21. Identify the species, those obey the 18 electron rule, from the following:

(a) A and B

(b) B and C

(c) C and D

(d) A and D

Ans. d

Sol.

Correct option is (d)

22. The following transformation

is an example of

(a) oxidative addition

(b) insertion

(c)

(d) reductive elimination

Ans. c

Sol.

Correct option is (c)

23. shows absorption bands at 8500, 15400, and 26000 cm^–1 whereas at 10750, 17500 and 28200 cm^–1. L and L´ are respectively.

(a)

(b)

(c)

(d)

Ans. d

Sol. has absorption bonds at 8500, 15400, 26000 cm^–1

at 10750, 17500 and 28200 cm^–1

as for first complex absorptions bands are at low energy i.e. it has weak splitting therefore, it has weak ligand and for II^nd complex high energy absorption bands corresponds to strong ligand. Thus, L is weak and L¹ is strong ligand.

Correct option is (d)

24. The number of microstates present in ³F term is

(a) 3

(b) 21

(c) 9

(d) 28

Ans. b

Sol. Number of microstate in ³F is calculated as (2S + 1) (2L + 1)

For F, L = 3

Hence, (3) (2 × 3 + 1) = 21

Correct option is (b)

25. fragment isolabal with a BH fragment is

(a) CpGe

(b) CpMn

(c) 5

(d) 6

Ans. d

Sol. B – H = 4e^– fragment isolobal fragment should have 14 electron.

Cp + Co = 5 + 9 = 14

Correct option is (d)

26. The number of metal – metal bonds in is

(a) 3

(b) 4

(c) 5

(d) 6

Ans. b

Sol. TVE = 18 + 16 + 22 + 8 = 64 = A

Correct option is (b)

27. Correct combination for orbitals in B₂ molecule is

(a)

(b)

(c)

(d)

Ans. b

Sol.

Correct option is (b)

28. The correct shape of ion on the basis of VSEPR theory is

(a) trigonal bipyramidal

(b) square pyramidal

(c) pentagonal planar

(d) see – saw

Ans. b

Sol.

Number of lone pair = 1

Structure is square pyramidal

Correct option is (b)

29. The number of P–S and P–P bonds in the compounds P₄S₃ are, respectively

(a) 6 and 3

(b) 4 and 3

(c) 3 and 6

(d) 6 and 2

Ans. a

Sol. Structure of P₄S₃ is

Number of P – S bond = 6

Number of P – P bond = 3

30. In the iodometric titration of sodium thiosulfate (Na₂S₂O₃) with acidic dichromate solution, 25 mL of 0.01 M dichromate requires 25 mL of 'x' M thiosulfate. The value of 'x' is

(a) 0.2

(b) 0.1

(c) 0.6

(d) 0.4

Ans. c

Sol. In acidic medium dichromate changes into Cr⁺⁺⁺ and in titration changes into (tetrathionate).

Now, milliequivalent of Cr₂O₇^2– = milli equivalent of S₂O₃^2–

Z₁M₁V₁ = Z₂M₂V₂

6 × 0.1 × 25 = 1 × x × 25 x = 0.6

Correct option is (c)

31. Decomposition temperature of CaCO₃ in thermogravimetric analysis will be highest in dynamic atmosphere of

(a) nitrogen

(b) synthesis gas

(c) 1 : 1 mixture of O₂ and CO

(d) water gas

Ans. c

Sol. Decomposition temperature of CaCO₃ in thermogravimetric analysis will be highest in dynamic atmosphere due to the presence of 1 : 1 mixture of O₂ and CO.

Correct option is (c)

32. On two sequential electron capture, ₅₆Ba¹³¹ will give

(a) ₅₄Xe¹³¹

(b) ₅₄Xe¹³⁰

(c) ₅₆Ce¹³¹

(d) ₅₆Ce¹³⁰

Ans. a

Sol.

Correct option is (a)

33. The compound which dissolves in POCl₃ to give a solution with highest chloride ion concentration, is

(a) Et₃N

(b) KCl

(c) FeCl₃

(d) SbCl₅

Ans. b

Sol.

The Cl^– ion produced in this reaction increases the concentration of Cl^– and it is not consumed by POCl₃.

Correct option is (a)

34. In the absence of bound globin chain, heme group on exposure to O₂ gives the iron –  oxygen species

(a)

(b)

(c)

(d)

Ans. a

Sol. Absence of globin chain following reaction is occur

Correct option is (a)

35. For monoionic complex the correct coordination number and geometry respectively, are

(a) 8 and hexagonal bipyramidal

(b) 5 and square pyramidal

(c) 8 and square antiprism

(d) 5 and trigonal bipyramidal

Ans. a

Sol. Co-ordinate number is 8 and geometry is hexagonal bipyramidal.

Correct option is (a)

36. Chelate effect is

(a) predominantly due to enthalpy change

(b) predominantly due to entropy change

(c) independent of ring size

(d) due to equal contribution of entropy and enthalpy change

Ans. b

Sol. Chelate effect if predominately due to entropy change

Correct option is (b)

37. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. d

Sol.

Correct option is (d)

38. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. b

Sol.

Correct option is (b)

39. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. a

Sol.

Correct option is (a)

40. Among the following, the compound that displays an IR band at 2150 cm^–1 is

(a)

(b)

(c)

(d)

Ans. c

Sol.

Correct option is (c)

41. In the ¹H NMR spectrum of myrtenal, the two methyl groups are expected to display signals at (chemical shift values in ppm)

(a) 1.35 (s, 3H) and 5.0 (s, 3H)

(b) 0.74 (s, 3H) and 1.33 (s, 3H)

(c) 1.22 (s, 6H)

(d) 0.70 (s, 6H)

Ans. b

Sol.

Correct option is (b)

42. Among the following, the compound(s) that can be classified as terpene derivative is(are)

(a) A and B

(b) A only

(c) B only

(d) B and C

Ans. c

Sol.

Correct option is (c)

43. The frontier orbital interactions involved in the formation of the carbocation intermediate in the reaction of isobutylene with HCl are

(a)

(b)

(c)

(d)

Ans. a

Sol.

Correct option is (a)

44. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. d

Sol.

Correct option is (d)

45. In the UN – visible absorption spectrum of an – unsaturated carbonyl compound, with increasing solvent polarity,

(a) transitions undergo hypsochromic shift, undergo bathochromic shift

(b) transitions undergo hypsochromic shift, undergo hypsochromic shift

(c) both and transitions undergo bathochromic shift

(d) both and transitions undergo hypsochromic shift.

Ans. a

Sol.

Correct option is (a)

46. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. a

Sol.

Correct option is (a)

47. In the following compound, the stereochemical descriptor for H_a and H_b is

(a) enantiotopic

(b) diastereotopic

(c) homotopic

(d) constitutionally heteretopic

Ans. b

Sol.

(P) and (Q) are diastereoisomers. Hence, H_a and H_b are diastereotopic.

Correct option is (b)

48. The correct statements are about the reaction of X and Y with NaNH₂ are

(A) X reacts faster than Y

(B) Y reacts faster than X

(C) X and Y behave as Lewis acids

(D) X is stronger Bronsted acid than Y

(a) A and C

(b) A and D

(c) B and C

(d) B and D

Ans. b

Sol.

Correct option is (b)

49. The correct statements about conformations X and Y of 2–butanone are

(A) X is more stable than Y

(B) Y is more stable than X

(C) Methyl groups in X are anti

(D) Methyl groups in Y are gauche

(a) A and D

(b) A and C

(c) B and C

(d) A, C and D

Ans. d

Sol.

In (Y) Gauche interaction, destabilies the conformation

Correct option is (d)

50. The correct order of magnitude of 'A values' for the given substituents in cyclohexane derivatives is

(a) Ph > CN > Me

(b) Me > Ph > CN

(c) CN > Me > Ph

(d) Ph > Me > CN

Ans. d

Sol.

The magnitude of 'A values' dependent on the steric hinderence.

The correct order is Ph > CH₃ > CN

Correct option is (d)

51. The correct order of pKa values for the compounds X, Y and Z is

(a) X > Y > Z

(b) Y > Z > X

(c) Z > X > Y

(d) Y > X > Z

Ans. c

Sol. Higher the pKa value lower will be the acidity.

But due to annelation effect the proton will be less acidic. Hence, higher pKa value.

Correct option is (c)

52. The following transformation proceeds through two consecutive electrocyclic processes, which are

(a) conrotatory and conrotatory

(b) disrotatory and conrotatory

(c) conrotatory and disrotatory

(d) disrotatory and disrotatory

Ans. c

Sol.

Correct option is (c)

53. The simultaneous eigenfunctions of angulra momentum operators L₂ and L_z are

(a) all of 2s, 2p_x, 2p_y and 2p_z orbitals

(b) only 2s, 2p_x and 2p_y orbitals

(c) only 2s and 2p_z orbitals

(d) only 2p_z orbital

Ans. c

Sol. 2s and 2p_z orbitals are eigen function of L₂ and L_z.

Correct option is (c)

54. An ideal gas is composed of particles of mass M in thermal equilibrium at a temperature T in one container. Another container contains ideal gas particles of mass 2M at a temperature 2T. The correct statement about the about two gases is

(a) average kinetic energy and average speed will be same in the two cases

(b) both the averages will be doubled in the second case.

(c) only the average kinetic energy will be doubled in the second case

(d) only the average speed will be doubled in the second case.

Ans. c

Sol.

So, it will be same in both. But average kinetic energy

So, it will be double is second case.

Correct option is (c)

55. The lowest energy term for the d₆ configuration is

(a) ²D

(b) ⁵D

(c) ¹P

(d) ¹D

Ans. b

Sol.

Correct option is (b)

56. If the rates of a reaction are R1 and R2 at concentrations C₁ and C₂ of a reactant respectively, the order of reaction, 'n' (assumping that the concentrations of all other reactants and T remain constant) with respect to the reactant is given by

(a)

(b)

(c)

(d)

Ans. a

Sol.

Correct option is (a)

57. Experimentally determined rate law for the chemical reaction,

The rate determining step consistent with the rat law is

(a)

(b)

(c)

(d)

Ans. c

Sol. Given rate is in accordance with the rate of NO₂F

Correct option is (c)

58. The symmetry point group of the most stable geometry of the following molecule is

(a) C₂

(b) C₁

(c) C_2v

(d) C_2h

Ans. a

Sol.

This molecule is chiral

Therefore, no plane, only one C₂.

Correct option is (a)

59. The eigenfunctions of the Hamiltonian of a harmonic oscillator are (where T and V are kinetic energy and potential energy operators, respectively)

(a) eigenfunctions of T as well as V

(b) eigenfunctions of T, but not of V

(c) eigenfunctions of V, but not of T

(d) eigenfunctions of neither T nor V

Ans. d

Sol. If commute i.e. then they have same set of eigen functions.

Correct option is (d)

60. In a potentiometric titration, the end point is characterised by

(a)

(b)

(c)

(d)

Ans. b

Sol.

Correct option is (b)

61. On titrating conductometrically a NaOH solution with a mixture of HCl and CH₃CO₂H solutions, plot of the volume of mixed acid added (b) in y-axis against the conductance (a) in x-axis is expected to look like

(a)

(b)

(c)

(d)

Ans. d

Sol. NaOH + (HCl + CH₃COOH)

First NaOH will react with HCl will use NaOH is completely titrated during this process conductance will decrease due to less of OH^– ions.

Second after that, as NaOH is completely consume. So, simple addition of mixture of HCl and CH₃COOH will only increase the conductance.

Correct option is (d)

62.

(a) pressure

(b) volume

(c) temperature

(d) heat capacity

Ans. b

Sol.

Correct option is (b)

63. In a cubic crystal, the plane [100] is equally inclined to the planes

(a) [010] and [011]

(b) [010] and [110]

(c) [001] and [101]

(d) [110] and [011]

Ans. a

Sol. Angle between (1 0 0) and (0 1 0)

And angle between (100) and (011)

64. The standard electrode potential E⁰ at a fixed temperature and in a given medium is dependent on

(a) only the electrode composition

(b) the electrode composition and the extent of the reaction

(c) the extent of the electrode reaction only

(d) the electrode reaction and the electrode composition

Ans. d

Sol. The standard electrode potential E₀ at a fixed temperature and in a given medium is dependent on the electrode reaction and the electrode composition.

Correct option is (d)

65. In a titration, the percentage uncertainties in the measured aliquot volume and the measured titre volume are ±x and ±y respectively. The percentage error in the calculated concentration of aliquot is

(a) x + y

(b) xy

(c)

(d)

Ans. d

Sol. In a titration, the percentage uncertainties in the measured aliquot volume and the measured titre volume are ±x and ±y respectively. The percentage error in concentration of aliquot is

Correct option is (d)

66. For an ideal gas at 300K

(a)

(b)

(c)

(d)

Ans. a

Sol. From Maxwell equation,

dU = TdS – PdV ...(i)

Divide the equation by (dV)_T at constant T

Correct option is (a)

67. The first excited state of hydrogen molecule is

(a)

(b)

(c)

(d)

Ans. d

Sol.

For half filled

Correct option is (d)

68. When river water containing colloidal clay flows into the sea, the major cause of silting is

(a) accumulation of sand at the bottom

(b) flocculation and coagulation

(c) decreased salinity of sea water

(d) micellization

Ans. b

Sol. When river water containing colloidal clay flows into the seac. The salt water induces flocculation and coagulation, and is a major cause of silting in estwers.

Correct option is (b)

69. Math the metal given in Column-A with its medicinal use as a compound in Column-B

Correct match is

(a) A – II; B – III; C – IV; D – I

(b) A – IV; B – II; C – I; D – III

(c) A – III; B – IV; C – I; D – II

(d) A – I; B – II; C – III; D – IV

Ans. c

Sol. Pt – complex used as anticoncern drug

Au – complex used as antiarthritis agent

Gd – complex used in MRI as contrast agent due to their strong paramagnetic behaviour

Li – complex used in Mariac depression

Correct option is (c)

70. At pH 10, tryptophan exists as

(a)

(b)

(c)

(d)

Ans. a

Sol. pH = 10 highly basic. So, there will be over all negative charge

Correct option is (a)

71. Complex shows red phosphorescene due to transition

(a)

(b)

(c)

(d)

Ans. c

Sol.

As phosphorescence is spin-forbidden transition and also occurs when electron comes from excited to ground state. Hence, transition is responsible for the phosphorescence.

Correct option is (c)

72. Choose the correct option for carbonyl cluoride with respect to bond angle and bond length

(a)

(b)

(c)

(d)

Ans. c

Sol. Structure of CoF₂ =

due to more space requirement for double bond the bond angle F–C–O will be higher that of F–C–F and also C–O bond order is larger than C-F hence, C–O bond length will be smaller than C–F.

Correct option is (c)

73. Which of the following react(s) with AsF₅ in liquid BrF₃?

(a) XeF₆ only

(b) XeF₆ and XeF₄

(c) XeF₆ and XeF₂

(d) XeF₄ and XeF₂

Ans. c

Sol.

Separation of XeF₄ from a mixture involves preferential complexation of XeF₂ and XeF₆ and XeF₄ is then removed in vaccum.

Correct option is (c)

74. Consider the following reactions:

Reactions which will give as a major product are:

(a) A and B

(b) C and D

(c) A and C

(d) B and D

Ans. b

Sol.

Correct option is (b)

75. The complex that shows orbital contribution to the magnetic moment, is

(a)

(b)

(c)

(d)

Ans. c

Sol. For orbital contribution the t_2g set should be unsymmetrically filled

Correct option is (c)

76. Among KF, SnF₄ and SbF₅, solute(s) that increase(s) the concentrations of is/are

(a) KF only

(b) KF and SnF₄

(c) SnF₄ and SbF₅

(d) KF, SnF₄ and SbF₅

Ans. a

Sol.

Correct option is (a)

77. Paramagnetic susceptibility of the order of 10^–6 cm³ mol^–1 observed for KMnO₄ is due to

(a) random spin alignment

(b) antiferromagnetic exchange interaction

(c) paramagnetic impurity

(d) temperature independent paramagnetism

Ans. d

Sol. In KMnO₄ all the electrons are paired. However, its paramagnetic susceptibility of the order of 10^–6 cm³ mol^–1 is due to the temperature independent paramagnetism.

Correct option is (d)

78. Correct order of M-C bond length of metallocenes (a–c)

(a) A > B > C

(b) B > C > A

(c) C > B > A

(d) A > C > B

Ans. b

Sol. As the electron goes to orbital of metallocene bond length increases as they are anologous eg set orbital of an octahedral complex.

Thus, in Ni(Cp)₂ there are highest number of electron in Hence, it has highest M–C bond length.

Correct option is (b)

79. A 100 mL solution of 2.5×10^–3 M in Bi(III) and Cu(II) each, is photometrically titrated at 745 nm with 0.1 M EDTA solution. Identify correct statements for this titration

(A) Total volume of EDTA solution used in 5 mL.

(B) 3 mL of EDTA is required to complex Bi(III) and 2 mL for Cu(II)

(C) 2.5 mL of EDTA is used for each metal ion

(D) First break in titraction curve is for Cu(II)

Correct statement are

(a) A and B

(b) A and C

(c) A, B and C

(d) B, C and D

Ans. b

Sol. Milimoles of each ion are 0.25 in solution.

On titrating with 0.1 MEDTA each ion will take 2.5 ml of EDTA..

So, total EDTA used = 2.5 + 2.5 = 5

Correct option is (b)

80. On continuous exposure of ¹⁰B sample to a slow neutron flux of 10¹⁶ m²s^–1, its 3% weight fraction disappears in 3 × 10⁷ s. Cross section for neutron capture (in bans) by ¹⁰B is

(a) 1000

(b) 3000

(c) 10,000

(d) 30,000

Ans. a

Sol. Neutron incident on ¹⁰B = 10¹⁶ m² sec^–1

t = 3 × 10⁷ sec

3% fraction disappearse i.e.

Correct option is (a)

81. The ¹H NMR spectrum of at 23⁰C consists of a sharp single line. The number of signals observed at low temperature (– 140⁰C) in its spectrum is

(a) 8

(b) 6

(c) 4

(d) 2

Ans. c

Sol.

Therefore, total protons 48 but in COT attached to the four carbon and all 4 protons show different environment at –140ºC because fluxonial behaviour slow down and hence, 4 signals will be observed.

Correct option is (c)

82. The g value for Ce³⁺ (4f¹) and Pre³⁺ (4f²) are, respectively

(a) 3/7 and 2/5

(b) 5/7 and 4/5

(c) 6/7 and 3/5

(d) 6/7 and 4/5

Ans. d

Sol. Lande's splitting factor g is calculated as

Correct option is (d)

83. The room temperature magnetic moment for a monomeric Cu(II) complex is greater than 1.73. This may be explained using the expression:

(a)

(b)

(c)

(d)

Ans. a

Sol. The complexes having T ground state have orbital contribution to magnetic moment however for complex having A or E ground state some time is slightly greater than is due to mixing of first excited state T (having same spin multiplicity to ground state) mixes with A or E due to spin coupling and there is calculated by

Correct option is (a)

84. The number of 3c – 2e bonds present in Al(BH₄)₃ is

(a) four

(b) three

(c) six

(d) zero

Ans. c

Sol.

Thus, number of 3c–2e^– bond is equal = 6

Correct option is (c)

85. The number of skeletal electrons present in the compounds C₂B₃H₅, C₂B₄H₆ and B₅H₉ are, respectively

(a) 10, 12 and 12

(b) 12, 14 and 14

(c) 10, 12 and 14

(d) 12, 14 and 12

Ans. b

Sol.

Each B–H unit give 2 electron. For cage bonding.

Hence, 5 (B – H) = 5 × 2 = 10 electron. Each hydrogen one electron,

Correct option is (b)

86. Identify correct statements for the EPR spectrum of VO(acac)₂ [with square pyramidal geometry at vanadium] at 77K [1(⁵¹V] = 7/2].

(A) It has two g values

(B) It has 8 lines only

(C) It has one g value

(D) It has two patterns of 8 lines each.

(a) A and D

(b) A and C

(c) B and C

(d) B and D

Ans. a

Sol.

For vanadium, I = 7/2. So, 2NI + 1 = 2(1) (7/2) + 1 = 8

So, in EPR spectrum it has two G values and it has two patterns of 8 lines each.

Correct option is (a)

87. The number of lines shown by the BH₃ part of the molecule Ph₃P. ¹¹BH₃ in the ¹H and ¹¹B NMR spectra are, respectively [I(¹¹B) = 3/2; I(³¹P)=¹/₂]

(a) 8 and 8

(b) 4 and 8

(c) 3 and 6

(d) 6 and 3

Ans. a

Sol. BH₃ part of the molecule Ph₃P. ¹¹BH₃

¹H NMR spectrum = (2N₁I + 1) × (2N₂I + 1) = (2×1×3/2 + 1) × (2×1 × ½ + 1) = 4×2 = 8

¹¹B spectrum = (2N₁I + 1) × (2N₂I + 1) = (2×3×½ + 1) × (2×1 × ½ + 1) = 4×2 = 8

Correct option is (a)

88. To record Mössbauer spectrum of Fe containing samples, a source 'X' is used. X after a nuclear transformation (Y), give γ– radiation used in Mössbauer spectroscopy.

(a)

(b)

(c) ⁵⁷Co, electron capture

(d) ⁵⁷Fe, electron capture

Ans. c

Sol. ⁵⁷Fe* can be prepared by electron capture process from radioactive

Correct option is (c)

89. Correct combination of number and size of rings present in a metal ion-porphine complex (including metal ion bearing chelate rings) is

(a) four 5 – membered and four 6 – membered

(b) two 5 – membered and four 6 – membered

(c) six 5 – membered and four 6 – membered

(d) five 5 – membered and four 6 – membered

Ans. a

Sol. Structure of m-porphyrine complex can be shown as

Here, (A) represents = five membered ring

(B) represents = six membered ring

Thus, these are four 5-membered ring and four-6-membered ring.

Correct option is (a)

90. In human body cis – platin hydrolyzes to diaqua complex and modifies the DNA structure by binding to

(a) N – atom of guanine base

(b) O – atom of cytosine base

(c) N – atom of adenine base

(d) O – atom of thymine base

Ans. a

Sol. cis-platin modifies the DNA structure by binding to N-7 nitrogen atom of guanine base.

Correct option is (a)

91. For fluxional Fe(CO)₅ (structure given below) in solution, the exchange of numbered CO groups will be between

(a) 2 and 5; 3 and 4

(b) 2 and 3; 4 and 5

(c) 2 and 3; 1 and 5

(d) 1 and 2; 4 and 5

Ans. a

Sol. During fluxional axial ligands are exchanged with equatorial ligand. Hence,

Correct option is (a)

92. In the following reaction sequence

where dtc = dithiocarbamate and tds = thiuramdisulfide.

(a) P – Et₂dtc ^–K⁺; R – Et₄tds; S – CpMo(Et₂dtc) (CO)₂

(b) P – Etdtc ^–K⁺; R – Et₃tds; S – CpMo(Et₃dtc) (CO)₂

(c) P – Et₄dtc ^–K⁺; R – Et₂tds; S – CpMo(Et₄dtc) (CO)

(d) P – Etdtc ^–K⁺; R – Ettdc; S – CpMo(Etdtc) (CO)

Ans. a

Sol.

Correct option is (a)

93. Reaction of Cr(CO)₆ with LiC₆H₅ gives A which reacts with [Me₃O][BF₄] to give B. The structures of A and B respectively, are

(a)

(b)

(c)

(d)

Ans. a

Sol.

Correct option is (a)

94. Heating a sample of results in the formation with eliminaton of 2 equivalents of CO. The Mo-Mo bond order in this reactions changes from

(a) 2 to 3

(b) 1 to 2

(c) 1 to 3

(d) 2 to 4

Ans. c

Sol. TVE = 10 + 12 + 12 = 34 = A

B = (n × 18) – A

B = (2×18)–34 = 36–34 = 2

Number of metal-metal bond

TVE = 10 + 12 + 8 = 30 = A

B = (n×18)–A = 36–30 = 6

Number of metal-metal bond

Correct option is (c)

95. A plausible intermediate involved in the self metathesis reaction of catalyzed by 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. d

Sol.

Correct option is (d)

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

(a)

(b)

(c)

(d)

Ans. c

Sol.

Correct option is (c)

98. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. a

Sol.

Correct option is (a)

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

(a)

(b)

(c)

(d)

Ans. a

Sol.

Correct option is (a)

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

(a)

(b)

(c)

(d)

Ans. a

Sol.

Correct option is (a)

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

(a)

(b)

(c)

(d)

Ans. b

Sol.

Correct option is (b)

102. The major product in the following reaction is

(a)

(b)

(c)

(d)

Ans. a

Sol.

Correct option is (a)

103. The correct reagent combination to effect the following reaction is

(a) (i) NaBH₄, CeCl₃, MeOH, 0⁰C; (ii) H₂,[Ir(COD) (py)P(Cy)₃]PF₆; (iii) Ph₃P, PhCO₂H, DEAD; (iv) LiAlH₄.

(b) (i) Li, liquid NH₃; (ii) H₂,[Ir(COD) (py)P(Cy)₃]PF₆; (iii) Ph₃P, PhCO₂H, DEAD; (iv) NaBH₄, CeCl₃, MeOH, 0⁰C

(c) (i) H₂, Pd/C; (ii) LIAlH₄, –78⁰C

(d) (i) H₂, Pd/C; (ii) Li, liquid NH₃

Ans. a

Sol.

Correct option is (a)

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

(a)

(b)

(c)

(d)

Ans. a

Sol.

Correct option is (a)

105. Structure of the intermediate A and major product B in the following reaction sequence are

(a)

(b)

(c)

(d)

Ans. a

Sol.

Correct option is (a)

106. The correct match for the following transformations P–S with the processes I-IV is

Reactions:

Processes:

(I) Diels – Alder

(II) Norrish Type – I

(III) Photocycloaddition followed by Diels – Alder

(IV) Norrish Type – II

(a) P – II; Q – IV; R – III; S – I

(b) P – II; Q – IV; R – I; S – II

(c) P – IV; Q – II; R – III; S – I

(d) P – IV; Q – II; R – I; S – III

Ans. c

Sol.

Correct option is (c)

107. The correct match for the reaction P – S with the names of cyclizations I – IV is

Reactions:

Name of Cyclization

(I) halocyclization

(II) Nazarov cyclization

(III) radical cyclization

(IV) electrocyclization

(a) P – IV; Q – I; R – II, S – II

(b) P – II; Q – I; R – IV; S – III

(c) P – IV; Q – II; R – III; S – I

(d) P – II; Q – I; R – III; S – IV

Ans. b

Sol.

Correct option is (b)

108. The correct structure of the intermediate, which leads to the product in the following reaction is

(a)

(b)

(c)

(d)

Ans. a

Sol.

Correct option is (a)

109. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. b

Sol.

Correct option is (b)

110. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. a

Sol.

Correct option is (a)

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

(a)

(b)

(c)

(d)

Ans. a

Sol.

Correct option is (a)

112. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. d

Sol.

Correct option is (d)

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

(a)

(b)

(c)

(d)

Ans. c

Sol.

Correct option is (a)

114. The correct structure of the compound, which shows following ¹³C NMR DEPT-135 data is ¹³C NMR DEPT–135: negative peaks at 30.2, 31.9, 61.8, 114.7 ppm; positive peack at 130.4 ppm

(a)

(b)

(c)

(d)

Ans. a

Sol. ¹³C NMR DEPT-135 shows positive CH₃ and CH and negative CH₂ signals.

Correct option is (a)

115. A compound displays the following spectral data. The correct structure of the compound is IR : 1690 cm^–1.

¹H NMR : 2.5 (s, 3H), 3.8 (s, 3H), 6.9 (d, J=8Hz, 2H), 7.8 (d, J = 8 Hz, 2H) ppm

¹³C NMR : 197, 165, 130, 129, 114, 56, 26 ppm

(a)

(b)

(c)

(d)

Ans. a

Sol. 6.9 (d, J = 8Hz, 2H); 7.8 (d, J = 8Hz, 2H)

Confirms the para-substitution pattern.

Since, J = 8Hz, value in aromatic region, 2 doublets of 4Hs.

Correct option is (a)

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

(a)

(b)

(c)

(d)

Ans. c

Sol.

Correct option is (a)

117. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. c

Sol.

Correct option is (c)

118. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. c

Sol.

Correct option is (c)

119. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. c

Sol.

Correct option is (c)

120. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. d

Sol.

Correct option is (d)

121. The signal – particle translational partition function (f) for an ideal gas in a fixed volume V depends on the thermal de – Broglie wavelength where

(a) n = 3

(b) n = 1

(c) n = – 1

(d) n = – 3

Ans. d

Sol. Single – particle translational partition function = de – Broglie wavelength

Correct option is (d)

122. 15 particles are distributed among 4 levels as shown in state I. Heat is given to the system and no work is done. The final state could be

(a) I

(b) III

(c) IV

(d) V

Ans. a

Sol. Since, only heat is given to system only number of particles in given states would change under the criterion of Boltzmann – Equilibrium.

Correct option is (a)

123. In an NMR spectrometer containing a 2.5T magnet, Larmor precession frequency of ¹H is 100 MHz. The radiofrequency used in this spectrometer has an associated magnetic field strength of 2.5 × 10^–4T. The duration of a 90⁰ pulse in this instrument is

(a) 25 × 10^–6 s

(b) 50 × 10^–6 s

(c) 25 ×10^–5 s

(d) 50 × 10^–5 s

Ans. a

Sol. A 90⁰ pulse width or pulse duration is the amount of time the pulse of energy is applied to the particular sample in order to flip all the spins into the X – Y plane.

The approximate field width of excitation is given by the formula, RF_field = 1 / (4*90⁰ pulse duration in sec.)

RF_field is frequency due associated field strength = 2.5 × 10^–4 × 42.57 MHz = 10642.5 Hz

Here, 42.57 MHz/T is gyromagnetic ratio for proton

Now, 1064.5 Hz = 1/(4*pulse duration in sec.)

Pulse duration = 1/(1064.5*4) sec. = 1/42570 sec. = 23.5 × 10^–6 sec.

Correct option is (a)

124. Upon application of a weak magnetic field, a line in the micirowave absorption spectrum of rigid rotor splits into 3 lines. The quantum number (J) of the rotational energy level from which the transition originates is

(a) 0

(b) 1

(c) 2

(d) 3

Ans. a

Sol. Under the section rule DM_j = 0, ± 1

Correct option is (a)

125. Phase diagram of a compound is shown below

The slpes of the lines OA, AC and AB are respectively. If melting point and of melting are 300K and 3 kJ mole^–1 respectively, the change in the volume on melting is

(a)

(b)

(c)

(d)

Ans. c

Sol.

Correct option is (c)

126. The figure below describes how a Carnot eingine works. It starts from the adiabatic compression step denoted by

(a) AB

(b) BC

(c) DC

(d) AD

Ans. b

Sol. BC is an adiabatic compression

Correct option is (b)

127. The point group obtained by adding symmetry operation to the point group C₄ is

(a) S₄

(b) C_4h

(c) D_2h

(d) D₄

Ans. b

Sol.

Correct option is (b)

128. For a particle of mass m confined in a rectangular box with sides 2a and a, the energy and degeneracy of the first excited state, respectively, are

(a)

(b)

(c)

(d)

Ans. a

Sol.

Correct option is (a)

129. The ionization energy of hydrogen atom in its ground state is approximately 1.3.6 eV. The potential energy of He⁺, in its ground state is approximately

(a) –54.4 eV

(b) –27.2 eV

(c) –13.6 eV

(d) –108.8 eV

Ans. d

Sol.

Correct option is (b)

130. The character table for the D₃ point group is provided below:

For this point group, the correct statement among the following is:

(a) it is possible to have a totally symmetric normal mode of vibration which is IR-active

(b) all IR-active normal modes are necessarily Raman inactive

(c) all Raman-active normal modes are necessarily IR-active

(d) it is possible to have a pair of IR-active normal modes that are degenerate.

Ans. d

Sol. From column '3' and column '4'

In front of 'E' irreducible representation, it is clear that there is a pair of I.R active momdes which are degenerate.

Degeneracy of 'E' = 2

Correct option is (d)

131. Suppose, are wavefunctions of an anharmonic oscillator and are wavefunctions of a harmonic oscillator with increasing order of energy. The subscripts denote vibrational quantum numbers in both the cases. Given.

the FORBIDDEN electric dipole (assumping the dipole operator is linear in normal coordinates) transition among the following is

(a)

(b)

(c)

(d)

Ans. a

Sol. are the wavefunctions of a harmonic oscillator in ground state, first excited state, second excited state respectively.

We know that for a harmonic oscillator will have even parity and will have odd parity.

Since, the dipole operator is linear, therefore electric dipole transition is forbidden if the initial and final state have same parity.

Since, have same parity.

So, transition is FORBIDDEN.

Correct option is (a)

132. If U is a function of V and T, is equal to are the internal pressure and the coefficient of thermal expansion, respectively)

(a) C_P

(b) C_V

(c)

(d)

Ans. d

Sol.

Correct option is (d)

133. The character table of C_3V point group is provided below, along with an additional reducible representation,

(a) A₁ + A₂ + 2E

(b) 2A₁ + 2E

(c) 2A₂ + 2E

(d) 2A₁ + 2A₂ + E

Ans. b

Sol.

Correct option is (b)

134. For the chemical reaction in aqueous solution,

the correct statement is

(a) increase of pressure increases the rate constant

(b) increase of dielectric constant increases the rate constant

(c) increase of ionic strength decreases the rate constant

(d) the entropy of activation is positive

Ans. b

Sol. Entropy of activation for bimolecular reaction is negative. Therefore, (d) is incorrect

Here, positive salt effect is observed. Therefore, increase of ionic strength increases the rate constant, Therefore option (c) is incorrect

Increase of Dielectric constant increases rate constant.

Correct option is (b)

135. If experimentally observed rate constant is greater than the maximum value of rate constant obtained using hard – sphere model of collision theory, then relation between the impact parameter (b) and sum of the radii of two reactants is

(a) b = r₁ + r₂

(b) b < r₁ + r₂

(c) b > r₁ + r₂

(d) b < r₁ + r₂

Ans. c

Sol. If experimental rate is greater than maximum rate using collision theory then impact parameter must be greater than sum of radii of reactants.

Correct option is (c)

136. Half – life t_1/2 for a third order reaction products, where C₀ is the initial concentration of C, will be

(a)

(b)

(c)

(d)

Ans. a

Sol. For a 3rd order reaction,

For half – life time,

Correct option is (a)

137. For a simple cubic lattice, the ratio between the unit cell length and the separation of two adjacent parallel crystal planes can NOT have a value of

(a) 5^1/2

(b) 7^1/2

(c) 11^1/2

(d) 13^1/2

Ans. b

Sol.

Correct option is (b)

138. Adsorption isotherm of three gases A, B and C are shown in the following figure, where is the percentage of surface coverage

The correct order of the extent of adsorption of these gases is

(a) A > B > C

(b) B > A > C

(c) C > A > B

(d) C > B > A

Ans. d

Sol.

Correct option is (d)

139. Choosing some Hamiltonian H and an orthonormal basis, a linear variation is carried out to get approximately energies With 2 basis function, one obtains Taking 3 basis functions, similarly three ordered energies are found. The relation which holds from the following is?

(a)

(b)

(c)

(d)

Ans. d

Sol. Correct option is (d)

140. Average value of momentum for the ground state of a particle in a 1 – d box is zero because

(a) [p, H] = 0

(b) V (potential) = 0

(c) H is hermitian

(d) the state is bound and stationary

Ans. d

Sol. For any bound state particle has equal and opposite value of momentum.

Correct option is (d)

141. For a hermitian operator A, which does NOT commute with the Hamiltonian H, let be an eigenfunction of A and be an eigenfunction of H. The correct statement regarding the average value of commutator of A with H is

(a)

(b)

(c)

(d)

Ans. d

Sol.

Correct option is (d)

142. The condensation of a hydroxy acid produces a polyster with the probability of linkage at both ends being p. The mole fraction of k–merchain formation is

(a)

(b)

(c)

(d)

Ans. c

Sol. In Flory Schulz distribution

Therefore, mole fraction of polymer with a chain length k is

Correct option is (c)

143. In simple molecular orbital theory of hydrogen molecule, bonding and anti – bonding molecular orbitals are constructed as linear combinations of atomic orbitals of two hydrogen atoms. Thes spatial part of a purely covalent singlet wavefunction is obtained by

(a)

(b)

(c)

(d)

Ans. c

Sol. Correct option is (c)

144. Two aqueous 1:1 electrolyte systems A and B are at different temperature T_A and T_B and C_A and C_B concentrations, respectively. Their Debye lengths will be equal if

(a)

(b)

(c)

(d)

Ans. a

Sol. Debye - Huckel screening legnths

Equation (i) = equation (ii)

Correction option is (a)

145. Aqueous solutions of NaCl, CaCl₂ and LaCl₃ show the following plots of logarithms of mean ionic activity coefficient vs molar concentration (c)

The correct option is then

(a) NaCl – C ; CaCl₂ – B ; LaCl₃ – C

(b) NaCl – A ; CaCl₂ – B ; LaCl₃ – C

(c) NaCl – A ; CaCl₂ – C ; LaCl₃ – B

(d) NaCl – C ; CaCl₂ – A ; LaCl₃ – B

Ans. b

Sol.

For same concentration of electrolyte will be smallest for NaCl and largest for LaCl₃. As negative sign is there so slope of NaCl will be least negative and slope of LaCl₃ will be most negative.

Correct option is (b)