21. Dinuclear anion has two bridging oxo groups. The geometry around each iodine is

(a) Octahedral

(b) Monocapped

(c) Square pyamidal

(d) Pentagonal bipyramidal

Ans. a

Sol. (a) The structure of is shown above-

22. Using a double beam UV – visible spectrophotometer, Beer's law fails for K₂Cr₂O₇ solution when

(a) Intensity of light source is changed

(b) Detector is not a photomultiplier tube

(c) Cuvette of 2 cm size is used

(d) pH is not kept same in all measurements.

Ans. d

Sol. (d) A double beam UV- Spectroscopy follows Beer-Lambert Law for different salt solution but this law is fails for K₂Cr₂O₇ solution when pH is not kept same in all measurements.

23. Trivalent lanthanide ion having isotropic magnetic susceptibility is –

(a) Eu³⁺

(b) Gd³⁺

(c) Yb³⁺

(d) Lu³⁺

Ans. b

Sol. (b) The ions having different magnetic field in different direction possess anisotropic magnetic susceptibility. In all given ion, only Gd⁺³ do not having any orbital contribution, so it has anisotropic magnetic susceptibility. The calculated and observed values of Gd⁺³ is 7.94.

24. The structure of CaB₆ is close to that of –

(a) Cesium chloride

(b) Nickel arsenide

(c) Rock salt

(d) Zink blende

Ans. a

Sol. (a) CaB₆ is cubic structure which is close to CsCl. The calcium atoms are arranged in simple cubic packing so that there are holes between groups of eight calcium atoms situated at the vartices of a cube. The simple cubic structure is expended by the octahedral B₆ groups and structure is a CsCl like packing of the calcium and hexaboride.

25. The correct of C – O bond length is –

(a)

(b)

(c)

(d)

Ans. b

Sol. (b) If species follows 18e^– rule then in that Metal – Carbon bond length will be weak where as carbon – oxygen bond length become strong.

26. Among the elements Zn, Ga, Ge and As, the one with the lowest first ionization energy is –

(a) As

(b) Zn

(c) Ga

(d) Ge

Ans. c

Sol. (c) The ionization energy decreases from top to bottom and increase left to right.

Zn²⁺ has stable and difficult to remove and e^– from an orbital so, Zn has IIIrd ionisation energy greater than 1st ionisation energy of so, Zn > Ga.

Therefor Ga is the lowest first ionisation energy.

27. The total degeneracy of the ground term of Co⁺² (high spin) in octahedal geometry is –

(a) 18

(b) 12

(c) 28

(d) 9

Ans. c

Sol. The total degeneracy of the ground term of Co^II (high spin) in Octahedral geometry is 28.

28. For the following reaction

The structure of the intermediate is

(a)

(b)

(c)

(d)

Ans. d

Sol. (d) Oxidative addition is not possible in 18 e^– species so, one CO must be dissociated and give 16 e–species and thus reaction is possible.

29. High spin complex of a 3d metal ion M has a magnetic moment of 2.9 B.M. in octahedral coordination environment and 4.1 B.M. in tetrahedral environment. The M ion is

(a) Co^III

(b) Ni^II

(c) Cu^II

(d) Co^II

Ans. b

Sol.

30. For electronic spectra K₂CrO₄ (A) and K₂MoO₄ (B) the correct combination is –

(a)

(b)

(c)

(d)

Ans. c

Sol. (c) Oxidation State of K₂CrO₄ and K₂MoO₄ is + 6i.e. higher oxidation state, so this exist as LM < T. downward in group decreases therefore Cr >> Mo.

31. Removal of an electron from NO molecule results in

A. An increase in the in the IR spectrum

B. An EPR active species

C. Electrons in HOMOs being closer to the oxygen than to nitrogen 2p orbital's

D. Electrons in HOMOs being closer to the nitrogen than to oxygen 2p orbital's

The correct answer is

(a) A only

(b) A and C

(c) B and D

(d) A, B and D

Ans. d

Sol. (d) Electronic Configuration of NO.

Bond order of NO⁺ > NO, thus and Electron in HOMO is being closer to the nitrogen than to oxygen 2p orbitals.

32. Consider the nature of solvents in column I and the corresponding for I₂ in various solvents given in column II. (for I₂ vapor l_max is 520 nm). Match column I with column II

(a) (a) – (i); (b) – (ii); (c) – (iii); (d) – (iv)

(b) (a) – (iii), (b) – (iv); (c) – (ii); (d) – (i)

(c) (a) – (i); (b) – (iii); (c) – (iv); (d) – (ii)

(d) (a) – (iv); (b) – (iii); (c) – (ii); (d) – (i)

Ans. c

Sol. (c) Strong donar gives minimum whereas non-donor gives same value due to no charge in i.e. 520 nm. Therefore strong donar gives 360 nm and non-donor 520 nm.

33. For the catalytic activity of Cu and Zn containing enzyme, superoxide dismutase, what is/are the correct statements (S)?

(A) Cu and Zn both are essential

(B) Only Cu is essential

(C) Zu is essential and Cu may be replaced by any other divalent metal atom

(D) Zn may be replaced by any other divalent metal atom

(a) A only

(b) C only

(c) D only

(d) B and D

Ans. d

Sol. (d) Structure of superoxide dismutase is shown below-

For catalytic activity only Cu is necessary but this is not blue-copper protein; and Zn may be replace by any other divalent metal atom.

34. Mass spectrum of a compound shows an [M+2] ion peak that is about 4% of M⁺. this indicates that the compound has one

(a) Fluorine

(b) Sulfur

(c) Bromine

(d) Chlorine

Ans. b

Sol. (b) Mass-Spectrum of a compound shows an [m + 2] ion peak this about that 4% of M⁺. This is indicate that is sulphur compound.

35. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. a

Sol. The oxidation of 2, 4 di-tert butyl phenol in the alkaline to the intensity blue coloured phenoxy-radical occur with potassium ferricynide.

The phonoxy radical reacts with oxygen as diradical form 4–4' linked forming yellow crystal.

36. For the following compounds, the correct order to reactivity towards nucleophilic acyl substitution is

(a) Acetyl chloride < methyl acetate < acetic anhydride < acetamide

(b) Acetamide < methyl acetate < acetic anhydride < acetyl chloride

(c) Acetamide < acetic anhydride < acetyl chloride < methyl acetate

(d) Methyl acetate < acetamide < acetic anhydride < acetyl chloride

Ans. b

Sol. (b) Mechanism of alyl solution this is based on addition elimination type driving force of this reaction less stabilised Carbonyl is converted more stabilised one

(More – REacti to less – Reactive)

37. The major product formed in the following reaction is –

(a)

(b)

(c)

(d)

Ans. d

Sol.

Li NH₃(1) Birch Reagent. Birch-Reagent more prefebly reduced

Less e dansity gp. Here e^– density order is–

38. The major product formed in the following reaction is –

(a)

(b)

(c)

(d)

Ans. b

Sol.

39. The major product in the following reaction is –

(a)

(b)

(c)

(d)

Ans. c

Sol. (c) MnO₂ used used as oxidation of allylic and benzylic alcohol selectively by a suspension of activated magnese (IV) dioxide MnO₂. Primary allylic alcohols are oxidised to aldehyde and cyclic allylic alcohols are oxidised to ketones.

40. The major product formed in the following reaction is –

(a)

(b)

(c)

(d)

Ans. a

Sol. (a) Anti-addition across an alkyne can be accomplished with lithium aluminium hydride using as propagylic alcohol and occure with very high stereo selectivity. Reduction in the presence of Sodium Mithoxide treated with iodine then the final product - 3 - iodollylic alcohol.

41. Number of signals observed in the ¹³C NMR spectrum of the following compound is –

(a) 4

(b) 5

(c) 6

(d) 9

Ans. c

Sol.

42. The structure of the product formed during the reaction of amino acid with ninhydrin is –

(a)

(b)

(c)

(d)

Ans. a

Sol.

43. The correct order to rate of sololysis in 80% ethanol at 25⁰C is–

(a) B > C > A

(b) A > B > C

(c) C > B > A

(d) C > A > B

Ans. d

Sol. (d) Solrolytic reaction is generally Substitution Reaction. It is type of Nucleophli substitution S_N1 or S_N2. Here S_N1 substitutiion Reaction in bridged bicyclic compound - at bridge reacts is very slow or does not take place. This is expected that planer intermediate can not be formed at bridge Head Carbon.

So greater the rigidity about the bridge head lower will bhe reactivity.

Order of Carbocation Stability.

44. IUPAC nomenclature of following propellane is –

(a) Tricyclo [1.1.1.0^2.4] pentane

(b) Tricyclo [1.1.0.1^1.3] pentane

(c) Tricyclo [1.1.1^1.3.0^1.5] pentane

(d) Tricyclo [1.1.1.0^1.3] pentane

Ans. d

Sol.

45. The correct statement about following species is –

(a) Both A and B are aromatic

(b) A is aromatic and B is antiaromatic

(c) A is nonaromatic and B is antiaromatic

(d) A is aromatic and B is homoaromatic

Ans. b

Sol.

46. The correct statement about the following compounds is –

(a) A is more stable than B

(b) B is more stable than A

(c) A and B are equally stable

(d) A and B are both locked conformations

Ans. c

Sol. (c) The Newmann projection of both A and B Compounds

In structure 'A' and B both methyl group exhibited same repulsion due to presence of an Alkyl group (R), although in 'A' methyl groups are Anti and in B, they are in Gauche form.

47. In the pure Raman rotational spectrum of ¹⁶O₂, whose electronic ground state is transitions to / from

(a) Even J levels are missing

(b) Odd J levels are missing

(c) All J levels appear

(d) None of the J levels appear

Ans. a

Sol. (a) In pure rotation Raman spectra of ¹⁶O₂. The transitions of Raman Spectra even J Levels are missing and odd J levels are obtained i.e., transition obtained at J = 1, 3, 5, 7 etc.

48. Elementary steps of a reaction are as follows

If steady state approximation is applicable to C, the rate of product formation in the reaction is

(a) Proportional to [A] [B]

(b) Proportional to [A]² [B]²

(c) Proportional to [A]^1/2 [B]^1/2

(d) Independent of [A] [B]

Ans. a

Sol. Explanation:

Given that,

Thus, we can write

Now applying SSA to 'C'

So, from equation (1),

Putting these value in equation (3),we get

Therefore,

49. The term symbol for the ground state of B₂ is

(a)

(b)

(c)

(d)

Ans. c

Sol. The molecular term symbol can be written as

Where is the total spin angular momentum quantum number, is the quantum numbers for total orbital anular momentum of the electrons about the internuclear axis 4 g/u subscript applies only to molecule with a centre of symmetry and (+)(–) super script applies only to

Electronic configuration of B₂ is

50. A Gaussian distribution has the functional from The variance of such distribution is

(a) a

(b) a²

(c) b

(d) b²

Ans. b

Sol. (b) The Gausium distribution function is defined as

Where b and a are two parameters, with a > 0.

Then we have,

Var (x) =

Using integral properties, we have,

51. The change in entropy for a reversible adiabatic process is

(a) Maximum

(b) Minimum

(c) Zero

(d) Positive

Ans. c

Sol. (c) The change of entropy for a reversible adiabatic process is zero.

If the reversible adiabatic process is done, then we know that the entropy of the universe can not be changing, and that heat flow from the gas to surroundings is zero.

From equation

52. The standard cell potential for the reaction

The Gibb's free energy change during the reaction is

(F = 96500 coulomb mol^–1)

(a) – 21.2 KJ mol^–1

(b) + 212 KJ mol^–1

(c) – 212 KJ mol^–1

(d) – 212 KJ mol^–1

Ans. c

Sol. The Gibb's free energy change during the reaction in terms of standard cell potential is.

53. If the unit of the rate constant of a reaction is L³mol^–3S^–1, the order of the reaction is

(a) 1

(b) 2

(c) 3

(d) 4

Ans. d

Sol. The unit of nth order reaction is,

= (mol/litre)^1-n sec.^–1

Thus comparting both side

The order of reaction is 4.

54. The lowest energy state of a ls¹2s¹ electronic configuration, according to hund's rule, is

(a) ³S₀

(b) ¹S₀

(c) ³S₁

(d) ¹S₁

Ans. c

Sol. The lowest electronic configuration is 1s¹ 2s¹

Thus, the term symbol is, = ³S₁

55. The commutator of with the Hamiltonian is

(a) 0

(b)

(c)

(d)

Ans. d

Sol.

(Because self operator are commute to each other)

56. A 5 g/L polymer solution is prepared with a polymer whose molar mass is 25 kg. the osmotic pressure (in atm) of this solution at 25⁰C is

(a) 0.002

(b) 0.05

(c) 0.005

(d) 0.008

Ans. c

Sol. Concentration of polymer solution = 5 gm/L

Molar mass of polymer (M) = 25 kg = 25000 gm

RT (given) = 2500 J.mol^–1

57. If all the lattice points of an FCC structure are occupied by uniform hard spheres that touch each other, the fraction of volume occupied is

(a)

(b)

(c)

(d)

Ans. a

Sol. (a) Suppose a FCC Unit cell with edge length of 'a', as shown in the figure.

According to figure,

The distance of closest approach = face digonal.

i.e. AC = 4r ...(1)

According to Solid Geometry

58. Origin of all colligative properties of a dilute solution is

(a) Volatility of solute molecule

(b) Interaction of solute – solvent molecules

(c) Zero enthalpy of mixing

(d) Entropy of mixing

Ans. d

Sol. (d) The colligative properties are determined by the number of particle in a solution and not on the molecular weight or size of the particles. The colligative properties of dilute solution ideally depend on change in the entropy of the solution on dissolving the solute, which is determined by the concentration of the solute molecules or ions but does not depend upon their structures. The rational for these colligative properties is the increase in entropy on mixing solutes with water.

59. The graph that represents the Langmuir adsorption Isotherm is

(a)

(b)

(c)

(d)

Ans. d

Sol. (d) The release of adsorbed gas is commonly described by a pressure relationship called Langmuir Isotherm. The Langmuir adsorption isotherm assumes that the gas attached to the surface of the coal or shale, and covers the surface as a single layer of gas. The typical formulation of Langmuir isotherm is

Where Cg is the gas content measured in scf.ton of coal for application to CBM reservoirs in the filed, the equation is modified to accound for ash and moisture content of the Coal

The Langmuir volume is the maximum amount of the gas that can be adsorbed Coal/Shale of infinite pressure. The following plot of a Langmuir isotherm demonstrate that gas content asymptotically approaches the Langmuir volume as pressure increases to infinity.

60. Correct match for the coenzymes is Column A with their function is Column B is

(a) P – i; Q – ii; R – iii

(b) P – iii; Q – i, R – ii

(c) P – iii; Q – ii; R – i

(d) P – ii; Q – i, R – iii

Ans. b

Sol. The correct match of the given coenzymes and their functions are shown in the following-

(P) NADH : This is used as a reducing agent to donate electrons.

(Q) FAD : The florin adenine dinucleotide (FAD) works as oxidising agent to except two electron and two protons to becomes FADH₂,

(R) COASA : It workss as acyl group transfer.

61. One of the product formed in the bis nickel complex catalyzed cyclo – dimerization of butadiene in the presence of PR₃ is compound A given below. Identify its precursor.

(a)

(b)

(c)

(d)

Ans. b

Sol. (b) The correct product in the reaction given is shown below.

62. The transformation are given in Column I and reagent in Column II. Match the items of Column I with those of Column II

(a) (a) – (i); (b) – (ii); (c) – (iii); (d) – (iv)

(b) (a) – (ii); (b) – (iii); (c) – (iv); (d) – (i)

(c) (a) – (iii); (b) – (ii); (c) – (i); (d) – (iv)

(d) (a) – (iii); (b) – (i); (c) – (iv); (d) – (ii)

Ans. b

Sol. in the presence of Na/Liq. NH₃ gives Ag. and Au in the presence of Liq. BrF₃ gives Ag [AuF₄]. Orthophosphoric acid in the presence of the presence of sulfuric and gives in the presence of gives

63. Consider the following statements for the oxygenation of hemocyanine:

(A) Oxidation state of both copper atoms changes by two

(B) It becomes intense blue from colorless

(C) Dioxygen is reduced to

(D) The bond forms between each oxygen and copper atoms.

The correct statements are:

(a) (A) and (C)

(b) (B) and (C)

(c) (A), (B) and (C)

(d) (B), (C) and (D)

Ans. d

Sol.

64. The correct increasing order of C–C bond length in the following molecules (A – D)

(a) (C) < (A) < (B)

(b) (A) < (B) < (C)

(c) (B) < (C) < (A)

(d) (C) < (B) < (A)

Ans. b

Sol.

CN is EWG (electron withdrawing group) and pull the e^– so C – C bond length increases.

Thus correct order is A < B < C.

65. Which of the following are NOT closo clusters?

(a) (C) and (C)

(b) (A) and (B)

(c) (A) and (C)

(d) (B) and (C)

Ans. d

Sol. The structure of B₄C₂H₈

C₂ = (BH)₂

B₄H₈

B₆H₁₀ = [BnHn]^4-

= [B₆H₆]^4–

= Nido

66. Identify the pair of molecules which are isoelecronic as well as isostructural from the following:

(a) B and C

(b) A and F

(c) A and D

(d) C and E

Ans. b

Sol. The complex have same no. of electrons called isoelectronic [Pd(PPh₃)₄]:

Pd = 10 e^–

67. For the following reaction, correct statements (S) is / are

(A) Oxidation state of iridium increases from I to II

(B) It is – hydride elimination reaction

(C) (I) and (II) both are diamagnetic

(D) It is migratory insertion reaction

The correct answer is

(a) (A) only

(b) (A) and (C)

(c) (C) and (D)

(d) (B), (C), and (D)

Ans. b

Sol.

Both are stable product having 16e^– and 18e^– respectively.

68. The reaction of with CH₃I gives compound A. The ¹H NMR spectrum of A shows two singlets in an integrated intensity ratio of 3:5. Compound A upon reaction with PPh₃ gives compound B. The ¹H NMR spectrum of B shows 3 sets of signals in an integrated intensity ratio of 3 : 5 : 15. Compounds A and B respectively, are

(a)

(b)

(c)

(d)

Ans. b

Sol. See the following reaction:

69. Identify the correct statements about the electro negativity of groups given below:

(A) CF₃ group has greater value than that of NF₂.

(B) NH₂ group has lower value than that of NF₂.

(C) OH group has greater value than that of NF₂.

(D) CH₃ and C₂H₅ groups have almost similar values

Correct answer is

(a) A, B and D

(b) B and C

(c) B, C and D

(d) B and D

Ans. d

Sol. The electronegativity of group depends upon % s- character and no. of highly electronegative atoms attached to the group central atom. In NH₂ and NF₂, the electronegativity of NF₂ is more than NH₂ because electronegativity of F is very high than 'H' so F-atom increases the electronegativity of NF₂ group. The OH group has greater value than NF₂ is not correct, because NF₂ have greater value than OH due to presence of 2F atoms.

The hydrocarbon (saturated) group have almost similar values, of electronegativity thus CH₃ and C₂H₅ have almost same electronegativity.

70. Height equivalent to theoretical plate (HETP) in gas – liquid chromatography depends significantly in which of the following?

(A) Temperature of column

(B) Velocity of carrier gas

(C) Packing of column

(D) Column material

Correct answer is

(a) A, B and C

(b) C and D

(c) B, C and D

(d) A and C

Ans. a

Sol. Height evuivalent of theoretical plate (HETP) in gas liquid chromatography depends on temperature, packing of column and velocity of carrier gas.

71. A binary fluoride (Z) of xenon combines with two moles of NaF to give a product which on hearting to 100⁰C affords compounds A. the alkaline hydrolysis of A gives perxenate salt. Z and A are, respectively,

(a) XeF₂ and XeF₄

(b) XeF₄ and XeF₆

(c) XeF₆ and XeF₄

(d) XeF₆ and XeF₆

Ans. d

Sol. See the following reaction

72. Match fluorescence colours given in column A with lanthanide ions given in Column B

Correct match is –

(a) (i) – (a); (ii) – (c); (iii) – (b); (iv) – (d)

(b) (i) – (d); (ii) – (c); (iii) – (b); (iv) – (a)

(c) (a) – (a); (ii) – (b); (iii) – (c); (iv) – (d)

(d) (a) – (c); (ii) – (b); (iii) – (d); (iv) – (a)

Ans. a

Sol. Colour of the following lanthanide is reported as,

73. Choose the correct set of statements for cis – platin.

(A) It can be prepared from K₂[PtCl₄].

(B) It can be prepared from [Pt(NH₃)₄Cl₂].

(C) In its preparation, the observed trans effect for Cl^– is greater than that of NH₃.

(D) In blood it stays in equilibrium with cis – [Pt(NH₃)₂Cl(H₂O)]⁺.

(E) In DNA strand, it binds to two adjacent cytosine bases.

The correct set is

(a) A, C and D

(b) A, C, D and E

(c) B, C and D

(d) B, C, D and E

Ans. a

Sol. In the cis – platin

It can be prepared by K₂ [PtCl₄] and its trans effect of Cl^– is greater than NH₃ and in blood it stays in equilibrium with cis–[Pt(NH₃)₂ Cl(H₂O)]⁺

74. In fission of ²³⁵U atom the energy released is 200 MeV. In one day fission of 1 kg ²³⁵U will given power (in M W) approximately)

(a) 550

(b) 650

(c) 950

(d) 1250

Ans. c

Sol. 1 kg of ²³⁵U cation

= (1000 / 235) × 6.023 × 10²³

= 25.62 × 10²³ atom

The energy released during each ²³⁵U fussuib reaction = 200 MeV

Assuming all the ²³⁵U – atom undergo fission the total energy released by the fission is =

= 25.62 × 10²³ × 200

= 5124 × 10²³ × MeV

= 5124 × 10²³ × 1.6 × 10^–13 Joules

= 8198.4 × 10¹⁰ Joules

Since 1 kg of 0.235 undergo fission over the course of a single day. The total powers released is–

= (8198.4 × 10¹⁰ Joules) / 86400 see

= 0.094889 × 10¹⁰ watts

= 948.88 MW

= 950 MW

75. The structure of (B) are made up of two MCl₄ units. For these structures, which statement is correct?

(a) (A) and (B) both have MCl₄ units eclipsed.

(b) (A) and (B) both have MCl₄ units staggered.

(c) (A) has both MCl₄ units staggered and (B) has both MCl₄ units eclipsed.

(d) (A) has both MCl₄ units esclipsed and (B) has both MCl₄ units staggered.

Ans. d

Sol. (A) or (D) [Re₂Cl₈]₂– exist in eclipsed form due to d-bond formation whereas [Os₂Cl₈]^2– exist as staggered form due to large size and it is not form

76. for the Waker process, pick the correct statement (s) from the following:

(A) Pd(II) is reduced to Pd(0) by Cu(i)

(B) Pd(0) is oxidized to Pd(II) by Cu(II)

(C) Cu(II) promotes the reductive elimination

correct answer is

(a) A and C

(b) B and C

(c) A and B

(d) B only

Ans. d

Sol. In the Wacker process Pd (0) is oxidised to Pd (II) whereas Cu (II) act as co-catalysed participate in the reaction.

77. Consider following statements:

I: AsCl₅ is thermally less stable than PCl₅.

II: Size of As is more than that of P.

Choose correct answer from the following

(a) Statements I and II are true and II is the correct explanation of I.

(b) Statements I and II are true but II is not the correct explanation of I.

(c) Statement I is true and statement II is false.

(d) Both the statements I and II are false.

Ans. b

Sol. AsCl₅ is thermally less stable than PCl₅ whereas size of 'As' is more than 'P' but these are no any relation between them.

78. Consider the following statements for Be₂Cl₄ (I), B₂Cl₄ (II) and Ga₂Cl₄ (III):

(A) There is an M – M (M = Be, B, Ga) bond in all.

(B) The oxidation state of Be, B and Ga is +2.

(C) The geometry around the central atom is planar for all.

(D) The geometry around the central atom is planar in I and II only.

The correct statement(s) is/are

(a) A, B and C

(b) A and B

(c) D only

(d) B, C and D

Ans. c

Sol. The geometry of the given compounds are shown below

The geometry around the central atom is planar only I and II.

79. Consider the following statements:

(A) Cr²⁺ is easier to oxidize than V²⁺ in the gas phase

(B) Cr²⁺ (aq) is a more powerful reducing agent than V²⁺ (aq).

(C) The rate of water exchange for Cr²⁺ (aq) is much faster than for V²⁺(aq).

The correct statements are

(a) A and B

(b) A and C

(c) B and C

(d) A, B and C

Ans. c

Sol. Reducing power of Cr²⁺ is greater than V²⁺ and rate of water exchange of Cr²⁺ = 10^–8 sec^–1 and V²⁺ = 1 – 10⁴ sec^–1, therefore Cr²⁺ is much faster than V²⁺.

80. Consider the statements A – D regarding equation I – III:

(A) Marcus equation is applicable to I and II.

(B) Marcus equation is applicable to II only.

(C) Equation I and II involve inner sphere electron transfer.

(D) Equation I and III involve inner sphere electron transfer.

The correct statements are:

(a) A and B

(b) B and C

(c) B and D

(d) C and D

Ans. c

Sol. Marcus equation is applicable here.

and in the option (I) and (III) involve inner sphere electron transfer.

81. The major formed in the following oxidation reaction is

(a)

(b)

(c)

(d)

Ans. b

Sol. See the following mechanism.

82. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. a

Sol. See the following mechanism.

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

(a)

(b)

(c)

(d)

Ans. a

Sol.

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

(a)

(b)

(c)

(d)

Ans. c

Sol.

85. Structure of the intermediate A and the final product B in the following reaction sequence are (dba = dibenzylidene acetone)

(a)

(b)

(c)

(d)

Ans. a

Sol.

86. Mechanism of the following transformation involves

(a) A [2+2] cycloaddition followed by 'con' rotatory electrocyclic ring opening

(b) A [4+2] cycloaddition followed by 'con' roratory elecrocyclic ring opening

(c) A [4+2] cycloaddition followed by cope rearrangement

(d) A [2+2] cycloaddition followed by 'dis' rotatory elecrocyclic ring opening

Ans. d

Sol. In case of photochemical (hv) only [2+2] cycloaddition allowed.

87. The correct statement about solvolysis using NaOAc/AcOH of following compounds is

(a) A reacts faster than B to give trans – 1, 2 – diacetoxycyclohexane

(b) B reacts faster than A to give trans – 1, 2 – diacetoxycyclohexane

(c) A reacts faster than B to give cis – 1, 2 – diacetoxycyclohexane

(d) B reacts faster than A to give cis – 1, 2 – diacetoxycyclohexane

Ans. a

Sol. See the following reaction.

88. The structure of the intermediate A and the major product B formed in the following reaction are

(a)

(b)

(c)

(d)

Ans. a

Sol. See the following mechanism.

89. A compound shows following spectral data:

¹H NMR: (d, J = 8 Hz, 2H), 6.6 (d, J = 8 Hz, 2H), 4.3 (q, J = 6 Hz, 2H), 4.0 (br s, 2H, D₂O exchangeable), 1.4 (t, J = 6 Hz, 3H) mass: m/z 165, 137, 120, 92

The correct structure of the compound is

(a)

(b)

(c)

(d)

Ans. a

Sol. First of all particle structure write –

Substituted benzene

90. The major allylic alcohol A and the ester B formed in the following reaction sequence are

(a)

(b)

(c)

(d)

Ans. b

Sol. See the following mechanism.

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

(a)

(b)

(c)

(d)

Ans. d

Sol. See the following mechanism.

92. Reaction of D – glucose with following reagents produces

Reagents: 1. Acetone, H⁺; 2. PDC; 3. (EtO)₂P(O)CH₂CO₂Et, NaH

(a)

(b)

(c)

(d)

Ans. a

Sol. D-Glucose exista as two form - from give rate of protection major.

93. For the following thermal [2+2] cycloaddition reaction, the correct statement about Transition state (TS) and preference for endo product formation is

(a)

(b)

(c)

(d)

Ans. b

Sol. In the following reaction.

The ratio of endo - exo product is given in the following table –

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

(a)

(b)

(c)

(d)

Ans. a

Sol. See the following mechanism.

95. The major products formed in the following photochemical reaction are

(a)

(b)

(c)

(d)

Ans. a

Sol. See the following mechanism.

96. Irradiation of either cis – or trans – stilbene at 313 nm results in the formation of a mixture of 93% cis and 7% trans olefin because.

(a) Trans – stilbene is more stable than cis – stilbene.

(b) The extinction coefficient of trans – stilbene is greater than cis – stilbene at exciting wavelength

(c) The transition state structures of cis and trans – stilbenes are different

(d) The triplet excited states of cis – and trans – stilbenes are at different energy levels.

Ans. b

Sol. Irradiation of either cis or trans - stilbene at 313 nm result in the formation of a mixture of a mixture of 93% cis and 7% trans olifine because, the excitation Cofficient of trans stilbene is greater than cis - stilbane at exciting wavelength.

97. The major heterocyclic compound formed in the following reaction is –

(a)

(b)

(c)

(d)

Ans. c

Sol. See the following mechanism.

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

(a)

(b)

(c)

(d)

Ans. a

Sol. See the following mechanism.

99. The correct order of rates for the following reaction is

(a) K₁ > K₂ and K₃ > K₄

(b) K₁ > K₂ and K₄ > K₃

(c) K₂ > K₁ and K₃ > K₄

(d) K₂ > K₁ and K₄ and K₃

Ans. b

Sol. Here two type reaction K₁ and K₂ follow elimination addition and K₃ and K₄ Addition elimination reaction.

* K₁ and K₂

Here the order of halogen reactivity 1 > Br > Cl > F and here elimination of halogen rate dermining step.

So K₁ > K₂

* K₂ and K₄

The EWG cleary influence the rate of reaction (EWG such as – CN, –CO etc) facilited substitution reaction.

Kinetic studies demonstrate that these reaction are second order. First order in nucleophile and first order aromatic substrate. The formation of the addition intermediate rate determining step (rds).

Order or reactivity of halogen F > Cl > Br > I. This is clearly suggested that stronger bond dipoles associates with the more electro negativity atom favour the addition step. Thus lowering the energy of activation of nucleophile addition step (rds).

So K₄ > K₃

100. The correct match of protons in Column A with the ¹H NMR chemical shift in Column B for the product of the following reaction is

(a) P – ii; Q – i; R – iii; S – iv

(b) P – i; Q – ii; R – iv; S – iii

(c) P – iv; Q – i; R – iii; S – ii

(d) P – ii; Q – iv; R – i; S – iii

Ans. a

Sol.

In this species an aromatic sextet is spread over Seven Caron as in the tropilium ion. The eight carbon is an sp³ carbon so can not take part is aromaticity. The NMR spectra show the presence of a diatropic Ring - Current Hb found at S = –0.3 and Ha = 5.1

H₁ & H₇ = 6.4

H₂ & H₇ = 8.5

101. Which of these is not a suitable unnormalized wave function for the excited 1s ^|2s^| electron configuration of the helium atom?

(a)

(b)

(c)

(d)

Ans. d

Sol. Electronic configuration of He in excited state is 1s^I 2s^I. The wave function of space is Antisymmetric and Spin part will be symmetric.

In these given option (4) is (1)2s(2) + 2s(1)1s(2)[2α(1)β(2)] is symmetric wave function and other are not symmetric wave function.

102. Two opposite sides (in the y – direction) of a square box of side L are slightly stretched. Consider the following four statements:

A. The point group changes from D_4h to D_2h.

B. The (1, 2) and (2, 1) energy levels remain doubly degenerate.

C. Both the energy levels are lowered and the energy of the (1, 2) level higher than that of the (2, 1) level.

D. Both the energy levels are lowered and the energy of the (1, 2) level lower than that of the (2, 1) level.

The two correct statements are:

(a) A and B

(b) A and C

(c) B and C

(d) A and D

Ans. d

Sol. From the given statement we can write y-axis

From the above energy level diagram E₂ < E₄ i.e. Both the energy level are lowered and the energy of the (1, 2) level is lower than that of the (2, 1) level.

103. Consider a model system of five non – interacting fermions in a single 3 – dimensional harmonic oscillator. The Hamiltonian of a single particle is where m is the mass of the particle, is the angular frequency, are the momentum operators. The ground state energy of the system of non – interacting fermions is

(a)

(b)

(c)

(d)

Ans. a

Sol.

104. A particle is in a state where are eigenfunctions of the Hamiltonian of the particle with eigenvalues E₁ and E₂, respectively. The average energy of the particle in the state is

(a)

(b)

(c)

(d)

Ans. a

Sol.

Since, this function is not normalized, using normalisation condition.

105. Which of the following statements on ground state perturbation theory, involving the zeroth order energy first order energy correction and second order energy correction is false?

(a) is the average value of perturbation operator with respect to the ground state of the zeroth order Hamiltonian.

(b) is necessarily negative.

(c) is necessarily negative.

(d) is an upper bound to the exact ground state energy.

Ans. b

Sol. First corrected energy is Ground state

= (+) ve value, (Never be negative)

Note: Only second corrected energy in ground state is always negative

Where m > n,

Second corrected energy in G.S.

106. Difference between activation energies of the reverse and forward steps of a reversible reaction is 9.21RT. If the pre-exponential factor of the forward reaction is double that of the reverse reaction at the same temperature, the equilibrium constant for the reaction at that temperature will be (In10 = 2.303)

(a) 1 × 10⁴

(b) 2 ×10⁴

(c) 1 ×10^–4

(d) 2 × 10^–4

Ans. b

Sol. From Arrhenius Equation:

Equilibrium Const.

107. For an enzyme – substrate reaction,

The slope and the intercept of the plot between are 10^–2 s and 10² M^–1s, respectively. If the value of K₁ will be close to (in units of M^–1s^–1) [r is the rate of the reaction and E₀ is the initial concentration of the enzyme)

(a) 1 × 10¹¹

(b) 1 × 10⁴

(c) 1 × 10⁸

(d) 1 × 10⁶

Ans. a

Sol.

108. Translation partition function of a D₂ molecule confined in a 100 cm² vessel at 25⁰C is

(a) 3.8 × 10²²

(b) 5.8 × 10²⁴

(c) 7.8 × 10²⁶

(d) 9.8 × 10²⁸

Ans. c

Sol. The translational partition function of D₂ molecule is given as,

109. The volume (cm³) of CO absorbed in charcoal (273 K) at two different pressures is given below

Assuming Langmuir isotherm, the maximum possible volume (cm³) CO that can be adsorbed is

(a) 50

(b) 100

(c) 150

(d) 200

Ans. 2

Sol.

110. The number of lines in EPR spectrum of CD₃ (I_D = 1) is

(a) 3

(b) 5

(c) 7

(d) 9

Ans. c

Sol. (c) The EPR Spectra of CD₃ (I_D = 1) is, gives as No. of EPR lines are equal to (2nI + 1)

Thus,

No. of EPR lines of CD3 = (2×3×1+1) = 7

111. A symmetric top molecule, among the following is

(a) Ethylene

(b) allene

(c) butatriene

(d) hexatriene

Ans. 2

Sol. The molecule can have following four type of tops, as

112. The allowed electronic transition in fluorine molecule is

(a)

(b)

(c)

(d)

Ans. c

Sol. (c) For allow transition, the conditions are,

113. Assuming harmonic approximation, the energy change for the reaction HCl + D₂ DCl + HD in cm^–1 is (the vibrational frequency data in cm^–1 is given in the table below),

(a) –258

(b) +258

(c) –129

(d) + 129

Ans. c

Sol. The reaction is,

114. The transition moment integral for a rotational transition between J = 1; M_j = 0 and J = 2; M_j = 0 states for a diatomic molecule along the z axis is proportional to

(a)

(b)

(c)

(d)

Ans. b

Sol. The transition moment integral for a rotational transition for a diatomic molecule along Z-axis is proportional to

115. One of the correct normalized sp² hybrid orbitals is

(a)

(b)

(c)

(d)

Ans. c

Sol. The three sp²-hybrid orbitals formed by combination of an s-orbital with two p-orbitals are comparessed as

Where i = 1, 2, 3

The coefficient ai, bi and ci (i = 1, 2, 3) can be determined as follows.

Since the three hybrids are the single s-orbital is considered to divide equally among them i.e.

One of the three hybrids say may be assigned any direction; Let if be the x-axis, then will be no contribution of P_y, i.e. c_i = 0

The normalisation condition required that

116. At 300 K, the thermal expansion coefficient and the isothermal compressibility of liquid water are 2 × 10^–5 bar^–1, respectively. (in Kbar) for water at 320 K and 1 bar will be

(a) 2.4

(b) 1.2

(c) 0.6

(d) 12.0

Ans. b

Sol. The thermal expansion coefficient

Isothermal compressibility,

Then

(–)ve Sign shows the direction only.

117. In the phase diagram of water, the solid – liquid boundary has a negative slope. The reason for this unusual behavior can be traced to decrease in

(a) Density of the system on melting

(b) Volume of the system on melting

(c) Entropy of the system on melting

(d) Enthalpy of the system on melting

Ans. 2

Sol. If the solid-liquid boundary of a phase diagram of water shows negative slope. The unusual behaviour of water phase diagram is due to decrease in volume of the system on melting.

118. The standard cell potential of cell was measured over a range of temperature, and the data was fited a the standard reaction entropy (JK^–1 mol^–1) at 298 K are

(a) – 9.65 and – 3.85

(b) – 3.84 and – 9.65

(c) – 18.3 and – 7.68

(d) – 7.68 and – 18.3

Ans. a

Sol. (a) A standard cell is given below Pt/H₂(g) | H Br (aq) | Ag Br (s) | Ag (s) The standard cell potential is given as.

119. The (002) plane of an elemental FCC crystal diffract X – rays at Bragg angle 90⁰. The density of the crystal is 4 × 10⁴ kg m^–3. The atomic weight of the elemental solid is

(a) 22

(b) 44

(c) 88

(d) 66

Ans. a

Sol.

120. A solution of Fe³⁺ is titrated potentiometrically using Ce³⁺ solution at 25⁰C. the emf (in V) of the redox system thus formed when, (i) 50% of Fe³⁺ and (ii) 80% of Fe³⁺ are titrated, would respectively be

(a) 0.734 and 0.77

(b) 0.77 and 0.385

(c) 0.77 and 0.734

(d) 0.385 and 0.367

Ans. c

Sol.