JEE Advanced Question Paper 2021 Chemistry Paper 2

Question 1. The reaction sequence(s) that would lead to o-xylene as the major product is(are).

Answer: (a, b)

Question 2. Correct option(s) for the following sequence of reactions is(are)

a. Q = KNO2, W = LiAlH4
b. R = benzenamine, V = KCN
c. Q = AgNO2, R = phenylmethanamine
d. W = LiAlH4, V = AgCN

Answer: (c, d)

Therefore, correct options are

Q = AgNO₂, R = phenylmethanamine

W = LiAlH₄, V = AgCN

Question 3. For the following reaction;

The rate of reaction is

a. The rate constant, k, of the reaction is 13.86 × 10^-4 s^-1.
b. Half-life of X is 50 s.
c. At 50 s, – d[X] / dt = 13.86 × 10^-3 mol L^-1 s^-1.
d. At 100 s, d[Y] / dt = 3.46 × 10^-3 mol L^-1 s^-1.

Answer: (b, c, d)

rate =

2X + Y → P

2 mole 1 mole

1 mole 0.5 mole 0.5 mole

– d[X] / dt = k₁[X] = 2k[X] ⇒ 2k = k₁

– d[Y] / dt = k₂[X] = 2k[X] ⇒ k₂ = k

2K = 1/50 In2

K = 1 / 100 In2 = 0.6 93 / 100 = 6.93 × 10^-3 × s^-1 = 50 sec

At 50 sec,

d[X] / dt = 2k[X] = 2 × 0.693 / 100 × 1

= 13.86 × 10^-3 mol L^-1 s^-1

At 100 sec

-d[Y] / dt = k₂[X] = k[X] × 0.693 / 100 × ½

(Concentration of X after 2 half-lives = ½ M)

= 3.46 × 10^-3 mol L^-1 s^-1

Question 4. Some standard electrode potentials at 298 K are given below:

Pb²⁺/Pb –0.13 V

Ni²⁺/Ni –0.24 V

C²⁺/Cd –0.40 V

Fe²⁺/Fe –0.44 V

To a solution containing 0.001 M of X²⁺ and 0.1 M of Y²⁺, the metal rods X and Y are inserted (at 298 K) and connected by a conducting wire. This resulted in the dissolution of X.

The correct combination(s) of X and Y, respectively, is(are)

(Given: Gas constant, R = 8.314 J K^-1 mol^-1, Faraday constant, F = 96500 C mol^-1)

a. Cd and Ni
b. Cd and Fe
c. Ni and Pb
d. Ni and Fe

Answer: (a, b, c)

X + Y²⁺ X²⁺ + Y

E = E^o + 0.06

(a) E° = –(– .4) + (– .24) = .16 > 0

(b) E° = –(– .4) + (– .44) = –.04 < 0 and E_cell = –0.04 + 0.06 = +0.02 > 0

(c) E° = –(– .24) + (– .13) = .11 > 0

(d) E° = –(– .24) + (– .44) = –.2 < 0

∴ E_cell = –0.2 + 0.06 = –0.14 < 0

∴ If E_cell > 0 then the cell construction is possible.

Question 5. The pair(s) of complexes wherein both exhibit tetrahedral geometry is(are) (Note: py = pyridine, Given: Atomic numbers of Fe, Co, Ni and Cu are 26, 27, 28 and 29, respectively)

a. [FeCl4]^– and [Fe(CO)4]^2–
b. [Co(CO)4]^– and [CoCl4]^2–
c. [Ni(CO)4] and [Ni(CN)4]^2–
d. [Cu(py)4]⁺ and [Cu(CN)4]^3–

Answer: (a, b, d)

In 3d¹⁰ electronic configuration, only sp³ hybridisation and tetrahedral geometry are possible.

Question 6. The correct statement(s) related to oxoacids of phosphorous is(are).

a. Upon heating, H₃PO₃ undergoes a disproportionation reaction to produce H₃PO₄ and PH₃.
b. While H₃PO₃ can act as a reducing agent, H₃PO₄ cannot.
c. H₃PO₃ is a monobasic acid.
d. The H atom of the P-H bond in H₃PO₃ is not ionizable in water.

Answer: (a, b, d)

In H₃PO₄, phosphorous is present in the highest oxidation state, i.e., +5. So H3PO4 cannot act as a reducing agent. Structure of H₃PO₃,

It is a dibasic acid.

H atom present in the P–H bond is not ionizable.

These P-H bonds are not ionisable to give H⁺ and do not play any role in basicity. Only those H atoms which are attached with oxygen in P-OH form are ionisable and cause the basicity. Thus, H₃PO₃ and H₃PO₄ are dibasic and tribasic, respectively as the structure of H₃PO₃ has two P – OH bonds and H₃PO₄ three.

Question Statement for Questions 7 and 8.

At 298 K, the limiting molar conductivity of a weak monobasic acid is 4 × 10² S cm² mol^–1. At 298 K, for an aqueous solution of the acid, the degree of dissociation is a and the molar conductivity is y × 10² S cm² mol^–1. At 298 K, upon 20 times dilution with water, the molar conductivity of the solution becomes 3y × 10² S cm² mol^–1.

Question 7. The value of α is _______.

Answer: (0.215)

Question 8. The value of y is _______.

Answer: (0.86)

Solution for Questions 7 and 8.

Molar conductivity of HX at infinite dilution

Molar conductivity of HX at conc. c₁ = y × 10² S cm² mol^-1

20 – 60α₁ = 9 – 9α₁

⇒ α₁ = 11/51 = 0.215

Y = 4α₁ = 0.086

Question Statement for Questions 9 and 10.

The reaction of x g of Sn with HCl quantitatively produced a salt. The entire amount of the salt reacted with y g of nitrobenzene in the presence of the required amount of HCl to produce 1.29 g of an organic salt (quantitatively).

(Use Molar masses (in g mol^–1) of H, C, N, O, Cl and Sn as 1, 12, 14, 16, 35 and 119, respectively).

Question 9. The value of x is ________.

Answer: (3.57)

Sn + HCl → SnCl₂

⇒ Moles of ammonium salt = 1.29 / 129 = 0.01

⇒ Moles of nitrobenzene = 0.01

No. of eq. of nitrobenzene = No. of eq. of SnCl₂

6 × (0.01) = 2 ×

⇒ n_Sn = 0.03

w_Sn = 0.03 × 119

x = 3.57

Question 10. The value of y is ________.

Answer: (1.23)

Solution of Question Nos. 9 and 10

Sn + HCl → SnCl₂

⇒ Moles of ammonium salt = 1.29 / 129 = 0.01

⇒ Moles of nitrobenzene = 0.01

⇒ y = 0.01 × Molar mass of nitrobenzene

= 0.01 × 123

= y = 1.23

Question Statement for Questions 11 and 12.

A sample (5.6 g) containing iron is completely dissolved in cold dilute HCl to prepare a 250 mL of solution. Titration of 25.0 mL of this solution requires 12.5 mL of 0.03 M KMnO4 solution to reach the endpoint. Number of moles of Fe²⁺ present in 250 mL solution is x × 10^–2 (consider complete dissolution of FeCl2). The amount of iron present in the sample is y% by weight.

(Assume: KMnO₄ reacts only with Fe²⁺ in the solution Use: Molar mass of iron as 56 g mol^–1)

Question 11. The value of x is ________.

a.swer: (1.875)

Question 12. The value of y is ________.

Answer: (18.75)

Solution of Question Nos. 11 and 12

8H⁺ + 5Fe²⁺ + MnO ^– → 5Fe³⁺ + Mn²⁺ + 4H₂O

For 25 ml,

meq of Fe²⁺ = meq of MnO ^–

= 12.5 × 0.03 × 5

For 250 ml,

mmoles of Fe²⁺ = 12.5 × 0.03 × 5 × 250 / 25

moles of Fe²⁺ = 18.75 / 1000 mol

= 18.75 × 10^–3 mol

= 1.875 × 10^–2 mol

x = 1.875

Weight of Fe²⁺ = 1.875 × 10^–2 × 56 = 1.05 g

% purity of Fe²⁺ y = 18.75%

= 1.05 / 5.6 ×100

Statement: The amount of energy required to break a bond is the same as the amount of energy released when the same bond is formed. In a gaseous state, the energy required for homolytic cleavage of a bond is called Bond Dissociation Energy (BDE) or Bond Strength. BDE is affected by the s-character of the bond and the stability of the radicals formed. Shorter bonds are typically stronger bonds. BDEs for some bonds are given below:

Question 13. The correct match of the C-H bonds (shown in bold) in Column J with their BDE in Column K is;

a. P – iii, Q – iv, R – ii, S – i
b. P – i, Q – ii, R – iii, S – iv
c. P – iii, Q – ii, R – i, S – iv
d. P – ii, Q – i, R – iv, S – iii

Answer: (a)

Order of stability of free radical

Q > P > R > S

Stability of free radical α 1 / Bond energy

∴ Order of bond energy :

S > R > P > Q

Question 14. For the following reaction,

the correct statement is;

a. Initiation step is exothermic with DH° = –58 kcal mol^–1
b. Propagation step involving ^·CH₃ formation is exothermic with DH° = –2 kcal mol^–1
c. Propagation step involving CH₃Cl formation is endothermic with DH° = +27 kcal mol^–1
d. The reaction is exothermic with DH° = –25 kcal mol^–1

Answer: (d)

Step (1) → Endothermic (bond breaking)

Step (2) → ∆H = 105 – 103 = 2 kcal/mol (Endothermic)

Step (3) → ∆H = 58 – 85 = –27 kcal/mol (Exothermic)

For complete reaction

∆H = 58 + 105 – 85 – 103

= –25 kcal/mol

Question Statement for Questions 15 and 16.

The reaction of K₃[Fe(CN)₆] with freshly prepared FeSO₄ solution produces a dark blue precipitate called Turnbull’s blue. The reaction of K₄[Fe(CN)₆] with the FeSO₄ solution in the complete absence of air produces a white precipitate X, which turns blue in the air. Mixing the FeSO₄ solution with NaNO₃, followed by slow addition of concentrated H₂SO₄ through the side of the test tube produces a brown ring.

Question 15. Precipitate X is

a. Fe₄[Fe(CN)₆]₃
b. Fe₄[Fe(CN)₆]
c. K₂Fe[Fe(CN)₆]
d. KFe[Fe(CN)₆]

Answer: (c)

Question 16. Among the following, the brown ring is due to the formation of

a. [Fe(NO)2(SO4)2]^2–
b. [Fe(NO)2(H2O)4]³⁺
c. [Fe(NO)4(SO4)2]
d. [Fe(NO)(H2O)5]²

Answer: (d)

Solution of Question Nos. 15 and 16

Question 17. One mole of an ideal gas at 900 K, undergoes two reversible processes, I followed by II, as shown below. If the work done by the gas in the two processes are the same, the value of

(U: internal energy, S: entropy, p: pressure, V: volume, R: gas constant)

(Given: molar heat capacity at constant volume, C of the gas is 5 / 2 R)

Answer: (10)

Process I is adiabatic reversible

Process II is a reversible isothermal process

Process I – (Adiabatic Reversible)

∆U / R = 450 – 2250

∆U = -1800 R

W_I = ∆U = -1800R

Process II – (Reversible Isothermal Process)

T1 = 900 K

Calculation of T₂ after the reversible adiabatic process

–1800R = nC_v(T₂ – T₁)

–1800R 1 × 5/2 R(T₂ – 900)

T₂ = 180 K

W_II = –nRT₂ In = W

–1 × R × 180ln v₃ / v₂ –1800R

ln v₃ / v₂10

Question 18. Consider a helium (He) atom that absorbs a photon of wavelength 330 nm. The change in the velocity (in cm s^-1) of the He atom after the photon absorption is_____.

(Assume: Momentum is conserved when the photon is absorbed.

Use: Planck constant = 6.6 × 10^-34 J s, Avogadro number = 6 × 1023 mol^-1, Molar mass of He = 4 g mol^-1)

Answer: (30)

Momentum of photon =

Momentum of 1 mole of He-atoms = m∆v

∴ m∆v = N_A × h / λ

∴ Change in velocity of He-atoms = 30 cm s^-1

Question 19. Ozonolysis of ClO2 produces oxide of chlorine. The average oxidation state of chlorine in this oxide is ____.

Answer: (6)

ClO₂ contains an odd electron and is paramagnetic. It reacts with ozone to give O₂ and Cl₂O₆.

2ClO₂ + 2O₃ → Cl₂O₆ + 2O₂

In Cl₂O₆, the average oxidation state of Cl is +6.