PYQ DEC 2011 - VedPrep

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

21. Identify which of the following operators is not hermitian ?

(a)

(b)

(c)

(d) x2

Ans. b

Sol. (Hermitian) and (Not hermitian)

For option (a),

It is hermitian.

For option (b),

It is Not hermitian.

Correct answer is (b)

22. The term symbol for the ground state of nitrogen atoms is

(a)

(b)

(c)

(d)

Ans. d

Sol.

23. PA and PB denote the populations of two energy states EA and EB , and EA > EB. The correct statement when the temperature T1 > T2 is

(a)

(b)

(c)

(d)

Ans. c

Sol. Population

Therefore at constant temperature lower the energy (E) higher will be population.

Also higher the energy value, higher will be the change in population ratio when temperature is changed from T2 to T1

24. The uncertainty in the NMR frequency of a compound in liquid state (relaxation time =1s) is 0.1 Hz. The uncertainty in the frequency (in Hz) of same compound in solid state (relaxation time = 10–4 s) is

(a) 10–4

(b) 100

(d) 10–3

Ans. c

Sol. From Heisenberg uncertaintity principle,

(where = life time or Relaxation time and = uncertainty in frequency)

From equation (ii) and (iii)

Correct answer is (c)

25. Which one of the following conductometric titrations will show a linear increase of the conductance with volume of the titrant added up to the break point and an almost constant conductance afterwards.

(a) A strong acid with a strong base

(b) A strong acid with a weak base

(d) A weak acid a weak base.

Ans. d

Sol.

When weak base is added to weak acid then salt is formed which is more dissocited w.r.t. weak acid, so conductance increases upto end point but after the end point only concentration of weak base increases which remains in almost undissociated form so does not lead to increases in conductance.

Correct answer is (d).

26. Flocculation value of K2SO4 is much less than that of KBr for Sol A. Floccultion value of CaCl2 is much less than that of NaCl for Sol B. Which of the following statements is correct?

(a) Sol A is negatively charged and Sol B is positively charged

(b) Both the sols are negatively charged.

(d) Both the sols are positively charged.

Ans. c

Sol. If charge on cation is high then its flocculation value will be low for an anionic sol and if charge on anion is high then its flocculation value will be low for a cationic Sol. As K2SO4has low flocculation value than KBr (due to SO42– has higher are charge than Br–) ; so sol. A is cationic in nature.

On sol B, CaCl2 has low flocculation value than NaCl due to higher the change on calcium ion than sodium ion. So sol B is negatively charged.

Correct answer is (c).

27. For a system of constant composition, the pressure (P) is given by.

(a)

(b)

(c)

(d)

Ans. b

Sol.

Correct answer is (b)

28. The value of d111 in a cubic crystal is 325.6 pm. The value of d333 is

(a) 325.6 pm

(b) 976.8 pm

(d) 625.6 pm

Ans. c

Sol.

Correct answer is (c)

29. The symmetry point group of ethane in its staggered conformation is

(a) C3v

(b) D3d

(d) S6

Ans. b

Sol. Correct answer is (b)

30. TFor the reaction C2H4 (g) + 3O2 (g) 2CO2(g) + 2H2O (l), The value of (in kJ) at 300 K and 1 bar is

(a) –5.0

(b) 0.0

(d) 5.0

Ans. a

Sol.

Correct answer is (a)

31. The sodium D lines are due to transitions. The splitting due to spin-orbit coupling in 2P state of the sodium atom is

(a)

(b)

(c)

(d)

Ans. b

Sol.

n = splitting due to spin orbital coupling =

Correct answer is (b)

32. The rate constant of a unimolecular reaction was 2.66 × 10–3 s–1 and 2.2 × 10–1 s–1 at T = 120 K and 360 K respectively. The rate constant (in s–1 units) at 240 K would be

(a) 2.4 × 10–2

(b) 2.4 × 10–1

(d) 1.8 × 10–3

Ans. a

Sol. If k1 and k2 are rate constant at temperature T1 and T2 respectively and is temperature coefficient, then these are related by the expression

Now at temperature 240 K, rate constant (k3)

Correct answer is (a)

33. For a potentiometric titration, in the curve of emf (E) vs volume (V) of the titrant added, the equivalence point is indicated by

(a)

(b)

(c)

(d)

Ans. c

Sol. In potentiometric titration the rate of change in slope of E vs V graph has maximum at the equivalence point

And as can be positive or negative. But will always be greater than zero.

Correct answer is (c)

34. The osmotic pressure of a polymer sample at different concentrations (c) was measured at T(K). A plot of versus c gave a straight line with slope (m) and intercept (c´). The number average molecular weight of the polymer is (R = gas constant).

(a)

(b)

(d) mRT

Ans. b

Sol. Osmotic pressure,

where C1 = concentration of polymer in mol L–1 unit and B = constant.

When concentration is expressed in terms of 'c' (gL–1) then, C1 = Mc

Correct answer is (b)

35. The concentration of a reactant undergoing decomposition was 0.1, 0.08 and 0.067 mol L–1 after 1.0, 2.0 and 3.0 hr respectively. The order of the reaction is

(a) 0

(b) 1

(d) 3

Ans. c

Sol.

Above data first with second order rate law expression for the reaction

From data (i) and (ii)

From data (i) and (iii)

As the value of k is constant so it is second order reaction

Correct answer is (c)

36. A particle is constrained in one dimensional box of length 2a with potential x < –a, x > a and V(x) = 0; –a < x < a. Energy difference between levels n = 3 and n = 2 is

(a)

(b)

(c)

(d)

Ans. d

Sol. For 1-D box, the energy is

Correct answer is (d)

37. In the 19F NMR spectrum of PF5, the number of signals and multiplicity, at room temperature are

(a) one, singlet

(b) one, doublet

(d) two, singlet

Ans. b

Sol. At room temperature due to rapid interconversion of axial and equatorial F atoms, all the fluorine will be equivalent. Hence 19F NMR of PF5 at room temperature will appear as doublet of 5F.

Correct answer is (b).

38. The correct statement regarding closo-{BnHn} species is

(a) it always has –2 charge

(b) it always has +2 charge

(d) It is more reactive than nido arachno-, and hypo-boranes

Ans. a

Sol. Closo BnHn species has always –2 charge.

Correct answer is (a)

39. Lewis acidity of BCl3, BPh3 and BMe3 with respect to pyridine follows the order

(a) BCl3 > BPh3 > BMe3

(b) BMe3 > BPh3 > BCl3

(d) BCl3 > BMe3 > BPh3

Ans. c

Sol.

B-attracts B-CH3 electron (Hyperconjugation)

B and C both are sp2-hybridised. No shifting of electrons (Resonance).

BPh3 > BMe3 > BCl3

Correct answer is (c)

40. Superoxide dismutase contains the metal ions

(a) Zn(II) and Ni(II)

(b) Cu(II) and Zn (II)

(d) Cu(II) and Fe(III)

Ans. b

Sol. Correct answer is (b)

41. The number of antibonding electrons in NO and CO according to MO theory are respectively.

(a) 1, 0

(b) 2, 2

(d) 2, 3

Ans. a

Sol. Energy level for NO,

Number of electrons in antibonding MO = 1

Energy level for CO,

No electron is present in antibonding MO.

So, number of antibonding electron in CO is zero

Correct answer is (a)

42. The correct combination of metal, number of carbonyl ligands and the charge for a metal carbonyl complex [M(CO)x]z– that satisfied the 18 electron rule is

(a) M = Ti, x = 6, z = 1

(b) M = V, x = 6, z = 1

(d) M = Mo, x = 5, z = 1

Ans. b

Sol. M = V, x = 6, z = 1

i.e.

Correct answer is (b).

43. Among the following pairs

Those in which the first ionization energies differ by more than 300 kJ mole–1 are:

(a) (1) and (3) only

(b) (1) and (2) only

(d) (3) and (4) only

Ans. b

Sol. (A) IE1 for O and S are 1314 and 1000 kJ mol–1.

difference = 1314 – 1000 = 314

(B) For N and P IE1 are 1402 and 1012.

difference = 1402 – 1012 = 390

difference = 1012 – 947 = 165 (less 300 kJ mol–1).

(D) Also for IE1 difference is less than 300 kJ mol–1.

Correct answer is (b)

44. The stable cyclopentadienyl complex of beryllium is

(a)

(b)

(c)

(d)

Ans. d

Sol. Because main group element follows octet rule.

2 + 5 + 1 = 8

Correct answer is (d)

45. The reaction between NH4Br and Na metal in liquid ammonia (solvent) results in the products

(a) NaBr, HBr

(b) NaBr, H2

(d) HBr, H2

Ans. b

Sol.

Correct answer is (b)

46. The material that exhibits the highest electrical conductivity among the following sulfur-nitrogen compounds is

(a) S4N4

(b) S7NH

(d) (SN)x

Ans. d

Sol. (SN)x because it is a giant cross linked molecule and have mobile electrons.

Correct answer is (d)

47. Uranium fluorides co-precipitate with

(a) CaF2

(b) AgF

(d) MgF2

Ans. a

Sol. Correct answer is (a)

48. The acid-base indicator (HIn) shows a colour change at pH 6.40 when 20% of it is ionized. The dissociation constant of the indicator is

(a) 9.95 × 10–8

(b) 3.95 × 10–6

(d) 6.0 × 10–8

Ans. a

Sol.

Correct answer (a)

49. The actual magnetic moment shows a large deviation from the spin-only formula in the case of

(a) Ti3+

(b) V3+

(d) Sm3+

Ans. d

Sol. Sm3+ due to spin-orbital coupling

Correct answer is (d)

50. The complex that absorbs light of shortest wavelength is

(a) [CoF6]3–

(b) [Co(H2O)6]3+

(d) [Co(OX)3]3– (OX = C2O42–)

Ans. c

Sol.

or lowest value of wavelength is absorbed

Correct answer is (c).

51. Two particles having speeds S1 and S2 have kinetic energies 1 and 2 MeV respectively; the relationship between S1 and S2 is

(a) S1 = 2S2

(b) S2 = 2S1

(c)

(d)

Ans. c

Sol.

Correct answer is (c)

52. Green coloured Ni(PPh2Et)2Br2, has a magnetic moment of 3.20 B.M. The geometry and the number of isomers possible for the complex respectively are

(a) Square planar and one

(b) tetrahedral and one

(d) tetrahedral and two

Ans. b

Sol. Ni(PPh2Et)2Br2;

Number of unpaired electrons = 2.

Thus this complex is tetrahedral. Tetrahedral complexes show only one isomer.

Magnetic moment corresponding to two unpaired electrons = 2.9 B.M. The hesh values 3.20 BM is due to spin-orbit coupling.

Correct answer is (b)

53. The chemiluminescence method for determining NO in environmental sample is based on formation of NO2* (excited) which is generally generated by reacting NO with

(a) O₂

(b)

(d)

Ans. a

Sol.

54. In the IR spectrum, carbonyl absorption band for the following compound appears at

(a) 1810 cm–1

(b) 1770 cm–1

(d) 1690 cm–1

Ans. b

Sol. C = O stretching frequency for simple ester 1750 – 1725 cm–1.

For six membered lactone = 1750 – 1735 cm–1

For five membered lactone = 1795 – 1760 cm–1

Correct answer is (b)

55. Among the following compounds, the formyl anion equivalent is

(a) acetylene

(b) nitromethane

(d) 1, 4-dithiane

Ans. b

Sol. Nitromethane is used as a synthetic equivalent for formyl anion.

Correct answer is (b)

56. In the following concerted reaction, the product is formed by a

(a) disrotatory electrocyclization

(b) disrotatory electrocyclization

(d) conrotatory electrocyclization

Ans. a

Sol. According to the given product stereochemistry, the product is formed by disrotatory electrocyclization

Correct answer is (a)

57. A suitable reagent combination for carrying out the following conversion is

(a) trimethyl orthoacetate and p-toluenesulfonic acid

(b) trimethyl ortho acetate and sodium hydroxide

(d) 2-methoxypropene and sodium hydroxide

Ans. c

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

Correct answer is (c)

58. The IUPAC name of the following compound is

(a) (R)-3-(prop-2-enyl) hex-5-ynoic acid

(b) (S)-3-(prop-2-enyl) hex-5-ynoic acid

(d) (S)-3-(prop-2-ynyl) hex-5-enoic acid

Ans. b

Sol. IUPAC name of above compound can be written using following keynotes

(1) carboxylic group has to be given highest priority

(2) keeping the chain tripple bond is parent chain.

(S)-3-(prop-2 enyl) hex-5-ynoic acid

Correct answer is (b).

59. In the mass spectrum of dodecahedrane (C20H20), approximate ratio of the peaks at m/z 260 and 261 is

(a) 1 : 1

(b) 5 : 1

(d) 20 : 1

Ans. b

Sol. In C20H20 due to 20 carbons, there will be 20 × 1.1% probability of one 13C in the molecule, where 1.1% is natural abundance of 13C with molecular ion, there will be 22% of M + 1 signal. Hence if 260 is 100% and 261 will be 22%.

60. The reaction given below proceeds through

(a)

(b)

(c)

(d)

Ans. a

Sol.

Correct answer is (a)

61. Among the following drugs, the anticancer agents is

(a) Captopril

(b) chloroquine

(d) ranitidine

Ans. c

Sol. Correct answer is (c)

62. The reaction that involves the formation of both C-C and C-O bonds is

(a) Diels-Alder reaction

(b) Darzen's glycidic ester condensation

(d) Beckmann rearrangement

Ans. b

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

Correct answer is (b)

63. Among A-C, the aromatic compound are

(a) A, B and C

(b) A and B only

(d) A and C only

Ans. c

Sol. Since in case of A the hydrogens present at 1st and 6th position repel each others. Hence planarity is distorts. While in case of B and C this type of planarity is maintained

Correct answer is (c)

64. In the following Markownikov addition reaction, the products A and B are

(a) homomers

(b) enantiomers

(d) regioisomers

Ans. c

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

Compound A and B are diastereomers due to difference in absolute configuration at carbon 2 and 3.

Correct answer is (c)

65. 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 answer is (a)

66. Among A-C, the compounds which can exhibit optical activity are

(a) A, B and C

(b) A and B only

(d) B and C only

Ans. c

Sol. Compound B can be represented in fisher projection as shown below

Here in fisher projection it contain plane of symmetry passes through mid of C2-C3. Hence it is optically inactive.

Correct answer is (c)

67. 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 answer is (a)

68. An organic compound (MF : C8H10O) exhibited the following 1H NMR spectral data : 2.5 (3 H,s), 3.8 (3H, s), 6.8 (2 H, d, J 8 = Hz), 7.2 (2H, d, J = 8 Hz) ppm. The compound among the choices, is

(a) 4-ethylphenol

(b) 2-ethylphenol

(d) 4-methylbenzyl alcohol

Ans. c

Sol. Characterization of organic compound can be done as

indicates para hetro disubstituted benzene ring.

Correct answer is (c)

69. With respect to electrophilic aromatic substitution, reactivity order of pyrrole, pyridine and indole is

(a) indole > pyrrole > pyridine

(b) pyrrole > pyridine > indole

(d) indole > pyridine > pyrrole

Ans. c

Sol. pyrrole > Indole > Pyridine

Pyrrole is considered to be centre where in HOMO is at higher energy level where as Pyridine is - sin k centre whose HOMO is at low energy level, and pyridine gives nucleophilic substitution reaction rather then electrophilic substitution reaction.

Correct answer is (c)

70. The most appropriate reagent suitable for the conversion of 2-octyne into trans-2-octene is

(a) zinc and acetic acid

(b) 10% Pd/C

(d) hydrazine hydrate

Ans. c

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

Correct answer is (c)

71. Consider a n-type semiconductor whose Ev = 0, Ec = 2.0 eV and Ed = 1.98 eV. The correct statement among the following is

(a) Ef = 1 eV and is independent of T

(b) Ef = 1.99 eV and remains independent of T

(d) Ef = 1.99 eV and decreases with increase of T.

Ans. d

Sol.

For fermi level lies exactly half any between the bottom or conduction band and the donor level. But as temperature is increased fermi level drops (i.e. Ef decreases).

Correct answer is (d)

72. Reaction of Fe(CO)5 with OH– leads to complex A which on oxidation with MnO2 gives B. Compounds A and B respectively are

(a) [HFe(CO)4]– and Fe3(CO)12

(b) [Fe(CO)5(OH)]– and Fe2(CO)9

(d) [HFe(CO)4]– and Fe2O3

Ans. a

Sol. Correct answer is (a)

73. For the reaction H2O(g) + C(graphite) CO(g) + H2O(g), the variation of energy parameter of the reaction over a large temperature range is shown below. The correct identification of the curves is given by

(a)

(b)

(c)

(d)

Ans. d

Sol.

As the number of mole of gaseous product is more than reactant

So there is overall increase in entropy

will increase with increase in temperature

Enthalpy of reaction is independent of temperature

So with increases of temperature decreases because

Correct answer is (d)

74. A Sodalite cage in zeolites is

(a) a truncated tetrahedron

(b) an icosahedron

(d) a dodecahedron

Ans. c

Sol. Correct answer is (c)

75. Two moles of a nonvolatile solute is dissolved in 48 mol of water and the resultant solution has a vapour pressure of 0.0392 bar at 300 K. If the vapour pressure of pure water at 300 K is 0.0400 bar, the activity coefficient of water in the solution is

(a) 0.96

(b) 0.98

(d) 1.02

Ans. d

Sol. From Roult's law, Relative lowering of vapour pressure = activity of solute

Correct answer (d)

76. The final product (s) of the reaction P(OR)3 + R´X is/are

(a) R´PO(OR)2 and RX

(b) [R´PO(OR)2]X

(d) ROR´ and P(OR)2 X

Ans. a

Sol.

Because P–O bond very strong affinity very high not easily break. So, R–O bond break.

Correct answer is (a)

77. 1 mol of CO2, 1 mol of N2 and 2 mol of O2 were mixed at 300 K. The entropy of mixing is

(a)

(b)

(c)

(d)

Ans. a

Sol. Since, we know that for the entropy of mixing is

Correct answer is (a)

78. For the eigenstates of the hydrogen atom, which of the following relations between the expectation value of kinetic energy (T) and potential (V) holds true?

(a)

(b)

(c)

(d)

Ans. b

Sol. For H-atom, virial theorem

(i) V = 2E (ii) V = –2T (iii) T = –E

Where, T = total energy

T = kinetic energy

V = potential energy

Correct answer is (b).

79. For the liquid vapour equilibrium of a substance at 1 bar and 400 is 8 × 10–3 bar K–1. If the molar volume in the vapour form is 200 L mol–1 and the molar volume in the liquid form is negligible, the molar enthalpy of vapourisation is (1.0 bar L = 100 J)

(a) 640 kJ mol–1

(b) 100 kJ mol–1

(d) 64 kJ mol–1

Ans. d

Sol.

Or,

= 640 × 100 J mol–1

= 64 kJ mol–1

Correct answer is (d)

80. The correct order of acidity among the following species is

(a)

(b)

(c)

(d)

Ans. b

Sol. Higher the charge and smaller size of the central metal cation increases the acidic strength of hydrated complex because it polarizes O–H band of H2O easily.

In [Sc(H2O)6]3+, Sc is in +3 oxidation state and has smallest size.

Therefore, in size of depositive ions increase in the order Ni2+ < Mn2+. Na+ has lowest oxidation state and largest size.

Correct answer is (b)

81. The Langmuir adsorption isotherm is given by where P is the pressure of the adsorabate gas. The Langmuir adsorption isotherm for a diatomic gas A2 undergoing dissociative adsorption is

(a)

(b)

(c)

(d)

Ans. d

Sol. For non-dissociative Langmuir adsorption,

Correct answer is (d)

82. The standard electrode potentials (Eº) of Fe3+ / Fe2+ and Fe2+ / Fe electrodes are +0.77 V and –0.44 V respectively at 300 K. The Eº of Fe3+ / Fe electrode at the same temperature is

(a) 1.21 V

(b) 0.33 V

(d) –0.04 V

Ans. d

Sol.

Adding above half cell reactions

Correct answer is (d)

83. Which of the following is true for the radial part of the hydrogen atom wavefunctions (n principal quantum number) and the nodes associated with them?

(a) The radial part of only s function is non-zero at the origin and has (n – 1) nodes.

(b) The radial part of s function is zero at the origin and has n number of nodes.

(d) The radial parts of all s functions are zero at the origin and have no nodes.

Ans. a

Sol. Number of radial nodes = n – l – 1

For s-function l = 0

Therefore, number of nodes = n – 0 – 1 = n – 1

Every function either s, p, d or f have non zero at the origin.

Correct answer is (a)

84. For non-degenerate perturbation theory for ground state, with as zeroth order energy, as the first order perturbation correction and E0 as the exact energy, which of the following is true?

(a)

(b)

(c)

(d)

Ans. c

Sol. According to "The variation principle" If any orbitary wavefunction is used to calculate the energy, the value calculated is never less the true energy.

Energy calculated by using an arbitary wavefunction and applying perturbation will always be greater than or equal to true energy.

Correct answer is (c)

85. Observe the following electronic transition of a diatomic molecule.

The allowed transitions are

(a) A and C only

(b) B and D only

(d) A, C and D only

Ans. c

Sol. For allowed transitions are

Correct answer is (b)

86. An excited triplet state wave function of hydrogen molecule with the electronic configuration has the following space part

(a)

(b)

(c)

(d)

Ans. c

Sol. According to the pauli exclusion principle, the wavefunction of fermions (here electrons) must be antisymmetric

For a triples state, spin wavefunctions can be which are symmetric in nature. As So, for to be antisymmetrical w.r.t. electron exchange, must be antisymmetric because for triplet state is symmetric so,

Correct answer is (c).

87. The NMR spectrum of AX3 exhibits lines at = 2.1 and 2.3 ppm (for X type protons) and = 4.1, 4.3, 4.5 and 4.7 ppm (for A type protons), measured from TMS with an instrument operating at 100 MHz. The chemical shift (in ppm) of A and X protons and coupling constant (in Hz) are respectively.

(a) 4.4, 2.2 and 20

(b) 2.2, 4.4 and 10

(d) 4.3, 2.1 and 20

Ans. a

Sol. In AX3 A proton spectra will be a quartet (Intensity ratio 1 : 3 : 3 : 1) at = 4.1, 4.3, 4.5 and 4.7 ppm and X proton spectra will be a doublet (Intensity ratio 1 : 1)

88. The character table of the C2v point group is given below:

The two function and (where pk is the p-orbital on the kth atom of cis-butadiene and is the molecular plane) belong to

(a) A1 and A2 respectively

(b) Both A2

(d) B1 and B2 respectively

Ans. c

Sol.

Both the molecular orbitals belongs to B2.

Correct answer is (c).

89. If denotes the characteristic temperature of rotation then the magnitude of (assume the bond lengths to be the same for all the molecules) is

(a) 2/3

(b) 3/2

(d) 9/8

Ans. c

Sol.

Correct answer is (c)

90. The overall reaction for the passage of 1.0 faraday of charge in the following cell

is given by (t denotes the transport numbers)

(a)

(b)

(c)

(d)

Ans. a

Sol. Let one faraday of electricity is withdrawn from the cell

The net effect is obtained by adding the above four changes, which gives

Correct answer is (a)

91. A system consisting of 4 identical and distinguishable particle, each possessing three available states of 1, 2 and 3 units, has 10 units of energy. The number of ways, W, in which these conditions are satisfied is

(a) 2

(b) 4

(d) 10

Ans. d

Sol. In order to have 10 unit total energy only 2 combinations of selecting particles are possible

(i) One particle with 1 unit energy and 3 particles with 3 unit energy each

Therefore, number of ways possible = 4C1 × 3C3 = 4 × 1 = 4 ways

(Because first particle can be selected in 4 ways next 3 particles from remaining 3-particles can be selected in only 1 way)

(ii) 2 particles with 2 unit energy each and 2 particles with 3 unit energy each

Therefore, number of ways possible = (selection of 2 particles out of 4-particles) × (selecting of 2 particles out of remaining 2-particles) = 4C2 × 2C2 = 6 × 1 = 6 ways

Therefore total number of ways = 4 + 6 = 10 ways.

Correct answer is (d)

92. The molar conductivities at infinite dilution for Na2SO4, K2SO4, KCl, HCl and HCOONa at 300 K are 260, 308, 150, 426 and 105 S cm–1 mol–2 respectively. Hence or formic acid in the same unit and at the same temperature is

(a) 381

(b) 405

(d) 531

Ans. b

Sol.

Correct answer is (b).

93. If the displacement vectors of all atoms in cis-butadiene are taken as the basis vectors the characters of the reducible representation of (molecular plane) and

(a) 30, 10, 30, 0

(b) 30, 0, 10, 0

(d) 30, 0, 20, 0

Ans. b

Sol. Basis set is displacement of vectors of all atoms = 3 N coordinates.

Correct answer is (b)

94. In least square fitting of a data set {XiVi} to the equation Y = A.X, the regression coefficient (A) is estimated by

(a)

(b)

(c)

(d)

Ans. b

Sol. For data set {Xi Yi}. When equation is Y = AX + B

Its deviatinos are token from the middle period so that the value of A and B can be directly written as

Correct answer is (b).

95. At any temperature for the following reaction (D and T are deuterium and tritium respectively) correct statement is

(a) A is fastest

(b) B is fastest

(d) All the above reactions have the same rate constant.

Ans. a

Sol. Heavier isotope contains stronger bonding. Stronger the bonding reaction rate is slow. But in lighter isotope bonding is weak, so reaction rate is faster.

Correct answer is (a)

96. An example of a relaxation method of measuring rates is:

(a) Spectroscopic monitoring of product concentration.

(b) Stopped flow technique

(d) Measurements of spectral line widths

Ans. c

Sol. When equilibrium reactions are disturbed from their equilibrium condition via either T-Jump or pH-Jump then measurement of relaxation time leads to the study of reaction rate.

Correct answer is (c).

97. The overall rate of the following complex reaction,

by steady state approximation would be

(a) K1K2k3[A]3[B]

(b) K2K1k3[A][B]3

(d) K1K2k3[A][B]

Ans. a

Sol. Overall reaction can be obtained by simple addition of mechanism steps,

So,

As A2 and C are not appearing in the overall reaction so they are intermediates. As equilibrium is fast so concentration of [A2] and [C] can be determined using equilibrium conditions.

From reaction (1), (2) and (3), we get

Correct answer is (a).

98. The vibrational energy levels, and of a diatomic molecule are separated by 2143 cm–1. Its anharmonicity is 14 cm–1. The values of (in cm–1) and first overtone (cm–1) of this molecule are respectively.

(a) 2143 and 4286

(b) 2157 and 4286

(d) 2171 and 4258

Ans. a

Sol.

First overtone more number is given by,

Correct answer is (a).

99. The addition polymerization of M (monomer) involves the following stages:

(I = initiator, R = free radical)

The rate constant for free radical formation is 2 × 10–3 s–1. The initial concentration of initiator is 10–3 mol dm–3. The overall rate of the reaction is 4 × 10–3 mol dm–3 s–1. Assuming steady state approximation for free radical, the kinetic chain length is:

(a) 2000

(b) 8 × 109

(d) 200

Ans. a

Sol. Kinetic chain length =

Correct answer is (a)

100. The electronic spectrum of [CrF6]3– shows three bands at 14,900 cm–1, 22400 cm–1 and 34,800 cm–1. The value of in this case is

(a) 5,500 cm–1

(b) 14,900 cm–1

(d) 34,800 cm–1

Ans. b

Sol. For d3 and d8, the lowest transition gives the value of

101. Among the following pairs, those in which both species have similar structures are:

(a) A and B only

(b) A and C only

(d) B, C and D only

Ans. b

Sol.

Correct answer is (b)

102. The number of metal-metal bonds in the dimers, [CpFe(CO)(NO)]2 and [CpMo(CO)3]2 respectively, are

(a) two and two

(b) two and three

(d) zero and one

Ans. d

Sol. [CpFe(CO)(NO)]2

Step-1: Determine the total valence electron (TVE) = A

A = [5 + 8 + 2 + 3] × 2 = [18] × 2 = 36

Step-2: B = (n × 18) – A, where n = number of metal.

B = (2 × 18) – 36 = 36 – 36 = 0

No metal-metal bond found.

[CpMo(CO)3]2

Step-1: A = [5 + 6 + 2 × 3] × 2 = [17] × 2 = 34

Step-2: B = (n × 18) – A = 2 × 18 – 34 = 36 – 34 = 2

Step-3: B/2 = gives the total number of M-M bonds in the complex

B = 2 and B/2 = 2/2 = 1

In this complex, the total number of M-M bonds is one.

Correct answer is (d).

103. The reduction of nitrogen to ammonia, carried out by the enzyme nitrogenase, needs,

(a) 2 electrons

(b) 4 electrons

(d) 8 electrons

Ans. d

Sol.

Correct answer is (d)

104. In the titration of 50 mL of 0.1 M HCl with 0.1 M NaOH using methyl orange as an indicator, the end point (color change) occurs as pH reaches 4.0. The titration error is:

(a) –0.2%

(b) –84.7%1

(d) +84.2%1

Ans. a

Sol. Correct answer is (a)

105. The styx code of B4H10 is

(a) 4120

(b) 4220

(d) 3203

Ans. c

Sol.

Correct answer is (c)

106. Match list 1 (compounds) with list II (structures), and select the correct answer using the codes using below

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

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

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

Ans. c

Sol. XeO4 = tetrahedral, BrF4– = Square Planar, SeCl4 = distorted Tetrahedral

Correct answer is (c)

107. In the trans-PtCl2L(CO) complex, the CO stretching frequency for L = NH3, pyridine, NMe3 decreases in the order.

(a) pyridine > NH3 > NMe3

(b) NH3 > pyridine > NMe3

(d) pyridine > NMe3 > NH3

Ans. a

Sol. Order of donor ability NMe3 > NH3 > Py

Order of Py > NH3 > NMe3

Correct answer is (a)

108. For the nuclear reactions.

(Given masses: 8Be = 8.005300, 4He = 4.002603 and The correct statement is

(a) (A) and (B) are both spontaneous fission processes.

(b) (A) is spontaneous fission but (B) is not

(d) Both (A) and (B) are not spontaneous fission processes.

Ans. b

Sol.

There is mass loss. Thus, energy evolve.

Therefore, exothermic

No mass loss, no energy released

Therefore, endothermic

Correct answer is (b)

109. A metal ion that replace manganese (II) ion in mangano-proteins without changing its function, is

(a) Fe(II)

(b) Zn(II)

(d) Cu(II)

Ans. c

Sol. Correct answer is (c)

110. In 57Fe* Mossbauer experiment, source of 14.4 keV (equivalent to 3.48 × 1012 MHz) is moved towards absorber at a velocity of 2.2 mm s–1. The shift in frequency of the source for this sample is:

(a) 35.5 MHz

(b) 25.5 MHz

(d) 15.5 MHz

Ans. b

Sol. Frequency shift where V = relatime velocity of source and observer

frequency of emitted radiation

Correct answer is (b)

111. Bayer's process involves

(a) Synthesis of B2H6 from NaBH4

(b) Synthesis of NaBH4 from borax

(d) Synthesis of B3N3H6 from B2H6

Ans. b

Sol.

Correct answer is (b)

112. A true statement about base hydrolysis of [Co(NH3)5Cl]2+ is

(a) It is a first order reaction

(b) The rate determining step involves the dissociation of chloride in [Co(NH3)4(NH2)Cl]+

(d) The rate determining step involves the abstraction of a proton from [Co(NH3)5Cl]2+

Ans. b

Sol. Substitution nucleophilic unimolecular conjugate base mechanism.

Correct answer is (b)

113. The catalyst involved in carrying out the metathesis of 1-butene to give ethylene and 3-hexene is

(a)

(b) Na2PdCl4

(d) RhCl(PPh3)3

Ans. a

Sol. Correct answer is (a).

114. The correct order of d-orbital splitting in a trigonal bipyramidal geometry is

(a)

(b)

(c)

(d)

Ans. d

Sol.

Correct answer is (d)

115. For the following outer sphere electron transfer reactions.

the rate constants are 10–6 M–1 s–1 and 8.2 × 102 M–1s–1 respectively. This difference in the rate constants is due to

(a) A change from high spin to low spin in Co* and high spin to low spin in Ru

(b) A change from high spin to low spin in Co* and low spin to high spin Ru*

(d) A change from low spin to high spin in Co* and high spin to low spin in Ru*

Ans. c

Sol.

Correct answer is (c)

116. The greater stability of compared to that of ((CH3)2CH – CH2 – )4Ti(B) is due to

(a) Hyperconjugation present in complex (A)

(b) -hydride elimination is not possible in complex (A)

(d) The stronger nature of Ti-C bond in complex (A)

Ans. b

Sol.

Correct answer is (b)

117. The coordination number and geometry of cerium in [Ce(NO3)6]2– are respectively,

(a) 6 and octahedron

(b) 6 and trigonal prism

(d) 12 and icosahedron

Ans. d

Sol. For Ce4+ ions NO3– behaves as bidentate ligand. Thus C.N. for [Ce(NO3)6]2– is 12. and shape corresponding to C.N. 12 is icosahedron.

Correct answer is (d)

118. A compound A having the composition FeC9H8O3 shows one signal at 2.5 ppm and another one around 5.0 ppm in its 1H NMR spectrum. The IR spectrum of this compound shows two bands around and 1680 cm–1. The compound follows the 18 electron rule of the following statements for A, the correct one is/are

(A) It has group.

(B) It has a terminal CO ligand

(D) It has Fe-H bond

(a) A and B only

(b) C only

(d) B and D only

Ans. a

Sol.

Correct answer is (a)

119. In bacterial rubredoxin, the number of iron atoms, sulfur bridges and cysteine ligands are

Fe atom Sulfer bridge Cysteine

(a) 4 4 4

(b) 2 2 4

(d) 1 0 4

Ans. d

Sol.

Correct answer is (d)

120. In the following reaction, the product formed and the mechanism involved are

(a) A is and is formed by addition-elimination mechanism

(b) A is and is formed by benzyne mechanism

(d) A is and is formed by SN2 displacement

Ans. a

Sol. This reaction is supposed to be proceeds through addition elimination mechanism as shown below.

Correct answer is (a)

121. An optically active compound enriched with R-enantiomer (60% ee) exhibited If the value of the sample is –135º, the ratio of R and S enantiomers would be

(a) R:S = 1:19

(b) R:S = 19:1

(d) R:S = 9:1

Ans. a

Sol. 60% enantiomeric excess of R-enantiomer gives +90º rotation.

Specific rotation of R enantiomer

Hence, specific rotation of R-enantiomer = +150º

specific rotation of S-enantiomer = –150º.

Now, for sample optical rotation = –135º

Enantiomeric excess of S isomer

Racemic part will be 10% (5% R-isomer, 5% S-isomer)

Total R = 5%, total S = 90 + 5 = 95%

Ratio R : S = 5 : 95 = 1 : 19

Correct answer is (a)

122. Match the amino acids with their structures:

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

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

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

Ans. d

Sol. Correct answer is (d)

123. Statement I: U(VI) is more stable than Nd(VI)

Statement II: The valence electrons in U are in 5f, 6d and 7s orbitals.

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

(b) Statement I and II are correct but statement II is not an explanation for Statement I

(d) Statements I and II both are incorrect

Ans. b

Sol. Correct answer is (b)

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

(a)

(b)

(c)

(d)

Ans. d

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

The last step in this reaction includes 1, 3 dipolar cycloaddition reaction also known as click reaction.

Correct answer is (d)

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

(a)

(b)

(c)

(d)

Ans. a

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

Correct answer is (a)

126. An organic compound having molecular formula C15H14O exhibited the following 1H and 13C NMR spectral data.

(a)

(b)

(c)

(d)

Ans. d

Sol. To solve this question we will calculate the DBE first

DBE = 9 indicates it may contain 2 benzene ring and one carbons group. Now lets look at NMR data

Two doublets in aromatic region indicates HOMO disubstituted benzene ring at ortho or hetero disubstituted benzene ring at para position.

Four 13C singlet in aromatic region confirms 4 types of aromatic carbon.

13C singlet at 190 indicates presence of

Hence, compound is p-heterodisubstituted.

Correct answer is (d)

127. Identify appropriate reagents A and B in the following reactions

(a) A = LiAlH4, B = BH3·Me2S

(b) A = BH3·Me2S, B = LiAlH4

(d) A = BH3·Me2S, B = LiBH4

Ans. c

Sol. LiAlH4 reduces COOEt and COOH both group

BH3, Me2S reduced only COOH group

LiBH4 reduced only COOEt group.

Chemical reaction involved in the above transformation can be illustrated as

Correct answer is (c)

128. The correct sequence of the reagent to be employed in the following transformation is

(a) (A) m – CPBA; (B) TsNHNH2; (C) AcOH; (D) H2, Pd / BaSO4

(b) (A) H2O2, NaOH (B) NH2NH2; (C) AcOH (D) H2, Pd / C

(d) (A) H2O2, NaOH; (B) TsNHNH2; (C) AcOH; (d) H2, Lindlar's catalyst

Ans. d

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

Correct answer is (d)

129. Reaction of 11.6 g of the aldehyde A with 462 mg of Wilkinson's catalyst provided 9.2 g of alkene B. The mol % of the catalyst used and the yield of the reaction, approximately are

(a) 1.0 mol% and 80%

(b) 1.0 mol% and 90%

(d) 0.2 mol % and 80%

Ans. b

Sol. This problem can be solved using mole concept

The use of catalyst in term of percentage mole can be calculated as

Correct answer is (b)

130. The major products A and B in the following reaction sequene are

(a)

(b)

(c)

(d)

Ans. c

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

Above shown process is an example of strecker synthesis used in the preparation of amino acids.

Correct answer is (c)

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

(a)

(b)

(c)

(d)

Ans. d

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

Correct answer is (d)

132. The major products A and B in the following raction sequence are

(a)

(b)

(c)

(d)

Ans. b

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

Correct answer is (b)

133. Appropriate 1H NMR chemical shifts for the protons A-D for the following compound are

(a) A–6.8; B–5.7; C–3.9; D–2.1 ppm

(b) A–6.8; B–5.7; C–2.1; D–3.9 ppm

(d) A–5.7; B–6.8; C–2.1; D–3.9 ppm

Ans. d

Sol. Position of classical shift for different hydrogen in this case can be decided by using the concept of resonance as shown below.

Correct answer is (d)

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

(a)

(b)

(c)

(d)

Ans. b

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

Correct answer is (b)

135. Citronellol A on oxidation with pyridinium chlorochromate (PCC) followed by treatment with aq. sodium hydroxide gives the product B (IR : 1680 cm–1); whereas oxidation with PCC in the presence of sodium acetate gives product C (IR: 1720 cm–1). Compound B and C are

(a)

(b)

(c)

(d)

Ans. a

Sol. PCC oxidise alcohol into carbonyl compound. Since the medium is acidic in PCC, so some side reaction may be takes place. To stop this side reaction normally buffer solution like sodium acetate is used.

Chemical reaction involved in the above transformation can be illustrated as

Correct answer is (a)

136. Match the following starting compounds with corresponding products in photochemical reactions:

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

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

(d) i-E, ii-A, iii-D

Ans. a

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

Correct answer is (a)

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

(a)

(b)

(c)

(d)

Ans. a

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

Correct answer is (a)

138. The major products A and B of the following reaction sequence are

(a)

(b)

(c)

(d)

Ans. a

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

Correct answer is (a)

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

(a)

(b)

(c)

(d)

Ans. b

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

Correct answer is (b)

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

(a)

(b)

(c)

(d)

Ans. b

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

Correct answer is (b)

141. The correct reagents for effecting the following reactions are

(a)

(b)

(c)

(d)

Ans. a

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

Sulphonium ylide cause CH2 group transfer to carbonyl double bond while sulphoxonium group transfer CH2 group to conjugated double bond on the other hand CH2I2, Zn|Cu causes CH2 group transfer to isolated double bond commonly known as simmon's smith reaction.

Correct answer is (a)

142. The major product A and B of the following reaction sequence are