Home / jee-main-27-july-2022-paper-shift-2_Chemistry

Chemistry Section-A

Question 1

The correct decreasing order of energy, for the orbitals having, following set of quantum numbers:
(A) n = 3, l = 0, m = 0
(B) n = 4, l = 0, m = 0
(C) n = 3, l = 1, m = 0
(D) n = 3, l = 2, m = 1
(A) (D) > (B) > (C) > (A)
(B) (B) > (D) > (C) > (A)
(C) (C) > (B) > (D) > (A)
(D) (B) > (C) > (D) > (A)
▶️ Answer/Explanation
Detailed solution

(A) n + l = 3 + 0 = 3

(B) n + l = 4 + 0 = 4

(C) n + l = 3 + 1 = 4

(D) n + l = 3 + 2 = 5

Higher n + l value, higher the energy & if same n + l value, then higher n value, higher the energy.

Thus : D > B > C > A.

✅ Answer: (A)

Question 2

Match List-I with List-II
List-I
(A) \( \Psi_{MO} = \Psi_A – \Psi_B \)
(B) \( \mu = Q \times r \)
(C) \( \frac{N_b – N_a}{2} \)
(D) \( \Psi_{MO} = \Psi_A + \Psi_B \)
List-II
(I) Dipole moment
(II) Bonding molecular orbital
(III) Anti-bonding molecular orbital
(IV) Bond order
(A) (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
(B) (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
(C) (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
(D) (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
▶️ Answer/Explanation
Detailed solution

(A) \( \Psi_{MO} = \Psi_A – \Psi_B \) → (III) ABMO

(B) \( \mu = Q \times r \) → (I) Dipole moment

(C) \( \frac{N_b – N_a}{2} \) → (IV) Bond order

(D) \( \Psi_{MO} = \Psi_A + \Psi_B \) → (II) BMO

✅ Answer: (C)

Question 3

The Plot of pH-metric titration of weak base \( NH_4OH \) vs strong acid HCl looks like:
▶️ Answer/Explanation
Detailed solution

Titration curve of \( NH_4OH \) vs HCl (WB + SA).

 

 

 

 

 

 

✅ Answer: (A)

Question 4

Given below are two statements:
Statement I: For KI, molar conductivity increases steeply with dilution.
Statement II: For carbonic acid, molar conductivity increases slowly with dilution.
In the light of the above statements, choose the correct answer from the options given below:
(A) Both Statement I and Statement II are true
(B) Both Statement I and Statement II are false
(C) Statement I is true but Statement II is false
(D) Statement I is false but Statement II is true
▶️ Answer/Explanation
Detailed solution

Statement I: KI is strong electrolyte thus almost constant on dilution.

Statement II: In weak electrolyte it increases, sharply.

 

 

 

 

 

✅ Answer: (B)

Question 5

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R)
Assertion (A) : Dissolved substances can be removed from a colloidal solution by diffusion through a parchment paper.
Reason (R) : Particles in a true solution cannot pass through parchment paper but the colloidal particles can pass through the parchment paper.
In the light of the above statements, choose the correct answer from the options given below:
(A) Both (A) and (R) are correct and (R) is the correct explanation of (A)
(B) Both (A) and (R) are correct but (R) is not the correct explanation of (A)
(C) (A) is correct but (R) is not correct
(D) (A) is not correct but (R) is correct
▶️ Answer/Explanation
Detailed solution

Assertion (A): Correct.

Reason(R): Incorrect.

Particles of true solution pass through parchment paper thus answer is (C).

✅ Answer: (C)

Question 6

Outermost electronic configurations of four elements A, B, C, D are given below:
(A) \( 3s^2 \)
(B) \( 3s^2 3p^1 \)
(C) \( 3s^2 3p^3 \)
(D) \( 3s^2 3p^4 \)
The correct order of first ionization enthalpy for them is:
(A) (A) < (B) < (C) < (D)
(B) (B) < (A) < (D) < (C)
(C) (B) < (D) < (A) < (C)
(D) (B) < (A) < (C) < (D)
▶️ Answer/Explanation
Detailed solution

(A) \( 3s^2 \) → Mg

(B) \( 3s^2 3p^1 \) → Al

(C) \( 3s^2 3p^3 \) → P

(D) \( 3s^2 3p^4 \) → S

\( P > S > Mg > Al \)

Half filled stability

Penetrating power of s-p.

\( C > D > A > B \).

✅ Answer: (B)

Question 7

An element A of group 1 shows similarity to an element B belonging to group 2. If A has maximum hydration enthalpy in group 1 then B is:
(A) Mg
(B) Be
(C) Ca
(D) Sr
▶️ Answer/Explanation
Detailed solution

\( Li^+ \) → Maximum hydration enthalpy in group 1 due to small size.

So ‘B’ is Mg.

 

 

 

 

 

✅ Answer: (A)

Question 8

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R)
Assertion (A) : Boron is unable to form \( BF_6^{3-} \)
Reason (R) : Size of B is very small.
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/Explanation
Detailed solution

Assertion (A): True

Reason (R): True but not correct explanation.

Correct explanation: Expansion of octet not possible for ‘B’.

✅ Answer: (B)

Question 9

In neutral or alkaline solution, \( MnO_4^- \) oxidises thiosulphate to:
(A) \( S_2O_7^{2-} \)
(B) \( S_2O_8^{2-} \)
(C) \( SO_3^{2-} \)
(D) \( SO_4^{2-} \)
▶️ Answer/Explanation
Detailed solution
\[ 8MnO_4^- + 3S_2O_3^{2-} + H_2O \rightarrow 8MnO_2 + 6SO_4^{2-} + 2OH^- \] 

✅ Answer: (D)

Question 10

Low oxidation state of metals in their complexes are common when ligands:
(A) have good π-accepting character
(B) have good σ-donor character
(C) are having good π-donating ability
(D) are having poor σ-donating ability
▶️ Answer/Explanation
Detailed solution

When metal is in low oxidation state then it forms complexes when ligands have good π-accepting character.

✅ Answer: (A)

Question 11

Given below are two statements:
Statement I : The non bio-degradable fly ash and slag from steel industry can be used by cement industry.
Statement II : The fuel obtained from plastic waste is lead free.
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/Explanation
Detailed solution

(I) Fly ash and slag from steel industry are utilised by cement industry.

(II) Fuel obtained from plastic waste has high octane rating. It contains no lead and it is known as green fuel.

Both statement (I) & (II) are correct.

✅ Answer: (A)

Question 12

The structure of A in the given reaction is:
▶️ Answer/Explanation
Detailed solution

 

 

 

 

 

 

✅ Answer: (B)

Question 13

Major product ‘B’ of the following reaction sequence is:
▶️ Answer/Explanation
Detailed solution

 

 

 

 

 

 

✅ Answer: (B)

Question 14

Match List-I with List-II.
List-I
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
List-II
(I) Gatterman Koch reaction
(II) Etard reaction
(III) Stephen reaction
(IV) Rosemnund reaction
Choose the correct answer from the options given below:
(A) (A)-(IV), (B)-(III), (C)-(II), (D)-(I)
(B) (A)-(I), (B)-(II), (C)-(III), (D)-(IV)
(C) (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
(D) (A)-(III), (B)-(II), (C)-(I), (D)-(IV)
▶️ Answer/Explanation
Detailed solution

 

 

 

 

 

 

 

 

✅ Answer: (A)

Question 15

Match List-I with List-II.

List-I (Polymer)
(A) Neoprene
(B) Teflon
(C) Acrilan
(D) Natural rubber

List-II (Monomer)
(I) Acrylonitrile
(II) Chloroprene
(III) Tetrafluoroethene
(IV) Isoprene

Choose the correct answer from the option given below:
(A) (A)-(II), (B)-(III), (C)-(I), (D)-(IV)
(B) (A)-(II), (B)-(I), (C)-(III), (D)-(IV)
(C) (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
(D) (A)-(I), (B)-(II), (C)-(III), (D)-(IV)
▶️ Answer/Explanation
Detailed solution

 

 

 

 

 

 

 

 

✅ Answer: (A)

Question 16

An organic compound ‘A’ contains nitrogen and chlorine. It dissolves readily in water to give a solution that turns litmus red. Titration of compound ‘A’ with standard base indicates that the molecular weight of ‘A’ is \(131 \pm 2\). When a sample of ‘A’ is treated with aq. NaOH, a liquid separates which contains N but not Cl. Treatment of the obtained liquid with nitrous acid followed by phenol gives orange precipitate. The compound ‘A’ is :
▶️ Answer/Explanation
Detailed solution

 

 

 

 

 

 

 

✅ Answer: (D) – 

Question 17

Match List-I with List-II
List-I
(A) Glucose + HI
(B) Glucose + Br\(_2\) water
(C) Glucose + acetic anhydride
(D) Glucose + HNO\(_3\)

List-II
(I) Gluconic acid
(II) Glucose pentacetate
(III) Saccharic acid
(IV) Hexane
Choose the correct answer from the options given below:
(A) (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
(B) (A)-(IV), (B)-(III), (C)-(II), (D)-(I)
(C) (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
(D) (A)-(I), (B)-(III), (C)-(IV), (D)-(II)
▶️ Answer/Explanation
Detailed solution

 

 

 

 

 

 

 

✅ Answer: (A)

Question 18

Which of the following enhances the lathering property of soap?
(A) Sodium stearate
(B) Sodium carbonate
(C) Sodium rosinate
(D) Trisodium phosphate
▶️ Answer/Explanation
Detailed solution

Rosin is added to soaps which forms sodium rosinate which lathers well.

✅ Answer: (C)

Question 19

Match List-I with List-II
List-I (Mixture)
(A) Chloroform & Aniline
(B) Benzoic acid & Napthalene
(C) Water & Aniline
(D) Napthalene & Sodium chloride

List-II (Purification Process)
(I) Steam distillation
(II) Sublimation
(III) Distillation
(IV) Crystallisation
Choose the correct answer from the options given below:
(A) (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
(B) (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
(C) (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
(D) (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
▶️ Answer/Explanation
Detailed solution

Separation Methods:

(A) Chloroform + Aniline → Different boiling points → (III) Distillation

(B) Benzoic acid + Napthalene → Different solubility → (IV) Crystallisation

(C) Water + Aniline → Immiscible liquids → (I) Steam distillation

(D) Napthalene + Sodium chloride → Napthalene sublimes → (II) Sublimation

✅ Answer: (D)

Question 20

Fe\(^{3+}\) cation gives a prussian blue precipitate on addition of potassium ferrocyanide solution due to the formation of:
(A) [Fe(H\(_2\)O)\(_6\)]\(_2\)[Fe(CN)\(_6\)]
(B) Fe\(_2\)[Fe(CN)\(_6\)]\(_2\)
(C) Fe\(_3\)[Fe(OH)\(_2\)(CN)\(_4\)]\(_2\)
(D) Fe\(_4\)[Fe(CN)\(_6\)]\(_3\)
▶️ Answer/Explanation
Detailed solution

Reaction: 4Fe\(^{3+}\) + 3[Fe(CN)\(_6\)]\(^{4-}\) → Fe\(_4\)[Fe(CN)\(_6\)]\(_3\)

This is the Prussian Blue precipitate.

✅ Answer: (D)

Section-B

Question 1

The normality of \( H_2SO_4 \) in the solution obtained on mixing 100 mL of 0.1 M \( H_2SO_4 \) with 50 mL of 0.1 M NaOH is ______ ×10\(^{-1}\) N. (Nearest Integer)
 
▶️ Answer/Explanation
Detailed solution

No. of equivalents of \( H_2SO_4 = 100 \times 0.1 \times 2 = 20 \)

No. of equivalents of NaOH = 50 × 0.1 = 5

No. of equivalents of \( H_2SO_4 \) left = 20 – 5 = 15

Total volume = 150 mL

⇒ 150 × x = 15

x = \(\frac{1}{10}\) = 0.1N = 1 × 10\(^{-1}\) N

✅ Answer: 1

Question 2

For a real gas at 25°C temperature and high pressure (99 bar) the value of compressibility factor is 2, so the value of Vander Waal’s constant ‘b’ should be ______ ×10\(^{-2}\) L mol\(^{-1}\) (Nearest integer) (Given R = 0.083 L bar K\(^{-1}\) mol\(^{-1}\))
 
▶️ Answer/Explanation
Detailed solution

For real gas under high pressure

\(Z = 1 + \frac{Pb}{RT} \quad \Rightarrow b = \frac{RT}{P}\)

\(= \frac{0.083 \times 298}{99}\)

\(= 0.25 \times 10^{-2} \, \text{L mol}^{-1}\)

✅ Answer: 25

Question 3

A gas (Molar mass = 280 g mol\(^{-1}\)) was burnt in excess O\(_2\) in a constant volume calorimeter and during combustion the temperature of calorimeter increased from 298.0 K to 298.45 K. If the heat capacity of calorimeter is 2.5 kJ K\(^{-1}\) and enthalpy of combustion of gas is 9 kJ mol\(^{-1}\) then amount of gas burnt is ______ g. (Nearest Integer)
 
▶️ Answer/Explanation
Detailed solution

Let x g is burnt

moles = \(\frac{x}{280}\)

heat released by \(\frac{x}{280}\) mole = 2.5 × 0.45 kJ

heat released by 1 mole = \(\frac{2.5 \times 0.45 \times 280}{x}\) kJ

\(\Delta H = \Delta U + \Delta n_gRT\)

\(\Delta H = \Delta U\)

\(9 = \frac{2.5 \times 280 \times 0.45}{x}\)

x = 35 g

✅ Answer: 35

Question 4

When a certain amount of solid A is dissolved in 100 g of water at 25°C to make a dilute solution, the vapour pressure of the solution is reduced to one-half of that of pure water. The vapour pressure of pure water is 23.76 mmHg. The number of moles of solute A added is ______. (Nearest Integer)
 
▶️ Answer/Explanation
Detailed solution

∵ Dilute solution given:

\[\frac{P^0 – P_s}{P^0} \sim \frac{\text{n-solute}}{\text{n-solvent}}\]

\[\frac{P^0 – \frac{P^0}{2}}{P^0} \sim \frac{\text{n-solute}}{\text{n-solvent}}\]

\[\text{n-solute} \sim \frac{\text{n-solvent}}{2} = \frac{100}{18 \times 2} = 2.78 \, \text{mol}\]

More accurate approach:

\[\frac{P^0 – P_s}{P_s} = \frac{\text{n-solute}}{\text{n-solvent}}\]

\[\frac{P^0 – \frac{P^0}{2}}{\frac{P^0}{2}} = \frac{\text{n-solute}}{\text{n-solvent}}\]

\[\text{n-solute} = \text{n-solvent} = \frac{100}{18} = 5.55 \, \text{mol}\]

✅ Answer: 3

Question 5

[A] → [B]
Reactant Product
If formation of compound [B] follows the first order of kinetics and after 70 minutes the concentration of [A] was found to be half of its initial concentration. Then the rate constant of the reaction is \( x \times 10^{-6} \, s^{-1} \). The value of \( x \) is ______. (Nearest Integer)
 
▶️ Answer/Explanation
Detailed solution

\( K = \frac{0.693}{t_{1/2}} = \frac{0.693}{70 \times 60} \)

\(= \frac{6930}{7 \times 6} \times 10^{-6}\)

\(= 165 \times 10^{-6} \, s^{-1}\)

✅ Answer: 165

Question 6

Among the following ores Bauxite, Siderite, Cuprite, Calamine, Haematite, Kaolinite, Malachite, Magnetic, Sphalerite, Limonite, Cryolite, the number of principal ores of iron is ______.
 
▶️ Answer/Explanation
Detailed solution

Bauxite — AlO\(_x\)(OH)\(_{3-x}\)(where \( 0 < x < 1 \))

✓ Siderite — FeCO\(_3\)

Cuprite — Cu\(_2\)O

Calamine — ZnCO\(_3\)

✓ Haematite — Fe\(_2\)O\(_3\)

Kaolinite — Al\(_2\)(OH)\(_x\)Si\(_2\)O\(_5\)

Malachite — CuCO\(_3\), Cu(OH)\(_2\)

✓ Magnetic — Fe\(_3\)O\(_4\)

Sphalerite — ZnS

✓ Limonite — Fe\(_2\)O\(_3\).3H\(_2\)O

Cryolite — Na\(_3\)AlF\(_6\)

✅ Answer: 4

Question 7

The oxidation state of manganese in the product obtained in a reaction of potassium permanganate and hydrogen peroxide in basic medium is ______.
 
▶️ Answer/Explanation
Detailed solution

\( 2KMnO_4 + 3H_2O_2 \rightarrow 2MnO_4 + 3O_2 + 2H_2O + 2KOH \)

✅ Answer: 4

Question 8

The number of molecule(s) or ion(s) from the following having non-planar structure is ______.
NO\(_3^-\) , H\(_2\)O\(_2\), BF\(_3\), PCI\(_3\), XeF\(_4\), SF\(_4\), XeO\(_3\), PH\(_4^+\), SO\(_3\), [Al(OH)\(_4\)]
 
▶️ Answer/Explanation
Detailed solution

SO\(_3\) — sp\(^2\) Planar

BF\(_3\) — sp\(^2\) Planar

NO\(_3^-\) — sp\(^2\) Planar

SF\(_4\) — sp\(^3\) Non-planar

H\(_2\)O\(_2\) — sp\(^3\) Non-planar

PCI\(_3\) — sp\(^3\) Non-planar

[Al(OH)\(_4\)] — sp\(^3\) Non-planar

XeF\(_4\) — sp\(^3\)d\(^2\) Planar

XeO\(_3\) — sp\(^3\) Non-planar

PH\(_4^+\) — sp\(^3\) Non-planar

✅ Answer: 6

Question 9

The spin only magnetic moment of the complex present in Fehling’s reagent is ______ B.M. (Nearest integer).
 
▶️ Answer/Explanation
Detailed solution

Fehling solution is a complex of Cu\(^{++}\)

Cu\(^{++}\) = 3d\(^9\)

No. of unpaired \( e^- = 1 \)

M.M = \(\sqrt{l(1+2)} = \sqrt{3} = 1.73 \, BM\)

✅ Answer: 2

Question 10

 
 
 
 
 
 
 
In the above reaction, 5 g of toluene is converted into benzaldehyde with 92% yield. The amount of benzaldehyde produced is ______ × \( 10^{-2} \) g. (Nearest integer)
 
▶️ Answer/Explanation
Detailed solution

 

 

 

moles = \(\frac{5}{92}\)

 

 

\(= \frac{5}{92} \times \frac{92}{100} = 5 \times 10^-2\)

 

 

\(=106\times5\times10^-2 = 5.3g\)

\(= 530\times 10^-2 g\)

 

 

✅ Answer: 530

Scroll to Top