1. Arrange the following in the decreasing order of acidity of the hydrogen indicated in italic

(a) (ii) > (iii) > (i) > (iv)

(b) (iv) > (ii) > (iii) > (i)

(c) (iv) > (iii) > (ii) > (i)

(d) (ii) > (iv) > (iii) > (i)

Ans. b

Sol. Conjugate base of strong acid is weak (i.e. stable), so stability of conjugate base of CH₃COCH₃, CH₃COCH₂COCH₃, CH₃COOCCH₂COOCH₃ and CH₃COCH₂NO₂ can be arranged as

has greater electron withdrawing ability than Therefore, (ii) has greater acidity than (iii).

Correct option is (b)

2. For the reaction shown below if the concentration of KCN is increased four times, the rate of the reaction will be :

(a) doubled

(b) increased four times

(c) unaffected

(d) halved

Ans. c

Sol. This reaction follows S_N1 – pathway due to living group is present at tertiary carbon atom.

The rate of S_N1 reaction reactant.

So, rate will be independent of nucleophilic concentration.

Correct option is (c)

3. Benzyl chloride is reacted with different nucleophiles shown below. Arrange them in the decreasing order of reactivity.

Nucleophiles :

(a)

(b)

(c)

(d)

Ans. b

Sol. in PhO^– and CH₃COO^– negative charge on oxygen is delocalized via resonance, but CH₃COO^– has more delocalization than PhO^–, so CH₃COO^– is less nucleophilic than PhO^–,

Correct option is (b)

4. The rate of nitration of the following aromatic compounds decreases in the order

(i) benzene (ii) pyridine (iii) thiophene (iv) toluene

(a) (iv) > (i) > (iii) > (ii)

(b) (iii) > (iv) > (i) > (ii)

(c) (iii) > (ii) > (i) > (iv)

(d) (ii) > (i) > (iv) > (iii)

Ans. b

Sol. The rate of electrophilic substitution reaction is depends on the electron density in ring. Thiophene has most electron density in the ring and pyridine has least. Hence, the order towards the electrophilic substitution will be

thiophene > toluene > benzene > pyridine

Correct option is (b)

5. The major product formed in the reaction on 1, 3-butadiene with bromine is

(a) BrCH₂CH(Br)CH = CH₂

(b) CH₂ = CH— CH₂CH₂Br

(c) CH₂ = C(Br)—C(Br) = CH₂

(d) BrCH₂CH = CHCH₂Br

Ans. d

Sol.

Correct option is (d)

6. The reaction of (+)2-iodobutane and NaI* (I* is radioactive isotope of iodine) in acetate was studied by measuring the rate of racemization (k_r) and the rate of incorporation of I* (k_i).

(+)CH₃CH(I)CH₂CH₃ + NaI* CH₃CH(I*)CH₂CH₃ + NaI

For the reaction, the relationship between k_r and k_i is:

(a) k_i = 2 × k_r

(b) k_i = (1/2) × k_r

(c) k_i = k_r

(d) k_i = (1/3) × k_r

Ans. b

Sol.

When mole of NaI* reacts with (I), then it causes formation of mole of (II) (i.e. inversion product), but mole of (I) and mole of (II) , give 1 mole of racemic mixture (I)

So rate of racemisation is twice of inversion.

Correct option is (b)

7.

In the scheme shown above (P), (Q), (R) and (S) are

(a) (P) = purine bases, (Q) = pyrimidine bases, (R) = nucleotides, (S) = nucleosides

(b) (P) = nucleosides, (Q) = nucleotides, (R) = pyrimidine bases, (S) = purine bases

(c) (P) = nucleosides, (Q) = nucleotides, (R) = (S) = purine bases

(d) (P) = nucleotides, (Q) = nucleosides, (R) = pyrimidine base, (S) = purine base

Ans. d

Sol. Correct option is (d)

8. The products obtained from the following reaction are:

(a)

(b)

(c)

(d)

Ans. b

Sol.

Correct option is (b)

9. The product(s) obtained in the following reaction is (are)

(a)

(b)

(c)

(d)

Ans. c

Sol. It is an example of claisen rearrangement which includes 3, 3-sigmatropic rearrangement and phenol tautomerism as shown below.

Correct option is (c)

10. Match the isoelectric point with the amino acids.

(a) (X)-(II), (Y)-(III), (Z)-(I)

(b) (X)-(III), (Y)-(I), (Z)-(II)

(c) (X)-(I), (Y)-(II), (Z)-(III)

(d) (X)-(II), (Y)-(I), (Z)-(III)

Ans. a

Sol. As the number of basic (–NH₂) units increases, the pI value increases.

As the number of acidic (–COOH) units increases, the pI value decrease.

Correct option is (a)

11. The compound having the highest melting point is:

(a) LiCl

(b) LiF

(c) LiI

(d) LiBr

Ans. b

Sol. LiF has highest lattice energy and also maximum ionic character, so it has highest melting point.

Correct option is (b)

12. The shape of SF₄ is

(a) tetrahedral

(b) trigonal bipyramidal

(c) square planer

(d) octahedral

Ans. b

Sol.

Correct option is (b)

13. The degree of hydration is expected to be maximum for

(a) Mg²⁺

(b) Na⁺

(c) Ba²⁺

(d) K⁺

Ans. a

Sol. (a) Degree of hydration is proportional to charge/radius; here Mg²⁺ has highest charge as well as smallest radius.

Correct option is (a)

14. The decreasing order of the first ionization energy of the following elements is:

(a) Xe > Be > As > Al

(b) Xe > As > Al > Be

(c) Xe > As > Be > Al

(d) Xe > Be > Al > As

Ans. c

Sol.

Correct option is (c)

15. The radioactive isotope used to locate brain tumors is

(a)

(b)

(c)

(d)

Ans. c

Sol. Correct option is (c)

16. The crystal field stabilization energy of high spin d⁷ octahedral complex is:

(a)

(b)

(c)

(d)

Ans. a

Sol. d⁷ in octahedral field split as shown below

Two electrons are paired in t_2g set of AOs. So correct value of CFSE

Correct option is (a)

17. The complex with the most colour among the following is:

(a) [FeF₆]^3–

(b) [MnCl₄]^2–

(c) [CoCl₄]^2–

(d) [CoF₆]^3–

Ans. c

Sol. Most colour (or maximum intensity) is observed in the order

Charge transfer > dⁿ – tetrahedral > dⁿ – octahedral (non-centrosymmetry) >

dⁿ – octahedral (centrosymmetry, > d⁵ - tetrahedral > d⁵ - octahedral (non-centrosymmetry) > d⁵ - octahedral (centrosymmetry)

Correct option is (c)

18. On addition of a solution of AgNO₃ to a solution of Na₂S₂O₃, it turns black on standing due to the formation of:

(a) Ag

(b) Ag₂S

(c) Ag₂S₂O₂

(d) Ag₂SO₄

Ans. b

Sol. (b) Chemical reaction occuring in above reaction can be represented as

Correct option is (b)

19. Among the following complexes,

(i) [Ru(bipyridyl)₃]⁺

(ii) [Cr(EDTA)]^–

(iii) trans-[CrCl₂(oxalate)₂]^3–

(iv) cis-[CrCl₂(oxalate)₂]^3–

the ones that show chirality are

(a) (i), (ii), (iv)

(b) (i), (ii), (iii)

(c) (ii), (iii), (iv)

(d) (i), (iii), (iv)

Ans. a

Sol. Out of (i), (ii), (iii), (iv) only (iii) has got plane of symmetry, so it is not chiral

Correct option is (a)

20. The electronic configurations that have orbital angular momentum contribution in octahedral environment are

(a) d¹ and high spin d⁴

(b) d¹ and d²

(c) d² and high spin d⁶

(d) high spin d⁴ and high spin d⁶

Ans. b

Sol. To have an orbital contribution, t_2g orbitals must not be symmetrically occupied (i.e. it should not have configuration).

Correct option is (b)

21. For an ideal solution formed by mixing of pure liquids A and B.

(a)

(b)

(c)

(d)

Ans. a

Sol. For an ideal solution,

Correct option is (a)

22. The relationship between the equilibrium constant K₁ for the reaction :

and the equilibrium constant K₂ for the reaction:

(a) 2K₁ = K₂

(b) K₁ = K₂²

(c) K₁ = K₂

(d) K₁² = K₂

Ans. d

Sol. For reaction,

Equilibrium constant K₁ can be calculated as

23. For H-like atoms, the ground state energy is proportional to

(a)

(b)

(c)

(d)

Ans. c

Sol.

Correct option is (c)

24.

(a)

(b)

(c)

(d)

Ans. d

Sol.

On integrating by parts,

Correct option is (d)

25. From the reaction aA products, the plot of versus time (t) gives a straight line. Order of the reaction is

(a) 0

(b) 1

(c) 2

(d) 3

Ans. c

Sol.

here vs t graph is straight line, so from question, n = 2

Correct option is (c)

26. The pH of a solution prepared from 0.005 mole of Ba(OH)₂ in 100 cc water is

(a) 10

(b) 12

(c) 11

(d) 13

Ans. d

Sol.

Correct option is (d)

27. For an electron whose x-positional uncertainity is 1 × 10¹⁰ m, the uncertainity in x-component of the velocity in ms^–1 will be of the order of (Data: m_e = 9 × 10^–31 kg, h = 6.6 × 10³⁴ Js)

(a) 10⁶

(b) 10⁹

(c) 10¹²

(d) 10¹⁵

Ans. a

Sol.

Correct option is (a)

28. For the following system in equilibrium,

the number of components, (C), phases (P) and degrees of freedom (F), respectively, are

(a) 2, 2, 2

(b) 1, 3, 0

(c) 3, 3, 2

(d) 2, 3, 1

Ans. d

Sol.

Phase = 3

Component (C) = N – E

N = 3 (No. of species)

E = 1 Equilibrium

C = 2

F = C – P + 2

= 2 – 3 + 2 = 1

Correct option is (d)

29. For the distribution of molecular velocities of gases, identify the correct order from following (where v_mp, v_av, v_rms are the most probable velocity, average velocity, root mean square velocity, respectively):

(a) v_rms, v_av, v_mp

(b) v_mp, v_rms, v_av

(c) v_av, v_rms, v_mp

(d) v_mp, v_av, v_rms

Ans. a

Sol. Expression representing value of most probable, average velocity and root mean square velocity are as shown below.

Correct option is (a)

30. Given that is:

(a) 1.21 V

(b) 0.33 V

(c) –0.036 V

(d) 0.036 V

Ans. c

Sol.

Correct option is (c)

31. Identify the major product(s) formed in the following reactions. Intermediates and reaction mechanisms need not be discussed.

(a)

(b)

(c)

(d)

Sol.

32. How may the following transformations be effected? Indicate the reagents/reaction conditions clearly in each step.

(a) (Not involving any functional group transformation of the COOH group in the starting material)

(b) (Using diethyl malonate as the only source of carbon)

Sol.

33. Suggest a suitable mechanism for each of the following reactions.

Sol. (a) Chemical reaction occuring in above transformation can be illustrated as

34. Rationalize the following observations using suitable mechanism.

(a) Nitratino of 4t-butyltoluene gives 4-nitrotoluene as one of the products.

(b) cis-1-t-butylcyclohexyltrimethylammonium hydroxide undergoes Hoffmann elimination to yield 4-t-butylcyclohexene whereas the trans isomer does not (use conformations) explain.

Sol. (a) Chemial reaction occuring in above transformation can be illustrated as

As shown in the above mechanism, electrophilic substitution of t-butyl carbocation by H⁺, takes place, because tertiary - butyl carbocation is quite stable, so it is completetively replaced by H⁺.

cis-4t-butylcyclohexyltrimethylammonium hydroxide undergoes Hoffmann elimination, because 'H' anti to trimethylammonium is present.

Here 'H' anti to leaving group is not present, so elimination is not possible.

(c) Chemical reaction occuring in statement given above can be illustrated as

35. (a) Suggest a chemical method for the separation of a mixture contain p-N, N-dimethylaminophenol and p-aminobenzoic acid and give a confirmatory test for phenol.

(b) Write the structures of X, Y and Z in the following

Sol. (a)

36. (a) Predict the hybridization and draw the structure of the following molecules based on VSEPR theory (i) I₃^– (ii) SO₃^2– (iii) P(CH₃)₃F₂

(b) Explain why PCl₅ exists and PH₅ does not.

Sol.

Central 'I' has 3 l.p. and 2 b.p., So, it is sp³d hybridized.

(ii) SO₃^2–

'S' has '4' electron clusters surrounding it, so it is sp³ hybridized.

'P' has 5 b.p. of electron-cluster surrounding it.

So, it is sp³d hybridized.

(b) As phosphorous contains vacant '3d' orbital, So, it can expand its valency from 3 to 5 via utilization of 3d orbital.

In case of PCl₅, overlapping of sp³d orbital of phosphorous with 3p orbital of chlorine is strong enough to make a stable compound, but sp³d(p) and 1s(H) overlap is not so effective, So, PH₅ does not exist.

37. (a) Write balanced equations for the formation of

(i) P₂O₇^–4 from PO₄^–3 (ii) [(H₂O)₄Fe(OH)₂Fe(OH₂)₄]⁴⁺ from [Fe(OH₂)₆]⁺³

(b) Which one of the two solutions has lower pH? Justify your answer.

(i) 0.1 M Fe(ClO₄)₂ or 0.1 M Fe(ClO₄)₃

(ii) 0.1 M Hg(NO₃)₂ or 0.1 M Zn(NO₃)₂

Sol. (a) Balanced equation for the formation of P₂O₇^–4 from P₂O₄^–3 can we written as

(ii) formation of [(H₂O)₄Fe(OH)₂Fe(OH₂)₄]⁴⁺ from [Fe(OH₂)₆]⁺³ can be written as

As Fe⁺³ is harder acid then Fe⁺², So [Fe(H₂O)₆]⁺³ prefers over [Fe(H₂O)₆]⁺² to get deprotonated.

So, that it can bind with OH^– (because OH^– is harder base than H₂O).

So, Fe(ClO₄)₃ is stronger acid in water and shows lower pH than Fe(ClO₄)₂.

(ii) Hg(NO₃)₂ in aq. medium exist as [Hg(H₂O)₆]⁺ and 2NO₃^– and Zn(NO₃)(aq) exists as [Zn(H₂O)₆]⁺² and 2NO₃^–.

As Zn⁺² is a harder lewis acid than Hg⁺².

So, [Zn(H₂O)₆]⁺² prefers to donate H⁺. So that Zn⁺² can get bind to OH^– (because OH^– is harder lewis base than H₂O).

So, Zn(NO₃)(aq) has lower pH.

38. (a) Between Co(H₂O)₆²⁺ and Cu(H₂O)₆²⁺, which has more distorted structure and why?

(a) (b) Calculate CFSE and spin only magnetic moment for the following complexes.

(i) [CoF₆]^3– (ii) [Fe(CN)₆]^3– (iii) [NiCl₄]^2–

Sol. (a) Crystal field splitting diagram of [Co(H₂O)₆]⁺² can be drawn as

39. (a) The radioactive element Ra (Z = 86) emits three alpha particles in succession. Deduce in which group the resulting element will be found?

(b) A radioisotope sample has an initial activity of 23 dis/min. After 1/2 h, the activity is 11.5 dis/min. How many atoms of the radioactive nuclide were present originally?

Sol.

40. (a) Write the products of the following reactions:

(b) Arrange BF₃, BCl₃ and BBr₃ in the increasing order of Lewis acidity and justice.

Sol. (a) (i) CH₃ – OH + I^– (ii) I – O^– + HCF₃ (iii) Mn(CO)₅I + C₂F₆ + NaI

(b) Increasing lewis acidity order:

BF₃ < BCl₃ < BBr₃

In back bonding most effective due to similar size and energy of 'p' orbitals.

back bonding less effective because difference in p-orbital energy and also bond length is large.

least back bonding, due to larger difference in size and energy of p orbital and also bond length is large.

41. Justify the following

(a) Considering CO₂ as an ideal gas, equipartition theorem products its total energy as 6.5 kT.

(b) for a process is the same whether the process takes place reversibly or irreversibly.

(c) The quantity equals the maximum non-expansion work done by a system in a constant temperaure-pressure process.

(d) At constant temperature and pressure, for a reversible phase change.

(e) Transition states cannot be isolated as independent chemical species.

Sol. (a) From Equipartition theorem,

Energy per translational or rotational degree of freedom

Energy per vibration degree of freedom = kT

CO₂ has 3 translational degree of freedom, 2 rotational degree of freedom and vibrational degree of freedom = (3 × 3 – 5) = 4.

Energy per molecule of CO₂ from euipartition theorem.

(b) If initial and final states achieved by any reversible and irreversible process are same, then will be same, because entropy is a state function. So, it is independent of path.

(c) Phase I Phase II, (Let be the enthalpy change from phase I to phase II).

Total entropy change for any process is given by,

If surrounding is at temperature 'T' (same as system temperature) and heat absorbed by the surrounding is q_rev.

Therefore, from equation (1) and (2),

(d) dG = VdP – SdT

At constant P, dP = 0 and at constant T, dT = 0

From equation (3)

So, at constant T and P, for reversible phase change.

(e) In transition state theory, a reaction is assumed to involve the attainment of an activated complex that goes on to product at an extremely rapid rate. The rate of decomposition of the activated complex about 6 × 10¹² s^–1 at room temperature.

42. The rate constant k for a second order reaction Products is expressed

where the concentration is in mol lit^–1, T is in absolute temperature and time is in minutes. The initial concentrations of both the reactants are 0.05 M. Calculate the activation energy and half life of the reaction at 27ºC.

(R = 2 cal K^–1 mol^–1)

Sol. For the second order reaction

At T = 300 K, k = 10¹⁰.

We have,

From (1) and (2)

E_a = 3000 × 2.303 × 2 cal/mol = 13818 cal/mol = 13.818 kcal/mol

Substituting

T = 300 K

A₀ = 0.05

half-life can be calculated

43. The equilibrium constant for the reaction

at 600°C is 1.00. If a mixture initially consisting of 1 mole of Fe₃O₄, 2 moles of CO, 0.5 mole of FeO and 0.3 mole of CO₂ is heated to 600°C at constant total pressure of 5 atmospheres, how many moles of each substance would be present at equilibrium?

Sol.

CO(g) = 2 – x = 1.15 mole

FeO(s) = 0.5 + 3x = 3.05 mole

CO₂(g) = 0.3 + x = 1.15 mole

44. (a) Use the time-independent Schrodinger equation to calculate the energy of a particle of mass "m" with V = 0 in the state in a cubical box of length "a".

(b) At 20ºC, the vapour pressure of two pure liquids X and Y which form an ideal solution are 70 torr and 20 torr respectively. If the mole fraction of X in solution is 0.5, find the mole fraction of X and Y in the vapour phase in equilibrium with the solution.

Sol.