Question 1: The stability of

Follows the order

a. I > II > III
b. II > I > III
c. II > III > I
d. III > II > I

Answer: (b)

Stability of carbocation increases with (+M) effect

Structure-II has +M effect of –NMe₂ group, hence maximum stability among the given set of cations. In structure-III due to steric inhibition of resonance effect –C⁺Me₂ moves out of the plane of the benzene ring hence stability decreases due to loss of conjugation.

Finally, the correct order should be: II > I > III.

Question 2: Among the following, the biodegradable polymer is:

a. Polylactic acid
b. Polyvinyl chloride
c. Bakelite
d. Teflon

Answer: (a)

Polylactic acid (PLA) is a biodegradable aliphatic polyester, PLA sometimes called polylactide, which can be produced by fermentation of renewable resources such as corn, cassava, potato and sugarcane.

Question 3: Among the following,

The compounds which can be reduced with formaldehyde and conc. aq. KOH are:

a. only II and V
b. only I and V
c. only II and III
d. only I, II and IV

Answer: (a)

These are examples of Cannizzaro reactions.

Aldehyde without α-H gives a Cannizzaro reaction and forms alcohol and salt of carboxylic acid.

Question 4: An organic compound that is commonly used for sanitizing surfaces is:

a. acetylsalicylic acid
b. chloramphenicol
c. aspartame
d. cetyltrimethylammonium bromide

Answer: (d)

Cetyltrimethylammonium bromide is a popular cationic detergent. Cationic detergents have germicidal properties and are expensive.

Question 5: The rates of reaction of NaOH with

follow the order:

a. II > I > III
b. II > III >I
c. I > III > II
d. III > II > I

Answer: (c)

I and III can react with NaOH but II do not react to room temperature. I and III give a reaction because at ortho and para position electron-withdrawing group is present.

Question 6: The most suitable reagent for the conversion of 2-phenylpropanamide into 1-phenylethylamine is:

a. H₂, Pd/C
b. Br₂, NaOH
c. LiAlH₄, Et₂O
d. NaBH₄, MeOH

Answer: (b)

Hoffmann bromamide degradation reaction.

Question 7: The compound X in the following reaction scheme is:

a. acetonitrile
b. methyl isocyanide
c. acetaldehyde
d. nitromethane

Answer: (a)

Question 8: A nucleus X captures a β particle and then emits a neutron and γ ray to form Y. X and Y are:

a. isomorphs
b. isotopes
c. isobars
d. isotones

Answer: (d)

_zX^A + _–1e⁰ →_Z–1Y^A → _Z–1X^A–1(Y) + ₀n¹ + γ

A = Mass number of X

Z = Atomic number of X

Number of neutrons in X = A – Z

Number of neutrons in Y = A – 1 – (Z – 1) = A – Z

i.e., X & Y are isotones (have the same number of neutrons)

Question 9: The boiling point (in ⁰C) of 0.1 molal aqueous solutions of CuSO₄ · 5H₂O at 1 bar is closest to:

[Given: Ebullioscopic (molal boiling point elevation) constant of water. K_b = 0.512 K Kg mol^–1]

a. 100.36
b. 99.64
c. 100.10
d. 99.90

Answer: (c)

∆T_b = iK_b .m

i = Van’t Hoff factor

CuSO₄→ Cu⁺² + SO₄^-2 (i = 2)

∆T_b = 2 × 0.512 × 0.1

∆T_b = 0.1024

T_sol– T⁰_A = ∆T_b

T_sol – 100 = 0.1024

T_sol = 100.1024 ⁰C

Question 10: A weak acid is titrated with a weak base. Consider the following statements regarding the pH of the solution at the equivalence point:

(i) pH depends on the concentration of acid and base.

(ii) pH is independent of the concentration of acid and base.

(iii) pH depends on the pK_a of acid and pK_b of the base.

(iv) pH is independent of the pK_a of acid and pK_b of the base.

The correct statements are:

a. only (i) and (iii)
b. only (i) and (iv)
c. only (ii) and (iii)
d. only (ii) and (iv)

Answer: (c)

For weak acid and weak base titration pH formula is

Take -ve log

pH = +½ (pK_w+ pK_a – pK_b)

pH = 7 +½ (pK_a – pK_b)

Question 11: Products are favoured in a chemical reaction taking place at a constant temperature and pressure.

Consider the following statements:

(i) The change in Gibbs energy for the reaction is negative.

(ii) The total change in Gibbs energy for the reaction and the surroundings is negative.

(iii) The change in entropy for the reaction is positive.

(iv) The total change in entropy for the reaction and the surroundings is positive.

The statements which are ALWAYS true are:

a. only (i) and (iii)
b. only (i) and (iv)
c. only (ii) and (iv)
d. only (ii) and (iii)

Answer: (b)

The reaction is taking place at constant temperature and pressure and products formation are favoured. i.e., the reaction is spontaneous.

ΔS_Total = ΔS_sys+ΔS_surr> 0

At constant T & P

ΔG_reaction< 0 (spontaneous)

Correct statements : (i) & (iv)

Question 12: A mixture of toluene and benzene forms a nearly ideal solution. Assume P_B⁰ and P_T⁰ to be the vapour pressures of pure benzene and toluene, respectively. The slope of the line obtained by plotting the total vapour pressure to the mole fraction of benzene is:

a. P⁰_B – P⁰_T
b. P⁰_T – P⁰_B
c. P⁰_B + P⁰_T
d. (P⁰_B -P⁰_T)/2

Answer: (a)

According to Raoult’s law

P_S = P_B + P_T

P_S = P_B⁰x_B + P_T⁰xT

P_S = P_B⁰x_B+ P_T⁰(1-x_B)

P_S = (P_B⁰ – P⁰_T)x_B + P⁰_T

y = mx + c

Question 13: Upon dipping a copper rod, the aqueous solution of the salt that can turn blue is:

a. Ca(NO₃)₂
b. Mg(NO₃)₂
c. Zn(NO₃)₂
d. AgNO₃

Answer: (d)

Cu + 2AgNO₃→Cu(NO₃)₂ + 2Ag

Blue

Order of reactivity:Ca> Mg>Zn> Cu>Ag

Hence Ag⁺ alone can oxidise Cu to Cu²⁺.

Question 14: Treatment of alkaline KMnO₄ solution with KI solution oxidizes iodide to:

a. I₂
b. IO₄^–
c. IO₃^–
d. IO₂^–

Answer: (c)

2KMnO₄ + H₂O + KI → 2MnO₂ + 2KOH + KIO₃

Question 15: If an extra electron is added to the hypothetical molecule C₂, this extra electron will occupy the molecular orbital:

a. π^*_2p
b. π_2p
c. σ^*_2p
d. σ_2p

Answer: (d)

Due to sp-mixing energy order of molecules have less than or equal to 14-proton is-

σ_1s<σ*_1s<σ_2s<σ*_2s<π_2p = π_2p<σ_2p<π*_2p = π*_2p<σ*_2p

Molecular orbital configuration of C₂-molecule – σ_1s²σ_1s*σ²_2s σ_2s*² π_2p² = π_2p²

C₂→ C₂^– (π_2p² = π_2p² σ_2p¹)

Question 16: Among the following, the square planar geometry is exhibited by

a. CdCl₄^2–
b. Zn(CN)₄^2–
c. PdCl₄^2–
d. Cu(CN)₄^3–

Answer: (c)

Pt⁺² and Pd⁺² – form a square planar and diamagnetic complex.

Question 17: The correct pair of orbitals involved in π-bonding between metal and CO in metal carbonyl complexes is:

a. metal d_xy and carbonyl π_x^*
b. metal d_xy and carbonyl π_x
c. metal

Question 18: The magnetic moment (in μB) of [Ni(dimethylglyoximate)₂] complex is closest to:

a. 5.37
b. 0.00
c. 1.73
d. 2.25

Answer: (b)

μB = √(n(n+2))BM

n = no. of unpaired electron

Question 19: A compound is formed by two elements M and N. Element N forms a hexagonal closed packed array with 2/3 of the octahedral holes occupied by M. The formula of the compound is:

a. M₄N₃
b. M₂N₃
c. M₃N₂
d. M₃N₄

Answer: (b)

N → HCP (Z = 6)

The number of atoms in HCP is 6. It has six octahedral voids.

M → ⅔ ×O.V.

⇒ ⅔ ×6 ⇒ 4 (Z = 4)

Formula is

M₄N₆ ⇒ M₂N₃

Question 20: If the velocity of the revolving electron of He⁺ in the first orbit (n = 1) is v. the velocity of the electron in the second orbit is:

a. v
b. 0.5v
c. 2v
d. 0.25v

Answer: (b)

V = 2.18×10⁶×(Z/n) m/sec

V ∝ 1/n

V₁/V₂ = n₂/n₁

V/V₂ = 2/1

V₂ = V/2 = 0.5V

Part II

Question 21: An organic compound X with molecular formula C₁₁H₁₄ gives an optically active compound on hydrogenation. Upon ozonolysis, X produces a mixture of compounds – P and Q. Compound P gives a yellow precipitate when treated with I₂ and NaOH but does not reduce Tollens’ reagent. Compound Q does not give any yellow precipitate with I₂ and NaOH but gives Fehling’s test. The compound X is:

Answer: (a)

gives Iodoform which is yellow ppt but it does not give Tollen’s reagent test due to the presence of ketone group. CH₃–CH₂–CH=O does not give an iodoform reaction but can give Tollen’s reagent test.

Question 22: The following transformation

can be carried out in three steps. The reagents required for these three steps in their correct order, are:

a. (i) NaBH₄; (ii) PCl₅ ; (iii) anh. AlCl₃
b. (i) NaBH₄; (ii) anh. AlCl₃; (iii) Zn(Hg)/HCl
c. (i) Zn(Hg)/HCl; (ii) SOCl₂ ; (iii) Anh. AlCl₃
d. (i) conc. H₂SO₄; (ii) H₂N-NH₂·H₂O; (iii) KOH, ethylene glycol, Δ

Answer: (c)

Question 23: In the following reaction

X and Y, respectively are :

Answer: (d)

Cumene hydroperoxide oxidation reaction.

Question 24: A two-dimensional solid is made by alternating circles with radius a and b such that the sides of the circles’ touch. The packing fraction is defined as the ratio of the area under the circles to the area under the rectangle with sides of length x and y.

The ratio r = b/a for which the packing fraction is minimized is closest to:

a. 0.41
b. 1.0
c. 0.50
d. 0.32

Answer: (a)

Area of rectangular = xy

= 2a(2a + 2b)

Area covered by circles = πa² + πb²

= π(a²+b²)

Packing fraction (PF) = (a² +b²)/(2a+2b)(2a)

PF = πa²(1+(b/a))²/4a²(1+b/a) = π(1+r²)/4(1+r) ; [r = b/a]

d(PF)/dr = π/4 [(1+r)2r – (1+r²)/(1+r)²] = 0 if P.F. is minimum.

∴2r + 2r² – 1 – r² = 0

r² + 2r -1 = 0

r = (-2±√(4+4))/2

= (-2+2√2)/2

= √2-1

≈ 0.41

Question 25: Consider a reaction that is first order in both directions

Initially only A is present, and its concentration is A₀. Assume A_t and A_eq are the concentrations of A at time “t” and at equilibrium, respectively. The time “t” at which A_t = (A₀ + A_eq)/2 is:

a. t = (ln(3/2))/ (k_f+k_b)
b. t = (ln (3/2))/(k_f–k_b)
c. t = (ln 2)/(k_f+k_b)
d. t = (ln 2)/(k_f–k_b)

Answer: (c)

-d[A]/dt = k_f[A] – k_b[B]

-d[A]/dt = k_f (A₀ – x_e) – k_b(x_e) = 0 [at equilibrium]

k_f(A₀ –x_e) = k_b(x_e)

k_b = k_f (A₀ – x_e)/x_e

+d[B]/dt = k_f(A₀ –x) – (k_f(A₀– x_e)/x_e) (x)

= x_ek_f(A₀-x) – k_f(A₀– x_e)x/x_e

+d[B]/dt = k_fA₀(x_e-x)/x_e

d[B]/(x_e-x) = k_f(A₀/x_e)dt

(x_e/A₀t) ln x_e/(x_e-x) = k_f ………(1)

From old relation

K_b = k_f(A₀-x_e)/x_e

K_bx_e = k_fA₀–k_fx_e

(k_b + k_f ) = k_ff A₀/x_e………..(2)

From equation (1) and (2)

(k_f + k_b) = (1/t) ln x_e/(x_e-x)

Now given data

⇒ (A₀ – x) = A_t

(A₀ – A_t) = x

⇒ A_t = (A₀+ A₀-x_e)/2 = (2A₀-x_e)/2

2A_t = 2A₀ – x_e

x_e = 2A₀ – 2A_t

(x_e – x) = 2A₀ – 2A_t – A₀ + A_t

(x_e – x) = (A₀ – A_t)

(k_f + k_b) = 1/t ln x_e/(x_e-x)

t = 1/(k_f + k_b) ln 2A₀-2A_t/(A₀-A_t)

= (1/(k_f+k_b)) ln2

Question 26: The reaction CaCO₃(s) ⇌ CaO(s) + CO₂ (g) is in equilibrium in a closed vessel at 298 K. The partial pressure (in atm) of CO₂(g) in the reaction vessel is closest to:

[Given: the change in Gibbs energies of formation at 298 K and 1 bar for

CaO (s) = –603.501 kJ mol^–1

CO₂ (g) = –394.389 kJ mol^–1

CaCO₃ (s) = –1128.79 kJ mol^–1

Gas constant R = 8.314 J K^–1 mol^–1]

a. 1.13 × 10^–23
b. 0.95
c. 1.05
d. 8.79 × 10²³

Answer: (a)

CaCO₃(s) ⇌ CaO(s) + CO₂ (↑)

ΔG⁰ = ∑∆G⁰_product– ∑∆G⁰_reactant

ΔG⁰= {ΔG⁰ [CaO(s)] + ΔG⁰ [CO₂ (↑)]} – {ΔG⁰ [CaCO₃(s)]}

ΔG⁰ = {–603.501 + (–394.389)] – [- 1128.79]

ΔG⁰= – 997.89 + 1128.79

ΔG⁰= 130.9 kJ/mole

ΔG⁰ = – 2.303 RT log K_p

130.9 = – 2.303 × 8.314 × 10^–3 × 298 log K_P

logK_P = – 22.94

K_P = Antilog (-22.94)

K_P = 1.13 × 10^–23

K_P = P_CO2= 1.13×10^–23

Question 27: ∆A container is divided into two compartments by a removable partition as shown below:

In the first compartment, n₁ moles of ideal gas He is present in a volume V₁. In the second compartment, n₂ moles of ideal gas Ne is present in a volume V₂. The temperature and pressure in both the compartments are T and P, respectively. Assuming R is the gas constant, the total change in entropy upon removing the partition when the gases mix irreversibly is:

a. n₁ R ln (v₁/(v₁+v₂)) + n₂R ln (v₁/(v₁+v₂))
b. n₁R ln ((v₁+v₂)/v₁) + n₂R ln (v₁+v₂)/v₁
c. (n₁+n₂) R ln n₁v₁/n₂v₂
d. (n₁+n₂)R ln n₂v₂/n₁v₁

Answer: (b)

Entropy is a state function

Entropy change ΔS = n C_Vln(T_f/T_i )+ nR ln(V_f/V_i)

For isothermal reversible process

∆S = nRln (V_f/V_i)

∆S_T = ∆S_{1st compartment} + ΔS_{IInd compartemnt}

∆S_T = n₁Rln (V₂+V₁)/V₁ + n₂R ln(V₂+V₁)/V₂

Question 28: Number of stereoisomers possible for the octahedral complexes [Co(NH₃)₃Cl₃] and [Ni(en)₂Cl₂], respectively, are:

[en = 1,2-ethylenediamine]

a. 2 and 4
b. 4 and 3
c. 3 and 2
d. 2 and 3

Answer: (d)

The possible number of stereoisomers of the above compound is 2.

The possible number of stereoisomers of the above compound is 3.

Question 29: When a mixture of NaCl, K₂Cr₂O₇ and conc. H₂SO₄ is heated in a dry test tube, a red vapour (X) is evolved. This vapour (X) turns an aqueous solution of NaOH yellow due to the formation of Y.

X and Y, respectively, are:

a. CrCl₃ and Na₂Cr₂O₇
b. CrCl₃ and Na₂CrO₄
c. CrO₂Cl₂ and Na₂CrO₄
d. Cr₂(SO₄)₃ and Na₂Cr₂O₇

Answer: (c)

This is a chromyl chloride test of Cl^– ion

4NaCl + K₂Cr₂O₇ + 6H₂SO₄ → 2CrO₂Cl₂(X) + 2KHSO₄ + 6H₂O + 4NaHSO₄

CrO₂Cl₂ + 4NaOH → Na₂CrO₄(Y) + 2NaCl + 2H₂O

X = CrO₂Cl₂

Y = Na₂CrO₄

Question 30: Sodium borohydride upon treatment with iodine produces a Lewis acid (X), which on heating with ammonia produces a cyclic compound (Y) and a colorless gas (Z). X, Y and Z are:

a. X = BH₃; Y = BH₃·NH₃; Z = N₂
b. X = B₂H₆; Y = B₃N₃H₆; Z = H₂
c. X = B₂B₆; Y = B₆H₆: Z = H₂
d. X = B₂H₆: Y = B₃N₃H₆: Z = N₂

Answer: (b)

2Na[BH₄] + I₂→ B₂H₆(X) + 2NaI + H₂

X = B₂H₆

Y = B₃N₃H₆

Z = H₂