JEE Main 2022 June 24 – Shift 2 Chemistry Question Paper with 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. 120 g of an organic compound that contains only carbon and hydrogen gives 330 g of CO₂ and 270 g of water on complete combustion. The percentage of carbon and hydrogen, respectively are

(A) 25 and 75

(B) 40 and 60

(C) 60 and 40

(D) 75 and 25

Answer (D)

Sol. Mass of organic compound = 120 g

Mass of CO₂ = 330 g

Moles of CO₂ = _330/44= 7.5

Mass of carbon = 7.5 × 12 = 90 gm

Percentage of C

Mass of H₂O = 270 g

Moles of H₂O = 270/18= 15

Mass of hydrogen = 15 × 2 = 30 gm

Percentage of H

2. The energy of one mole of photons of radiation of wavelength 300 nm is (Given h = 6.63 × 10^–34Js, N_A = 6.02 × 10²³ mol^–1, c = 3 × 10⁸ ms^–1)

(A) 235 kJ mol^–1

(B) 325 kJ mol^–1

(C) 399 kJ mol^–1

(D) 435 kJ mol^–1

Answer (C)

Sol. Wavelength of radiation = 300 nm

Photon energy

= 6.63 × 10^–19 J

Energy of 1 mole of photons

= 6.63 × 10^–19 × 6.02 × 10²³ × 10^–3

= 399 kJ

3. The correct order of bond orders of C₂^2-, N₂^2-, O₂^2- is, respectively

(A) C₂^2-< N₂^2-< O₂^2-

(B) O₂^2-< N₂^2-< C₂^2-

(C) C₂^2-< O₂^2-< N₂^2-

(D) N₂^2-< C₂^2-< O₂^2-

Answer (B)

Sol.

4. At 25°C and 1 atm pressure, the enthalpies of combustion are as given below :

The enthalpy of formation of ethane is

(A) +54.0 kJ mol^–1

(B) –68.0 kJ mol^–1

(C) –86.0 kJ mol^–1

(D) +97.0 kJ mol^–1

Answer (C)

Sol.

Enthalpy of formation of C₂H₆(g) = –86.0 kJ mol^–1

5. For a first order reaction, the time required for completion of 90% reaction is ‘x’ times the half life of the reaction. The value of ‘x’ is (Given: In 10 = 2.303 and log 2 = 0.3010)

(A) 1.12

(B) 2.43

(C) 3.32

(D) 33.31

Answer (C)

Sol.

For a first order reaction,

Time for 90% conversion,

6. Metals generally melt at very high temperature. Amongst the following, the metal with the highest melting point will be

(1) Hg

(2) Ag

(3) Ga

(4) Cs

Answer (B)

Sol. Melting points of the given metals

Hg : –38.83° C

Ag : 961.8° C

Ga : 29.76° C

Cs : 28.44° C

∴ Metal having highest melting point is Ag.

7. Which of the following chemical reactions represents Hall-Heroult Process?

(1) Cr₂O₃ + 2Al → Al₂O₃ + 2Cr

(2) 2Al₂O₃ + 3C → 4Al + 3CO₂

(3) FeO + CO → Fe + CO₂

(4)

Answer (B)

Sol. Hall-Herault process is used for the extraction of aluminium by electrolysis molten Al₂O₃

2Al₂O₃ + 3C → 4Al + 3CO₂

8. In the industrial production of which of the following, molecular hydrogen is obtained as a byproduct?

(1) NaOH

(2) NaCl

(3) Na metal

(4) Na₂CO₃

Answer (A)

Sol. Molecular hydrogen is produced as a byproduct in the industrial production of NaOH by electrolysis of aq NaCl solution

NaCl → Na⁺ + Cl^–

Cathode : 2H₂O + 2e → H₂ + 2OH^–

Anode : 2Cl^– → Cl₂ +2e

NaOH is crystallised from the remaining part of electrolyte.

9. Which one of the following compounds is used as a chemical in certain type of fire extinguishers?

(1) Baking soda

(2) Soda ash

(3) Washing soda

(4) Caustic Soda

Answer (A)

Sol. Baking soda (NaHCO₃) is used in certain type of fire extinguishers because it decomposes at high temperature to produce CO₂ which extinguishes fire

10. PCl₅ is well known, but NCl₅ is not. Because,

(1) nitrogen is less reactive than phosphorous

(2) nitrogen doesn’t have d-orbitals in its valence shell.

(3) catenation tendency is weaker in nitrogen than phosphorous.

(4) size of phosphorous is larger than nitrogen

Answer (B)

Sol. PCl₅ is well known but NCl₅ is not because nitrogen does not have vacant d-orbitals in its valence shell. So, nitrogen cannot expand its octet. On the other hand phosphorus has vacant d-orbitals in its valence shell which enables it to expand its octet.

11. Transition metal complex with highest value of crystal field splitting (Δ₀) will be

(A) [Cr(H₂O)₆]³⁺

(B) [Mo(H₂O)₆]³⁺

(C) [Fe(H₂O)₆]³⁺

(D) [Os(H₂O)₆]³⁺

Answer (D)

Sol. Crystal field splitting (Δ₀) for octahedral complexes depends on oxidation state of the metal as well as to which transition series the metal belongs. For the same oxidation state, the crystal field splitting (Δ₀) increases as we move from 3d → 4d → 5d. Cr³⁺ and Fe³⁺ belong to 3d series, Mo³⁺ belongs to 4d series and Os³⁺ belongs to 5d series. Therefore crystal field splitting (Δ₀) is highest for [Os(H₂O)₆]³⁺.

12. Some gases are responsible for heating of atmosphere (green house effect). Identify from the following the gaseous species which does not cause it.

(A) CH₄

(B) O₃

(C) H₂O

(D) N₂

Answer (D)

Sol. Among the given gases, the green house gases which are responsible for heating the atmosphere are CH₄, water vapour and ozone. Nitrogen is not a green house gas.

13. Arrange the following carbocations in decreasing order of stability.

(A) A > C > B

(B) A > B > C

(C) C > B > A

(D) C > A > B

Answer (Bonus)

Sol. The given carbocations are

Carbocation (A) is stabilised by hyperconjugation due to 4 α hydrogen atoms. Carbocation (C) is also stabilised by hyperconjugation due to 4 α hydrogen atoms but destabilised by –I effect of O-atom. Carbocation (B) is most stable as it is stabilised by resonance.

∴ Correct decreasing order of stability is

B > A > C

None of the given options is correct.

14. Given below are two statements.

Statement I: The presence of weaker π-bonds make alkenes less stable than alkanes.

Statement II: The strength of the double bond is greater than that of carbon-carbon single bond.

In the light of the above statements, choose the correct 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. The π-bond present is alkenes is weaker than σ-bond present in alkanes. That makes alkenes less stable than alkanes. Therefore, statement-I is correct.

Carbon-carbon double bond is stronger than Carbon-carbon single bond because more energy is required to break 1 sigma and 1 pi bond than to break 1 sigma bond only. Therefore, statement-II is also correct.

15. Which of the following reagents/reactions will convert ‘A’ to ‘B’?

(A) PCC oxidation

(B) Ozonolysis

(C) BH₃, H₂O₂/^–OH followed by PCC oxidation

(D) HBr, hydrolysis followed by oxidation by K₂Cr₂O₇.

Answer (C)

Sol.

The first step involves addition of H₂O to alkene according to anti-markownikoff’s rule while the second step involves oxidation of 1° alcohol to aldehyde.

16. Hex-4-ene-2-ol on treatment with PCC gives ‘A’ on reaction with sodium hypoiodite gives ‘B’, which on further heating with soda lime gives ‘C’. The compound ‘C’ is

(A) 2-pentene

(B) Proponaldehyde

(C) 2-butene

(D) 4-methylpent-2-ene

Answer (C)

Sol.

17. The conversion of propan-1-ol to n-butylamine involves the sequential addition of reagents. The correct sequential order of reagents is

(A) (i) SOCl₂ (ii) KCN (iii) H₂/Ni, Na(Hg)/C₂H₅OH

(B) (i) HCl (ii) H₂/Ni, Na(Hg)/C₂H₅OH

(C) (i) SOCl2 (ii) KCN (iii) CH₃NH₂

(D) (i) HCl (ii) CH₃NH₂

Answer (A)

Sol.

18. Which of the following is not an example of a condensation polymer?

(A) Nylon 6,6

(B) Dacron

(C) Buna-N

(D) Silicone

Answer (C)

Sol. Nylon 6, 6 is a condensation polymer of hexamethylene diamine and adipic acid

Dacron is a condensation polymer of terephthalic acid and ethylene glycol.

Buna-N is an addition polymer of 1, 3-butadiene and acrylonitrile

Silicone is a condensation polymer of dialkyl silanediol.

19. The structure shown below is of which well-known drug molecule?

(A) Ranitidine

(B) Seldane

(C) Cimetidine

(D) Codeine

Answer (C)

Sol. The given structure is that of cimetidine which is well known antacid.

20. In the flame test of a mixture of salts, a green flame with blue centre was observed. Which one of the following cations may be present?

(A) Cu²⁺

(B) Sr²⁺

(C) Ba²⁺

(D) Ca²⁺

Answer (A)

Sol. Cupric salts give green flame with blue centre. The colour of other salts are

Sr²⁺ Crimson red

Ca²⁺ Brick red

Ba²⁺ Green

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. At 300 K, a sample of 3.0 g of gas A occupies the same volume as 0.2 g of hydrogen at 200 K at the same pressure. The molar mass of gas A is ______ g mol^–1. (nearest integer) Assume that the behaviour of gases as ideal.

(Given: The molar mass of hydrogen (H₂) gas is
2.0 g mol^–1).

Answer (45)

Sol. V₁, Volume of 0.2 g H₂ at 200 K

V₂, Volume of 3.0 g of gas A at 300 K

V₁ = V₂ (Given)

∴ M = 45 g mol^–1

2. A company dissolves ‘x’ amount of CO₂ at 298 K in 1 litre of water to prepare soda water. X = _____
× 10^–3 g. (nearest integer)

(Given: partial pressure of CO₂ at 298 K = 0.835 bar.

Henry’s law constant for CO₂ at 298 K = 1.67 kbar.

Atomic mass of H, C and O is 1, 12, and 6 g mol^–1, respectively)

Answer (1221)

Sol. According to Henry’s law, partial pressure of a gas is given by

P_g = (K_H) X_g

where X_g is mole fraction of gas in solution

Mass of CO₂ in 1 L water = 1221 × 10^–3 g

3. PCl₅ dissociates as

5 moles of PCl₅ are placed in a 200 litre vessel which contains 2 moles of N₂ and is maintained at 600 K. The equilibrium pressure is 2.46 atm. The equilibrium constant K_p for the dissociation of PCl₅ is ______ × 10^–3. (nearest integer)

(Given: R = 0.082 L atm K^–1 mol^–1; Assume ideal gas behaviour)

Answer (1107)

Sol.

Initial 5

moles

Equilibrium 5 – x x x

moles

Number of moles of N₂ = 2

Equilibrium pressure = 2.46 atm

On solving, x = 3

= 1107 × 10^–3 atm

4. The resistance of a conductivity cell containing 0.01 M KCl solution at 298 K is 1750 Ω. If the conductivity of 0.01 M KCl solution at 298 K is
0.152 × 10^–3 S cm^–1, then the cell constant of the conductivity cell is _______ × 10^–3 cm^–1.

Answer (266)

Sol. Molarity of KCl solution = 0.1 M

Resistance = 1750 ohm

Conductivity = 0.152 × 10^–3 S cm^–1

Conductivity

∴ Cell constant = 0.152 × 10^–3 × 1750 = 266 × 10^–3 cm^–1

5. When 200 mL of 0.2 M acetic acid is shaken with 0.6 g of wood charcoal, the final concentration of acetic acid after adsorption is 0.1 M. The mass of acetic acid adsorbed per gram of carbon is ________g.

Answer (2)

Sol. Mass of wood charcoal = 0.6 g

Initial moles of acetic acid = 0.2 × 0.2 = 0.04

Final moles of acetic acid = 0.1 × 0.2 = 0.02

Moles of acetic acid adsorbed = 0.04 – 0.02

= 0.02

Mass of acetic acid adsorbed per gm of charcoal

6. (a) Baryte, (b) Galena, (c) Zinc blende and
(d) Copper pyrites. How many of these minerals are sulphide based?

Answer (3)

Sol. Baryte BaSO₄

Galena PbS

Zinc blende ZnS

Copper pyrites CuFeS₂

Of the given minerals, only 3 are sulphide based.

7. Manganese (VI) has ability to disproportionate in acidic solution. The difference in oxidation states of two ions it forms in acidic solution is _____.

Answer (3)

Sol. Manganese (VI) disproportionates in acidic medium as

Difference in oxidation states of Mn in the products formed = 7 – 4 = 3

8. 0.2 g of an organic compound was subjected to estimation of nitrogen by Duma’s method in which volume of N₂ evolved (at STP) was found to be 22.400 mL. The percentage of nitrogen in the compound is _____. [nearest integer]

(Given : Molar mass of N₂ is 28 g mol^–1, Molar volume of N₂ at STP : 22.4L)

Answer (14)

Sol. Mass of organic compound = 0.2 g

Volume of N₂ gas evolved at STP = 22.4 mL

Mass of N₂ gas evolved

= 0.028 g

Percentage of nitrogen in the compound

⁹

Consider the above reaction. The number of π electrons present in the product ‘P’ is_________.

Answer (2)

^Sol.

The given reaction undergoes nucleophilic substitution by SN2 mechanism at room temperature

∴ No. of π electrons present in P = 2

10. In alanylglycylleucylalanyvaline, the number of peptide linkages is __________.,

Answer (4)

Sol. The given pentapeptide is

ALA – GLY – LEU – ALA – VAL

It has 4 peptide linkages.