1. Molecular shape of SOCl₂ is

(a) Square planar

(b) Trigonal pyramidal

(c) Triangular planar

(d) T-shape

Ans. b

Sol.

Correct option is (b)

2. Number of three-centre two-electron (3e-2e) bonds present in diborane is:

(a) 2

(b) 4

(c) 6

(d) 8

Ans. a

Sol.

2(3c – 2e) bonds are present in diborane

Correct option is (a)

3. The lattice energy of LiF calculated from Born-Lande equation – 1000 kJ mol_–1. Assume that for both LiF and MgO the Madelung constants, interionic distances and Born exponents have the same value. The lattice energy of MgO in kJ mol_–1 is:

(a) – 4000

(b) – 2000

(c) 2000

(d) 4000

Ans. d

Sol. From Born-Lande equation

because 'n' is always greater than one.

Correct option is (d)

4. The compound formed by dissolving elemental gold in aqua regia is:

(a) AuCl

(b) AuNO₃

(c) H[AuCl₄]

(d) H[Au(NO₃)₄]

Ans. c

Sol.

Correct option is (c)

5. Number of moles of ions produced by complete dissociation of one mole of Mohr's salt in water is

(a) 3

(b) 4

(c) 5

(d) 6

Ans. c

Sol. Iron (II) ammonium sulfate (NH₄)₂Fe(SO₄).6H₂O ions produced by complete dissociation of Mohr salt

Correct option is (c)

6. The tetrachloro complexes of Ni(II) and Pd(II) respectively, are (atomic numbers of Ni and Pd are 28 and 46 respectively)

(a) diamagnetic and diamagnetic

(b) paramagnetic and paramagnetic

(c) diamagnetic and paramagnetic

(d) paramagnetic and diamagnetic

Ans. d

Sol. [NiCl₄]_2– Paramagnetic

Pairing occurs in case of Pd sine it has greater nuclear charge leading to quater interaction to the ligands.

Correct option is (d)

7. The total number of steps involved and number of beta particles emitted in the spontaneous decay of respectively, are

(a) 8 and 6

(b) 14 and 6

(c) 6 and 8

(d) 14 and 8

Ans. b

Sol.

Let 'x' = number of alpha-particles emitted

y = No. of beta-particles emitted.

Conserving the number of nucleons-

238 = 206 + 4x + y(0)

4x = 238 – 206 = 32

x = 8

Now, 92 = 82 + 8(2) + y(–1)

92 = 82 + 16 – y

y = 98 – 92 = 6

Number of beta particles = 6

Total steps = 8 + 6 = 14

Correct option is (b)

8. A filter paper moistioned with ammonical sodium nitroprusside solution turns violet on contact with a drop of alkaline Na₂S solution. The violet colour is due to the formation of

(a) [Fe(SCN)₅(NO)]_1–

(b) [Fe(SCN)₅(NO)]_2–

(c) [Fe(CN)₅(NOS)]_3–

(d) [Fe(CN)₅(NOS)]_4–

Ans. d

Sol.

Correct option is (d)

9. The species/compounds that are aromatic among the following are

(a) R and S

(b) P and Q

(c) Q and S

(d) P and S

Ans. d

Sol.

n = 0 (following Huckel Rule)

Hence aromatic,

1. Planar 2. Fully conjugated 3. 4n + 2 = 14

n = 3 (following Huckel rule)

Hence, it is aromatic.

Correct option is (d)

10. The major product obtained in the reaction below is

(a)

(b)

(c)

(d)

Ans. c

Sol.

11. The rates of acetolysis for the following norbornyl derivatives are in the order

(a) R > Q > P

(b) Q > R > P

(c) P > R > Q

(d) R > P > Q

Ans. d

Sol.

Removal of the tosyl group (the rate determining step) is subject to strong anchimeric assistance by the double bond and forms non-classical carbocation. Hence acetolysis is fastest in case of R. Among P and Q.

Rate of acetolysis for P is faster than Q due to participation of π-electrons of two allylic 1-6, and 4-5 bonds. Rate of acetolysis R > P > Q

Correct option is (d)

12. The Haworth projection for of D-glucose is

(a)

(b)

(c)

(d)

Ans. c

Sol.

–OH which is right will be downward

–OH which is left will be upward.

Correct option is (c)

13. The complementary DNA sequence of the given DNA 5'-G-A-A-T-T-C-3' is:

(a) 5'-C-T-T-A-A-G-3'

(b) 5'-C-U-U-A-A-G-3'

(c) 3'-C-T-T-A-A-G-5'

(d) 3'-G-A-A-T-T-C-5'

Ans. c

Sol. Complementary DNA sequence of

5´– G – A – A – T – T – C – 3´ 3´ – C – T – T – A – A – G – 5´

This is called base pairing in DNA

A (adenine) pair with T (thymine)

G (guanine) pair with C (cytosine)

Correct option is (c)

14. The order of nucleophilicity of the following anions in a S_N2reaction is:

(a) Q > R > S > P

(b) Q > P > R > S

(c) Q > R > P > S

(d) P > S > R > Q

Ans. b

Sol. Among P, R and S conjugate base of the weakest acid is the strongest nucleophile

Between P and Q, is better nucleophilic than

(Large size of anion better nucleophile)

Correct option is (b)

15. The pair of conformation that has maximum energy difference is:

(a)

(b)

(c)

(d)

Ans. a

Sol. Chair form is the most stable and half chair is the least stable. hence maximum energy difference is between half chair and chair form.

Chair < twist boat < boat < half chair.

[Chair < half chair]

Correct option is (a)

16. The major mono-sulfonation product of is

(a)

(b)

(c)

(d)

Ans. b

Sol.

among I, II and III (I and III) are unstable so II is the more appropriate choice.

Correct option is (b)

17. Electrophilic nitrations of the following compounds follow the trend

(a) S > R > P > Q

(b) R > S > P > Q

(c) R > P > S > Q

(d) P > S > R > Q

Ans. b

Sol.

Among P, Q, R and S, NHCOCH₃ and –C₂H₅ groups are ring activators and –Cl, SO₃H are ring deactivators.

Resonance effect dominates over hyper-conjugation hence is the more powerful activator.

between –Cl and –SO₃H, –SO₃H is the powerful deactivator.

P > Q

Correct option is (b)

18. The compounds those would not respond to tests of both nitrogen and sulfur with sodium fusion extracts are

(a) I and III

(b) III and IV

(c) I and IV

(d) II and IV

Ans. a

Sol.

Organic compounds containing C, H, N, S give test with sodium fusion extracts. For test of Nitrogen atom presence of carbon is necessary, since it is the test for

(I) has no carbon atom hance it would not give test with sodium fusion extracts. (III) has no S atom. So it would not give test of Sulfur. I and III is appropriate choice.

Correct option is (a)

19. The correct epimeric pair of the following is

(a) P and Q

(b) R and Q

(c) Q and S

(d) R and S

Ans. a

Sol. Two diastereomers that differ in the configuration around one only stereogenic centre are called epimers.

So, the configuration is changed at the only C₃ carbon. Hence it is called epimer and epimer is also called diastereomer.

Correct option is (a)

20. shown below is a

(a) diterpene having two isoprene units

(b) triterpene having three isoprene units

(c) triterpene having four isoprene units

(d) Sesquiterpene having three isoprene units.

Ans. d

Sol.

d-farmesene has 3 isoprene unit hence it is a sesquiterpene. (Molecular formula = (C₁₅H₂₄))

Correct option is (d)

21. For the equilibrium the equilibrium constant, K_P is expressed as

(a)

(b)

(c)

(d)

Ans. c

Sol.

Correct option is (c)

22. The average speed of H₂, N₂ and O₂ gas molecules is in the order

(a) H₂ > N₂ > O₂

(b) O₂ > N₂ > H₂

(c) H₂O₂ > N₂

(d) N₂ > O₂ > H₂

Ans. a

Sol.

Correct option is (a)

23. The enthalpy of vaporization is zero at

(a) Boyle temperature

(b) critical temperature

(c) Inversion temperature

(d) boiling temperature

Ans. b

Sol. The heat of vaporization diminishes at the critical temperature because above the critical temperature the liquid and vapour phase no longer co-exist.

Correct option is (b)

24. The half-life of any zero-order reaction is:

(a) independent of concentration

(b) Proportional to inverse of concentration

(c) proportional to concentration

(d) proportional to square to the concentration

Ans. c

Sol. for n_th order reaction

Correct option is (c)

25. The molality of (NH₄)₂SO₄ solution that has the same ionic strength as 1 mol kg_–1 solution of KCl is

(a)

(b)

(c)

(d)

Ans. a

Sol.

Correct option is (a)

26. The standard enthalpy of formation at 1 bar and 300 K for the formation of CF₂ClCF₂Cl(g) from its constituent elements in the standard state is –900 kJ mol_–1. Given R = 8.3 J K_–1 mol_–1, the standard internal energy of formation at the same pressure and temperature is

(a) –905 kJ mol_–1

(b) –895 kJ mol_–1

(c) 895 kJ mol_–1

(d) 905 kJ mol_–1

Ans. b

Sol.

Formation of CF₂ClCF₂Cl(g)

2C(s) + 2F₂(g) + Cl₂ (g) = CF₂ClCF₂Cl(g)

= –900 kJ / mole + 5kJ / mole = –895 kJ/mole

Correct option is (b)

27. The percent transmittance of a solution having absorbance (optical density) 1.0 is

(a) 1

(b) 10

(c) 50

(d) 99

Ans. b

Sol. A = –log T; –A = log T, 10_–A = T

Correct option is (b)

28. The matrix which transforms

(a)

(b)

(c)

(d)

Ans. d

Sol.

Correct option is (d)

29. A concentration cell with two hydrogen electrodes at two different pressures is depicted as

The potential (E_cell) of the cell is

(a)

(b)

(c)

(d)

Ans. d

Sol.

Correct option is (d)

30. An aqueous solution containing 1 g L_–1 of a polymer exerts osmotic pressure of 4 torr at 300 K. Given R = 0.082 L atm, the molar mass (g mol_–1) of the polymer is

(a) 4500

(b) 4564

(c) 4674

(d) 4800

Ans. c

Sol. Given that mass of solute, w = 1.0 g L

Volume, V = 1 ltr.

= 4674 g mole_–1

Correct option is (c)

31. (a) identify the most acidic compound from the following: CH₃ – CH₃, CH₂ = CH₂ and CH CH, and justify your answer. Draw overlap of the orbitals to show bonding in the most acidic compound using the concept of hybridization.

(b) Write a balanced chemical equation to represent acid-base reaction of orthoboric acid in water. Addition of ethylene glycol to aqueous orthoboric acid enhances its acidity. Explain the above statement using appropriate chemical equation.

Sol. (a) CH CH is the most acidic compound among ethane and ethylene.

The acidity of H–A increases as the percent S-character of the A_– increases.

The higher the percent S-character of the hybrid orbital, the closer the lone pair is held to the nucleus, and the more stable the conjugate base.

(b) orthoboric acid in water:

Addition of ethylene glycol of aqueous orthoboric acid forms chelated complex compound and generates H₊ ions.

32. (a) Draw the unit cell structure of NaCl. Calculate the limiting radius ratio of any ionic solid having NaCl like structure.

(b) Give molecular formula and structure of the compound formed by reaction of Be(OH)₂ with acetic acid.

Sol. (a) Unit cell of NaCl

Cl_– = F.C.C arrangement

Na₊ = All octahedral voids.

Limiting Radius Ratio:

Let a = Edge length

33. (a) The spin-only magnetic moments of K₃[Fe(oxalate)₃] are 5.91 respectively. Write down their ligand field electronic configuration. Justify your answer. Atomic number of Fe and Ru are 26 and 44 respectively.

(b) Draw the structures of Arrange them in the increasing order of O–N–O bond angles.

Sol.

In both the complexes ligand is same but one complex is high spin and another complex is low spin. Fe and Ru belongs to same group. On moving down the group crystal field splitting energy increases. hence Ru complex is a low spin complex.

increasing order of ONO bond angles

34. (a) Show with labels the splitting of d-orbitals in an octahedral ligand field. Calculate the CFSE of (i) high spin d₆ and (ii) low spin d₆ metal ions in octahedral field.

(b) Schematically represents orbital overlaps in metal carbonyls. Show the correct signs of the lobes.

Sol.

35. (a) A coordination compound is composed of one Co(III), one chloride, one sulfate and four molecules of ammonia. The aqueous solution of the compound gives no precipitate when combined with aqueous BaCl₂, while a white precipitate is formed with aqueous AgNO₃ solution. Draw its structure and explain the observations with chemical equations.

(b) Draw the structure of dimethylglyoxime (DMGH₂) and its Ni(II) complex formed in aqueous ammonia.

Sol. (a) Complex is [Co(NH₃)₄SO₄]Cl

No reaction. since SO_42– is present in the coordination sphere.

Cl_– is present in the ionic sphere

36. (a) Write the structure of E, F and G in the following scheme of reactions.

(b) Identify the structure of H and I in the following synthetic transformation

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

37. (a) Complete the following reaction sequence with appropriate structures of J, K and L.

(b) Identify the structures of M and N in the following synthetic transformation

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

(b) Chemical transformation involved in above chemical reaction can be illustrated as

38. (a) In the following reaction scheme, write the structure of O, P and Q

(b) Given below are structure of some natural products. Identify them as vitamin A, B₆, C and D and classify them according to their classes (isoprenoid, alkaloid, carbohydrate and steroid)

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

39. (a) Write the appropriate structures for R, S and T in the following scheme.

(b) Choose the correct stereoisomer between U and V that would furnish W on controlled hydrolysis. Write the stable conformation of W.

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

In both U and V, V is the correct choice. Since in V both groups are anti to each other. (Suitable condition for SN₂)

Stable conformation of W due to presence of three membered ring. The cyclohexane ring is deformed chair.

40. The mechanism of isomerization of cyclobutene (CB) to 1, 3-butadiene (BD) is as follows.

(a) Show that the rate law is

(b) The apparent first-order rate constant, At the CB concentration of 1 × 10_–5 mol dm_–3, the value of k_app reaches 50% of its limiting value obtained at very high concentrations of CB. Evaluate the ratio

Sol.

At very high concentration of CB,

41. (a) The molar conductance of 0.012 mol dm_–3 aqueous solution of chloroacetic acid is 100 cm₂ mol_–1. The ion conductance of chloroacetate and H₊ ions are 50 W_–1 cm₂ mol_–1 and 300 cm₂ mol_–1, respectively. Calculate (i) degree of dissociation and pK_a of chloroacetic acid, and (ii) H₊ ion concentration in the solution.

(b) Sketch the conductivity versus concentration of base curves for the titration of aqueous solutions of acetic acid (i) with NaOH, and (ii) with NH₄OH.

Sol. (a) Concentration of chloroacetic acid = 0.012 mol dm_–3

(b) (1) Weak acid with strong base:

CH₃COOH with NaOH

(i) Weak acid (CH₃COOH) has low conductance than its salt (CH₃COONa). Due to its lower dissociation w.r.t. salt. So, addition of base increase conductance (region ab in graph)

(ii) The reson for initial decrease in conductance (region (Oa)) is because small addition of base leads to common anion i.e. CH₃COO_– ion formation which represses the dissociation of CH₃COOH which gives highly conducting H₊ ions.

(iii) After the end point there is higher rate of increases in conductance because Na₊ + OH_– has higher conductance than Na₊ + CH₃COO_– (region bc in graph).

(2) Weak acid with weak base:

CH₃COOH + NH₄OH CH₃COO_– + NH₄₊ + H₂O

(i) Weak acid (CH₃COOH) has low conductance than its salt (CH₃COONa). Due to its lower dissociation w.r.t. salt. So, addition of base increase conductance (region ab in graph).

(ii) The reason for initial decrease in conductance (region(Oa)) is because small addition of base leads to common anion i.e. CH₃COO_– ion formation which represses the dissociation of CH₃COOH which gives highly conducting H₊ ions.

(iii) The region b´c´ of graph is due to weak nature of base. So, conductance remains almost constant with the addition of weak base after end point.

42. A solution of a free particle Schrodinger equation is

(a) Derive expressions for energy 'E' and momentum 'p' of the particle.

(b) Using the above relations, show that the wavelength

Sol.

(Solution is e_ikx = cos kx + i sin kx for x-direction)

Particle is free so potential is constant. Since the particle is not subjected to any external force. Its potential energy assume zero. There is no boundary condition and no restriction.

The function e_+ikx are eigen function of the linear momentum operator with eigenvalues Expression for momentum So, momentum is

(b) As we know from quantum mechanics,

We have derived above that,

43. (a) Sketch the temperature composition phase diagram at 1 atm pressure for the ethanol-water system.

(i) Label all the areas in the diagram.

(ii) Indicate the temperature at which the composition of the vapour is same as that of the liquid.

(iii) What is the degree of freedom at the corresponding composition?

(b) Estimate the pressure necessary to melt ice at –10ºC if the molar volume of liquid water is 18.01 mL and molar volume of ice is 19.64 mL. The entropy change for the melting process is 16.3 J K_–1. Assume that the molar volumes and entropy change remain constant in this temperature range. [100 J = 1 L bar].

Sol. (a) T Vs composition for C₂H₅OH – H₂O

(ii) The temperature at which composition of liquid and vapour is same is 351.2K. This is known as Azeotropic mixture

(iii) Degree of freedom of azeotropic composition, at constant pressure in liquid region.

F = C – P + 1

= 2 – 1 + 1 = 2

In vapour region, F = C – P + 1 = 2 – 1 + 1 = 2

At liquid – vapour equilibrium

F = C – P + 1 = 2 – 2 + 1 = 1

(b) From Clapyeron equation,

44. (a) (i) Show that for 'n' moles of a Vander waals gas,

(ii) Can a gas that obeys the equation of state p(V – nb) = nRT be liquefied? Explain.

(b) Consider ideal mixing of 2 moles of toluene and 2 moles of benzene at 1 atm and 300 K. Calculate the values of for the process. (ln 2 = 0.69)

Sol.

(ii) If a gas obey the equation of state P(V-nb) = nRT

This gas difficult to liquify because this gas donot contain n₂a/V₂m part because 'a' is responsible for attraction. if 'a' is not present gas molecule not attract, gas not liquify and 'z' value always greater than 1 and it increase with increase of P.

For example: Hydrogen and helium, the value of 'a' is extremely small for these gases as they are difficult to liquify. Thus we have the equation of state as P(Vm-b) = RT, obtained from vander waals equation by ignoring the term a/V₂ m. Hence, 'z' is always greate than 1 and it increase with increasing of P.