JEE Advanced Question Paper 2022 Chemistry Paper 1

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JEE Advanced 2022 Chemistry Question Paper 1 with Solutions

SECTION – 1 (Maximum marks : 24)

∙ This section contains EIGHT (08) questions.

∙ The answer to each question is a NUMERICAL VALUE.

∙ For each question, enter the correct numerical value of the answer using the mouse and the on-screen virtual numeric keypad in the place designated to enter the answer. If the numerical value has more than two decimal places, truncate/round-off the value to TWO decimal places.

∙ Answer to each question will be evaluated according to the following marking scheme:

Full Marks : +3 ONLY if the correct numerical value is entered;

Zero Marks : 0 In all other cases.

1. 2 mol of Hg(g) is combusted in a fixed volume bomb calorimeter with excess of O₂ at 298 K and 1 atm into HgO(s). During the reaction, temperature increases from 298.0 K to 312.8 K. If heat capacity of the bomb calorimeter and enthalpy of formation of Hg(g) are 20.00 kJ K^–1 and 61.32 kJ mol^–1 at 298 K, respectively, the calculated standard molar enthalpy of formation of HgO(s) at 298 K is X kJ mol^–1. The value of |X| is _______.

Answer (90.39)

Sol. 2Hg(g) + O₂(g) ⎯→ 2HgO(s)

Heat capacity of calorimeter = 20 kJ K^–1

Rise in temperature = 14.8 K

Heat evolved = 20 × 14.8 = 296 kJ

ΔH° = ΔU° + Δn_gRT

= –296 – 3 × 8.3 × 298 × 10^–3

≃ –303.42 kJ

2. The reduction potential (E⁰, in V) of

Answer (0.77)

Sol. (1)

(2)

(3)

Adding (1), (2) and (3),

= (–5.04 – 2.42 + 2.06) F

–7F E° = –5.4F

E° = 0.77 V

3. A solution is prepared by mixing 0.01 mol each of H₂CO₃, NaHCO₃, Na₂CO₃, and NaOH in 100 mL of water. pH of the resulting solution is _______.

Answer (10.02)

Sol. First acid base reaction between H₂CO₃ and NaOH takes place.

In the final solution, we have 0.01 mole Na₂CO₃ and 0.02 moles of NaHCO₃.

Here, we have a buffer of NaHCO₃ and Na₂CO₃.

= 10.32 – log2

= 10.32 – 0.3

= 10.02

∴ pH = 10.02

4. The treatment of an aqueous solution of 3.74 g of Cu(NO₃)₂ with excess KI results in a brown solution along with the formation of a precipitate. Passing H₂S through this brown solution gives another precipitate X. The amount of X (in g) is ________.

Answer (00.32)

Sol.

5. Dissolving 1.24 g of white phosphorous in boiling NaOH solution in an inert atmosphere gives a gas Q.
The amount of CuSO₄ (in g) required to completely consume the gas Q is _______.

Answer (2.38)

^Sol.

As NaOH is present in excess. So, amount of phosphine formed is 0.01 mole (as P₄ is limiting)

6. Consider the following reaction.

On estimation of bromine in 1.00 g of R using Carius method, the amount of AgBr formed (in g) is ________.

Answer (01.50)

Sol. 2P + 3Br₂ → 2PBr₃

Number of moles in 1 gm of (R) = 1/250

Number of moles of AgBr formed from (R) = 2/250

7. The weight percentage of hydrogen in Q, formed in the following reaction sequence, is ________.

Answer (1.31)

Sol.

Formula of compound = C₆H₃N₃O₇

Molar Mass of compound = (12 × 6 + 3 + 14 × 3 + 16 × 7) g

= 229 g

8. If the reaction sequence given below is carried out with 15 moles of acetylene, the amount of the product D formed (in g) is __________.

The yields of A, B, C and D are given in parentheses.

Answer (136.00)

Sol.

Molecular formula of D is C₈H₈O₂

Molar mass of D is (12 × 8 + 8 × 1 + 16 × 2) = 136 g

∴ Mass of D is 136

SECTION – 2 (Maximum marks : 24)

∙ This section contains SIX (06) questions.

∙ Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four option(s) is(are) correct answer(s).

∙ For each question, choose the option(s) corresponding to (all) the correct answer(s).

∙ Answer to each question will be evaluated according to the following marking scheme:

Full Marks : +4 ONLY if (all) the correct option(s) is(are) chosen;

Partial Marks : +3 If all the four options are correct but ONLY three options are chosen;

Partial Marks : + 2 If three or more options are correct but ONLY two options are chosen, both of which are correct;

Partial Marks : +1 If two or more options are correct but ONLY one option is chosen and it is a correct option;

Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered);

Negative Marks : –2 In all other cases.

9. For diatomic molecules, the correct statement(s) about the molecular orbitals formed by the overlap of two 2p_z orbitals is(are)

(A) σ orbital has a total of two nodal planes.

(B) σ* orbital has one node in the xz-plane containing the molecular axis.

(C) π orbital has one node in the plane which is perpendicular to the molecular axis and goes through the center of the molecule.

(D) π* orbital has one node in the xy-plane containing the molecular axis.

Answer (A, D)

Sol.

10. The correct option(s) related to adsorption processes is(are)

(A) Chemisorption results in a unimolecular layer.

(B) The enthalpy change during physisorption is in the range of 100 to 140 kJ mol^–1.

(C) Chemisorption is an endothermic process.

(D) Lowering the temperature favours physisorption processes.

Answer (A, D)

Sol. (A) First statement is correct as chemisorption results in a unimolecular layer and physisorption result in a multimolecular layer.

(B) Second statement is incorrect as enthalpy change during physisorption is of the range of (20 – 40) kJ mol^–1.

(C) Chemisorption is an exothermic process with (80 – 240) kJ mol^–1 as the enthalpy of adsorption.

(D) Lowering the temperature results in increase in the extent of physisorption.

Hence (A) and (D) are correct.

11. The electrochemical extraction of aluminium from bauxite ore involves

(A) the reaction of Al₂O₃ with coke (C) at a temperature > 2500°C.

(B) the neutralization of aluminate solution by passing CO₂ gas to precipitate hydrated alumina (Al₂O₃.3H₂O).

(C) the dissolution of Al₂O₃ in hot aqueous NaOH.

(D) the electrolysis of Al₂O₃ mixed with Na₃AlF₆ to give Al and CO₂.

Answer (B, C, D)

Sol. (A) The reduction of Al₂O₃ with coke (C) at a temperature > 2500°C is not carried out due to the formation of carbides.

(B) It is correct as neutralisation of aluminate solution is done by passing CO₂ gas to precipitate hydrated alumina.

(C) Reaction of powdered one is carried out with hot concentrated NaOH at 473 K – 523 K and 35 – 36 bar pressure.

(D) Electrolysis of Al₂O₃ is done mixed with Na₃AlF₆ to produce Al and CO₂. It is a correct statement.

12. The treatment of galena with HNO₃ produces a gas that is

(A) paramagnetic

(B) bent in geometry

(C) an acidic oxide

(D) colorless

Answer (A, D)

Sol. PbS + dil. HNO₃ → Pb(NO₃)₂ + S + NO + H₂O

NO is paramagnetic due to the presence of unpaired electrons. It is a neutral oxide. It is colourless.

Hence, (A) and (D) are correct statements.

13. Considering the reaction sequence given below, the correct statement(s) is(are)

(A) P can be reduced to a primary alcohol using NaBH₄.

(B) Treating P with conc. NH₄OH solution followed by acidification gives Q.

(C) Treating Q with a solution of NaNO₂ in aq. HCl liberates N₂.

(D) P is more acidic than CH₃CH₂COOH.

Answer (B, C, D)

Sol.

14. Considering the following reaction sequence,

the correct option(s) is(are)

Answer (A, B, C)

Sol.

→ P may be → H₂/Pd, ethanol; Sn/HCl

→ R may be → NaNO₂/HCl; HNO₂

→ U may be → (i) H₃PO₂, (ii) KMnO₄ – KOH, Δ or (i) CH₃ – CH₂ – OH, (ii) KMnO₄ – KOH, Δ

SECTION – 3 (Maximum marks : 12)

∙ This section contains FOUR (04) Matching List Sets.

∙ Each set has ONE Multiple Choice Question.

∙ Each set has TWO lists: List-I and List-II.

∙ List-I has Four entries (I), (II), (III) and (IV) and List-II has Five entries (P), (Q), (R), (S) and (T).

∙ FOUR options are given in each Multiple Choice Question based on List-I and List-II and ONLY ONE of these four options satisfies the condition asked in the Multiple Choice Question.

∙ Answer to each question will be evaluated according to the following marking scheme:

Full Marks : +3 ONLY if the option corresponding to the correct combination is chosen;

Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered);

Negative Marks : −1 In all other cases.

15. Match the rate expressions in LIST-I for the decomposition of X with the corresponding profiles provided in
LIST-II. X_s and k are constants having appropriate units.

(A) I→ P; II→ Q; III → S; IV → T

(B) I→ R; II→ S; III → S; IV → T

(C) I→ P; II→ Q; III → Q; IV → R

(D) I→ R; II→ S; III → Q; IV → R

Answer (A)

Sol. (I)

Case-1: [X] >> X_s; [X] + X_s ≈ [X]

I → P, S

Case-2: [X] << X_s; [X] + X_s ≈ X_s

∴ I → Q, T

Case-3: [X] ≈ X_s

In this case curve-R given in List-II will match.

∴ I → P, Q, R, S, T (The graph of half-life should start from origin)

(II)

∵ [X] << X_s

∴ X_s + [X] ≈ X_s

∴ II → Q, T

(III)

∵ [X] >> X_s

∴ X_s + [X] ≈ [X]

∴ III → P, S

(IV)

∵ [X] >> X_s

∴ X_s + [X] ≈ [X]

∴ IV → Q, T

16. LIST-I contains compounds and LIST-II contains reactions.

Match each compound in LIST-I with its formation reaction(s) in LIST-II, and choose the correct option

(A) I → Q; II → P; III → S; IV → R

(B) I → T; II → P; III → Q; IV → R

(C) I → T; II → R; III → Q; IV → P

(D) I → Q; II → R; III → S; IV → P

Answer (D)

Sol.

I → Q

II → R

III → S

IV → P

Option (D) is correct.

17. LIST-I contains metal species and LIST-II contains their properties.

[Given: Atomic number of Cr = 24, Ru = 44, Fe = 26]

Match each metal species in LIST-I with their properties in LIST-II, and choose the correct option

(A) I → R, T; II → P, S; III → Q, T; IV → P, Q

(B) I → R, S; II → P, T; III → P, Q; IV → Q, T

(C) I → P, R; II → R, S; III → R, T; IV → P, T

(D) I → Q, T; II → S, T; III → P, T; IV → Q, R

Answer (A)

Sol. (I) [Cr(CN)₆]^4–:

Cr²⁺ = [Ar] 3d⁴ 4s⁰

It is d²sp³ hybridised as CN^– is a strong field ligand.

(II) [RuCl₆]^2–:

Ru⁴⁺ = [Kr] 4d⁴ 5s⁰

t_2g set contains four electron.

(III) [Cr(H₂O)₆]²⁺:

Cr²⁺ = [Ar] 3d⁴ 4s⁰

It has four unpaired electrons as H₂O is weak field ligand.

So, its μ = 4.9 B.M.

(IV) [Fe(H₂O)₆]²⁺:

Fe²⁺ = [Ar]3d⁶ 4s⁰

It has four unpaired electrons, its μ = 4.9 B.M.

18. Match the compounds in LIST-I with the observations in LIST-II, and choose the correct option.

(A) I → P, Q; II → S; III → Q, R; IV → P

(B) I → P; II → R, S; III → R; IV → Q, S

(C) I → Q, S; II → P, T; III → P; IV → S

(D) I → P, S; II → T; III → Q, R; IV → P

Answer (D)

Sol. (I) Aniline:

Since it contains both carbon and nitrogen so its sodium fusion extract with boiling FeSO₄, followed by acidification with conc. H₂SO₄ gives Prussian blue colour.

Fe²⁺ + 6CN^– → [Fe(CN)₆]^4–

I – (P, S)

(II) o-Cresol:

(III) Cysteine:

Since it has both, sulphur and nitrogen, so its sodium fusion extract will give blood red colour with Fe³⁺ and it has carboxylic group so it will give effervescence with NaHCO₃.

(IV) Caprolactam:

Its sodium fusion extract will give Prussian blue colour on boiling with FeSO₄ followed by acidification with conc. H₂SO₄.

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