GATE Chemistry 2007
Previous Year Question Paper with Solution.
1. The rate of sulphonation of benzene can be significantly enhanced by the use of
(a) a mixture of HNO3 and H2SO4
(b) cons. H2SO4
(c) a solution of SO3 in H2SO4
(d) SO3
Ans. (b)
Sol. Rate of sulphonation of benzene can be significantly enhanced by the use of conc. H2SO4. Concentrated sulphuric acid contains trace of SO3 due to slight dissociation of the acid.
Correct option is (b)
2. The reaction,
is an example of a
(a) Birch reduction
(b) Clemmenson reduction
(c) Wolff – Kishner reduction
(d) hydride reduction
Ans. (a)
Sol. Birch Reaction :
Correct option is (a)
3. The major product (X) of the monobromination reaction is
(a)
(b)
(c)
(d)
Ans. (d)
Sol. NBS + AIBN NBS in the presence of AIBN gives free radical Br
Correct option is (d)
4. Benzene can not be iodinated with I2 directly. However, in presence of oxidants such as HNO3, iodination is possible. The electrophile formed in this case is
(a) [I+]
(b) I*
(c)
(d)
Ans. (a)
Sol. I2 + HNO3 + H+
I+ + NO2 + H2O
Correct option is (a)
5. Classify the following species as electrophiles (E) and nucleophiles (N) in routine organic synthesis
SO3 Cl+ CH3NH2 H3O+ BH3 CN–
(a) E = SO3, Cl+, BH3; N = CH3NH2, H3O+, CN–
(b) E = Cl+, H3O+; N = SO3, CH3NH2, BH3, CN–
(c) E = Cl+, H3O+, BH3; N = SO3, CH3NH2, H3O+, CN–
(d) E = SO3, Cl+, H3O+; BH3; N = CH3NH2, CN–
Ans. (d)
Sol. Electrophiles - Electron difficient species
Nucleophiles- electron rich species
Electrophiles-SO3, Cl+, H3O+, BH3
Nucleophile – CH3NH2,
Correct option is (d)
6. The major product obtained upon treatment of compound X with H2SO4 at 80ºC is:
(a)
(b)
(c)
(d)
Ans. (c)
Sol.
Correct option is (c)
7. BaTi[Si3O9] is a class of
(a) ortho silicate
(b) cyclic silicate
(c) chain silicate
(d) sheet silicate
Ans. (b)
Sol. BaTi (Si3O9) = Ba2+Ti4+ (Si3O9)6–
Thus, it is an example of cyclic silicate.
Correct option is (b)
8. The ground state term for V3+ ion is
(a) 3F
(b) 2F
(c) 3P
(d) 2D
Ans. (a)
Sol. V3+ is a d2 system with electronic arrangement in ground state as
L = +2 +1 = 3 corresponds to F state
Spin multiplicity, SM = 2S + 1
= 2(1) + 1 = 3.
Ground state term = SML = 3F
Correct option is (a)
9. In photosynthesis, the predominant metal present in the reaction centre of photosystem II is
(a) Zn
(b) Cu
(c) Mn
(d) Fe
Ans. (c)
Sol. In photosynthesis, Mn is the predominant metal present in the reaction centre of photosystem –II.
Correct option is (c)
10. The octahedral complex / complex ion which shows both facial and meridonial isomers is
(a) Triglycinatocobalt (III)
(b) Tris(etihylenediamine) cobalt (III)
(c) Dichlorodiglycinatocobalt (III)
(d) Trioxalactocobaltate (III)
Ans. (a)
Sol.
Correct option is (a)
11. Zn is carbonic anhydrase is co – ordinated by three histidine and one water molecule. The reaction of CO2 with this enzyme is an example of
(a) electrophilic addition
(b) electron transfer
(c) nucleophilic addition
(d) electrophilic substitution
Ans. (a)
Sol. Zn is carbonic anhydrase, the CO2 undergoes electrophilic addition reaction
Corret option is (a)
12. The difference in the measured and calculated magnetic moment (based on spin – orbit coupling) is observed for
(a) Pm3+
(b) Eu3+
(c) Dy3+
(d) Lu3+
Ans. (b)
Sol. The difference is observed and calculated magnetic moment observed in Eu3+ ion
Correct option is (b)
13. For a redox reaction, the
observed in cyclic voltametry at hanging mercury drop electrode is –650 mV vs. SCE. The expected value for
is
(a) –708 mV
(b) –679 mV
(c) –650 mV
(d) –621 mV
Ans. (b)
Sol.
(EP)anodic observed in cyclic voltametry at hanging mercury drop electrode is –650 mV vs SCE.
Correct option is (b)
14. The dimension of Planck constant is (M, L and T denote mass, length and time respectively)
(a) ML3T–2
(b) ML3T–1
(c) M2L–1T–1
(d) M–1L2T–2
Ans. (b)
Sol.
h =
E or = force × distance
= mass × acceleration × distance
= kg × ms–2 × m
Unit of h = = kgm2 s–1
Dimension of h is ML2T–1
Correct option is (b)
15. For a homonuclear diatomic molecule, the bonding molecular orbital is
(a) of lowest energy
(b) of second lowest energy
(c) of lowest energy
(d) of lowest energy
Ans. (d)
Sol. Energy order of molecular orbital be given as
Correct option is (d)
16. The selection rules for the appearance, of P branch in the rotational – vibrational absorption spectra of a diatomic molecule within rigid rotor – harmonic oscillator model are
(a)
(b)
(c)
(d)
Ans. (c)
Sol. For p branch in the rotational vibrational spectra
Correct option is (c)
17. The S2 operation on a molecule with the axis of rotation as the z axis, moves a nucleus at (x, y, z) to
(a) (–x, –y, x)
(b) (x, –y, – z)
(c) (– x, y, – z)
(d) (– x, – y, – z)
Ans. (d)
Sol. S2 operation on a molecule with axis of rotation as the z-axis inversion occur
(–x –y –z)
Correct option is (d)
18. The expression which represents the chemical potential of the ith sp species (µi) in a mixture is
(a)
(b)
(c)
(d)
Ans. (a)
Sol. Chemical potential is
Correct option is (a)
19. Which of the following statements is NOT correct for a catalyst?
(a) It increases the rate of a reaction
(b) It is not consumed in the course of a reaction
(c) It provides an alternate pathway for the reaction
(d) It increases the activation energy of the reaction
Ans. (d)
Sol. Catalyst decreases the activation energy of the reaction.
Correct option is (d)
20. The value of the rate constant for the gas phase reaction is 38 dm3 mol–1 s–1 at 300 K. The order of the reaction is
(a) 0
(b) 1
(c) 2
(d) 3
Ans. (c)
Sol. Rate constant = 38 dm3 mol–1 s–1
Unit = dm3 mol–1 s–1 is second order
Correct option is (c)
21. Boric acid in aqueous solution in presence of glycerol behaves as a strong acid due to the formation of
(a) an anionic metal – chelate
(b) borate anion
(c) glycerate ion
(d) a charge transfer complex
Ans. (a)
Sol.
Correct option is (a)
22. Match the compounds in List I with the cooresponding structure / property given in List II
List – I List – II
A. (Ph3P)3 RhCl (i) Spinel
B. LiC6 (ii) Intercalation
C. PtF6 (iii) Oxidising agent
D. Ni3S4 (iv) Catalyst for alkene hydrogenation
(a) A – iii, B – i, C – ii, D – iv
(b) A – iv, B – ii, C – iii, D – i
(c) A – iii, B – ii, C – i, D – iv
(d) A – iv, B – iii, C – ii, D – i
Ans. (b)
Sol. Rh(PPh3)3 Cl Catalyst for alkene hydrogenation
LiC6
Intercalation compound
PtF6
Oxidising agent, Ni3S4
Spinel
Correct option is (b)
23. W(CO)6 reacts with MeLi to give an intermediate which upon treatment with CH2N2 gives compound X. X is represented as
(a) WMe6
(b) (CO)5 W– Me
(c) (CO)5 W = C (Me) OMe
(d)
Ans. (c)
Sol.
Correct option is (c)
24. Considering the quadrupolar nature of M – M bond in [Re2 Cl8]2–, the M – M bond order in [Re2Cl4 (PMe2Ph)4]+ and [Re2Cl4 (PMe2Ph)4] respectively are
(a) 3.0 and 3.0
(b) 3.0 and 3.5
(c) 3.5 and 3.5
(d) 3.5 and 3.0
Ans. (d)
Sol. [Re2Cl4 (PMe2Ph)4]+
Step (1) : Let the oxidation state of 'R' atom is x
2x – 4 = + 1 2x = + 5
Re+2 = 4d5 Re3+ = 4d4
Step (2) : There fore, number of d electron on Re+2 and Re+3 ions
= 5 + 4 = 9
Step (3) : Electronic configuration in cluster ion is
Similarly [Re2Cl4(PMe2Ph)4] Step (1) 2x – 4 = 0
2x = + 4
x = +2
Re+2 = 4d5 Re+2 = 4d5
Step (2) = 5 + 5 = 10
Step (3) =
Corret option is (d)
25. A student recorded a polarogram of 2.0 mM Cd2+ solution and forgot to add KCl solution. What type of error do you expect in his results?
(a) Only migration current will be observed
(b) Only diffusion current will be observed
(c) Both migration current as well as diffusion current will be observed
(d) Both catalytic current as well as diffusion current will be observed
Ans. (b)
Sol. A student recorded a polarogram of 2.0 mM Cd2+ solution and forgot to add KCl solution that is the supporting electrolyte, then migration current is supressed so only diffusion current will be observed.
Correct option is (b)
26. The separation of trivalent lanthanide ions, Lu3+, Yb3+, Dy3+, Eu3+ can be effectively done by a cation exchange resion using ammonia o – hydroxy isobutyrate as the eluent. The order in which the ions willl be separated is
(a) Lu3+, Yb3+, Dy3+, Eu3+
(b) Eu3+, Dy3+, Yb3+, Lu3+
(c) Dy3+, Yb3+, Eu3+, Lu3+
(d) Yb3+, Dy3+, Lu3+, Eu3+
Ans. (a)
Sol. In lanthanoids, as the atomic number increases, size of lanthanide ion decreases. O-hydroxylsobutyrate ion forms a stable complex with smaller ion. The correct order of trivalent lanthanide ion is separation by
Lu3+ > Yb3+ > Dy3+ > Eu3+
Correct option is (a)
27. Arrange the following metal complexes in order of their increasing hydration energy
(P) [Mn(H2O)6]2+
(Q) [V(H2O)6]2+
(R) [Ni(H2O)6]2+
(S) [Ti(H2O)6]2+
(a) P < S < Q < R
(b) P < Q < R < S
(c) Q < P < R < S
(d) S < R < Q < P
Ans. (a)
Sol. Observed H.E = Calculated H.E + CFSE
CFSE for Mn2+ complex ion is zero that means H.E is least for [Mn(H2O)6]2+, CFSE for Ni2+ is maximum amongst all that mean hydration energy for [Ni(H2O)6]2+ is maximum.
Correct option is (a)
28. In the complex, the IR stretching frequency appears at 1857 cm–1 (strong) and 1897 cm–1 (weak). The valence electron count and the nature of the M-CO bond respectively are
(a) 16 e–, bridging
(b) 17 e–, bridging
(c) 18 e–, terminal
(d) 18 e–, briding
Ans. (c)
Sol. According to neutral atom method of 18 electron count (TVE)
= 2 Ni = 20 electron
= 10 electron
2CO = 4 electron
TVE = 34 electron
B = n × 18 – TVE where, n = number of metal in cluster
= 2 × 18 – 34
= 36 – 34 = 2
= number of M–M bond
= 1
= 1857 cm–1 suggest ter min al carbonyl
As for M–CO terminal = 2120 – 1850 cm–1
Per 'Ni' atom valence electron = 18 electron
Correct option is (c)
29. The correct classification of [B5H5]2–, B5H9 and B5H11 respectively is
(a) closo, arachno, nido
(b) arachno, closo, nido
(c) closo, nido, arachno
(d) nido, arachno, closo
Ans. (c)
Sol. [B5H5]–2
[BnHn]–2
closo
[B5H9] [B5H5]–4
[BnHn]–4
Nido
[B5H11] [B5H5]–6
[BnHn]–6
Arachano
Correct option is (c)
30. The compounds X and Y in the following reaction are
(a) X = (Et)2P(S)SH; Y = (Et)2P(S)Cl
(b) X = (EtO)2P(S)SH; Y = (EtO)2P(S)Cl
(c) X = (EtO)2PSH; Y = (EtO)2PCl
(d) X = (Et)3PO; Y = (Et)3PCl
Ans. (b)
Sol.
Correct option is (b)
31. Consider the reactions
(A) [Cr(H2O)6]2+ + [CoCl(NH3)5]2+
[Co(NH3)5(H2O)]2+ + [CrCl(H2O)5]2+
(B) [Fe(CN)6]4– + [Mo(CN)8]3–
[Fe(CN)6]3– + [Mo(CN)8]4–
Which one of the following is the correct statement?
(i) Both involve an inner sphere mechanism
(ii) Both involve an outer sphere mechanism
(iii) Reaction A follows inner sphere and reaction B follows outer sphere mechanism
(iv) Reaction A follows outer sphere and reaction B follows inner sphere mechanism
(a) i
(b) ii
(c) iv
(d) iii
Ans. (d)
Sol. Reaction,
[Cr(H2O)6]2+ + [CoCl(NH3)5]2+
[Co(NH3)5(H2O)]2+ + [CrCl(H2O)5]2+
follows inner sphere mechanism because it involves exchange of ligand. While reaction,
[Fe(CN)6]4– + [Mo(CN)8]3–
[Fe(CN)6]3– + [Mo(CN)8]4–
follows outer sphere mechanism because it involves only transfer of charge.
Correct option is (d)
32. The pair of compounds having the same hybridization for the central atom is
(a) XeF4 and [SiF6]2–
(b) [NiCl4]2– and [PtCl4]2–
(c) Ni(CO)4 and XeO2F2
(d) [Co(NH3)6]3+ and [Co(H2O)6]3+
Ans. (a, d)
Sol. [Co(NH3)6]3+ and [Co(H2O)6]3+ both are low spin complexes. They involve inner 3d orbital in hybridization. hence they have same hybridization i.e. d2sp3 and XeF4 and [SiF6]2– have hybridisation sp3d2.
Correct option is (a, d)
33. In the reaction shown below, X and Y respectively are
(a) [Mn(CO)4]2–, [CH3C(O)Mn(CO)5]–
(b) [Mn(CO)4]2–, [CH3C(O)Mn(CO)5]
(c) [Mn(CO)5]–, [ClMn(CO)5]
(d) [Mn(CO)5]2–, [ClMn(CO)5]–
Ans. (d)
Sol. [Co(NH3)6]3+ and [Co(H2O)6]3+ both are low spin complexes. They involve inner 3d orbital in hybridization. Hence, they have same hybridization i.e., d2sp3
Correct option is (d)
34. The Lewis acid character of BF3, BCl3 and BBr3 follows the order
(a) BF3 < BBr3 < BCl3
(b) BCl3 < BBr3 < BF3
(c) BF3 < BCl3 < BBr3
(d) BBr3 < BCl3 < BF3
Ans. (c)
Sol. Back donation of lone pair
Due to small size of fluorine, fluorine has tendency to back donate its lone pair of electron to B. Due to electron donation, electron deficiency of B is reduced. Such tendency to back donation of lone pair decreases as the size of halide increases.
Correct order of lewis acidity is
BBr3 > BCl3 > BF3
Correct option is (c)
35. The compound which shows L M charge transfer is
(a) Ni(CO)4
(b) K2Cr2O7
(c) HgO
(d) [Ni(H2O)6]2+
Ans. (a)
Sol. Metal in low oxidation state and ligand must have empty orbitals and will shows M
L charge transfer.
Correct option is (a)
36. The reaction of [PtCl4]2– with NH3, gives rise to
(a) [PtCl4(NH3)2]2–
(b) trans-[PtCl2(NH3)2]
(c) [PtCl2(NH3)4]
(d) cis-[PtCl2(NH3)2]
Ans. (d)
Sol.
Correct option is (d)
37. Zeise's salt is represented as
(a) H2PtCl6
(b) [PtCl4]2–
(c) [ZnCl4]2–
(d)
Ans. (d)
Sol.
Correct option is (d)
38. The catalyst used in the conversion of ethylene to acetaldehyde using Wacker process is
(a) HCo(CO4)
(b) [Pd(Cl4)]2–
(c) V2O5
(d) TiCl4 in the presence of Al(C2H5)3
Ans. (b)
Sol. Wecker's process
Correct option is (b)
39. The temperature of 54 g of water is raised from 15ºC to 75ºC at constant pressure. The change in the enthalpy of the system (given that Cp.m of water = 75 JK–1mol–1) is:
(a) 4.5 kJ
(b) 13.5 kJ
(c) 9.0 kJ
(d) 18.0 kJ
Ans. (b)
Sol. = 75 – 15 = 60°
w = 54 g of H2O = 3 moles of H2O
At constant pressure, = qp
qp = n Cv,m
or
= 3 × 75 × 60 = 13500 J =
Correct option is (b)
40. The specific volume of liquid water is 1.001 mL g–1 and that of ice is 1.0907 mL g–1 at ºC. If the heat of fusion of ice at this temperature is 333.88 J g–1, the rate of change of melting point of ice with pressure in deg atm–1 will be
(a) –0.0075
(b) 0.0075
(c) 0.075
(d) –0.075
Ans. (a)
Sol.
Now, 1 atm = 1.01325 × 105 Pa or
Hence = –0.0074 atm–1 K or –0.0074 atm–1 °C.
Correct option is (a)
41. Given that E0(Fe3+, Fe) = –0.04 V and E0(Fe2+, Fe) = –0.44 V, the value of E0(Fe3+, Fe2+) is:
(a) 0.76 V
(b) –0.40 V
(c) –0.76 V
(d) 0.40 V
Ans. (a)
Sol.
From Eq. (1) & Eq. (2) gives required equation
–nFE = –n1F E1 + n2F E2
–F (nE) = –F(n1 E1 – n2E2)
E = 0.76
Correct option is (a)
42. For the reaction P + Q + R S, experimental data for the measured initial rates is given below.
The order of the reaction with respect to P, Q and R respectively is:
(a) 2, 2, 1
(b) 2, 1, 2
(c) 2, 1, 1
(d) 1, 1, 2
Ans. (c)
Sol. From experiment (1) and (2), concentration of P is double and rate increase in four times. So, order is 2 with respect to P.
From experiment (2) and (3), concentration of Q is fourth and rate is four times. So, order is 1 with respet ot Q.
Experiment (1) and (4),
Correct option is (c)
43. Sucrose is converted to a mixture of glucose and fructose in a pseudo first order process under alkaline conditions. The reaction has a half life of 28.4 min. The time required for the reduction of a 8.0 mM sample of sucrose to 1.0 mM is
(a) 56.8 min
(b) 170.4 min
(c) 85.2 min
(d) 227.2 min
Ans. (c)
Sol. t1/2 = 28.4 min
K = = 0.0244 min–1
[A]0 = 8 mM; [A] = 1.0 mM; In= Kt; t = 85.2 min
Correct option is (c)
44. The reaction, 2NO(g) + O2(g) 2NO2(g)
proceeds via the following steps
The rate of this reaction is equal to
(a) 2kb[NO][O2]
(b) (kakb[NO]2[O2])/(ka + kb[O2])
(c) 2kb[NO]2[O2]
(d) ka[NO]2[O2]
Ans. (b)
Sol. 2NO(g) + O2(g) 2NO2(g)
Mechanism :
r = Kb [N2O2] [O2]
Applying SSA on [N2O2]
Rate of formation = Rate of consumption
Ka [NO]2 = K'a [N2O2] + Kb [N2O2] [O2]
Correct option is (b)
45. 40 millimoles of NaOH are added to 100 mL of a 1.2 M HA and Y M NaA buffer resulting in a solution of pH 5.30. Assuming that the volume of the buffer remains unchanged, the pH of the buffer (KHA = 1.00 × 10–5) is
(a) 5.30
(b) 5.00
(c) 0.30
(d) 10.30
Ans. (b)
Sol. Vsolution = 100 mL [HA] = 1.2 A [NaA] =
The given solution is buffer solution
nHA = 1.2 × 100 = 120 m mol
nNaA = y × 100 = 100y mmol
On adding NaOH = 40 mmol,
HA + NaOH NaA + H2O
Initial amount 120 40 0 0
After reaction 120–40 0 40
= 80
In solution, nHA = 80 mmol; nNaA = (100y + 40) mmol
In the beginning nNaA = 100y = 100 × 1.2 = 120 mm
Correct option is (b)
46. The entropy of mixing of 10 moles of 10 moles of helium and 10 moles of oxygen at constant temperature and pressure, assuming both to be ideal gas, is:
(a) 115.3 JK–1
(b) 5.8 Jk–1
(c) 382.9 JK–1
(d) 230.6 JK–1
Ans. (a)
Sol.
Correct option is (a)
47. The ionisation potential of hydrogen atom is 13.6 eV. The first ionisation potential of a sodium atom, assuming that the energy of its outer electron can be represented by a H – atom like model with an effective nuclear charge of 1.84, is
(a) 46.0 eV
(b) 11.5 eV
(c) 5.1 eV
(d) 2.9 eV
Ans. (c)
Sol.
Correct option is (c)
48. The quantum state of a particle moving in a circular path in a plane is given by
When a perturbation H1 = P cos is applied (P is a constant), what will be the first order correction to the energy of the mth state
(a) 0
(b)
(c)
(d)
Ans. (a)
Sol.
Correct option is (a)
49. The correct statement (s) among the following is/are
(i) The vibrational energy levels of a real diatomic molecule are equally spaced.
(ii) At 500K, the reaction A B is spontaneous when
= 18.83 kJ mol–1 and
= 41.84 J K–1 mol–1.
(iii) The process of fluorescence involves transition from a singlet electronc state to another singlet electronic state by absoprtion of light.
(iv) When a constant P is added to each of the possible energies of a system, its entropy remains unchanged.
(a) Only i
(b) Only ii
(c) Both i and iii
(d) Both ii and iv
Ans. (c)
Sol. (I) Correct statements are
Gap equally spaced
(II) In fluorescence transition from a singlet electronic state to another singlet electronic state by absorption of light
Correct option is (c)
50. Assuming H2 and HD molecules having equal lengths, the ratio of the rotational partition functions of these molecules, at temperature above 100 K is
(a) 3/8
(b) 3/4
(c) 1/2
(d) 2/3
Ans. (a)
Sol. We know that rotational partition function
Correct option is (a)
51. N non – interacting molecules are distributed among three non – degenerate energy levels and
at 100 K. If the average total energy of the system at this temperature is
the number of molecules in the system is:
(a) 1000
(b) 1503
(c) 2354
(d) 2987
Ans. (a)
Sol.
We know
Correct option is (a)
53. The rate constant of two reactions at temperature T are k1(T) and k2(T) and the corresponding activation energies are E1 and E2 with E2 > E1. When temperature is raised from T1 and T2, which one of the following relations is correct?
(a)
(b)
(c)
(d)
Ans. (d)
Sol. E2 > E1
At same temperature K2 < K1
On increasing the temperature K2 increase more as compared to K1
So,
Correct option is (d)
54. The number of degrees of freedom for a system consisting of NaCl(s), Na+(aq) and Cl– (aq) at equilibrium is
(a) 2
(b) 3
(c) 4
(d) 5
Ans. (a)
Sol. NaCl Na+ + Cl–1
C = 2 – 1 = 1
F = C – P + 2
F = 2 – 2 + 2 = 2
Therefore, F = 2
Correct option is (a)
55. Match the structure in List – I with their correct names in List – II.
List – I List – II
(P) (i) 3 – methyl furan
(Q) (ii) Imidazole
(R) (iii) 5 – hydroxybenzothiazole
(S) (iv) 2 – amino piperidine
(T) (v) 2 – amino morpholine
(a) P – i, Q – ii, R – v, S – iii, T – iv
(b) P – ii, Q – iii, R – iv, S – v, T – i
(c) P – iii, Q – iv, R – v, S – i, T – ii
(d) P – iv, Q – v, R – i, S – ii, T – iii
Ans. (a)
Sol.
Correct option is (a)
56. The result of the reduction of either (R) or (S) 2 – methylcyclohexanone, in separate reactions, using LiAlH4 is that the reduction of
(a) The R enantiomer is stereoselective
(b) The R enantiomer is stereospecific
(c) The S enantiomer is stereospecific
(d) Both the R and S enantiomers is stereoselective
Ans. (d)
Sol.
Reduction of (R) or (S)-2-methylcyclohexanone by LilH4 is a stereoselective reaction. Reduction of carbonyl group by LiAlH4 is possible from both side (above and below the plane). The major product (select) fromed from less hindered side.
Correct option is (d)
57. The increasing order of basicity among the following is
(a) Y < X < Z
(b) Y < Z < X
(c) X < Z < Y
(d) X < Y < Z
Ans. (c)
Sol.
Due to +M effect high electron density on nitrogen atom.
So, more basic than (X)
(X)
Lone pair present on nitrogen atom delocalization with ring so less as compare to (Z)
(Y)
Due to SIR effect most basic in nature.
So correct order of basicity, X < Z < Y.
Correct option is (c)
58. In the reaction,
if the concentration of both the reactants is doubled, then the rate of the reaction will
(a) remain unchanged
(b) quadruple
(c) reduce to one fourth
(d) double
Ans. (d)
Sol.
Reaction proceed through SN1 mechanism
Rate (reactant)
Rate of reaction depends upon the concentration of reactant only.
If the concentration of reactant is doubled then the rate of reaction will be double.
Correct option is (d)
59. Match the structure in List – I with the coupling constant [1HJ(Hz)] given in List – II
List – I List – II
(1) (i) ~ 1Hz
(2) (ii) ~ 10 Hz
(3) (iii) ~ 15 Hz
(a) 1 – (i), 2 – (ii), 3 – (iii)
(b) 1 – (ii), 2 – (iii), 3 – (i)
(d) 1 – (iii), 2 – (ii), 3 – (i)
(d) 1 – (iii), 2 – (i), 3 – (ii)
Ans. (b)
Sol.
Correct option is (b)
60. Phenol on reaction with formaldehyde and dimethyl amine mainly gives
(a)
(b)
(c)
(d)
Ans. (a)
Sol. Mannich reaction
More stable due to intramolecular H-bonding
Correct option is (a)
61. The mono protonation of adenine (X) in acidic solution
mainly occurs at
(a) position 1
(b) position 2
(c) position 3
(d) either position 4 or 5
Ans. (b)
Sol. The monoprotonation of adenine in acidic solution occurs at second position due to stablization of positive charge.
Correct option is (b)
62. In the following reaction,
(X) and (Y) respectively are
(a) 1:CH2 and cis 1, 2 – dimethylcyclopropane
(b) 3:CH2 and cis 1, 2 – dimethylcyclopropane
(c) 1:CH2 and a mixture of cis / trans 1, 2 – dimethylcyclopropane
(d) 3:CH2 and a mixture of cis / trans 1, 2 – dimethylcyclopropane
Ans. (d)
Sol. Photosensitized through benzophenone, produced triplet carbene (3:CH2)
Correct option is (d)
63. The major products obtained upon treating a mixture of
with a strongly acidic solution of H2SO4 is
(a)
(b)
(c)
(d)
Ans. (a)
Sol.
Correct option is (a)
64. Match the observed principal absoprtions in the visible spectrum shown List – I with the bond show this absorption in List – II
(a) 1 – (i), 2 – (ii), 3 – (iii), 4 – (iv)
(b) 1 – (i), 2 – (iii), 3 – (ii), 4 – (iv)
(c) 1 – (ii), 2 – (i), 3 – (iv), 4 – (iii)
(d) 1 – (iv), 2 – (ii), 3 – (iii), 4 – (i)
Ans. (a)
Sol.
Corret option is (a)
65. Among the isomers C10H14 shown,
the isomer that can be identified uniquely by mass spectrometry alone is:
(a) P
(b) Q
(c) R
(d) S
Ans. (b)
Sol.
Correct option is (b)
66. The direction of rotation of the following thermal electrocyclic ring closures.
respectively is:
(a) Disrotory, disrotatory, disrotatory
(b) Conrotatory, conrotatory, conrotatory
(c) Disrotatory, disrotatory, conrotatory
(d) Disrotatory, conrotatory, disrotatory
Ans. (a)
Sol.
Correct option is (a)
67. The molecules (s) that exist an meso structure (s).
is/are:
(a) Only M
(b) Both K and L
(c) Only L
(d) Only K
Ans. (d)
Sol.
For meso, there must be more than one asymmetric centre and plane of symmetry should be present
Correct option is (d)
68. Stereochemical descriptors for the atoms labeled Ha and Hb in the structure.
respectively are:
(a) X – homotopic, Y – enantiotopic and Z – diastereotopic
(b) X – enantiotopic, Y – homotopic and Z – diastereotopic
(c) X – diastereotopic, Y – homotopic and Z – enantiotopic
(d) X – homotopic, Y – diastereotopic and Z – enantiotopic
Ans. (c)
Sol.
Other method,
(I) In case I, Ha and Hb are diastereotopic.
(II) In case II, C2 symmetry present so they are homotopic.
(III) In case III. Ha and Hb reflect through the plane, so they are enantiotopic.
Correct option is (c)
69. Treatment of the pentapeptide Gly-Arg-Phe-Ala-Ala, in separate experiments, with the enzymes Trypsin, Chemotrypsin and Carboxypeptidase A respectively, gives :
(a) Gly-Arg + Phe-Ala-Ala; Gly-Arg-Phe+Ala-Ala; Gly-Arg-Phe-Ala+Ala
(b) Gly-Arg-Phe+Ala-Ala; Gly-Arg-Phe+Ala-Ala; Gly-Arg-Phe-Ala+Ala
(c) Gly-Arg+Phe-Ala-Ala; Gly-Arg-Phe-Ala+Ala; Gly-Arg-Phe+Ala-Ala
(d) Gly-Arg + Phe-Ala-Ala; Gly-Arg-Phe+Ala-Ala; Gly+ Arg-Phe- Ala + Ala
Ans. (a)
Sol. (i) Enzyme trypsin cuts the peptide chain from c-terminal side of Arg, Cys
(ii) Chemotrysin cuts peptide chain from c-terminal side of Phe, Trp, Tyr
(iii) Carboxypeptidase cuts peptide chain from c-terminal side.
Correct option is (a)
70. Hordenine (x), an alkaloid, an alkaloid, undergoes Hoffmann degradation to give compound (Y).
(Y) on treatment with alkaline permanganate (Z). Y and Z respectively are
(a)
(b)
(c)
(d)
Ans. (d)
Sol.
Correct option is (d)
Common data for Q. 71, Q. 72, Q. 73:
Trans 1, 2 – difluoroethylene molecule has a 2 – fold rotational axis, a symmetry plane perpendicular to the rotational axis and an inversion centre.
71. The number of distinct symmetry operations that can be performed on the molecule is:
(a) 2
(b) 4
(c) 6
(d) 8
Ans. (b)
Sol.
Molecule having C2h point group. Distinct symmetry operation E C2
i
Correct option is (b)
72. The number of irreducible representations of the point group of the molecule is:
(a) 1
(b) 2
(c) 3
(d) 4
Ans. (d)
Sol. Number of irreducible representation is 4,
Ag, Bb, Au, Bu
Correct option is (d)
73. If two H atoms of the above molecule are also replaced by F atoms, the point group of the resultant molecule will be
(a) Ci
(b) C2h
(c) C2v
(d) D2h
Ans. (d)
Sol.
Correct option is (d)
Common Data for Q. 74 and Q.75
Reactivity of ary 1 amines towards electrophilic aromatic substitution is much higher than that of aliphatic amines. Hence differential reactivity of the amino group is desirable in many reactions.
74. The compound which on reacting with aniline will NOT form an acetanilide is
(a)
(b)
(c)
(d)
Ans. (b)
Sol.
Correct option is (b)
75. Aniline can be distinguished from methylamine by its reaction with
(a) p – toluene sulphonyl chloride / KOH
(b) (i) NaNO2/HCl, 0 – 5ºC (ii) alkaline – napthol
(c) Sn/HCl
(d) acetyl chloride
Ans. (b)
Sol. (1) Aniline ract with HNO2 (NaNO2 + HCl) at 0 – 5°C to form arenediazonium salt which couple with an alkaline solution of -naptol to form orange or red coloured azo dye.
(2) Methyl amine react with HNO2 at 0 – 5°C to form alcohol with birsk effervescence due to quantitative evolution of N2 gas.
CH3 – NH2 + HONO CH3 – OH + H2 + H2O
Correct option is (b)
76. In the reaction,
Compound X is
(a)
(b)
(c)
(d)
Ans. (b)
Sol.
Same ozonolysis (mechanism) on remaining both double bond observed product is
Correct option is (b)
77. Oxidation of X with chromic acid gives
(a)
(b)
(c)
(d)
Ans. (a)
Sol.
Correct option is (a)
Linkage Answer Type Q. 78 and Q. 79:
78. In the reaction,
Compound X is
(a) Adenine
(b) Xanthine
(c) 2, 6 – diaminopurine
(d) Adenosine
Ans. (d)
Sol.
Correct option is (d)
79. Compound X on treatment with conc. HCl gives
(a) Uric acid
(b) Adenine
(c) Hypoxanthine
(d) Guanine
Ans. (b)
Sol.
Correct option is (b)
80. The reaction of ammonium chloride with BCl3 at 140ºC at followed by treatment with NaBH4 gives the product X. The formula of X is
(a) B3N3H3
(b) B3N3H6
(c) B3N3H12
(d) [BH – NH]n
Ans. (b)
Sol. 3NH4Cl + 3BCl3
B3N3H3Cl3 + 9HCl
2B3N3H6 + 6NaCl + 3B2H6
Correct option is (b)
81. Which of the following statement(s) is/are true for X?
(I) X is not isoelectronic with benzene
(II) X undergoes addition reaction with HCl.
(III) Electrophilic substitution reaction on X is much faster than that of benzene
(IV) X undergoes polymerization at 90ºC
(a) (i) and (ii)
(b) (ii) only
(c) (ii) and (iii)
(d) (i) and (iv)
Ans. (b)
Sol. B3N3H6 is isoelectronic with benzene and hence had been called inorganic benzene.
Polar species like HCl, attack the double bond between N and B, i.e, why borazine in contranst to benzene, readily undergoes addition reaction.
Electrophillic substitution reaction on borazine slower than benzene.
Only (II)
Correct option is (b)
82. Consider a particle of mass m moving in a one – dimensional box under the potential V = 0 for 0 < x < a and outside the box. When the particle is in its lowest energy state the average momentum (< px > of the particle is
(a)
(b)
(c)
(d)
Ans. (a)
Sol.
Correct option is (a)
83. The uncertainty in the momentum of the particle in its lowest energy state is:
(a)
(b)
(c)
(d)
Ans. (c)
Sol.
Correct option is (c)
Linkage Answer Type Q. 84 and Q. 85:
84. In the mixture obtained by mixing 25.0 mL 1.2 × 10–3 M MnCl2 and 35.0 mL of 6.0 × 10–4 M KCl solution, the concentrations (M) of Mn2+, K+ and Cl– ions respectively are
(a)
(b)
(c)
(d)
Ans. (c)
Sol.
nKCl = 35 × 6 × 10–4
210 × 10–4 = 0.021 mmol
= 0.021 mmol = 0.021 mmol
Correct option is (c)
85. The activity (M) of Mn2+ ion in the above solution is
(a) 1.0 × 10–4
(b) 2.0 × 10–4
(c) 3.0 × 10–4
(d) 4.0 × 10–4
Ans. (d)
Sol. Activity of Mn2+ ion in the above solution is
a = a = = 0.8 × 5 × 10–4 = 4 × 10–4
Correct option is (d)