PAPER : CSIR-UGC-NET/JRF June 2017 · 2017. 12. 5. · 6 PAPER : CSIR-UGC-NET/JRF June 2017 57. The v = 0 to 1 vibration-rotation spectrum of a diatomic molecule exhibits transitions

1PAPER : CSIR-UGC-NET/JRF June 2017

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

PART-B

21. Which one of the following pairs has two magic numbers for closed nuclear shells ?(a) 8, 10 (b) 10, 20 (c) 50, 82 (d) 82, 130

22. Identify the correct statement(s) for phosphorimetric measurement from the following :A. It is done after a time delay when fluorescence, if present becomes negligibleB. Immobilization of analytic increases phosphorescenceC. Phosphorescence decreases in the presence of heavy atoms(a) A only (b) A and B (c) A and C (d) B and C

23. Choose the isoelectronic pair among the following :A. 6[V(CO) ]B. 5

5 5[Cu( C H ) (CO)] C. 4[Co(CO) ]

D. 3 2[IrCl(CO) (PPh ) ](a) A and B (b) B and C (c) C and D (d) A and D

24. An organometallic fragment that is isolobal to 3CH+ is(a) 5[Fe(CO) ] (b) 5[Mn(CO) ] (c) 5[Cr(CO) ] (d) 3[Ni(CO) ]+

25. The calculated and observed magnetic moments (in B.M.) of aqua complex of a lanthanide ion are 0and ~3.5, respectively. The lanthanide ion is(a) Pm3+ (b) Pr3+ (c) Eu3+ (d) Sm3+

26. The compound that gives a basic solution in HF is :(a) AsF5 (b) PF5 (c) BF3 (d) BrF3

27. Based on VSEPR theory, the predicted shapes of [XeF5]– and BrF5 respectively, are

(a) pentagonal planar and square pyramidal(b) square pyramidal and trigonal bipyramidal(c) trigonal bipyramidal and square pyramidal(d) square pyramidal and pentagonal planar

28. Both potassium and sulfuric acid form intercalation compounds with graphite. The graphite layersare(a) reduced in both the cases(b) oxidized in both the cases(c) oxidized in the case of potassium and reduced in the case of sulphuric acid(d) reduced in the case of potassium and oxidized in the case of sulphuric acid

29. The resonance Raman stretching frequencies (in cm–1) of the bound O2 species in oxy-hemerthyrinand oxy-hemoglobin, respectively, are(a) ~850 and 1100 (b) ~750 and 850 (c) ~850 and 850 (d) ~1100 and 850

30. CdS, HgS and BiI3, are coloured due to(a) L M charge transfer transitions(b) d d electronic transitions(c) M L charge transfer transitions(d) combination of L M charge transfer and d d electronic transitions

2 PAPER : CSIR-UGC-NET/JRF June 2017


31. The relative rates of water exchange for the hydrated complexes of (1) Ni2+, (2) V2+ and (3) Cr3+

ions follows the trend(a) (1) > (2) > (3) (b) (1) < (2) < (3) (c) (1) > (2) < (3) (d) (1) < (2) > (3)

32. Consider the following sulfur donor atom bearing bidentate ligand where X and name of ligands aregiven in following columns :

XS

S

X Ligand nameA. NR2 I. DithiocarbonateB. OR II. DithiocarbamateC. O– III. XanthateD. SR IV. ThioxanthateCorrect match of entries given in two columns is(a) A-II, B-III, C-I, D-IV (b) A-III, B-II, C-IV, D-I(c) A-I, B-II, C-III, D-IV (d) A-IV, B-I, C-II, D-III

33. In vitro reaction of an excess of O2 with free heme B in aqueous medium the end product is(a) hematin (b) [ 2O Fe(III) -protoporphyrin-IX](c) heme B(O2) (d) oxoferrylprotoporphyrin-IX cation radical

34. 13C NMR spectrum of DMSO-d6 gives a signal at 39.7 ppm as a(a) singlet (b) triplet (c) quintet (d) septet

35. Following reaction is an example of

S

1.

2. t-BuOKS

(a) Ramberg-Backlund reaction (b) [2, 3]-sigmatropic shift(c) [3, 3]-sigmatropic shift (d) Pummerer rearrangement

36. Among the following, the synthetic equivalent of acetyl anion is

(a)O

H3C Cl(b) 3CH CN (c)

S

S

(d) 3 2 2CH CH NO

37. Following reaction is an example of

S

S

+

NC CN

NC CN

heatS

S

NC CNCN

CN

(a) [3 + 2] cycloaddition (b) [4 + 2] cycloaddition(c) [6 + 2] cycloaddition (d) [8 + 2] cycloaddition



38. The major product of the following reaction isO

H+

(a) (b) (c)

O

(d)O

39. The most stable conformation of 2-fluoroethanol is

(a)

H

H

F

OH

HH(b)

H

F HOH

HH(c)

H

F

H

OH

HH(d)

F

H HOH

HH

40. The IUPAC name of the following compound is

B

H

(a) 9-borabicyclo[3.3.1]nonane (b) 1-borabicyclo[3.3.1]nonane(c) 9-borabicyclo[3.3.0]octane (d) 1-borabicyclo[3.3.0]octane

41. The correct match of natural products in Column-I with their biosynthetic precursors in Column-IIisColumn-I Column-II

A.

Me

Me Me

OH I. L-Lysine

B.

N

NH II. L-Ornithine

III.Farnesyl



IV. Geranyl pyrophosphate(a) A-IV, B-I (b) A-IV, B-II (c) A-III, B-I (d) A-III, B-II

42. The correct order of pKa values for the following species is

(a) 3 2 2 2 2PhNH i-Pr NH Ph NH + + + (b) 2 2 3 2 2Ph NH PhNH i-Pr NH + + +

(c) 2 2 2 2 3i-Pr NH Ph NH PhNH + + + (d) 3 2 2 2 2PhNH Ph NH i-Pr NH + + +

43. Among the following, the natural product that is a steroid and contains an , -unsaturated ketone is(a) estrone (b) prostaglandin (c) cortisone (d) morphine

44. The major product formed in the following reaction is

N

Li, liq. NH3

(a)N

(b)NH

(c)N

(d)N

45. The major product formed in the sodium ethoxide mediated reaction between benzophenone andethyl chloroacetate is

(a)O

PhPh

O

Cl

(b)

PhEtO

PhCOOEt

Cl

(c)COOEt

O

PhPh

(d)

PhCl

PhCOOEt

OH


NBS (2 equiv)hv, CCl4

reflux

(a)

BrBr

(b)

BrBr

(c)

Br

Br

(d)

Br

Br

47. Consider a particle in its ground state confined to a one-dimensional box in the interval (0, 8). The

probability of finding it between 2 24.0 and 4.0 is close to ( is sufficiently small so that thewavefunction can be taken as a constant in this interval).

(a) 4

(b) 3

(c) 2

(d)



48. Which of the functions below is a common eigenfunction of ddx

and 2

2ddx

operators ?

(a) cos x (b) kx (c) eix (d) 2e x

49. A one-component system with the associated phase diagram (see the figure) is not possible because

Pres

sure

Temperature

B C

D

A

OPhase

Phase Phase

Phase

(a) OB has a negative slops (b) OC has a positive slope(c) Both OB and OC are linear (d) OB, OC and OD cannot all coexist, given OA

50. A phase transition process is always(a) isothermal – isoentropic (b) isochoric – isothermal(c) isobaric – isochoric (d) isothermal – isobaric

51. The correct statement for any cyclic thermodynamic process is

(a) 0dq (b) 0dw (c) 0dU (d) 0Vdq 52. Metallic silver crystallizes in face-centred-cubic lattice structure with a unit cell of length 40 nm.

The first order diffraction angle of X-ray beam from (2, 1, 0) plane of silver is 30º. The wavelengthof X-ray used is close to(a) 11 nm (b) 18 nm (c) 25 nm (d) 32 nm

53. If the pre-exponential factor in Arrhenius equation is 12 11.6 10 s , the value of the rate constant atextremely high temperature will be close to(a) 12 11.6 10 s (b) 12 14.2 10 s (c) 9 12.4 10 s (d) 6 11.2 10 s

54. In kinetic study of a chemical reaction, slopes are drawn at different times in the plot of concentra-tion of reactants versus time. The magnitude of slopes with increase of time(a) remains unchanged (b) increases(c) decreases (d) increases and decreases periodically

55. The electrochemical cell potential (E), after the reactants and products reach equilibrium, is (E0 isthe standard cell potential and n is the number of electrons involved)(a) 0 /E E nF RT (b) 0 /E E RT nF (c) 0E E (d) 0E

56. For the electronic configuration 2 2 41 2 2s s p , two of the possible term symbols are 1S and 3 p . TheTheremaining term is(a) 1D (b) 1F (c) 3D (d) 3F



57. The v = 0 to 1 vibration-rotation spectrum of a diatomic molecule exhibits transitions for

0 , 1 , 1R R P and 2P lines at 2241, 2254, 2216 and 2203 cm–1, respectively. From this data,we can conclude that the molecule(a) has rigid rotation and harmonic vibration (b) has anharmonic vibration(c) has rotational-vibrational interaction (d) is affected by nuclear spin-statistics

58. Consider aqueous solutions of two compounds A and B of identical concentrations. The surfacetension of the solution of A is smaller than that of pure water while for B it is greater than that of purewater under identical conditions. From this one infers that(a) surface concentration of A is smaller than its bulk concentration(b) surface concentration of B is larger than its bulk concentration(c) surface concentration of A is larger than that of B(d) surface concentration of A is smaller than that of B

59. For a monodisperse polymer, the number-average molar mass nM and weight-average molar mass

wM are related according to

(a) w nM M (b) w nM M (c) w nM M (d) logw nM M

60. An intense purple colour (Plasmon band) is exhibited by a colloid consisting of spherical(a) silver particles of 10 mm diameter (b) silicon particles of 5 mm diameter(c) gold particles of 5 nm diameter (d) iron particles of 3 mm diameter

PART-C

61. Choose the correct statement for magnitude of threshold energy of an endoergic nuclear reactionbetween stationary nucleus and a moving projectile.(a) It is greater than ‘ Q ’ of nuclear reaction.(b) It has to be more than kinetic energy of a projectile.(c) It is less than ‘ Q ’ of nuclear reaction.(d) It has to be equal to kinetic energy of a projectile.

62. Identify correct statements from the following:A. Area of differential thermal analysis peak is proportional to amount of sample.B. Area of differential thermogravimetric analysis curve is proportional to mass loss.C. Phase transition cannot be studied with differential scanning calorimetry.D. Simulatneous determination of two metal ions is possible with thermogravimetric analysis.Answer is(a) A, B and C (b) A, B and D (c) B, C and D (d) A, C and D

63. Consider following statements for fission of 235U with thermal neutrons.A. The % of nuclei undergoing unsymmetrical fission is maximum.B. In each fission, one thermal neutron is produced.C. Magnitude of energy released per fission is of the order 200 MeVCorrect statement(s) is/are(a) A and B (b) A and C (c) B and C (d) C only

64. Addition of two electrons to the bismuth cluster 35Bi results in a change of structure type from

(a) closo to nido (b) nido to arachno (c) closo to arachno (d) arachno to hypho



65. Reaction of 5Na Mn CO with 2 2H C CHCH Cl gives A along with NaCl. Photolysis of com-

pound A results in compound B together with elimination of CO. the correct structural formulationsof compounds A and B are respectively,

(a)

CO

MnOCCO

OCCO

CO

MnOCCO

CO

A B

(b)Mn

COCOCOOC

OCMn

COCOCOOC

A B

OC

(c)

CO

Mn

CO

CO

CO

CO

Mn

CO

CO

MnOCCO

CO

BA

(d)Mn

COCO

COCOMn

CO

COB

Mn

CO COOC

A

COOC

66. A copper(II) complex having distorted octahedral geometry shows an absorption band at 625 nm.Given spin-orbit coupling of the complex as –625 cm–1, the eff (in B.M.) is(a) 1.73 (b) 1.81 (c) 1.63 (d) 1.93

67. Match items in column A with items in column B:

+5 3 2 6 3

2 6 43

6 3+

3 4+

3 2 4

Column A Column BI: SbF BrF BrF SbF A. Lewis acid behaviour of BrF

II: BrF SbF Ag BrF B. Lewis base behaviour of BrFAg SbF 2BrFIII: KF BrF K BrF C. Self ionisationIV: 2BrF BrF BrF D. Neutralisatio

n

The correct answer is(a) I-(A); II-(B); III-(C); IV-(D) (b) I-(B); II-(D); III-(C); IV-(A)(c) I-(C); II-(D); III-(B); IV-(A) (d) I-(B); II-(D); III-(A); IV-(C)

68. Mössbauer spectrum of complex 2 2Fe 1,10 phenanthroline NCS shows two lines at 300 K,

four lines at 186 K, and again two lines at 77 K. This can be attributed toA. change in the coordination mode of NCSB. change in the spin-state of ironC. cis-trans isomerisationD. change in metal-ligand bond distancesThe correct statements are(a) A and B (b) B and C (c) A and C (d) B and D

69. 3 2R Ge on photolysis gives a radical which shows ESR spectrum. The ESR signals carrying the

signature of 73Ge I 9/2 are in terms of(a) Nine lines (b) Ten lines (c) Two lines (d) One line



70. Mass fragment of +IrCl in mass spectrometry shows three mass peaks at m/z = 226, 228, and 230.

Given that natural abundances of 191 193 35Ir, Ir, Cl and 37Cl are 37%, 63%, 76%, and 24% respec-tively, the intensities of the mass peaks are in the order(a) 49.5 : 100 : 26.6 (b) 100: 49.5: 26.6 (c) 26.6: 100: 49.5 (d) 26.6: 49.5: 100

71. The 31 1P H NMR spectrum of 4 4 4 2 42,2,6,6-N P Cl NMe is expected to show(a) two triplets (b) two doublets(c) one doublet and one triplet (d) one quartet and one doublet

72. The number of bonding molecular orbitals and the number of available skeletal electrons in 26 6B H ,

respectively, are(a) 7 and 14 (b) 6 and 12 (c) 18 and 12 (d) 11 and 14

73. The compound 2 2N F has two isomers. Choose the correct option from the following:

(a) both isomers possess v plane

(b) both isomers possess h plane

(c) one isomer has h plane while the other has a v plane

(d) none of them have a h plane

74. Consider the following statements for metallothioneins:A. the contain about 30% cysteine residuesB. they prefer to bind soft metal ions such as Cd(II), Hg(II) and Zn(II)C. they are involved in electron transfer reactionsD. they are low molecular weight proteinsCorrect statements are(a) A, B and C (b) A, B and D (c) A, C and D (d) B and C

75. Consider the following statements for deoxy-hemerythrin and deoxy-hemocyanin:A. they are involved in O2 transport in biological systemsB. they contain two metal ions in their active siteC. active site metal centres are bridged by amino acid residuesD. they prefer to bind only one O2 per active siteThe correct statements are(a) A, B and D (b) A, C and D (c) B, C and D (d) A and C

76. Consider the following statements for octahedral complexes, (a) 36CrF , (b) 3

3Cr ox

and (c)

3+

3Cr en :

A. their d d transitions are at 14900, 17500, and 21800 cm–1, respectivelyB. their spin-only magnetic moments are sameC. two of them have optical isomersD. all of them show Jahn-Teller distortionThe correct statements are(a) A, B, and C (b) A, C, and D (c) B, C, and D (d) B and D



77. Addition of 4NaBH to 5 66 6Cp Fe C H

will give

(a) 52

Cp Fe H

(b) 5 66 6Cp Fe H C H

(c) 5 66 6Cp Fe C H (d) 5 6

6 7Cp Fe C H

78. The eff of 2 2 3Fe S CNEt changes with temperature with the involvement of two electronic

states. The states are

(a) low spin 22gT and high-spin 6

1gA . (b) low spin 11gA and high-spin 3

2gT

(c) low spin 2gE and high-spin 6

1gA . (d) low spin 22gT and high-spin 4

1gT .

79. Match the items in the three columns.

max

2 36 2

3 36 2

33 2

Absorption maxComplex (column 1) Color (column 2) ( ,nm) (column 3)A. Ni H O NO I. Blue X. 675B. Ni NH NO II. Green Y. 565C. Ni en NO III. Violet Z. 615

The correct answer is(a) A-II-X; B-I-Z; C-III-Y (b) A-I-X; B-II-Y; C-III-Z(c) A-III-Y; B-I-Z; C-II-X (d) A-I-X; B-II-Z; C-III-Y

80. Identify the product in the reaction between

IrCOPh3P

PPh3Cl and 3CH I going at room temperature via 2NS mechanism

(a) IrCOPh3P

PPh3Cl

CH3

I

(b) IrIPh3P

PPh3Cl

CH3

CO

(c) IrIPh3P

PPh3OC

CH3

Cl

(d) IrCOPh3P

ClPh3P

CH3

I

81. The major products A and B formed in the following reactions sequence are

Ph CHO A B

i. n-BuLi (2 equiv.) THF, -78 ºC

ii. CO2iii. H3O+

Ph3PCBr4Zn

(a) Ph Br

Br PhA = B =

CO2H

(b) Ph BrA = B = Ph CO2H

(c) PhA = B = PhBr CO2H

(d) PhA = B = PhBr CO2H

Br CO2H



82. The intermediate A and product B formed in the following reaction sequence are

H

CO2H

NO

SH

DCC, DMAPC6H6, reflux

AC6H6, reflux

n-Bu3SnD AlBN

B

(a)

H

O

SN

OH

O

DA = B =

(b)

H

O

SN

OH

DA = B =

(c)

H

O

O N

SH

A = B =

D O O

(d)

H

O

O N

SH

DA = B =

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

OO O

NH1.

2. CH2N2Et2OA

AlCl3CH2Cl2

B

(a)

MeO2C

NO

NO

MeO2C

A = B =

(b)

MeO2C

NO

NO

MeO2C

A = B =

(c) NO NO

A = B =MeO2CH H

MeO2C



(d) NO

A = B =MeO2C NO

MeO2C


OO

t-BuOBr CO2Me

O

i NaH 0 ºC

iiA

F3C

O

OHCH2Cl2THF, RT

B

(a)OO

CO2Met-BuO

OO

O

HO CO2MeA = B =

(b)

OO

CO2Met-BuO

OO

Ot-BuO

CO2MeA = B =

(c)

OO

t-BuO

OCO2HA = B =

O

CO2MeHO2C

(d)

OO

CO2Me

t-BuOA = B =BrOH

OO

CO2Me

O

85. The major product formed in the following reaction isH

OHH

1. NaH, CS2, Mel

2. n-dodecane heat

(a)

H

(b)

H

H

(c)

H

H

(d)

H

H



86. The correct combination of reagents to effect the following reaction is

OH

CHO

OH

HO

(a) A. 3POCl , pyridine; B. AgOAc ; C. 4LiAlH

(b) A. 4NaBH ; B. 3 2Ph P, DEAD, PhCO H

(c) A. 3 2Ph P, DEAD, PhCO H ; B. 4LiAlH

(d) A. PCC ; B. L-selectride


Ph CO2EtAlCl3

CH2Cl2, 0 ºC

i. O3

iii. H3O+A B

ii. NaBH4

(a)Ph

CO2EtO

HH

O

Ph

OH

A = B =

(b)

O

HH

O

A = B =

H

H

Ph

CO2Et

HOPh

(c)Ph

CO2EtO

HH

O

Ph

OH

A = B =

(d) A = B =

Ph

CO2EtO

OH Ph

O

88. The correct combination of reagents A and B to effect following transformations are

H

OH

HOH

B A OH

H

OHH



(a) 4 2 2A cat. OsO , NMO; B = i. I , PhCO Ag, ii. aq. NaOH

(b) 4 2 2 2A alkaline KMnO ; B = i. I , PhCO Ag, H O, ii. aq. NaOH

(c) 2 2 4A = I , PhCO Ag, ii. aq. NaOH; B = cat. OsO , TMEDA, NMO

(d) 4A = i. m-CPBA, ii. aq. NaOH; B = alkaline KMnO


O OH

(i-PrO)4TiL-(+)-DETt-BuOOH

CH2Cl2, -20 ºCMol. Sieves 4 Å

A

t-BuSHNaOH

B

(a) O OH

OO OH

OH

SBu-t

A = B =

(b) O OH

OO OH

OH

SBu-t

A = B =

(c) O OH

OO SBu-t

OH

OH

A = B =

(d) O OH

OO SBu-t

OH

OH

A = B =


O

OMeDDQ

(1 equiv.)A

Phl(OAc)2 BMeOH

(a)OH

A =O

B =

OMe

(b)OH

A =O

B =

OMe

(c)OH

A =O

B =

O

(d) A =O

B =O

OMe

OAc



91. The specific rotation D for (S)–(+)–2–butanol is 10º mL/g dm. The observed optical rotation

obs of a sample composed of a mixture of (R)- and (S)-2-butanol is –0.45º. If the cell path lengthis 0.6 dm and the concentration of 2-butanol in the sample is 0.15 g/mL, the percentages of (R) and(S) enantiomers in the sample are(a) (R) = 25%, (S) = 75% (b) (R) = 40%, (S) = 60%(c) (R) = 60%, (S) = 40% (d) (R) = 75%, (S) = 25%


CH2Br

1. i. Mg, ii. CO2, iii. H3O+

2. (COCl)2

3. Et3N

(a)O

(b)

Cl

O

O

(c) O

O

Cl

(d)

O

93. Following reaction involves

NH N

OH CHO

p-TSAC6H6, reflux

N

N

O

(a) Claisen followed by Mannich reaction (b) aza-Cope followed by Mannich reaction(c) Claisen followed by aza-aldol reaction (d) aza-Cope followed by aza-aldol reaction

94. The intermediate A and the major product B formed in the following reaction is

IPd(OAc)2

Ph3P

Ag2CO3[A] B

(a)

H PdLn H

A = B =

(b)

H PdLn

A = B =

(c)

H PdLn

A = B =

H



(d)

H PdLn

A = B =


N

O Oi. Bu2BOTf

Et3N

ii. PhCHO iii. LiOH, H2O2

H2O

(a) HO

O OH

Ph (b) HO

O OH

Ph (c) HO

O OH

Ph (d) HO

O OH

Ph


N

COOMe

PhMe OTBDMS

O+

175 ºCtoluene

(sealed tube)

(a) N

OTBDMSOMe

MeOOC

Ph

(b) N

OTBDMSOMe

MeOOC

Ph

(c) N

OTBDMSOMe

Ph

COOMe (d) N

OTBDMSOMe

Ph

COOMe

97. The most stable conformation for the following compound isHO

OH

(a)OH

OH(b)

OH HO

(c)

OH OH

HH

(d)

H

OHHO

H• •



98. The correct structure of the compound based on the following characteristic spectral data isIR: 1736 cm–1

1H NMR: 3.59 (s, 3H), 3.32 (t, 2H), 2.25 (t, 2H), 1.85-1.75 (m, 2H), 1.73-1.62 (m, 2H)13C NMR: 174.0, 51.0, 32.9, 32.9, 32.8, 31.0, 23.0

(a) BrOEt

O(b) Br

O

O

(c)Br OMe

O(d)

OMe

OBr

99. The major product formed in the reaction of D-glucose with 2ZnCl in MeOH is a methylglucopyranoside (A or B). The structure of this product and the molecular orbital interaction presentbetween ring-oxygen and the anomeric C-O bond responsible for its stability, respectively, are

O

OH

HO

HOHO

OMeO

OH

HO

HOHO

OMeA B

(a) A and n (b) A and n (c) B and n (d) B and n

100. Among the following correct statement for nucleic acids is(a) Uracil is present in DNA(b) Uracil is present in RNA(c) Phosphorylation in RNA is at 2’ and 5’ positions(d) Normally three hydrogen bonds stabilize A-T base pair

101. The figure below depicts an adsorption isotherm of O2 on charcoal at 90 K.

0 200 400P/torr

100

200

v(cm

/g)

3

At a pressure 25 torr, only 10% of charcoal sites are occupied by O2. Therefore, the ratio of adsorp-tion to desorption rate constants (in torr–1) is close to(a) 0.003 (b) 0.004 (c) 0.006 (d) 0.015

102. Polonium is the only metal known to exist in a simple cubic lattice form. The density of polonium at0ºC is measured to be 10.00 g/cm3. The atomic radius of polonium would then be (assume the massof a polonium atom = 2.7 × 10–22 g)(a) 1.1 Å (b) 1.9 Å (c) 1.5 Å (d) 2.3 Å

103. The specific conductance of a solution is 1 10.176 cm . If the cell constant is 0.255 cm–1, the

conductance 1 of that solution is(a) 1.449 (b) 0.690 (c) 0.045 (d) 0.431



104. Photochemical decomposition of HI takes place with the following mechanismHI + hv (I ) H + Ia

21H + HI H Ik

22I + I + M I Mk

Considering hydrogen (H) and iodine (I) atoms as intermediates, the rate of removal of HI is(a) aI /2 (b) aI (c) a2I (d) 2

aI

105. In an enzyme-catalysed reaction

1 2

1E + S ES E + Pk k

k

4 12 3.42 10 sk . If 2 3

0 1 10 mol dmE , the magnitude of maximum velocity and turn overnumber using Michaelis-Menten kinetics are(a) 2 3 1 4 13.42 10 mol dm s ; 3.42 10 s (b) 6 3 1 4 13.42 10 mol dm s ; 3.42 10 s

(c) 4 3 1 6 13.42 10 mol dm s ; 3.42 10 s (d) 4 3 1 2 13.42 10 mol dm s ; 3.42 10 s

106. Arrhenius equations for two chemical reactions are: 1 2/ /1 1 2 2,E RT E RTk A e k A e . If 1 2E E , then at

a given temperature T,

(a) 1 1

2 2

k Ak A

(b) 2 2

1 1

k Ak A

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

107. The fugacity of a real gas is less than the pressure (P) of an ideal gas at the same temperature (T) onlywhen (Tb is the Boyle temperature of the real gas)(a) high , bP T T (b) low , bP T T (c) high , bP T T (d) low , bP T T

108. For the reaction 2 2 21H O g H g O g2

, the equilibrium constant pK depends on the degree

of dissociation 1 and total pressure P as

(a) 2pK P (b) 3/2 1/2

pK P (c) 1/2 3/2pK P (d) 2

pK P

109. The minimum work required by an engine to transfer 5 J of heat from a reservoir at 100 K to one at300 K is(a) 5 J (b) 10 J (c) 15 J (d) 20 J

110. The correct relation involving symmetry operations

(a) 24 2S S (b) 2xz yz C x

(c) 3 34 4S C (d) 3

6 2S S

111. A polydisperse polymer sample has ten molecules of molar mass 20,000 g mol–1 and fifteen mol-

ecules of molar mass 10,000 g mol–1. The number-average molar mass 1g mol nM of the sampleis(a) 13,000 (b) 14,000 (c) 15,000 (d) 16,000



112. Consider a system of three particles which can occupy energy levels with energy 0, and 2, such

that the total energy 4E . Cases A, B and C correspond to spin 12 fermions, spin 0 bosons, and

classically distinguishable particles, respectively. The correct ordering of entropy is(a) A B CS S S (b) B A CS S S (c) C B AS S S (d) C A BS S S

113. For a point group, an incomplete character table is given below with one irreducible representationmissing

3

1

E 2C 3A 1 1 1

E 2 1 0

v

The Mulliken symbol and characters of the missing representation are(a) 1 1 1 1A (b) 1 1 1 1B (c) 2 1 1 1A (d) 2 1 1 1B

114. Given below is a specific vibrational mode of 3BCl with and denoting movements of therespective atoms above and below the plane of the molecule respectively. The irreducible represen-tation of the vibrational mode and its IR / Raman activity are

3 3 2 32 2 2

1

22 2

1

2

2 3 2 31 1 1 1 1 1 ,1 1 1 1 1 12 1 0 2 1 0 , ,1 1 1 1 1 11 1 1 1 1 12 1 0 2 1 0 , ,

h b v

z

x y

D E C C SA x y zA RE x y x y xyAA zE R R xz yz

Cl Cl

B

Cl

(a) 2A ; neither IR nor Raman active (b) E ; both IR and Raman active

(c) 1A ; Raman active (d) 2A ; IR active

115. The first excited state 21/2P of fluorine lies at an energy of 400 cm–1 above the ground state 2

3/2P .

The fraction of Fluorine atoms in the first excited state at 1420 cmBk T is close to

(a) 1

1 e (b) 1

2 e (c) 1

1 4e(d)

11 2e



116. The two limiting wavefunctions of the ground state of +2H molecular ion, as the internuclear separa-

tion R goes to (i) (infinity) and (ii) 0 (zero) are ( a b1s ,1s are 1s-orbital wave functions of hydrogen

atoms a and b in +2H , and He1s is the wave function of the 1s orbital of He )

(a) (i) a1s r ; (ii) b1s r (b) (i) b1s r ; (ii) a1s r

(c) (i) a 1 b 21s r 1s r ; (ii) He 1 He 21s r 1s r (d) (i) a b1s r 1s r ; (ii) He1s r

117. For a certain magnetic field strength, a free proton spin transition occurs at 700 MHz. Keeping the

magnetic field strength constant the 14 N nucleus will resonate at 145.6 and N 0.4g p g

(a) 700 MHz (b) 400 MHz (c) 200 MHz (d) 50 MHz

118. The first electronic absorption band maximum of a polar and relatively rigid aromatic moleculeappears at 310 nm but its fluorescence maximum in acetonitrile solution appears with a large Stokesshift at 450 nm. The most likely reason for the Stokes shift is(a) large change in molecular geometry in the excited state(b) increase in dipole moment of the molecule in the excited state(c) decrease in polarizability of the molecule in the excited state(d) lowered interaction of the excited molecule with polar solvent

119. The un-normalized radial wave function of a certain hydrogen atom eigenstate is 26 exp /3r r r .A possible angular part of the eigenstate is(a) 35cos 3cos (b) 23cos 1 (c) cos (d) 1

120. Given a trial wave function 1 1 2 2t C C , and the Hamiltonian matrix elements, *1 1 0H dv ,

*1 2 2.5H dv , *

2 2 12.0 H dv , the variationally determined ground state energy is(a) –0.52 (b) –0.50 (c) 12.50 (d) 12.52

20 SOLVED PAPER : CSIR-UGC-NET/JRF June 2017


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

PART–B21. 2, 8, 20, 28, 50, 82 and 126 are magic number.

Correct option is (c)22. Phosphorescence has been observed form a wide variety of compounds and is differentiated from

fluorescence by long lived emission of light after extinction of the excitation source i.e. it done afterfluorescence, if parsists.Phosphorescence increases when heavy atoms like iodine silver and lead etc. are added.Phosphorescence increases, when analyte used become immobile.

Hence, both (A) and (B) are correct.Correct option is (b)

23. The species having same number of electrons are called isoelectronic.

6A V CO 5 12 17 e

55 5B Cu C H CO 11 5 2 18 e

4C Co CO 9 8 1 18 e

3 2D IrCl CO PPh 9 1 2 4 16 e Correct option is (b)

24. Two fragments are isolobal if the number, symmetry properties, approximate energy and shape oftheir frontier molecular orbitals and the number of electrons in them are similar not identical butsimilar.

3CH 4 3 1 6 ; 5Cr CO 6 10 16 Correct option is (c)

25. 3 6Eu f

+3 +2 +1 0 –1 –2

L = 6–3 = 3 J = |L–S| = 0S = 3

–3

Hence, 1 0µ g J J Correct option is (c)

26. 5 6 2AsF 2HF AsF H F

5 6 2PF 2HF PF H F

3 4 2BF 2HF BF H F

3 2 2BrF HF BrF HF BrF3 acts as base as it donate F– in HF.

Hence, correct option is (d)

21SOLVED PAPER : CSIR-UGC-NET/JRF June 2017


27. [XeF5]–

F F

F

F

FXe

••

••

–

Shape = Pentagonal planar

[BrF5]

F

F F

F

Br

F

••

Shape = square pyramidal

Correct option is (a)

28. K graphite graphite K

2 4 4H SO graphite graphite HSO

Correct option is (d)

29. In oxyhemerythrin 1850 cmO O

In oxyhemoglobin 11100 cmO O


30. In CdS, HgS and BiI3 colour is due to LMCT as Cd2+ and Hg2+ has d10 configuration. Hence, no d-dtransition. Also, no MLCT as ligand is not -acceptor..Hence, correct option is (a).

31. Change in CFSE on going for Oh to five coordinate intermediate for ions is as follows.

2 3 2.00V d

3 3 2.00Cr d

2 8 2.00Ni d


32. NR2 Dithiocarbamate

OR XanthateO– DithiocarbonateSR ThioxanthateCorrect option is (a)



33. PFe + O2

IIPFeO2 + PFe

II IIPFe — O — Fe — PIII III

2PFe = OIVPFe

II

PFe—O—FePIII III

Hematin


34.D3C

S

O

CD313C NMR n = 3; I = 1 (for D)Multiplicity = 2nI + 1 = 2×3×1 + 1 = 6 + 1 = 7 (septet)


35.

S S

Ht-BuOK

S

2

1 2 3[2, 3] S.T.R.

S

I

1

Correct option is (b)

36.

H3C CH

H

NO2Base

H3C CH NO2

Nitroethane

H C Cacetyl anion

Hence, nitroethane anion is synthetic equivalent of acetyl anion.Correct option is (d)

37.S

S

NC

NC

CN

CN

[8+2]

cyclo-addition S

S

NC CN

CN

CN




38.

O

H+

OH HO

RingContraction

Oringexpansion


39.

H

F HO

HH

H

Most stable due to intramolecular hydrogen bonding


40.

BH9

87

6 5 4 3

12

9-borabicyclo [3, 3, 1] nonane


41.

N

NH

Anabasine

It is a pyridine and piperidine alkaloid found in the tree tobacco plant. Its principal industrial use isan insecticide.

Lysine serves as the precursor of piperidine ring of anabasine.

OPP

Gernyl pyrophosphate is the biosynthetic precursor of (+)–menthol.Correct option is (a)



42. Basicity order for their conjugate base isi

2 2 2Pr NH PhNH Ph NH Hence, pKa value for their conjugate acid

i2 3 2 2

a

Pr NH PhNH Ph NH

pK

Hence, correct option is (b)

43.

O

O

H

H HCortisone

OH

O

OH

Correct option is (c)

44.N

Li, liq. NH3

(Birch reduction)N

H H

H

N

e–

N

+H+

N

H

•e–

NH

•

N

H

H+




45.

O

Ph Ph+ Cl CO2Et

O

Ph PhCl

CO2Et OPh

PhCO2Et

Darzen's glycidic estersynthesis


46. NBS + HBr Br2hv 2Br•

(Limited supply)

H

+ Br•

•

–Br•Br Br

Br

• BrBrBr Br •

+ Br•

Trans-1, 2-dibromoacenaphthene

Below the plane attack

H

+ Br•

– HBr


47. The probability of finding the particle lies between x to x + dx is given by

2p x dx x dx

4,x dx

2 2 2sin sinrn x n xP dx dx

22 sin n x

2 22 2 1 4sin sin8 8

n x

21 sin4 2 4


48. ix ixd e i edx

therefore, eigen function



22

2ix ixd e i e

dx therefore, eigen function


49. In one component system 4 phases carnot simultaneously exist at single point.Correct option is (d)

50. (i) Phase transition cannot be isoentropic

solid liquidSo, entropy is going to change during phase transition(ii) Phase transition cannot be isochoric as

solid gas or liquid gasVolume is going to change(iii) Volume is going to changeAs phase transition takes place at constant temperature and pressureCorrect option is (d)

51. For any cyclic process cyclic integral of state function is zero.So, 0dU as U is a state functionCorrect option is (c)

52. 2 2 22 sina n

h k

40 12 125

18 nm Correct option is (b)

53.a

RTk A e

at very high temperature T

00

aR Ak A e k A e k k A

e

Correct option is (a)54. As the time increases slope will decreases.

So, correct option is (c)

55. At equilibrium E is zero, because G is zero at equilibrium.G nFE

if 0G So, E = 0Correct option is (d)

56. 1s2, 2s2, 2p4

Number of microstate for P4 = 6! 15

4! 6 4 !

1s = 1×1 = 1 3 3 3 9p Thus, remaining microstate are 5 which comes from 1D = 1×5 = 5Correct option is (a)



57. Since, due to vibrational-rotational interaction P and R lines are obtained and molecule behave as adiatomic vibrating rotor.

21 11

2 2e e eE BJ J x

Correct option is (c)58. Greater the surface tension, the greater the surface concentration.


59. P.D.I. = 1n

MM

(for monodisperse)

w nM M Correct option is (b)

60. Intermediate colours been understood and correlated to the formation of intermediate nanostructuresbefore the formation of the final gold nanoparticles. Specifically, TEM images have shown that afterfew seconds of citrate addition, gold nanowires 5 to 8 nm are formed, which are responsible for thedark purple colour. Beyond a certain threshold, the nanowires disintegrate into nanoparticles, andthe solution turns ruby-blue.Correct option is (c)

PART–C61. We have the relation between threshold energy and ‘Q’ of a reaction, as

a ATh

A

M MK QM

Here, Ma = mass of moving projectile or bombarding particle

MA = mass of stationary nucleusAlso, for endoergic reaction, Q > 0Therefore, magnitude of threshold energy

a ATh

A

M MK QM

1a A

A

M MM

ThK Q

Correct option is (a)62. Correct statement are

• Area of different thermal analysis peak is proportional to amount of sample• In the thermogravimetric analysis area of curve is proportional to mass loss• In thermogravimetric analysis two metal ion simultaneously determine.Correct option is (b)

63. In fission of U235 with thermal neutrons(1) On the average, each fission of a U235 nucleus produces about 2.5 free neutrons(2) Each fission produce 200 MeV energy(3) Fission always occure in assymmetric fashion, it means number of nuclei undergoing unsym-metrical fission is maximum.Hence, correct statement is (1) and (3)Correct option is (b)



64. 35Bi 5 5 3 22

Hence, (4n + 2), closoOn addition of 2 electron it become (4n + 4), nidoCorrect option is (a)

65. 2 25Na Mn CO CH CH CH Cl Mechanism:

MnCOOC

CO

COOC

(A)

+ NaCl –COhv

MnCOOC

CO

COOC

(B)upon heating 1 allyl converted into 3 allylwith evolution ofCO gas

[Mn(CO)5]– + CH2 = CH – CH2 – Cl


66. µeff = µspin 1

77 11 1 1010 cm cm

625 625

For Cu2+ Term is 2D, hence, 2 , 1625 cm

µeff = 1

7 1 72 625 cm 2 625 6251.73 1 625 1.73 1

10 cm 10

72 625 6251.73 1 1.73 1 0.078 1.73 1.078

10

effµ 1.864


67. 5 3 2 6 3SbF BrF BrF SbF Lewis base behaviour of BrF

2 6 4 6 3BrF SbF Ag BrF Ag SbF 2BrF Neutralization

3 4 3KF BrF K BrF Lewis acid behaviour of BrF

3 2 42BrF BrF BrF self ionization




68. 2 2Fe 1,10-phenanthroline NCS

77K Low spin

–4 –2 0 42

186K

HS300K

• Since NCS is ambidentate ligand therefore change in co-ordinate mode• Change in the spin-state of iron at high temperatureCorrect option is (a*)

69. h •3 32R Ge 2R Ge in •

3R Ge radical

multiplicity = (2NI + 1) = 2×1×92 + 1 = 10 lines

73 9Ge I2


70. 191 193 35 37Ir : Ir Cl : Cl

Natural abundance37% : 63% 76% : 24%

x y a b

1 13.7 6.3 7.6 2.4x y a b

3.7 7.6 6.3 7.6 3.7 2.4 6.3 2.4xa ya xb yb

28.12 47.88 8.88 15.122 2 4

xa ya xb ybM M M M

28.12 56.76 2 15.12 4M M M

28.12 1.76 56.76 1.76 2 15.12 1.76 4M M M

49.49 100 2 26.6 4M M M

49.5 100 2 26.6 4M M M

: 2 : 449.5 : 100 : 26.6M M M




71.

N

PMe2N Cl

NP

P

(a)

Cl

Me2N NP

N

ClNMe2

NMe2

Cl(b)

(b)

31 1a

1P H 2NI 1 2 2 1 3 triplet2

31 1b

1P H 2NI 1 2 2 1 3 triplet2


72. 26 6B H has 6 + 1 = 7 binding molecular orbital

26 6 2 6 12B H B H e

for 2 negative charge, hence 12 2 14e e e Correct option is (a)

73. N2F2 has two isomers,

N N

F F

N N

F

F

1C2 + 2v 1C2 + 1h


74. Metallothioneins are cystein rich, low molecular weight protein. Due to soft sulfur centre they preferto bind.Correct option is (b)

75. Both deoxyhemerythrin and deoxy hemocyanin are O2 transport protein in biological system. Bothcontain two metal ion at active site and they bind only O2 per active site.Correct option is (a)

76. As the strength of ligand increases energy for d-d transition increases.Strength of ligand order is 2F OX en .All complex have same number of unpaired electron. Hence, all have same spin only magneticmoment.Correct option is (a)



76. 3

3Cr ox

and 3

3Cr en

will show optical isomer. As t2g level is electronically non-degener--

ate. Hence, there will be no Jahn-Teller distortion.Correct option is (a)

77.

Fe

+

NaBH4 Fe

H

H 5 5

5 5 6 7Fe C H C H

5 56 6 5 5Fe C H C H


78. In 2 2 3Fe S CNEt oxidation state of Fe = +3

Fe3+ (low spin)

2T2g

Fe3+ (high spin)

6A1g

;


79. A Ni H O NO B Ni NH NO C Ni en NO2 3 3 3 36 32 6 2 2increases0

Thus, order for energy absorption is 675 (A) < 615 (B) < 565 (C)Complementary colour of blue Orange, green Red, Violet YellowEnergy order for absorbed light is, yellow > orange > redThus, A absorbs red hence it is green

B absorbs orange hence it is blueC absorbs yellow hence it is violet

Thus, correct option is (a)

80. IrCO

Cl

Ph3P

PPh3

CH3I (polar)SN2

anti addition

IrCO

Cl

Ph3P

PPh3

CH3

IMechanism:

Ph3P

Cl PPh3

CO

Ir

•• Ph3P

Cl PPh3

CO

IrCH3I

CH3

I

Ph3P

Cl PPh3

CO

Ir

CH3

I




81. PhCHO Ph3P, CBr4, Zn

Corey-Fuchs reaction Ph

H

Br

Br

n–Bu–Li+Ph

Br

n–Bu–Li+PhLi

(i)(ii) CO2(iii) H3O+

PhCOOH


82.

H

COOH +

NHS

O

DCC, DMAPC6H6, reflux

H

C

O

O

S

H

DBu3SnD, AlBN

Bu3Sn•

–CO2+

Bu3SnS


83.O

O O + NH

NO

HO2CCH2N2

methylation

NO

EtO2C

HAlCl3

ene reaction

NO

MeO2C


84.BuO

O

H

O

t-Na+H– BuO

O O

t-

BrO

CO2MeBuO

O O

O

CO2Me

O

O O

O

CO2Me

t-

H

But-HO

O O

O

CO2Me

H

O

H

OOH2

CO2Me

HO

OHO CO2Me

OH+




85.

OH H

H

Me–I Pyrolyticsyn-elimination

H

H

O H

H

S C S

NaH

CH3 H

H

O H

H

C

SMe

S

H

H


86.

CHO

OH

Ph3P, DEAD, PhCO2H

CHO

O C Ph

O

LiAlH4

OH

OH


87.+

Ph

EtO2C

(4+2)Cycloaddition

CO2Et

PhO3/NaBH4

CO2Et

Ph

HO

H3O+HH

Ph

OH

O O

H

O

H H

OH

Ph

O

HO


88.KMnO4/OH–

OH

OH

(i) I2, PhCO2Ag, H2O (woodward reagent)

(ii) Aq. NaOHOH

OH




89. OOH

L–(+)–DETepoxidation

below the plane

OO

O

t-Bu-SH , NaOH O

O

OH

OS-tBu

OH

OH t-Bu-S

Product (A)

Product (B)

OHH O

O

O

above the plane attack


90.

O

OMe

DDQ1 equiv.

Deprotection of –OH group

OH

Ph IOAc

OAc+

OI

OAc

PhO

OMeMeOH

–OAc–PhI–OAc


91. D for S–(+)–2-butanol = 10º mL/g dm

obs for mixture of R and S 2-butanol = – 0.45ºIf l = 0.6 dm; C = 0.15 g/mL

0.45 –5º0.6 0.15

T

C

pure

5º 100 50% excess of R10º

mixtee

5050 75%2

R

25%S Correct option is (d)



92.

Br

(i) Mg

MgBr

(ii) CO2

(iii) H3O+

C

OH

O

(COCl)2

C

Cl

OH

Et3N

C

C

O

O

(2+2)addition


93.

N

OH

H

N

C

O

HN

N

OH

3 2

32

11

[3,3] S.T.R.Aza-cope reagent

N

NO H

(Mannich reaction)

N

N

O


94. 3 32 4Pd OAc PPh Pd PPh

Active form is 3 2Pd PPh or PdL2 (where L = PPh3)This reaction is an example of intramolecule Heck reaction(1) Oxidative Addition :

I4

3

2

1

PdL2Pd

IL2

(2) Olefin insertion (Syn addition):

Pd4

3

2

1

L2 I

32

4

1

PdL2H

HH elimination

SYN

Olefinic insertionSYN addition

H




95.N

O OBBu21

1

22

z-enolatesyn

N

O O

Me

Ph

OH

LiOH, H2O2

H2O2(Hydrolysis)HO Ph

O OH

Me

Bu2BOTfN

O O

Et3N

PhCHO


96. N

CO2Me

Ph

175ºCN

Ph

CO2Me

N

Ph

CO2Me

O

OTBDMS

(1, 3-Dipole)

1, 3-DipolarCyclo-addition

N

O

Me

CO2Me

OTBDMS

Ph


97.

OH

OH

Highly unstable due to 1, 3-diaxial interaction, sterichinderance and Gaucheinteraction

O O

Twist boat

H H

Hydrogen bonding




98.Br O

O

Me

3.59 , 3s H suggest option (c) and (d)

3.32, , 2 only possible in option (c)

2.25, , 2t Ht H


99. Anomeric effect present in compound (B)

O

HOHO

OH

OHOMe

••

n

interaction

Correct option is (c)100. In case of RNA-uracil is present. Inspite of thymine-DNA


101.1 0.1 1 0.0044

1 1 0.1 25k

P


102. 3Zma

1/31/3 223

31 2.7 10 3 10 3

10Zm ga cm Å

gcm

And 1.52ar Å


103. k GA

0.176 0.255G 0.1760.255

G 176255

G

10.6901 G Correct option is (b)

104. Rate of removal of HI = 1 0k H HI I ... (1)

SSA on •H •1aI k H HI

Therefore, (1) ar ICorrect option is (c)



105. 4 2 2 3 1max 2 0 3.42 10 1 10 3.42 10 seck E mol dm

And T.O.N. = 4 12 3.42 10k s


106./1 1

1

E RTk eA

... (1)

/2 2

2

E RTk eA

... (2)

/2 1 1 2

2 1

21

E E RTk A eA k

2 1

1 21k A

k A 1 1

2 2

A kA k


107. Fugacity of gas is less than P when attractive forces are dominant. It happens at low P and whenT < Tb.Correct option is (b)

108.

2 2 212

'' 1 '2

H O g H g O g

PP P

1/2

1/21/2

' '2 '

' 1 1 2P

P Pk P

P

12T

PP P P

2TPP P P 1

2 2 12

TT T

PPP P P P P

1/23/2

1/22 11

2

TP

Pk

1 so 1 1 and 1 1

2

3/2 1/2

2T

PPk

so 3/2P Tk P




109. The given engine is working as refrigerator as it is transfering heat from sink (lower temperature) tosource (higher temperature). So, efficiency of refrigerator/coefficient of performance

1

2 1

100 100 1'300 100 200 2

TT T

output' 'input

cq

1 5 102

J

Correct option is (b)110. 3 3 3

6 6 2 2'S C C a S Correct option is (d)

111.10 20000 15 10000 14, 000

10 15nM


112. no. of arrangements no. of arrangements no. of arrangementsC A Bclassical fermions bosons

C A BS S S Correct option is (d)

113. Character table of C3V points group and complete table is3

1

2

2 31 1 11 1 12 1 0

vE CAAE


114. The given vibrational is not symmetric w.r.t. hCorrect option is (d)

115.

2 12 22 2 2 2 2 2

2 21 2 22 1 1 11 2 1 2400420

400 400420 420

2 1 11 2

4 2 2

E EE E

E E EE Ei

N g e g e N g N g eN NPN N gg e g e g eg g e

ee

e e


116.

u

g

1s 1sa b

H2 H2+

u

g

1s 1sa b

2 12 He 1s ; He 1s

2 a bH : r 1s r 1s r



HeHe : r 0 1s r


117.

p

N N

v g pv g

700 5.6 700 0.4 500.4 5.6N

N

MHz MHzv MHzv


118. The first electronic absorption band maximum of a polar and relatively rigid aromatic moleculeappears at 310 nm but its fluorescence maximum in acetonitrile solution appears with a large stokesshift at 450 nm. The reason for stokes shift increasing dipole moment of the molecule in the excitedstate.Correct option (b)

119. From radial part 2 36 expr

r r

Minimum power of 1 1r So, 1Corresponding 1 possible angular part of eigen state is cosFor cos , 1 Correct option is (c)

120.11 11 12 12

21 21 22 220

H ES H ESH ES H ES

0 2.50

2.5 12E

E

12 6.25 0E E

2

2

12 6.25 0

12 6.25 0

E E

E E

The root of this equation is 0.50E 0.50E


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PAPER : CSIR-UGC-NET/JRF June 2017 · 2017. 12. 5. · 6 PAPER : CSIR-UGC-NET/JRF June 2017 57. The v = 0 to 1 vibration-rotation spectrum of a diatomic molecule exhibits transitions