KVPY-SA 2017 Chemistry Paper with Solutions

Question 1: The structure of 3-methylpent-2-ene is

Answer: (A)

Question 2: The stability of carbanions

follows the order

a. III < IV < I < II
b. I < II < IV < III
c. III < II < I < IV
d. IV < III < II < I

Answer: (C)

And more the +I effect/hyperconjugation effect less will be the stability of carbanion.

Therefore 1° > 2° > 3° (stability of carbanion)

Hence overall order IV > I > II > III.

Question 3: In the following reaction

The major product is

Answer: (D)

Question 4: In the reaction of 1-bromo-3-Chlorocyclobutane with two equivalents of sodium in ether, the major product is

Answer: (D)

Question 5: The order of basicity of

in water is

a. IV < III < I < II
b. II < I < IV < III
c. IV < I < III < II
d. II < III < I < IV

Answer: (C)

More the tendency of donating lone pair more will be the basicity.

More the lone pair involved in resonance less will be the basicity.

More electronegativity, less basicity.

Hence overall order of basicity II > III > I > IV.

Question 6: The first ionisation energy of Na, B, N and O atoms follows the order

a. B < Na < O < N
b. Na < B < O < N
c. Na < O < B < N
d. O < Na < N < B

Answer: (B)

Na →1s² 2s² 2p⁶ 3s¹→ will attain noble gas configuration after removing one e^– → least I.E.

B → 1s² 2s² 2p¹

N → 1s² 2s² 2p³ → half- filled stabilized → highest I.E.

O → 1s² 2s² 2p⁴

In between B and O

Along the period size ↓ I.E. ↑.

Na < B < O < N

Question 7: Among P₂O₅, As₂O₃, Sb₂O₃ and Bi₂O₃ the most acidic oxide is

a. P₂O₅

b. As₂O₃

c. Sb₂O₃

d. Bi₂O₃

Answer: (A)

⁺⁵P₂O₅

⁺³As₂O₃

⁺³Sb₂O₃

⁺³Bi₂O₃

More the oxidation number more will be the tendency to accept lone pair of e^– → more will be the acidic character.

OR

Down the group, metallic character increase, hence the basic character of oxides increases. Hence P₂O₅ is the most acidic oxide.

Question 8: Among K, Mg, Au and Cu, the one which is extracted by heating its ore in the air is

a. K
b. Mg
c. Au
d. Cu

Answer: (D)

2CuFeS₂ + 2SiO₂ + 4O₂ → Cu₂S + 2FeSiO₃ + 3SO₂

Cu₂S + O₂ → 2Cu + SO₂

Question 9: The metal ion with the total number of electrons same as S_2– is

a. Na⁺

b. Ca²⁺

c. Mg²⁺

d. Sr²⁺

Answer: (B)

Given:-

S_2– → 18 e^–

From the options,

Na⁺ → 10e^–

Ca²⁺ → 18e^–

Mg²⁺ → 10e^–

Sr²⁺ → 36 e^–

Ca²⁺ and S^2– have the same number of electrons.

Question 10: X g of Ca [atomic mass = 40] dissolves completely in concentrated HCl solution to produce 5.04 L of H₂ gas at STP. The value of X is closest to

a. 4.5
b. 8.1
c. 9.0
d. 16.2

Answer: (C)

Apply mole – mole analysis:-

1 mole of Ca produce → 1 mole of H₂(g)

Given: [x / 40] = 0.225

= 9.0

Question 11: A 20 g object is moving with a velocity of 100 ms^–1. The de Broglie wavelength (in m) of the object is [Planck’s constant h = 6.626 × 10^–34 J s]

a. 3.313 × 10^–34

b. 6.626 × 10^–34

c. 3.313 × 10^–31

d. 6.626 × 10^–31

Answer: (A)

Given, m = 20g, v = 100 ms^–1

De Broglie wavelength → λ = h / mv = [6.626 × 10^–34 J s] / [20 * 10^-3 kg * 100ms^-1]

Units:

m → kg

v → m / s

h → Js

then λ → (m)

λ = 3.313 × 10^–34 m

Question 12: In a closed vessel at STP, 50 L of CH₄ is ignited with 750 L of air (containing 20% O₂). The number of moles of O₂ remaining in the vessel on cooling to room temperature is closest to

a. 5.8
b. 2.2
c. 4.5
d. 6.7

Answer: (B)

1CH₄ + 2O₂ → CO₂ + 2H₂O

Air → 750L

20% O₂ → [20 * 750] / [110] = 150L

Hence, CH₄ is the limiting reagent (consumed completely).

After the reaction

0 150 – 100 50L 100L

= 50L

Volume of O₂ remained = 50L

Moles of O₂ remained = 50L / 22.4L = 2.2moles

Question 13: CO₂ is passed through lime water. Initially, the solution turns milky and then becomes clear upon continued bubbling of CO₂. The clear solution is due to the formation of

a. CaCO₃

b. CaO
c. Ca(OH)₂

d. Ca(HCO₃)₂

Answer: (D)

Ca(OH)₂ + CO₂ → CaCO₃ + H₂O

(turn milky) → because of the formation of CaCO₃ ppt

CaCO₃ + H₂ O

Clear solution because of the formation of soluble Ca(HCO₃)₂

Question 14: The maximum number of electrons that can be filled in the shell with the principal quantum number n = 3 is

a. 18
b. 9
c. 8
d. 2

Answer: (A)

n = 3

Then possible subshell =

Maximum electrons occupied → 2 + 6 + 10 = 18 electrons

Question 15: The atomic radii of Li, F, Na and Si follow the order

a. Si > Li > Na > F
b. Li > F > Si > Na
c. Na > Si > F > Li
d. Na > Li > Si > F

Answer: (D)

→ Size of 3^rd period element > size of 2^nd element period

→ along the period → size ↓

Overall order

Na > Li > Si > F

Question 16: The reaction of an alkene X with bromine produces a compound Y, which has 22.22% C, 3.71% H and 74.07% Br. The ozonolysis of alkene X gives only one product. The alkene X is: [Given: atomic mass of C = 12; H = 1; Br = 80]

a. Ethylene
b. 1-butene
c. 2-butene
d. 3-hexene

Answer: (C)

Hence empirical formula of Y = C₂H₄Br

On Ozonolysis of X → it gives only one product.

Hence the correct option is C (2 – butene).

Question 17: In the following reaction

respectively, are

Answer: (B)

Question 18: KMnO₄ reacts with H₂O₂ in an acidic medium. The number of moles of oxygen produced per mole of KMnO₄ is

a. 2.5
b. 5
c. 1.25
d. 2

Answer: (A)

Add both reactions

2MnO₄^– + 5H₂O₂ + 8H⁺ → 2Mn²⁺ + 5O₂ + 8H₂O

From the reaction

2 mole of KMnO₄ produce → 5 moles of O₂

1 mole of KMnO₄ produce → 2.5 moles of O₂

Hence mole of O₂ produced / mole of KMnO₄ = 2.5

Question 19: The photoelectric behaviour of K, Li, Mg and Ag metals is shown in the plot below. If the light of wavelength 400 nm is incident on each of these metals, which of them will emit photoelectrons?

a. K
b. K and Li
c. K, Li and Mg
d. K, Li, Mg and Ag

Answer: (B)

Incident light λ = 400nm.

Energy of incident light = 12400 eV / 4000A^o = 3.1 eV

If the energy of incident light ≥ work function then photoelectron will be ejected. From the graph,

In the case of K, Li→ work function < 3.1 eV: hence photoelectrons will be emitted.

But in case of Mg, Ag → work function > 3.1 eV: hence no photoelectrons will be emitted.

Question 20: A piece of metal weighing 100 g is heated to 80°C and dropped into 1 kg of cold water in an insulated container at 15°C. If the final temperature of the water in the container is 15.69°C. If the final temperature of the water in the container is 15.69°C, the specific heat of the metal in J/g °C is

a. 0.38
b. 0.24
c. 0.45
d. 0.13

Answer: (C)

For metal,

M = 100 g

T₁ = 80°C

Heat loss by metal = Heat gain by cold water

– m₁S₁ (T_f – T₁) = m₂S₂ (T_f – T₂)

– 100S₁ (288.69 – 353) = 1000 × 4.2 (288.69 – 288)

S₁ = 0.45 J/g°C

(Specific heat of metal)