JEE Main 2022 June 27 – Shift 1 Chemistry Question Paper with Solutions

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JEE Main 2022 27th June Morning Shift Chemistry Question Paper and Solutions

SECTION – A

Multiple Choice Questions: This section contains 20 multiple choice questions. Each question has 4 choices (1), (2), (3) and (4), out of which ONLY ONE is correct.

Choose the correct answer :

1. Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A) : At 10°C, the density of a 5 M solution of KCl [atomic masses of K & Cl are 39 & 35.5 g mol^–1 respectively], is ‘x’ g ml^–1. The solution is cooled to –21°C. The molality of the solution will remain unchanged.

Reason (R) : The molality of a solution does not change with temperature as mass remains unaffected with temperature.

In the light of the above statements, choose the correct answer from the options given below.

(A) Both (A) and (R) are true and (R) is the correct explanation of (A).

(B) Both (A) and (R) are true but (R) is not the correct explanation of (A).

(C) (A) is true but (R) is false.

(D) (A) is false but (R) is true.

Answer (A)

Sol. Density = ‘x’ gm ml^–1

∴ molality,

≃ 8 m (Assuming x = 1)

∴ ΔT_f = iK_fm

Assuming complete dissociation of salt (100%)

(i = 2)

ΔT_f = 2 × 1.86 × 8 ≃ 29.76

Hence, the solution does not freeze at –21°C. This means that molality of the solution won’t change as

x ≥ 1.

Statement (II) is also correct as molality is mass dependent and hence, does not change with temperature. However, as solvents are not mentioned, statement (I) can also be incorrect.

2. Based upon VSEPR theory, match the shape (geometry) of the molecules in List-I with the molecules in List-II and select the most appropriate option.

List-I List-II

(Shape) (Molecules)

(A) T-shaped (I) XeF₄

(B) Trigonal planar (II) SF₄

(C) Square planar (III) CIF₃

(D) See-saw (IV) BF₃

(A) (A)-(I), (B)-(II), (C)-(III), (D)-(IV)

(B) (A)-(III), (B)-(IV), (C)-(I), (D)-(II)

(C) (A)-(III), (B)-(IV), (C)-(II), (D)-(I)

(D) (A)-(IV), (B)-(III), (C)-(I), (D)-(II)

Answer (B)

Sol.

(Shape) (Molecules)

(A) T-shaped (III) ClF₃

(B) Trigonal planar (IV) BF₃

(C) Square planar (I) XeF₄

(D) See-saw (II) SF₄

Hence, (B) is the correct option.

3. Match List-I with List-II

List-I

(A) Spontaneous process

(B) Process with ΔP = 0, ΔT = 0

(C) ΔH_reaction

(D) Exothermic Process

List-II

(I) ΔH < 0

(II) ΔG_T,P < 0

(III) Isothermal and isobaric process

(IV) [Bond energies of molecules in reactants] – [Bond energies of product molecules]

Choose the correct answer from the options given below :

(A) (A)-(III), (B)-(II), (C)-(IV), (D)-(I)

(B) (A)-(II), (B)-(III), (C)-(IV), (D)-(I)

(C) (A)-(II), (B)-(III), (C)-(I), (D)-(IV)

(D) (A)-(II), (B)-(I), (C)-(III), (D)-(IV)

Answer (B)

Sol. Correct match is

1. Spontaneous (II) ΔG_T,P < 0 process (Constant temperature and pressure condition)
2. Process with ΔP = 0, ΔT = 0 (III) Isothermal and isobaric process

(C) ΔH_reaction (IV) [Bond energies of molecules in reactants – bond ene0072gies of product molecules]

(D) Exothermic process (I) ΔH < 0

Hence, the correct option is (B).

4. Match List-I with List-II

List-I List-II

(A) Lyophilic colloid (I) Liquid-liquid colloid

(B) Emulsion (II) Protective colloid

(C) Positively charged (III) FeCl₃ + NaOH

colloid

(D) Negatively charged (IV) FeCl₃ + hot water

colloid

Choose the correct answer from the options given below :

(A) (A)-(II), (B)-(I), (C)-(IV), (D)-(III)

(B) (A)-(III), (B)-(I), (C)-(IV), (D)-(II)

(C) (A)-(II), (B)-(I), (C)-(III), (D)-(IV)

(D) (A)-(III), (B)-(II), (C)-(I), (D)-(IV)

Answer (A)

Sol. Correct match of List-I and List-II is:

1. Lyophilic colloid (II) Protective colloid
2. Emulsion (I) Liquid-liquid colloid

(C) Positively charged (IV) FeCl₃ + hot water colloid (It forms a positively charged sol of Fe(OH)₃)

(D) Negatively charged (III) FeCl₃ + NaOH colloid (Negatively charged colloid is formed due to adsorption of OH^– ions on Fe(OH)₃)

Hence, the correct option is (A).

5. Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A) : The ionic radii of O^2– and Mg²⁺ are same.

Reason (R) : Both O^2– and Mg²⁺ are isoelectronic species.

In the light of the above statements, choose the correct answer from the options given below.

(A) Both (A) and (R) are true and (R) is the correct explanation of (A).

(B) Both (A) and (R) are true but (R) is not the correct explanation of (A).

(C) (A) is true but (R) is false.

(D) (A) is false but (R) is true.

Answer (D)

Sol. Correct order of ionic radii:

O^–2 > Mg⁺²

This is because among isoelectronic species, the size of anions are greater than the size of cations. Statement (II) is correct as both O^–2 and Mg⁺² are isoelectronic.

6. Match List-I with List-II.

List-I List-II

(A) Concentration of (I) Aniline

Gold ore

(B) Leaching of alumina (II) NaOH

(C) Froth stabiliser (III) SO₂

(D) Blister copper (IV) NaCN

Choose the correct answer from the options given below.

(A) (A)-(IV), (B)-(III), (C)-(II), (D)-(I)

(B) (A)-(IV), (B)-(II), (C)-(I), (D)-(III)

(C) (A)-(III), (B)-(II), (C)-(I), (D)-(IV)

(D) (A)-(II), (B)-(IV), (C)-(III), (D)-(I)

Answer (B)

Sol. List-I List-II

(A) Concentration of (IV) NaCN

Gold ore

(B) Leaching of alumina (II) NaOH

(C) Froth stabiliser (I) Aniline

(Aniline and cresols

are used as froth

stabilisers in froth

floatation process)

(D) Blister copper (III) SO₂ (During self reduction process used in the formation of blister copper SO₂ gas is evolved)

Hence (B) is most appropriate option.

7. Addition of H₂SO₄ to BaO₂ produces:

(A) BaO, SO₂ and H₂O

(B) BaHSO₄ and O₂

(C) BaSO₄, H₂ and O₂

(D) BaSO₄ and H₂O₂

Answer (D)

Sol.

Hence, the correct option is (D)

8. BeCI₂ reacts with LiAIH₄ to give:

(A) Be + Li[AICI₄] + H₂

(B) Be + AIH₃ + LiCI + HCI

(C) BeH₂ + LiCI + AICI₃

(D) BeH₂ + Li[AICI₄]

Answer (C)

Sol.

The above reaction using LiAlH₄ is an important preparation method for production of hydrides.

9. Match List-I with List-II.

List-I List-II

(Si-Compounds) (Si-Polymeric/other

Products)

(A) (CH₃)₄Si (I) Chain Silicone

(B) (CH₃)Si(OH)₃ (II) Dimeric Silicone

(C) (CH₃)₂Si(OH)₂ (III) Silane

(D) (CH₃)₃Si(OH) (IV) 2D-Silicone

Choose the correct answer from the options given below:

(A) (A)-(III), (B)-(II), (C)-(I), (D)-(IV)

(B) (A)-(IV), (B)-(I), (C)-(II), (D)-(III)

(C) (A)-(II), (B)-(I), (C)-(IV), (D)-(III)

(D) (A)-(III), (B)-(IV), (C)-(I), (D)-(II)

Answer (D)

Sol. List-I List-II

(Si-Compounds) (Si-Polymeric/other

Products)

(A) (CH₃)₄Si (III) Silane

(B) (CH₃)Si(OH)₃ (IV) 2D-Silicone

(C) (CH₃)₂Si(OH)₂ (I) Chain Silicone

(D) (CH₃)₃Si(OH) (II) Dimeric Silicone

2D-Silicone

10. Heating white phosphorus with conc. NaOH solution gives mainly:

(A) Na₃P and H₂O

(B) H₃PO and NaH

(C) P(OH)₃ and NaH₂PO₄

(D) PH₃ and NaH₂PO₂

Answer (D)

Sol. P₄(white) + NaOH

→ PH₃ + NaH₂PO₂ + H₂O

11. Which of the following will have maximum stabilization due to crystal field?

(A) [Ti(H₂O)₆]³⁺

(B) [Co(H₂O)₆]²⁺

(C) [Co(CN)₆]^–3

(D) [Cu(NH₃)₄]²⁺

Answer (C)

Sol. The given complexes are:

CN^– is the strongest ligand among the given complexes CFSE value for the [Co(CN)₆]^–3 complex will be highest as it has d⁶ configuration with a CFSE value of –2.40 Δ₀ + 2P, where P represents pairing energy and Δ₀ represents splitting energy in octahedral field.

The value of Δ₀ is high for cyanide complexes.

12. Given below are two Statements:

Statement I: Classical smog occurs in cool humid climate. It is a reducing mixture of smoke, fog and sulphur dioxide.

Statement II: Photochemical smog has components, ozone, nitric oxide, acrolein, formaldehyde, PAN etc.

In the light of the above statements, choose the most appropriate answer from the options given below.

(A) Both Statement I and Statement II are correct.

(B) Both Statement I and Statement II are incorrect.

(C) Statement I is correct but Statement II is incorrect.

(D) Statement I is incorrect but Statement II is correct

Answer (A)

Sol. (I) Classical smog occurs in cool humid climate. It is a reducing mixture of smoke, fog and sulphur dioxide. This is a correct statement.

(II) This statement is also based on fact and is a correct statement.

13. Which of the following is structure of a separating funnel?

Answer (A)

Sol. The diagram is option (A) clearly represents separating funnel which is used to separate two immiscible liquids.

14. ‘A’ and ‘B’ respectively are:

(A) 1-methylcyclohex-1, 3-diene & cyclopentene

(B) Cyclohex-1, 3-diene & cyclopentene

(C) 1-methylcyclohex-1, 4-diene & 1-methylcyclo-pent-ene

(D) Cyclohex-1, 3-diene & 1-methylcyclopent-1-ene

Answer (D)

Sol.

(B) should be 1-methylcyclopent-1-ene.

15. The major product of the following reaction is:

Answer (A)

Sol.

Rate of S_N2 >S_N2 (AR)

16. Which of the following reactions will yield benzaldehyde as a product?

(A) (B) and (C)

(B) (C) and (D)

(C) (A) and (D)

(D) (A) and (C)

Answer (C)

Sol.

17. Given below are two statements:

Statement-I : In Hofmann degradation reaction, the migration of only an alkyl group takes place from carbonyl carbon of the amide to the nitrogen atom.

Statement-II : The group is migrated in Hofmann degradation reaction to electron deficient atom.

In the light of the above statements, choose the most appropriate answer from the options given below:

(A) Both Statement I and Statement II are correct

(B) Both Statement I and Statement II are incorrect

(C) Statement I is correct but Statement II is incorrect

(D) Statement I is incorrect but Statement II is correct

Answer (D)

Sol. Hofmann bromamide degradation

In this degradation, the migration of the alkyl/aryl group occurs to the electron deficient nitrogen (nitrene).

Statement (I) is not absolutely correct as it mentions only the alkyl group, whereas migration of aryl groups may also occur depending on migratory aptitude.

Statement (II) is correct as migration occurs to electron deficient atom.

18. Match List-I with List-II

List-I List-II

(Polymer) (Used in)

(A) Bakelite (I) Radio and television cabinets

(B) Glyptal (II) Electrical switches

(C) PVC (III) Paints and Lacqures

(D) Polystyrene (IV) Water pipes

Choose the correct answer from the options given below:

(A) (A) –(II) (B) – (III), (C) – (IV), (D) – (I)

(B) (A) –(I) (B) – (II), (C) – (III), (D) – (IV)

(C) (A) –(IV) (B) – (III), (C) – (II), (D) – (I)

(D) (A) –(II) (B) – (III), (C) – (I), (D) – (IV)

Answer (A)

Sol. List-I List-II

(Polymer) (Used in)

(A) Bakelite (II) Electrical switches

(B) Glyptal (III) Paints and Lacqures

(C) PVC (IV) Water pipes

(D) Polystyrene (I) Radio and television cabinets

Therefore, the correct option is (A).

19. L-isomer of a compound ‘A’ (C₄H₈O₄) gives a positive test with [Ag(NH₃)₂]⁺. Treatment of ‘A’ with acetic anhydride yields triacetate derivative. Compound ‘A’ produces an optically active compound (B) and an optically inactive compound (C) on treatment with bromine water and HNO₃ respectively. Compound (A) is:

Answer (A)

Sol.

When(A) is heated with acetic anhydride, acetylation occurs and –OH group is replaced by and hence, triacetate is formed.

20. Match List-I with List-II

List-I

(C) C₁₇H₃₅COO^–Na⁺+Na₂CO₃+Rosinate

(D) CH₃(CH₂)₁₆COO(CH₂CH₂O)CH₂CH₂OH

List-II

(I) Dishwashing power

(II) Toothpaste

(III) Laundry soap

(IV) Hair conditional

Choose the correct answer from the options given below:

(A) (A) –(III) (B) – (II), (C) – (IV), (D) – (I)

(B) (A) –(IV) (B) – (II), (C) – (III), (D) – (I)

(C) (A) –(IV) (B) – (III), (C) – (II), (D) – (I)

(D) (A) –(III) (B) – (IV), (C) – (I), (D) – (II)

Answer (B)

Sol.

_{→ Cationic detergent}

(Hair conditioner)

(B)

→ Toothpaste

(Anionic detergent)

(C) C₁₇H₃₅COO^–Na⁺+Na₂CO₃+Rosinate

→ Laundry soap

(D) CH₃(CH₂)₁₆COO(CH₂CH₂O)CH₂CH₂OH

→ Dishwashing powder

SECTION – B

Numerical Value Type Questions: This section contains 10 questions. In Section B, attempt any five questions out of 10. The answer to each question is a NUMERICAL VALUE. For each question, enter the correct numerical value (in decimal notation, truncated/rounded-off to the second decimal place; e.g. 06.25, 07.00, –00.33, –00.30, 30.27, –27.30) using the mouse and the on-screen virtual numeric keypad in the place designated to enter the answer.

1. Metal deficiency defect is shown by Fe_0.93O. In the crystal, some Fe²⁺ cations are missing and loss of positive charge is compensated by the presence of Fe³⁺ ions. The percentage of Fe²⁺ ions in the Fe_0.93O crystals is ______. (Nearest integer)

Answer (85)

Sol. Fe_0.93O

Let the number of O^–2 ions be 100

and the number of Fe⁺² ions be X

The number of Fe⁺³ ions be (93 – X)

∴ X(2) + (93 – X)3 = 200

279 – X = 200

X = 79

∴ % of Fe⁺² ions

≃ 85%

2. If the uncertainty in velocity and position of a minute particle in space are, 2.4 × 10^–26 (m s^–1) and 10^–7 (m), respectively. The mass of the particle in g is _____. (Nearest integer)

(Given : h = 6.626 × 10^–34 Js)

Answer (22)

Sol.

≥ 0.02198 kg

∴ Mass of the particle ≃ 22 g

3. 2 g of a non-volatile non-electrolyte solute is dissolved in 200 g of two different solvents A and B whose ebullioscopic constants are in the ratio of 1 : 8. The elevation in boiling points of A and B are in the ratio

Answer (8)

Sol. ΔTb = k_bm

4.

In an experiment, 2.0 moles of NOCl was placed in a one-litre flask and the concentration of NO after equilibrium established, was found to be 0.4 mol/ L. The equilibrium constant at 30°C is _____ × 10^–4.

Answer (125)

Sol.

t = 0 2

t = t_eq 2 – 0.4 0.4 0.2

5. The limiting molar conductivities of NaI, NaNO₃ and AgNO₃ are 12.7, 12.0 and 13.3 mS m² mol–1, respectively (all at 25°C). The limiting molar conductivity of Agl at this temperature is ____ mS m² mol–1.

Answer (14)

Sol.

= 12.7 + 13.3 – 12.0

= 26 – 12

= 14 mS m² mol^–1

6. The rate constant for a first order reaction is given by the following equation :

The activation energy for the reaction is given by ______ kJ mol^–1. (In nearest integer)

(Given : R = 8.3 J K^–1 mol^–1)

Answer (166)

Sol.

∴

E_a = 2 × 10⁴ × 8.3

= 166 kJ/mol

7. The number of statement(s) correct from the following for Copper (at. no. 29) is/are ______.

(A) Cu(II) complexes are always paramagnetic

(B) Cu(I) complexes are generally colourless

(C) Cu(I) is easily oxidized

(D) In Fehling solution, the active reagent has Cu(I)

Answer (3)

Sol. (A) Cu(II) complexes are always paramagnetic as they have one unpaired electron due to d⁹ configuration of Cu(II)

(B) Cu(I) complexes are generally colourless due to d¹⁰ configuration.

(C) Cu(I) is easily oxidised to Cu⁺² in aqueous solution

2Cu⁺ → Cu⁺² + Cu

Cu⁺¹ disproportionates to Cu⁺² and Cu

(

In Fehling’s solution, active reagent has Cu(II) which is reduced to Cu(I) on reaction with aldehydes.

Hence (D) statement is incorrect.

8. Acidified potassium permanganate solution oxidises oxalic acid. The spin-only magnetic moment of the manganese product formed from the above reaction is ______ B.M.

(Nearest Integer)

Answer (6)

Sol.

Mn⁺² has 5 unpaired electrons

∴ Spin only magnetic moment

9. Two elements A and B which form 0.15 moles of A₂B and AB₃ type compounds. If both A₂B and AB₃ weigh equally, then the atomic weight of A is _____ times of atomic weight of B.

Answer (2)

Sol. Mole of A₂B = moles of AB₃

A + 3B = 2A + B

2B = A

Atomic weight of A is 2 times that of B.

10. Total number of possible stereoisomers of dimethyl cyclopentane is _______.

Answer (Bonus)

Sol. Position of methyl groups not mentioned.

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