PYQ 2014 - VedPrep

GATE Chemistry 2014
Previous Year Question Paper with Solution.

Q.1 – Q.25 : Carry ONE mark each.

1. The maximum non-PV work that a system can perform at constant P is

(a)

(b)

(c)

(d)

Ans. (b)

Sol. Maximum non-PV work is equal to

Correct option is (b)

2. Consider the reaction:

The unit of the thermodynamic equilibrium constant for the reaction is

(a) mol L^–1

(b) L mol^–1

(d) dimensionless

Ans. (d)

Sol.

Thermodynamic equation constant is dimensionless.

If equilibrium constant is k_p

Correct answer is (d)

3. The number of IR active vibrational normal modes of CO₂ is _________

Ans. (3)

Sol. Number of IR active vibrational normal mode of CO₂

4. The number of C₂ axis in CCl₄ is __________

Ans. (3)

Sol. Number of C₂ axis in CCl₄ is 3

5. The value of the magnetic quantum number of a p_x orbital is

(a) –1

(b) 0

(d) undefined

Ans. (4)

Sol. The value of magnetic quantum number of p_x orbitial is undefined first of all fix the direction of applied magnetic field. Then decide the value of magnetic quantum number.

Correct option is (d)

6. The molecular partition function for a system in which the energy levels are equispaced by

(a)

(b)

(c)

(d)

Ans. (d)

Sol.

Since, energy level are equi-spaced,

if we consider ground state

This is an exponential series, which is equal to

7. A monoatomic gas, X, adsorbed on a surface, Langmuir adsorption isotherm. A plot of the fraction of surface coverage, against the concentration of the gas [X], for very low concentration of the gas, is described by the equation

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

8. At a given temperature and pressure, the ratio of the average speed of hydrogen gas to that of helium gas is approximately _________.

Ans. (1.4 to 1.5)

Sol.

Correct answer is 1.4 to 1.5.

9. An example of nido-borane from the following is

(a) B₄H₁₀

(b) B₆H₁₀

(d) B₈H₁₄

Ans. (b)

Sol.

Correct answer is (b)

10. The geometries of Ni(CO)₄ and [NiCl₄]^2–, respectively are

(a) tetrahedral and square planar

(b) square planar and tetrahedral

(d) square planar and square planar

Ans. (c)

Sol. In Ni(CO)₄, Ni is present as Ni⁰.

Ni⁰ = [Ar]3d⁸ 4s² 4p⁰

CO being strong field ligand pair up the s-electrons with d-electrons making the s-orbital empty for hybridisation. i.e., Ni in the presence of CO = [Ar]3d¹⁰ 4s⁰ 4p⁰

Thus,

Thus, the geometry of Ni(CO)₄ is tetrahedral and it is a diamagnetic complex.

Further,

in [NiCl₄]^2–, Ni is present as Ni²⁺.

Cl^– being a weak field ligand is not capable to pair up the unpaired d-electrons. So, the available orbitals for it are only 4s and 4p-orbitals.

Ni in the presence of Cl^– = [Ar] 3d⁸ 4s⁰ 4p⁰

Thus, it is also a tetrahedral complex.

Correct option is (c)

11. The number of S–S bonds in H₂S₅O₆ is _________

Ans. 4

Sol.

12. In atomic absorption spectroscopy, the atomization process utilizes

(a) flame

(b) electric field

(d) electron beam

Ans. (a)

Sol. The atomic absorption spectroscopy, the atomization process utilizes flame (flame atomizer).

Correct option is (a)

13. At room temperature, the number of singlet resonances observed in the ¹H NMR spectrum of Me₃CC(O)NMe₂ (N N–dimethyl pivalamide) is ________

Ans. 3

Sol.

At room temperature

Restricted rotational three singlet

Correct answer is (3)

14. Amongst the following, the metal that does NOT form homoleptic polynuclear carbonyl is

(a) Mn

(b) Fe

(d) Co

Ans. (c)

Sol. Cr metal is unable to form polynuclear metal carbonyl complexes due to steric hindrance.

Correct option is (c)

15. The reaction of [Cp₂TaMe₂]I (Cp = C₅H₅^–) with NaOMe yields.

(a) [Cp₂Ta(OMe)₂]I

(b) [Cp₂Ta(Me)OMe]I

(d) Cp₂Ta(OMe) = CH₂

Ans. (c)

Sol.

Correct answer is (c)

16. The complexes [Co(H₂O)₄Cl₂]NO₂ and [Co(H₂O)₄Cl(NO₂)]Cl are

(a) linkage isomers

(b) positional isomers

(d) optical isomers

Ans. (c)

Sol. The given two complexes give different ions when subjected to ionisation.

Thus, these are called ionisation isomers.

Correct option is (c)

17. The major product of the following reaction is

(a)

(b)

(c)

(d)

Ans. (c)

Sol.

Correct answer is (c)

18. Amongst the following, the structure of guanosine is

(a)

(b)

(c)

(d)

Ans. (d)

Sol.

19. The correct order of IR stretching frequency of the C=C in the following olefins is

(a) I > II > III

(b) II > III > I

(d) III > I > II

Ans. (c)

Sol. Increasing in the size decreasing IR (C = C) streching frequency)

(III) > (II) > I

Correct option is (c)

20. The correct order of the solvolysis for the following chlorides in acetic acid is

(a) II > I > III

(b) III > II > I

(d) I > III > II

Ans. (b)

Sol.

Reaction proceed through S_N1 mechanism, formation of 2° carbocation.

Reaction occurs through NGP, formation of four and Six membered ring produce formation of four membered ring in less feasible which is not stable product. Also, –I effect of O-atom destabilised the carbocation.

Reaction proceed through NGP formation five, five ring membered ring formation.

So correct order of solvolysis

III > II > I

Correct option is (b)

21. Formation of the product in the following photochemical reaction involves

(a) rearrangement

(b) Paterno-Buchi reaction

(d) Norrish type I reaction

Ans. (a)

Sol.

22. The correct order of stability for the following conformations of cyclohexane is

(a) I > II > III

(b) I > III > II

(d) III > I > II

Ans. (b)

Sol. Twist boat form is free from flagpole interaction. Boat form is less stable than twist boat form due to presence of flagpole interaction. Half chair is less stable due to very steric occur. Half chair contain angular strain and eclipsing effect. But boat form contain 1, 4 flagpole interaction and edipsing effect, so less stable. But in twist boat form both facts decrease. So twist boat forms is more stable than boat form.

Correct option is (b)

23. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. (c)

Sol.

24. The overall yield (in %) for the following reaction sequence is _______

Ans. (57.6)

Sol. Step I Reaction (A) is 100%, react with reagent observed yield is 90%.

Step II 90% yield of B react with reagent, observed yield is 72%

Step II 72% yield react with reagent C observed yield is 57.6

Finally observed yield = 57.6

Correct option is (57.6)

25. The most suitable reagent combination to effect the following conversion is

(a) (i) NaH, CS₂, then MeI; (ii) Bu₃SnH, AlBN, C₆H₆, Reflux

(b) (i) I₂, PPh₃, imidazole; (ii) H₂, 10% Pd-C, AcOH, high pressure

(d) (i) MsCl, pyridine, DMAP; (ii) LiAlH₄, THF, reflux

Ans. (a)

Sol.

Q.26 – Q.55: Carry TWO marks each.

26. is a proposed hydrogenic wavefunction, where Z = atomic number, r = radial distance from the nucleus, azinuthal angle, N is a constant. The INCORRECT statement about

(a) in the xy-plane

(b) two radial nodes are present in

(d) the size of the orbital decreases with increase in atomic number

Ans. (b)

Sol.

The power of one angular node

two radical node not prasent on eV radial node Z = r/6

The size of the orbital decrease with increase in active number

Correct option is (b)

27. The van der waals constant a and b of CO₂ are 3.64 L² bar mol^–2 and 0.04 L mol^–1, respectively. The value of R is 0.083 bar dm³ mol^–1 K^–1. if one mole of CO₂ is confined to a volume of 0.15L at 300K, then the pressure (in bar) exerted by the gas, is ________.

Ans. (60 – 66)

Sol. We know that van der Waals' equation for 1 mole of a gas is

On substituting values, we get

28. A plot of osmotic pressure against concentration (gL^–1) of a polymer is constructed. The slope of the plot

(a) increases with increase in temperature

(b) increases with increase in molar mass of the polymer

(d) decreases with increase in temperature.

Ans. (a)

Sol.

The slope of the plot increases with increase in temperature

Correct option is (a)

29. A platinum electrode is immersed in a solution containing 0.1 M Fe²⁺ and 0.1 M Fe³⁺. Its potential is found to be 0.77V against SHE. Under standard conditions and considering activity coefficients to be equal to unity, the potential of the electrode, when the concentration of Fe³⁺ is increased to 1 M, is _____

Ans. (0.8291)

Sol.

E = 0.8291V

Correct option is (0.8291).

30. Molybdenum crystallizes in a bcc structure with unit cell dimensions of 0.314 nm. considering the atomic mass of molybdenum to be 96, its density (in kg m^–3) is __________

Ans. (10296.7)

Sol.

Correct answer is (10296.7).

31. The ratio of molecules distributed between two states is 9.22 × 10⁶ at 300K. the difference in energy (in kJ mol^–1) of the two states is _________.

Ans. (39.98)

Sol.

16.03 × 1.38 × 10^–25 × 3 J × 6.023 × 10²³

399.8 × 10² J = 39.98 KJ/mole

Correct answer is (39.98)

32. A Carnot engine operates at 55% efficiency. If the temperature of reject steam is 105ºC, then the absolute temperature of input steam is __________

Ans. (840)

Sol.

33. Of the following plots, the correct representation of chemical potential against absolute temperature (T) for a pure substance is (S, L and g denote solid, liquid and gas phases, respectively)

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

34. The enthalpy of fusion of ice at 273K is 6.01 kJ mol^–1 and the enthalpy of vaporization of water at 273K is 44.83 kJ mol^–1. The enthalpy of sublimation (in kJ mol^–1) of ice at 273K, is ________

Ans. (50.84)

Sol.

= 6.01 + 44.83

35. Suppose are two hybrid orbitals:

The angle (in degrees) between them is ________

Ans. (179)

Sol.

36. BCl₃ and NH₄Cl were heated at 140ºC to give compound X, which when treated with NaBH₄ gave another compound Y. Compounds X and Y are

(a) X = B₃N₃H₃Cl₃ and Y = B₃N₃H₆

(b) X = B₃N₃H₉Cl₃ and Y = B₃N₃H₆

(d) X = B₃N₃Cl₆ and Y = B₃N₃H₆

Ans. (a)

Sol.

37. The number of microstates in term ¹G is

(a) 13

(b) 11

(d) 7

Ans. (c)

Sol. (2S + 1) = 1 S = 0

l = 4[for G]

Microstate = (2S + 1) (2l + 1)

1 × (2 × 4 + 1) = (9)

Correct option is (c)

38. The set of protons (underlined) in CH₃CH₂CH₂OCH₃ that would exhibit different splitting patterns in high (500 MHz) and low (60 MHz) field ¹H NMR, is

(a)

(b)

(c)

(d)

Ans. (b)

Sol.

The underlined proton shows different signal in NMR at 500 MHz and 60 mHz

Correct option is (b)

39. Amongst the following, the complex ion that would show strong Jahn-Teller distortion is

(a) [Cr(H₂O)₆]²⁺

(b) [Ti(H₂O)₆]³⁺

(d) [Fe(H₂O)₆]²⁺

Ans. (a)

Sol. The compounds having unsymmetrically filled e_g orbitals show strong J.T.D.

40. Amongst the following, the metal carbonyl species having the highest stretching frequency is

(a) [Mn(CO)₆]⁺

(b) Cr(CO)₆

(d) [Fe(CO)₄]^2–

Ans. (a)

Sol. The stretching frequency bond order positive oxidation state.

Or we can say greater the negative charge on metal greater will be back bonding and hence lesser will be new CO.

Correct option is (a)

41. The correct order of thermal stability for the given compounds is

(a) TiMe₄ > Ti(CH₂CMe₃)₄ > TiEt₄

(b) TiEt₄ > Ti(CH₂CMe₃)₄ > TiMe₄

(d) Ti(CH₂CMe₃)₄ > TiMe₄ > TiEt₄

Ans. (d)

Sol. The stability of compound decreases due to presence of Transition metal possess vaccant vaccant d-orbital which involve in kinetically facile reaction such as

The driving force for elimination reaction is the formation of stronger M – H bond than M-alkyl bond.

Correct option is (d)

42. Amongst the following, the complex ion that is expected to show the highest magnetic moment at room temperature is

(a) [Ni(CN)₄]^2–

(b) [Fe(CN)₆]^3–

(d) [Co(CN)₆]^3–

Ans. (b)

Sol. dsp²-square planar zero unpaired electron.

unpaired electron is one and t_2g is unsymmetrically filled so orbital contribution.

43. MnCr₂O₄ is

(a) normal spinel with total CFSE of –15.5 Dq

(b) inverse spinel with total CFSE of –15.5 Dq

(d) inverse spinel with total CFSE of –24 Dq

Ans. (c)

Sol. MnCr₂O₄ is normal spinel because Mn²⁺ having CFSE less than Cr³⁺ in Oh site.

Total CFSE = CFSE of Mn²⁺ + 2CFSE of Cr³⁺

Correct option is (c)

44. Mg²⁺ is preferred in photosynthesis by chlorphyll because

(a) it has strong spin-orbit coupling

(b) it has weak spin-orbit coupling

(d) it binds strongly with chlorophyll

Ans. (d)

Sol. In photosynthesis, Mg²⁺ present in chlorophyll. Because it binds strongly with chlorophyll.

Correct option is (d)

45. In Monsanto acetic acid process shown below, the role of HI is

(a) to convert CH₃OH to a stronger nucleophile (CH₃O^–)

(b) to reduce the Rh(I) catalyst to a Rh(0) species

(d) to convert CH₃OH to CH₃I

Ans. (d)

Sol. In monsanto acetic and process role of HI is to convert CH₃OH to CH₃I

46. Formation of the ketone H from the diazoketone I involves

(a) generation of carbene and a [2, 3]-sigmatropic rearrangement

(b) generation of carbene and an electrocyclic ring closing reaction

(d) generation of ketene and a [3, 3] sigmatropic rearrangement

Ans. (d)

Sol.

47. The major products X and Y formed in the following reaction sequence are

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

Correct answer is (a)

48. The major products X and Y formed in the following reactions are

(a)

(b)

(c)

(d)

Ans. (b)

Sol.

49. The major product X and Y formed in the following reaction sequence are

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

Correct answer is (a)

50. The product of the following reaction gave 6 line ¹³C NMR spectrum with peaks at 46, 37, 33 ppm. The structure of the product is

(a)

(b)

(c)

(d)

Ans. (c)

Sol.

Correct answer is (c)

51. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. (c)

Sol.

Correct answer is (c)

52. The major products X and Y formed in the following reaction sequence are

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

Correct answer is (a)

53. The major products X and Y formed in the following reaction sequence are

(a)

(b)

(c)

(d)

Ans. (b)

Sol.

Correct answer is (b)

54. Given the fact that 1, 3-butadiene has a UV absorption of 217nm, the absorption wavelength (in nm) for the conjugated system shown below is ______________