PYQ DEC 2013 - VedPrep

CSIR NET CHEMISTRY (DEC-2013)
Previous Year Question Paper with Solution.

21. L – DOPA is used for the treatment of

(a) Tuberculosis

(b) Parkinson's disease

(d) Cancer

Ans. b

Sol. L – DOPA is used for Parkinsons' disease.

Correct answer is (b)

22. In the IR spectrum of p – nitrophenyl acetate, the carbonyl absorption band appears at

(a) 1660 cm^–1

(b) 1700^–1

(d) 1770 cm^–1

Ans. d

Sol.

p – nitrophenoxy group will be strong electron withdrawing group for carbonyl so, C = O stretching frequency will increase from normal ester to 1770 cm^–1.

Correct answer is (d)

23. The major product formed in the reaction of styrene with an excess of lithium in liquid ammonia and t – butyl alcohol is:

(a)

(b)

(c)

(d)

Ans. d

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

Correct option is (d)

24. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. c

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

Correct option is (c)

25. For estrone, among the statements A–C, the correct ones are

A. It is a steroidal hormone

B. It has two hydroxyl groups

C. It has one ketone and one hydroxyl groups

(a) A, B and C

(b) A and B

(d) B and C

Ans. c

Sol. The structure of estrone is

Correct answer is (c)

26. An organic compound having the molecular formula C₁₀H₁₄ exhibited two singlets in the ¹H NMR spectrum and three signals in the ¹³C NMR spectrum. The compound is

(a)

(b)

(c)

(d)

Ans. a

Sol.

¹H NMR : Singlet 12H, singlet 2 H

¹³C NMR : CH₃ Signal, CH signal, C signal.

Correct answer is (a)

27. Amongst the following, the compound which has the lowest energy barrier for the cis – trans isomerisation is:

(a)

(b)

(c)

(d)

Ans. c

Sol.

Among all options, option (c) has gain aromaticity in both the rings. Therefore, the lowest energy barries for cis – trans isomerisation is option (c)

28. The IUPAC name of the compound given below is

(a) (2E, 4E) – 3 – chlorohexa – 2, 4 – diene – 1, 6 – diol

(b) (2Z, 4E) – 3 – chlorohexa – 2, 4 – diene – 1, 6 – diol

(d) 4(2E, 4Z) – 3 – chlorohexa – 2, 4 – diene – 1, 6 – diol

Ans. b

Sol.

Correct answer is (b)

29. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. c

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

Correct option is (c)

30. The constituent amino acids present in the following dipeptide, respectively, are

(a) (R) – aspartic acid and (S) – lysine

(b) (S) – aspartic acid and (R) – lysine

(d) (S) – glutamic and (S) – arginine

Ans. a

Sol. Constituent amino acid present in above dipeptide can be shown as

Correct option is (a)

31. A suitable organocatalyst for enantioselective synthesis of Wieland – Miescher ketone (A) is

(a) (–) – proline

(b) (+) – menthone

(d) (+) – BINOL

Ans. a

Sol. This reaction is an example of Robinson annulation. In this reaction we required a base and for the synthesis of optically active compound (A) we will have to use chiral base among all the option only protein is naturally occuring base.

Correct answer is (a)

32. For acylation with acetic anhydride / trienthylamine and oxidation with chromium trioxide of the transand cis – alcohols A and B, the correct statement is

(a) A undergoes acylation as well as oxidation faster than B

(b) B undergoes acylation as well as oxidation faster than A

(d) B undergoes acylation faster than A, whereas A undergoes oxidation faster than B

Ans. c

Sol.

'A' undergoes acylation faster while as 'B' undergoes oxidation faster.

Correct answer is (c)

33. The two benzylic hydrogens H_A and H_B in the compounds I and II, are

(a) diastereotopic in I and enantiotopic in II

(b) diastereotopic in II and enantiotopic in I

(d) diastereotopic in both I and II

Ans. b

Sol.

Correct answer is (b)

34. The following reaction proceeds through a

(a) 1, 3 – sigmatropic rearrangement

(b) 2, 3 – sigmatropic rearrangement

(d) 3, 5 – sigmatropic rearrangement

Ans. c

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

Correct option is (c)

35. The number of nodes present in the highest occupied molecular orbital of 1, 3, 5 – hexatriene in its ground state is

(a) One

(b) Two

(d) Four

Ans. b

Sol.

[Here nodes are represented by dot (•)]

There are two node present in the highest occupied molecular orbital HOMO of 1, 3, 5 hexatrine.

Correct option is (b)

36. Deuterium kinetic isotope effect for the following reaction was found to be 4.0. Based on this information, mechanism of the reaction is

(a) E₁

(b) E₂

(d) free radical

Ans. b

Sol. Kinetic isotope effect is seen in slow step. In case of E₂ – reaction, 'H' or 'D' is involved.

Correct answer is (b)

37. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. b

Sol. Correct answer is (b)

38. The bond order of the metal - metal bond in the dimeric complex is

(a) 4.0

(b) 3.5

(d) 2.5

Ans. b

Sol.

Correct option is (b)

39. The reaction on FeCl₃.6H₂O with SOCl₂ yields.

(a) FeCl₂(s), SO₂(g) and HCl (g)

(b) FeCl₃(s), SO₂(g) and HCl (l)

(d) FeCl₃(s), SO₂(g) and HCl (g)

Ans. d

Sol.

Correct answer is (d)

40. Patients suffering from Wilson's disease have

(a) Low level of Cu – Zn superoxide dismutase

(b) High level of Cu – Zn superoxide dismutase

(d) High level of cooper – storage protein, ceruloplasmin

Ans. c

Sol. Wilson's disease is caused by excess accumulation of Cu in the body.

Correct answer is (c)

41. High does of dietary supplement ZnSO₄ for the cure of Zn deficiency

(a) reduces myoglobin

(b) increases iron level in blood

(d) reduces copper, iron and calcium levels in body

Ans. d

Sol. Correct option is (d)

42. Which of the following in NOT suitable as catalyst for hydroformylation?

(a) HCo(CO)₄

(b) HCo(CO₃)PBu₃

(d) H₂Rh(PPh₃)₂Cl

Ans. d

Sol. and These all are the catalyst for hydroformylation process.

Correct answer is (d)

43. Commonly used scintillator for measuring radiation is

(a) NaI(AI)

(b) Nal(TI)

(d) CsI(AI)

Ans. b

Sol. Correct answer is (b)

44. A sample of aluminium ore (having) no other metal) is dissolved in 50 mL of 0.05 M EDTA. for the titration of unreacted EDTA, 4 mL of 0.05 M MgSO₄ is required. The percentage of Al in the sample is:

(a) 27

(b) 31

(d) 40

Ans. b

Sol. Correct answer is (b)

45. In a cluster, H₃CoRu₃(CO)₁₂, total number of electrons considered to be involved in its formation is

(a) 57

(b) 60

(d) 72

Ans. b

Sol.

3 + 9 + 8 × 3 + 12 × 2 = 60 electron.

Correct answer is (b)

46. Among the following, the correct acid strength thrend is represented by

(a)

(b)

(c)

(d)

Ans. c

Sol.

Reference – Shriver Atkin (P. No. 123 Bronsted acid base concept)

Correct answer is (c)

47. Among the molten alkali metals, the example of an immiscible pair (in all proportions) is

(a) K and Na

(b) K and Cs

(d) Rb and Cs

Ans. c

Sol. Large difference between the size of Li and Cs. So, it is difficult to get a solid solution of these two metals.

Correct answer is (c)

48. Among the following, an example of a hypervalent species is

(a) BF₃.OEt₂

(b) SF₄

(d) Sb₂S₃

Ans. c

Sol. More than 8 electrons in the valence shell of P in

Correct answer is (c)

49. An octahedral metal ion M²⁺ has magnetic moment of 4.0 B.M. The correct combination of metal ion and d-electron configuration is given by

(a)

(b)

(c)

(d)

Ans. a

Sol.

Correct answer is (a)

50. According to VSEPR theory, the geometry (with lone pair) around the central iodine in I₃⁺ and I₃^– ions respectively are

(a) tetrahedral and tetrahedral

(b) trigonal bipyramidal and trigonal bipyramidal

(d) tetrahedral and octahedral

Ans. c

Sol. Number of valence electrons in central atom are 8 while in are 10 and hence the geometries are tetrahedral and trigonal bipyramidal.

Correct answer is (c)

51. Treatment of CIF₃ with SbF₅ leads to the formation of a/an

(a) polymeric material

(b) covalent cluster

(d) lewis acid-base addcut

Ans. c

Sol.

SbF₅ is a very strong F – acceptro and leads to the formation of ionic compound.

Correct answer is (c)

52. The reason for the chemical inertness of gaseous nitrogen at room temperature1 is best given by its

(a) high bonding energy only

(b) electronic configuration

(d) high bond energy and HOMO–LUMO gap

Ans. d

Sol.

High bond energy and HOMO – LUMO gap.

Correct answer is (d)

53. Two tautomeric forms of phosphorus acid are

(a)

(b)

(c)

(d)

Ans. a

Sol. H₃PO₃ is a dibasic and reducing in nature.

Correct answer is (a)

54. The correct thermodynamics relation among the following is

(a)

(b)

(c)

(d)

Ans. a

Sol.

Correct option is (a)

55. The boiling point of a solution of non-volatile solid is higher than that of the pure solvent, It always indicates that

(a) the enthalpy of the solution is higher than that of the pure solvent.

(b) the entropy of the solution is higher than that of the pure solvent.

(d) the internal energy of the solution is higher than that of pure solvent.

Ans. b

Sol. Correct answer is (b)

56. According to Arrhenius equation (K = rate constant and T = temperature)

(a) In K decreases lineraly with 1/T

(b) In K decreases lineraly with T

(d) In K increases lineraly with T

Ans. a

Sol.

Hence, ln K decreases lineraly as l/T increases.

Correct answer is (a)

57. The angle at which the first order Bragg reflection is observed from (110) plane in a simple cubic unit cell of side 3.238Å, when chromium radiation of wavelength 2.29Å is used, is

(a) 30⁰

(b) 45⁰

(d) 90⁰

Ans. a

Sol.

Correct answer is (a)

58. The orbital with two radiaal and two angular nodes is

(a) 3p

(b) 5d

(d) 8d

Ans. b

Sol. For radial node = n – 1 – 1 = 5 – 2 – 1 = 2

Angular node = l = 2

Correct answer is (b)

59. Michael Faraday observed that the colour of colloidal suspensions of gold nanoparticles changes with the size of the nanoparticles. This is because

(a) Gold forms complex with the solvent.

(b) Band gap of gold changes with size of the nanoparticle.

(d) Colloidal suspensions diffract light

Ans. b

Sol. Correct answer is (b)

60. The energy 2s and 2p orbitals is the same for

(a) Li

(b) Li²⁺

(d) H^–

Ans. b

Sol.

Li has one electron in 2s and no electron in 2p hence the energies of 2s and 2p orbitals will be different.

Li⁺⁺ has only one electron in ls¹ hence will have no screening effect therefore, 2s and 2p will have same energies.

Correct answer is (b)

61. If a homonuclear diatomic molecule is oriented along the Z-axis, the molecular orbital formed by linear combination of p, orbitals of the two atoms is

(a)

(b)

(c)

(d)

Ans. c

Sol. If a homonuclear diatomic molecule is oriented the z – axis, then linear combination of p_z orbitals of two atoms form – orbitals.

The molecular orbitals formed by linear combination of p_x or p_y orbitals of two atoms is

Correct answer is (c)

62. A reaction contains a mixture of N₂, H₂ and NH₃ in equilibrium (Kp = 3.75 atm^–2). If sufficient. He is introduced into the reactor to double the total pressure, the value of K_P at the new equilibrium would be

(a) 0.94 atm^–2

(b) 3.75 atm^–2

(d) 15.00 atm^–2

Ans. b

Sol. k_p changes only by change in temperature.

Correct answer is (b)

63. The volume of a gas absorbed on a solid surface is 10.0 ml, 11.0 ml, 12.2 ml, 14.5 ml and 22.5 ml at 1.0, 2.0, 3.0, 4.0 and 5.0 atm, pressure, respectively. These data are best represented by

(a) Gibb's isotherm

(b) Langmuir isotherm

(d) BET isotherm

Ans. d

Sol. When we plot graph it increasing gradually and multilayered is formed. No monolayer is cormed.

Correct option is (d)

64. A compound of M and X atoms has a cubic unit cell. M atoms are at the corners and body centre position and X atoms are at face centre positions of the cube. The molecular formula of the compound is

(a) MX

(b) MX₂

(d) M₂X₃

Ans. d

Sol.

Correct answer is (d)

65. When Frenkel defects are created in an otherwise perfect ionic crystal, the density of the ionic crystal

(a) increases

(b) decreases

(d) oscillates with the number of defects

Ans. c

Sol. In case of Frenkel defect, ions are not missing from the lattice. They just displace into interstitial sites. Also, no extra ion come from outside into the lattice. Therefore, density reamisn same.

Correct answer is (c)

66. The molecule in which the bond order increases upon addition of an electron is

(a) O₂

(b) B₂

(d) N₂

Ans. b

Sol.

If an electron is added to B₂, then will go bonding to – orbital.

Correct answer is (b)

67. In a potentiometric titration, the end point is obtained by observing

(a) change in colour

(b) jump in potential

(d) increase in turbidity

Ans. b

Sol. Correct answer is (b)

68. Electrolysis of an aqueous solution of 1.0 M NaOH results in

(a) Na at the cathode and O₂ at the anode

(b) H₂ at the cathode and O₂ at the anode

(d) O₂ at the cathode and H₂ at the anode.

Ans. b

Sol.

Correct answer is (b)

69. The cell voltage of Daniel cell is 1.07 V. If reduced potential of is 0.34 V, the reduction potential of is

(a) 1.141 V

(b) – 1.41 V

(d) –0.73 V

Ans. d

Sol.

Correct answer is (d)

70. In the mechanism of reaction, H₂+Br₂ 2HBr, the first stepis

(a) dissociation of H₂ into H• radicals

(b) dissociation of Br₂ into Br• radicals

(d) reaction of Br• radical with H₂

Ans. b

Sol. In this reaction, the first step is chain initiation step, in which the bromine molecule acquires energy as a result of collision with another bromine molecule to dissociate into two Br atoms.

Correct answer is (b)

71. For an electronic configuration of two non-equivalent electronics which of the following terms is not possible?

(a)

(b)

(c)

(d)

Ans. d

Sol. For two non – equivalent

Correct option is (d)

72. Consider a two – dimensional harmonic oscillator with potential energy If are the eigensolutions and are the eigenvalues of harmonic oscillator problem in x and y direction with potential respectively, the wave function and eigenvalues of the above two – dimensional harmonic oscillator problem are

(a)

(b)

(c)

(d)

Ans. c

Sol.

Correct option is (c)

73. The quantum mechanical virial theorem for a general potential is given by where T is the kinetic energy operator and < > indicates expectation value. This leads to the following relation between the expectation value of kinetic energy and potential energy for a quantum mechanical harmonic oscillator problem with potential

(a)

(b)

(c)

(d)

Ans. a

Sol.

Correct option is (a)

74. Consider a particle in a one dimensional box of length 'a' with the following potential

Starting with the standard particle in a box hamiltonian as the zeroth order Hamiltonian and the potential of V₁ from 'a/2' to 'a' as a perturbation, the first – order energy correction to the ground state is

(a) V₁

(b) V₁/4

(d) V₁/2

Ans. d

Sol. Particle in a box hamiltonian as zeroth order hamiltonian and potential v_i from a/2 to a.

Correct option is (d)

75. The most porbable value of 'r' for an electron in 1s orbital of hydrogen atom is

(a) a₀/2

(b) a₀

(c)

(d)

Ans. b

Sol.

Correct option is (b)

76. The angular momentum operator is

(a)

(b)

(c)

(d)

Ans. b

Sol.

Correct option is (b)

77. The molecule with the smallest rotation partition function at any temperature among the following is

(a)

(b)

(c)

(d)

Ans. b

Sol. Since rotational partition function is given by

Hence, two factor should be considered

In present case, has smallest (reduced mass) also having

Therefore, will have lowest rotational partition function.

Correct option is (b)

78. Both NaCl and KCl crystallize with the FCC structure. However, the X-ray powder diffraction pattern of NaCl corresponds to be FCC structure whereas, that of KCI corresponds to simple cubic structure. This is because

(a) K⁺ and Cl^– are isoelectronic

(b) Na⁺ and Cl^– are isoelectronic

(d) KCl has anti – site defects.

Ans. a

Sol. Since, K⁺ and Cl^– are isoelectric, some of the peaks are missing entirely. The scattering intensities for X – rays are directly related to the number of electrons in the atom. Light atoms scatter X – rays weakly, while heavy atoms scatter X – rays more effectively.

Correct answer is (a)

79. Consider the cell:

E_cell= 1.71V at 25⁰C for the above cell. The equilibrium constant for the reaction:

at 25⁰C would be close to

(a) 10²⁷

(b) 10⁵⁴

(d) 10⁴⁰

Ans. b

Sol.

Correct option is (b)

80. The molecule that has the smallest diffusion coefficient in water is

(a) glucose

(b) fructose

(d) surcrose

Ans. d

Sol. Diffusion coefficient D

which measures the mobility of the ion due to its thermal energy.

Diffusion is the process which involves the migration of a component in solution down a gradient of itw own concentration.

Diffusion coefficient in water of sucrose is smallest because its mobility is very low. It contains high concentration of ion. so, migration of component in solution very small.

Correct answer is (d)

81. Metallic gold crystallizes in FCC structure with unit cell dimension of 4.00 Å. The atomic radius of gold is

(a) 0.866Å

(b) 1.414Å

(d) 2.000Å

Ans. b

Sol. We know that, for FCC atomic radius is given by

Correct answer is (b)

82. A first order gaseous reaction is 25% complete in 30 minutes at 227⁰C and in 10 minutes at 237⁰C. The activation energy of the reaction is closest to (R = 2 cal K^–1 mol^–1)

(a) 27 kcal mol^–1

(b) 110 kcal mol^–1

(d) 5.5 kcal mol^–1

Ans. c

Sol. From Arrhenius, equation, we have

Correct answer is (c)

83. In the reaction between NO and H₂ the following data are obtained

The orders with respect to H₂ and NO are

(a) 1 with respect to NO and 2 with respect to H₂

(b) 2 with respect to NO and 1 with respect to H₂

(d) 2 with respect to NO and 2 with respect to H₂

Ans. b

Sol. where a is order w.r.t. NO & b is order w.r.t. H₂

From experiment I, on doubling pressure of NO rate becomes 4 times

Therefore, a = 2

From experiment II, on doubling pressure of H₂ rate becomes almost double.

Therefore, b = 1

Correct option is (b)

84. The energy for a single electron excitation in cyclopropenium cation in Hückel theory is

(a)

(b)

(c)

(d)

Ans. c

Sol.

Correct option is (c)

85. The atomic masses of fluorine and hydrogen are 19.0 and 1.0 amu, respectively (1 amu = 1.67×10^–27 kg). The bond length of HF is 2.0Å. The moment of inertia of HF is

(a) 3.2×10^–47 kg m²

(b) 6.4×10^–47 kg m²

(d) 4.8×10^–47 kg m²

Ans. b

Sol. F atomic mass = 19.0 amu.

H atomic mass = 1 amu

H – F bond length = 2.0 Å

Correct option is (b)

86. The masses recorded when a substance is weighed 4 times are 15.8, 15.4, 15.6 and 16.0 mg. The variance (square of the standard deviation) is closest to

(a) 0.02

(b) 0.05

(d) 0.20

Ans. b

Sol.

Correct option is (b)

87. The transition that is allowed by x-polarized light in trans-butadiene is

(The character table for C_2h is given below)

(a)

(b)

(c)

(d)

Ans. b

Sol. If direct product of ground and excited state transforms according to the x, y and z axiz then the transition will be allowed.

and B_u transforms as per ^x–axis. Hence, B_g to A_u transition will be x – polarized.

Correct option is (b)

88. The heat capacity of 10 mol of an ideal gas at a certain temperature in 300 JK^–1 at constant pressure. The heat capacity of the same gas at the same temperature and at constant volume would be

(a) 383 JK^–1

(b) 217 JK^–1

(d) 466 JK^–1

Ans. b

Sol.

Correct answer is (b)

89. The Maxwell's relationship derived from the equation dG = VdP – SdT is

(a)

(b)

(c)

(d)

Ans. c

Sol. dG = VdP = SdT ...(i)

In an isobaric – isothermal process

dP = 0 and dT = 0

dG = 0 or G = constant

Thus Gibb's function is that physical quantity which remains constant in reversible isobaric process

From equation (i)

G = G (P, T)

From equation (i) and (ii)

And since dG is a percent differential

Correct option is (c)

90. The chemical potential of the i^th component is defined as

(a)

(b)

(c)

(d)

Ans. d

Sol. The chemical potentials are thus the rate of change of free energy per mole at constant temperature and constant pressure.

and constant temperature pT - 0 and constant pressure dP=0

Correct answer is (d)

91. Work (w) involved in isothermal reversible expansion from V_i to V_r of n moles of an ideal gas is

(a)

(b)

(c)

(d)

Ans. a

Sol. Work done is reversible expansion process for an ideal gas

If gas expend from volume, V_i to V_f if process is isothermal then we may integrate it from V_i to V_f,

Correct answer is (a)

92. The limiting molar conductivities of NaCl, Nal and RbI are 12.7, 10.8 and 9.1 mS m²mol^–1, respectively. The limiting molar conductivity of RbCl would be

(a) 32.6 mS m² mol^–1

(b) 7.2 mS m² mol^–1

(d) 11.0 mS m² mol^–1

Ans. d

Sol.

Correct answer is (d)

93. The number of ways in which four molecules can be distributed in two different energy levels is

(a) 6

(b) 3

(d) 8

Ans. c

Sol. Number of energy levels (g_i) = 2

Number of molecules (Ni) = 4

The number of ways of distribution

(since, it is not given either particles are distinguishable or indistinguishable, therefore, we will consider it as distinguishable)

Correct option is (c)

94. An element exists in two crystallographic modifications with FCC and BCc structures. the ratio of the densities of the FCC and BCC modifications in terms of the volumes of their unit cells (V_FCC and V_BBC) is

(a) V_BBC : V_FCC

(b) 2V_BCC : V_FCC

(d)

Ans. b

Sol.

Correct option is (b)

95. Given The resonance frequency of a proton in magnetic field of 12.6 T is close

(a) 60 MHz

(b) 110 MHz

(d) 780 MHz

Ans. c

Sol.

Correct answer is (c)

96. In Mössbauer experiment, a source emitting at had to be move towards absorber at 2.2 mm s^–1 for resonance. The shift in the frequency between the source and the absorber is

(a) 15.0 MHz

(b) 20. 0 MHz

(d) 30.0 MHz

Ans. c

Sol.

Correct answer is (c)

97. Among the following, the correct combination of complex and its color is

(a) Complex : Color : Red

(b) Complex : Color : Orange

(d) Complex : Color : Yellow

Ans. c

Sol. Order of ligand is spectro chemical series

From above it is clear that CN^– will produce highest energy difference, followed by F^–, Cl^–, SCN^–. As colour of complex is complementary colour of absorbed light.

Thus absorbs orange colour radiation which complementary colour is blue therefore colour of is blue.

The other combination according to energy are mismatched. Hence, only option (c) is correct.

Correct answer is (c)

98. In a specific reaction, hexachlorocyclotriphosphazene, N₃P₃Cl₆ was reacted with a metal fluoride to obtain mixed halo derivatives namely N₃P₃ClF(A), N₃P₃Cl₄F₂(B), N₃P₃Cl₂F₄(D), N₃P₃CIF₅(E). Compositions among these which can give isomeric products are

(a) A, B and C

(b) B, C and D

(d) E, A and B

Ans. b

Sol. (A) N₃P₃Cl₅F = only one isomer

Correct option is (b)

99. Xenon forms several fluorides and oxofluorides which exihibit acidic behaviour. The correct sequence of descending Lewis acidity among the given species is represented by

(a)

(b)

(c)

(d)

Ans. a

Sol. Relative acidic strength of xenon fluorides follows order

XeF₆ > XeO₂F₄ > XeO₄ > XeOF₄ > XeF₄ > XeO₂F₂ > XeO₃

This order depends upon

(i) number of lone pair (ii) number of 'F' atoms

Correct answer is (a)

100. Number of isomeric derivaties possible for the neutral closo – carborane, C₂B₁₀H₁₂ is

(a) three

(b) two

(d) six

Ans. a

Sol.

C₂B₁₀H₁₂ is show three isomeric derivative which is depend on the temperature.

Correct answer is (a)

101. For higher boranes 3c – 2e 'BBB' bond may be a part of their structures. In B₅H₉, the number of such electron deficient bond(s) present is/are

(a) four

(b) two

(d) one

Ans. d

Sol.

Correct option is (d)

102. In the atomic absorption spectroscopic estimation of Fe(III) using O₂/H₂ flame, the absorbance decreases with the addition of

(a)

(b)

(d) Cl^–

Ans. b

Sol. Chemical reactions between analyte and matrix that prevent the atomization of the analyte. One common one is that Sulfate or phosphates anions will form non – volatile salts with Fe⁺³, so its signal decreases. Can be treated in two ways 1. Releasing agents – can add to metal to prevent formation of interference complexes. this this example EDTA or 8 – hydroxyquino line.

Correct option is (b)

103. In a polarographic estimation, the limiting currents were 0.15, 4.65, 9.15 and 27.15 when concentration (mM) of Pb(II) were 0, 0.5 1.0 and 3.0 respectively. An unknown solution of Pb(II) gives a limiting current of Concentration of Pb(II) in the unknown is

(a) 1.355 mM

(b) 1.408 mM

(d) 1.500 mM

Ans. d

Sol. Limiting current (id)

Limiting current proportional to concentration

Correct option is (d)

104. The gases SO₂ and SO₃ were reacted separately with CIF gas under ambient conditions. The major products expected from the two reactions respectively, are

(a) SOF₂ and CIOSO₂F

(b) SOF₂ and SO₂F₂

(d) SO₂CIF and CIOSO₂F

Ans. d

Sol.

Correct answer is (d)

105. The correct statement regarding terminal/bridging CO groups in solid Co₄(CO)₁₂ and Ir₄(CO)₁₂ is

(a) both have equal number of bridging CO groups

(b) number of bridging CO groups in Co₄(CO)₁₂ is 4

(d) the number of bridging CO groups in Ir₄(CO)₁₂ is zero

Ans. d

Sol.

Correct answer is (d)

106. On reducing Fe₃(CO)₁₂ with an excess of sodium, a carbonylate ion is formed. The iron is isoelectronic with

(a)

(b)

(c)

(d)

Ans. b

Sol.

Correct answer is (b)

107. The correct statement for ozone is

(a) It absorbs radiations in wavelength region 290 – 320 nm.

(b) It is mostly destroyed by NO radical in atmosphere

(d) Its concentration near poles is high due to its paramagnetic nature.

Ans. a

Sol. (i) Ozone is diamagnetic in nature

(ii) It is non toxic even at 1ppm level

(iii) It is not destroyed by no radial in atmosphere

(iv) It absorbs radiations in wave length region 290 – 320 nm

Correct answer is (a)

108. Among the following clusters,

H is encapsulated in

(a) A only

(b) B only

(d) A and B only

Ans. a

Sol.

Correct answer is (a)

109. The solid state structure of aluminum fluoride is

(a)

(b)

(c)

(d)

Ans. c

Sol. A1F₃ has octahedral arrangement in 3 – dimethyl structure which causes high methyl point of A1F₃ in comparison to A1C1₃, A1Br₃ and AlI₃. AlCl₃ dimer has layered structure.

Correct answer is (c)

110. Oxidised form of enzyme catalase (structure A); prepared by the reaction of with H₂O₂, has green color because

(a) Oxidation sate of iron changed from Fe^III to Fe^IV.

(b) Porphyrin ring is oxidized by one electron

(d) Fe^IV is coordinated with anionic tyrosinate ligand in axial position.

Ans. a

Sol. Catalase is a tetramer of four polypeptide chains covering 50 amino acids.

The colour arises as Fe(III) changes into Fe(IV).

Correct answer is (a)

111. The reactive position of nicotinamide adenine dinucleotide (NAD) in biological redox reactions is

(a) 2 – position of the pyridine ring

(b) 6 – position of the pyridine ring

(d) 5 – position of the pyridine ring

Ans. c

Sol.

Therefore the 4 – position of the pyridine ring is reactive position in the biological reaction.

Correct answer is (c)

112. The electrophiel Ph₃C⁺ reacts with to give a product A. The product A is formed because

(a) Fe is oxidised

(b) Alkyl is substituted with Ph₃C

(d) Alkyl is converted to alkene

Ans. d

Sol.

5 + 8 + 4 + 1 = 18 electron

Correct answer is (d)

113. Substitution of L with other ligands will be easiest for the species

(a)

(b)

(c)

(d)

Ans. c

Sol. In case of (c) the chances of conversion of haptacity from is more due to partial bond fixation.

Correct option is (c)

114. Among the following, the correct statement is

(a) CH is isolobal to Co(CO)₃

(b) CH₂ is isolobal to Ni(CO)₂

(d) CH₂ is isolobal to Mn(CO)₄

Ans. a

Sol.

ans so on..........

Correct answer is (a)

115. MnCr₂O₄ is likely to have a normal spinel structure because

(a) Mn²⁺ will have a LFSE in the octahedral site whereas the Cr³⁺ will not

(b) Mn is +2 oxidation state and both the Cr and in +3 oxidation state

(d) Cr³⁺ will have a LFSe in the octahedral site whereas the Mn²⁺ ion will not

Ans. d

Sol.

Correct answer is (d)

116. The ground state forms of Sm³⁺ and Eu³⁺ respectively, are

(a) ⁷F₀ and ⁶H_5/2

(b) ₆H_5/2 and ⁷F₀

(d) ⁷F₆ and ²F_7/2

Ans. b

Sol.

Correct answer is (b)

117. The orbital interactions shown below represent

(a) CH₃ – Al interactions in Al₂(CH₃)₆

(b) B – H interactions in B₂H₆

(d) CH₃CH₂– Mg interactions in EtMgBr.(OEt²)²

Ans. c

Sol.

The diagram clearly indicates the four centered – two electron interaction (4c – 2e). This takes place in The sp³ hybrid orbital is of carbon while the three s – orbitals are of three surrounding lithium atoms.

Correct option is (c)

118. Compounds K₂Ba[Cu(NO₂)₆] (A) and Cs₂Ba[Cu(NO₂)₆] (B) exhibit tetragonal elongation and tetragonal compression, respectively. The unpaired electron in A and B are found respectively, in orbitals,

(a)

(b)

(c)

(d)

Ans. b

Sol.

Correct option is (b)

119. Reaction of Ph₂PCH₂CH₂PPh₂ with [RhC₁(CO)₂]₂ in a 2:1 molar ratio gives a crystalline solid A. The IR spectrum of complex A shows v_co at 1985 cm^–1. The ³¹P(¹H) NMR spectrum of A consists of two doublets of doublets of equal intensities (¹⁰³Rh is 100% abundant and I=¹/₂). The structure of complex A is

(a)

(b)

(c)

(d)

Ans. a

Sol.

IR band at 1985 cm^–1 indicates terminal CO. Both P atoms are non – equivalent. Each P atom will show doublet of doublet with

Correct option is (a)

120. The most appropriate structure for the complex is

(a)

(b)

(c)

(d)

Ans. c

Sol. bonding in weaker than that of Pt – P bonding. When Pt – S and Pt – P bond are trans to each other then Pt – S become weaker therefore in such a situation SCN ligand tends to form bonding through M atoms of ligand as M do not form bonding. Hence, the most probable structure will be

Correct option is (c)

121. The major product formed in the following 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)

122. 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

since lindlar's catalyst convert

alkyne to cis alkene

Correct option is (b)

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

(a)

(b)

(c)

(d)

Ans. d

Sol.

Correct option is (b)

124. The most suitable reagent combination of A – C, required in the following conversions are

(a) A = Li/liq. NH₃; B = NaBH₄, CeCL₃, 7H₂O; C = H₂, (Ph₃P)₃RhCl

(b) A = Li/liq. NH₃; B = NaBH₄, CeCL₃, 7H₂O; C = H₂, 10% Pd/C

(d) A = NaBH₄, CeCl₃, 7H₂O; B = Li/liq. NH₃; C = H₂, 10% Pd/C

Ans. a

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

Correct option is (a)

125. The major product B formed in the following reaction sequence and overall yield of its formation are

(a)

(b)

(c)

(d)

Ans. a

Sol.

Therefore option (b) and (d) and not possible

Since, the reaction is not happen at the stereocentre, therefore stereochemistry of the reactant will preserved in the product.

Correct option is (a)

126. An organic compound (C₈H₁₀O₂), which does not change the color of ferric chloride solution, exhibited the following ¹H NMR spectral data. 7.3 (1H, t, J = 8 Hz), 7.0 (1H, d, J = 8 Hz), 6.95 (1H, s), 6.9 (1H, d, J = 8 Hz) 5.3 (1H, br, s, D₂O exchangeable), 4.6 (2H, s), 3.9 (3H, s). Structure of the compound is

(a)

(b)

(c)

(d)

Ans. b

Sol.

Correct option is (b)

127. Methyl 4 – oxopentanoate exhibited signals at 208, 172, 51, 37, 32 and 27 ppm in its ¹³C NMR spectrum. The signals due to the methoxy, Cl, C4 and C5 carbons are

(a) OMe – 32; Cl – 208; C4 – 172; C5 – 51

(b) OMe – 51; Cl – 208; C4 – 172; C5 – 32

(d) OMe – 51; Cl – 172, C4 – 208; C5 – 32

Ans. d

Sol.

OMe = 52 is due to attachement of carbon to O.

C – 5 CH₃ is due to attachment of carbon to carbonyl group.

C – 4 is aliphatic carbonyl centre so above 200.

C – 1 is ester carbonyl group so below 200.

Correct answer is (d)

128. In the following reaction, the intermediate and the major product A are

(a)

(b)

(c)

(d)

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 sulfuric acid mediated rearrangement of the sesquiterpene saritonin A is

(a)

(b)

(c)

(d)

Ans. b

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

This is a dienone – phenol type rearrangement.

Correct option is (b)

130. In the following transformation, the reagent A and the major product B, respectively, are

(a)

(b)

(c)

(d)

Ans. d

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

Correct answer is (d)

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

(a)

(b)

(c)

(d)

Ans. a

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

Correct option is (a)

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

(a)

(b)

(c)

(d)

Ans. a

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

Correct option is (a)

133. 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)

134. The peptide A on reaction with 1 – fluoro – 2, 4 – dinitrobenzene followed by exhaustive hydrolysis gave phenylalanine, alanine, serine and N – (2, 4 – dinitrophenyl) glycine. On the other hand, peptide A after two cycles of Edman degradation gave Phe – Ser as the product. The structure of the peptide A is

(a) Phe – Ser – Ala – Gly

(b) Phe – Ser – Gly – Ala

(d) Ala – Gly – Phe – Ser

Ans. c

Sol. 2, 4 – dinotrofluorobenzene is said to be Sanger reagent and it will react with N – terminal aminoacid. The DNB – derviative of glycene indicates that, glycene is present at the N – terminal side in the peptide.

i.e.

Correct option is (c)

135. The compound (B) (labeled) is precursor for biosynthesis of the natural product A. the labeled carbons in the product A are

(a) C1, C3, C5 and Me

(b) C2, C4, C6 and Me

(d) C1, C3, C5 and COOH

Ans. c

Sol. Correct answer is (c)

136. The major product formed in the following 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)

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

(a)

(b)

(c)

(d)

Ans. a

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

Correct option is (a)

138. 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 option is (b)

139. The conditions A-B, required for the following pericyclic reactions are

(a)

(b)

(c)

(d)

Ans. b

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

Correct option is (b)

140. The number of electrons participating and the pericyclic mode in the following reaction are

(a) 4 and conrotatory

(b) 4 and disrotatory

(d) 6 and disrotatory

Ans. d

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

Correct option is (d)

141. Stereoselective reduction of the dione A with a chiral reducing agent provides the corresponding diol B in 100% diastereoselectivity and 90% ee favoring R, R configuration. The composition of the product is