21.    The HOMO (highest occupied molecular orbital) to LUMO (lowest unoccupied molecular orbital) electronic transition responsible for the observed colours of halogen molecules (gas) is

(a)

(b)

(c)

(d)

Ans. a

Sol. In halogen molecules, thetransitions are responsible for their colours as evident from the following figure

Correct option is (a)

22.    In the hydrolysis to transif the leaving group is chloride, the formation of cis product is the least, when A is

(a)

(b)

(c)

(d)

Ans. a

Sol. The -donor ligands such as Cl^–, Br^–, I^–, N-bonded NCS etc. stabilize the TBP intermediate and favour the stereochemical change. The trans-forms tendency to isomers in the order

Correct option is (a)

23.    The expected number of ¹⁹F NMR spectral lines, including satellites, for is [Abundance of

(a) two

(b) twenty one

(c) three

(d) one

Ans. c

Sol.

Pentagonal Planar

as 74% Xe are NMR inactive hence a singlet of five fluorine and due to 26% nuclei a doublet of satellite.

Hence, total number of signal is 3.

Correct option is (c)

24.    The expected H–H–H bond angle in is

(a) 180⁰

(b) 120⁰

(c) 60⁰

(d) 90⁰

Ans. c

Sol.

Each bond angle is 60º.

Correct option is (c)

25.    The number of bridging ligand (s) and metal – metal bond(s) present in the complex(obeys 18 – electron rule), respectively, are

(a) 0 and 1

(b) 2 and 1

(c) 3 and 1

(d) 1 and 2

Ans. c

Sol. T.V.E. = 8 × 2 + 5 × 2 + 2 × 2 + 4 = 34 = A

B = (n × 18) – A = 36 – 34 = 2

Correct option is (c)

26. The oxidation state of gold in the following complex is

(a) 0

(b) 1

(c) 2

(d) 3

Ans. c

Sol. Oxidation state of gold in the complex is 2

Correct option is (c)

27.    The rate of alkene coordination to [PtCl₄]^2– is highest for

(a) norbornene

(b) ethylene

(c) cyclohexene

(d) 1-butene

Ans. a

Sol. When the norbornene is strained molecule because the hybridization is sp² when the metal attached with norbornene back -bonding takes place and due to this back -bonding the hyberdization changes to sp² – sp³ and hence relief to strain

Correct option is (a)

28.    The nephelauxetic parameteris highest for

(a) Br^–

(b) Cl^–

(c) CN^–

(d) F^–

Ans. d

Sol. According to Nephelauxetic series

F^–> H₂O > NH₃>en> NCS^–> Cl^– ~ CN^–> Br^–

The highest value of is for F^–

Correct option is (d)

29.    Thetransition in the electronic spectrum of [Cr(NH₃)₆]³⁺ occurs nearly at

(a) 650 nm

(b) 450 nm

(c) 350 nm

(d) 200 nm

Ans. a

Sol. Thetransition in the electronic spectrum of [Cr(NH₃)₆]³⁺ occurs nearly at 650 nm.

Correct option is (a)

30.    In the catalytic hydration of CO₂ by carbonic anhydrase, CO₂ first interacts with

(a) OH group of the active site of the enzyme and then with zinc

(b) H₂O of the active site of the enzyme and then with zinc

(c) zinc of the active site of the enzyme and then with OH group

(d) zinc of the active site of the enzyme and then with H₂O

Ans. a

Sol. In carbonic anhydrase OH group first attack on CO₂ and then CO₂intract with Zn.

Correct option is (a)

31. For the reaction,

The highest value of , when X^– is

(a) OCl^–

(b) F^–

(c) Cl^–

(d)

Ans. c

Sol. Due to more acidity of HCl it will undergo more ionization. Hence, concentration of [Cl^–]_aq will be maximum.

Correct option is (c)

32.    The correct statement for d.c. polarography is

(a) E_1/2 is concentration dependent

(b) Dropping mercury electrode is a macro electrode

(c) Limiting current is equal to diffusion current

(d) A large excess of supporting electrolyte eliminates migration current

Ans. d

Sol. In d.c. polarography supporting electrolyte is taken in excess, so that all migration current is carried by supporting electrolyte. Thus electroactive species remain free from migration current

1. E_1/2 does not depend on concentration of electroactive species.

2. DME is a micro electrode.

3. limiting current is sum of diffusion current and residual current

i_l = i_d + i_r

Correct option is (d)

33. Saturation factor in neutron activation analysis is

(A = induced radioactivity; = neutron flux; = effective nuclear cross section; N = no of target atoms; = decay constant)

(a)

(b)

(c)

(d)

Ans. a

Sol. Saturation factor in neutron activation analysis.

Activity equation,

saturation factor

Saturation factor

Where, A = number of decays per second

N = number of atoms of the target isotopes

= activation cross section

= neutron flux

= decay constant

t_irr = irradiation time

Correct option is (a)

34.    The primary analytical method (not using a reference) is

(a) inductively coupled plasma emission spectrometry

(b) energy dispersive X-ray fluorescence spectrometry

(c) anodic stripping voltammetry

(d) isotopic dilution mass spectrometry

Ans. d

Sol. Primary analytical method is isotopic dilution mass spectrometry (not using a reference).

Correct option is (d)

35.    The number of inorganic sulphur (or sulphide) atoms present in the metalloprotein active sites of rubredoxin, 2-iron ferredoxin and 4iron ferredoxin, respectively, are

(a) 0, 2 and 4

(b) 2, 4 and 3

(c) 0, 4 and 2

(d) 0, 2 and 3

Ans. a

Sol.

Correct option is (a)

36.    The metal iodide with metallic lustre and high electrical conductivity is

(a) NaI

(b) CdI₂

(c) LaI₂

(d) BiI₃

Ans. c

Sol. LaI₂ exists as La³⁺(2I^–)(e^–)

The electron present is responsible for metallic cluster and high electrical conductivity.

Correct option is (c)

37. The correct order of the bond dissociation energies for the indicated C-H bond in following compounds is

(a) C > B > A

(b) A > B > C

(c) A > C > B

(d) C > A > B

Ans. d

Sol.

More s% character, stronger will be the bond.

More strong bond required more bond dissociation energy. So, the correct order of bond dissociation energy, C > B > A.

Correct option is (d)

38. The correct order of the acidity for the following compounds is

(a) B > C > A

(b) C > B > A

(c) B > A > C

(d) C > A > B

Ans. c

Sol. Cyclic are more acidic than open chain

As the ring decreases, acidity increases

(B) > (A) > (C) Least acidic

Correct option is (c)

39. The correct statement about the following compound is

(a) compound is chiral and has P configuration

(b) compound is chiral and has M configuration

(c) compound is achiral as it possesses C₂-axis of symmetry

(d) compound is achiral as it possesses plane of symmetry

Ans. a

Sol.

Helical chirality, P-configurations. Since, clockwise path from front side.

Correct option is (a)

40. Methyl groups in the following compound are

(a) homotopic

(b) diasterotopic

(c) enantiotopic

(d) constitutionally heterotopic

Ans. a

Sol.

Correct option is (a)

41. Among the structures given below, the most stable conformation for the following compound is

(a)

(b)

(c)

(d)

Ans. c

Sol.

All three methyl groups on same side (above). But in option (a) : 1, 3-diaxial interaction occurs.

So, it is less stable conformation. It undergoes ring flipping to give more stable conformation as shown in option (c)

Correct option is (c).

42. Molecular orbital interactions involved in the first step of the following reaction is

(a)

(b)

(c)

(d)

Ans. a

Sol.

Correct option is (a)

43. The major product formed in the dinitration of 4-bromotoluene is

(a)

(b)

(c)

(d)

Ans. b

Sol.

Correct option is (b)

44.    The correct order of the rate constants for the following series of reactions (Z = CF₃/CH₃/OCH₃) is

(a) CF₃> CH₃> OCH₃

(b) CF₃> OCH₃> CH₃

(c) OCH₃> CF₃> CH₃

(d) CH₃> OCH₃> CF₃

Ans. a

Sol.

z = more electron withdrawing group, stabalize the carbanion, more will be rate constant.

Electron Withdrawing Effect : CF₃> CH₃>OMe

Correct option is (a)

45.    ¹H NMR spectrum of a mixture of benzene and acetonitrile shows two singlets of equal integration. The molar ratio of benzene: acetonitrile is

(a) 1 : 1

(b) 2 : 1

(c) 1 : 2

(d) 6 : 1

Ans. c

Sol. C₆H₆ : CH₃ – CN

6H 3H

1 equivalent : 2 equivalent

6H 6H

1 : 1 (Integration ratio)

Correct option is (c)

46.    The compound which shows IR frequencies at both 3314 and 2126 cm^–1 is

(a) CH₃(CH₂)₄CH₂SH

(b) CH₃(CH₂)₄CH₂C=N

(c) CH₃(CH₂)₄CH₂C = C – H

(d) CH₃(CH₂)₂C = C(CH₂)₂CH₃

Ans. c

Sol.

Correct option is (c)

47.    Number of signals present in the proton decoupled ¹³C NMR spectrum of the following compound is

(a) four

(b) six

(c) eight

(d) ten

Ans. a

Sol.

Molecule have plane of symmetry. So, in ¹³C NMR four signal.

Correct option is (a)

48. The most stable product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. a

Sol.

Correct option is (a)

49. The major product in the following reaction is

(a)

(b)

(c)

(d)

Ans. b

Sol.

Correct option is (b)

50. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. d

Sol.

Simmons Smith reaction.

Simmons Smith reagent reacts on isolated double bond, to form a cyclopropane.

Correct option is (d)

51.    Correct characteristics of the functional groups of adenine in DNA base pair are

(a) N(3) is a hydrogen bond acceptor and C(6)NH₂ is a hydrogen bond donor

(b) N(1) is a hydrogen bond acceptor and C(6)NH₂ is a hydrogen bond donor

(c) Both N(3) and C(6)NH₂ are hydrogen bond acceptors

(d) Both N(1) and C(6)NH₂ are hydrogen bond acceptors

Ans. b

Sol.

Correct option is (b)

52.    ¹H NMR spectrum of an organic compound recorded on a 500 MHz spectrometer showed a quartet with line positions at 1759, 1753, 1747, 1741 Hz. Chemical shift and coupling constant (Hz) of the quartet are

(a) 3.5 ppm, 6 Hz

(b) 3.5 ppm, 12 Hz

(c) 3.6 ppm, 6 Hz

(d) 3.6 ppm, 12 Hz

Ans. a

Sol.

Chemical shift

Coupling constant J = 1747 – 1741 = 6 Hz of difference between any two adjacent peaks (coupling constant is not change in Hertz).

Correct option is (a)

53.    The weight of the configuration with two up and three down spins in a system with five spinparticles is

(a) 120

(b) 60

(c) 20

(d) 10

Ans. d

Sol.

Correct option is (d)

54.    For a reaction with an activation energy of 49.8 kJ mol^–1, the ratio of the rate constants at 600 K and 300 K, (k₆₀₀/k₃₀₀), is approximately (R = 8.3 J mol^–1 K^–1)

(a) ln (10)

(b) 10

(c) 10 + e

(d) e¹⁰

Ans. d

Sol.

Correct option is (d)

55.    Covariance is defined by the relationGiven the arbitrary constants A, B andwill be zero only when

(a)

(b)

(c)

(d)

Ans. c

Sol. Covariance shows the tendency in the linear relationship between the variables.

Correct option is (c)

56.    Each void in a two dimensional hexagonal close-packed layer of circles is surrounded by

(a) six circles

(b) three circles

(c) four circles

(d) twelve circles

Ans. b

Sol. From figure it is clear that each void is surrounded by three circles.

Correct option is (b)

57.    The ionic mobilities ofrespectively. The transport numbers ofare, respectively

(a) 0.545 and 0.455

(b) 0.455 and 0.545

(c) 0.90 and 0.910

(d) 0.910 and 0.090

Ans. a

Sol.

Correct option is (a)

58.    The ionic strength of a solution containing 0.008 M AlCl3 and 0.005 M KCl is

(a) 0.134 M

(b) 0.053 M

(c) 0.106 M

(d) 0.086 M

Ans. b

Sol.

Correct option is (b)

59.    The correct normalized wavefunction for one of the sp² hybrid orbitals is

(a)

(b)

(c)

(d)

Ans. c

Sol. For normalised wave function,

Correct option is (c)

60.    The correct statement in the context of NMR spectroscopy is

(a) static magnetic field is used to induce transition between the spin states

(b) magnetization vector is perpendicular to the applied static magnetic field

(c) the static magnetic field is used to create population difference between the spin states

(d) static magnetic field induces spin-spin coupling

Ans. c

Sol. In NMR spectroscopy the static magnetic field is used to create population difference between the spin states

Correct option is (c)

61.    The parameter which always decreases during a spontaneous process at constant S and V, is

(a) U

(b) H

(c) C_p

(d) q

Ans. a

Sol. dU = +TdS – PdV

If S & V are constant, then equation (1) becomes

(dU)_{S, V} < 0

Criteria of spontaniety.

Correct option is (a)

62.    Triple point pressure of substances A, B, C and D are 0.2, 0.5, 0.8 and 1.2 bar, respectively. The substance which sublimes under standard conditions on increasing temperature is

(a) A

(b) B

(c) C

(d) D

Ans. d

Sol. Standard condition means 1 bar pressure

For A

So at 1 bar transition will be solidliquid. So number sublimation

For B

Same explanation as above

Correct option is (d)

63.    According to the transition state theory, the plot with slope equal to is

(a) ln k vs. T

(b)

(c)

(d)

Ans. c

Sol.

Correct option is (c)

64.    The transition that belongs to the Lyman series in the hydrogen-atom spectrum is

(a)

(b)

(c)

(d)

Ans. b

Sol. For allowed transition,and for Lyman Series Transition is from higher level to n = 1

Correct option is (b)

65.    The molecule that possesses S₄ symmetry element is

(a) ethylene

(b) allene

(c) benzene

(d) 1, 3 – butadiene

Ans. b

Sol.

Correct option is (b)

66.    Vibrations of diatomic molecules are usually modelled by a harmonic potential. If the? potential is given by x², the correct statement is

(a) force is 2x and force constant is 2

(b) force is –2x and force constant is 2

(c) force is 2x and force constant is –1

(d) force is –2x and force constant is –1

Ans. b

Sol. Given: V = x²

Correct option is (b)

67.    When 1 × 10^–5 of a fatty acid (M = 602.3 g/mol) was placed on water as a surface film, a monomolecular layer of area 100 cm² was formed on compression. The cross-sectional area (in Ų) of the acid molecule is

(a) 50

(b) 100

(c) 150

(d) 200

Ans. b

Sol. Total surface area = Number of molecules × Area covered by 1 mole (A₁)

Correct answer is (b)

68.    Mark – Gouwink equation is used for the determination of

(a) number-average molar mass

(b) weight-average molar mass

(c) viscosity-average molar mass

(d) z-average molar mass

Ans. c

Sol.

Correct answer is (c)

69.    Many properties of nanoparticles are significantly different than the corresponding bulk material due to

(a) smaller band gap of nanoparticles compared to bulk

(b) higher heterogeneity of the nanoparticle solutions

(c) larger ratio of surface area to volume of the nanoparticles compared to the bulk

(d) smaller ratio of surface?area to volume of the nanoparticles compared to the bulk

Ans. c

Sol. Many properties of nano-particles are significantly different than the corresponding bulk material because large ratio of surface area to volume of nano-particles in compare to the bulk.

Correct option is (c)

70. The correct match for the following is

(a) (i) – (A); (ii) – (C); (iii) – (E)

(b) (i) – (E); (ii) – (B); (iii) – (A)

(c) (i) – (D); (ii) – (C); (iii) – (A)

(d) (i) – (E); (ii) – (B); (iii) – (D)

Ans. b

Sol. Camphor Terpene

Insulin Hormone

Keratin Structural Protein

Correct option is (b)

71.    Consider the following statements for KC₈:

(A) It is paramagnetic,

(B) It has eclipsed layer structure,

(C) Its electrical conductivity is greater than that of graphite. The correct answer is

(a) A and B

(b) A and C

(c) B and C

(d) A, B and C

Ans. d

Sol. 8C(graphite) + K⁺(am) + e^–(am) [K (am)]⁺ [C8]^– (s)

to the presence of unpaired electrons in band of graphite it is paramagnetic and its electrical conductance increases. It has eclipsed layer structure

Correct option is (d)

72.    Among the following, choose the correct products that are formed in the reaction of S₂Cl₂ with ammonia in CCl₄:

NH₄Cl (A), S₄N₄ (B), S_g (C), and S₃N₃Cl₃ (D).

(a) A, B and C

(b) A, B and D

(c) B, C, and D

(d) A, C and D

Ans. a

Sol.

Correct option is (a)

73.    For [Ce(NO₃)₄(OPPh₃)₂], from the following

A. Its aqueous solution is yellow-orange in colour

B. Coordination number of Ce is ten

C. It shows metal to ligand charge transfer

D. It is diamagnetic in nature

the correct answer is

(a) A and B

(b) A and C

(c) A, B and D

(d) B, C and D

Ans. c

Sol.

N_D unpaired electrons.

It is diamagnetic and its colour is due to LMCT. Its C.N. = 10

Correct option is (c)

74. Consider the following statements, I and II:

I: [Rh(CO)₂I₂]^– catalytically converts CH₃I and CO to CH₃COI

II: [Rh(CO)₂I₂]^– is diamagnetic in nature

The correct from the following is

(a) I and II are correct and II is an explanation of I

(b) I and II are correct and II is not an explanation of I

(c) I is correct and II is incorrect

(d) I and II are incorrect

Ans. b

Sol.

I and II are correct but II is not correct explanation of I.

Correct explanation : Monsanto process occurs via oxidative reaction, migratory insertion and reductive elimination reaction for that the starting compound must be square planar.

Correct option is (b)

75.    In a direct isotopic dilution method for determination of phosphate, 2 mg of ³²PO₄^3– (specific activity 3100 disintegration s^–1 mg^–1) was added to 1g of a sample solution. The 30 mg of phosphate isolated from it has an overall activity of 3000 disintegration s^–1. The % mass of PO₄^3– in the sample is

(a) 30

(b) 6

(c) 9

(d) 15

Ans. b

Sol. In 30 mg sample the overall activity is 3000 s^–1. So, mass of ¹²PO₄^–2 is

Correct answer is (b)

76.    Consider the following statements for [FeO₄]^4–.

A. It is paramagnetic

B. It has T_d symmetry

C. Adopts distorted square planar geometry

D. Shows approximately D2d symmetry

The correct answer is

(a) A, B and C

(b) A, C and D

(c) A and D

(d) A and B

Ans. c

Sol. Jahn Teller distorted away from the ideal tetrahedral towards a flatterred structure

It is paramagnetic due to unpaired electron D_2d symmetry.

Correct option is (c)

77.    The geometry of [ReH₉]^2– is

(a) monocapped square antiprism

(b) monocapped cube

(c) tricapped trigonal prism

(d) heptagonal bipyramid

Ans. c

Sol. Tricapped trigonal prismatic

Atoms 1 to 6 are the prism atoms, 7 to 9 are H-atom are equatorial hydrogen.

Correct option is (c)

78.    The reaction between PI₃, PSCl₃ and zinc powder gives P₃I₅ as one of the products. The solution state ³¹P NMR spectrum of P₃I₅ shows a doublet and a triplet The correct structure of P₃I₅ is

(a)

(b)

(c)

(d)

Ans. c

Sol. Correct structure is

. One c₂ axis and one plane of symmetry exist in this molecules.

So, two types of phosphorous atoms

Correct option is (c)

79. Some molecules and their properties in liquid ammonia are given in columns A and B respectively. Match column A with column B

The correct match is

(a) (A) – (i); (B) – (ii); (C) – (iii); (D) – (iv)

(b) (A) – (ii); (B) – (iii); (C) – (iv); (D) – (i)

(c) (A) – (iii); (B) – (iv); (C) – (i); (D) – (ii)

(d) (A) – (iv); (B) – (iii); (C) – (ii); (D) – (i)

Ans. d

Sol. solvolysis and disproportionation of Cl₂ in NH₃

undergoes disproportionation in liquid NH₃

CH₃COOH acts as strong acid in NH₃

Molecules that do not behave as acid in water, may behave as weak acids in NH₃.

Correct option is (d)

80.    The spectroscopic ground state term symbols for the octahedral aqua complexes of Mn(II), Cr(III) and Cu(II), respectively, are

(a) ²H, ⁴F and ²D

(b) ⁶S, ⁴F and ²D

(c) ²H, ²H and ²D

(d) ⁶S, ⁴F and ²P

Ans. b

Sol. is a weak ligand, No pairing occurs.

Correct answer is (b)

81. From the following transformations,

A. Epoxidation of alkene

B. Diol dehydrase reaction

C. Conversion of ribonucleotide – to – deoxyribonucleotide

D. 1, 2 – carbon shift in organic substrates

those promoted by coenzyme B₁₂ are

(a) A and B

(b) B, C and D

(c) A, B and D

(d) A, B and C

Ans. b

Sol. Co-enzyme B₁₂ catalyses dehydration, 1, 2-carbon shift reaction.

Hence, Correct option is (b)

82. Match the items in column A with the appropriate items in column B

The correct answer is:

(a) (A) – (ii); (B) – (iii); (C) – (v); (D) – (iv)

(b) (A) – (ii); (B) – (iii); (C) – (iv); (D) – (i)

(c) (A) – (ii); (B) – (iii); (C) – (v); (D) – (vi)

(d) (A) – (iii); (B) – (v); (C) – (vi); (D) – (ii)

Ans. c

Sol. Metallothionines Cystein rich protein

Plastocyanin Electron transfer

Ferritin Iron storage

Chemotherapy Carboplatin

Correct option is (c)

83.    For OH^– catalysed S_N1 conjugate base mechanism of [Co(NH₃)₅Cl]²⁺, the species obtained in the first step of the reaction is/are T

(a) [Co(NH₃)₅(OH)]²⁺ + Cl^–

(b) [Co(NH₃)₄(NH₂)Cl]⁺ + H₂O

(c) [Co(NH₃)₄(NH₂)]²⁺ + Cl^–

(d) [Co(NH₃)₅Cl(OH)]⁺ only

Ans. b

Sol.

Correct answer is (b)

84. Match the species in column X with their properties in column Y

Column X Column Y

(a) Heme A (i) oxo-bridged Mn₄ cluster

(b) water spliting enzyme (ii) tetragonal elongation

(c) [Mn(H₂O)₆]²⁺ (iii) predominantly electronic transitions

(d) [Cr(H₂O)₆]²⁺ (iv) d d spinforbidden transitions

(v) tetragonal compression

Ans. b

Sol. Heme A Iron porphyrin Colour due to water splitting enzyme contains transition 0 × 0 bridged Mn₄ cluster

Correct option is (b)

85.    According to isolobal analogy, the right set of fragments that might replace Co(CO)₃ in [Co₄(CO)₁₂] is

(a) CH, BH and Mn(CO)₅

(b) P, CH and

(c) Fe(CO)₄, CH₂ and SiCH₃

(d) BH, SiCH₃ and P

Ans. b

Sol. Co(CO)₃

9 + 6 = 15 (for 18 electron require 3 electron)

Correct answer is (b)

86.    According to Wade's rules, the correct structural types of and are

(a) closo and nido

(b) nido and arachno

(c) closo and aracarachno

(d) nido and nido

Ans. b

Sol.

Correct answer is (b)

87.    The correct geometry of

(a) octahedron

(b) pentagonal pyramid

(c) trigonal prism

(d) monocapped square pyramid

Ans. c

Sol.

TVE = 9 × 6 + 4 + 15 × 2 + 2 = 54 + 4 + 2 + 30 = 90

Hence, it has trigonal prism geometry.

Correct option is (c)

88.    The final product(s) of the reaction of arachno borane, B₄H₁₀ with NMe₃ is/are

(a)

(b)

(c)

(d) 4

Ans. a

Sol.

Correct answer is (a)

89. Product A in the following reaction is

(a)

(b)

(c)

(d)

Ans. c

Sol.

Correct answer is (c)

90.    Treatment of Fe(CO)₅ with 1, 3 – butadiene gives B that shows two signals in its ¹H NMR spectrum. B on treatment with HCl yields C which shows four signals in its ¹H NMR spectrum. The compound C is

(a)

(b)

(c)

(d)

Ans. b

Sol.

Compound (C) has show four ¹H NMR signal

Correct option is (b)

91.    In the following redox reaction with an equilibrium constant K = 2.0 × 10⁸,

the self exchange rates for oxidant and reductant are 5.0 M^–1s^–1 and 4.0 × 10³ M^–1s^–1, respectively. The approximate rate constant (M^–1s^–1) for the reaction is

(a) 3.16 × 10⁶

(b) 2.0 × 10⁶

(c) 6.32 × 10⁶

(d) 3.16 × 10⁴

Ans. b

Sol. k₁₂ = (k₁₁K k₂₂f)^1/2 = (5 × 4 × 10³ × 2 × 10⁸ × 1.0)^1/2 = 20 × 10⁵ = 2.0 × 10⁶

Correct option is (b)

92.    The correct statement for a Fischer carbene complex is

(a) the carbene carbon is electrophilic in nature

(b) metal exists in high oxidation state

(c) metal fragment and carbene are in the triplet states

(d) CO ligands destabilize the complex

Ans. a

Sol. Fischer carbene carbon is electrophilic in nature due to strong -acceptor ligands

Correct option is (a)

93.    The acidic solution containing trimethylamine (A), dimethylamine (B) and methyl amine (C) (pK_a of cations 9.8, 10.8 and 10.6, respectively) was loaded on a cation exchange column. The order of their elution with a gradient of increasing pH > 7 is

(a) A < C < B

(b) B < C < A

(c) B < A < C

(d) C < B < A

Ans. a

Sol. (A) trimethylamine pK_a 9.8

(B) dimethylamine pK_a 10.8

(C) methyl amine pK_a 10.6

pH > 7 is a basic pH and amine with lower pK_a value will give the proton easily, and will elute at last.

Correct option is (a)

94. For complex A, deuterationof NH protons does not alter the EPR spectrum. The number of hyperfine lines expected in the EPR spectrum of A is

(a) 20

(b) 12

(c) 60

(d) 36

Ans. a

Sol. Number of hyperfine line (due to Cu and nitrogen)

Correct option is (a)

95.    The numbers of triangular faces in square antiprism, icosahedron and tricapped trigonal prism (capped on square faces), respectively, are

(a) 8, 20 and 14

(b) 8, 20 and 12

(c) 10, 12 and 14

(d) 10, 12 and 12

Ans. a

Sol. Number of triangular anti-prism are 8, in icosahedron it is 20 and in tricapped trigonal prism capped on square faces) it is 14

Correct answer is (a)

96.    Number of lines in the ¹⁹F NMR spectrum of F₂C(Br) – C(Br)Cl₂ at – 120⁰C assumping it a mixture of static conformations given below, are

(a) One

(b) Two

(c) Four

(d) five

Ans. d

Sol.

Both fluorine in same environment.

Correct answer is (d)

97. The correct statement for the reactants A, B to give products C, D is

(a) A gives C and B give D

(b) A gives D and B gives C

(c) A and B give identical amounts of C and D

(d) A and B give D

Ans. c

Sol.

Correct answer is (c)

98. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. c

Sol.

Correct answer is (c)

99. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. b

Sol.

Correct answer is (b)

100. The compound that exhibits following spectral data is

(a)

(b)

(c)

(d)

Ans. c

Sol.

Spectral edata confirms the Para-pattern, so option (a) and (b) ruled out. Options (c) and (d) may be correct but in option (d), the methyl are non-equivalent due to resonating structure, exhibit different signal. But in option (c), two methyl group are chemical equivalent and exhibits a singlet of 6-Hs.

Correct option is (c)

101. The major product in the following reaction is

(a)

(b)

(c)

(d)

Ans. d

Sol.

102. The major product formed in the following reaction is

NIS : N – iodosuccinimide

(a)

(b)

(c)

(d)

Ans. b

Sol.

Correct answer is(b)

103. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. b

Sol.

Correct option is (b)

104. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. a

Sol.

Correct answer is (a)

105. Correct sequence of reagents for the following conversion is

(a) (i) K₂CO₃, (ii) HC CCOCH₃, (iii) Br₂, (iv) NaBH₄

(b) (i) NaBH₄, (ii) HC CCOCH₃, (iii) Br₂, (iv) K₂CO₃

(c) (i) HC CCOCH₃, (ii) K₂CO₃, (iii) Br₂, (iv) NaBH₄

(d) (i) Br₂, (ii) HC CCOCH₃, (iii) K₂CO₃, (iv) NaBH₄

Ans. a

Sol.

Correct option is (a)

106. The major product in the following reaction is

(a)

(b)

(c)

(d)

Ans. b

Sol. This is Julia Olefinations reactions, under the given set of reaction E-alkene is formed. The detail mechanism is shown below.

Correct answer is (b)

107. For the four reactions given below, the rates of the reactions will vary as

(a) 1 > 2 and 3 > 4

(b) 2 > 1 and 3 > 4

(c) 2 > 1 and 4 > 3

(d) 1 > 2 and 4 > 3

Ans. d

Sol.

–R effect of this group, increase stability of intermediate.

In that case electron withdrawing group increase the rate of reaction.

EWG containing carbocation intermediate decrease the stability.

Correct answer is (d)

108. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. a

Sol.

Correct answer is (a)

109. The correct sequence of pericyclic reactions involved in the following transformation is

(a) (i) ene reaction, (ii) [2, 3]-sigmatropic shift, (iii) [3, 3]-sigmatropic shift

(b) (i) ene reaction, (ii) [3, 3]-sigmatropic shift, (iii) [1, 3]-sigmatropic shift

(c) (i) [2, 3]-sigmatropic shift, (ii) ene reaction, (iii) [1, 3]-sigmatropic shift

(d) (i) [1, 3]-sigmatropic shift, (ii) [2, 3] sigmatropic shift, (iii) [3, 3]-sigmatropic shift

Ans. a

Sol.

Correct answer is (a)

110. The intermediate that leads to the product in the following transformation is

(a)

(b)

(c)

(d)

Ans. b

Sol.

Correct answer is (b)

111. Product(s) of the following reaction is (are) [* – indicates isotopically labelled carbon]

(a)

(b)

(c)

(d)

Ans. c

Sol.

Correct answer is (c)

112. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. a

Sol. Coupling reaction

Correct option is (a)

113. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. d

Sol.

Correct option is (d)

114. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. d

Sol.

Correct option is (d)

115. Correct match for the products of the reactions in Column–A with the properties in Column – B is

(a) (i) – P, (ii) – S, (iii) – R, (iv) – Q

(b) (i) – P, (ii) – R, (iii) – Q, (iv) – S

(c) (i) – Q, (ii) – R, (iii) – S, (iv) – P

(d) (i) – S, (ii) – Q, (iii) – R, (iv) – P

Ans. a

Sol.

Correct option is (a)

116. The correct starting compound A in the following reaction is

(a)

(b)

(c)

(d)

Ans. b

Sol.

Correct answer is (b)

117. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. b

Sol.

Correct option is (b)

118. The major product formed in the following reaction is

.

(a)

(b)

(c)

(d)

Ans. a

Sol. Barton reaction:

Correct option is (a)

119. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. b

Sol.

Correct answer is (b)

120. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. d

Sol. DDQ used for deprotection of benzylether through one electron oxident process. As the number of OMe increases on the aryl ring, reactivity towards deprotection increases. So,

Correct answer is (d)

121.    A constant of motion of hydrogen atom in the presence of spin-orbit coupling is

(a)

(b) s

(c)

(d)

Ans. c

Sol.

For constant of motion C, we must have

Correct option is (c)

122.    The orbital degeneracy of the level of a oneelectron atomic system with Z = 5 and energy is

(a) 1

(b) 5

(c) 25

(d) 36

Ans. c

Sol.

Therefore, orbtial degeneracy = n² = 25.

Correct option is (c)

123.    If we write a normalized wavefunction is also normalized when

(a) is hermitian

(b) is anti-hermitian

(c) is unitary

(d) is any linear operator

Ans. c

Sol.

124.    The ground state of a certain system with energy is subjected to a perturbation V, yielding a first-order correction If E₀ is the true ground – state energy of the perturbed system, the inequality that always holds is

(a)

(b)

(c)

(d)

Ans. *

Sol. According to Perturbation theory, "The energy of perturbed system is equal or greater than the standard system."

Correct option is (*)

125.    The spatial part of an excited state b of hydrogen molecule is proportional to Using LCAO – MO expansion of in terms of ls – atomic orbitals, one can infer that this wavefunction has

(a) only ionic parts

(b) only covalent parts

(c) both ionic and covalent parts

(d) neither ionic nor covalent parts

Ans. b

Sol.

Correct option is (b)

126.    The highest molecular orbitals for an excited electronic configuration of the oxygen molecule are A possible molecular term symbol for oxygen with this electronic configuration is

(a)

(b)

(c)

(d)

Ans. a

Sol. The arrangements for configurations are

Thus one of possible molecular term symbol is

Correct answer is(a)

127.    For H₂O molecule, the electronic transition from the ground state to an excited state of B₁ symmetry is

(a) not allowed

(b) allowed with polarisation

(c) allowed with polarisation

(d) allowed with polarisation

Ans. b

Sol.

128.    The pair of symmetry point groups that are associated with only polar molecules is

(a)

(b)

(c)

(d)

Ans. d

Sol. The polar point group are

Correct option is (d)

129.    The rotational constant and the fundamental vibrational frequency of HBr are, respectively, 10 cm^–1 and 2000 cm^–1. The corresponding values for DBR approximately are

(a) 20 cm^–1 and 2000 cm^–1

(b) 10 cm^–1 and 1410 cm^–1

(c) 5 cm^–1 and 2000 cm^–1

(d) 5 cm^–1 and 1410 cm^–1

Ans. d

Sol.

Correct answer is (d)

130.    Among the following, both microwave and rotational Raman active molecule is

(a) CH₄

(b) N₂O

(c) C₂H₄

(d) CO₂

Ans. b

Sol. N₂O is microwave and rotational Raman active molecule.

Correct option is (b)

131.    In a 200 MHz NMR spectrometer, a molecule shows two doublets separated by 2 ppm. The observed coupling constant is 10 Hz. The separation between these two signals and the coupling constant in a 600 MHz spectrometer will be, respectively

(a) 600 Hz and 30 Hz

(b) 1200 Hz and 30 Hz

(c) 600 Hz and 10 Hz

(d) 1200 Hz and 10 Hz

Ans. d

Sol.

So, at 600 MHz, 2ppm = 1200 Hz. Hence, correct option is (d)

132.    The equation of state for one mole of gas is given by P(V – b) = RT, where b and R and constants. The value of

(a) V – b

(b) b

(c) 0

(d)

Ans. b

Sol. dH = TdS + UdP

This is T.E.S. II

Now for given gas

Constant P on differentiating with respect to T

Correct option is (b)

133.    The volume change in a phase transition is zero. From this, we may infer that the phase boundary is represented by

(a)

(b)

(c)

(d)

Ans. a

Sol. According to Clapyeron equation

Correct option is (a)

134.    The partial derivative is equal to

(a)

(b)

(c)

(d)

Ans. a

Sol.

Now, according to reciprocal theorem of partial derivatives

Putting these values in equation (1), we get

Correct option is (a)

135.    If the energies of a bare proton aligned alonge and against an external static magnetic field (B_z) are respectively, then the ratio of probabilities of finding the proton along and against the magnetic field is

(a)

(b)

(c)

(d)

Ans. d

Sol.

Rate of probability of finding the proton along end against magnetic field

Correct option is (d)

136.    Partition function of a one – dimensional oscillator having equispaced energy levels with energy spacing equal to k_BT and zero ground state energy is

(a) e

(b)

(c)

(d)

Ans. c

Sol. Since, zero point energy = 0

Correct option is (c)

137. A reaction goes through the following elementary steps

Assuming that steady state approximation can be applied to C, on doubling the concentration of A, the rate of production of D will increase by (assuming k₂[A]<<k_–1[C])

(a) 2 times

(b) 4 times

(c) 8 times

(d)

Ans. d

Sol. Rate of production of D

Applying SSA on 'C',

Correct option is (d)

138. The rate of an acid-catalyzed reaction in aqueous solution follows the rate equation

If k₁₆ and k₄ are rate constants for the reactions at ionic strength of 16 mo1L^–1 and 4 molL^–1, respectively, in terms of Debye – Hückel constant (B = 0.51), is

(a) 4B

(b) 8B

(c) 10B

(d) 12B

Ans. d

Sol.

Correct answer is (d)

139. For two reactions,

According to the collision theory, the ratio of squares of pre – exponential factors of reactions 2 (A₂) and 1(A₁) at the same temperature,

(a) 4/5

(b) 5/5

(c) 5/3

(d) 3/5

Ans. a

Sol. Pre-exponential factor in collision Theory,

Correct option is (a)

140.    If the specific conductances of a sparingly (1 : 1) salt (MW = 200 g mol^–1) in its saturated aqueous solution at 25⁰C and that of water are 1.5 × 10^–3 ohm^–1 and 1.5 × 10^–5 ohm^–1 dm^–1, respectively, and the ionic conductances for its cation and anion at infinite dilution are 0.485 and 1.0 ohm^–1 dm^–1, respectively, the solubility (in g L^–1) of the salt in water at 25⁰C is

(a) 1 × 10^–6

(b) 1 × 10^–3

(c) 2 × 10^–1

(d) 2 × 10^–4

Ans. c

Sol.

Correct option is (c)

141.

The formation constant of the complex is approximately

(a) 1 × 10⁵

(b) 1 × 10⁷

(c) 1 × 10⁹

(d) 1 × 10¹²

Ans. c

Sol.

on reversing 2^nd reaction and adding it to reaction 1

Equilibrium constant of above cell reaction is formation constant

But values in equation given are of oxidation potential

Correct option is (c)

142. The molar conductivity ( Ʌ ) vs. concentration (c) plot of sodium dodecylsulfate in water is expected to look like

(a)

(b)

(c)

(d)

Ans. d

Sol. The molar conductance of anionic surfactant of the type Na⁺R^– (sodium dodecylsulfate in water) in water is plotted against the square root of the normality of the solution. The curve obtained, instead of being the smoothly decreasing curve characteristic of ionic electrolytes of this type, has a sharp break in it, at low concentrations. This sharp break in the curve accompanied by reduction in the conductance of the solution, indicating a sharp increase in the mass per unit charge of the material in solution, is interpreted as evidence for the formation of micelles at that point from the unassociated molecules of surfactant with part of the charge of the micelle neutralized by associated counter ions. The concentration at which this phenomenon occurs is called the critical micelle concentration (CMC)

The effect of concentration of electrolyte is given by

For homologous ionic surfactant

Correct option is (d)

143.    The values obtained from X-ray powder diffraction pattern of a solid are 2x, 4x 6x, 8x where x is equal to 0.06. The wavelength of X – ray used to obtain this pattern is 1.54Å. The until cell and the until cell length, respectively are

(a) BCC, 3.146 Å

(b) FCC, 3.146 Å

(c) SCC, 6.281 Å

(d) BCC, 1.544 Å

Ans. a

Sol.

Correct option is (a)

144. Distribution of molar masses in a typical polymer sample is shown below.

The A, B and C represent

(a)

(b)

(c)

(d)

Ans. b

Sol.

Correct option is (b)

145.    Two bound stationary states, 1 and 2, of a one – electron atom, with E₂ > E₁ (E is the total energy) obey the following statement about their kinetic energy (T) and potential energy (V)

(a) T₂ > T₁ ; V₂ > V₁

(b) T₂ > T₁ ; V₂ < V₁

(c) T₂ < T₁ ; V₂ > V₁

(d) T₂ = T₁ ; V₂ > V₁

Ans. c

Sol.

Correct option is (c)