IIT-JEE-Main-05042024-Chemistry-Paper-With-Solution-Morning

CHEMISTRY PAPER WITH SOLUTION

SECTION- A

Question 61

The incorrect postulates of Dalton’s atomic theory are :

(A) Atoms of different elements differ in mass.
(B) Matter consists of divisible atoms.
(C) Compounds are formed when atoms of different element combine in a fixed ratio.
(D) All the atoms of a given element have different properties including mass.
(E) Chemical reactions involve reorganisation of atoms.

Choose the correct answer from the options given below:

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

▶️ Answer/Explanation

Ans. (4)

Sol.

(B), (D) are not correct.

Question 62

The following reaction occurs in the Blast furnace where iron ore is reduced to iron metal:
Fe₂O₃(s) + 3CO(g) ↔ 2Fe(l) + 3CO₂(g)
Using Le-chatelier’s principle, predict which one of the following will not disturb the equilibrium.

Addition of Fe₂O₃
Addition of CO₂
Removal of CO
Removal of CO₂

▶️ Answer/Explanation

Ans. (1)

Sol.

When a solid (Fe₂O₃) is added, it does not affect the equilibrium position.

Question 63

Identify compound (Z) in the following reaction sequence.

$\ce{2-nitrophenol}$
$\ce{2,6-dinitrophenol}$
$\ce{2,4,6-trinitrophenol}$
$\ce{4-nitrophenol}$

▶️ Answer/Explanation

Ans. (3)

Sol.

X: sodium phenoxide
Y: phenol
Z: 2,4,6-trinitrophenol (picric acid).
The correct compound Z is (3).

Question 64

Given below are two statements: One is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): Enthalpy of neutralisation of strong monobasic acid with strong monoacidic base is always –57 kJ mol^–1
Reason (R): Enthalpy of neutralisation is the amount of heat liberated when one mole of H⁺ ions furnished by acid combine with one mole of OH^– ions furnished by base to form one mole of water.
In the light of the above statements, choose the correct answer from the options below:

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

▶️ Answer/Explanation

Ans. (2)

Sol.

Both are true and (R) explains (A). For strong acid + strong base neutralisation, enthalpy is always –57 kJ/mol by the given definition.

Question 65

The statement(s) that are correct about the species O^2–, F^–, Na⁺ and Mg²⁺:

(A) All are isoelectronic.
(B) All have the same nuclear charge.
(C) O^2– has the largest ionic radii.
(D) Mg²⁺ has the smallest ionic radii.

Choose the most appropriate answer:

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

▶️ Answer/Explanation

Ans. (4)

Sol.

All are isoelectronic; O^2– has largest, Mg²⁺ smallest radii. Nuclear charge is different.

Question 66

For the compounds:
(A) H₃C–CH₂–O–CH₂–CH₂–CH₃
(B) H₃C–CH₂–CH₂–CH₂–CH₃
(C) CH₃–CH₂–C(=O)–CH₂–CH₃
(D) H₃C–CH–CH₂–CH₂–CH₃ (with –OH on 2nd C)
The increasing order of boiling point is:

(A) < (B) < (C) < (D)
(B) < (A) < (C) < (D)
(D) < (C) < (A) < (B)
(B) < (A) < (D) < (C)

▶️ Answer/Explanation

Ans. (2)

Sol.

Alcohol (–OH) > acid/ether > alkane
Boiling point: hydrogen bonding > high polarity > low polarity > non-polar.

Question 67

Given below are two statements:
Statement I: In group 13, the stability of +1 oxidation state increases down the group.
Statement II: The atomic size of gallium is greater than that of aluminium.
Choose the most appropriate answer:

Statement I is incorrect but Statement II is correct
Both Statement I and Statement II are correct
Both Statement I and Statement II are incorrect
Statement I is correct but Statement II is incorrect

▶️ Answer/Explanation

Ans. (4)

Sol.

Due to poor shielding by d and f electrons, lower oxidation state is more stable down the group; atomic size of Al > Ga.

Question 68

Number of σ and π bonds present in ethylene molecule is respectively:

3 and 1
5 and 2
4 and 1
5 and 1

Solution:

Ethylene is
H σ

H—C ≡πσ—C—H
σ
So it has 5 σ bonds and 1 π bond.
Ans. (4)

Question 69

Identify ‘A’ in the following reaction:

CH₃CH(OH)CH₃
CH₃CH₂CH₃
CH₃CH=N–NH₂
C₂H₅CH=N–NH₂

▶️ Answer/Explanation

Ans. (2)

Sol.

Question 70

The reaction at cathode in the cells commonly used in clocks involves.

reduction of Mn from +4 to +3
oxidation of Mn from +3 to +4
reduction of Mn from +7 to +2
oxidation of Mn from +2 to +7

▶️ Answer/Explanation

Ans. (1)

Sol.

In the cathode reaction, manganese (Mn) is reduced from the +4 oxidation state to the +3 state.

Question 71

Which one of the following complexes will exhibit the least paramagnetic behaviour?
[Atomic number, Cr = 24, Mn = 25, Fe = 26, Co = 27]

[Co(H₂O)₆]²⁺
[Fe(H₂O)₆]²⁺
[Mn(H₂O)₆]²⁺
[Cr(H₂O)₆]²⁺

▶️ Answer/Explanation

Ans. (1)

Sol.

Complex	Number of unpaired e^–	μ = √n(n + 2) B.M.
[Co(H₂O)₆]²⁺	3	3.87
[Fe(H₂O)₆]²⁺	4	4.89
[Mn(H₂O)₆]²⁺	5	5.92
[Cr(H₂O)₆]²⁺	4	4.89

Least paramagnetic behaviour: [Co(H₂O)₆]²⁺

Question 72

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): Cis form of alkene is found to be more polar than the trans form.
Reason (R): Dipole moment of trans isomer of 2-butene is zero.

In the light of the above statements, choose the correct answer:

Both (A) and (R) are true but (R) is NOT the correct explanation of (A)
(A) is true but (R) is false
Both (A) and (R) are true and (R) is the correct explanation of (A)
(A) is false but (R) is true

▶️ Answer/Explanation

Ans. (3)

Sol.

Dipole moment is a vector quantity and for a compound the net dipole moment is the vector sum of all dipoles. Hence, dipole moment of cis form is greater than trans form.

$\mu:\mathrm{CH}_3\overset{\displaystyle{}}{C=C}\mathrm{CH}_3$
Cis (μ > 0) > Trans (μ = 0)

Question 73

Given below are two statements:
Statement I: Nitration of benzene involves the following step –
H
|
H–O–NO₂ ⇌ H₂O + NO₂⁺
Statement II: Use of Lewis base promotes the electrophilic substitution of benzene.
In the light of the above statements, choose the most appropriate answer:

Both Statement I and Statement II are incorrect
Statement I is correct but Statement II is incorrect
Both Statement I and Statement II are correct
Statement I is incorrect but Statement II is correct

▶️ Answer/Explanation

Ans. (2)

Sol.

In nitration of benzene, concentrated H₂SO₄ and HNO₃ generate the electrophile NO₂⁺ in steps:
H₂SO₄ + HNO₃ ⇌ HSO₄^– + H–O–NO₂⁺
HSO₄^– + H₂O + NO₂⁺
Lewis acids—not Lewis bases—promote the formation of electrophiles.

Question 74

The correct order of ligands arranged in increasing field strength.

Cl^– < OH^– < Br^– < CN^–
F^– < Br^– < I^– < NH₃
Br^– < F^– < H₂O < NH₃
H₂O < OH^– < CN^– < NH₃

▶️ Answer/Explanation

Ans. (3)

Sol.

Experimental order: Br^– < F^– < H₂O < NH₃

Question 75

Which of the following gives a positive test with ninhydrin?

Cellulose
Starch
Polyvinyl chloride
Egg albumin

▶️ Answer/Explanation

Ans. (4)

Sol.

Ninhydrin test is a test of amino acids. Egg albumin contains protein which is a natural polymer of amino acids and will show positive ninhydrin test.

Question 76

The metal that shows highest and maximum number of oxidation states is:

▶️ Answer/Explanation

Ans. (2)

Sol.

Mn shows the highest oxidation state (Mn⁺⁷) in 3d series metals.

Question 77

An organic compound has 42.1% carbon, 6.4% hydrogen and the remainder is oxygen. If its molecular weight is 342, then its molecular formula is:

C₁₁H₁₈O₁₂
C₁₂H₂₀O₁₂
C₁₄H₂₀O₁₀
C₁₂H₂₂O₁₁

▶️ Answer/Explanation

Ans. (4)

Sol.

Only C₁₂H₂₂O₁₁ has 42.1% carbon, 6.4% hydrogen and 51.5% oxygen.

Question 78

Given below are two statements:
Statement I: Bromination of phenol in solvent with low polarity such as CHCl₃ or CS₂ requires Lewis acid catalyst.
Statement II: The Lewis acid catalyst polarises the bromine to generate Br⁺.
In the light of the above statements, choose the correct answer:

Statement I is true but Statement II is false
Both Statement I and Statement II are true
Both Statement I and Statement II are false
Statement I is false but Statement II is true

▶️ Answer/Explanation

Ans. (4)

Sol.

Phenol is a highly activated compound which can undergo bromination directly with bromine, without any Lewis acid.

Question 79

Molar ionic conductivities of divalent cation and anion are 57 S cm² mol^–1 and 73 S cm² mol^–1 respectively. The molar conductivity of solution of an electrolyte with the above cation and anion will be:

65 S cm² mol^–1
130 S cm² mol^–1
187 S cm² mol^–1
260 S cm² mol^–1

▶️ Answer/Explanation

Ans. (2)

Sol.

Λ⁺²_C = 57 S cm²mol^–1
Λ⁺²_A = 73 S cm²mol^–1
Λ_Solution = 57 + 73 = 130 S cm²mol^–1

Question 80

The number of neutrons present in the more abundant isotope of boron is ‘x’. Amorphous boron upon heating with air forms a product, in which the oxidation state of boron is ‘y’. The value of x + y is …

▶️ Answer/Explanation

Ans. (4)

Sol.

More abundant isotope is B¹¹ [Number of neutrons = 6, x = 6]
B + O₂ → B₂O₃ (oxidation state +3, y = 3)
x + y = 6 + 3 = 9

SECTION-B

Question 81

The value of Rydberg constant (R_H) is 2.18×10^–18 J.
The velocity of electron having mass 9.1×10^–31 kg in Bohr’s first orbit of hydrogen atom = ……… × 10⁵ ms^–1 (nearest integer)

▶️ Answer/Explanation

Ans. (22)

Sol.

V = 2.18 × 10⁶ × (Z/n)
For hydrogen, n = 1, Z = 1
V = 2.18 × 10⁶ × 1/1 = 2.18 × 10⁶ ms^–1 = 21.8 × 10⁵ ≈ 22 × 10⁵ ms^–1 (nearest integer)

Question 82

In a borax bead test under hot condition, a metal salt (one from the given) is heated at point B of the flame, resulted in green colour salt bead. The spin-only magnetic moment value of the salt is ……….. BM (Nearest integer)
[Given atomic number of Cu = 29, Ni = 28, Mn = 25, Fe = 26]

▶️ Answer/Explanation

Ans. (6)

Sol.

Fe³⁺ will give green coloured bead when heated at point B. Number of unpaired e^– in Fe³⁺ is 5; μ = 5.92; nearest integer = 6.

Question 83

The heat of combustion of solid benzoic acid at constant volume is –321.30 kJ at 27°C. The heat of combustion at constant pressure is (–321.30 – xR) kJ, the value of x is ……………

▶️ Answer/Explanation

Ans. (150)

Sol.

ΔH = ΔU + Δn_gRT
Δn_g = –0.5;
ΔH = –321.30 – (1/2 × 300) = –321.30 – 150R
So x = 150.

Question 84

Total sum of oxygen atoms in Product A and Product B are ………

▶️ Answer/Explanation

Ans. (14)

Sol.

Product A contains 6 O atoms, Product B contains 8 O atoms (picric acid), so total is 14.

Question 85

The spin only magnetic moment value of the ion among Ti²⁺, V²⁺, Co³⁺ and Cr²⁺, that acts as strong oxidising agent in aqueous solution is ………… BM (Nearest integer).
(Given atomic numbers: Ti : 22, V : 23, Cr : 24, Co : 27)

▶️ Answer/Explanation

Ans. (5)

Sol.

Strong oxidising agent = Co³⁺
No. of unpaired e^– in Co³⁺ (3d⁶) = 4
μ = √n(n+2) = √24 = 4.89 ≈ 5

Question 86

During Kinetic study of reaction 2A + B → C + D, the following results were obtained:

A[M]	B[M]	initial rate of formation of D
0.1	0.1	6.0 × 10^–3
0.3	0.2	7.2 × 10^–2
0.3	0.4	2.88 × 10^–1
0.4	0.1	2.40 × 10^–2

Based on the above data, overall order of the reaction is ……….

▶️ Answer/Explanation

Ans. (3)

Sol.

x (order wrt A) = 1, y (order wrt B) = 2 → overall: 1 + 2 = 3.

Question 87

An artificial cell is made by encapsulating 0.2 M glucose solution within a semipermeable membrane. The osmotic pressure developed when the cell is placed within a 0.05 M solution of NaCl at 300 K is ______ × 10^–1 bar. (Nearest Integer)
[Given: R = 0.083 L bar mol^–1 K^–1]
Assume complete dissociation of NaCl.

▶️ Answer/Explanation

Ans. (25)

Sol.

Question 88

The number of halobenzenes from the following that can be prepared by Sandmeyer’s reaction is ….

▶️ Answer/Explanation

Ans. (2)

Sol.

Only bromobenzene and chlorobenzene can be prepared by Sandmeyer’s reaction.

Question 89

In the Lewis dot structure for NO₂^–, total number of valence electrons around nitrogen is …….

▶️ Answer/Explanation

Ans. (8)

Sol.

Number of valence electrons around N atom = 8.

Question 90

9.3 g of pure aniline is treated with bromine water at room temperature to give a white precipitate of product ‘P’. The mass of product ‘P’ obtained is 26.4 g. The percentage yield is ……….. %.

▶️ Answer/Explanation

Ans. (80)

Sol.

93 g aniline produces 330 g of tribromoaniline.
9.3 g aniline → 33 g tribromoaniline.
% yield = 26.4/33 × 100 = 80%

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