GATE Chemistry 2012
Previous Year Question Paper with Solution.
1. In the proton decoupled 13C NMR spectrum of 7 – norbornanone, the number of signals obtained is
(a) 7
(b) 3
(c) 4
(d) 5
Ans. (b)
Sol.
total number of signal in 13C NMR = 3
Correct option is (b)
2. Identify the most probable product in the given reaction
(a)
(b)
(c)
(d)
Ans. (c)
Sol.
3. In the cyclization reaction given below, the most probable product formed is
(a)
(b)
(c)
(d)
Ans. (c)
Sol.
Correct option is (c)
4. If and
are the uncertainties in the y – coordinate and the y component of the momentum of a particle respectively, then, according to uncertainty principle
is:
(a)
(b)
(c)
(d)
Ans. (d)
Sol. From Heisenberg uncertainty principle
In y coordinate,
Correct option is (d)
5. The average length of a typical – helix comprised of 10 amino acids is
(a) 10Å
(b) 15 Å
(c) 36 Å
(d) 54 Å
Ans. (b)
Sol. 3.6 amino acids ................... 5.4Å
Correct option is (b)
6. Number of thymine residues in a 5000 kb DNA containing 23% guanine residues is:
(a) 2.70 × 106
(b) 2.70 × 107
(c) 1.35 × 106
(d) 1.35 × 107
Ans. (c)
Sol.
Correct option is (c)
7. Show below is a Hammett plot obtained for the reaction
The change in slope of the plot indicates that
(a) The reaction does not follow linear free energy relationship
(b) Electrons are being withdrawn from the transition state in the mechanism
(c) Electrons are being donated to the transition state in the mechanism
(d) The mechanism of the reaction is changing
Ans. (d)
Sol. = (–4.4), electrons flow away from the aromatic ring in the rate-determining step.
= (+2.5), electrons flows towards the aromatic ring in the rate-determining step.
So, the mechanism of the reaction is changing.
Correct option is (d)
8. The ratio of relative intensities of the two molecular ion peaks of methyl bromide in the mass spectrum is:
(a)
(b)
(c)
(d)
Ans. (c)
Sol. In CH3Br molecule Br having two isotope 79Br and 81Br
CH3Br79 : CH3Br81
M+ : (M + 2)+ = 1 : 1
Correct option is (c)
9. A disaccharide that will not given Benedict's test and will not form osazone is
(a) maltose
(b) lactose
(c) cellobiose
(d) sucrose
Ans. (d)
Sol.
Correct option is (d)
10. Choose the allowed transition
(a)
(b)
(c)
(d)
Ans. (c)
Sol. Selection rules for electronic transition for diatomic molecules are
Correct option is (c)
11. The angular part of the wavefunction for the electron in hydrogen atom is proportional to . The values of the azimuthal quantum number l and the magnetic quantum number (m) are respectively.
(a) 2 and 2
(b) 2 and – 2
(c) 3 and 2
(d) 3 and – 2
Ans. (c)
Sol. Angular part
Compare with
m = 2
Function is 'f' orbital l = 3
= 3 m = 2
Correct option is (c)
12. Let and
denote the wavefunctions of the 2px and 2pz orbitals of carbon, respectively, and
and
represent the wavefunction of the and orbitals of oxygen, respectively. If c1 and c2 are constants used in linear combinations and the CO molecule is oriented along the z axis than, according to molecular orbital theory, the
– bonding molecular orbital has a wavefunction given by
(a)
(b)
(c)
(d)
Ans. (d)
Sol. Since the co-molecule is oriented along the z-axis, combination of and
result in the formation of
donating orbitals while combination of either and
or
and
result in the formation of
bonding orbital.
The wave function of bonding molecular orbital
Correct option is (d)
13. The bond that gives the most intense band in the infrared spectrum for its stretching vibrations is
(a) C – H
(b) N – H
(c) O – H
(d) S – H
Ans. (c)
Sol. Intensity is proportional to the polarity of bond more polar bond more intense
more polar
Correct option is (c)
14. If xA and xB are the respective mole fractions of A and B in a ideal solution of the two and TA, TB, T are the fusion temperatures of pure A, and pure B and the ideal solution respectively, then
(a)
(b)
(c)
(d)
Ans. (d)
Sol.
correct option is (d)
15. For a reaction involving two steps given below
assume that the first step attains equilibrium rapidly. The rate of formation of P is proportional to
(a) [G]1/2
(b) [G]
(c) [G]2
(d) [G]3/2
Ans. (d)
Sol.
Putting [H] in equation (i)
Correct option is (d)
16. A metal chelate that can be used for separation and quantitative analysis of aluminium ion by gas chromatography is
(a) EDTA
(b) ethylene glycol
(c) dinonyl phthalate
(d) trifluoroacetylacetone
Ans. (d)
Sol. Trifluoroacetylacetone can be used for separation and quantitative analysis of aluminium ion by gas chromatography.
Correct option is (d)
17. The enthalpies of hydration of Ca2+, Mn2+ and Zn2+ follow the order
(a) Mn2+ > Ca2+ > Zn2+
(b) Zn2+ > Ca2+ > Mn2+
(c) Mn2+ > Zn2+ > Ca2+
(d) Zn2+ > Mn2+ > Ca2+
Ans. (d)
Sol. The electronic configuration of the given ions is as
Ca2+ = 1s2 2s2 2p6 3s2 3p6
Mn2+ = 1s2 2s2 2p6 3s2 3p6 3d5
Zn2+ = 1s2 2s2 2p6 3s2 3p6 3d10
Thus, the size of Zn2+ ion is smallest and of Ca2+ is largest among the three due to increase in effective nuclear charge with increase in atomic number. Smaller the size of ion, higher is the hydration enthalpy. Thus the order of hydration enthalpy is
Correct option is (d)
18. The number of terminal carbonyl groups present in is
(a) 2
(b) 5
(c) 6
(d) 3
Ans. (c)
Sol. Fe2(CO)9
Correct option is (c)
19. Among the following substituted silanes, the one that gives cross – linked silicon polymer upon hydrolysis is
(a)
(b)
(c)
(d)
Ans. (b)
Sol.
Correct option is (b)
20. The plot of versus T (where
is molar magnetic susceptibility and T is the temperature) for a paramagnetic complex which strictly follows Curie equation is:
(a)
(b)
(c)
(d)
Ans. (c)
Sol. Plot of magnetic susceptibility versus T for paramagnetic
Correct option is (c)
21. Among the following donors, the one that forms most stable adduct with the Lewis acid is:
(a) 4 – methylpyridine
(b) 2, 6 – dimethylpyridine
(c) 4 – nitropyridine
(d) 2, 6 – di – tert – butylpyridine
Ans. (a)
Sol.
More the electron density on nitrogen, more stable adduct will form with lewis acid, but in case of (b) and (d), crowding increases so stability decreases.
Correct option is (a)
22. The complex with inverse-spinel structure is
(a) Co3O4
(b) Fe3O4
(c) MgAlO4
(d) Mn3O4
Ans. (b)
Sol. In mixed oxide
Fe3+ is d5 system with zero CFSE in oh void. Whereas Fe2+ is a d6 with CFSE in oh void. So preferentially occupy octahedral site to give inverse spinel.
Correct option is (b)
23. The IUPAC nomenclature of Na[PCl6] is
(a) sodium hexachlorophosphine(V)
(b) sodium hexachlorophosphate(V)
(c) sodium hexachlorophosphine
(d) sodium hexachlorophosphite(V)
Ans. (b)
Sol. In Na[PCl6], let the oxidation state of P is x.
+1 + x + (–1)6 = 0
x – 5 = 0
x = +5
The IUPAC name of Na[PCl6] is sodium hexachlorophosphate (V).
Correct option is (b)
24. An intermediate formed during the hydroformylation of olefins using as catalyst is
(a)
(b)
(c)
(d)
Ans. (d)
Sol. Catalytic Cycle of Hydroformylation
Co2 (CO)8 HCo (CO)4
Correct option is (d)
25. The order of polarity of and is:
(a)
(b)
(c)
(d)
Ans. (b)
Sol.
Correct option is (b)
26. From a carboxymethyl – cellulose column at pH 6.0, arginine, valine and glutamic acid will elute in the order
(a) arginine, valine, glutamic acid
(b) arginine, glutamic acid, valine
(c) glutamic acid, arginine, valine
(d) glutamic acid, valine, arginine
Ans. (d)
Sol. pI
Arginine 10.8 positively charged cation
Valine 6.0 neutral
Glutamic Acid 3.2 negatively charged anion
At a pH equal to a proteins pI, the protein will carry no net charged.
At a pH below the pI, the protein wil carry a net positive charged.
If the buffer pH is raised above a proteins pI, it will carry a net negative charged.
Carboxymethyl – cellulose column is a catioin exchanger column so cation will be eluted last. So, here arginine with positive charged will be eluted at last and order of elution : glutamic acid, valine, arginine.
Correct option is (d)
27. Symmetry operations of the four C2 axes perpendicular to the principle axis belong to the same class in the point group (s)
(a) D4
(b) D4d
(c) D4h
(d) D4h and D4d
Ans. (b)
Sol.
Symmetry operator of the four C2 axis perpendicular to the principal axis belong to the same class in D4d
Correct option is (b)
28. At 298K, the EMF of the cell
is 0.7530V. The standard potential of the calomel electrode is 0.2802V. If the liquid junction potential is zero, the pH of the solution is:
(a) 4.7
(b) 7.4
(c) 8.0
(d) 12.7
Ans. (c)
Sol.
E = 0.2802 + 0.0591 pH
Correct option is (c)
29. The wavefunction of a 1 – D harmonic oscillator between and
is given by
. The value of N that normalizes the function
is:
(a)
(b)
(c)
(d)
Ans. (c)
Sol.
Correct option is (c)
30. Consider the reaction,
The molecular diamesters of H2 and C2H4 are 1.8Å and 3.6Å respectively. The pre – exponential factor in the rate constant calculated using collision theory in is approximately. (For this reaction at 300 K,) where the symbols have their usual meanings)
(a) 2.5 × 108
(b) 2.5 × 1014
(c) 9.4 × 1017
(d) 9.4 × 1023
Ans. (a)
Sol. Consider the reaction, H2 + C2H4 C2H6
The molecular diameter of H2 and C2H4 = 1.8Å and 3.6Å.
The pre exponential factor in the rate constant calculated using collision Theory is (m3 per mole per sec).
A = NA = 1.11 × 1027 m mole–1 sec–1
k2 =
P = 1
k2 =
= 25.408 × 10–20 1027 = 25.408 × 107 m3 mol–1 s–1
k2 = 2.54 × 108 m3 mol–1 s–1
Correct option is (a)
31. The molecular partition function of a system is given by where the symbols have their usual meanings.
The heat capacity at constant volume for this system is
(a) 3R
(b) 6R
(c) 9R / 2
(d) 3R / 2
Ans. (a)
Sol.
Correct option is (a)
32. Consider the phase diagram given below.
At the intersection point Q the phases that are in equilibrium are
(a) Solid A, solid B and solid AB2
(b) Solid A, solid AB2 and liquid
(c) Solid B, solid AB2 and liquid
(d) Solid A, solid B, solid AB2 and liquid
Ans. (c)
Sol.
So, at point Q. Solid AB2, B and liquid are in equilibrium
Correct option is (c)
33. Identify the product from the following reaction
(a)
(b)
(c)
(d)
Ans. (a)
Sol.
In the case of (A)
Hydroboration occurs in syn number on double bond. Hydrogen present on axial position due to less 1-3 diaxial interaction than CH3, which present on equatorial position.
Correct option is (a)
34. The product from the following reaction is
(a)
(b)
(c)
(d)
Ans. (a)
Sol.
Correct option is (a)
35. The acid catalyzed cyclization of 5 – ketodecan – 1, 9 – diol is given below
The most predominant sprioketal is
(a)
(b)
(c)
(d)
Ans. (a)
Sol.
Following stereochemistry is possible with this spiroketal structure.
Correct option is (a)
36. For a face centered cubic lattice, the Miller indices for the first Bragg's peak (smallest Bragg angle) are
(a) 002
(b) 111
(c) 001
(d) 110
Ans. (b)
Sol. For FCC h, k, l should be all either even or all odd.
{considering first order of reflection}
will be minimum for (1 1 1)
Correct option is (b)
37. For the titration of a 10 mL (aq) solution of of 0.001 M Na2EDTA is required to reach the end point. The concentration of CaCO3 (assume molecular weight of CaCO3 = 100) is
(a)
(b)
(c)
(d)
Ans. (d)
Sol. n factor for both CaCO3 and Na2Edta = 2
(CaCO3) n1M1V1 = n2M2V2 (Na2Edta)
2 × M1 × 10 = 2 × 0.001 × 2
M1 = 0.001 / 5 = 2 × 10–4 Mol/L
M1 = 2 × 10–4 × 100 / 1000 g/mL (molecular weight of CaCO3 = 100)
M1 = 2 × 10–5 g/mL
Correct option is (d)
38. In the reaction
the production formed is
(a)
(b)
(c)
(d)
Ans. (a)
Sol.
Correct option is (a)
39. In the reaction given below, identify the product
(a)
(b)
(c)
(d)
Ans. (c)
Sol.
Correct option is (c)
40. Consider the following pairs of complexes
[CoF(NH3)5]2+ and [Cr(OH2)6]2+
[Co(NH3)5(OH2)]3+ and [Cr(OH2)6]2+
[Co(NH3)6]3+ and [Cr(OH2)6]2+
[CoI(NH3)5]2+ and [Cr(OH2)6]2+
The electron transfer rate will be fastest in the pair
(a) [CoF(NH3)5]2+ and [Cr(OH2)6]2+
(b) [Co(NH3)5(OH2)]3+ and [Cr(OH2)6]2+
(c) [Co(NH3)6]3+ and [Cr(OH2)6]2+
(d) [CoI(NH3)5]2+ and [Cr(OH2)6]2+
Ans. (d)
Sol. In all the reactant one of the reactant is [Cr(H2O)6]2+ which is very labile complex.
As the size of halide increase bridging tendency increases. Hence the rate of electron transfer also increases.
Correct option is (d)
41. The extent of Mossbauer quadrupole splitting of iron follows the order
(a)
(b)
(c)
(d)
Ans. (a)
Sol.
Electronic -spherical ligand spherical
FeCl2 . 4H2O > K2 [Fe(CN)5 NO] > FeCl3 . 6H2O
Correct option is (a)
42. Hemoglobin is an oxygen carrying protein. The correct statement about oxy – hemoglobin is that
(a) the metal is low – spin in + 3 oxidation state while dioxygen is in form
(b) the metal is high – spin in + 3 oxidation state while dioxygen is in form
(c) the metal is low – spin in + 3 oxidation state while dioxygen is in neutral form
(d) the metal is high – spin in + 3 oxidation state while dioxygen is in neutral form
Ans. (a)
Sol. In oxyhemoglobin the metal is low-spin in +3 oxidation state while dioxygen is in O2–.
Correct option is (a)
43. If a mixture of NaCl, conc. H2SO4 and K2Cr2O7 is heated in a dry test tube, a red vapour (P) is formed. This vapour (P) dissolves in aqueous NaOH to form a yellow solution, which upon treatment with AgNO3 forms a red solid (Q). P and Q are, respectively
(a)
(b)
(c)
(d)
Ans. (d)
Sol.
Correct option is (d)
44. For the following reaction
At 298 K, the equilibrium constant is:
(a) 10500
(b) 10338
(c) 1038
(d) 10833
Ans. (b)
Sol. 2MnO4– + 5H2C2O4 + 6H+ 2Mn+2 + 8H2O + 10CO2
E° (MnO4–/Mn+2) = 1.51V
E° (CO2/H2C2O4) = – 0.49
Total number of electron = 10
= –RT In K
–nFE = – 2.303 RT log K
Correct option is (b)
45. The ground states of high-spin octahedral and tetrahedral Co(II) complexes are, respectively
(a) 4T2g and 4A2
(b) 4T1g and 4A2
(c) 3T1g and 4A2
(d) 4T1g and 3A1
Ans. (b)
Sol.
Correct option is (b)
46. The INCORRECT statement about Zeise's salt is:
(a) Zeise salt is diamagnetic
(b) The oxidation state of Pt in Zeise's salt is +2
(c) All the Pt – Cl bond lengths in Zeise's salt are equal
(d) C – C bond length of ethylene moiety in Zeise's salt is longer than that of free ethylene molecule
Ans. (c)
Sol. Zeise's salt :
Bond length of (a) is greater than (b). Hence, all the bond Zeise's salt are unequal.
Correct option is (c)
47. The number of possible isomers for the square planar mononuclear complex [(NH3)2M(CN)2] of a metal M is
(a) 2
(b) 4
(c) 6
(d) 3
Ans. (c)
Sol. The possible isomers of [(NH3)2M(CN)2] are
Correct option is (c)
Statement
Consider the reaction sequence shown below:
TsCl = p – toluenesulfoynyl chloride
48. The oxidant X used in step 1 is
(a) CrO3
(b) OsO4
(c) NaIO4
(d) m – CPBA followed by NaOH
Ans. (b)
Sol. For syn addition OsO4 used.
Correct option is (b)
Statement
Consider the reaction sequence shown below:
TsCl = p – toluenesulfoynyl chloride
49. The product is
(a)
(b)
(c)
(d)
Ans. (d)
Sol.
Correct option is (d)
50. In the above reaction, X = Cl, Br or I. Based on the graph, identify the alkyl halides (R – X) as S1 S2 and S3
(a) S1 = R – Cl, S2 = R – Br and S3 = R – 1
(c) Sl = R – I, S2 = R – Br and S3 = R – Cl
(c) S1 = R – Cl, S2 = R – I and S3 = R – Br
(d) SI = R – I, S2 = R – Cl and S3 = R – Br
Ans. (b)
Sol. Since is a better leaving group, then Br– is than Cl–. So, S1 have very low activation energy and S3 have very high activation energy.
Correct option is (b)
51. Identify product P1 and its yield relative to P2
(a) P1 is M and is the major product
(c) P1 is N and is the minor product
(c) P1 is N and is the major product
(d) P1 is M and is the minor product
Ans. (d)
Sol.
Correct option is (d)
Linked Answer Question
Statement for Linked Answer for Q. 52 and Q. 53
A 20491 cm–1 laser line was used to excite oxygen molecules (mad of 16O only) to obtain the rotational Raman spectrum. The resulting rotational Raman spectrum of oxygen molecule has the first Stokes line at 20479 cm–1.
52. The rotational constant (usually denoted as B) for the oxygen molecule is
(a) 1.2 cm–1
(b) 2.0 cm–1
(c) 3.0 cm–1
(d) 6.0 cm–1
Ans. (a)
Sol. A 20491 cm–1 laser line was used to excite oxygen molecule because molecule made 16O only because molecule have spin zero. In the spin zero molecule even J line are missing in Raman spectrum. e.g. O2, CO2.
Rotational constant B = 20491 – 20479 = 12
10 B = 12
B = 1.2 cm–1
53. The next rotational Stokes line is expected at
(a) 20467 cm–1
(b) 20469 cm–1
(c) 20471 cm–1
(d) 20475 cm–1
Ans. (b)
Sol. The next rotational stoke line is first stokes line in 20479
Second line difference is 8B
B = 1.2
8 × 1.2 = 9.6
So, second stock line is 20479 – 9.6 = 20469.7 cm–1.
Correct option is (b)
Statement for Linked Answer for Q. 54 and Q. 54
Huckel molecular orbital theory can be applied to the allene radical
54. The secular determinant (where and E have their usual meanings) is given by
(a)
(b)
(c)
(d)
Ans. (a)
Sol.
Secular determine
Correct option is (a)
55. The possible value of E are
(a)
(b)
(c)
(d)
Ans. (a)
Sol. Possible value of E
x(x2 – 1) – 1 (x – 0) + 0
x(x2 – 1) – 1 (x–0) + 0 x = 0
x3 – x – x = x3 – 2x = 0 x = ±
Put value of 'x' in equation (i)
Correct option is (a)