PYQ JUNE 2011 - VedPrep

CSIR NET CHEMISTRY (JUNE - 2011)
Previous Year Question Paper with Solution.

21. According to crystal field theory, Ni²⁺ can have two unpaired electrons in

(a) Octahedral geometry only

(b) Square-planar geometry only

(d) Both octahedral and tetrahedral geometry.

Ans. d

Sol.

Correct option is (d)

22. [Ni(CN)₄]^2– and [NiCl₄]^2– complex ions are

(a) Both diamagnetic

(b) Both paramagnetic

(d) Antiferromagnetic and diamagnetic respectively

Ans. c

Sol.

Correct option is (c)

23. Which of the following spectroscopic techniques will be useful to distinguish between M-SCN and M-NCS binding modes?

(a) NMR

(b) IR

(d) Mass

Ans. b

Sol.

Correct option is (b)

24. Which of the following compound show a charge-transfer band?

(a) Lanthanum nitrate

(b) Ceric ammonium nitrate

(d) Copper (II) sulphate pentahydrate

Ans. b

Sol. Charge transfer spectra are possible due to transfer of electron from ligand to metal. This is more probable if the metal is in a high oxidation state or ligand has reducing properties. Intense colour of ceric ammonium nitrate, (NH₄)₂[Ce(NO₃)₆] is due to charge transfer from bidentate NO₃^– to Ce⁺⁴ which has C.N. = 12.

Correct option is (b)

25. Among SF₄, BF₄^–, XeF₄ and ICl₄^– the number of species having two lone pair of electrons on the central atom according to VSEPR theory is

(a) 2

(b) 3

(d) 0

Ans. a

Sol. Structure of above compounds can be represented as

Correct answer is (a)

26. The False statement for a polarographic measurement procedure is:

(a) O₂ is removed

(b) Dropping mercury electrode is working electrode.

(d) Residual current is made zero by adding supporting electrolyte.

Ans. d

Sol. (a) O₂ must be removed as it give its own polarogram, due to following reactions,

(b) Dropping mercury electrode is the working electrode

(c)

(d) Residual current need not to be made zero.

Correct option is (d)

27. The ligand system present in vitamin B₁₂ is:

(a) Porphyrin

(b) Corrin

(d) Crown ether

Ans. b

Sol. Vitamin B₁₂ molecule is built-around a corrin ring containing cobalt (III) atom.

Correct option is (b)

28. Which one of the following exhibits rotational spectra?

(a) H₂

(b) N₂

(d) CO₂

Ans. c

Sol.

Correct option is (c)

29. In Ziegler-Natta catalysis the commonly used catalysed system is

(a) TiCl₄, Al(C₂H₅)₃

(b)

(d) TiCl₄, BF₃

Ans. a

Sol. Ziegler-Natta catalyst is Al(C₂H₅)₃ + TiCl₄

Correct answer is (a)

30. Oxidation occurs very easily in case of

(a)

(b)

(c)

(d)

Ans. b

Sol.

Oxidation of gives an 18 electron compound, so easier to oxidise.

Correct option is (b)

31. Complex in which organic ligand is having only with metal is

(a) W(CH₃)₆

(b)

(c)

(d)

Ans. a

Sol.

Correct option is (a)

32. In the molecules H₂O, NH₃ and CH₄

(a) The bond angles are same

(b) The bond distances are same.

(d) The shapes are same

Ans. c

Sol.

Correct option is (c)

33. The correct order of stability of difluorides is:

(a) GeF₂ > SiF₂ > CF₂

(b) CF₂ > SiF₂ > GeF₂

(d) CF₂ > GeF₂ > SiF₂

Ans. b

Sol.

Stabilisation via back-bonding. Due to innert pair effect stability of +2 oxidation state increases down the group.

Correct option is (b)

34. The number of possible isomers for [Ru(bpy)₂Cl₂] is (bpy = 2,2´-bipyridine)

(a) 2

(b) 3

(d) 5

Ans. b

Sol. Isomers of [Ru(bpy)₂Cl₂] can be represented as

II and III are enantiomers.

Correct option is (b)

35. The species ¹⁹Ne and ¹⁴C emit a positron and respectively. The resulting species formed are respectively

(a) ¹⁹Na and ¹⁴B

(b) ¹⁹F and ¹⁴N

(d) ¹⁹F and ¹⁴B

Ans. b

Sol.

Correct option is (b)

36. Cis and trans complexes of the type [PtA₂X₂] are distinguished by

(a) Chromyl chloride test

(b) Carbylamine test

(d) Ring test

Ans. c

Sol. Kurnakov test is used to distinguish between cisplatin and transplatin, by using thiourea as a reagent.

Correct option is (c)

37. The term symbol of a molecule with electronic configuration

(a)

(b)

(c)

(d)

Ans. b

Sol. Given electronic configuration of the molecule is

Only valence shell is considered while writing term symbol.

Reflection under plane containing nuclie = (–) × (+) = (–).

and u × u = g

Correct option is (b)

38. A process is carried out at constant volume and at constant entropy. It will be spontaneous if:

(a)

(b)

(c)

(d)

Ans. c

Sol. From Clausius inequality,

At constant volume, dw = – PdV = 0

Therefore, from first law of thermodynamics,

At constant entropy ds = 0,

Therefore, from equation (1), we get

Here equality holds only for reversible process.

Correct option is (c)

39. The half life of a zero order reaction is given by (k = rate constant):

(a)

(b)

(c)

(d)

Ans. a

Sol.

On integration and putting proper limit we get,

Correct option is (a)

40. For an aqueous solution at 25ºC, the Debye-Huckel limiting law is given by

(a)

(b)

(c)

(d)

Ans. c

Sol. For an aqueous solution at 25ºC, the Debye-Huckel limiting law is

Correct option is (c)

41. The microwave spectrum of a molecule yields three rotational constants. The molecule is

(a) Prolate symmetric top

(b) Sphereical top

(d) Oblate symmetric top

Ans. c

Sol. For an asymmetric top,

Therefore, Total energy (E) = E_x + E_y + E_z

So, above equation will give three rotational constants, B_x, B_y and B_z.

Correct option is (c)

42. The Q band in the vibrational spectrum of acetylene is observed in the

(a) C–C stretching mode

(b) C–H symmetric stretching mode

(d) C–H antisymmetric stretching mode.

Ans. c

Sol. Q-band (or PQR contour) is obtained for the bending vibration of C₂H₂ when its linearity is lost.

Correct option is (c)

43. The Stark splitting for a given field is larger for a molecule AX as compared to BX. Which one of the following is true?

(a)

(b)

(c)

(d)

Ans. b

Sol. The degeneracy associated with the quantum number (M_J) (the orientation of the rotation in space) is partly removed when an electric field is applied on a polar molecule (e.g. HCl, NH₃, etc)

This splitting of states by an electric field is called Stark Effect.

For any linear molecule (e.g. AX or BX) in electric field (E), the energy of the state with quantum number J and M_J is given by

here energy depends on (square of dipole moment), so large (dipole moment) will cause larger splitting.

Correct option is (b)

44. The adsorption of a gas on a solid surface exhibits the following isotherm. Which one of the following statements is true?

(a) Heat of adsorption is independent of coverage

(b) Adsorption is multilayer

(d) Heat of adsorption varies exponentially with coverage

Ans. a

Sol. For Langmuir adsorption (Monolayer, only),

At Langmuir assumed that binding at a site has no influence on the properties of neighbouring sites; this means that enthalpy of adsorption is independent of coverage.

Correct option is (a)

45. In a chemical reaction, A(s) + B(g) C(g), the total pressure at equilibrium is 6 atm. The value of equilibrium constant is:

(a) 1/2

(b) 9

(d) 36

Ans. c

Sol.

K_p = 1 is possible when, partial pressure of B (P_B) is equal to partial pressure of C (P_C),

i.e. P_B = P_C = 3 atm and P_Total = P_B + P_C = 6 atm.

Correct option is (c)

46. A molecule, AX, has a vibrational energy of 1000 cm^–1 and rotational energy of 10 cm^–1. Another molecule. BX, has a vibrational energy of 400 cm^–1 and rotational energy of 40 cm^–1. Which one of the following statements about the coupling of vibrational and rotational motion is true?

(a) The coupling is stronger in BX

(b) The coupling is stronger in AX

(d) There is no coupling in both AX and BX

Ans. a

Sol. As per B.O. approximation we can consider the two different motions independent if their energy difference is large. In other words if energy difference of two different motions is large, the coupling between them will be minimum.

In case of AX the energy difference of vibrational and rotational energy is large compared to BX. Hence coupling will be stronger in BX.

Correct answer is (a)

47. At room temperature, which molecule has the maximum rotational entropy?

(a) H₂

(b) O₂

(d) N₂

Ans. b

Sol. The entropy of a compound is related to the number of ways in which the molecules can be distributed among different energy states; the greater the number; the large is the degree of disorder, i.e. larger is the entropy.

According to Boltzmann distribution, larger, the energy difference w.r.t. ground state lower will be population in exited state.

so higher the value of B, lower will be population in different exited state (i.e. lower disorder or entropy)

Correct option is (b)

48. The normalized hydrogen atom 1s wavefunction is given by and energy is –0.5 au. If we use a normalized wavefunction of the above form with the average value of energy of the ground state of hydrogen atom is:

(a) Greater than –0.5 au

(b) Equal to –0.5 au

(d)

Ans. c

Sol. True normalised wave function for 1s orbital of H-atom is given by

So, –0.8 < – 0.5.

Correct option is (c)

49. A constant of motion is defined by the equation:

(a) [H, A] = 0

(b) <[H, A]> = 0

(d)

Ans. b

Sol. A is constant of motion (i.e., does not change with time) if

If is time independent then, for to be constant of motion,

Note: If then also we can say A is a constant of motion.

Correct option is (b)

50. The hermitian conjugate of operator d/dx, called is actually equal to

(a) –d/dx

(b) d/dx

(d) –i(d/dx)

Ans. a

Sol. Hermitian conjugate or adjoint of an operator and is defined as

Correct option is (a)

51. An ideal gas expands by following an equation PV^a = constant. In which case does one expect heating?

(a) 3 > a > 2

(b) 2 > a > 1

(d) –1 < a < 0

Ans. c

Sol. For ideal gas, Joule-Thomson coefficient . So we can not produce cooling or heating by expansion under isoenthalpic process.

Under the given process,

PV^a = constant ... (1)

For ideal gas PV = nRT ... (2)

Replaing P from equation (2) in (1), we get

TV^a–1 = constant

a must be less than 1. Also expansion causes decrease in pressure in gas, this is possible for a > 0, as PV^a = constant

Correct option is (c)

52. If y² = 4x and 0.1% error is incurred for x, the percentage error involved in y will be

(a) 0.4

(b) 0.025

(d) 0.05

Ans. d

Sol.

Correct option is (d)

53. The configuration at the two stereocentres in the compound given below are

(a) 1R, 4R

(b) 1R, 4S

(d) 1S, 4S

Ans. a

Sol.

Correct option is (a)

54. The two compounds given below are

(a) Enantiomers

(b) Identical

(d) Regioisomers

Ans. b

Sol.

Correct option is (b)

55. A suitable catalyst for bringing out the transformation given below is:

(a) BF₃.Et₂O

(b) NaOEt

(d) Dibenzoyl peroxide

Ans. a

Sol. Chemical reaction involved in the above transformation can be illustrated as

Correct option is (a)

56. Thermolysis of allyl phenyl ether generates

(a) o-allylphenol only

(b) o- and p-allylphenols

(d) m-allylphenol only

Ans. a

Sol. Chemical reaction involved in the above transformation can be illustrated as

Correct option is (a)

57. The major product formed in the reaction given below is:

(a)

(b)

(c)

(d)

Ans. a

Sol. Chemical reaction involved in the above transformation can be illustrated as

Correct option is (a)

58. The most suitable reagent for the following transformation is:

(a) LiAlH₄

(b) NH₂NH₂/KOH

(d) Li/liq. NH₃

Ans. d

Sol. Chemical reaction involved in the above transformation can be illustrated as

Correct option is (d)

59. The intermediate involved in the reaction given below is:

(a) Free radical

(b) Carbocation

(d) Carbene

Ans. d

Sol. Chemical reaction involved in the above transformation can be illustrated as

Correct option is (d)

60. In the most stable conformation of trans-1-t-butyl-3-methylcyclohexane, the substituents at C-1 nd C-3, respectively are

(a) Axial and equatorial

(b) Equatorial and equatorial

(d) Axial and axial

Ans. c

Sol. 1, 3-trans-disubstitution = a, e substitution

At C – 1 - equatorial -t-Bu; At C–3 - axial -methyl group.

Larger substituent always prefer equatorial position. Hence, correct conformation can be represented as

Correct option is (c)

61. Among the carbocations given below

(a) A is homoaromatic, B is antiaromatic and C is aromatic

(b) A is aromatic, B is antiaromatic and C is homoaromatic

(d) A is homoaromatic, B is aromatic and C is antiaromatic

Ans. a

Sol.

(A) Homoaromatic character

(A romatised by bypassing the sp³-carbon atom)

Correct option is (a)

62. The order of carbonyl stretching frequency in the IR spectra of ketone, amide and anhydride is:

(a) Anhydride > amide > ketone

(b) Ketone > amide > anhydride

(d) Anhydride > ketone > amide

Ans. d

Sol.

Correct option is (d)

63. In the mass spectrum of the compound given below, during the the order of preferential loss of groups is:

(a) Me > C₃H₇ > Et

(b) C₃H₇ > Et > Me

(d) Et > C₃H₇ > Me

Ans. b

Sol.

Longer the alkyl chain, more stable is free radical.

Correct option is (b)

64. The reaction given below is an example of

(a) 1, 3-sigmatropic hydrogen shift

(b) 1, 3-sigmatropic methyl shift

(d) 1, 5-sigmatropic methyl shift

Ans. c

Sol. Chemical reaction involved in the above transformation can be illustrated as

Correct answer is (c)

65. The concerted photochemical reaction between two olefins leading to a cyclobutane ring is:

(a)

(b)

(c)

(d)

Ans. b

Sol.

Correct option is (b)

66. Addition of BH₃ to a carbon-carbon double bond is

(a) anti-Markovinikov syn addition

(b) anti-Markovnikov anti addition

(d) Markovnikov anti addition

Ans. c

Sol. Chemical reaction involved in the above transformation can be illustrated as

Correct option is (c)

67. The absorption at in the UV spectrum of acetone is due to

(a)

(b)

(c)

(d)

Ans. b

Sol. which can be differentiated by low

Correct option is (b)

68. The reaction given below is an example of

(a) E₂-elimination

(b) E₁-elimination

(d) E₁CB-elimination

Ans. d

Sol. Chemical reaction involved in the above transformation can be illustrated as

Correct option is (d)

69. The suitable reagent for the following conversion is

(a) m-CPBA

(b) H₂O₂/AcOH

(d) H₂O₂/NaOH

Ans. d

Sol. Chemical reaction involved in the above transformation can be illustrated as

Correct option is (d)

70. The relative rates of solvolysis of iodides A-C are

(a) C > A > B

(b) C > B > A

(d) B > A > C

Ans. a

Sol.

Correct option is (a)

71. Alkali metal superoxides are obtained by the reaction of

(a) Oxygen with alkali metals in liquid ammonia

(b) Water with alkali metals in liquid ammonia

(d) H₂O₂ with alkali metals in liquid ammonia.

Ans. a

Sol. Alkali metals in liq. NH₃ act as source of electrons and are supposed to be a good one-electron reducing agents.

M + (x + y)NH₃ [M(NH₃)_x]⁺ + [e(NH₃)y]^–

These solvated electrons can reduce O₂ molecule to superoxide ion.

[e(NH₃)y]^– + O₂ O₂^– + yNH₃

These superoxide ions can combine with solvated metal ion to give Alkali metal superoxides.

[M(NH₃)_x]⁺ + O₂^– MO₂ + xNH₃

Correct option is (a)

72. H₂O₂ reduces

(A) [Fe(CN)₆]^3–

(B) KIO₄

(D) SO₃^2–

(a) A and B only

(b) B and C only

(d) B and D only

Ans. b

Sol. H₂O₂ oxidise, Fe⁺² to Fe⁺³,NH₂OH to HNO₃, SO₃^2– to SO₄^2–

H₂O₂ reduce KMnO₄, KIO₄, Ce(SO₄)₂ [All are stronger O.A. than H₂O₂]

Note: In the question, only I⁺⁷ and Ce⁺⁴ are in their highest oxidation states, so the only possibility of reduction by H₂O₂ exists.

Correct option is (b)

73. Match List-I (compounds) with List-II (application) and select the correct answer using the codes given below the lists.

(a) A-ii, B-i, C-iv, D-iii

(b) A-i, B-ii, C-iv, D-iii

(d) A-iii, B-i, C-ii, D-iv

Ans. a

Sol.

Correct option is (a)

74. Among the following statements, identify the correct ones for complexes of lanthanide (III) ion.

(A) Metal-ligand bond is significantly ionic.

(B) Complexes rarely show isomerism

(D) The magnetic moments are not accounted even approximately by spin only value for majoritiy of lanthanides.

(a) A, B and D only

(b) A, B and C only

(d) A and D only

Ans. a

Sol. (A) Correct, As 4f-electrons are too well shielded to interact, so they are like inert gas-type ion, like those of alkaline earth metals, that attracts ligands only by overall electrostatic forces.

(B) Correct, Lanthanides, owing to the unavailability of orbitals for bonding, high basicity and rather larger size of Ln⁺³, form few complexes and mostly with oxygen or oxygen plus nitrogen chelating ligands such as hydroxy acids EDTA etc. So, they rarely show isomerism and also lability of ligands make the isolation of isomers is difficult.

(C) Incorrect, C.N. more than 8 are common, due to their larger size and smaller chelating ligand like, NO₃^–, SO₄^2– etc.

e.g. [Ce(NO₃)₆]^2–, C • N = 12

(D) Correct, As 4f electrons are well shielded from external fields by the overlying 5s and 5p electrons. Thus the magnetic effect of orbital motion is can not be neglected.

Correct option is (a)

75. According to VSEPR theory, the molecule/ion having ideal tetrahedral shape is:

(a) SF₄

(b) SO₄^2–

(d) SO₂Cl₂

Ans. b

Sol.

Correct option is (b)

76. The complex [Mn(H₂O)₆]²⁺ has very light pink colour. The best reason for it is

(a) The complex does not have a charge transfer transition

(b) d-d transitions here are orbital forbidden but spin allowed.

(d) d-d transitions here are both orbital forbidden and spin forbidden.

Ans. d

Sol. For a d⁵ octahedral complex [Mn(H₂O)₆]²⁺, all transitions are Laporte forbidden (orbital forbidden) as well s spin forbidden. Absorptions associated with doubly forbidden transitions are extremely weak, so it is very lightly coloured.

Correct answer is (d)

77. The highest occupied MO in N₂ and O₂⁺ respectively are (take x-axis as internuclear axis)

(a)

(b)

(c)

(d)

Ans. a

Sol. X-axis as internuclear axis

Correct answer is (a)

78. The correct order of LMCT energies is

(a) MnO₄^– < CrO₄^2– < VO₄^3–

(b) MnO₄^– > CrO₄^2– > VO₄^3–

(d) MnO₄^– < CrO₄^2– > VO₄^3–

Ans. a

Sol. In the given case ligand in same, so higher the formal oxidation state of the metal, lower will be LMCT, So, order is MnO₄^– < CrO₄^2– < VO₄^3–

Correct answer is (a)

79. Carboxypeptidase contains:

(a) Zn(II) and hydrolyses CO₂

(b) Zn(II) and hydrolyses peptide bonds

(d) Mg(II) and hydrolyses peptide bonds.

Ans. b

Sol. Carboxypeptidase enzyme contains Zn⁺² ion and it cleaves (hydrolyses) the carboxy terminal amino acid from a peptide chain.

Correct answer is (b)

80. In the EPR spectrum of tetragonal Cu(II) complex, when the unpaired electron resides in the orbital

(a) d_xy

(b)

(c)

(d) d_xz

Ans. b

Sol. gives the evidence that Cu⁺² ion is present in tetragonally distorted octahedral field.

Correct option is (b)

81. The oxidative addition and reductive elimination steps are favoured by

(a) Electron rich metal centres

(b) Electron deficient metal centers

(d) Electron rich and electron deficient metal centers respectively

Ans. d

Sol. Oxidation of metal is easier for electron rich systems, so electron rich metals undergo oxidative addition and as reduction is easier for electron-deficient metal, so it undergoes reductive elimination.

Correct answer is (d)

82. Identify the order according to increasing stability of the following organometallic compounds,

TiMe₄, Ti(CH₂Ph)₄, Ti(i-Pr)₄ and TiEt₄.

(Me = methyl, Ph = phenyl, i-Pr = isopropyl, Et = ethyl)

(a) Ti(CH₂Ph)₄ < Ti(i-Pr)₄ < TiEt₄ < TiMe₄

(b) TiEt₄ < TiMe₄ < Ti(i-Pr)₄ < Ti(CH₂Ph)₄

(d) TiMe₄ < TiEt₄ < Ti(i – Pr)₄ < Ti(CH₂Ph)₄

Ans. c

Sol. Ti(i – Pr)₄ < TiEt₄ < TiMe₄ < Ti(CH₂Ph)₄. This is correct stability order because Ti(CH₂Ph)₄ has no elimination but Ti(i – Pr)₄ has more β-elimination. More elimination less stable.

Correct option is (c)

83. Among the metals, Mn, Fe, Co and Ni the ones those would react in its native form directly with CO giving metal carbonyl compounds are:

(a) Co and Mn

(b) Mn and Fe

(d) Ni and Co

Ans. c

Sol.

Correct answer is (c)

84. The molecule with highest number of lone pairs and has a linear shape based on VSEPR theory is:

(a) CO₂

(b) I₃^–

(d) NO₂⁺

Ans. b

Sol.

Correct option is (b)

85.

A 100 ml of solution is 1080 mg with respect to Ag⁺ and 635 mg with respect to Cu²⁺. If 0.1 mg Ag⁺ left in the solution is considered to be the complete deposition of Ag⁺, the cathode potential, so that no copper is deposited during the process is

(a) 0.16 V

(b) 0.84 V

(d) –0.16 V

Ans. c

Sol.

So, we will have to fix cathode potential above 0.31 volt so that no. Cu is deposited at the cathode.

We can not reduce the potential upto 0.203 V. So, that all Ag⁺ has been deposited.

Correct option is (c)

86. In the H₂Ru₆(CO)₁₈ cluster, containing 8-coordinated Ru centers, the hydrogen atoms are

(a) Both terminal

(b) One terminal and the other bridging

(d) Both bridging between three Ru centers.

Ans. d

Sol.

86-electron cluster H₂Ru₆(CO)₁₈ displays a distorted octahedral metal geometry.

Here each hydride ligand is connected to 3-Ru atoms.

Correct option is (d)

87. In the hydroformylation reaction, the intermediate CH₃CH₂CH₂Co(CO)₄:

(a) Forms are acyl intermediate CH₃CH₂CH₂COCo(CO)₃

(b) Forms an adduct with an olefin reactant

(d) Eliminates propane.

Ans. a

Sol. During hydroformylation the intermediate CH₃ – CH₂ – CH₂ – Co(CO)₄ gets transformed to acyl intermediate CH₃CH₂CH₂ – COCo(CO)₃.

Correct option is (a)

88. Statement I: The sizes of Zr and Hf are similar

Statement II: Size of Hf is affected by lanthanide contraction.

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

(b) Statement I and II are correct but II is not a correct explanation of I

(d) Statements I and II both are incorrect

Ans. a

Sol. As we move from 2nd row transition element (Zr) to 3rd row transition element (Hf), size should increase, but due to lanthanide contraction offset the increase, so Zr and Hf have almost similar size.

Correct option is (a)

89. Consider the compounds, (A) SnF₄ (B) SnCl₄ and (C) R₃SnCl. The nuclear quadrupole splitting are observed for

(a) A, B and C

(b) A and B only

(d) A and C only

Ans. c

Sol. SnF₄ is ionic, here Sn^–4 = 5sº, so no quadrupole splitting.

SnCl₄ i covalent, and Sn is sp³ hybridised. So there is S-electron density in the molecule.

quadrupole splitting is possible

R₃SnCl is covalent, and Sn is sp³, so there is S-electron density in the molecule.

quadrupole splitting is possible.

Correct option is (c)

90. Consider two redox pairs

(1) Cr(II)/Ru(III) (2) Cr(II)/Co(III)

The rate of acceleration in going from a outer sphere to a inner sphere mechanism is lower for (1) relative to (2). Its correct explanation is:

(a) HOMO/LUMO are respectively

(b) HOMO/LUMO are respectively

(d) HOMO/LUMO are respectively

Ans. b

Sol.

transition which is responsible for its slower acceleration in going from outer sphere mechanism to inner sphere mechanis.

Correct option is (b)

91. The correct value of isomer shift (in Mossbauer spectra) and its explanation for Fe(II)-TPP and Fe(III) TPP respectively from the following are:

(TPP = tetraphenylporphyrinate)

(A) 0.52 mms^–1

(B) 0.45 mms^–1

(D) Decrease in s electron density

(a) (A) and (D); (B) and (C)

(b) (A) and (C); (B) and (C)

(d) (B) and (D); (A) and (C)

Ans. b

Sol. Both Fe(II) TPP and Fe(III) – TPP are square planar (because of tetradentate-planar, porphyrine type ligand), and both Fe(II) and Fe(III) are dsp² hybridised, so there is s-electron density, which is responsible for their quadrupole moment and existence of mossbauer spectra.

So both Fe(II)-TPP and Fe(III)-TPP show increase in s-electron density, but Fe(II) has higher s-electron density. So Fe(II) has isomer higher shift (0.52 mms^–1) than Fe(III) (0.45 mms^–1).

Correct option is (b)

92. In IR spectrum of [Co(CN)₅H]^3– the Co-H stretch is observed at 1840 cm^–1. The (Co-D) stretch in [Co(CN)₅D]^3– will appear at nearly

(a) 1300 cm^–1

(b) 1400 cm^–1

(d) 1600 cm^–1

Ans. a

Sol.

Correct option is (a)

93. For the complex

(A) [Ni(H₂O)₆]²⁺

(B) [Mn(H₂O)₆]²⁺

(D) [Ti(H₂O)₆]³⁺

the ideal octahedral geometry will not be observed in

(a) A and D

(b) C and D

(d) D only

Ans. d

Sol. Due to Jahn-Teller distortion

Correct answer is (d)

94. Among the following, the number of anhydrides of acids ae

CO, NO, N₂O, B₂O₃, N₂O₅, SO₃ and P₄O₁₀.

(a) 3

(b) 4

(d) 6

Ans. b

Sol.

Correct option is (b)

95. For a given nuclear fission reaction of ²³⁵U

the amount of energy (in kJ/mol) released during this process is (given ²³⁵U = 235.0439 amu, ¹⁴²Ba = 141.9164 amu, ⁹¹Kr = 90.9234 amu, neutron = 1.00866 amu)

(a) 3.12 × 10¹²

(b) 2.8 × 10¹¹

(d) 1.68 × 10¹⁰

Ans. d

Sol.

E = [(235.0435 + 1.00866) – (141 × 9164 + 90.9234 + 3 × 1.00866)] × 1.67 × 10^–27 × (3 × 10⁸)² J

Correct option is (d)

96. The decomposition of gaseous acetaldehyde at T(K) follows second order kinetics. The half-life of this reaction is 400 s when the initial pressure is 250 Torr. What will be the rate constant (in Torr^–1 s^–1 and half-life (in s) respectively, if the initial pressure of the acetaldehyde is 200 Torr at the same temperature)?

(a) 10⁵ and 500

(b) 10^–5 and 400

(d) 10^–5 and 500

Ans. d

Sol.

Correct option is (d)

97. For an enzyme catalyzed reaction, a Lineweaver-Burk plot gave the following data:

slope = 40 s

intercept = 4 (mmol dm^–3 s^–1)^–1.

If the initial concentration of enzyme is 2.5 × 10^–9 mol dm^–3, what is the catalytic efficiency (in dm^–3 mol^–1 s^–1) of the reaction?

(a) 10⁵

(b) 10⁶

(d) 10⁴

Ans. c

Sol.

Correct option is (c)

98. A hydrogenic orbital with radial function of the form corresponds to

(a)

(b)

(c)

(d)

Ans. b

Sol. Given for any H-orbital

Radial function has form

From the solution of Schrodinger equation for H-atom,

Correct option is (b)

99. For an assembly of molecules (molar mass = M) at temperature T, the standard deviation of Maxweller's speed is approximately

(a)

(b)

(c)

(d)

Ans. a

Sol.

Correct option is (a)

100. The unperturbed energy levels of a system are The second order correction to energy for the ground state in pressure of the perturbation V for which V₁₀ = 2, V₂₀ = 4 and V₁₂ = 6 has been found to be

(a) 6

(b) 0

(d) –8

Ans. a

Sol. Second order perturbation for nth state is given by

Correct option is (a)

101.

Its reducible components are

(a) E + 2A₁ + 2A₂

(b) 2E + A₁ + A₂

(d) E² + 2A₁

Ans. a

Sol.

Correct option is (a)

102. Refer to the character table of the point group C_3v given above. Find which of the following transition is forbidden

(a)

(b)

(c)

(d)

Ans. d

Sol. The direct product of two energy state should transform according to x, y or z axis for electronic transition.

A₁ × A₁ = A₁ = transition with z axis = allowed

A₁ × E = E = Transform with x, y axis = allowed

A₂ × E = E = Transform with x, y axis = allowed

A₁ × A₂ = A₂ = Not transform x, y or z = forbidden

Correct option is (d)

103. The electronic configuration for gadalonium (Gd) is [Xe]4f⁷5d¹6s², where as that of Gd²⁺ is

(a) [Xe]4f⁵5d6s²

(b) [Xe]4f⁶6s²

(d) [Xe]4f⁷5d¹

Ans. d

Sol. ₆₄Gd = [Xe]4f⁷ 5d¹ 6s², Gd⁺² [Xe]4f⁷5d¹ 6s⁰

Removal of electrons takes place from the outermost orbital only.

Correct option is (d)

104. The possible J values for ³D term symbol are

(a) 2

(b) 3

(d) 5

Ans. b

Sol. For ³D

2S + 1 = 3, 2S = 3 – 1 for Term, S, P, D, F

2S = 2, S = 1 Value of L = 0, 1, 2, 3

S = 1, L = 2

J can have values from |L + S| ............ |L – S|

|2 + 1| ........... |2 – 1|

3 2 1

Correct option is (b)

105. The energy levels for cyclobutadiene are The delocalization energy in this molecule is

(a) 0

(b)

(c)

(d)

Ans. a

Sol.

106. The variation of equilibrium constant (K) of a certain reaction with temperature (T) is ln k = given R = 8.3 Jk^–1 mol^–1, the values of are

(a) 166 kJ mol^–1 and 24.9 Jk^–1 mol^–1

(b) 166 kJ mol^–1 and –24.9 Jk^–1 mol^–1

(d) –166 kJ mol^–1 and 24.9 Jk^–1 mol^–1

Ans. c

Sol.

Correct option is (c)

107. The chemical potential of component 1 in a solution of binary mixture is + RT ln p₁, when p₁ is the partial pressure of component 1 vapour phase. The standard state is:

(a) Independent of temperature and pressure

(b) Depends on temperature and pressure both

(d) Depends on pressure only

Ans. c

Sol.

Correct option is (c)

108. Debey Huckel screening length is a measure of size of diffuse ion cloud around an ion, provided at 298 K, which of the following values of is true for a 0.03 molal solution for Na₂SO₄ in water

(a)

(b)

(c)

(d)

Ans. a

Sol.

Correct option is (a)

109. If the ratio of composition of oxidized and reduced species in electrochemical cell, is given as the correct potential difference will be

(a)

(b)

(c)

(d)

Ans. a

Sol. For oxidation reduction half-cell

Oxisised state (O) + ne^– Reduced state (R)

From modified nernst equation (in terms of formal potential Eº)

Correct option is (a)

110. If the equilibrium constants for the reactions 1 and 2

are k₁ and k₂, the equilibrium constant for the reaction

(a) k₁ + k₂

(b) k₁ – k₂

(d) k₁/k₂

Ans. c

Sol.

Correct option is (c)

111. The virial expansion for a real gas can be written in either of the following forms:

(a) PV/RT

(b) RT/PV

(d) RT

Ans. d

Sol. The virial expression for a real gas can be written as

Correct option is (d)

112. A certain system of noninteracting particles has the single-particle partition has the single-particle partition function where A is some constant. The average energy per particle will be

(a)

(b)

(c)

(d)

Ans. a

Sol.

Correct option is (a)

113. Observe the following aqueous solutions of same compound. All the measurements are made at same wavelength and same temperature.

Solution A: The transmittance of 0.1 mol dm^–3 using 1 cm cell is 0.5.

Solution B: The optical density 0.5 mol dm^–3 is measured using 1 mm cell.

Solution C: The transmittance of this solution is 0.1.

The optical density of these solutions follow the order

(log 20 = 1.3010; log 30 = 1.4771, log 50 = 1.6990)

(a) A > B > C

(b) B > C > A

(d) C > A > B

Ans. d

Sol.

Therefore, optical density order is C > A > B.

Correct option is (d)

114. The rotational constant of ¹⁴N₂ is 2 cm^–1. The wave number of incident radiation in a Raman spectrometer is 20487 cm^–1.What is the wave number of first scattered Stokes line (in cm^–1) of ¹⁴N₂?

(a) 20479

(b) 20475

(d) 20495

Ans. b

Sol. In Raman spectra, Stokes lines are separated by 4 B cm^–1 and have wave number value less than the incident radiation, but the first Stoke's line is observed at

6B = 6 × 2 cm^–1 = 12 cm^–1

Therefore, first Stoke line wave number = wave number of incident – 12 cm^–1 radiation.

= (20487 – 12) cm^–1 = 20475 cm^–1

Correct option is (b)

115. For a certain particle encoutering a barrier, the tunneling probability is approximately e^–10. If the mass is halved and width of the barrier (rectangular) doubled, approximate value of the tunneling probability will be

(a)

(b)

(c)

(d)

Ans. b

Sol. Given, Tunelling probability, T = e^–2kL = e^–10

Correct option is (b)

116. An operator A is defined as Which one of the following statements is true?

(a) A is a Hermitian operator

(b) is an antihermitian operator

(d) is Hermitian, but is antihermitian.

Ans. c

Sol.

Correct option is (c)

117. Isothermal which has fractional coverage, linearly, dependent on pressure at low pressures but almost independent at high pressure is called

(a) BET isotherm

(b) Langmuir isotherm

(d) Temkin isotherm

Ans. b

Sol. IN Langmuir isotherm, Fractional coverage

Correct option is (b)

118. A one dimensional crystal of lattice dimension 'a' is metallic. If the structure is distorted in such a way that the lattice dimension is enhanced to '2a'.

(a) The electronic structure remains unchanged

(b) The width of conduction band decreases and a band gap is generated

(d) The width of the conduction band remains unchanged.

Ans. b

Sol. Width of conduction band

Correct option is (b)

119. For a H₂molecule, the ground state wavefunction is refers to the space part and to the spin part. Given that the form of would be

(a)

(b)

(c)

(d)

Ans. c

Sol. From Pauli Principle, "When the labels of any two identical fermions are exchanged, the total wavefunction changes sign".

As electrons are Fermions, So

Correct option is (c)

120. There are several types of mean molar masses for polymer and they are dependent on experimental methods like:

(1) Osmometry (2) Light scattering (3) Sedimentation

Correct relation between mean molar masses and experimental method is:

(a)

(b)

(c)

(d)

Ans. c

Sol.

Correct option is (c)

121. An organic compound (C₇H₁₂O₂) exhibited the following data in the ¹H NMR spectrum.

7.10(1H, d t, J = 16 and 7.2 Hz), 5.90 (1H, d t, J = 16 and 2 Hz),

4.1 (2H, q, J = 7.2 Hz), 2.10(2H, m), 1.25(3H, t, J = 7.2 Hz),

0.90 (3H, t, J = 7.2 Hz) ppm.

The compound, among the choices given below, is:

(a)

(b)

(c)

(d)

Ans. a

Sol. C₇H₁₂O₂. Two important data will differentiate the structure

(i) J value cis and trans alkene

(ii) Position of oxygen in ester.

In the given data J value of define protons is 16.5 that indicates trans structure.

One quartet of 2H at 4.1 ppm confirms structure (a).

Correct option is (a)

122. In the broad band decoupled ¹³C NMR spectrum, the number of signals appearing for the bicyclooctane A-C respectively, are

(a) Five, four and eight

(b) Three, two and five

(d) Three, two and eight

Ans. a

Sol.

Correct option is (a)

123. In the mass spectrum of dichlorobenzene the ratio of the peaks at m/z 146, 148 and 150, is:

(a) 1:1:1

(b) 3:3:1

(d) 9:6:1

Ans. d

Sol. Natural ratio of Cl³⁵ and Cl³⁷ is 3 : 1

Ratio = 9 : 6 : 1

Correct option is (d)

124. The major compound X formed in the following reaction exhibited a strong absorption at v_max 1765 cm^–1 in the IR spectrum. The structure of X is:

(a)

(b)

(c)

(d)

Ans. a

Sol. Chemical reaction involved in the above transformation can be illustrated as

Correct option is (a)

125. The correct order of acidity of the following compound A-C is

(a) B > C > A

(b) C > B > A

(d) A > B > C

Ans. c

Sol.

Correct option is (c)

126. The major product formed in the reaction sequence is

(a)

(b)

(c)

(d)

Ans. d

Sol. Chemical reaction involved in the above transformation can be illustrated as

Correct option is (d)

127. For the following allylic oxidation reaction, the appropriate statement, among the choices given below, is

(a) Suitable reagent is KMnO₄ and the major product is A.

(b) Suitable reagent is KMnO₄ and the major product is B.

(d) Suitable reagent is SeO₂ and the major product is B

Ans. c

Sol. For Allylic oxidation of alkene SeO₂ is the most important reagent. The oxidation are believed to involve an ene reaction between the alkene and the hydrated form of the oxide, followed by [2-3]-sigmatropic rearrangement of the resulting allylseleninic acid and final hydrolysis of the Se(II) ester of the allylic alcohol.

In the case of endo arrangement there is strong electronic repulsion between and lone pair of electron on oxygen which makes it less stable as compared to exo in which this repulsion is minimum. Thus major product will be (A)

Correct option is (c)

128. The intermediate A and the major product B in the following conversion are

(a) A is carbocation and B is

(b) A is a carbanion and B is

(d) A is benzyne and B is

Ans. d

Sol. Chemical reaction involved in the above transformation can be illustrated as

Correct option is (d)

129. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. d

Sol. Chemical reaction involved in the above transformation can be illustrated as

Correct option is (d)

130. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. b

Sol. Chemical reaction involved in the above transformation can be illustrated as

Correct option is (b)

131. The major product formed in the reaction of glucose with benzaldehyde and p-TSA is

(a)

(b)

(c)

(d)

Ans. d

Sol. Chemical reaction involved in the above transformation can be illustrated as

Correct option is (d)

132. Papaverine on oxidation with potassium permanganate gives a ketone, which on fusion with potassium hydroxide gives

(a)

(b)

(c)

(d)

Ans. b

Sol. When papaverine is oxidised with hot conc. permangnate it is broken down into small fragments for example, veratric acid, metahemipinic acids, pyridine-2, 3, 4-tricarboxylic acid and 6, 7-dimethoxyisoquinoline-1-carboxylic acid.

Correct option is (b)

133. The major product formed on nitration (HNO₃/H₂SO₄) of uridine followed by reduction with tin and HCl is

(a)

(b)

(c)

(d)

Ans. c

Sol. Chemical reaction involved in the above transformation can be illustrated as

Correct option is (c)

134. In the following reaction sequence, the correct structures for the major products X and Y are

(a)

(b)

(c)

(d)

Ans. b

Sol. Chemical reaction involved in the above transformation can be illustrated as

Correct option is (b)

135. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. a

Sol. Chemical reaction involved in the above transformation can be illustrated as

Correct option is (a)

136. Match the following

(a) A-iii, B-ii, C-iv

(b) A-iv, B-iii, C-ii

(d) A-ii, B-iii, C-iv

Ans. d

Sol. I. Axis of chirality present in the following compound.

(i) Allene (ii) Biphenys (iii) Bipyrroles (iv) Spiranes

(vi) Hemispiranes

II. Plane of chirality present in the following compound

(i) Ansa compound (ii) Paracyclophanes (iii) Transcyclooctene

III. Helical chirality present in

Correct option is (d)

137. The gauche interaction values for Me/Me, Me/Br and Br/Br are 3.3, 0.8 and 3.0 kJ/mol, respectively. Among the following, the most stable conformation of 2, 3-dibromobutane is:

(a)

(b)

(c)

(d)

Ans. b

Sol. Correct option is (b)

138. The major product formed in the reaction of (S)-1, 2, 4-butanetriol with 3-pentanone in the presence of a catalytic amount of p-TSA is:

(a)

(b)

(c)

(d)

Ans. c

Sol. Chemical reaction involved in the above transformation can be illustrated as

Correct option is (c)

139. The major product formed in the following reaction is:

(a)

(b)

(c)

(d)

Ans. b

Sol. Chemical reaction involved in the above transformation can be illustrated as

Correct option is (b)

140. The major product formed in the following transformation is

(a)

(b)

(c)

(d)

Ans. c

Sol. Chemical reaction involved in the above transformation can be illustrated as

Correct option is (c)

141. The major product formed in the following transformation is

(a)

(b)

(c)

(d)

Ans. d

Sol. Chemical reaction involved in the above transformation can be illustrated as

Correct option is (d)

142. The structure of the major products X and Y in the following transformation are

(a)

(b)

(c)

(d)

Ans. c

Sol. Chemical reaction involved in the above transformation can be illustrated as

Correct option is (c)

143. Match the following

(a) A-i, B-ii, C-iii

(b) A-ii, B-iii, C-iv

(d) A-iii, B-iv, C-i

Ans. d

Sol. Correct answer is (d)

144. Consider the following reaction:

In an experiment, 1.99 g of bromide A on reaction with ethanolic potassium hydroxide gave 1.062 g of a mixture of the olefins B and C. If the ratio of olefins B:C formed is 2:1, the yields for their formation, respectively are

(a) 60 and 30%

(b) 50 and 25%

(d) 54 and 27%

Ans. a

Sol.

Molecular weight of A = 199 g mol^–1

Molecular weight of B = 118 g mol^–1

According to the above balanced equation 199 of A will give product = 118 g

Thus 1.99 will gives the product

But the yield is 1.062 g

Percentage yield =

Since the ratio of B and C is 2 : 1

Therefore, yield of B = 60% and C = 30%

Correct option is (a)

145. An organic compound A(C₈H₁₆O₂) on treatment with an excess of methylmagnesium chloride generated two alcohols B and C, whereas reaction of A with lithium aluminium hydride generated only a single alcohol C. Compound B on treatment with an acid yielded on olefin (C₆H₁₂), which exhibited only a singlet at 1.6 ppm in the ¹H NMR spectrum. The compound A is