PYQ 2010 - VedPrep

GATE Chemistry 2010
Previous Year Question Paper with Solution.

1. The Lewis acidity of BF₃ is less than BCl₃ even though fluorine is more electronegative than chlorine. It is due to

(a) stronger 2p (B) – 2p (F) – bonding

(b) stronger 2p (B) – 2p (F) – bonding

(d) stronger 2p (B) – 2p (Cl) – bonding

Ans. (b)

Sol. In Boron Halides back donation takes place between boron and halogen This is

Back donation which is maximum in B–F it is while in B–Cl it is hence weaker in B–Cl. Therefore, electron deficiency of boron is compensated by 'F' more as compare to 'Cl'. Henc BF₃ is weaker lewis acid than BCl₃.

Correct option is (b)

2. Pyrozenes are a class of silicate minerals, which exhibit a polymeric chain structure, as shown below

Its simplest repeat unit is

(a) [SiO₄]^4–

(b) [SiO₃]^2–

(d) [Si₄O₁₁]^6–

Ans. (b)

Sol. Single chain silicate are called pyroxenes and are formed by sharing of oxygen atoms on two corners of each tetrahedron. They have general formula of if n = 1, [SiO₃]^2–

Correct option is (b)

3. Among the following pentachlorides the one which does not exist due to the 'inertpair effect' is

(a) PCl₅

(b) BiCl₅

(d) AsCl₅

Ans. (b)

Sol. As we move downward in group the inertness of ns² electron pair increases (called inert pair effect). This IPE will be maximum in last element of group. Hence BiCl₅ is least stable.

Correct option is (b)

4. Band theory predicts that magnesium is an insulator. However, in practice it acts as a conductor due to

(a) presence of filled 3s orbital

(b) overlap of filled 2p and filled 3s orbital

(d) presence of unfilled 3p orbital

Ans. (c)

Sol. The electric configuration of Mg–3s²3p°(Ne). The energy gap between 3s and 3p is small hence they the 3s orbital and can overlap with 3p orbital band. So, pertubation from filled valence band to empty conduction band can take place, making magnesium conductor.

Correct option is (c)

5. The number of 'framework electron pairs' present in the borane cluster [B₁₂H₁₂]^2– is:

(a) 10

(b) 11

(d) 13

Ans. (c)

Sol. Framework electron pairs are calculated by wade rules as

FEP =

Where, B = number of boron atoms

H = number of hydrogen atoms

C = number f carbon atom

x = negative charge

n = B + C

Correct option is (c)

6. The reaction between [PdCl₄]^2– and C₂H₄ produces a new compound. Compare to free C₂H₄, the C – C bond order of the product is:

(a) between 1 and 2

(b) less than 1

(d) greater than 2

Ans. (a)

Sol.

The C–C bond of ethelene become large when coordinate with Pd because, the electron density from metal to shift in to orbital of ethylene.

Correct option is (a)

7. Among the following pair of metal ions present in Nature, the first one functions as an electron transfer agent and the second one catalyzes the hydrolysis reactions. The correct pair is

(a) Fe and Zn

(b) Mg and Fe

(d) Ca and Cu

Ans. (a)

Sol. Fe act as electron transfer agent in many reaction. Zn act as catalyst in hydrolysis reaction specially enzymatic reaction by carboxypeptidase.

Correct option is (a)

8. Structurally nickellocene is similar to ferrocene. Nickellocene attains stability due to the formation of

(a) a monocation

(b) a dication

(d) a dianion

Ans. (b)

Sol. Nickellocene (Cp₂Ni) is 20e^– species. The electronic configuration of .

Nickellocence are less stable and easily oxidised or loss two electron from antebonding to give 18 electron ion.

Correct option is (b)

9. The absolute configurations for compounds X and Y, respectively are

(a) R, S

(b) S, R

(d) S, S

Ans. (b)

Sol.

Correct option is (b)

10. In the reaction,

the major product [X] is

(a)

(b)

(c)

(d)

Ans. (c)

Sol.

Correct option is (c)

11. Among the following, a pair of resolvable configurational enantiomers is given by

(a) cis – 1, 2 – dimethylcyclohexane

(b) cis – 1, 3 – dimethylcyclohexane

(d) trans – 1, 3 – dimethylcyclohexane

Ans. (d)

Sol. (1) Cis : 1-2 dimethylohexane

In Cis 1-2 dimethylcyclohexane, one4 enantiomers is converted into another enantiomers by Ring fiping. So they are non resolvable.

(2) Cis 1-3 dimethylcyclohexane

(3) Cis 1, 4 dimethyl cyclohexane

(4) Trans 1,3 dimethyl cyclohexane.

One can't converted by Ring fliping into other. So they are resolvable.

Correct option is (d)

12. In the reaction,

the major product [X] is:

(a)

(b)

(c)

(d)

Ans. (b)

Sol.

One of the form of trans 2-amino cyclo hexanol.

Since, the conformation e, e is more stable.

Correct option is (b)

13. The decreasing order of isoelectric point for the following α – amino acids is

Lysine Alanine Glutamic acid

(I) (II) (III)

(a) I > II > III

(b) II > I > III

(d) I > III > II

Ans. (a)

Sol. Isoelectric point of basic aminoacid ~ 9 – 10

Isoelectric point of acidic amino acid ~ 3 – 3.5

Isoelectric point of neutral amino acid ~ 5.5 ~ 6

So, Lysine basic amino acid

Alanine neutral amino acid

Glutamic acid acidic amino acid

So, the order is I > II > III

Correct option is (a)

14. The decreasing order of the reactivity of the following compounds towards electrophiles is

(a) II > I > III

(b) II > III > I

(d) I > II > III

Ans. (d)

Sol.

Pyrrole is more activated because the electron density present on the nitrogen involves in the aromatic sextet and increases the electron density in the ring whereas thiophene have vacant d-orbital. So, it withdraw some% of electron density towards itself and also overlaping of 2p of C and 3p of 'S' atom is slightly poor than that of 2p of 'C' and 2p of 'N' of pyrrole. So, less electron density available in thiophene ring. Wheareas in case of pyridine 'N' atom reduce the electron density.

Pyrrole > Thiophene > Pyridine

Correct option is (d)

15. In the reaction,

(a)

(b)

(c)

(d)

Ans. (a)

Sol. It is an example of ene reaction.

Correct option is (a)

16. The decreasing order of acidity of marked H of the following molecules is

(a) I > II > III

(b) III > I > II

(d) II > I > III

Ans. (c)

Sol.

As the value of pKa decreases acidity increases

As the stability of conjugate bas increases acidity increases.

So, the order of acidity is III > II > I.

Correct option is (c)

17. The decreasing order of nucleophilicity for the following anions is

(a)

(b)

(c)

(d)

Ans. (c)

Sol. Nucleophilicity depends on donor ability.

As the stability of negative charge increases its nucleophilic behaviour decrease.

Correct order of nucleophilicity

Correct option is (c)

18. The molar entropy of crystalline CO at absolute zero is

(a) Zero

(b) –Rln 2

(d) 2Rln 2

Ans. (a)

Sol. S = k In w

S = k In (arrangement)^NA

S = k. NA In (2)

Correct option is (a)

19. For an ideal gas

(a)

(b)

(c)

(d)

Ans. (b)

Sol.

Correct option is (b)

20. Among W (work), Q (heat), U (internal energy) and S (entropy)

(a) W and U are path functions but Q and S are state functions.

(b) W and S are path functions but Q and U are state functions.

(d) W and Q are path functions but U and S are state functions.

Ans. (d)

Sol. Internal energy u and entropy S are state function as they depend only on the initial and final state of the system. They follow Euler thorem and for this cyclic integral is zero.

But W and U are path and depend only the path followed not follow Euler theorem cyclic integral is not zero.

Correct option is (d)

21. For eigen functions and os particle in a 1 – D box of length

(a) is normalized but not orthogonal to

(b) is normalized but not orthogonal to

(d) is neither normalized nor orthogonal to

Ans. (c)

Sol.

Correct option is (c)

22. The bond order of C₂ molecule is

(a) 0

(b) 1

(d) 3

Ans. (c)

Sol. According to MOT the electronic configuration of C₂ molecule is

Bond order = = 2

Correct option is (c)

23. Sulfur can exist in four phases. The possible number of triple points is

(a) 1

(b) 2

(d) 4

Ans. (d)

Sol.

Possible number of triple points is 4

Three are 3 stable triple point (B, C and D) and one is metastable triple point (x).

Correct option is (d).

24. The standard reduction potentials at 298 K for single electrodes are given below:

Electrode Electrode Potential (volt)

Mg²⁺ / Mg – 2.34

Zn⁺² / Zn – 0.76

Fe⁺² / Fe – 0.44

From this we can infer that

(a) Zn can reduce both Mg²⁺ and Fe²⁺

(b) Fe can reduce both Mg²⁺ and Zn²⁺

(d) Mg can reduce Zn²⁺ but not Fe²⁺

Ans. (c)

Sol. R.P = Fe > Zn > Mg

Metals with lower standard reduction potential can displace the metal with higher standard reduction potential from their solution. Mg with lower standard reduction potential (–2.34) can reduce both Zn²⁺ and Fe⁺² ions.

Correct option is (c)

25. For the pair of reactions given below

If at a particular temperature, K_P1 and K_P2 are the equilibrium constants for reactions (i) are (ii) respectively, then

(a) K_P1 = 2K_P2

(b)

(c)

(d)

Ans. (c)

Sol.

Correct option is (c)

26. According to VSEPR model, the shape of [XeOF₅]^– is

(a) Octahedral

(b) Trigonal bipyramidal

(d) pentagonal monopyramidal

Ans. (d)

Sol.

Correct option is (d)

27. The number of unpaired electron(s) present in the species [Fe(H₂O)₅(NO)]²⁺ which is formed during 'brown ring test' is

(a) 2

(b) 3

(d) 5

Ans. (b)

Sol. In [Fe(H₂O)₅(NO)]²⁺, Fe is in +1 oxidation state with five H₂O neutral ligands and one unipositive NO⁺ ligand. During the formation of the complex, NO gives its unpaired electron to Fe²⁺ and change it into Fe⁺.

Fe = [Ar] 3d⁶ 4s²

Fe⁺ = [Ar] 3d⁷ 4s⁰

Correct answer is (b)

28. Fe₃O₄ and Co₃O₄ are metal oxides having spinel structure. Considering their CFSE, the correct statement regarding their structure is

(a) both have normal spinel structure

(b) both have inverse spinel structure

(d) Fe₃O₄ has a inverse and Co₃O₄ are normal spinels

Ans. (d)

Sol. In mixed oxide

Fe⁺³ is d⁵ system with zero cfse in octahedral void whereas Fe⁺² is a d⁶ with –0.4 cfse value in octahedral void so preferentially occupy octahedral site to give inverse spinel.

In mixed oxide

Co⁺³ is a d⁶ system with configuration in strong field octahedral O^2– voids to give max CFSE of –2.4 and hence this oxide is normal spinel.

Correct option is (d)

29. The mechanism of the reaction between [Fe(CN)₆]^4– and [Fe(bpy)₃]³⁺ (bpy = 2,2'-bipyridine) is

(a) outer-sphere electron-transfer

(b) inner-sphere electron-transfer

(d) ligand exchange followed by electron-transfer

Ans. (a)

Sol. The mechanism of the reaction between [Fe(CN)₆]^4– and [Fe(bpy)₃]³⁺ is outer sphere electron transfer.

Correct option is (a)

30. The d-d absorption band of [Fe(H₂O)₆]²⁺ is split due to

(a) presence of octahedral geometry

(b) static Jahn-Teller distortion

(d) Presence of trigonal bipyramidal geometry

Ans. (c)

Sol. Dynamic Jahn-Teller distortion is exhibited when there is a small distortion.

In [Fe(H₂O)₆]²⁺, distortion is small, since degeneracy occurs in t_2g orbitals, so it shows dynamic Jahn-Teller distortion.

Correct option is (c)

31. The crystal field symbol for the ground state of [Mn(CN)₆]^4– i

(a) ²T_2g

(b) ¹A_1g

(d) ⁶A_1g

Ans. (a)

Sol. [Mn(CN)₆]^4–

Mn = 3d⁵4s²

Mn²⁺ = 3d⁵

The crystal field symbol for ground state [Mn(CN)₆]^4– is ²T_2g.

Correct option is (a)

32. In the following reactions:

the reagent / conditions X and Y are

(a) X = BF₃; Y = heating at 1250ºC

(b) X = NaF; Y = heating at 25ºC

(d) X = CF₃SO₃H; Y = H₂SO₄

Ans. (b)

Sol.

Correct option is (b)

33. [CoCl₄]^2– is a blue coloured complex. Controlled – treatment of this complex with watger generates two isomeric light pink coloure complexes of composition [Co(H₂O)₄Cl₂]. Identify the correct point groups for [CoCl₄]^2– and two isomeric complexes [Co(H₂O)₄Cl₂].

(a) D_2h and (C_2v and C_2h)

(b) T_d and (C_2v and D_4h)

(d) T_d and (C_2v and C_4v)

Ans. (b)

Sol. [CoCl₄]^–2 sp³ tetrahedral geometry point group Td

[Co(H₂O)₄Cl₂] two isomers

Correct option is (b)

34. In the reaction

the major product [X] is:

(a)

(b)

(c)

(d)

Ans. (b)

Sol.

Correct option is (b)

35. In the reaction,

the major product [X] is:

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

It is an example of oxy cope rearrangement.

Correct option is (a)

36. In the following reaction sequence

(a)

(b)

(c)

(d)

Ans. (c)

Sol.

Correct option is (c)

37. In the reaction,

the major products, [X] and [Y], respectively, are

(a)

(b)

(c)

(d)

Ans. (b)

Sol.

Correct option is (b)

38. In the reaction,

the major product [X] is

(a)

(b)

(c)

(d)

Ans. (b)

Sol.

Correct option is (b)

39. In the reaction sequence

the major products, [X] and [Y] respectively, are

(a)

(b)

(c)

(d)

Ans. (d)

Sol.

Correct option is (d)

40. The change in entropy when two moles of Argon gas are heated at constant volume from 300 K to 500 K is:

(a) –12.74 J K^–1 mole^–1

(b) – 6.37 J K^–1 mole^–1

(d) 12.74 J K^–1 mole^–1

Ans. (c)

Sol.

= 3 × 8.314 × 2.303 log 5 – log 3

= 3 × 8.314 × 2.303 [0.6990 – 0.4771] = 3 × 8.314 × 2.303 × 0.2219

For 2 moles, = 12.74 J/K

For 1 mole = = 6.37 J/K – mole

Correct option is (c)

41. At any temperature T, fugacity coefficient is given by

where Z is the compressibility factor. The fugacity coefficient of a real gas governed by equations of state, P (V – b) = RT with'b' a constant is given by

(a)

(b)

(c)

(d)

Ans. (d)

Sol.

Correct option is (d)

42. The specific rate constant of decomposition of compound is represented by

The activation energy of decomposition for this compound at 300 K is

(a) 24 kcal/mole

(b) 12 kcal/mole

(d) 12 cal/mole

Ans. (a)

Sol. In k = 5.0 – ...(i)

From Arrhenius equation

In k = In A – ...(ii)

Comparing both equation, we get

E_a = 12000 × 2 = 24.00 kcl/mol

Correct option is (a)

43. The commutator [x³, p_x] is equal to

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

Correct option is (a)

44. An electron of mass 'm' is confined to a one dimensional box of length 'b'. If it makes a radiative transition from second excited state to the ground state, the frequency of the photon emitted is

(a)

(b)

(c)

(d)

Ans. (c)

Sol.

Correct option is (c)

45. The point group of ClF₃ molecule and its corresponding number of irreducible representation are respectively.

(a) C_3v and 4

(b) C_2v and 4

(d) C_2v and 3

Ans. (b)

Sol.

For C_2v point group, four irreducible representation

Correct option is (b)

46. The most populated rotational state for HCl(B = 8.5 cm^–1) at 300 K is:

(a) 2

(b) 3

(d) 7

Ans. (b)

Sol.

Correct option is (b)

47. The ratio of life times of two states that gives rise to line widths of 1.0 cm^–1 and 0.2 cm^–1 respectively is:

(a) 1 : 2

(b) 1 : 5

(d) 5 : 1

Ans. (b)

Sol.

T₁ : T₂ = 1.5

Lifetime ratio

Correct option is (b)

Common Data for Q. 48 and Q. 49

A six – coordinate transition – metal complex is ESR and Mossbauer active. The effective magnetic moment of this complex is – 5.9 B.M.

48. The metal – ion along with its oxidation state and the number of unpaired electron present are

(a) Fe(II) and 4

(b) Mn(II) and 5

(d) Fe(III) and 5

Ans. (d)

Sol. The effective magnetic moment ~ 5.9 B.M. means it has 5 unpaired electron, Mn(II) and Fe(III) both contains five unpaired electron and both are ESR active (due to presence of unpaired electrons) in 3d-series only Fe, Ni, Zn are Mossbauer active elements. So, Fe(III) is the element which are ESR and Mossbauer active

Correct option is (d)

49. The complex is

(a) [Mn(H₂O)₆]²⁺

(b) [Fe(CN)₆]^3–

(d) [Fe (H₂O)₆]³⁺

Ans. (d)

Sol. [Fe (H₂O)₆]³⁺

Fe = 3d⁶ 4s²

Fe³⁺ = 3d⁵

Five unpaired electron

Correct option is (d)

Common Data for Q. 50 and Q. 51

An organic compound [X](C₁₂H₁₆O₃) exhibits the following spectral data IR : – 1720 cm^–1

¹H NMR : 2.35 (s, 6H), 3.30 (s, 3H), 3.83 (t, 2H), 4.42 (t, 2H), 7.07 (s, 1H), 7.58 (s, 2H)

The compound [X] with an excess of MeMgBr gives a 1 : 1 mixture of compounds [Y] and [Z]. The compound [Z] exhibits the following ¹H NMR data: 2.0 (bs, 1H), 3.30 (s, 3H), 3.56 (t, 2h), 3.70 (t, 2H)

50. The compound [X] is:

(a)