KVPY-SX 2017 Chemistry Paper with Solutions

Question 1: The major product formed in the following reaction is

Answer: (a)

In the above molecule, lone pair of nitrogen is in conjugation with benzene as well as a carbonyl group. Therefore, the +M effect of the ‘NH’ group on benzene will be weak. The only mono substitution will take place.

Question 2: Among the α-amino acids – threonine, tyrosine, methionine, arginine and tryptophan, those which contain an aromatic group in their side chain are

a. threonine and arginine
b. tyrosine and tryptophan
c. methionine and tyrosine
d. arginine and tryptophan

Answer: (b)

Question 3: The number of stereoisomers possible for the following compound is CH₃ – CH = CH – CH(OH) – CH₃

a. 1
b. 2
c. 3
d. 4

Answer: (d)

n = 2

Number of possible stereoisomers = 2ⁿ = 2² = 4

G.I. → Can be cis/trans.

The chiral centre → can be R/S.

Therefore possible combination

Cis – R

Cis – S

Trans – R

Trans – S

Question 4: In electrophilic aromatic substitution reactions of chlorobenzene, the ortho/para directing ability of chlorine is due to its

a. Positive inductive effects (+I)
b. negative inductive effect (–I)
c. Positive resonance effect (+R)
d. negative resonance effect (–R)

Answer: (c)

Question 5: Among the following,

the anti-aromatic compounds are

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

Answer: (b)

Question 6: Upon reaction with CH₃MgBr followed by protonation, the compound that produces ethanol is

a. CH₃CHO
b. HCOOH
c. HCHO
d. (CHO)₂

Answer: (c)

Question 7: Which of the following is NOT an oxidation-reduction reaction?

a. H₂ + Br₂ → 2HBr
b. NaCl + AgNO₃→NaNO₃ + AgCl
c. 2Na₂S₂O₃ + I₂ → Na₂S₄O₆ + 2NaI
d. Cl₂ + H₂O → HCl + HOCl

Answer: (b)

Question 8: The thermal stability of alkaline earth metal carbonates – MgCO₃, CaCO₃, SrCO₃ and BaCO₃, follows the order

a. BaCO₃ > SrCO₃ > CaCO₃ > MgCO₃
b. CaCO₃ > SrCO₃ > BaCO₃ > MgCO₃
c. MgCO₃ > CaCO₃ > SrCO₃ > BaCO₃
d. SrCO₃ > CaCO₃ > MgCO₃ > BaCO₃

Answer: (a)

Mg²⁺

Ca²⁺

Sr²⁺

Ba²⁺

Down the group, the size of cation increases and CO₃^2– is also large in size.

And large cation + large anion → more stability.

Hence, more comparable the size of cation and anion, more will be the thermal stability of alkaline earth metal carbonates.

Hence, BaCO₃ > SrCO₃ > CaCO₃ > MgCO₃.

Question 9: When a mixture of diborane and ammonia is heated, the final product is

a. BH₃
b. NH₄BH₄
c. NH₂NH₂
d. B₃N₃H₆

Answer: (d)

Borazole or Inorganic benzene

Question 10: Among the following metals, the strongest reducing agent is

a. Ni
b. Cu
c. Zn
d. Fe

Answer: (c)

According to electrochemical series,

Lesser the value of E⁰_RP, higher will be the tendency to get oxidized itself and reduce others and hence stronger will be the reducing agent.

Question 11: The molecule which is NOT hydrolysed by water at 25°C is

a. AlCl₃
b. SiCl₄
c. BF₃
d. SF₆

Answer: (a)

a. AlCl₃ + 3H₂O → Al(OH)₃ ↓ + 3HCl

(Al has vacant p-orbital where H₂O can attack.)

b. SiCl₄ + 4H₂O → Si(OH)₄ + 4HCl

(Si has vacant d-orbital where H₂O can attack.)

c. BF₃ + 3H₂O → B(OH)₃ + 3HF

(B has vacant p-orbital where H₂O can attack.)

d. SF₆ + H₂O → (At 25^o C) No reaction

Sulphur has vacant d-orbital but due to steric hindrance created by 6F, SF₆ cannot be hydrolysed at 25°C.

Question 12: Among the following compounds, the one which does NOT produce nitrogen gas upon heating is

a. (NH₄)₂Cr₂O₇
b. NaN₃
c. NH₄NO₂
d. (NH₄)₂(C₂O₄)

Answer: (d)

(NH₄)₂Cr₂O₇ → N₂ + Cr₂O₃ + H₂O

2NaN₃ → 2Na + 3N₂

NH₄NO₂ → N₂ + H₂O

(NH₄)₂C₂O₄ → 2NH₃ + H₂C₂O₄

Question 13: Chlorine has two naturally occurring isotopes, ³⁵Cl and ³⁷Cl. If the atomic mass of Cl is 35.45, the ratio of the natural abundance of ³⁵Cl and ³⁷Cl is closest to

a. 3.5 : 1
b. 3 :1
c. 2.5 : 1
d. 4 : 1

Answer: (b)

Atomic mass of Cl = 35.45 = Average atomic mass

Average atomic mass = [n₁M₁ + n₂M₂] / [n₁ + n₂]

35.45 = [n₁ * 35 + n₂ * 37] / [n₁ + n₂]

35.45n₁ + 35.45n₂ = 35n₁ + 37n₂

0.45n₁ = 1.55n₂

n₁ / n₂ = 1.55 / 0.45 ≈ 3 / 1

3 : 1

Question 14: The reaction C₂H₆(g) → C₂H₄(g) + H₂(g) is at equilibrium in a closed vessel at 1000 K. The enthalpy change (ΔH) for the reaction is 137.0 kJ mol^–l. Which one of the following actions would shift the equilibrium to the right?

a. Decreasing the volume of the closed reaction vessel
b. Decreasing the temperature at which the reaction is performed
c. Adding an inert gas to the closed reaction vessel
d. Increasing the volume of the closed reaction vessel

Answer: (d)

C₂H₆(g) → C₂H₄(g) + H₂(g); ΔH = +ve

a. Volume ↓ → Reaction will shift in the direction where the number of moles ↓ according to Le-chatelier principle → backward direction.

b. T ↓ → Reaction will shift in a direction where the temperature will increase (backwards) as the reaction is endothermic.

c. Addition of inert gas at constant volume (no effect on equilibrium).

d. Volume ↑ → Reaction will shift in a direction where the number of moles will increase i.e. forward direction.

Correct option is (d).

Question 15: The enthalpy (H) of an elementary exothermic reaction A -> B is schematically plotted against the reaction coordinate. The plots in the presence and absence of a catalyst are shown in dashed and solid lines, respectively. Identify the correct plot for the reaction.

Answer: (a)

For exothermic reaction

ΔH = E_P – E_R < 0

E_B – E_A < 0

E_B < E_A

Catalyst does not change the initial and final position of the reaction. It will only decrease the activation energy, hence increase the rate of reaction.

Question 16: Mg(OH)₂ is precipitated when NaOH is added to a solution of Mg²⁺. If the final concentration of Mg²⁺ is 10^–10M, the concentration of OH^–(M) in the solution is [Solubility product for Mg (OH)₂ = 5.6 × 10^–12]

a. 0.056
b. 0.12
c. 0.24
d. 0.025

Answer: (c)

Mg(OH)_2(s) ⇌ Mg²⁺ + 2OH^–

Equilibrium s 2s

K_sp = [Mg²⁺][OH^–]² = 5.6 × 10^–12

10^–10 [OH^–]² = 5.6 × 10^–12

Question 17: A constant current (0.5 amp) is passed for 1 hour through

(i) aqueous AgNO₃,

(ii) aqueous CuSO₄ and

(iii) molten AlF₃, separately.

The ratio of the mass of the metals deposited on the cathode is [M_Ag, M_Cu, M_Al are molar masses of the respective metals]

a. M_Ag: 2 M_Cu: 3 M_A1
b. M_Ag: M_Cu: M_A1
c. 6 M_Ag: 3 M_Cu: 2 M_A1
d. 3 M_Ag: 2M_Cu: M_A1

Answer: (c)

Question 18: A reaction has an activation energy of 209 kJ mol^–1. The rate increases 10–fold when the temperature is increased from 27°C to X °C. The temperature X is closest to [Gas constant, R = 8.314 J mol^–1K^–1]

a. 35
b. 40
c. 30
d. 45

Answer: (a)

𝛆_a = 209 kJmol^–1

Given:

T₁ = 27°C = 300 K

T₂ = x°C = (x + 273)K

K₂ / K₁ = 10

Apply

log (K₂ / K₁) = [𝛆_a / 2.303 R] [(1 / T₁) – (1 / T₂)]

1 = log 10 = {[209 * 10³] / [2.303 * 8.314]} {[(1 / 300) – (1 / (x + 273)]}

x + 273 = 308.4 K

x = 35.4°C ≈ 35°C

Question 19: A mineral consists of a cubic close-packed structure formed by O^2–ions, where half the octahedral voids are occupied by Al3+ and one-eighth of the tetrahedral voids, are occupied by Mn²⁺. The chemical formula of the mineral is

a. Mn₃Al₂O₆
b. MnAl₂O₄
c. MnAl₄O₇
d. Mn₂Al₂O₅

Answer: (b)

No. of atoms in ccp = N

Then, No. of O.V. = N

No. of T.V. = 2N

No. of O^–2 per unit cell = 8 × (1 / 8) + 6 × (1 / 2) = 4

No. of Al⁺³ per unit cell = 4 × (1 / 2) = 2

No. of Mn⁺² per unit cell = 8 × (1 / 8) = 1

MnAl₂O₄

Question 20: For a 4p orbital, the numbers of radial and angular nodes, respectively, are

a. 3, 2
b. 1, 2
c. 2, 4
d. 2, 1

Answer: (d)

4p

n = 4, l = 1

No. of Radial nodes = n – l – 1 = 4 – 1 – 1 = 2

No. of Angular nodes = l = 1

Question 21: In the following reaction sequence

X and Y are

Answer: (b)

Anti-markovnikov’s rule

Question 22: In the following reactions

X and Y are

Answer: (a)

Question 23: Which of the following alkenes can generate optically active compounds upon hydrogenation?

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

Answer: (c)

Question 24: When heated in air, brown copper powder turns black. This black powder would turn brown again when heated with

a. CO

b. O₂

c. H₂

d. NH₃

Answer: (c)

Question 25: The geometry and magnetic property of [NiCl₄]^2–, respectively, are

a. Tetrahedral, paramagnetic
b. tetrahedral, diamagnetic
c. Square planar, paramagnetic
d. square planar, diamagnetic

Answer: (a)

Cl^– is a weak ligand

No pairing of the electron will take place.

• sp³ hybridisation, tetrahedral

• 2 unpaired electrons → paramagnetic

Question 26: Among

(i) [Cr(en)₃]³⁺,

(ii) trans-[Cr(en)₂Cl₂]⁺,

(iii) Cis-[Cr(en)₂Cl₂]⁺

(iv) [Co(NH₃)₄Cl²]⁺

the optically active complexes are

a. i and ii
b. i and iii
c. ii and iii
d. ii and iv

Answer: (b)

Question 27: ²²⁷Ac has a half-life of 22 years with respect to radioactive decay. The decay follows two parallel paths: ²²⁷Ac → ²²⁷Th and ²²⁷Ac → ²²³Fr. If the percentage of the two daughter nuclides are 2.0 and 98.0, respectively, the decay constant (in year^–1) for ²²⁷Ac → ²²⁷Th path is closest to

a. 6.3 × 10^–2
b. 6.3 × 10^–3
c. 6.3 × 10^–1
d. 6.3 × 10^–4

Answer: (d)

k₁ / k₂ = 2 / 98

t½ = 22 years

k = 0.693 / t_½ = 0.693 / 22 year = k₁ + k₂

0.693 / 22 = k₁ + (98 / 2) k₁

k₁ = 6.3 × 10^–4

Question 28: A system consisting of 1 mol of an ideal gas undergoes a reversible process, A → B → C → A (schematically indicated in the figure below). If the temperature at the starting point A is 300 K and the work done in the process B → C is 1 L atm, the heat exchanged in the entire process in L atm is

a. 1.0
b. 0.0
c. 1.5
d. 0.5

Answer: (d)

Given: n = 1 mole, T_A = 300 K, ω_BC = 1 L-atm

ω_AB = –Pext.(V₂ – V₁) = –1(1.5 – 1) = –0.5 L-atm

(Isobaric process)

ω_CA = 0 (dV = 0)

(Isochoric process)

ω_total = ω_AB + ω_BC + ω_CA

= –0.5 + 1 + 0 = 0.5 L-atm

ΔU calculation

For complete cyclic process ABC → ΔU = 0 (As U → is a state function) According to first law,

ΔU = q + w = 0

q = –w = –0.5 L-atm

Heat exchanged during the process = 0.5 L-atm.

Correct option is (D).

Question 29: A mixture of toluene and benzene boils at 100°C. Assuming ideal behaviour, the mole fraction of toluene in the mixture is closest to [Vapour pressures of pure toluene and pure benzene at 100°C are 0.742 and 1.800 bar respectively. 1 atm = 1.013 bar]

a. 0.824
b. 0.744
c. 0.544
d. 0.624

Answer: (b)

P°_toluene = 0.742 bar

P°_benzene = 1.800 bar

Mixture boils at 100°C → at 1 atm = 1.013 bar

P_T = 1.013 bar = P_AX_A + P_BX_B

P_T = P^o_t X_t + P^o_b X_b

1.013 = 0.742 X_t + 1.8 (1 – X_t)

X_t = 0.744

X_b = 1 – 0.744 = 0.256

X_t = mole fraction of toluene.

Question 30: A two-dimensional solid pattern formed by two different atoms X and Y is shown below. The black and white squares represent atoms X and Y, respectively. The simplest formula for the compound based on the unit cell from the pattern is

a. XY₈
b. X₄Y₉
c. XY₂
d. XY₄

Answer: (a)

The unit cell of the above pattern will consist of 8 white squares and 1 black square i.e. it will form the centre unit cell.

No. of white square Y = 8

No. of black square X = 1

Formula XY₈.