CSIR NET CHEMISTRY (DEC-2017)
Previous Year Question Paper with Solution.
21. Among the following nuclear reactions of thermal neutrons, the
cross section is highest for
(a)
(b)
(c)
(d)
Ans. d
Sol. The cross – section can be interpreted as the effective 'target area' that a
nucleus interacts with an incident neutron. Larger the effective area, greater the
probability for reaction.
In reaction, one neutron react with
to give two neutron, which have more tendency to react with other
radioactive nuclei. Thus, increases effective area for reaction.
Correct option is (d)
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
ERROR fo
(c) cobalt(II) vs eriochrome black T
Invalid do
(d) nickel (II) vs dimethylglyoxime
Ans. a
Sol. Oxalic acid versus potassium permaganante:
End – point is colorless pink – purple
(b) Iron (II) versus 1, 10 – phenanthroline:
End – point is colorless Reddish – orange
(d) Nickel (II) versus Dimethylglyoxime:
End – point is colorless Red
(c) Cobalt (II) versus Eriochrome Black – T:
End – point is Blue – green Blue
Since, color change at end – point is not significant for option (c) which is
requirement of the spectrophotometric titration. Therefore, color mointoring
method is not suitable in this case.
Correct option is (c)
23. The first ionization energy is the lowest for
(a) Br
(b) Se
(c) P
(d) As
Ans. b
Sol. 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. Among special with pyramidal shape is/are?
(a)
(b)
(c)
(d)
Ans. a
Sol.
Correct option is (a)
25. The role of BF3 as an industrial plymerization catalyst is to generate
(a) carbanion
(b) carbocation
(c) organic radical
(d) cation radical
Ans. b
Sol.
Correct option is (b)
26. For the following complexes, the increasing order of magnetic
moment (spin only value) is
(a) D < A < B < C
(b) C < A < D < B
(c)
(d)
Ans. a
Sol.
Hence, A < B < C = D
Correct option is (d)
27. The correct statement for cytochrome c is
(a) It is a non – heme protein
(b) The coordination number of iron in cytochrome c is five
(c) It is a redox protein and an electron carrier
(d) It can store or carry dioxygen
Ans. c
Sol. 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
oxygen
Correct option is (c)
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
Ans. d
Sol.
Correct option is (d)
29. The IR spectrum of Co(CO)4H shows bands at 2121, 2062, 2043 and
1934 cm–1. The VCo-D (in cm–1) expected in the spectrum of Co(CO)4D is
(a) 2111
(b) 1396
(c) 2053
(d) 1910
Ans. b
Sol.
Correct option is (b)
30. In trigonal prismatic ligand field, the most stabilized d orbital is
(a)
(b)
(c)
(d)
Ans. b
Sol. In trigonal prismatic ligand field, energy of 'd' orbital is
Hence, the most stablized 'd' orbital is dxy
Correct option is (b)
31. The most unstable complex on the basis of electro–neutrality
principle among the following is
(a)
(b)
(c)
(d)
Ans. b
Sol. According to electroneutrality principle, each atom in any stable substance
has a charge close to zero. This means if central atom has more positive charge then
it would be stabilized by ligand having high negative charge density i.e. more
electronegative donor atom.
In case of , nitrogen is sp3 hybridized hence it is least
electronegative hence least stable complex.
Correct option is (b)
32. The band in the electronic spectrum of 12 appearing at 520 nm will
undergo maximum blue shift in
(a) water
(b) hexane
(c) benzene
(d) methanol
Ans. d
Sol. Electronic absorption band of I2 vapours appears at 520 nm i.e.
It solution is prepared with a solvent that able to form a charge transfer
complex (CTC) like aromatic solvents, ethers, alcohols etc. This absorption
band shifted towards UV range i.e. blue shift. More the donation of electrons
from solvent to the of more will be the blue shift.
Here, option(d) methanol has highest electrons donating capacity (i.e. most
basic)
Note: Methanol has low ionization potential (10.85 eV) than water (12.59 eV),
therefore methanol is better electron donar than water.
Correct option is (d)
33. Mismatch among the following is
(a) Sharp transition and fluorescene 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
Ans. c
Sol. Spin – orbit coupling is larger for the heavier atoms (Inner transition) and very
small for lighter atoms.
Correct option is (c)
34. Among the following, the compound that gives base peak at m/z 72
in the El mass spectrum is
(a)
(b)
(c)
(d)
Ans. a
Sol.
Correct option is (a)
35. The among molecule has
(a) plane of symmetry
(b) R configuration
(c) S configuration
(d) centre of symmetry
Ans. b
Sol.
Correct option is (b)
36. The following natural product Entcrodiol is a
(a) terpene
(b) steroid
(c) lignan
(d) alkaloid
Ans. c
Sol.
Enterodiol is a lignan, formed by the action of intestinal bacteria on lignan
precursors found in plants.
Correct option is (c)
37. The correct order of basicity for the following heterocycles is
(a) A > C > B
(b) C > A > B
(c) C > B > A
(d) B > A > C
Ans. c
Sol.
Lesser is the availability of lone pair for donation, lesser is the basic behaviour.
Correct option is (c)
38. The kinetic product formed in the following reaction is
(a)
(b)
(c)
(d)
Ans. d
Sol.
Correct option is (d)
39. Among the structure given below, the one that corresponds to the
most stable conformation of compound A is
(a)
(b)
(c)
(d)
Ans. a
Sol. Hydrogen bonding and no 1, 3 – diaxial interaction
Correct option is (a)
40. According to Frontier Molecular Orbital (FMO) Theory, the Highest
Occupied Molecular Orbital (HOMO) of hexatriene in the following reaction
is
(a)
(b)
(c)
(d)
Ans. d
Sol. HOMO of hexatriene in hv condition is molecular orbital has three
nodes.
No. of nodes.
Correct option is (d)
13
41. The number of signals observed in the proton decoupled C NMR
spectrum of the following compound is
(a) Five
(b) Six
(c) Ten
(d) Thirteen
Ans. a
Sol.
42. The correct order of stability of the following carbocations is
(a) A > C > B
(b) B > C > A
(c) C > A > B
(d) C > B > A
Ans. d
Sol.
Order of stability of carbocation C > B > A
Correct option is (d)
43. An optically pure organic compound has specific rotation of +400.
The optical purity of the sample that exhibits specific rotation of +320 is
(a) 8%
(b) 12%
(c) 20%
(d) 80%
Ans. d
Sol.
Correct option is (d)
44. The major product formed in the following reaction is
(a)
(b)
(c)
(d)
Ans. c
Sol.
Correct option is (c)
45. Correct match of the compounds in Column P with the IR stretching
frequencies (cm–1) in Column Q is
(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–B
Ans. c
Sol.
Correct option is (c)
46. The organic compound that displays following data is
(a)
(b)
(c)
(d)
Ans. a
Sol.
Correct option is (a)
47. The molecule with a C2 axis of symmetry among the following is
(a) BH2Cl
(b) CH3Cl
(c) NH2Cl
(d) HOCl
Ans. a
Sol.
Once C2 passing through 'B' and 'Cl' atoms, two hydrogen interchange their
position but remain same molecule.
Correct option is (a)
48. The molecule that will show Raman spectrum, but not IR spectrum,
among the following is
(a) H2
(b) HCl
(c) BrCl
(d) CS2
Ans. a
Sol. 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. Boron in CBl3 has
(a) sp hybridization
(b) sp2 hybridization
(c) sp3 hybridization
(d) no hybridization
Ans. b
Sol.
Correct option is (b)
50. The number of degencrate spatial orbitals of a hydrogen-like atom
with principal quantum number n = 6 is
(a) 1
(b) 6
(c) 72
(d) 36
Ans. d
Sol. Orbital degencracy = n2 = 62 = 36
Correct option is (d)
51.
If then which of the following
necessarily holds:
(a)
(b)
(c)
(d)
Ans. b
Sol.
Correct option is (b)
52.
The correct statement among the following is ( is a hermitian
operator)
(a)
(b)
(c)
(d)
Ans. b
Sol. which is a Hermitian operator and be operand
Therefore, eign value is positive.
Correct option is (b)
53. If the atoms/ions in the crystal are taken to be hard spheres
touching each other in the until cell, then the fraction of volume occupied
in the body centered cubic structure is
(a)
(b)
(c)
(d)
Ans. d
Sol. Packing fraction
Correct option is (d)
54. Repeated measurements of Pb in a lake water sample gave 3.2, t.2
and 7.2 ppb of Pb. Standard deviation in the measurement of Pb is
(a) 2 ppb
(b) 4ppb
(c) 0 ppb
(d)
Ans. a
Sol.
Correct option is (a)
55. The stability of lyophobic colloids is a consequence of the
(a) electrical double layer at the surface of the particles.
(b) van der Walls force between the particles
(c) small particle size
(d) shape of the particles
Ans. a
Sol. The stability of lyophobic colloids is a consequence of electrical double layer
at the surface of the particles
Correct option is (a)
56. The equivalent conductance at infinite dilution of a strong electrolyte
can be obtained from the plot of
(a)
(b)
(c)
(d)
Ans. b
Sol. For strong electrolyte, can be obtained from the plot of
Correct option is (b)
57.
The number – average molar mass for a monodisperse
polymer is related to the weight – average molar mass by the
relation
(a)
(b)
(c)
(d)
Ans. d
Sol.
Correct option is (d)
58.
For a sequence of consecutive reaction, the
concentration of I would be, by steady state approximation.
(a)
(b)
(c)
(d)
Ans. d
Sol. Rate of formation of I = Rate of deformation of I
Correct option is (d)
59. Enthalpy is equal to
(a)
(b)
(c)
(d)
Ans. b
Sol.
Correct option is (b)
60. The structure of ribonucleoside uridine is
(a)
(b)
(c)
(d)
Ans. b
Sol.
Uridine is glycosylated pyrimidine – along containing uracil attached to ribose
ring (ribofuranose) via a – N1 – glycosidic bond
Correct option is (b)
61. The peak area of differential thermal analysis curve is proportional
to one or more of the following:
A. mass loss
B. mass of the sample
C. heat of decomposition/phase change
The correct answer is
(a) A only
(b) B only
(c) A and C
(d) B and C
Ans. d
Sol. The peak area of differential thermal analysis is propportional to
(i) mass of the sample
(ii) heat of decompositional phase change
Correct option is (d)
62. To determine the bond parameters at 250C, 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
Ans. b
Sol. Electron differaction is used to get the bond parameters of gaseous samples.
Correct option is (b)
63. Match lanthanides in Colum I and with their properties in Column II
(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)
Ans. a
Sol. (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 colour
Correct option is (a)
64. Among the following species isolobal to CH2 are
A. CpCr(CO)2
B. CpCu
C. Ni(CO)2
D. Cr(CO)4
E. Fe(CO)4
(a) A, C and E
(b) B, C and D
(c) B, C and E
(d) A, B and D
Ans. c
Sol. Number of valence electron in CH2 = 6
Hence, fragment having 16 electron will be isolobal to this fragment.
Correct option is (c)
65. Choose the incorrect statement for the phosphomolybdate anion,
(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)
Ans. c
Sol. is less basic. It is slightly less basic than
Correct option is (c)
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 transitions
C. are stable
D. Some of actinides do not have radioactive isotopes
The correct answer is
(a) A and C
(b) B and C
(c) A, B and C
(d) B, C and D
Ans. a
Sol. (A) Due to more expansion of 5f orbital actinoids has more tendency to
release electron hence have more 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) and are stable
(D) All the actinoids have radioactive isotopes
Correct option is (a)
67. According to Bent's rule, for p – block elements, the correct
combination of geometry around the central 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
Ans. a
Sol. According to Bent's Rule, 'more electro negative element occupy axial position
in trigonal bipyramidal geometry.
Correct option is (a)
68. Allred – Rochow electronegativity of an element is
A. directly proportional to the effective nuclear charge
B. directly proportional to the covalent radius
C. inversely proportional to the square of the covalent radius
D. directly proportional to the square of the effective nuclear charge
The correct answer is
(a) A and B
(b) A and C
(c) B and C
(d) A and D
Ans. b
Sol. According to Allred – Rochow, electronegativity can be given as
where, r = covalent radius.
Correct option is (b)
69. Br2 with propanonc forms a charge transfer complex and l2 forms
triiodidc anion with I–. This implies 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 acid
Ans. b
Sol. In there is intraction between of 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. In the complex [Pd (L-L) (Me) (Ph], the bisphosphine (L-L) that does
not allow reductive elimination of PhMe, is
(a)
(b)
(c)
(d)
Ans. d
Sol. According to reductive elimination reaction mechanism cis – complex can give
the product but trans – 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 flexible methylene (–CH2–) groups.
Therefore, reductive elimination not occured.
Correct option is (d)
71. In the reaction given below, the bisphosphine (P–P) that is in
effective for transfer hydrogenation reaction is
(a) Diphenylphosphinomethane
(b) 1, 2–Diphenylphosphinoethane
(c) 1, 3–Diphenylphosphinopropane
(d) 1, 4–Diphyenylphosphinobutane
Ans. d
Sol. 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.
Correct option is (d)
72. For high spin and low spin d6 octahedral complexes (ML6), the
generally observed spin allowed transitions, respectively, are
(a) two and one
(b) one and two
(c) zero and one
(d) two and two
Ans. b
Sol. For d6 (high spin) there is one spin allowed transition from
In case of d6 low spin commonly two spin allowed transitions are observed.
there are additional spin allowed transitions at higher energy but they are
marked by allowed transitions. Hence, are not observed.
Correct options is (b)
73. The reactions given below,
are examples of
(a) disproportionation only
(b) disproportionation (A) and solvation (B)
(c) solvation (A) and disproportionation (B)
(d) solvalysis as well as disproportionation
Ans. d
Sol.
Above both reactions are example of solvolysis as well as disproportionation
Correct option is (d)
74. According to Wade's rules, the culster 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
Ans. b
Sol. thus it follow 4n + 4. Hence, it has nido structure
Correct option is (b)*
75. 31P
Assuming the expected NMR spectrum of
(a)
(b)
(c)
(d)
Ans. c
Sol.
Intensity ratio: 1 : 1 : 1 : 1 : 3 : 3 : 3 : 3 : 3 : 3 : 3 : 3 : 1 : 1 : 1 : 1
Correct option is (c)
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 (tetrahedra, high-spin)
(d) (trigonal prismatic, low-spin) and (tetrahedral, high-spin)
Ans. a
Sol.
Correct option is (a)
77. The active site structure for oxy-hemerythrin is:
(a)
(b)
(c)
(d)
Ans. c
Sol. In oxyhemerythrin, both the Fe are present as Fe(III) and there is presence of
hydrogen bonding and dioxygen bind to only one Fe.
Correct option is (c)
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 hepatcoordinated 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
Ans. a
Sol. Mechanism of base hydrolysis can be given as
Correct option is (a)
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
Ans. d
Sol. Cubane like ferredoxin is Fe4S4
It has four inorganic sulfide which can be removed by treatment with acid.
Correct option is (d)
80. Considering the ambidentate behaviour of thiocyanate ion, the most
stable structure among the following is
(a)
(b)
(c)
(d)
Ans. a
Sol.
Correct option is (a)
81. Major product of the following reaction is
(a)
(b)
(c)
(d)
Ans. c
Sol.
Correct option is (c)
82. Major product in the following reaction is
(a)
(b)
(c)
(d)
Ans. a
Sol.
Correct option is (a)
83. Major product A and B of the following reaction sequence are
(a)
(b)
(c)
(d)
Ans. c
Sol.
Correct option is (c)
84. The major product formed in the following reaction is
(a)
(b)
(c)
(d)
Ans. c
Sol.
Correct option is (c)
85. Major products A and B of the following reaction sequence are
(a)
(b)
(c)
(d)
Ans. b
Sol.
Correct option is (b)
86. The major product formed in the following reaction is
(a)
(b)
(c)
(d)
Ans. d
Sol.
Correct option is (d)
87. Correct sequence of reagents (i) – (iii) required for the conversion of
A to B is
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
Ans. c
Sol.
Correct option is (c)
88. The following reaction involves
(a) [1, 2] sigmatropic rearrangement
(b) [2, 3] sigmatropic rearrangement
(c) [3, 3] sigmatropic rearrangement
(d) C – H insertion reaction
Ans. b
Sol.
Correct option is (b)
89. Correct sequence of steps involved in the following transformation is
(a) Michael addition, aldol condensation, syn-elimination, keto-enol
tautomerism
(b) aldol condensation, electrocyclic ring closing, syn-elimination,
dehydrogenation
(c) Michael addition, Claisen condensatin, anti-elemination, keto-enol
tautomerism
(d) Robinson annulation, dehydrogenation, anti-elimination
Ans. a
Sol.
Correct option is (a)
90. The major products A and B in the following reaction sequence are
(a)
(b)
(c)
(d)
Ans. a
Sol.
Correct option is (a)
91. The major product formed in the following reaction sequence is
(a)
(b)
(c)
(d)
Ans. b
Sol.
Correct option is (b)
92. The number of optically active steroisomers possible for
(a) Two
(b) Four
(c) Six
(d) Eight
Ans. a
Sol. The Number of optically active stereoisomer are two
Correct option is (a)
93. The correct match of the circled protons in Column P with 1H NMR
chemical shift ( δ ppm ): Column Q is
(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–A
Ans. b
Sol.
Correct option is (b)
94. The major product formed in the following reaction sequence is
(a)
(b)
(c)
(d)
Ans. b
Sol.
Correct option is (b)
95. For the successful synthesis of peptide linkage leading to the
product A, the side chain of the amino acid B should have
(a) XH = –OH
(b) XH = –(CH2)4NH
(c) XH = – p – (C6H4)OH
(d) XH = – SH
Ans. d
Sol.
Correct option is (d)
96. The major products A and B in the following reaction sequence are
(a)
(b)
(c)
(d)
Ans. a
Sol.
Correct option is (a)
97. The intermediate A and the major product B in the following reaction
are
(a)
(b)
(c)
(d)
Ans. a
Sol.
Correct option is (a)
98. The correct intermediate which leads to the product in the following
reaction in
(a)
(b)
(c)
(d)
Ans. c
Sol.
Correct option is (c)
99. In the following transformation, the mode of electrocyclization A and
the major product B are
(a)
(b)
(c)
(d)
Ans. b
Sol.
Correct option is (b)
100. The major product A and B in the following reaction sequence are
(a)
(b)
(c)
(d)
Ans. b
Sol.
Correct option is (b)
101. The correct statement about the symmetry of the eigenfunctions of
a quantum of 1–D harmonic oscillator is
(a) All the eigenfunctions are only even functions, because the potential is an
even function.
(b) All the eigenfunctions 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.
Ans. d
Sol. Eigen functions are even for even values of vibrational quantum number (n)
and odd for odd values of n.
Correct option is (d)
102. The correct statement about the difference of second and first
excited state energies of a particle in 1–D, 2–D square and 3–D cubic
boxes with same length for each, is
(a)
(b)
(c)
(d)
Ans. c
Sol.
Correct option is (c)
103. A one – dimensional quantum harmonic oscillator is perturbed by a
potential The first order correction to the energy for the ground state
(DE(1)) is
(a)
(b)
(c)
(d)
Ans. c
Sol. For odd power of perturbed potential, is always '0'
Correct option is (c)
104. The normal mode of ethylene represented, by the figure below, is
(a) only IR active
(b) only Raman active
(c) both IR and Raman active
(d) neither IR nor Raman active
Ans. b
Sol. Only Raman active, (symmetric vibration, change in dipole moment is zero
during vibration)
Correct option is (b)
105. The pair that contains a spherical top and a symmetric top, among
the following, is
(a)
(b)
(c)
(d)
Ans. c
Sol.
Correct option is (c)
106. A part of the character table of a point group (of order 4) is given
below.
The four characters of are, respectively
(a) 1, 1, –1, –1
(b) 2, 0, 0, 1
(c) 1, i, i, 1
(d) 1, –i, i, –1
Ans. a
Sol. According to GOT theorem, any two IR must be orthogonal to each other.
Correct option is (a)
107. The electronic transition energy from in propenyl radical is
4.8 eV. Within the frame work of Huckel theory, the transitions energy
from would be
(a) 2.4 cV
(b) 4.8 cV
(c) 9.6 cV
(d) 14.4 cV
Ans. c
Sol.
Correct option is (c)
108. The g – factors of 1H and 13C are 5.6 and 1.4 respectively. For the
same value of the magnetic field strength, if the 1H resonates at 600 MHz,
the 13C would reasonate at
(a) 2400 MHz
(b) 600 MHz
(c) 150 MHz
(d) 38 MHz
Ans. c
Sol.
Correct option is (c)
109. The term symbol for the ground state of a metal ions is 3P2. The
residual entropy of a crystal of a salt of this metal ion at 0 K is
(a)
(b)
(c)
(d)
Ans. c
Sol.
Correct option is (c)
110. In stretching of a rubber band.
Which of the following relations in true?
(a)
(b)
(c)
(d)
Ans. a
Sol.
Similarly differentiating equation (1), w.r.t. L at constant P and T
Correct option is (a)
111. For 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
Ans. c
Sol. For distinguishable particles,
Correct option is (c)
112. One mole of an ideal gas undergoes a cyclic process (ABCDA) starting
from point A through 4 reversible steps as shown in the figure. Total work
done in the process is
(a)
(b)
(c)
(d)
Ans. d
Sol.
Correct option is (d)
113. If the specific conductance of an electrolyte solution is
and cell constant is 0.25 cm–1, the conductance of the solution is
(a)
(b)
(c)
(d)
Ans. c
Sol.
Correct option is (c)
114. The predicted electromotive force (emf) of the electrochemical cell
(a) –0.850 V
(b) +0.044 V
(c) +0.0850V
(d) –0.044 V
Ans. b
Sol.
Correct option is (b)
115. A polymer has the following molar mass distribution
The calculated number average molar mass of the polymer is
(a) 5200
(b) 5600
(c) 5800
(d) 6000
Ans. b
Sol.
Correct option is (b)
116. The separaton 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
Ans. a
Sol.
Correct option is (a)
117. The slope and intercept obtained from (1/Rate) against (1/substrate
concentration) of an enzyme catalyzed reaction are 300 and 2×105,
respectively. The Michaelis–Menten constants of the enzyme in this
reaction is
(a) 5 × 106 M
(b) 5 × M–6 M
(c) 1.5 × 103M
(d) 1.5 × 10–3M
Ans. d
Sol.
Correct option is (d)
118. The pressure inside (Pin)a sperical cavity with a radius r formed in a
liquid with surface tension is related to the external pressure (Pout) as
(a)
(b)
(c)
(d)
Ans. b
Sol. For a bubble, surface tension counteracts internal pressure
Correct option is (b)
119. Reaction between A and B is carried out for different initial
concentrations and the corresponding half-life times are measured. The
data listed in the table:
The rate can be represented as
(a)
(b)
(c)
(d)
Ans. c
Sol. From entry (1) and (2), order w.r.to, B = 1
And from entry (3) and (4), order w.r.to. A = 2
Correct option is (c)
120. The polot of the rate constant vs. ionic strength of the reaction
follows the line (refer to the figure)
(a) I
(b) II
(c) III
(d) IV
Ans. d
Sol.
Thus, plot having negative slope
Correct option is (d)
JOIN VEDPREP LIVE APP
Your Name
Your Email
Phone Number
City / Town
Submit Query
LOCATE US COMPANY COURSES ACADEMIES GET IN
TOUCH
72 Mall Road, About IIT JAM Physics
Near GTB Contact CSIR NET Maths
Nagar, Metro Blog GATE Biology
Station Gate Terms And Previous Year
No. 3, Conditions Papers
Kingsway Privacy Policy
camp GTB Shipping
Nagar, New Policy
Delhi -110009 Refund Policy
+91
9136597244 /
+91
9821304611
© 2023 VedPrep
