PYQ DEC 2015 - VedPrep

CSIR NET CHEMISTRY (DEC-2015)
Previous Year Question Paper with Solution.

21. The biological function of cytochrome P₄₅₀ and myoglobin are, respectively

(a) oxidation of alkene and O₂ storage

(b) O₃ transport and O₂ storage

(d) electron carrier and O₂ transport

Ans. a

Sol. Cytochrome P₄₅₀ function as monooxygenase and catalyses the insertion of oxygen in substrate

Myoglobin is found in tissue and store the oxygen in tissue transported by Haemoglobin.

Correct option is (a)

22. Deoxy – hemocyanin is

(a) heme protein and paramagnetic

(b) colorless and diamagnetic

(d) blue colored and diamagnetic

Ans. b

Sol. In deoxyhaemocyanin Cu is in Cu(I) state i.e. have d¹⁰s⁰ configuration therefore it colourless and diamagnetic. Diamagnetism is due absence of unpaired electrons and colourless is due to no d–d transition as all the orbitals are filled.

Correct option is (b)

23. The oxidizing power of , and follows the order

(a)

(b)

(c)

(d)

Ans. a

Sol.

All have +6 oxidation state but due to smaller size on going from Cr to Fe. Fe has very high charge density. Hence, it has very high tendency to accept electron. Hence, strongest oxidising agent.

Correct option is (a)

24. Using crystal field theory, identify from the following complex ions that show same (spin only) values

(a) A and B

(b) B and C

(d) A, B and C

Ans. c

Sol.

Since, A and C has same number of unpaired electron. Hence, they have same magnetic moment. As A and C are 3d – block metal with weak ligand. Hence, they are high spin while Ir being 5d – metal is low spin. Because 5d has greater splitting power than 3d.

Correct option is (c)

25. The W – W bond order in
is

(a) three

(b) two

(d) zero

Ans. d

Sol.

Correct option is (d)

26. The correct statement for Mn – O bond lengths in is

(a) all bonds are equal

(b) four bonds are longer than two others

(d) they are shorter than the Mn–O bond in

Ans. a

Sol.

As all the levels are electronically non – degenerate. Hence, No Jahn – Teller distortion. Therefore, all the Mn – O bond length will be equal.

27. For the reaction of with PMe₃, the main intermediate is

(a)

(b)

(c)

(d)

Ans. b

Sol.

Example of migratory insertion reaction in which alkyl group inser to CO. A new ligand create acycl a vaccancy is created which is occupied by new ligand.

Correct option is (b)

28. Identify the complex ions in sequential order when ferroin is used as an indicator in the titration of iron (II) with potassium dichromate, (phen = 1, 10 – phenathroline)

(a)

(b)

(c)

(d)

Ans. a

Sol.

In presence of oxidising agent like changes it colours from red to blue hence, they are used in redox titration.

Correct option is (a)

29. The structures of XeF₂ and XeO₂F₂ respectively are

(a) bent, tetrahedral

(b) linear, square planar

(d) bent, see –saw

Ans. c

Sol.

Therefore, G = Geometry = trigonal bipyramidal (TBP)

Hence, the shape = linear

Correct option is (c)

30. Spin motion of which of the following gives magnetic moment

(A) electron, (B) poton (C) neutron

(a) A and B

(b) B and C

(d) A, B and C

Ans. d

Sol. Correct option is (d)

31. Correct statement for coulometry is

(a) it is based on Faraday's law of electrolysis

(b) it is a type of voltammetry

(d) it uses ion selective electrode

Ans. a

Sol. Columetry is based on Faraday's law of electrolysis

Correct option is (a)

32. The order of increasing Bronsted acidity for boron hydrides is

(a) B₅H₉ < B₆H₁₀ < B₁₀H₁₄

(b) B₁₀H₁₄ < B₅H₉ < B₆H₁₀

(d) B₁₀H₁₄ < B₆H₁₀ < B₅H₉

Ans. a

Sol. Acidic strength of Borane depends upon

(i) number of Boron atoms

(ii) decapping

Hence, the correct order of acidity is B₅H₉ < B₆H₁₀ < B₁₀H₁₄

Therefore, correct option is (a)

33. Among the following, species expected to show fluxional behaviour are

(a) B and C

(b) B and D

(d) A and D

Ans. b

Sol. In IF₇ and Fe(CO)₅, the rearrangement is so fast that axial and equatorial legands can not be differential by spectroscopy). The molecules in which rearrangement leads to configuration which are chemically equivalent are called fluxional molecule.

Correct option is (b)

34. The ring size and the number of linked tetrahedral present in are, respectively

(a) 6 and 6

(b) 12 and 6

(d) 6 and 12

Ans. b

Sol. The structure of is

Hence, the number of linked tetrahedrals = 6

The size of ring = 12

Therefore, Correct option is (b)

35. The molecule C₃O₂ has a linear structure. This compound has

(a)

(b)

(c)

(d)

Ans. a

Sol. The structure of C₃O₂ (carbon suboxide) is

Correct option is (a)

36. The metallic radii are abnormally high for which of the following pairs?

(a) Eu, Yb

(b) Sm, Tm

(d) Nd, Ho

Ans. a

Sol. The Eu and Yb donate two electrons in their metallic bonding unlike other lanthanoids which donate three electrons.

This is due to stable configurations

Hence, they show abnormally larger sizes than other lanthanoids.

Eu 1.99Å, Yb = 1.94Å

Correct option is (a)

37. Identify two enantiomers among the following compounds

(a) A and B

(b) A and C

(d) C and D

Ans. d

Sol.

Correct option is (d)

38. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. d

Sol.

Correct option is (d)

39. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. c

Sol.

Correct option is (c)

40. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. b

Sol.

Correct option is (b)

41. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. a

Sol. Correct option is (a)

42. The major product A and B in the following reactions are

(a)

(b)

(c)

(d)

Ans. c

Sol.

Correct option is (c)

43. D-Mannose upon refluxing in acetone with CuSO₄ and H₂SO₄ gives

(a)

(b)

(c)

(d)

Ans. c

Sol.

Correct option is (c)

44. The major product formed by photochemical reaction of (2E, 4Z, 6E) – decatriene is

(a)

(b)

(c)

(d)

Ans. a

Sol.

Correct option is (a)

45. The correct statement about the following reaction is that

(a)
and the reaction proceeds through carbene intermediate.

(b) and the reaction proceeds through nitrene intermediate.

(d) and the reaction proceeds through Norrish type I path.

Ans. a

Sol.

Correct option is (a)

46. The structure of the compounds that matches the ¹H NMR data given below is

¹H NMR (DMSO – d₆):
7.75 (dd, J = 8.8, 2.4 Hz, 1H), 7.58 (d, J = 2.4 Hz, 1H), 6.70 (d, J = 8.8 Hz, 1H), 6.50 (broad s, 2H), 3.80 (s, 3H).

(a)

(b)

(c)

(d)

Ans. b

Sol.

Correct option is (b)

47. Correctly matched structure and carbonyl stretching frequency set is

COLUMN – A COLUMN – B

(a) P – Y, Q – Z, R – X

(b) P – Y, Q – X, R – Z

(d) P – X, Q – Z, R – Y

Ans. a

Sol.

Correct option is (a)

48. The number of chemical shift non – equivalent protons expected in ¹H NMR spectrum of – pinene is

(a) 7

(b) 8

(d) 10

Ans. d

Sol.

Correct option is (d)

49. In the mass spectrum of 1, 2 – dichloroethane, approximate ratio of peaks at m/z values 98, 100, 102 will be

(a) 3 : 1 : 1

(b) 9 : 6 : 1

(d) 1 : 2 : 1

Ans. b

Sol.

Correct option is (b)

50. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. d

Sol.

Correct option is (d)

51. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. a

Sol.

Correct option is (a)

52. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. c

Sol.

Correct option is (c)

53. The concentration of a reactant R varies with time for two different reactions as shown in the following plots:

The order of these two reactions I and II, respectively are

(a) zero and one

(b) one and zero

(d) two and zero

Ans. c

Sol. c_t vs t straight line is observed for '0' order and
vs t straight line is observed for second order.

Correct option is (c)

54. For a simple cubic crystal lattice, the angle between the [2 0 1], plane and the xy plane is

(a) less than 30⁰

(b) between 30⁰ and 45⁰

(d) greater than 60⁰

Ans. d

Sol.

Correct option is (d)

55.

(a)

(b)

(c)

(d)

Ans. b

Sol. Rate of formation of B – Rate of deformation of B

Correct option is (b)

56. If the reduced mass of a diatomic molecule is doubled without changing its force constant, the vibrational frequency of the molecule will be

(a)
times the original frequency

(b) times the vibrational frequency

(d) unchanged

Ans. b

Sol.

times the original frequency.

Correct option is (b)

57. The standard deviation of speed for Maxwell's distribution satisfies the relation

(a)

(b)

(c)

(d)

Ans. b

Sol.

Correct option is (b)

58. The value of for the reaction is

(a) –3RT

(b) +3RT

(d) –RT

Ans. a

Sol.

Correct option is (a)

59. If the pressure p(system) is greater than the p(surroundings), then

(a) work is done on the system by the surroundings

(b) work is down on the surroundings by the systems

(d) internal energy of the system increases

Ans. b

Sol.

Process of expansion takes place. Work is done by the system on the surrounding.

Correct option is (b)

60. The different non – zero operators satisfy the reaction

(a)

(b)

(c)

(d)

Ans. d

Sol.

Correct option is (d)

61. The degeneracy of an excited state of a particle in 3 – dimensional cubic box with energy 3 times its ground state energy is

(a) 3

(b) 2

(d) 4

Ans. a

Sol.

Correct option is (a)

62. of a reaction is equal to slope of the plot of

(a)

(b)

(c)

(d)

Ans. d

Sol. According to Gibbs – Helmholtz equation

At absolute zero, i.e. when T = 0, of reaction is equal to the slope of the plot of T) versus 1 / T

Correct option is (d)

63. The correct form for a simple Langmuir isotherm is

(a)

(b)

(c)

(d)

Ans. c

Sol.

Correct option is (c)

64.

(a) depend only on stoichiometry

(b) depend only on specific identify of the electrolyte

(d) are mainly dependent on specific identify of the electrolyte and stoichiometry, respectively

Ans. d

Sol. Kohlrausch's constant is specified for uni – univalent electrolyte (stoichiometry) and strong electrolyte (specific identify).

Correct option is (d)

65. The correct expression for the product are the number – average and weight average molar masses, respectively, of a polymer] is

(a)

(b)

(c)

(d)

Ans. b

Sol.

Correct option is (b)

66. The concentration of a MgSO₄ solution having the same ionic strength as that of a 0.1 M Na₂SO₄ solution is

(a) 0.05 M

(b) 0.067 M

(d) 0.133 M

Ans. c

Sol.

Correct option is (c)

67. sp hybrid orbitals are of the form (2s and 2p_z are normalised individually). The coefficients of the normalized form of the above sp hybrid orbitals are

(a)

(b)

(c)

(d)

Ans. a

Sol.

Correct option is (a)

68. The correct statement among the following is

(a) N₂ has higher bond order than N₂⁺ and hence has larger bond length compared to N₂⁺.

(b) N₂⁺ has higher bond order than N₂ and hence has larger bond length compared to N₂.

(d) N₂ has lower bond order than N₂⁺ and hence has lower dissociation energy compared to N₂⁺ energy.

Ans. c

Sol. The Mo configuration of N₂ and N₂⁺ are

Correct option is (c)

69. The formation constant for the complexation of M⁺ (M = Li, Na, K and Cs) with cryptand, C222 follows the order

(a)

(b)

(c)

(d)

Ans. a

Sol. The stability (formation constant for complexation of cryptate complex depends upon)

(1) size of cavity

(2) size of metal cation

Hence, k⁺ ion will form most stable complex with cryptand - 222.

Correct option is (a)

70. The correct match for the compounds in column A with the description in column B is

(a) P – Y, Q – Z, R – X

(b) P – Z, Q – X, R – Y

(d) P – X, Q – Z, R – Y

Ans. b

Sol.

(b) P – Z, Q – X, R – Y

Correct option is (b)

71. The resonance Raman stretching frequency of O₂ is 1580. The
for O₂ in bound oxy – hemoglobin is close to

(a) 1600

(b) 1900

(d) 1100

Ans. d

Sol. In oxyhaemoglobin Fe is in +3 oxidation state and O₂ is in in O₂^–

As oxygen accept electron in orbital. Hence, bond get reduced and frequency decreases. Since, after binding to Fe in Hb bond between O–O become weaker comparison to free O–O bond in O₂.

Therefore, Force constant k decreases and

Stretching frequency would decrease

1100cm^–1 is

Correct option is (d)

72.

The correct match is

(a) A – VI, B – I, C – IV and D – III

(b) A – V, B – I, C – IV and D – III

(d) A – V, B – VI, C – II and D – IV

Ans. a

Sol. Hb it functions as O₂ transfer and contain Fe as active centre.

Cytochrome. b it functions as electron transfer and contain Fe

Vitamin B₁₂
it is responsible for group transfer reaction and contain cobalt.

Haemocyanin it is a non – haem, non – iron, copper containing protein and functions as O₂ transport system.

Correct option is (a)

73. Pick the correct statements about Atomic Absorption Spectrometry (AAS) from the following

(A) Hg lamp is not a suitable source for AAS

(B) Graphite furnace is the best atomizer for AAS

(D) AAS is better than ICP – AES for simultaneous determination of metal ions.

Correct answer is

(a) A, B and C

(b) B, C and D

(d) D, A and B

Ans. a

Sol. Inductively coupled plasma atomic spectroscopy (ICP – AES) is an analytic technique used for detection for trace metals and it is better than AAS for simultaneous determination of metal ions. Hence, option (d) is correct.

In Atomic Absorption Spectroscopy mercury lamp is not suitable cource but hollow cathode lamp (HCl) are the most common radiation source in Atomic Absorption Spectroscopy and atomizers most commonly used are flame and electro thermal (graphite). Solid sample is used in determination.

Correct option is (a)

74. Identify radioactive capture from the following nuclear reactions.

(a)

(b)

(c)

(d)

Ans. b

Sol. The reaction which involve evolution of radiation are called radioactive capture reaction.

Correct option is (b)

75. The calibration curve in spectrofluorimetric analysis becomes non–linear when

(a) molecular weight of analyte is high

(b) intensity of light source is high

(d) molar absorptivity of analyte is high

Ans. c

Sol. Fluorescence intensity is typically directly proportional (.inear) to concentration. There are however factor that affect tis linear relationship. When concentration is too high, light can not pass through the sample to cause excitation, thus very high concentration can have very low Fluorescence and thus calibration curves become non – linear.

Correct option is (c)

76.
is deep purple in color whereas is colorless. This is due to greater energy required for

(a) d–d transitions in the Re compound compared to the Mn compound

(b) d–d transitions in the Mn compound compared to the Re compound

(d) charge transfer from O to Mn compared to O to Re

Ans. c

Sol.

E₂ > E₁ hence, energy required for transfer of electron from oxygen to Re falls in UV region. Hence,
colourless while in due to less energy difference it falls in visible region. Hence, coloured. Also as both Mn and Fe have +7 oxidation state i.e. they have no d – electron hence. no d – d transition and colour arises due to LMCT.

Correct option is (c)

77.
shows fluxional behavior. The ¹H NMR spectrum of this compound when it is in the non – fluxional state shows

(a) one signal

(b) two signals in the intensity ratio of 4 : 1

(d) five signals of equal intensity.

Ans. c

Sol.

H_c = central proton

H_a = is anti with respect to Hc

H_s = syn with respect to Hc

H_a, H_s, H_e environment are different new intensity ratio 2 : 2 : 1. When allyl is non – fluxional.

Correct option is (c)

78. The number of lone pair(s) of electrons on the central atom in are respectively

(a) 2, 0 and 1

(b) 1, 0 and 0

(d) 2, 1 and 0

Ans. c

Sol.

Hence, correct option is (c)

Note: The lone pair of
is stereochemically inactive and the of XeF₆ is boderline of stereochmically active and stereochmically inactive condition.

Correct option is (c)

79. Consider the following reaction:

The number of possible isomers for [A] is

(a) 4

(b) 3

(d) 5

Ans. b

Sol.

Correct option is (b)

80. Using Wade's rules predict the structure type of

(a) nido

(b) closos

(d) hypho

Ans. b

Sol.

Correct option is (b)

81. Among the following complexes

(a) A and B

(b) C and B

(d) A and C

Ans. d

Sol.

Thus, A and C are chiral and B is achiral. Hence, correct option is (d)

82. Mössbauer spectrum of a metal complex gives information about

(A) oxidation state and spin state of metal

(B) types of ligands coordinated to metal

(D) geometry of metal

Correct answer is

(a) A and C

(b) B and C

(d) B and D

Ans. c

Sol. Mössbauer spectrum of a complex gives information about

(1) oxidation state and spin state of metal

(2) types of ligands co – ordinate to metal define by quardurpole splitting on the behalf of electronic environment of ligand symmetry.

(3) Also information about geometry of metal

Correct option is (c)

83. For uranocene, the correct statement(s) is/are

(A) oxidation state of uranium is '+4'

(B) it has cyclooctatetraenide ligands

(D) it has '–2' charge

Correct answer is

(a) A and B

(b) B and C

(d) B only

Ans. a

Sol. Structure of Uranocane

Hence, oxidation state of U = +4

Thus in uranocene U is in +4 oxidation state and have cyclooctatetraenide ligand.

Correct option is (a)

84. The final products of the reaction of carbonyl metalates with respectively, are

(a)

(b)

(c)

(d)

Ans. a

Sol.

Correct option is (a)

85. The correct statement about the substitution reaction of to give is,

(a) it obeys first order kinetics

(b) its rate is proportional to the concentration of both the reactants

(d) its rate is dependent only on the concentration of

Ans. a

Sol.

The reaction proceed through dissociative pathway and rate is dependent only concentration of substrates. As substate do not have acidic hydrogen. Hence, it does not undergo S_N1CB mechanism.

Correct option is (a)

86. Aqueous effects one electron reduction of giving compound Y. Compound Y undergoes rapid hydrolysis. Y is,

(a)

(b)

(c)

(d)

Ans. a

Sol.

Correct option is (a)

87. The reaction of BCl₃ with NH₄Cl gives product A which upon reduction by NaBH₄ gives product B. Product B upon reacting with HCl affords compound C, which is

(a)

(b)

(c)

(d)

Ans. a

Sol.

Correct option is (a)

88. The number of valence electrons provided by fragment towards cluster bonding is

(a) 1

(b) 14

(d) 2

Ans. d

Sol. Number of electron involving in metal cluster = T.V.E – 12

T.V.E in Ru(CO)₃ = 8 + 6 = 14 (T.V.E = total valence electron)

Hence, number of electron provided for cluster bonding = 14 – 12 = 2

Correct option is (d)

89. Choose the correct statements about Tanabe – Sugano diagrams

(A) E/B is plotted against

(B) The zero energy is taken as that of the lowest term

(D) Two terms of the same symmetry upon increases of ligand field strength bend apart from each other.

Correct answer is

(a) A and B

(b) A and C

(d) A, B, C and D

Ans. c

Sol. Tanabe – Sugano are useful in interpretation of spectra of both high spin and low spin complexes of d² – d⁸ metal cation.

In Tanabe – Sungano, the energy of excited sate (expressed as E/B) are plotted against ligand field strength (expressed as

Zero energy is taken for the lowest term and also two form of same symmetry never cross each other and they bent for apart from each other due to repulsion.

Correct option is (c)

90. Which of the following statements are TRUE for the lanthanides?

(A) the observed magnetic moment of Eu³⁺ at room temperature is higher than that calculated from spin – orbit coupling

(B) Lanthanide oxides are predominantly acidic in nature.

(D) Lanthanide (III) ions can be separated by ion exchange chromatography

Correct answer is

(a) A and D

(b) A and B

(d) (d) B and C

Ans. a

Sol. (a) There is no good agreement between the calculated and experimental magnetic moments for Eu³⁺ and Sm³⁺ ions because of law value (~300 cm^–1) of spin orbit coupling constant. Therefore, the ground state and the first excited states become equality populated, due to the thermal motion become equally populated due to thermal motion. Hence, the observed magnetic moment of Eu³⁺ at room temperature is higher than that calculated from, spin – orbit coupling.

(b) Lanthanide oxides are predominantly basic in nature.

(d) Since the chemical properties of the tripositive lanthanide ions (due to their similar size) they can not be separated easily by chemical method. Instead they can be separated by ion exchange chromatography.

Correct option is (a)

91. The intermediate and the final major product of photolysis of Z.

From the following:

are

(a) A and D

(b) B and D

(d) A and C

Ans. b

Sol.

Correct option is (b)

92. Reaction of with results in A without loss of CO. Compound A, on heating to 120⁰C loses a CO ligand to give B, which does not have a Mn–Mn bond. Compound B reacts with pyridine to give 2 equivalents of C. Compounds A, B and C from the following respectively, are

(a) II, V and IV

(b) II, III and IV

(d) II, V and III

Ans. a

Sol.

Correct option is (a)

93. The approximate positions of v_CO bands (cm^–1) in the solid – state infrared spectrum and the Fe–Fe bond order in
(non – centrosymmetric) respectively, are

(a) (2020, 1980, 1800) and one

(b) (2020, 1980 1800) and two

(d) (2143) and one

Ans. a

Sol.

In this complex M – M bond order = 1 and two types of CO's terminal as well as m² – bridging. So, v_C–O band lies in both range

Correct option is (a)

94. Protonated from of ZSM – 5 catalyzes the reaction of ethene with benzene to produce ethylbenzene. The correct statement for this catalytic process is

(a) alkyl carbocation is formed

(b) carbanion is formed

(d) vinyl radical is formed

Ans. a

Sol. Correct option is (a)

95. Three electronic transitions at 14900, 22700 and 34400 cm^–1 are observed in the absorption spectrum of
value (in cm^–1) and the corresponding transition are

(a)

(b)

(c)

(d)

Ans. b

Sol.

Since, have no fix energy therefore, they will not provide accurate value of

Thus, energy difference between will be correspond to

96. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. c

Sol.

Correct option is (c)

97. The following transformation involves sequential

(a) Claisen rearrangement – Cope rearrangement – ene reaction

(b) Cope rearrangement – Claisen rearrangement – ene reaction

(d) ene reaction – Claisen rearrangement – Cope rearrangement

Ans. b

Sol.

Correct option is (b)

98. The major product formed in the following reaction sequence is

(a)

(b)

(c)

(d)

Ans. b

Sol.

Correct option is (b)

99. The major products A and B in the following reaction sequence are

(a)

(b)

(c)

(d)

Ans. c

Sol.

Correct option is (c)

100. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. d

Sol. Correct option is (d)

101. The major products A and B in the following reaction sequences are

(a) A = D – threose ; B = D – glucose

(b) A = D – etythrose; B = D – glucose + D – mannose

(d) A = D – tartaric acid; B = D – glucose

Ans. b

Sol.

Correct option is (b)

102. The major products A and B in the following reaction sequence are

(a)

(b)

(c)

(d)

Ans. d

Sol.

Now, stablized phosphorous ylide forms Z-alkene from aldehyde.

Correct option is (d)

103. The major products A and B in the following reaction sequence are

(a)

(b)

(c)

(d)

Ans. a

Sol.

Correct option is (a)

104. The major products A and B in the following reactions sequence are

(a)

(b)

(c)

(d)

Ans. a

Sol.

Correct option is (a)

105. The major products A and B in the following reaction sequence are

(a)

(b)

(c)

(d)

Ans. d

Sol.

Correct option is (d)

106. The major product formed in the following reaction in

(a)

(b)

(c)

(d)

Ans. d

Sol.

Correct option is (d)

107. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. b

Sol.

Intramolecular Heck – coupling reaction.

Correct option is (b)

108.

(a)

(b)

(c)

(d)

Ans. b

Sol.

Correct option is (b)

109. The major products A and B in the following reaction sequence are

(a)

(b)

(c)

(d)

Ans. c

Sol.

Correct option is (c)

110. The correct reagent combination to effect the following transformation is

(a) A = NaBH₄, BF.OEt₂; B = MeMgBr (2.5 equiv.), THF then H₃O⁺

(b) A = BH₃. THF; B = MeLi (2.5 equiv.), THF then H₃O⁺

(d) A = (i) Me₃Al, MeNHOMe, (ii) MeMgBr, THF then H₃O⁺; B = LiAlH₄, THF

Ans. c

Sol.

Correct option is (c)

111. The mechanism and the product formed in the following reaction, respectively, are

(a)

(b)

(c)

(d)

Ans. d

Sol.

Correct option is (d)

112. A Concerted [1, 3]– sigmatropic rearrangement took place in the reaction shown below. The structure of the resulting product is

(a)

(b)

(c)

(d)

Ans. c

Sol. A concerted [1, 3] sigmatropic shift.

[1, 3]- allyl shift under thermally condition occurs antrafacially and hence stereochemistry changes (inversion in configuration occurs)

Hence, stereochemistry of (D) and (OAc) will become same / cis.

Correct option is (c)

113. The major product A and B in the following reaction sequence are

(a)

(b)

(c)

(d)

Ans. a

Sol.

Correct option is (a)

114. The major product formed in the following reaction sequence is

(a)

(b)

(c)

(d)

Ans. d

Sol.

Correct option is (d)

115. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. b

Sol.

Correct option is (b)

116. The major products A and B in the following reaction sequence are

(a)

(b)

(c)

(d)

Ans. c

Sol.

Correct option is (c)

117. The major product of the following reaction is

(a)

(b)

(c)

(d)

Ans. c

Sol.

Correct option is (c)

118. The major product A and B in the following reaction sequence are

(a)

(b)

(c)

(d)

Ans. b

Sol.

Correct option is (b)

119. The major products A and B in the following reaction sequence are

(a)

(b)

(c)

(d)

Ans. c

Sol.

Correct option is (c)

120. The major product of the following reaction is

(a)

(b)

(c)

(d)

Ans. c

Sol.

Correct option is (c)

121. According to the transition state theory, one of the vibrations in the activated complex is a loose vibration. The partition function for this loose vibration is equal to (k_B is the Boltzmann's constant and h is the Planck's constant)

(a)

(b)

(c)

(d)

Ans. d

Sol.

Correct option is (d)

122. Possible term symbol(s) of the excited states of Na atom with the electronic configuration is / are

(a)

(b)

(c)

(d)

Ans. b

Sol.

Correct option is (b)

123. For a simple cubic crystal, X–ray diffraction shows intense reflections for angles which are assigned to [1 0 1] and [1 1 1] planes, respectively. The ratio is

(a) 1.5

(b) 1.22

(d) 0.67

Ans. c

Sol.

Correct option is (c)

124. Stability of lyophobic dispersions is determined by

(a) inter – particle electric double layer repulsion and intra – particle van der walls attraction.

(b) inter – particle electric double layer attraction and intra – particle van der waals repulsion.

(d) inter – particle excluded volume attraction and intra – particle van der waals repulsion.

Ans. a

Sol. Stability of lyophobic dispersions is determined by inter – particle electric double layer repulsion and intra – particle van der waals attraction.

Correct option is (a)

125. A certain 2 – level system has stationary state energies E₁ and E₂ (E₁ < E₂) with normalized wave function respectively. In the presence of a perturbation V, the second – order correction to the energy for the first state will be

(a)

(b)

(c)

(d)

Ans. c

Sol. The second order correction to n^th state

Correct option is (c)

126. The ¹H NMR frequency at 1.0 T is 42.4 MHz. If the gyromagnetic ratios of ¹H and ¹³C are 27 × 10⁷ and 6.75 × 10⁷ T^–1S^–1, respectively, what will be the ¹³C frequency at 1.0 T?

(a) 10.6 MHz

(b) 169.9 MHz

(d) 21.3 MHz

Ans. a

Sol.

Equation (i) divided by equation (ii),

Correct option is (a)

127. 10 mL aliquots of a mixture of HCl HNO₃ are titrated are titrated conductometrically using a 0.1M NaOH and 0.1M AgNO₃ separately. The titre volumes are V₁ and V₂ mL, respectively. The concentration of HNO₃ in the mixture is obtained from the combination.

(a) V₁ – V₂

(b) 2V₁ – V₂

(d) 2V₂ – V₁

Ans. a

Sol. First titration:

10 mL of HCl + HNO₃ with NaOH

Reaction,

Suppose in 10 ml aliquots n₁ mole of HCl are present, and n₂ moles of HNO₃ are present. (Number of moles of HCl = number of moles of NaOH + no. of moles of HNO₃)

n₂ + n₁ = 0.1 V₁

Second titration:

Number of moles of HCl = number of moles of AgNO₃

n₁ = 0.1V₂

Number of moles of HNO₃ only

So, (V₁ – V₂) will give number of moles of HNO₃. So, from it well get conc.

Correct option is (a)

128. Given that at 25⁰C, E⁰ corresponding to the electrode reaction

(a) 0.75V

(b) 1.05V

(d) 1.95V

Ans. d

Sol. Reaction – 1:

Reaction – 2:

Reaction – 3: On reversing reaction – 2

On adding reaction (1) and reaction (3), we get

Correct option is (d)

129. The standard EMF of the cell

(a) increases with T

(b) decreases with T

(d) decreases with [HCl]

Ans. b

Sol.

As entropy of reaction decreases because gaseous is being consumed. So,

decreases with increase in temperature.

Correct option is (b)

130. The molecule with the smallest rotational constant (in the microwave spectrum) among the following is

(a)

(b)

(c)

(d)

Ans. c

Sol.

I for ClC = CF is highest

Therefore, B would be smallest for CIC = CF

Correct option is (c)

131. The spectrosopic technique that can distinghuish unambiguously between trans – 1, 2 – dichloroethylene and cis – 1, 2 – dichloroetylene without any numerical calculation is

(a) microwave spectroscopy

(b) UV – visible spectroscopy

(d) – ray spectroscopy

Ans. a

Sol.

These two compound are distinguish on the behalf of dipolemoment by microwave spectroscopy without any calculation.

Correct option is (a)

132. The ground state electronic configuration of C₂ using all electron is

(a)

(b)

(c)

(d)

Ans. d

Sol. C₂ = valence electron = 12

Correct option is (d)

133.
for an enzyme catalyzed reaction are respectively. The rate of the reaction when the substrate concentration is 1.0 × 10^–6 M is

(a)

(b)

(c)

(d)

Ans. b

Sol.

Correct option is (b)

134. The first order rate constant for a unimoleclar gas phase reaction A products that follows Lindemann mechanism is 2.0 s^–1 at p_A = 1 atm and 4.0 s^–1 at p_A = 2 atm. The rate constant for the activation step is

(a)

(b)

(c)

(d)

Ans. b

Sol.

Correct option is (b)

135. The molecule diborane belongs to the symmetry point group

(a)

(b)

(c)

(d)

Ans. d

Sol.

Correct option is (d)

136. Though a constant shift of energy levels of a system changes the partition function, the properties that do not change are

(a) average energy, entropy and heat capacity

(b) average energy and entropy

(d) entropy and heat capacity.

Ans. d

Sol. Entropy and heat capacity

Since, entropy change in partition function is compensated by change in total energy and in heat capacity differentiation with respect to temperature lead to cancellation of constant shift in energy.

Correct option is (d)

137. The vibrational frequency of a homo – nuclear diatomic molecule is v. The temperature at which the population of the first excited state will be half that of the ground state is given by

(a)

(b)

(c)

(d)

Ans. b

Sol.

Correct option is (b)

138. The irreducible representations of The Raman active modes of trans–1, 3 – butadiene belong to the irreducible representations

(a) A_g and B_g

(b) A_g and A_u

(d) B_g and B_u

Ans. a

Sol. Raman active vibrations transform according to quadractic functions of x, y and z which are A_g and B_g.

Correct option is (a)

139. The symmetry – allowed atomic transition among the following is

(a)

(b)

(c)

(d)

Ans. b

Sol.

Since, for allowed transition (atomic),

Correct option is (b)

140. The average end – to – end distance of a random coil polymer 10⁶ monomers (in units of segment length) is

(a) 10⁶

(b) 10⁵

(d) 10³

Ans. d

Sol. Average end – to – end distance

Correct option is (d)

141. The reversible expansion of 1.0 mol of an ideal gas is carried out from 1.0 L to 4.0 L under isothermal condition at 300K. for this process is

(a)

(b)

(c)

(d)

Ans. c

Sol.

Correct option is (c)

142. A temperature – dependence of the vapour pressure of solid A can be represented by log and that of liquid A by log The temperature of the triple point of A is

(a) 200 K

(b) 300 K

(d) 500 K

Ans. a

Sol. At triple point vapour pressure of solid A and liquid A are same. So,

Correct option is (a)

143. The non – spontaneous process among the following is

(a) vapourisation of superhead water at 105⁰C and 1 atm pressure

(b) expansion of a gas into vacuum freezing

(d) freezing of water at 0⁰C and 1 atm pressure.

Ans. d

Sol. Option (a), (b) and (c) talks about the processes which are spontaneous processes.

Option (a) and (c) can be seen through phase diagram of water and (b) is free expansion which is spontaneous.

Option (d) talks about an equilibrium process.

So, this is a non - spontaneous process.

Correct option is (d)

144. The radial part of a hydrogenic wave function is given as This function is then indentifiable as

(a) 2s

(b) 3p

(d) 5f

Ans. b

Sol. In radial part of wave function highest power of r denote the value of

Here and radial node

Correct option is (b)

145. A normalised state is constructed as a linear combination of the ground state and the first excited state of some harmonic oscillator with energies 1/2 and 3/2 units, respectively. If the average energy of the state is 7/6, the probability of finding will be