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Previous Years’ Question

TIFR | IIT-JAM | BHU | HCU

DEC. 2017

CSIR-UGC-NET JUNE 2018

CHEMICAL SCIENCES

PAPER WITH SOLUTION

1PAPER : CSIR-UGC-NET/JRF Dec. 2017

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PAPER : CSIR-UGC-NET/JRF Dec. 2017CHEMICAL SCIENCES BOOKLET-[A]

PART-B

21. Among the following nuclear reactions of thermal neutrons, the cross section is highest for

(a) 235 1 235 192 0 92 0U n U n (b) 235 1 236

92 0 92U n U

(c) 235 1 232 492 0 92 2U n Th He (d) 235 1 94 140 1

92 0 36 56 0U n Kr Ba 2 n

22. Spectrophotometric monitoring is not suitable to determine the end point of titration of(a) oxalic acid vs potassium permanganate (b) iron(II) vs 1, 10-phenanathroline(c) cobalt(II) vs eriochrome black T (d) nickel (II) vs dimethylglyoxime

23. The first ionization energy is the lowest for(a) Br (b) Se (c) P (d) As

24. Among 3 3ClO , XeO and 3SO , species with pyramidal shape is/are?

(a) 3 e 3ClO and X O (b) e 3 3X O and SO (c) 3 3ClO and SO (d) 3SO

25. The role of BF3 as an industrial plymerization catalyst is to generate(a) carbanion (b) carbocation (c) organic radical (d) cation radical

26. For the following complexes, the increasing order of magnetic moment (spin only value) is

A. 36TiF B. 36CrF C. 36MnF D. 36CoF

(a) D < A < B < C (b) C < A < D < B (c) B A D C (d) A B C D 27. The correct statement for cytochrome c is

(a) It is a non-heme protein(b) The coordination numer of iron in cytochrome c is five(c) It is a redox protein and an electron carrier(d) It can store or carry dioxygen

28. Geometries of SNF3 and XeF2O2, respectively, are(a) square planar and square planar (b) tetrahedral and tetrahedral(c) square planar and trigonal bipyramidal (d) tetrahedral and trigonal bipyramidal

29. The IR spectrum of Co(CO)4H shows bands at 2121, 2062, 2043 and 1934 cm–1. The Co D (incm–1) expected in the spectrum of Co(CO)4D is(a) 2111 (b) 1396 (c) 2053 (d) 1910

30. In trigonal prismatic ligand field, the most stabilized d orbital is

(a) 2zd (b) xyd (c) xzd (d) yzd

31. The most unstable complex on the basis of electro-neutrality principle among the following is

(a) 32 6Al OH

(b) 33 6Al NH

(c) 36AlF (d) 33 6Al NCCH

32. The band in the electronic spectrum of l2 appearing at 520 nm will undergo maximum blue shift in(a) water (b) hexane (c) benzene (d) methanol

2 PAPER : CSIR-UGC-NET/JRF Dec. 2017


33. Mismatch among the following is(a) Sharp transition and fluorescence in lanthanides(b) Broad bands and d-d transitions(c) Very high spin-orbit coupling and transition elements(d) Charge transfer and molar absorptivity of the order of 104 L mol–1 cm–1

34. Among the following, the compound that gives base peak at m/z 72 in the El mass spectrum is

(a)

O

(b)O

(c)O

(d)

O

35. The following molecule hasCO2H

H

H

HO2C(a) plane of symmetry (b) R configuration (c) S configuration (d) centre of symmetry

36. The following natural product Enterodiol is a

OHOH

HO

HO(a) terpene (b) steroid (c) lignan (d) alkaloid

37. The correct order of basicity for the following heterocycles is

NH

A

NN

H

B

N

NH

C(a) A > C > B (b) C > A > B (c) C > B > A (d) B > A > C

38. The kinetic product formed in the following reaction isOH

conc. H2SO4

(a)OH

SO3H

(b)

OH

SO3H

(c)OH

HO3S

(d)

OHSO3H



39. Among the structures given below, the one that corresponds to the most stable conformation ofcompound A is

O

OOH

A

(a) OO

OH

Me(b) O

O OHMe

(c)OO

Me

OH (d) OO

Me

OH

40. According to Frontier Molecular Orbital (FMO) Theory, the Highest Occupied Molecular Orbital(HOMO) of hexatriene in the following reaction is

hv

(a) (b)

(c) (d)

41. The number of signals observed in the proton decoupled 13C NMR spectrum of the following com-pound is

(a) Five (b) Six (c) Ten (d) Thirteen42. The correct order of stability of the following carbocations is

MeO

MeO

MeO

(A) (B) (C)

(a) A > C > B (b) B > C > A (c) C > A > B (d) C > B > A43. An optically pure organic compound has specific rotation of +40°. The optical purity of the sample

that exhibits specific rotation of +32° is(a) 8% (b) 12% (c) 20% (d) 80%



44. The major product formed in the following reaction isO O

KOHMeOH

(a)

O

(b)

O

(c)

O

(d)

O45. Correct match of the compounds in Column P with the IR stretching frequencies (cm–1) in Column Q

isColumn P Column Q

I

O

A 1865

II O

O

B 1770

IIIOO O

C 1745

(a) I - B; II - C; III - A (b) I - C; II - A; III - B (c) I - C; II - B; III - A (d) I - A; II - C; III - B46. The organic compound that displays following data is

1HNMR (400 MHz): 7.38 (d), 7.25 (d), 1.29 (s) ppm

(a)

Br

(b)

Br

(c)

Br

(d)Br

47. The molecule with a C2 axis of symmetry among the following is(a) BH2Cl (b) CH3Cl (c) NH2Cl (d) HOCl

48. The molecule that will show Raman spectrum, but not IR spectrum, among the following is(a) H2 (b) HCl (c) BrCl (d) CS2

49. Boron in BCl3 has(a) sp hybridization (b) sp2 hybridization (c) sp3 hybridization (d) no hybridization



50. The number of degenerate spatial orbitals of a hydrogen-like atom with principal quantum number n= 6 is(a) 12 (b) 6 (c) 72 (d) 36

51. If ˆ ˆ, 0A B and ˆ ˆ, 0A C , then which of the following necessarily holds:

ˆ ˆˆ, and are operatorsA B C

(a) ˆˆ, 0B C (b) ˆ, 0A BC (c) ˆ , 0B AC (d) ˆ , 0C AB

52. The correct statement among the following is ( A is a hermitian operator)

(a) The eigenvalues of 2A can be negative.

(b) The eigenvalues of 2A are always positive

(c) No eigenfunction of A is an eigenfuncation of 2A .

(d) The eigenvalues of 2A can be complex.53. If the atoms/ions in the crystal are taken to be hard spheres touching each other in the unit cell, then

the fraction of volume occupied in the body centered cubic structure is

(a) 3 (b)26

(c) 6

(d)38

54. Repeated measurements of Pb in a lake water sample gave 3.2, 5.2 and 7.2 ppb of Pb. Standarddeviation in the measurement of Pb is(a) 2 ppb (b) 4 ppb (c) 0 ppb (d) 2 2 ppb

55. The stability of lyophobic colloids is a consequence of the(a) electrical double layer at the surface of the particles.(b) van der Waals force between the particles.(c) small particle size.(d) shape of the particles.

56. The equivalent conductance at infinite dilution of a strong electrolyte (0) can be obtained from theplot of

(a) vs. C (b) vs. C (c) vs. C2 (d)1vs.C

57. The number-average molar mass nM for a monodisperse polymer is related to the weight-average

molar mass wM by the relation

(a)3

wn

MM (b)4

wn

MM (c) 2n wM M (d) n wM M

58. For a sequence of consecutive reactions, 1 2k kA I P the concentration of I would be, bysteady state approximation.

(a) 1k A (b) 1 2k k A (c) 1 2k k A (d) 1

2

k Ak

59. Enthalpy is equal to

(a) i iTS PV u n (b) i iTS u n (c) i iu n (d) i iPV u n



60. The structure of ribonucleoside uridine is

(a)N

NH

O

OO

OH

HO (b)N

NH

O

OO

OH

HO

OH

(c)N

NH

OO

OH

HOO

(d)N

NH

OO

OH

HO

OH

O

PART-C

61. The peak area of differential thermal analysis curve is proportional to one or more of the following:A. mass lossB. mass of the sampleC. heat of decomposition/phase changeThe correct answer is(a) A only (b) B only (c) A and C (d) B and C

62. To determine the bond parameters at 25°C, electron diffraction is generally unsuitable for both(a) O3 and NO2 (b) Sulfur and dry ice (c) NO2 and sulfur (d) O3 and dry ice

63. Match lanthanides in Column I with their properties in Column IIColumn I Column IIA. Lu (i) Reagent in oxidation state IVB. Eu (ii) Ml2 of metallic lustreC. Ce (iii)Diamagentic M(III)D. Tb (iv)Pink in oxidation state IIICorrect match is(a) A-(iii), B-(ii); C-(i); D-(iv) (b) A-(ii), B-(iii); C-(iv); D-(i)(c) A-(iv), B-(ii); C-(i); D-(iii) (d) A-(iii), B-(ii); C-(iv); D-(i)

64. Among the following species isolobal to CH2 areA. CpCr(CO)2 B. CpCu C. Ni(CO)2 D. Cr(CO)4E. Fe(CO)4(a) A, C and E (b) B, C and D (c) B, C and E (d) A, B and D

65. Choose the incorrect statement for the phosphomolybdate anion, [PMO12O40]3–.

(a) It has a Keggin structure.(b) Phosphorus is in +5 oxidation state.(c) It is extremely basic.(d) It forms crystalline precipitates with [R4N]+ (R = bulky alkyl or aryl group)



66. Consider the following statement(s) for actinides (An):A. Oxidation states greater than +3 are more frequent in An compared to lathanides (Ln)B. Some An(III) ions show d-d transitionsC. UO2

2+ and PuO22+ are stable

D. Some of actinides do not have radioactive isotopesThe correct answer is(a) A and C (b) B and D (c) A, B and C (d) B, C and D

67. According to Bent’s rule, for p-block elements, the correct combination of geometry around thecentral atom and position of more electro-negative substituent is(a) Trigonal bipyramidal and axial (b) Triogonal bipyramidal and equatorial(c) Square pyramidal and axial (d) Square pyramidal and basal

68. Allred-Rochow electronegativity of an element isA. directly proportional to the effective nuclear chargeB. directly proportional to the covalent radiusC. inversely proportional to the square of the covalent radiusD. directly proportional to the square of the effective nuclear chargeThe correct answer is(a) A and B (b) A and C (c) B and C (d) A and D

69. Br2 with propanone forms a charge transfer complex and l2 forms triiodide anion with I–. This im-plies that(a) both Br2 and I2 act as bases (b) both Br2 and I2 act as acids(c) Br2 acts as an acid and I2 acts as a base (d) Br2 acts as a base and I2 acts an an acid

70. In the complex [Pd(L-L)(Me)(Ph)], the bisphosphine (L-L) that does not allow reductive elimina-tion of PhMe, is

(a) P

P

(b)

P

P

Ph PhPh

Ph

(c) MeOP

PMeO

(d)

PPPh Ph Ph

Ph



71. In the reaction given below, the bisphosphine (P-P) that is in effective for transfer hydrogenationreaction is

CpRu(P–P)H + N

Ph

BF44CH3CNCD2Cl2

CpRu(P–P)(CH3CN)]+BF4- + N

Ph(a) Diphenylphosphinomethane (b) 1, 2-Diphenylphosphinoethane(c) 1, 3-Diphenylphosphinopropane (d) 1, 4-Diphyenylphosphinobutane

72. For high spin and low spin d6 octahedral complexes (ML6), the generally observed spin allowedtransitions, respectively, are(a) two and one (b) one and two (c) zero and one (d) two and two

73. The reactions given below,A. Cl2 + 2H2O HOCl + H3O

+ + Cl–B. Cl2 + 2NH3 NH2Cl + NH4

+ + Cl–are examples of(a) disproportionation only (b) disproportionation (A) and solvation (B)(c) solvation (A) and disproportionation (B) (d) solvalysis as well as disproportionation

74. According to Wade’s rules, the cluster type and geometry of [Sn9]4-, respectively, are

(a) closo and tricapped trigonal prismatic (b) nido and monocapped square-antiprismatic(c) arachno and heptagonal bipyramidal (d) closo and monocapped square antiprismatic

75. Assuming 1 1PH PBJ J , the expected 31P NMR spectrum of H3P: 111BCl3 [for 11B, I = 3/2] is

(a) (b)

(c) (d)

76. The geometry around Cu and its spin state for K3CuF6 and KCuL2, [H2L = H2NCONHCONH2],respectively are:(a) (octahedral, high-spin) and (square planar, low-spin)(b) (octahedral, low-spin) and (square planar, low-spin)(c) (trigonal prismatic, high-spin) and (tetrahedral, high-spin)(d) (trigonal prismatic, low-spin) and (tetrahedral, high-spin)



77. The active site structure for oxy-hemerythrin is:

(a)

O

Fe(III)

O

Fe(III)N(His)

N(His)OO

O

HO

GluC

O

(His)N

(His)N

Asp

(b)

O

Fe(II)

O

Fe(III)N(His)

N(His)OO

O

HO

GluC

O

(His)N

(His)N

Asp

(c) Fe(III) Fe(III)N(His)

N(His)OO

O

O

GluC

O

(His)N

(His)N

Asp

(His)N

HO

O

(d) Fe(II) Fe(III)N(His)

N(His)OO

O

O

GluC

O

(His)N

(His)N

Asp

(His)N

HO

O

78. Consider the following statements with respect to the base hydrolysis of [CoCl(NH3)5]2+ to

[Co(NH3)5(OH)]2+.A. One of the ammonia ligands acts as a Bronsted acid.B. The entering group is water.C. A heptacoordinated Co3+ species is an intermediate.The correct statement(s) is/are(a) A and B (b) A and C (c) B and C (d) C only

79. The number of inorganic sulfides in cubane like ferredoxin and their removal method, respectively,are(a) eight and washing with an acid (b) four and washing with a base(c) eight and washing with a base (d) four and washing with an acid

80. Considering the ambidentate behaviour of thiocyanate ion, the most stable structure among the fol-lowing is

(a)P P

Pd

NCS SCN

(b)P P

Pd

SCN NCS



(c)P P

Pd

SCN NCS

(d)P P

Pd

NCS SCN81. Major product of the following reaction is

O

O

1. LDA, TBDMSCl

2. CO2Meheat

then H2O

(a) HO2CCO2Me

(b) HO2CCO2Me

(c) HO2C

CO2Me

(d) HO2C

CO2Me

82. Major product in the following reaction is

O

1. NH2OH

2. AcCl

(a) CN (b) CN

(c) CN (d) CN

83. Major products A and B of the following reaction sequence are

1. TsNHNH2, AcOH

2. baseO

O

A H2, Pd/CaCO3

Pb(OAc)2B



(a)O

A =O

B = (b) A = B =

OO

(c)O

A = B =O

(d) A = B =

OO

84. The major product formed in the following reaction is

CH2Cl2/MeLi

NH

(a)N

Cl

(b)N

Cl

(c)

N

(d)

N

Cl

85. Major products A and B of the following reaction sequence are

CO2H l2

NaHCO3A B

DBU

(a)A =

OO

I

H

B =O

O

H

(b) A =O

O

I

H

B =O

O

H

(c) A =O

O

I

H

B =O

O

H



(d) A =O

O

I

H

B =O

O

H86. The major product formed in the following reaction is

BnOO

S CuCN

1.

2. NH4Cl/H2O

2-

2Li+

(a)S

BnO OH(b)

BnO OH

(c)OH

BnOS

(d)OH

BnO

87. Correct sequence of reagents (i)-(iii) required for the conversion of A to B is

H2N COOH

N NH NO

(A) (B)A. Thionyl chloride, B. 4-Chloropyridine C. Piperidine(a) A, B and C (b) A, C and B (c) B, A and C (d) C, A and B

88. The following reaction involves

O N2

O

O

ORh2(OAc)4 (cat)

(a) [1, 2] sigmatropic rearrangement (b) [2, 3] sigmatropic rearrangement(c) [3, 3] sigmatropic rearrangement (d) C-H insertion reaction

89. Correct sequence of steps involved in the following transformation is

O

SOPhO

KOHOH

(a) Michael addition, aldol condensation, syn-elimination, keto-enol tautomerism(b) aldol condensation, electrocyclic ring closing, syn-elimination, dehydrogenation(c) Michael addition, Claisen condensation, anti-elimination, keto-enol tautomerism(d) Robinson annulation, dehydrogenation, anti-elimination



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

heat

O

NEt2A

H2O

AcOHB

(a)

O

A =

NEt2

O

B =COOH

(b)

O

A =

NEt2

O

B =

NEt2

O

(c)

O

A =

NEt2

O

B =COOH

(d)

O

A =

NEt2

O

B =

NEt2

O

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

3. PCl3N

1. H2O2, AcOH2. HNO3, H2SO4

(a)

N

NO2

(b)

N

NO2

(c)N

Cl

(d)

N

Cl

92. The number of optically active steroisomers possible forCH3–CH(OH)–CH(OH)–CH(OH)–CH3 is(a) two (b) four (c) six (d) eight



93. The correct match of the circled protons in Column P with 1H NMR chemical shift ( ppm) inColumn Q is

Column-P Column-Q

(I) H

Br

(A) 6.72

(II)

O

O

H

(B) 16.4

(III) H

(C) –0.61

(a) I-A; II-B; III-C (b) I-B; II-A; III-C (c) I-B; II-C; III-A (d) I-C; II-B; III-A94. The major product formed in the following reaction sequence is

OHO

HO

OH

1. MeOH/H+

2. TBDMSCl (1 equiv)

Et3

N

3. PDC4. AcOH/H

2O

(a) TBDMSO OH

OOH

OH

(b)TBDMSO

O OH

O

(c)O

O OTBDMS

O

H

(d)O

O OH

TBDMSO

H



95. For the successful synthesis of peptide linkage leading to the product A, the side chain of the aminoacid B should have

SPhN

OHpeptide CO2Me

NH2

HX+

NN

OHpeptide

H

CO2Me

XH

(B)

XN

OHpeptide

H2N CO2Me

pH = 7

(A)

(a) XH = – OH (b) XH = –(CH2)4NH (c) XH = – p – (C6H4)OH (d) XH = –SH96. The major products A and B in the following reaction sequence are

OTBDMS

SiMe3

OH

1. H2SO42. TBAF

A

Ti(PrO)4(+)-DET

t-BuOOHCH2Cl2-20°C

B

(a) OHA = OHB =O

(b) OHA = OHB =O

(c)OH

A = B =OH

O

(d)OH

A = B =OH

O

97. The intermediate A and the major product B in the following reaction are

OMe

Br1. LDA (1 equiv.)

MeO2C CO2Me

[A] B

2.

(a)

OMe

CO2Me

CO2Me

A =

OMe

CO2MeB =

CO2Me



(b)

OMe

CO2Me

CO2Me

A =

OMe

CO2Me

B =CO2Me

(c)

OMe

A =

OMe

B =CO2Me

CO2Me

CO2Me

CO2Me

(d)

OMe

A =

OMe

CO2Me

B =CO2Me

CO2Me

CO2Me

98. The correct intermediate which leads to the product in the following reaction is

OH

O

NH2

NaNO2

HBrOH

O

Br

(a) OH

O

N2Br(b)

O

O

(c)O

O

(d)N

O

N

99. In the following transformation, the mode of electrocyclization A and the major product B are

heat

A

H

H

cat. OsO4NMO B

(a) A = (4n) e–, dis

H

H

B =

OH

OH



(b) A = (4n + 2) e–, dis

H

H

B =

OH

OH

(c) A = (4n) e–, con

H

H

B =

OH

OH

(d) A = (4n + 2) e–, con

H

H

B =

OH

OH

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

then H2On-Bu O

BO

H

A(Ph3P)2PdCl2

KOH

Br

B

(a) A = n-BuB(OH)2

B = n-Bu

(b) A = n-BuB(OH)2

B = n-Bu

(c) A =n-Bu B(OH)2 B =

n-Bu

(d) A =n-Bu B(OH)2 B =

n-Bu

101. The correct statement about the symmetry of the eigenfunctions of a quantum of 1-D harmonicoscillator is(a) All the eigenfunctions are only even functions, because the potential is an even function.(b) All the eigenfunction are only odd functions, although the potential is an even function.(c) The eigenfunctions have no odd-even symmetry.(d) All the eigenfunctions are either odd or even functions, because the potential is an even function.

102. The correct statement about the difference of second and first excited state energies (E) of a par-ticle in 1-D, 2-D square and 3-D cubic boxes with same length for each, is(a) E (1 – D box) = E (2 – D box) = E (3 – D box)(b) E (1 – D box) > E (2 – D box) > E (3 – D box)(c) E (1 – D box) > E (2 – D box) = E (3 – D box)(d) E (1 – D box) < E (2 – D box) < E (3 – D box)



103. A one-dimensional quantum harmonic oscillator is perturbed by a potential x3. The first order cor-rection to the energy for the ground state (E(1)) is(a) E(1) > 0 but < 1 (b) E(1) < 0 (c) E(1) = 0 (d) E(1) > 2

104. The normal mode of ethylene represented, by the figure below, is

HH

H H

(a) only IR active (b) only Raman active(c) both IR and Raman active (d) neither IR nor Raman active

105. The pair that contains a spherical top and a symmetric top, among the following, is

(a) 4 2 2,CH CH Cl (b) 2 2 3,CH Cl CH Cl (c) 3 4,CH Cl CH (d) 4 3 4,CH C CH

106. A part of the character table of a point group (of order 4) is given below.

1 2 3

1

2

3

4

1 1 1 11 1 1 11 1 1 1? ? ? ?

E X X X

The four characters of 4 are, respectively(a) 1, 1, –1, –1 (b) 2, 0, 0, 1 (c) 1, i, i, 1 (d) 1, –i, i, –1

107. The electronic transition energy from 1 2 in propenyl radical is 4.8 eV. Within the frame work

of Huckel theory, the transitions energy from 1 3 would be(a) 2.4 eV (b) 4.8 eV (c) 9.6 eV (d) 14.4 eV

108. The g-factors of 1H and 13C are 5.6 and 1.4 respectively. For the same value of the magnetic fieldstrength, if the 1H resonates at 600 MHz, the 13C would resonate at(a) 2400 MHz (b) 600 MHz (c) 150 MHz (d) 38 MHz

109. The term symbol for the ground state of a metal ion is 3P2. The residual entropy of a crystal of a saltof this metal ion at 0 K is(a) kB ln 1 (b) kB ln 3 (c) kB ln 5 (d) kB ln 7

110. In stretching of a rubber band,dG = V dp – SdT + f dLWhich of the following relations in true?

(a), ,p T p L

S fL T

(b), ,p T p L

S fL V

(c), ,p T p L

S VL T

(d), ,p T T L

S fL p



111. Four distinguishable molecules are distributed in energy levels E1 and E2 with degeneracy of 2 and 3,respectively. Number of microstates, with 3 molecules in energy level E1 and one in energy level E2,is(a) 4 (b) 12 (c) 96 (d) 192

112. One mole of an ideal gas undergoes a cyclic process (ABCDA) starting from point A through 4reversible steps as shown in the figure. Total work done in the process is

A

B C

D

V2

V1

T1 T2

Volu

me

Temperature

(a) 21 2

1

VR T TV

(b) 21 2

1

VR T TV

(c) 21 2

1ln VR T T

V (d) 2

2 11

ln VR T TV

113. If the specific conductance of an electrolyte solution is 0.2 –1 cm–1 and cell constant is 0.25 cm–1,the conductance of the solution is(a) 1.25 –1 (b) 1.0 –1 (c) 0.8 –1 (d) 2.0 –1

114. The predicted electromotive force (emf) of the electrochemical cell 2 2/ 0.01 | 0.01 /Fe s Fe aq M Cd aq M Cd s is

2 20 0

/ /0.447 V and 0.403 V

Fe Fe Cd CdE E

(a) –0.850 V (b) +0.044 V (c) +0.0850 V (d) –0.044 V115. A polymer has the following molar mass distribution

1Number of molecules Molar mass (g.mol )50 500075 6000

The calculated number average molar mass nM of the polymer is(a) 5200 (b) 5600 (c) 5800 (d) 6000

116. The separation of the (123) planes of an orthorhombic unit cell is 3.12 nm. The separation of (246)and (369) planes are, respectively,(a) 1.56 nm and 1.04 nm (b) 1.04 nm and 1.56 nm(c) 3.12 nm and 1.50 nm (d) 1.04 nm and 3.12 nm

117. The slope and intercept obtained from (1/Rate) against (1/substrate concentration) of an enzymecatalyzed reaction are 300 and 2 × 105, respectively. The Michaelis-Menten constants of the enzymein this reaction is(a) 5 × 106 M (b) 5 × M–6 M (c) 1.5 × 103 M (d) 1.5 × 10–3 M



118. The pressure inside (Pin) a spherical cavity with a radius r formed in a liquid with surface tension isrelated to the external pressure (Pout) as

(a) 2in outP P

r

(b) 2in outP P

r

(c) in outP Pr

(d) in outP Pr

119. Reaction between A and B is carried out for different initial concentrations and the correspondinghalf-life times are measured. The data listed in the table:

0 1/20Entry sec1 500 10 602 500 20 603 10 500 604 20 500 30

tB MA M

The rate can be represented as

(a) k A B (b) 2k A (c) 2k A B (d) 2k A B

120. The plot of the rate constant vs. ionic strength of the reaction 2A B follows the line (refer to thefigure)

I

II

III

IV

log

k

I

0

(a) I (b) II (c) III (d) IV

21SOLVED PAPER : CSIR-UGC-NET/JRF Dec. 2017


SOLVED PAPER : CSIR-UGC-NET/JRF Dec. 2017CHEMICAL SCIENCES BOOKLET-[A]

PART–B

21. The cross-section can be interpreted as the effective ‘target area’ that a nucleus interacts with anincident neutron. Larger the effective area, greater the probability for reactionIn reaction, 235 1 94 140 1

92 0 36 56 0U n Kr Ba 2 n , one neutron react with 23592 U to give two neu-

tron, which have more tendency to react with other radioactive nuclei. Thus, increases effective areafor reaction.Correct option is (d)

22. (a) Oxalic acid versus potassium permanganate:

End-point is colorless maxpink-purple 525 nm

(b) Iron (II) versus 1, 10-phenanthroline:

End-point is colorless Reddish-orange max 512 nm

(d) Nickel (II) versus Dimethylglyoxime :

maxEnd-point is colorless Red 445 nm

(c) Cobalt (II) versus Eriochrome Black-T:

max max

End-point is Blue-green Blue

8 660 600

Free EBT Co II +EBT complex

pH nm nm

Since, color change at end-point is not significant for option(c) which is requirement of the spectro-photometric titration. Therefore, color monitoring method is not suitable in this case.Correct option is (c)

23. • Generally, ionisation energy increases on moving from left to right in a period in periodic table.• Ionisation energy decreases as we move down in given group• Ionisation energy will be more for full filled and half filled orbital.Correct option is (b)

24. 33

7 6 3 1 26ClO 3 1 4 sp Tetrahedral8 8

Cl

OOO

= Pyramidal shape

••

33

8 6 3 26XeO 3 1 4 sp Tetrahedral8 8

22 SOLVED PAPER : CSIR-UGC-NET/JRF Dec. 2017


Xe

OOO

= Pyramidal shape

••

23

6 6 3 24SO 3 sp Trigonal planar8 8

S

O

OOCorrect option is (a)

25. n + BF3 n BF3

H2O + BF3 H+ + HOBF3cation formation carbocationCorrect option is (b)

26. (A) 362 2

3 1 0

TiF

Ti 3d 4s

Ti 3d 4s

eg

t2g

1 unpaired electron

d1

(B) 36

5 1

3 3 0

CrF

Cr 3d 4s

Cr 3d 4s

eg

t2g

3 unpaired electron

d3

(C) 365 2

3 4

MnF

Mn 3d 4s

Mn 3d

eg

t2g

4 unpaired electron

d4

(D) 36

7 2

3 6 0

CoF

Co 3d 4s

Co 3d 4s

t2g

4 unpaired electron

eg

d6

Hence, A < B < C = DCorrect option is (d)



27. Cytochrome-C shows redox reaction and it is an essential component of electron transport chain,whereas it is carriers of one electron and does not bind oxygenCorrect option is (c)

28. 33

6 5 7 3 32SNF 4 sp Td8 8

32 2

8 6 2 7 2 34XeO F 4 1 5 sp d TBP8 8

S

N

F FF

Xe

OF

FO

••

Tetrahedral Trigonal bipyramidal

Correct option is (d)

29.1vµ

1934 19342 1367.7 1396

2

Co H Co DCo D

Co D Co H Co D

µµ

Correct option is (b)

30. In trigonal prismatic ligand field, energy of ‘d’ orbital is

2 2 2yz xz xyz x yd d d d d

Hence, the most stablized ‘d’ orbital is xydCorrect option is (b)

31. According to electroneutrality principle, each atom in any stable substance has a charge close tozero. This means if central atom has more positive charge then it would be stabilized by ligandhaving high negative charge density i.e. more electronegative donor atom.

In case of 33 6Al NH

, nitrogen is sp3 hybridized hence it is least electronegative hence

least stable complex.Correct option is (b)

32. Electronic absorption band of I2 vapours appears at 520 nm i.e. max 2 of I gas 520 nm If solution is prepared with a solvent that able to form a charge transfer complex (CTC) like aromaticsolvents, ethers, alcohols etc. This absorption band shifted towards UV range i.e. blue shift. Morethe donation of electrons from solvent to the * (LUMO) of I2 more will be the blue shift.

Here, option(d) methanol has highest electron donating capacity (i.e. most basic)

max 2I Hexane 523nm Red shift

max 2I Benzene 502nm Blue shift

max 2I Methanol 440 nm Blue shift

max 2I Water 460 nm Blue shift

Note: Methanol has low ionization potential (10.85 eV) than water (12.59 eV), therefore methanolis better electron donar than water.Correct option is (d)



33. Spin-orbit coupling is larger for the heavier atoms (Inner transition) and very small for lighter atoms.Correct option is (c)

34.

OH

H

M.C.O

+O

m/z = 72

• •H

•

Correct option is (a)

35.H

HOOC

(4)

(3)

COOH

H

(1)

(2)(R) Configuration


36. HO

HO

OH

OH

Enterodiol is a lignan, formed by the action of intestinal bacteria on lignan precursors found inplants.Correct option is (c)

37.

N

H

Pyrrole

NN

H

Pyrazole

N

N

H

Imidazole

H+ H+ H+

WeakBase

N

H H

Strongerconjugate

acidN

NH

H

N

NH

H

pKa = –4.5 pKa = 2.5 pKa = 7.1

N

N

H

H

apK basicity

Lesser is the availability of lone pair for donation, lesser is the basic behaviour.Correct option is (c)



38.

OS

O

HO OH2

OS

O

O

OS

O

HO OH

H2O

H

OH

OS

O

OO

HO3S HH

SO3H

OHH

Correct option is (d)39.

OOMe

OH

Hydrogen bonding and no 1, 3-diaxial interaction


40. HOMO of hexatriene in hv condition is 4 . In 4 molecular orbital has three nodes.

6

5

4

3

2

10

1.

2.

3.

4.

5.No. of nodes

LUMO

HOMO

HOMO

LUMO

hv




41.

c

a

d

c

a

b

ee

b c

d

cb

a

a

13 Total 5-signalC NMR


42.

MeO MeO

MeO

+

Carbocation stabilizedthrough (8 hydrogen)

hyperconjugation

Carbocation stabilizedthrough (6 hydrogen)

hyperconjugation and allylicdouble bond

Carbocation stabilizedthrough (6 hydrogen)hyperconjugation andnp-vacant orbital

resonance stabilization

Order of stability of carbocation C > B > ACorrect option is (d)

43.

T

T

of sample 32Optical purity 100 100 80%40 of pure enantiomer


44.

O O

H HOH

KOH, MeOH

O O

H

IntramolecularAldol reaction

OH

H OOH

E1cB

O

OH

O

Correct option is (c)



45. (I)

O

Cyclic KetoneIR = 1745 cm–1

(II) O

O

Cyclic EsterIR = 1770 cm–1

(C) (B)

(III)

AnhydrideIR = 1865 cm–1

(A)

OO O


46.

Br

Plane of symmetry

Br

C C

C

H

H

HH

H

HH

H HH

H

H

H

= 7.38 (d)

= 7.25 (d)

singlet 1.29 ppm

a a

b b

c

c

c


47. B

Cl

H H

Once C2 passing through ‘B’ and ‘Cl’ atoms, two hydrogen interchange their position but remain same molecule.

Correct option is (a)48. H2 a homo diatomic molecule will have not change in dipole moment. Thus, it will be IR inactive but

since H2 having polarisability in a Raman spectra. Thus, it will be Raman active.Correct option is (a)

49.B

Cl

ClClsp2 [hybridisation]

(3 + 3×7)/8 = 3 sp2Correct option is (b)

50. Orbital degeneracy = n2 = 62 = 36Correct option is (d)

51. ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ ˆA, BC A, B C B A,C 0 ˆ ˆ ˆˆA, B 0 & A,C 0


52. e.g. Let ˆ ˆ xA p which is a Hermitian operator and ikxe be operand

2

2 2 2 22

ˆ ˆ ˆ;

ikx ikx ikxA A e k A e k e

x

Therefore, eigen value is positive.Correct option is (b)



53. Packing fraction =334 4 3 3 32 3.14

3 3 4 8 4

r aZ ra


54. 3.2 5.2 7.2 5.2

3

x ppb ppb

2 2 2

2 2

3.2 5.2 5.2 5.2 7.2 5.21 3 1

2 0 2 8 22 2

Lx xs ppb

N

ppb ppb ppb

Correct option is (a)55. The stability of lyophobic colloids is a consequence of electrical double layer at the surface of the

particlesCorrect option is (a)

56. For strong electrolyte, 0 can be obtained from the plot of vs cCorrect option is (b)

57.MwP.D.I.= =1Mn for Monodisperse sample

Mw Mn Correct option is (d)

58. Rate of formation of I = Rate of deformation of I

11 2

2

kk A k I I Ak

Correct option is (d)59. i iG u n

G H TS H G TS i iH u n TS


60.

N

NH

O

OO

HO

OH OHUridine is glycosylated pyrimidine-analog containing uracil attached to ribose ring (ribofuranose)via a 1, N glycosidic bond Correct option is (b)



PART–B61. The peak area of differential thermal analysis is propportional to

(i) mass of the sample(ii) heat of decompositional phase changeCorrect option is (d)

62. Electron diffraction is used to get the bond parameters of gaseous samples.Correct option is (b)

63. (a) Lu(III) = [Xe] 4f145d06s0, hence diamagnetic(b) EuI2 has metallic lustre(c) Ce(IV) used as oxidising agent(d) Tb(III)-f8 has pink colourCorrect option is (a)

64. Number of valence electron in CH2 = 6

Hence, fragment having 16 electron will be isolobal to this fragment.

2A CpCr CO 5 6 4 15

B CpCu 5 11 16

2C Ni CO 10 4 14

4D Cr CO 6 8 14 ; 4E Fe CO 8 8 16


65. 312 40PMo O is less basic. It is slightly less basic than 4ClO


66. (A) Due to more expansion of 5f orbital actinoids has more tendency to release electron hence havemore tendency to show oxidation state greater than +3 compared to Lanthanoids(B) Actinoids (III) show either f-d or f-f transition, but no d-d transition(C) 2

2UO and 22PuO are stable

(D) All the actinoids have radioactive isotopesCorrect option is (a)

67. According to Bent’s Rule, ‘more electro negative element occupy axial position in trigonal bipyramidalgeometry.Correct option is (a)

68. According to Allred-Rochow, electronegativity can be given asAR eff

20.3590 Z 0.744

r where, r = covalent radius.

Correct option is (b)69. In 3I

, there is intraction between * of I2 and non-bonding electron of I–. In case propanone and Br2,

there in interaction between * of Br2 and non-bonding electron of propanone.Thus, both I2 and Br2 behave as acid.Correct option is (b)



70. According to reductive elimination reaction mechanism cis-complex can give the product buttrans-complex is not suitable for reaction and does not give reductive elimination product. Since,ligand given in option (d) forms a trans-complex with Pd(II) due to the presence of flexiblemethylene (–CH2–) groups. Therefore, reductive elimination not occured.

P PPd

Me

PhPh

PhPh

Ph(trans)

Actually formed but does not give reaction

P PPd

Me Ph

PhPh

PhPh

(cis)This can give the reaction but not formed

Correct option is (d)71. As the chelate ring size increases (in case of (d)), the energy of LUMO of complex cation

CpRu(P–P)+ (which formed after H– transfer) decreases and makes the complex cation better H–

acceptor and therefore decreasing the rate of H– transfer from CpRu(P–P)H.

P PRu

Ph

Ph

Ph

PhCp H

CH3CNP P

Ru

Ph

Ph

Ph

PhCp NCCH3

+

+ H–

(d)High steric crowding due tolarge chelating ring

Low energy LUMOso, better H– acceptor

No reaction

Correct option is (d)72. For 6d high spin there is one spin allowed transition from 5 5

2g gT EIn case of d6 low spin commonly two spin allowed transitions are observed.

1 11g 1g

1 11g 2g

A T

A T

there are additional spin allowed transitions at higher energy but they are marked by allowed transi-tions. Hence, are not observedCorrect option is (b)

73. 2 2 3A Cl 2H O HOCl H O Cl

2 3 2 4B Cl 2NH NH Cl NH Cl Above both reactions are example of solvolysis as well as disproportionationCorrect option is (d)

74. 49Sn 9 4 4 40 , thus it follow 4n + 4. Hence, it has nido structure

Correct option is (b)*



75. 113 3H P BCl

31 2 1HP NMR NI for 13 2 3 1 4 quartet2

H

1: 3: 3 :1

31 2 1BP NMR NI for 32 1 1 4 quartet2

B

1:1:1:1

1 3 3 1

B B B B

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

3H

Intensity ratio : 1 : 1 : 1 : 1 : 3 : 3 : 3 : 3 : 3 : 3 : 3 : 3 : 1 : 1 : 1 : 1Correct option is (c)

76. 33 6 6K CuF 3k CuF , oxidation state of Cu3+ = 3d8

Thus, Cu3+ 8d Oh

KCuL2 oxidation state of Cu = 1 + x–4 = 0 x 3

3 8Cu d

Hence, square planar low spin

C N

HN

C N

O

O H

N C

N C

NH

O

O

CuK+

–H H

HCorrect option is (a)

77. In oxyhemerythrin, both the Fe are present as Fe(III) and there is presence of hydrogen bonding anddioxygen bind to only one Fe.Correct option is (c)

78. Mechanism of base hydrolysis can be given as

2

3 3 25 4Co Cl NH OH Co Cl NH NH

2slow3 2 3 2RDS4 4

intermediate C.N. 5

Co Cl NH NH Co NH NH Cl

22 fast

3 2 2 34 5Co NH NH H O Co NH OH




79. Cubane like ferredoxin is Fe4S4It has four inorganic sulfide which can be removed by treatment with acid.

8H 84 4 4 2Fe S Fe 4H S


80.

P

Pd

P

Ph

Ph

Ph

Ph

NCS SCN

76º

Small P–Pd–P angle, less steric hindrance effective Pd–SCN– bonding

P

Pd

P

Ph

Ph

Ph

Ph

SCN NCS

85º

Increase in P–Pd–P angle,increase in steric hindrance. No effective bonding from the P atom, the S atom of SCN–

ligand can form bond but due to steric hindrance SCN–

binds through N atom.

P

Pd

P

Ph

Ph

Ph

Ph

SCN NCS

89º

P–Pd–P bond angle, further increasewith increase in size of chelate ring andhence steric hindrance increases. No effective bonding from P, the S atom of SCN– canform the bond but steric hindrance favoursthe Pd-N bonding


81.

O

O

HH

N

LiLDA

O

OLi

TBDMS Cl

O

OTBDMS

MeO2C

+

heat(4+2)

cycloaddition

O

OTBDMS

32

2

1

1CO2Me

TBDMSO

OH2OCO2Me

HOOC

CO2Me

3

[3,3]sigmatropic


82.

O

H2N OH

OH

NH OH

N OH

Ac–Cl

N OAc

Backmannfragmentation

CN

H

CN




83.

OH

O O

N

NH TsBase

O

N

NTs

O

(Lindlar catalyst)

H2, Pd/CaCO3

Pd(OAc)2

O

O

O AcOH

Product (A)Product (B)

TsNHNH2


84.

CH

H

Cl

Cl

C

H

Cl•• (Carbene)

N

CH

Cl

• •

N

C

Cl

H

NN

H

Li+Me–

Me–Li+

Li

Carbene insertion

Li

–LiCl

(Product)


85.C

OH

O

I2

O

O HI

NaHCO3

O

OI

O

I

H

H

O

Product (A)

O

I

O

O

O

H

DBU

Base

Product (B)

DBU

*DBU = [1, 8-Diazabicycle [5.4.0] undec-7-ene]


86. BnOO Cu

CN

SNH4Cl/H2O

BnO

OH2–




87.HN C

O

OHS

O

ClCl

H2N C

O

OH

N

Cl

N

NH

C

O

ClN

NH

NH

C

O

HN N

NH

N

HN CO

Cl

N

Cl


88. O

O

N2

Rh2(OAc)4(cat.)

O

O

••+ N2

••••

O

O

32

1

1

2[2, 3 STR]

O

O


89.S

O

O

Ph

H KOH

OH

S

O

O

Ph OMichael addition

SOPh

O OH

H

KOH

SOPh

O O

SOPh

O OKOH

IntramolecularAldol

Reaction

O

PhOS

OH

O

S

OH

Ph OH

syn eliminationH

O

keto enolTautomerism

OH

H




90.

O

NEt2••

O

C

NEt2O

NEt2

H+

O

NEt2

OH2

O

O

OH2

O

OH

OH3O+

O

OH

O

–NEt2Product (A)

Product (B)


91. N

H2O2/AcOH

N

O

HNO3/H2SO4

O N O

N

H NO2

O

N

NO2

O : PCl3

N

NO2


92. The number of optically active stereoisomer are two

CH3

H OH

H OH

CH3

HO H

CH3

HO H

HO H

CH3

H OH

Optically activeCorrect option is (a)



93. H

Br

(I)

4n = 12 e–

in delocalization

Therefore,anti-aromatic inner proton will be highly deshielded i.e. paratropicanisotropy = 16.4 ppm

O

O

H

= 6.72 ppmvinylic proton with EWG

(II)H

(III)

4n+2 = 14 e–

in delocalizationTherefore, aromatic inner proton will be highly shielded i.e. diatropic anisotropy = – 0.61


94.

O

HO

HOOH2MeOH/H+ O

HO

HOOMe

TBDMS-Cl(1eq.) Et3N

O

HO

TBDMSOOMe

PDC

O

O

TBDMSOOMe

AcOH/H2O

O

O

HOOH

O

HO

HOOH

H+

O

HO

HO MeOH


95.

Peptide

O

SPhN

H

NH2

CO2MeHS

Peptide

O

SN

H H2N CO2Me

Peptide

O

NHN

H

CO2Me

SH


96.OH

SiMe3

OTBDMSH2SO4

OH2

SiMe3

OTBDMS

Anti-elimination

OHTBAF

Deprotection of (–OH)

Ti(Pro)4, (+)–DETt-BuOOH, CH2Cl2, –20ºC

OHO




97.

OMe

Br

H

LDA

OMe

CO2Me

CO2Me

OMe

C

C

OMe

O

OMeO

OMe

CO2Me

OMeC

O

OMe

CO2Me

CO2Me

H3O+

workup

1 equivalent

OMe

CO2Me

OMeC

O

(A)

(B)


98. OH

NH2

O

NaNO2OH

N2

O

NGP

O

O

Br

HBrOH

O

BrBr


99.

H

H

heat

H

H

cat. OsO4

NMO

H

H

OH

OH

6 e–

(cis-hydroxylation fromless hindered face)


100. n BuO

BO

H

then H2On-Bu

B(OH)2

Br

(Ph3)2PdCl2, KOHn-Bu


101. Eigen functions are even for even values of vibrational quantum number (n) and odd for odd valuesof n.Correct option is (d)



102. In 1-D box, 2

258

hEm and in 2-D box

2

238

hEm

And in 3-D box, 2

238

hEm


103. For odd power of perturbed potential, 1E is always ‘0’


104. Only Raman active, (symmetric vibration, change in dipole moment is zero during vibration)


105. Spherical top molecule 4CH Perfect geometry

a b cI I I

Symmetrical top molecule 3 a b cCH Cl I 0, I I Correct option is (c)

106. According to GOT theorem, any two IR must be orthogonal to each other.Correct option is (a)

107.

3 = – 2

2 =

1 = + 2

1 3 1 22 2 4.8 9.6E E eV

Correct option is (c)108. 0 0E g B h g B

g

600 5.6 600 1501.4 4

H HC

C C C

g MHz MHzg

Correct option is (c)109. 3

2PDegeneracy = (2×2+1) = 5

B eS k nq For electron ground state, e eq g

5B e BS k ng k n Correct option is (c)

110. dG VdP SdT fdL ... (1)Differentiating equation w.r.t. T at constant P, L

,P L

G ST

Similarly differentiating equation (1), w.r.t. L at constant P and T



,T P

G F Since, G is a stable function,

, ,, ,P L T PT P P L

G GL T T L

Or,, ,T P P L

S FL T

Correct option is (a)111. For distinguishable particles,

1ni

ii

gW Nn

; 1 3 10

0 1

0 1

2 3 84 4 4 3 24 4 963 1 3 2

n ng gWn n


112.

T1 T2

V2

V1

B C

DA

Volume

total AB BC CD DAw w w w w Process AB is isothermal,

21

1AB

Vw nRT nV

Process BC is isochoric,0BCw

Process CD is isothermal,

1 22 2

2 1CD

V Vw nRT n nRT nV V

Process DA is isochoric

0DAw

2 21 2

1 1

22 1

1

totalV Vw nRT n nRT nV V

VnR T T nV




113. 10.2; 0.80.25

AR G k G

A


114. 0 0 0.0591 0.01log2 0.01cell right leftE E E

0.403 0.447cellE ; 0.044cellE VCorrect option is (b)

115.1 1 2 2

1 2

N M N M 50 500 75 6000Mn 5600N N 50 75


116. 123246

d 3.12d 1.56 nm2 2

; 123369

d 3.12d 1.04 nm3 3


117.m

max

kslope 300r

... (1)

5

max

1Intercept 2 10r

...(2)

3m

1k 1.5 10

2


118. For a bubble, surface tension counteracts internal pressure

int2

outP Pr

int2

outP Pr


119. From entry (1) and (2), order w.r.to, B = 1And from entry (3) and (4), order w.r.to. A = 2Correct option is (c)

120. 2 productA B

Slope = 2 2 2 1 4 ve salt effect A BAZ Z A AThus, plot having negative slopeCorrect option is (d)

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