PYQ 2013 - VedPrep

GATE Chemistry 2013
Previous Year Question Paper with Solution.

1. The point group symmetry of is

(a) D_2h

(b) C_2h

(d) D_2d

Ans. (d)

Sol. CH₂ = C = CH₂

point group (D_2d point group)

Correct option is (d)

2. Two trial wave function give ground state energies E₁ and E₂, respectively, for the microscopic particle in a 1 – D box by using the variation method. If the exact ground state energy is E₀, the correct relationship between E₀, E₁ and E₂ is:

(a)

(b)

(c)

(d)

Ans. (c)

Sol. A trial wave function that can be expressed as a linear combination of 2 variational parameter leads to variational minimum energy E₀ < E₂ < E₁

Correct option is (c)

3. The ground state energies of H atom and H₂ molecule –13.6 eV and –31.7 eV, respectively. The dissociation energy of H₂ is ___________ eV.

Ans. (4.5)

Sol. H₂ + dissociation energy = 2 h atom

–31.7eV + dissociation energy = 2 × (–13.6)

Dissociation energy = + 31.7 – 27.2 = 4.5

4. A 2 L vessel containing 2g of H₂ gas at 27ºC is connected to a 2L vessel containing 176 g of CO₂ gas at 27ºC. Assuming ideal behaviour of H₂ and CO₂, the partial pressure of H₂ at equilibrium is _______ bar.

Ans. (6.2325)

Sol. V_total = 4L

5. Consider the reaction, at equilibrium,

The equilibrium can be shifted towards the forward direction by

(a) Increasing the amount of carbon in the system

(b) Decreasing the volume of the system

(d) Increasing the temperature of the system

Ans. (c)

Sol. According to L.C. principles, On decreasing pressure shifts in the direction in which the number of gaseous molecules increases.

Correct option is (c)

6. A sparingly soluble electrolyte M₂X ionizes as

The solubility product (K_sp), molal solubility (S) and molal activity coefficient are related by

(a)

(b)

(c)

(d)

Ans. (d)

Sol.

When actually in the form of ion is ksp

7. For the first order consecutive reaction, under steady state approximation to [Q], the variation of [P], [Q] and [R] with time are best represented by

(a)

(b)

(c)

(d)

Ans. (c)

Sol. The concentration of the component which is present in steady practicaly remains constant except at the very begining and at the very end.

Correct option is (c)

8. At 273 K and 10 bar, the langmuir adsorption of a gas on a solid surface gave the fraction of surface coverage as 0.01. The Langmuir adsorption isotherm constant is ___________ bar^–1.

Ans. (0.001)

Sol.

Correct option is (0.001)

9. Conversion of boron trifluoride to tetrafluoroborate accompanies

(a) Increase in symmetry and bond elongation

(b) Increase in symmetry and bond contraction

(d) Decrease in symmetry and bond elongation

Ans. (a)

Sol. In BF₃, 'B' is sp³ hybridised and there is B–F back donation, due to which B–F bond order increases and hence bond length decreased. B–F bond length is 130 pm.

Also BF₃ is trigonal planar.

In BF₄^– (tetra fluoro borate) 'B is sp³ hybridised and there is no back donation. Hence B–F bond order is one. [BF₄]^– perfect Td.

Correct answer is (a)

10. The correct statement with respect to the bonding of the ligands, Me₃N and Me₃P with the metal ions Be²⁺ and Pd²⁺ is,

(a) The ligands bind equally strong with both the metal ions as they are dicationic

(b) The ligands bind equally strong with both the metal ions as both the ligands are pyramidal

(d) The binding is stronger for Me₃N with Pd²⁺ and Me₃P with Be²⁺

Ans. (c)

Sol. According to HSAB concept (Pearson theory)

Hard acid has a tendency to combine with hard base in order to give the stable product. Hence bond is stronger.

Correct answer is (c)

11. A crystal has the lattice parameters The crystal system is

(a) Tetragonal

(b) Monoclinic

(d) Orthorhombic

Ans. (d)

Sol. A crystal has the lattice parameters The crystal system is orthorhombic.

Correct option is (d)

12. The by – product formed in the characteristic reaction is

(a) CO

(b) MeOH

(d) MeCONH₂

Ans. (b)

Sol.

Fischer carbene, Carbene carbon is electrophilic in nature

Correct option is (b)

13. The catalyst and co – catalyst used in Wacker process, respectively, are

(a) PdCl₂ and Cu

(b) CuCl₂ and [PdCl₄]^2–

(d) [PdCl₄]^2– and CuCl₂

Ans. (d)

Sol. The [PdCl₄]^2– (cat) can be removed by reaction with CuCl₂ (Co Catalyst)

Correct answer is (d)

14. Oxymyoglobin Mb(O₂) and oxhyhemoglobin Hb(O₂)₄, respectively, are

(a) Paramagnetic and paramagnetic

(b) Diamagnetic and diamagnetic

(d) Diamagnetic and paramagnetic

Ans. (b)

Sol. Oxymyglobin Mb(O₂)

Due to antiferromagnetic coupling it is diamagnetic in nature.

Correct option is (b)

15. Hapticity of cycloheptatriene in is __________

Ans. 6

Sol. Mo(C₇H₈) (CO)₃

18electron

To follow stable 18 electron rule the heptacity of C₇H₈

Correct option is (6)

16. The number of oxygen molecule (s) that a molecule of hemerythrin can transport is ________

Ans. 1

Sol.

Thus, number of dioxygen molecule transported by one hemerythrin is one

Correct answer is (1)

17. The maximum number of stereoisomers possible for the compound given below is __________

Ans. 4

Sol.

Correct answer is 4

18. The correct sequence of the amino acids present in the tripeptide given below is

(a) Val – Ser – Thr

(b) Val – Thr – Ser

(d) Leu – Thr – Ser

Ans. (a)

Sol.

So, the correct sequence of the tripeptide is val–ser–thr

Correct option is (a)

19. Among the compounds given in the options (a) – (b), the one that can be used as a formyl anion equivalent (in the presence of a strong base) is:

(a) ethylene

(b) nitroethane

(d) 1, 4 – dithiane

Ans. (c)

Sol.

Correct answer is (c)

20. The major product formed in the reaction given below is:

(a)

(b)

(c)

(d)

Ans. (d)

Sol.

Correct answer is (d)

21. The major product formed in the reaction given below is:

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

Correct answer is (a)

22. The pericyclic reaction given below is an example of

(a) [1, 3] – sigmatropic shift

(b) [1, 5] – sigmatropic shift

(d) [3, 3] – sigmatropic shift

Ans. (d)

Sol. It is an example of 3, 3-sigmatropic shift.

Correct option is (d)

23. The major product formed in the reaction of quinoline with potassium amide (KNH₂) in liquid ammonia is:

(a)

(b)

(c)

(d)

Ans. (b)

Sol. It is an example of chichibabin reaction

Correct answer is (b)

24. The number of signals that appear in the proton decoupled ¹³C NMR spectrum of benzonitrile (C₇H₅N) is _________.

Ans. (5 to 5)

Sol.

So, the pronton decopled ¹³C NMR is 5.

Correct answer is (5 to 5)

25. Among the compound given in the option (a) to (d), the one that exhibits a sharp band at around 3300 cm^–1 in the IR spectrum is:

(a) 1, 2 – butadiene

(b) 1, 3 – butadiene

(d) 2 – butyne

Ans. (c)

Sol.

26. In the metathesis reaction given below, 4.32 g of the compound X was treated with 822 mg of the catalyst Y to yield 2.63 g of the product Z. The mol% of the catalyst Y used in this reaction is ___________ [Atomic weights of Ru = 101; P = 31; Cl = 35.5]

Ans. (5)

Sol. 4.32 of compound (X)

The mol% of the catalyst Yused in this reaction

= 0.05 × 100% = 5%

0.02 >> 10^–3, it can be neglected.

Correct option is (5)

27. An organic compound Q exhibited the following spectral data:

IR: 1760 cm^–1

Compound Q is

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

Correct answer is (a)

28. The major product formed in the Beckmann rearrangement of the compound given below is:

(a)

(b)

(c)

(d)

Ans. (d)

Sol.

Correct answer is (d)

29. The major product formed in the reaction given below is

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

(Example of Pinacole-Pinacole rearrangement)

30. The major product formed in the reaction given below is

(a)

(b)

(c)

(d)

Ans. (d)

Sol.

Correct answer is (d)

31. The major product (s) formed in the reaction sequence given below is/are

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

Correct answer is (a)

32. Match the compound in the Column – I with photochemical reactions that they can undergo given in the Column – II:

(a) (i) – (q); (ii) – (s); (iii) – (p)

(b) (i) – (r); (ii) – (p); (iii) – (s)

(d) (i) – (r); (ii) – (q); (iii) – s)

Ans. (b)

Sol.

33. is an eigen function on the operator The corresponding eigen value is

(a) +4

(b) –4

(d) –2

Ans. (b)

Sol. Eigen function

For eigen value function operate by operator

Eigenvalue = –4

Correct option is (b)

34. The infrared spectrum of HCl gas shows an absorption band centered at 2885 cm^–1. The zero point energy of HCl molecule under hamonic oscillator approximation is:

(a) 2.866 × 10^–22J

(b) 2.8665 × 10^–20J

(d) 5.7330 × 10^–20J

Ans. (b)

Sol. For harmonic oscillator every band gap is

35. For the reaction given the values, = 84 J K^–1,

(a) –11.08 kJ

(b) + 11.08 kJ

(d) + 13.55 kJ

Ans. (a)

Sol.

= 13.955 kJ – 298 × 0.84 = 13.955 kJ – 25.032 kJ = –11.08 kJ

Correct option is (a)

36. The change in enthalpy when 3 mol of liquid benzene transforms to the vapour state at its boiling temperature (80ºC) and at 1 bar pressure is ____________ kJ.

Ans. (92.13)

Sol.

For 3 mole = 92.1 3 kJ

Correct option is (92.13)

37. The moment of iertia of a homonuclear diatomic molecular is 7.5 × 10^–45 kg m². Its rotational partition function at 500 K is___________.

Ans. (4649.5)

Sol. I = 7.5 × 10^–45 kgm^–2

38. For a reaction of the type the correct rate expression is ([X]₀ and [X] corresponds to the concentration of X at time t = 0 and t = t, respectively)

(a)

(b)

(c)

(d)

Ans. (b)

Sol.

Correct option is (b)

39. The temperature dependence of partition are as follows:

According to the Conventional Transition State Theory (CTST), the temperature dependence of the Arrhenius pre – exponential factor for a reaction of the type given below is

linear molecule + linear molecule non – liner linear transition state

(a) T^–1

(b) T⁰

(d) T²

Ans. (a)

Sol.

Arrhenius pre-exponential factor

Correct answer is (a)

40. Decarbonylation reaction of [cis – (CH₃CO) Mn (¹³CO)(CO)₄] yields X, Y and Z, where X = [(CH₃) Mn (CO)₅]; Y = [cis – (CH₃) Mn (¹³CO) CO₄]; [Z = trans – (CH₃) Mn (¹³CO) (CO)₄]

The molar ratio of the products (X : Y : Z) in this reaction is

(a) 1 : 1 : 1

(b) 1 : 2 : 1

(d) 2 : 1 : 1

Ans. (b)

Sol.

Correct answer is (b)

41. According to polyhedral electron count rule, the structure of Rh₆(CO)₁₆ is:

(a) closo

(b) nido

(d) hypo

Ans. (a)

Sol. Rh₆ (CO)₁₆

Polyhedral electron cont (PEC) = 86 – 6 × 12 = 14

42. The increasing order of melting points of the halides NaCl, CuCl and NaF is:

(a) CuCl < NaCl < NaF

(b) NaF < NaCl < CuCl

(d) CuCl < NaF < NaCl

Ans. (a)

Sol. In NaCl CuCl

In NaF and NaCl, 'Cl^–' is of bigger size than F^– hence Cl^– will be easily polarisable than F^–. Due to this NaCl will have more polarisation, more co-valent character (Fajan rule) and lower melting point.

Therefore, overall order of melting point will be

CuCl < NaCl < NaF

Correct option is (a)

43. The correct electronic configuration and spin only magnetic moment of Gd³⁺ (atomic number 64) are

(a) [Xe]4f⁷ and 7.9 BM

(b) [Xe]4f⁷ and 8.9 BM

(d) [Rn]5f⁷ and 7.9 BM

Ans. (a)

Sol. Electronic configuration of Gd³⁺ = [Xe]4f ⁷

Correct option is (a)

44. Among the following octahedral complexes, the one that has the highest enthalpy of hydration is

(a) [Ca(H₂O)₆]²⁺

(b) [Mn(H₂O)₆]²⁺

(d) [Cr(H₂O)₆]²⁺

Ans. (c)

Sol. [V(H₂O₆)]²⁺ has V²⁺ metal having 3d³ electronic configuration due to which it has maximum negative value of CFSE out of all. Hence posses maximum enthalpy of hydration.

Greater the negative value of CFSE greater will be its hydration energy.

Correct option is (c)

45. A metal crystallizes in the face – centered cubic lattice parameter of 4.20Å. The shortest atom to atom contact distance in the lattice is

(a)

(b)

(c)

(d)

Ans. (b)

Sol.

46. Polarographic method of analysis to obtain individual amounts of Cu²⁺ and Cd²⁺ in a given mixture of the two ions (Cu²⁺ and Cd²⁺) is achieved by measuring their

(a) half – wave potentials

(b) migration currents

(d) diffusion currents

Ans. (d)

Sol. Polarography method of analysis to obtain individual amounts of Cu⁺² and Cd^–2 in a given mixture of the two ions (Cu²⁺ and Cd⁺²) is achioeved by measuring their diffusion currents.

Ilkovic equation = id = 708 n D^1/2 m^2/3 t^1/6 C

Correct option is (d)

47. The ground state term of [Ni(H₂O)₆]²⁺ is

(a)

(b)

(c)

(d)

Ans. (c)

Sol.

Orgel diagram of d⁸– Oh compound

Correct option is (c)

Common Data for Q. 48 and Q.49:

N, N – Dimethylformamide (DMF) gives different patterns of signals for the methyl protons whenits ¹H NMR spectrum is recorded at different temperatures.

48. Match the patterns of the NMR signals givenin the Column – I with temperatures given in the Column – II.

Column – I Column – II

(i) Two singlets, for three protons each, at 2.87 and 2.97 ppm (x) 25ºC

(ii) One sharp singlet for six protons at 2.97 ppm (y) 120ºC

(iii) One broad signal for protons (z) 150ºC

(a) (i) – (x); (ii) – (y); (iii) – (z)

(b) (i) – (x); (ii) – (z); (iii) – (y)

(d) (i) – (z); (ii) – (y); (iii) – (x)

Ans. (b)

Sol.

at normal 25°C two singlet for three proton

at 120°C one broad signal for six proton

at 150°C one sharp singlet for six proton

Correct option is (b)

49. Based on the above data, the calculated difference in the frequencies of the two methyl singlets, if the spectrum is recorded on a 300 MHz spectrometer, is ___________ Hz.

Ans. (30)

Sol.

Common Data for Q. 50 and Q. 51:

Heating a mixture of ammonium chloride and sodium tetrahydridoborate gives one liquid product (X), along with other products under ambient conditions.

50. Compound X is:

(a)