, pKa value for their conjugate acid
Hence, correct option is (b)
43. Among the following, the natural product that is a steroid and contains an 
unsaturated ketone is
(a) estrone
(b) prostaglandin
(c) cortisone
(d) morphine
Ans. c
Sol. 
Correct option is (c)
44. The major product formed in the following reaction is
(a) 
(b) 
(c) 
(d) 
Ans. b
Sol. 
Correct option is (b)
45. The major product formed in the sodium ethoxide mediated reaction between benzophenone and ethyl chloroacetate is
(a) 
(b) 
(c) 
(d) 
Ans. c
Sol. 
Correct option is (c)
46. The major product formed in the following reaction is
(a) 
(b) 
(c) 
(d) 
Ans. b
Sol. 
Correct option is (b)
47. Consider a particle in its ground state confined to a one – dimensional box in the interval (0, 8). The probability of finding it between 
is close to 
is sufficiently small so that the wavefunction can be taken as a constant in this interval).
(a) 
(b) 
(c) 
(d) 
Ans. a
Sol. The probability of finding the particle lies between x to x + dx is given by
Correct option is (a)
48. Which of the function below is a common eigenfunction of 
operators?
(a) cos x
(b) kx
(c) eis
(d) 3 – x2
Ans. c
Sol. 
Correct option is (c)
49. A one – component system with the associated phase diagram (see the figure) is not possible because
(a) OB has a negative slops
(b) OC has a positive slope
(c) Both OB and OC are linear
(d) OB, OC and OD cannot all coexist, given OA
Ans. d
Sol. In one component system 4 phases carnot simultaneously exist at single point.
Correct option is (d)
50. A phase transition process is always
(a) isothermal – isoentropic
(b) isochoric – isothermal
(c) isobaric – isochoric
(d) isothermal – isobaric
Ans. d
Sol. (i) Phase transition cannot be isoentropic
So, entropy is going to change during phase transition
(ii) Phase transition cannot be isochoric as
Volume is going to change
(iii) Volume is going to change
As phase transition takes place at constant temperature and pressure
Correct option is (d)
51. The correct statement for any cyclic thermodynamic process is
(a) 
(b) 
(c) 
(d) 
Ans. c
Sol. For any cyclic process cyclic integral of state function is zero.
Correct option is (c)
52. Metallic silver crystallizes in face – centred – cubic lattice structure with a unit cell of length 40 nm. The first order diffraction angle of X – ray beam from (2, 1, 0) plane of silver is 300. The wavelength of X – ray used is close to
(a) 11 nm
(b) 18 nm
(c) 24 nm
(d) 32 nm
Ans. b
Sol. 
Correct option is (b)
53. If the pre – exponential factor in Arrhenius equation is 1.6 × 1012 s–1, the value of the rate constant at extremely high temperature will be close to
(a) 1.6 × 1012 s–1
(b) 4.2 × 1012 s–1
(c) 2.4 × 109 s–1
(d) 1.2 × 106 s–1
Ans. a
Sol. 
Correct option is (a)
54. In kinetic study of a chemical reaction, slopes are drawn at different times in the plot of concentration of reactants versus time. The magnitude of slopes with increase of time
(a) remains unchanged
(b) increases
(c) decreases
(d) increases and decreases periodically
Ans. c
Sol. As the time increases slope will decreases.
So, correct option is (c)
55. The electrochemical cell potential (E), after the reactants and products reach equilibrium, is (E0 is the stantdard cell potential and n is the number of electrons involved)
(a) E = E0 + nF/RT
(b) E = E0 – RT/nF
(c) E = E0
(d) E = 0
Ans. d
Sol. At equilibrium E is zero, because 
is zero at equilibrium.
So, E = 0
Correct option is (d)
56. For the electronic configuration ls22s22p4, two of the possible term symbols are 1S and 3p. The remaining term is
(a) 1D
(b) 1F
(c) 3D
(d) 3F
Ans. a
Sol. 1s2, 2s2, 2p4
Number of microstate for 
Thus, remaining microstate are 5 which comes from 1D = 1 × 5 = 5
Correct option is (a)
57. The v = 0 to 1 vibration – rotation spectrum of a diatomic molecule exhibits transitions for R (0), R (1), P (1) and P (2) lines at 2241, 2254, 2216 and 2203 cm–1, respectively. From this data, we can conclude that the molecule
(a) has rigid rotation and harmonic vibration
(b) has anharmonic vibration
(c) has rotational – vibrational interaction
(d) is affected by nuclear spin – statistics
Ans. c
Sol. Since, due to vibrational rotational interaction P and R lines are obtained and molecule behave as a diatomic vibrating rotor.
Correct option is (c)
58. Consider aqueous solutions of two compounds A and B of identical concentrations. The surface tension of the solution of A is smaller than that of pure water while for B it is greater than that of pure water under identical conditions. From this one infers that
(a) surface concentration of A is smaller than its bulk concentration
(b) surface concentration of B is larger than its bulk concentration
(c) surface concentration of A is larger than that of B
(d) surface concentration of A is smaller than that of B
Ans. c
Sol. Greater the surface tension, the greater the surface concentration
Correct option is (c)
59. For a monodisperse polymer, the number – average molar mass 
and weight – average molar mass 
are related according to
(a) 
(b) 
(c) 
(d) 
Ans. b
Sol. 
Correct option is (b)
60. An intense purple colour (Plasmon band) is exhibited by a colloid consisting of spherical
(a) silver particles of 10 mm diameter
(b) silicon particles of 5 mm diameter
(c) gold particles of 5 nm diameter
(d) iron particles of 3 mm diameter
Ans. c
Sol. Intermediate colours been understood and correlated to the formation of intermediate nanostructures before the formation of the final gold nanoparticles. Specifically, TEM images have shown that after few seconds of citrate addition, gold nanowires 5 to 8 nm are formed, which are responsible for the dark purple colour. Beyond a certain threshold, the nanowires disintegrate into nanoparticles, and the solution turns ruby-blue.
Correct option is (c)
61. Choose the correct statement for magnitude of threshold energy of an endoergic nuclear reaction between stationary nucleus and a moving projectile.
(a) It is greater than '|Q|' of nuclear reaction.
(b) It has to be more than kinetic energy of a projectile.
(c) It is less than '|Q|' of nuclear reaction
(d) It has to be equal to kinetic energy of a projectile.
Ans. a
Sol. We have the relation between threshold energy and 'Q' of a reaction, as
Here, Ma = mass of moving projectile or bombarding particle
MA = mass of stationary nucleus
Also, for endoergic reaction, Q > 0
Therefore, magnitude of threshold energy
Correct option is (a)
62. Identify correct statements from the following.
A. Area of differential thermal analysis peak is proportional to amount of sample.
B. Area of differential thermogravimetric analysis curve is proportinal to mass loss.
C. Phase transition cannot be studied with differential scanning calorimetry.
D. Simulatneous determination of two metal ions is possible with thermogravimetric analysis.
Answer is
(a) A, B and C
(b) A, B and C
(c) B, C and D
(d) A, C and D
Ans. a
Sol. Correct statement are
Area of different thermal analysis peak is proportional to amount of sample
In the thermogravimetric analysis area of curve is proportional to mass loss
In thermogravimetric analysis two metal ion simultaneously determine.
Correct option is (a)
63. Consider following statements for fission of 235U with thermal neutrons.
A. The % of nuclei undergoing unsymmetrical fission is maximum.
B. In each fission, one thermal neutron is produced.
C. Magnitude of energy released per fission is of the order 200 MeV
Correct statement(s) is/are
(a) A and B
(b) A and C
(c) B and C
(d) C only
Ans. b
Sol. In fission of U235 with thermal neutrons
(1) On the average, each fission of a U235 nucleus produces about 2.5 free neutrons
(2) Each fission produce 200 MeV energy
(3) Fission always occure in asymmetric fashion, it means number of nuclei undergoing unsymmetrical fission is maximum.
Hence, correct statement is (1) and (3)
Correct option is (b)
64. Addition of two electrons to the bismuth cluster 
results in a change of structure type from
(a) closo to nido
(b) nido to arachno
(c) closo to arachno
(d) arachno to hypho
Ans. a
Sol. 
Hence, (4n + 2), closo
On addition of 2 electron it become (4n + 4), nido
Correct option is (a)
65. Reaction of 
with 
gives A along with NaCl. Photolysis of compound A results is compound B together with elimination of CO, the correct structural formulations of compounds A and B are respectively.
(a) 
(b) 
(c) 
(d) 
Ans. a
Sol. 
Mechanism:
Correct option is (a)
66. A copper (II) complex having distorted octahedral geometry shows an absorption band at 625 nm. Given spin – orbit coupling of the complex as –625 cm–1, the 
is
(a) 1.73
(b) 1.81
(c) 1.63
(d) 1.93
Ans. b
Sol. 
Correct option is (b)
67. Math items in column A with items in column B:
The correct answer is
(a) I – (A); II – (B); III – (C); IV – (D)
(b) I – (B); II – (D); III – (C); IV – (A)
(c) I – (C); II – (D); III – (B); IV – (A)
(d) I – (B); II – (D); III – (A); IV – (C)
Ans. d
Sol. 
Correct option is (d)
68. Mössbauer spectrum of complex [Fe (1, 10 – phenanthroline)2 (NCS)2] shows two lines at 300 K, four lines at 186 K, and again two lines at 77 K. This can be attributed to
A. change in the coordination mode of NCS
B. change in the spin – state of iron
C. cis – trans isomerisation
D. change in metal – ligand bond distances
The correct statements are
(a) A and B
(b) B and C
(c) A and C
(d) B and D
Ans. a*
Sol. [Fe(1, 10-phenanthroline)2(NCS)2]
Since NCS is ambidentate ligand therefore change in co-ordinate mode
Change in the spin-state of iron at high temperature
Correct option is (a*)
69. (R3Ge)2 on photolysis gives a radical which shows ESR spectrum. The ESR signals carrying the signature of 73Ge(I = 9/2) are in terms of
(a) Nine lines
(b) Ten lines
(c) Two lines
(d) One line
Ans. b
Sol. 
Correct option is (b)
70. mass fragment of [IrCl]+ in mass spectrometry shows three mass peaks at m/z = 226, 228 and 230. Given that natural abundances of 191Ir, 193Ir, 35Cl and 37Cl are 37%, 63%, 76% and 24% respectively, the intensities of the mass peaks are in the order
(a) 49.5 : 100 : 26.6
(b) 100 : 49.5 : 26.6
(c) 26.6 : 100 : 49.5
(d) 26.6 : 49.5 : 100
Ans. a
Sol. 191Ir : 193 Ir + 35Cl : 37Cl
37% : 63% + 76% : 24%
Natural abundance x y a b
(3.7x + 6.3y)1(7.6a + 2.4b)1
= 3.7 × 7.6xa + 6.3 × 7.6ya + 3.7 × 2.4xb + 6.3 × 2.4yb
= 28.12M + 56.76 (M + 2) + 15.12 (M + 4)
= 28.12 × 1.76M + 56.76 × 1.76 (M + 2) + 15.12 × 1.76 (M + 4)
= 49.49M + 100(M + 2) + 26.6(M + 4)
= 49.5M + 100(M + 2) + 26.6 (M + 4)
M : M + 2 : M + 4
49.5 : 100 : 26.6
Correct option is (a)
71. The 31P {1H} NMR spectrum of 2, 2, 6, 6 – N4P4Cl4 (NMe2)4 is expected to show
(a) two triplets
(b) two doubles
(c) one doublet and one triplet
(d) one quartet and one doublet
Ans. a
Sol. 
Correct option is (a)
72. The number of bonding molecular orbitals and the number of available skeletal electrons in [B6H6]2–, respectively, are
(a) 7 and 14
(b) 6 and 12
(c) 18 and 12
(d) 11 and 14
Ans. a
Sol. 
has 6 + 1 = 7 binding molecular orbital
[B6H6]2– B – H = 2 × 6 = 12 e–
for 2 negative charge, hence 12e– + 2e– = 14e–
Correct option is (a)
73. The compound N2F2 has two isomeris. Choose the correct option from the following:
(a) both isomers passes 
plane
(b) both isomers posses 
plane
(c) one isomer has 
plane while the other has 
plane
(d) none of them have a 
plane
Ans. c
Sol. N2F2 has two isomers.
Correct option is (c)
74. Consider the following statements for metallothioneins:
A. the contain about 30% cysteine residues
B. they prefer to bind soft metal ions such as Cd(II), Hg(II) and Zn (II)
C. they are involved in electron transfer reactions
D. they are low molecular weight proteins
Correct statements are
(a) A, B and C
(b) A, B and C
(c) A, C and D
(d) B and C
Ans. b
Sol. Metallothioneins are cystein rich, how molecular weight protein. Due to soft sulfur centre they prefer to bind.
Correct option is (b)
75. Consider the following statements for deoxy – hemerythrin and deoxy – hemocyanin:
A. they are involved in O2 transport in biological systems
B. they contain two metal ions in their active site
C. active site metal centres are bridged by amino acid residues
D. they prefer to bind only one O2 per active site
The correct statements are
(a) A, B and D
(b) A, C and D
(c) B, C and D
(d) A and C
Ans. a
Sol. Both deoxyhemerythrin and deoxy hemocyanin are O2 transport protein in biological system. Both contain two metal ion at active site and they bind only O2 per active site.
Correct option is (a)
76. Consider the following statements for octahedral complexes (a) [CrF6]3–, (b) [Cr(ox)3]3– and (c) [Cr(en)3]3+:
A. their 
transitions are at 14900, 17500 and 21800 cm–1, respectively
B. their spin – only magnetic moments are same
C. two of them have optical isomers
D. all of them show Jahn – Teller distortion
The correct statements are
(a) A, B and C
(b) A, C and D
(c) B, C and D
(d) B and D
Ans. a
Sol. [Cr(ox)3]3– and [Cr(en)3]3+ will show optical isomer. As t2g level is electronically non-degenerate. Hence, there will be no Jahn-Teller distortion.
Correct option is (a)
77. Addition of NaBH4 to 
will give
(a) 
(b) 
(c) 
(d) 
Ans. d
Sol. 
Correct option is (d)
78. The 
changes with temperature with the involvement of two electronic states. The states are
(a) low spin 2T2g and high – spin 6A1g.
(b) low spin 1A1g and high – spin 3T2g.
(c) low spin 2Eg and high – spin 6A1g.
(d) low spin 2T2g and high – spin 4T1g.
Ans. a
Sol. In [Fe(S2CNEt2)3] oxidation state of Fe = +3
Correct option is (a)
79. Match the items in the three columns.
The correct answer is
(a) A – II – X; B – I – Z; C – III – Y
(b) A – I – X; B – II – Y; C – III – Z
(c) A – III – Y; B – I – Z; C – II – X
(d) A – I – X; B – II – Z; C – III – Y
Ans. a
Sol. 
Thus, order for energy absorption is 675(A) < 615(B) < 565 (C)
Complementary colour of blue 
Energy order for absorbed light is, yellow > orange > red
Thus, A absorbs red hence it is green
B absorbs orange hence it is blue
C absorbs yellow hence it is violet
Thus, correct option is (a)
80. Identify the product in the reaction between 
and CH3I going at room temperature via Sn2 mechanism
(a) 
(b) 
(c) 
(d) 
Ans. a
Sol. 
Correct option is (a)
81. The major products A and B formed in the following reactions sequence are
(a) 
(b) 
(c) 
(d) 
Ans. a
Sol. 
Correct option is (a)
82. The intermediate A and product B formed in the following reaction sequence are
(a) 
(b) 
(c) 
(d) 
Ans. d
Sol. 
Correct option is (d)
83. The major products A and B formed in the following reaction sequence are
(a) 
(b) 
(c) 
(d) 
Ans. d
Sol. 
Correct option is (d)
84. The major products A and B formed in the following reaction sequence are
(a) 
(b) 
(c) 
(d) 
Ans. a
Sol. 
Correct option is (a)
85. The major product formed in the following reaction is
(a) 
(b) 
(c) 
(d) 
Ans. c
Sol. 
Correct option is (c)
86. The correct combination of reagents to effect the following reaction is
(a) A. POCI3, pyridine; B. AgOAc; C. LialH4
(b) A. NaBH4 ; B. Ph3P, DEAD, PhCO2H
(c) A. Ph3P, DEAD, PhCO2H ; B. LiAlH4
(d) A. PCC ; B. L – selectride
Ans. c
Sol. 
Correct option is (c)
87. The major products A and B formed in the following reactions sequence are
(a) 
(b) 
(c) 
(d) 
Ans. a
Sol. 
Correct option is (a)
88. The correct combination of reagents A and B to effect following transformations are
(a) A = cat. OsO4, NMO; B = i. I2, PhCO2Ag, ii. aq. NaOH
(b) A = alkaline KMnO4; B = i, I2, PhCO2Ag, H2O, ii. aq. NaOH
(c) A = I2, PhCO2Ag, ii. aq. NaOH; B = cat. OsO4, TMEDA, NMO
(d) A = i, m – CPBA, ii. aq. NaOH; B = alkaline KMnO4
Ans. b
Sol. 
Correct option is (b)
89. The major products A and B formed in the following reaction sequence are
(a) 
(b) 
(c) 
(d) 
Ans. c
Sol. 
Correct option is (c)
90. The major products A and B formed in the following reaction sequence are
(a) 
(b) 
(c) 
(d) 
Ans. b
Sol. 
Correct option is (b)
91. The specific rotation 
for (S) – (+) – 2 butanol is 100 Ml/g dm. The observed optical rotation 
of a sample composed of a mixture of (R) – and (S) – 2 – butanol is –0.450. If the cell path length is 0.6 dm and the concentration of 2 – butanol in the sample is 0.15 g/mL, the percentages of (R) and (S) enantiomers in the sample are
(a) (R) = 25%, (S) = 75%
(b) (R) = 40%, (S) = 60%
(c) (R) = 60%, (S) = 40%
(d) (R) = 75%, (S) = 25%
Ans. d
Sol. 
Correct option is (d)
92. The major product formed in the following reaction is
(a) 
(b) 
(c) 
(d) 
Ans. d
Sol. 
Correct option is (d)
93. Following reaction involves
(a) Claisen followed by Mannich reaction
(b) aza–Cope followed by Mannich reaction
(c) Claisen followed by aza– aldol reaction
(d) aza–Cope followed by aza–aldol reaction
Ans. b
Sol. 
Correct option is (b)
94. The intermediate A and the major product B formed in the following reaction is
(a) 
(b) 
(c) 
(d) 
Ans. a
Sol. 
Active form is Pd(PPh3)2 or PdL2 (where L = PPh3)
This reaction is an example of intramolecule Heck reaction
(1) Oxidative Addition:
(2) Olefin insertion (Syn addition):
Correct option is (a)
95. The major product formed in the following reaction is
(a) 
(b) 
(c) 
(d) 
Ans. b
Sol. 
Correct option is (b)
96. The major product formed in the following reaction is
(a) 
(b) 
(c) 
(d) 
Ans. c
Sol. 
Correct option is (c)
97. The most stable conformation for the following compound is
(a) 
(b) 
(c) 
(d) 
Ans. c
Sol. 
Correct option is (c)
98. The correct structure of the compound based on the following characteristic spectral data is IR : 1736 cm–1
1H NMR : 
3.59 (s, 3H), 3.32 (t, 2H), 2.25 (t, 2H), 1.85 – 1.75 (m, 2H), 1.73 – 1.62 (m, 2H)
13C NMR : 
174.0, 51.0, 32.9, 32.9, 32.8, 31.0, 23.0
(a) 
(b) 
(c) 
(d) 
Ans. c
Sol. 
Correct option is (c)
99. The major product formed in the reaction of D-glucose with ZnCl2 in MeOH is a methyl glucopyranoside (A or B). The structure of this product and the molecular orbital interaction present between ring – oxygen and the anomeric C – O bond responsible for its stability, respectively, are
(a) 
(b) 
(c) 
(d) 
Ans. c
Sol. Anomeric effect present in compound (B)
Correct option is (c)
100. Among the following reaction statement for nucleic acids is
(a) Uracil is present in DNA
(b) Uracil is present in RNA
(c) Phosphorylation in RNA is at 2' and 5' positions
(d) Normally three hydrogen bonds stabilize A – T base pair
Ans. b
Sol. In case of RNA-Uracil is present. Inspite of thymine-DNA
Correct option is (b)
101. The figure below depicts an adsorption isotherm of O2 on charcoal at 90 K.
At a pressure 25 torr, only 10% of charcoal sites are occupied by O2. Therefore, the ratio of adsorption to desorption rate constants (in torr–1) is close to
(a) 0.003
(b) 0.004
(c) 0.006
(d) 0.015
Ans. b
Sol. 
Correct option is (b)
102. Polonium is the only metal known to exist in a simple cubic lattice form. The density of polonium at 0°C is measured to be 10.00 g/cm3. The atomic radius of polonium would then be (assume the mass of a polonium atom = 2.7 × 10–22 g)
(a) 1.1 Å
(b) 1.9 Å
(c) 1.5 Å
(d) 2.3 Å
Ans. c
Sol. 
Correct option is (c)
103. The specific conductance of a solution is 0.176 
cm–1. If the cell constant is 0.255 cm–1, the conductance (
) of the solution is
(a) 1.449
(b) 0.690
(c) 0.045
(d) 0.431
Ans. b
Sol. 
Correct option is (b)
104. Photochemical decomposition of HI takes place with the following mechanism
Considering hydrogen (H) and iodine (I) atoms as intermediates, the rate of removal of HI is
(a) Ia/2
(b) Ia
(c) 2Ia
(d) 
Ans. c
Sol. Rate of removal of HI = k1[H][HI] + Ia ... (1)
Applying SSA on
Correct option is (c)
105. In an enzyme – catalysed reaction
mol dm–3, the magnitude of maximum velocity and turn over number using Michaelis – Menten kinetics are
(a) 
(b) 
(c) 
(d) 
Ans. a
Sol. 
And T.O.N. = k2 = 3.42 × 104 s–1
Correct option is (a)
106. Arrhenius equations for two chemical reactions are: 
then at a given temperature T,
(a) 
(b) 
(c) 
(d) 
Ans. a
Sol. 
Correct option is (a)
107. The fugacity of a real gas is less than the pressure (P) of an ideal gas at the same temperature (T) only when (Tb is the Boyle temperature of the real gas)
(a) high P, T < Tb
(b) low, P, T < Tb
(c) high P, T > Tb
(d) low P, T > Tb
Ans. b
Sol. Fugacity of gas is less than P when attractive forces are dominant. it happens at low P and when T < Tb.
Correct option is (b)
108. For the reaction 
the equilibrium constant Kp depends on the degree of dissociation 
and total pressure P as
(a) 
(b) 
(c) 
(d) 
Ans. b
Sol. 
Correct option is (b)
109. The minimum work required by an engine to transfer 5 J of heat from a reservoir at 100 K to one at 300 K is
(a) 5 J
(b) 10 J
(c) 15 J
(d) 20 J
Ans. b
Sol. The given engine is working as refrigerator as it is transfering heat from sink (lower temperature) to source (higher temperature). So, efficiency of refrigerator/coefficient of performance.
Correct option is (b)
110. The correct relation involving symmetry operations
(a) 
(b) 
(c) 
(d) 
Ans. d
Sol. 
Correct option is (d)
111. A polydisperse polymer sample has ten molecules of molar mass 20,000 g mol–1 and fifteen molecules of molar mass 10,000 g mol–1. The number – average molar mass 
of the sample is
(a) 13,000
(b) 14,000
(c) 15,000
(d) 16,000
Ans. b
Sol. 
Correct option is (b)
112. Consider a system of three particles which can occupy energy levels with energy 0, 
such that the total energy 
Cases A, B and C correspond to spin 
fermions, spin 0 bosons, and classically distinguishable particles, respectively. The correct ordering of entropy is
(a) SA > SB > SC
(b) SB > SA > SC
(c) SC > SB > SA
(d) SC > SA > SB
Ans. d
Sol. 
Correct option is (d)
113. For a point group, an incomplete character table is given below with one irreducible representation missing
The Mulliken symbol and characters of the missing representation are
(a) 
(b) 
(c) 
(d) 
Ans. c
Sol. Character table of C3V points group and complete table is
Correct option is (c)
114. Given below is a specific vibrational mode of BCl3 with 
denoting movement of the respective atoms above and below the plane of the molecule respectively. The irreducible representation of the vibrational mode and its IR / Raman activity are
(a) 
(b) 
(c) 
(d) 
Ans. d
Sol. 
Correct option is (d)
115. The first excited state 
of fluorine lies at an energy of 400 cm–1 above the ground state 
The fraction of Fluorine atoms in the first excited state 
is close to
(a) 
(b) 
(c) 
(d) 
Ans. d
Sol. 
Correct option is (d)
116. The two limiting wavefunctions of the ground state of 
molecular ion, as the internuclear separation R goes to (i) 
(infinity) and (ii) 0 (zero) are (lsa, lsb are ls – orbital wave functions of hydrogen atoms a and b in 
and lsHe is the wave function of the ls orbital of He+)
(a) (i) 1sa (r); (ii) 1sb (r)
(b) (i) 1sb (4); (ii) 1sa (r)
(c) (i) 1sa (r1) 1sb (r2); (ii) 1sHe (rl) 1sHe (r2)
(d) (i) 1sa (r) + 1sb (r) ; (ii) 1sHe (r)
Ans. d
Sol. 
Correct option is (d)
117. For a certain magnetic field strength, a free proton spin transition occurs at 700 MHz. Keeping the magnetic field strength constant the 14N nucleus will resonate at 
(a) 700 MHz
(b) 400 MHz
(c) 200 MHz
(d) 50 MHz
Ans. d
Sol. 
Correct option is (d)
118. The first electronic absorption band maximum of a polar and relatively rigid aromatic molecule appears at 310 nm but its fluorescene maximum in acetonitrile solution appears with a large Stokes shift at 450 nm. The most likely reason for Stokes shift is
(a) large change in molecular geometry in the excited state
(b) increase in dipole moment of the molecule in the excited state
(c) decrease in polarizability of the molecule in the excited state
(d) lowered interaction of the excited molecule with polar solvent
Ans. b
Sol. The first electronic absorption band maximum of a polar and relatively rigid aromatic molecule appears at 310 nm but its fluorescence maximum in acetonitrile solution appears with a large stokes shift at 450 nm. The reason for stokes shift increasing dipole moment of the molecule in the excited state.
Correct option is (b)
119. The un – normalized radial wave function of a certain hydrogen atoms eigenstate is (6r – r2) exp (– r/3). A possible angular part of the eigenstate is
(a) 
(b) 
(c) 
(d) 1
Ans. c
Sol. 
Corresponding 
possible angular part of eigen state is 
Correct option is (c)
120. Given a trial wave function 
and the Hamiltonian matrix elements, 
the variationally determined ground state energy is
(a) –0.52
(b) –0.50
(c) 12.50
(d) 12.52
Ans. b
Sol. 
–12E + E2 – 6.25 = 0
E2 – 12E – 6.25 = 0
The root of this equation is E = –0.50
E = –0.50
Correct option is (b)