JEE Main 2022 June 28 – Shift 2 Chemistry Question Paper with Solutions

JEE Main 2022 June 28 Shift 2 Chemistry question paper with solutions can be accessed here online and in the PDF format. All the JEE main 2022 question papers are solved by our experts to help students verify their answers, and analyse their performance in the examination. Students are advised to solve multiple JEE main question papers to improve their problem-solving skills, and focus on time management in the exam. You can also get the JEE main answer keys for the chemistry papers on our website to verify your answers, whenever you begin to solve the question papers.

JEE Main 2022 June 28th Chemistry Shift 2 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. Compound A contains 8.7% Hydrogen, 74% Carbon and 17.3% Nitrogen. The molecular formula of the compound is,

Given : Atomic masses of C, H and N are 12, 1 and 14 amu, respectively.

The molar mass of the compound A is 162 g mol^–1.

(A) C₄H₆N₂

(B) C₂H₃N

(C) C₅H₇N

(D) C₁₀H₁₄N₂

Answer (D)

Sol.

Empirical Formula = C₅H₇N

Empirical formula mass = 81 g

n × 81 = 162

n = 2

Hence, molecular formula is C₁₀H₁₄N₂.

2. Consider the following statements :

(A) The principal quantum number ‘n’ is a positive integer with values of ‘n’ = 1, 2, 3, ….

(B) The azimuthal quantum number ‘l’ for a given ‘n’ (principal quantum number) can have values as ‘l’ = 0, 1, 2, …n

(C) Magnetic orbital quantum number ‘m_l’ for a particular ‘l’ (azimuthal quantum number) has (2l + 1) values.

(D) ±1/2 are the two possible orientations of electron spin.

(E) For l = 5, there will be a total of 9 orbital

Which of the above statements are correct?

(A) (A), (B) and (C)

(B) (A), (C), (D) and (E)

(C) (A), (C) and (D)

(D) (A), (B), (C) and (D)

Answer (C)

Sol. Possible values of l for a given ‘n’ = 0,1,2 …(n – 1)

For l = 5, total orbitals = 2l + 1

= 2(5) + 1 = 11 orbital

Hence A, C and D are correct statements

3. In the structure of SF₄, the lone pair of electrons on S is in.

(A) Equatorial position and there are two lone pair – bond pair repulsions at 90º

(B) Equatorial position and there are three lone pair – bond pair repulsions at 90º

(C) Axial position and there are three lone pair – bond pair repulsion at 90º

(D) Axial position and there are two lone pair – bond pair repulsion at 90º

Answer (A)

Sol. SF₄ → sp³d hybridisation.

The lone pair of electrons on S is in equatorial position and there are two lone pair-bond pair repulsions at 90°.

4. A student needs to prepare a buffer solution of propanoic acid and its sodium salt with pH 4. The ratio of

Given : K_a(CH₃CH₂COOH) = 1.3 × 10^–5

(A) 0.03

(B) 0.13

(C) 0.23

(D) 0.33

Answer (B)

Sol.

From Henderson equation

5. Match List-I with List-II :

Choose the correct answer from the options given below:

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

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

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

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

Answer (C)

Sol. (A) Negatively charged sol CdS sol

(B) Macromolecular colloid Starch

(C) Positively charged sol Fe₂O₃.xH₂O

(D) Cheese A gel

6. Match List-I with List-II:

List-I(Oxide) List-II (Nature)

(A) Cl₂O₇ (I) Amphoteric

(B) Na₂O (II) Basic

(C) Al₂O₃ (III) Neutral

(D) N₂O (IV) Acidic

Choose the correct answer from the options given below:

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

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

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

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

Answer (B)

Sol. (A) Cl₂O₇ → Acidic

(B) Na₂O → Basic

(C) Al₂O₃ → Amphoteric

(D) N₂O → Neutral

Oxides of metals are basic in nature whereas oxides of non metals are acidic in nature. N₂O is a neutral oxide.

7. In the metallurgical extraction of copper, following reaction is used :

FeO + SiO₂ → FeSiO₃

FeO and FeSiO₃ respectively are.

(A) Gangue and flux

(B) Flux and slag

(C) Slag and flux

(D) Gangue and slag

Answer (D)

Sol. FeO + SiO₂ → FeSiO₃

Gangue Slag

8. Hydrogen has three isotopes: protium (¹H), deuterium (²H or D) and tritium (³H or T). They have nearly same chemical properties but different physical properties. They differ in

(A) Number of protons

(B) Atomic number

(C) Electronic configuration

(D) Atomic mass

Answer (D)

Sol. ¹H, ²D and ³H have same atomic number but their atomic masses are different.

Isotopes have same atomic number i.e. same number of protons.

9. Among the following, basic oxide is:

(A) SO₃

(B) SiO₂

(C) CaO

(D) Al₂O₃

Answer (C)

Sol. Since, oxides of metals are basic in nature. Hence CaO is a basic oxide

SO₃ and SiO₂ are acidic oxides and Al₂O₃ is a amphoteric oxide.

10. Among the given oxides of nitrogen; N₂O, N₂O₃, N₂O₄ and N₂O₅, the number of compound/(s) having N – N bond is:

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Sol.

N₂O, N₂O₃ and N₂O₄ contain N — N bond

11. Which of the following oxoacids of sulphur contains ‘‘S’’ in two different oxidation states?

(A) H₂S₂O₃

(B) H₂S₂O₆

(C) H₂S₂O₇

(D) H₂S₂O₈

Answer (A)

Sol. In H₂S₂O₃, sulphur exhibits two different oxidation states +6 and –2.

12. Correct statement about photo-chemical smog is:

(A) It occurs in humid climate.

(B) It is a mixture of smoke, fog and SO₂.

(C) It is reducing smog.

(D) It results from reaction of unsaturated hydrocarbons.

Answer (D)

Sol. Photochemical smog occurs in warm, dry and sunny climate. The main components of photochemical smog result from the action of unsaturated hydrocarbons and nitrogen oxides.

This is an oxidising smog.

13. The correct IUPAC name of the following compound is:

(A) 4-methyl-2-nitro-5-oxohept-3-enal

(B) 4-methyl-5-oxo-2-nitrohept-3-enal

(C) 4-methyl-6-nitro-3-oxohept-4-enal

(D) 6-formyl-4-methyl-2-nitrohex-3-enal

Answer (C)

Sol.

4-methyl-6-nitro-3-oxohept-4-enal

14. The major product (P) of the given reaction is (where, Me is –CH₃)

Answer (C)

Sol.

15.

In the above reaction ‘A’ is

Answer (C)

Sol.

16. Isobutyraldehyde on reaction with formaldehyde and K₂CO₃ gives compound ‘A’. Compound ‘A’ reacts with KCN and yields compound ‘B’, which on hydrolysis gives a stable compound ‘C’. The compound ‘C’ is

Answer (C)

Sol.

17. With respect to the following reaction, consider the given statements:

(A) o-Nitroaniline and p-nitroaniline are the predominant products.

(B) p-Nitroaniline and m-nitroaniline are the predominant products.

(C) HNO₃ acts as an acid.

(D) H₂SO₄ acts as an acid.

Choose the correct option.

(A) (A) and (C) are correct statements.

(B) (A) and (D) are correct statements.

(C) (B) and (D) are correct statements.

(D) (B) and (C) are correct statements.

Answer (C)

Sol.

Hence, H₂SO₄ acts as an acid.

18. Given below are two statements, one is Assertion (A) and other is Reason (R).

Assertion (A): Natural rubber is a linear polymer of isoprene called cis-polyisoprene with elastic properties.

Reason (R): The cis-polyisoprene molecules consist of various chains held together by strong polar interactions with coiled structure.

In the light of the above statements, choose the correct one 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 (C)

Sol.

The cis-polyisoprene molecule consists of various chains held together by weak van der Waals interactions and has a coiled structure.

Hence assertion is true but reason is false.

19. When sugar ‘X’ is boiled with dilute H₂SO₄ in alcoholic solution, two isomers ‘A’ and ‘B’ are formed. ‘A’ on oxidation with HNO₃ yields saccharic acid whereas ‘B’ is laevorotatory. The compound ‘X’ is :

(A) Maltose

(B) Sucrose

(C) Lactose

(D) Starch

Answer (B)

Sol.

D-(–)-Fructose is a laevorotatory compound.

20. The drug tegamet is:

Answer (C)

Sol. Tegamet (cimetidine) is

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. 100 g of an ideal gas is kept in a cylinder of 416 L volume at 27°C under 1.5 bar pressure. The molar mass of the gas is ________ g mol^–1. (Nearest integer).

(Given : R = 0.083 L bar K^–1 mol^–1)

Answer (4)

Sol. From combined gas law,

PV = nRT

M = 4 g/mol

2. For combustion of one mole of magnesium in an open container at 300 K and 1 bar pressure, Δ_CH^Θ = –601.70 kJ mol^–1, the magnitude of change in internal energy for the reaction is ______ kJ. (Nearest integer)

(Given : R = 8.3 J K^–1 mol^–1)

Answer (600)

Sol.

ΔU = –601.70 + 1.245

ΔU ≃ –600 kJ

Magnitude of change in internal energy is 600 kJ.

3. 2.5 g of protein containing only glycine (C₂H₅NO₂) is dissolved in water to make 500 mL of solution. The osmotic pressure of this solution at 300 K is found to be 5.03 × 10^–3 bar. The total number of glycine units present in the protein is _______.

(Given : R = 0.083 L bar K^–1 mol^–1)

Answer (330)

Sol. Since,

π = icRT

Molar mass of protein = 24751.5 g/mol

Number of glycine units in protein

= 330

4. For the given reactions

Sn²⁺ + 2e^– → Sn

Sn⁴⁺ + 4e^– → Sn

the electrode potentials are;

Answer (16)

Sol.

5. A radioactive element has a half life of 200 days. The percentage of original activity remaining after 83 days is ________. (Nearest integer)

(Given : antilog 0.125 = 1.333, antilog 0.693 = 4.93)

Answer (75)

Sol.

Hence, percentage of original activity remaining after 83 days is 75%.

6. [Fe(CN)₆]^4–

[Fe(CN)₆]^3–

[Ti(CN)₆]^3–

[Ni(CN)₄]^2–

[Co(CN)₆]^3–

Among the given complexes, number of paramagnetic complexes is_______.

Answer (2)

Sol.

7. (a) CoCl₃⋅4 NH₃, (b) CoCl₃⋅5NH₃, (c) CoCl₃.6NH₃ and (d) CoCl(NO₃)₂⋅5NH₃. Number of complex(es) which will exist in cis-trans form is/are______.

Answer (1)

Sol. CoCl₃ ⋅ 4NH₃ ⇒ [Co(NH₃)₄Cl₂]Cl

CoCl₃ ⋅ 5NH₃ ⇒ [Co(NH₃)₅Cl]Cl₂

CoCl₃ ⋅ 6NH₃ ⇒ [Co(NH₃)₆]Cl₃

Only [Co(NH₃)₄Cl₂] can show geometrical isomerism. Hence can exist in cis-trans form.

8. The complete combustion of 0.492 g of an organic compound containing ‘C’, ‘H’ and ‘O’ gives 0.793 g of CO₂ and 0.442 g of H₂O. The percentage of oxygen composition in the organic compound is_____.[nearest integer]

Answer (46)

Sol. C_xH_yO_z + O₂ → CO₂ + H₂O

weight of oxygen = 0.492 – (0.216 + 0.05)

= 0.226 g

9. The major product of the following reaction contains______bromine atom(s).

Answer (1)

^Sol.

10. 0.01 M KMnO₄ solution was added to 20.0 mL of 0.05 M Mohr’s salt solution through a burette. The initial reading of 50 mL burette is zero. The volume of KMnO₄ solution left in burette after the end point is _____ml. [nearest integer]

Answer (30)

Sol. Meq of oxidising agent = Meq of reducing agent

0.01 × 20 × 5 = 0.05 × V × 1

Volume required = 20 ml

Since initial volume of KMnO₄ in burette is 50 ml. Hence volume of KMnO₄ left in the burette after end point is 30 ml.

Download PDF of JEE Main 2022 June 28 Shift 2 Chemistry Paper & Solutions

JEE Main 2022 Question Paper with Solutions – June 28 Shift 2

JEE Main 2022 June 28 Shift 2 Question Paper – Physics Solutions

JEE Main 2022 June 28 Shift 2 Question Paper – Chemistry Solutions

JEE Main 2022 June 28 Shift 2 Question Paper – Maths Solutions