PYQ 2011 - VedPrep

IIT JAM Chemistry 2011
Previous Year Question Paper with Solution.

1. The pair of semimetals in the following is:

(a) Al, Si

(b) Ge, As

(d) Ca, B

Ans. c

Sol.

Correct option is (c)

2. The most probable oxidation states for both Cr and Mo are

(a) +2, +3, +4

(b) +2, +3, +5

(d) +3, +4, +5

Ans. c

Sol. +2 oxidation state, CrX₂, MoX₂(where X = F, Cl, Br, I)

+3 oxidation state, CrX₃, MoX₃, CrO₃

+6 oxidation state, CrO₃, CrO₄^2–, MoO₃

Correct option is (c)

3. The correct order of acidic character is:

(a) Al₂O₃ > MgO > SiO₂ > P₄O₁₀

(b) P₄O₁₀ > Al₂O₃ > MgO > SiO₂

(d) SiO₂ > P₄O₁₀ > Al₂O₃ > MgO

Ans. c

Sol. Oxides of metal are basic in character and also higher the electropositive character of metal, higher will be basic character of the oxide, so MgO is more basic than Al₂O₃.

Oxides of non metals are acidic character.

Correct option is (c)

4. The pair of amphoteric oxides is:

(a) VO, Cr₂O₃

(b) V₂O₃, Cr₂O₃

(d) V₂O₅, CrO₃

Ans. c

Sol. Amphoteric oxides get dissoved in both acids as well as bases.

VO dissolves in acids but not in bases.

V₂O₃ dissolves in acids but not in bases.

VO₃ dissolves in acids as well as bases, so it is an amphoteric oxide.

V₂O₅ dissolves in acids as well as bases, so it an amphoteric oxide.

Cr₂O₃ dissolves in acids as well as bases, so it is an amphoteric oxide.

CrO₃ dissolves only in bases, but not in acids.

Correct option is (c)

5. In the structure of B₄O₅ (OH)₄^2–

(a) All four B atoms are trigonal planar

(b) One B atom is tetrahedral and the other three are trigonal planar

(d) Two B atoms are tetrahedral and the other two are trigonal planar

Ans. d

Sol.

Boron with –ve charges, are tetrahedral and Boron with no charge are trigonal planar.

Correct option is (d)

6. The pH of an aqueous solution of Al³⁺ is likely to be

(a) Neutral

(b) Acidic

(d) Highly basic

Ans. b

Sol.

As above equilibrium reaction generates H₃O⁺, So aq. solution of Al⁺³ is acidic.

Correct option is (b)

7. Hydrolysis of (CH₃)₂ SiCl₂ and CH₃SiCl₃ leads to

(a) Linear chain and cross-linked silicones, respectively

(b) Cross-linked and linear chain silicones, respectively

(d) Cross-linked silicones only

Ans. a

Sol. Hydrolysis of (CH₃)₂SiCl₂ and CH₃SiCl₃ can be illustrated by following chemical reactions.

Correct option is (a)

8. The oxide that has the inverse spinel structure is

(a) FeCr₂O₄

(b) MnCr₂O₄

(d) Fe₂CoO₄

Ans. d

Sol. A compound AB₂O₄ (where A is in +2 and B in +3 o.s.), has inverse spinel structure if ccp lattice is made by oxide ions and

(i) A is occupying th of octahedral site.

(ii) Half of B is occupying th of octahedral site.

(iii) Rest half of of 'B' is occupying octahedral site.

Tetra hedral site is occupied by metal ion (B⁺³) which has lower C.F.S.E.

In Fe₂CoO₄, Fe⁺³ has zero C.F.S.E. So, one of Fe⁺³ occupies tetrahedral site and it acquires inverse spinel structure.

Correct option is (d)

9. The transition metal monoxide that shows metallic conductivity is:

(a) NiO

(b) MnO

(d) CoO

Ans. c

Sol. 3d metal mono-oxide like TiO and VO have high electrical conductivities that decrease with increasing temperature.

As in these compound conduction band is formed by the overlap of t_2g orbitals of metal ions in neighbouring octahedral sites that are oriented towards each other. The radial extension of the d-orbitals of these early d-block elements is greater than for element later in period, hence a band result from their overlap this band is only partly filled. Hence these oxide so metallic conductivity.

However in MnO, FeO, CoO and NiO, no such type of band formation result due to smaller radial extension of d-orbital, and they behaves as semiconductor.

[overlap of the d_zx orbitals in TiO to give a t_2g band. In perpendicular direction the d_yx and d_zx orbitals overlap in an identical manner]

Correct option is (c)

10. The metal that is extracted by the reduction method is:

(a) Al

(b) Au

(d) Mg

Ans. c

Sol. Metal Extraction method

Al Electrolysis of Al₂O₃ in moltel Na₃AlF₆.

Au Leading with CN^– first, so soluble complex [Au(CN)₂]^– is formed then reduction with Zn.

Hg Direct reduction of HgS by heat alone (HgS + O₂ Hg + SO₂)

Mg Electrolysis of fused MgCl₂ with added KCl.

Correct option is (c)

11. The most viscous liquid is

(a) Water

(b) Methanol

(d) Glycerol

Ans. d

Sol. Due to 3 O – H groups there will be large number of intermolecular H–bonding possible, which in turn increases the viscosity, as cross linked polymer like network is formed.

Correct option is (d)

12. In ammonical buffer, oxine (8-hydroxyquinoline) forms yellow precipitate with

(a) Mg(II)

(b) Ca(II)

(d) Sr(II)

Ans. a

Sol.

Correct option is (a)

13. Addition of an aqueous solution of Fe(II) to potassium hexacyanochromate (III) produces a brick-red colured complex, which turns dark green at 100ºC. The dark green complex is

(a) Fe₄[Cr(CN)₆]₃

(b) KFe[Cr(CN)₆]

(d) Fe[Cr(CN)₆]

Ans. c

Sol.

Correct option is (c)

14. In the following equation X is:

(a)

(b)

(c)

(d)

Ans. a

Sol.

Correct option is (a)

15. Based on the principle of equipartition of energy, the molar heat capacity of CO₂ at constant volume C_v,m is:

(a) 3.5 R

(b) 6R

(d) 9R

Ans. c

Sol. For CO₂, translational degree of freedom = 3

rotational degree of freedom = 2 (because molecule is linear)

and vibrational degree of freedom = 3N – 5 = 3 × 3 – 5 = 4 (because CO₂ is linear)

Therefore, total internal energy of CO₂ molecule, is

Correct option is (c)

16. One mole of a van der waals gas undergoes reversible isothermal transformation from an initial volume V₁ to a final volume V₂. The expression for the work done is:

(a)

(b)

(c)

(d)

Ans. d

Sol.

Correct option is (d)

17. The scalar product of two vectors u and v, where

(a) – 10

(b)

(c)

(d) 10

Ans. a

Sol. Given two vectors,

Scalar product, u . v = (2.1) + (3.1) + (–5.3) = 2 + 3 – 15 = –10

Correct option is (a)

18. The minimum concentration of silver ions that is required that is required to start the precipitation of Ag₂S(K_sp = 1 × 10^–51) in a 0.1 M solution of S^2– is

(a) 1 × 10^–49 M

(b) 1 × 10^–50 M

(d) 1 × 10^–25 M

Ans. d

Sol.

So, for the precipitation of Ag₂S, [Ag⁺] must be greater or equal to 10^–25 M.

Correct option is (d)

19. Identify the correct statement regarding Einstein's photoelectic effect

(a) The number of electrons ejected depends on the wavelength of incident radiation.

(b) Electron ejection can occur at any wavelength of incident radiation.

(d) The kinetic energy of the ejected electrons is independent of the wavelength of incident radiation.

Ans. c

Sol.

= threshold wavelength (wavelength above which electrons cannot be ejected)

K.E. is dependent on l_incident.

Number of ejected electrons depends on intensity of incident radiation only.

Correct option is (c)

20. The hydrolysis constant (K_h) of NH₄Cl is 5.6 × 10^–10. The concentration of H₃O⁺ in a 0.1 M solution of NH₄Cl at equilibrium is

(a)

(b)

(d) 2.8 × 10^–5

Ans. a

Sol.

As k_h is very small. So, neglecting

(Removing negative sign, as concentration can not be – ve)

[neglecting the first term of numerator].

Correct option is (a)

21. The acid dissociation constant (K_a) for HCOOH, CH₃COOH, CH₂ClCOOH and HCN at 25ºC are 1.8 × 10^–4, 1.8 × 10^–5, 1.4 × 10^–3 and 4.8 × 10^–10, respectively. The acid that gives highest pH at the equivalence point when 0.2 M solution of each acid is titrated with a 0.2 M solution of sodium hydroxide is:

(a) HCOOH

(b) CH₃COOH

(d) HCN

Ans. d

Sol. For any acid base titration,

Now, NaA can undergo hydrolysis and will give,

As Conc. of NaA is same in all cases, so higher pk_a (lower k_a) will have higher pH.

For HCN pH will be highest.

Correct option is (d).

22. For an ideal gas undergoing reversible Carnot Cycle, the plot of enthalpy (H) versus entropy (S) is:

(a)

(b)

(c)

(d)

Ans. b

Sol. (b) In adiabatic process, q_rev = 0

Correct option is (b)

23. Hybridization of the atoms indicated with the asterisk (*) in the following compounds sequantly are

(a) sp², sp², sp³, sp²

(b) sp², sp³, sp³, sp²

(d) sp², sp², sp³, sp³

Ans. a

Sol. Whenever there is possibility of resonance, then l.p. of oxygen align itself in a orbital such that it can participate in resonance (i.e. sidewide overlap of orbitals) is possible, which causes change in hybridisation of oxygen from sp³ to sp².

Correct option is (a)

24. The Cahn-Ingold-Prelog (CIP) priorities of the group and the absolute configuration (R/S) of the following compounds are

(a) CH₂OH > CH(CH₃)₂ > CH = CH₂ > CH₃ and S

(b) CH₂OH > CH = CH₂ > CH(CH₃)₂ > CH₃ and S

(d) CH₂OH > CH(CH₃)₂ > CH = CH₂ > CH₃ and R

Ans. b

Sol. Since 4th group (CH₃) is present at above the plane. So, absolute configuration is inverted.

Correct option is (b)

25. The optically active stereoisomer of the following compound is

(a)

(b)

(c)

(d)

Ans. b

Sol. Options given in (a), (c) and (d) has plane of symmetry, so they are optically inactive

Correct option is (b)

26. The correct relationship within each pair of the natural products is:

(a) Camphor – terpene; insulin – protein; nicotine – alkaloids; streptomycin – carbohydrate

(b) Camphor – terpene; insulin – carbohydrate; nicotine – alkaloid; streptomycin – lipid

(d) Camphor – carbohydrate; insulin – protein; nicotine – alkaloid; streptomycin – terpene.

Ans. a

Sol. Correct option is (a)

27. The correct sequence of relationship between the compounds of the following pairs i-iv is

(a) Identical, enantiomers, diastereomers and structural isomers.

(b) Enantiomers, identical, structural isomers and diastereomers.

(d) Identical, identical, diastereomers and structural isomers.

Ans. c

Sol. Correct option is (c)

28. The INCORRECT statement in the following is:

(a) The nucleobase pairs are aligned perpendicular to the helical axis in DNA.

(b) RNA contains uracil and thymine, but DNA contains only thymine.

(d) All enzymes are proteins, but all proteins are not necessarily enzymes.

Ans. b

Sol. RNA contains adenine guanine, uracil and cytosine, but DNA contains adenine, guanine, thymine and cytosine.

Correct option is (b)

29. The product P and Q in the following reactions, respectively, are

(a)

(b)

(c)

(d)

Ans. c

Sol. Chemical transformation involved in above chemical reaction can be illustrated as

Correct option is (c)

30. The major product in the following reaction is

(a)

(b)

(c)

(d)

Ans. c

Sol. Chemical transformation involved in above chemical reaction can be illustrated as

Correct option is (c)

31. (a) In the following reactions, identify X, Y and Z.

(b) Draw the structure of S₄N₄H₄ and N₄S₄F₄.

Sol. (a) Complete chemical reaction occurring in above chemical transformate can be shown as

32. (a) The magnetic moment of [Fe(phen)₂(NCS)₂] varies with temperature. The magnetic moments at 200 K and 50 K are 4.9 B.M. and 0 B.M., respectively. Write the d-electron configurations of Fe at both temperatures and give reason for the observed change in the magnetic moment.

(phen = 1, 10-phenanthroline)

(b) PCl5 exists as a discrete covalent molecule in the gaseous state, but is ionic in the solid state.

Draw the structures of PCl5 in gaseous and solid states.

Sol. (a) Magnetic moment of [Fe(phen)₂(NCS)₂] varies with temperature from 4.9 BM to 0 BM due to spin crossover from high spin to low spin state of Fe⁺² in octahedral field.

A combination with tetrahedral and octahedral structural units.

33. In the following equilibrium and reactions, identify species B to E.

Write the balanced chemical equation for the conversion of C to E.

Sol. Complete chemical reaction occurs in above chemical transformation can be shown as

34. (a) Identify species A and C in the following:

Write the balanced chemical equation for the conversion of A and A³⁺.

Hint: C on the dilution with water gives B

(b) Draw the structures of X and Y in the following reactions.

Sol.

35. (a) The molar conductances at infinite dilution for BaCl₂, KCl, K₂SO₄ and Cl^– are 280, 150, 300 and 76 W^–1 m² mol^–1, respectively. Calculate the transport number of Ba²⁺ in BaSO₄ solution at infinite dilution.

(b) If 4 moles of a MX₂ salt in 1 kg of water raises the boiling point of water by 3.2 K. Calculate the degree of dissociation of MX₂ in the solution.

Sol. (a) Transport number at infinite dilution for cation or anion is given by

36. (a) For the reaction the plot of ln[R] versus time (t) gives a straight line with a negative slope. The half life for the reaction is 3 minutes.

(ln 2 = 0.693, ln 0.1 = –2.303)

(i) Derivative the expression for t_1/2.

(ii) Calculate the slope of the straight line

(iii) Calculate the time required for the concentration of R to decrease to 10% of its initial value.

(b) Shown below is the Jablonski diagram that describes various photophysical processes. The solid arrows represent radiative transitions and the wave arrow represents a non-radiative transition.

(i) Name the photophysical pathways X, Y and Z.

(ii) Which of the radiative decays is faster?

Sol. (a) (i) For the reaction,

The plot of ln[R] vs t is straight line with –ve slope.

Above reaction is a first order reaction

On rearranging and then integrating we get,

Where [R]₀ = initial concentration of reactant.

[R] = Concentration of R at time t

(ii) Among radiative decays (Fluorescence and (ISC) phosphorescence), Fluorescence is faster because this transition is spin allowed, while Phosphorescence is spin forbidden.

37. (a) (i) Given that derive the expression for the temperature dependence of the cell potential (E) in terms of the change in entropy

(ii) For a cell reaction, E(at 25ºC) = 1.26 V, n = 2 and = –96.5 J K^–1 mol^–1. Calculate E at 85ºC by assuming to be independent of temperature. (F = 96500 C mol^–1).

(b) The phase diagram for the lead-antimony system at a certain pressure is given below.

(i) Identify the phases and components in region I and region II.

(ii) Calculate the number of degrees of freedom (variance) at point M.

Sol.

Substituting value from equation (1) in equation (2).

Temperature dependence of cell potential in terms of entropy

(ii) From relation derived, in part (ii), we have

(i) Region I has solid (Sb) and melt (Sb + Pb).

Components(c) = 2 (Sb and Pb)

Phases (p) = 2 (Solid and melt)

Region II has two solid phases having composition either Sb(solid) + Eutectic (solid)

Or Pb(solid) + Eutectic (solid), depending upon the mole fraction of Pb less than or more than eutectic composition composition mole fraction of Pb.

[i.e. (i) When X_Pb < X_{Pb, euectic} then Sb(s) + eutectic (s) phase.

(ii) When X_Pb > X_{Pb, eutectic} then Pb(s) + Eutectic (s) phase]

So, in region II,

C = 2 P = 2

(ii) From phase rule, F = C – P + 2

but at constant pressure,

F = C – P + 1

At point M, C = 2 [Sb and Pb]

P = 2 [Solid (Pb) and Liquid (Pb + Sb)]

F = 2 – 2 + 1

F = 1

Therefore, variance or degree of freedom at M = 1

38. (a) One mole of an ideal gas initially at 300 K and at a pressure of 10 atm undergoes adiabatic expansion.

(i) Reversibly and

(ii) Irreversibly against a constant external pressure of 2 atm until the final pressure becomes equal to the external pressure.

Calculate for (i) and (ii). For (ii), express the final answer in terms of R. Given: Molar heat capacity at constant volume C_v,m = 3R/2.

(b) For the following equilibrium at 300ºC.

Calculate K_p when N₂O₂ is 30% dissociated and the total pressure is 2 bar.

Sol. (a) From Clausius inequality

equality holds for reversible process i.e. when q = q_rev, then

(i) For reversible adiabatic process

q = q_rev = 0

(ii) For adiabatic irreversible process,

q = 0

As entropy is a state function, so it is independent of path followed (i.e. reversible or irreversible) if initial and final states are same.

Net entropy change can be written as,

39. (a) The Maxwell probability distribution of molecular speeds for a gas is:

where 'v' is the speed, 'm' the mass of the gas molecule and k the Boltzmann constant.

(i) Use F(v) to show that the most probable speed v_mp is given by the expression.

(ii) Use R = 8 J K^–1 mol^–1 in the above expression to calculate the v_mp for CH₄(g) at 127ºC.

(b) The wavefunction of a quantum state of hydrogen atom with principal quantum number n = 2 is:

(i) Identify the values of quantum numbers l and m and hence the atomic orbital.

(ii) Find where the radial node of the wavefunction occurs.

Sol. (a) From Maxwell probability distribution, the number of molecules having velocity between v and v + dv is,

So, velocity (v) at which F(v) will be maximum is said to be most probable velocity,

at v = v_m,p, F(v) is maximum

For F(v) to be maximum,

(ii) For CH₄(g)

M = 16 × 10^–3 kg/mol

T = 127º = 400 K

R = 8 J k^–1 mol^–1.

(i) As the above function is independent of angle (probability density) is also independent of angle. This is only possible for s-orbital (which is spherical)

and must have different sign (i.e. + ve or –ve for r < r_node and r > r_node)

40. (a) Write the possible substitution products in the following reactions. Indicate the types of mechanisms (S_N1/S_N2) that is/are operative in each reaction.

(b) Write the elimination products A to C in the following reaction. Identify the major product

Sol.

Only S_N2 mechanism is operating here because.

is a strong nucleophilic.

(b) DMF is an aprotic polar solvent in which nucleophilic does not get solvated.

(d) Allylic group increases the electropositive character (because, sp² carbon is more electronegative than sp³ carbon), so nucleophile can attach faster.