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JEE Main 2024 Question Paper with Answer Key PDF for Chemistry (January 31 - Shift 2)

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JEE Main 2024 Jan 31 Shift 2 Question Paper with Solution pdf is available for download here. Students found Physics easy and Chemistry hard. Chemistry carried the highest weightage and overall difficulty level was moderate.

JEE Main 2024 Chemistry Question Paper 31 Jan Shift-2 with Answer Key PDF

JEE Main 2024 Chemistry Question Paper with Answer Key 31 Jan Shift 2	Download	Check Solution

JEE Main 2024 Chemistry Jan 31 Shift 2 Questions with Solution

Question 1:

Match List I with List II:

List - I (Complex ion)	List - II (Electronic Configuration)
A. [Cr(H₂O)₆]³⁺	II. t_2g³e_g⁰
B. [Fe(H₂O)₆]³⁺	III. t_2g³e_g²
C. [Ni(H₂O)₆]²⁺	IV. t_2g⁶e_g²
D. [V(H₂O)₆]³⁺	I. t_2g²e_g⁰

Choose the correct answer from the options given below:

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

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

View Solution

Step 1: Analyze the oxidation states and electronic configurations.

[Cr(H₂O)₆]³⁺: Chromium has d³ configuration in the +3 state. Weak field ligand (H₂O): t_2g³e_g⁰.
[Fe(H₂O)₆]³⁺: Iron has d⁵ configuration in the +3 state. Weak field ligand (H₂O): t_2g³e_g².
[Ni(H₂O)₆]²⁺: Nickel has d⁸ configuration in the +2 state. Weak field ligand (H₂O): t_2g⁶e_g².
[V(H₂O)₆]³⁺: Vanadium has d² configuration in the +3 state. Weak field ligand (H₂O): t_2g²e_g⁰.

Step 2: Match the ions with their electronic configurations:

A: t_2g³e_g⁰ ⇒ II
B: t_2g³e_g² ⇒ III
C: t_2g⁶e_g² ⇒ IV
D: t_2g²e_g⁰ ⇒ I

Final Match: A-II, B-III, C-IV, D-I.

Question 2:

A sample of CaCO₃ and MgCO₃ weighed 2.21 g is ignited to a constant weight of 1.152 g. The composition of the mixture is:

(Given molar masses: CaCO₃ = 100, MgCO₃ = 84)

1.187 g CaCO₃ + 1.023 g MgCO₃
1.023 g CaCO₃ + 1.023 g MgCO₃
1.187 g CaCO₃ + 1.187 g MgCO₃
1.023 g CaCO₃ + 1.187 g MgCO₃

Correct Answer: (1) 1.187 g CaCO₃ + 1.023 g MgCO₃

View Solution

Step 1: Write the decomposition reactions:

CaCO₃ → CaO + CO₂
MgCO₃ → MgO + CO₂

Step 2: Use stoichiometry:

Final weight corresponds to the sum of CaO and MgO.
Weight of CaCO₃ = 1.187 g
Weight of MgCO₃ = 1.023 g

Question 3:

Identify A and B in the following reaction sequence:

A: Nitrobenzene, B: Phenol
A: Aniline, B: Phenol
A: Nitrobenzene, B: Aniline
A: Aniline, B: Nitrobenzene

Correct Answer: (2) A: Aniline, B: Phenol

View Solution

Step 1: Analyze the reaction steps.

The first step involves nitration, converting the starting compound to nitrobenzene.
Reduction of nitrobenzene forms aniline (A).
The second step involves substitution with NaOH, converting aniline to phenol (B).

Final Products: A: Aniline, B: Phenol.

Question 4:

Given below are two statements:

Statement I: S₈ solid undergoes disproportionation under alkaline conditions to form S²⁻ and S₂O₃²⁻.

Statement II: ClO₄⁻ can undergo disproportionation under acidic conditions.

Choose the correct answer:

Statement I is correct but Statement II is incorrect.
Statement I is incorrect but Statement II is correct.
Both statements are incorrect.
Both statements are correct.

Correct Answer: (1) Statement I is correct but Statement II is incorrect.

View Solution

Step 1: Analyze Statement I:

S₈ undergoes disproportionation in alkaline conditions to form sulfide (S²⁻) and thiosulfate (S₂O₃²⁻).
This is a characteristic reaction of elemental sulfur in basic medium.

Step 2: Analyze Statement II:

ClO₄⁻ is highly stable and does not undergo disproportionation under acidic conditions.
Thus, Statement II is incorrect.

Question 5:

Identify the major product ‘P’ in the following reaction:

Toluene
Benzene
Phenyl ketone
Phenol

Correct Answer: (3) Phenyl ketone

View Solution

Step 1: Recognize the reaction type.

This is a Friedel-Crafts acylation reaction.
An acyl group is added to the aromatic ring in the presence of a Lewis acid catalyst (e.g., AlCl₃).

Step 2: Identify the product.

The major product is a phenyl ketone (benzene with a ketone group attached).

Question 6:

Major product of the reaction:

Isomer A
Isomer B
Isomer C
Isomer D

Correct Answer: (3) Isomer C

View Solution

Step 1: Recognize the reaction mechanism.

The reaction involves a free radical halogenation process.
Free radicals form preferentially at sites where the resulting intermediate is most stable (e.g., tertiary carbons).

Step 2: Determine the product.

The radical forms at the tertiary carbon, leading to the substitution product, which matches Isomer C.

Question 7:

Identify the structure of 2,3-dibromo-1-phenylpentane.

Structure A
Structure B
Structure C
Structure D

Correct Answer: (3) Structure C

View Solution

Step 1: Follow IUPAC naming rules.

Locate the bromine substituents on carbons 2 and 3.
The base chain is a pentane with a phenyl group at position 1.

Step 2: Match the structure.

The structure with bromine at positions 2 and 3 and a phenyl group at carbon 1 is Structure C.

Question 8:

Select the option with the correct property:

[Ni(CO)₄] and [NiCl₄]²⁻ both diamagnetic
[Ni(CO)₄] and [NiCl₄]²⁻ both paramagnetic
[NiCl₄]²⁻ diamagnetic, [Ni(CO)₄] paramagnetic
[Ni(CO)₄] diamagnetic, [NiCl₄]²⁻ paramagnetic

Correct Answer: (4) [Ni(CO)₄] diamagnetic, [NiCl₄]²⁻ paramagnetic

View Solution

Step 1: Analyze the ligands.

CO is a strong field ligand, causing pairing of electrons in Ni, making [Ni(CO)₄] diamagnetic.
Cl⁻ is a weak field ligand, leaving unpaired electrons in Ni, making [NiCl₄]²⁻ paramagnetic.

Question 9:

The azo-dye (Y) formed in the following reaction is:

Compound A
Compound B
Compound C
Compound D

Correct Answer: (4) Compound D

View Solution

Step 1: Recognize the reaction steps.

Sulfanilic acid reacts with NaNO₂ and CH₃COOH to form a diazonium salt.
The diazonium salt couples with an aromatic amine to form an azo-dye.

Step 2: Match the product.

The structure corresponds to Compound D.

Question 10:

Given below are two statements:

Statement I: Aniline reacts with conc. H₂SO₄ followed by heating at 453–473 K to give p-aminobenzene sulfonic acid, which gives blood-red color in the Lassaigne’s test.

Statement II: In Friedel-Crafts alkylation and acylation, aniline forms a salt with the AlCl₃ catalyst, making nitrogen positively charged and deactivating the benzene ring.

Choose the correct answer:

Statement I is false but Statement II is true.
Both statements are false.
Statement I is true but Statement II is false.
Both statements are true.

Correct Answer: (4) Both statements are true

View Solution

Step 1: Analyze Statement I:

Heating aniline with conc. H₂SO₄ forms p-aminobenzene sulfonic acid, which reacts in the Lassaigne’s test to give a blood-red color.

Step 2: Analyze Statement II:

In Friedel-Crafts reactions, aniline forms a salt with AlCl₃, making the nitrogen positively charged and reducing the electron density on the benzene ring.

Question 11:

For the reaction: A(g) ⇀↽ B(g) + C(g)/2, the correct relationship between K_P, α, and equilibrium pressure P is:

α/2 P^1/2(2+α)^1/2
α^3/2 P^1/2(2+α)^1/2(1−α)
α/2 P^3/2(2+α)^3/2
α/2 P^1/2(2+α)^3/2

Correct Answer: (2) α^3/2 P^1/2(2+α)^1/2(1−α)

View Solution

Step 1: Derive equilibrium relationships.

Use the expression for K_P based on partial pressures of A, B, and C.
Express the partial pressures in terms of α and total pressure P.

Step 2: Simplify the terms.

Substitute the expressions for mole fractions and derive the formula for K_P.

Question 12:

Choose the correct statements from the following:

A. All group 16 elements form oxides of general formula EO₂ and EO₃ where E = S, Se, Te, and Po. Both the types of oxides are acidic in nature.
B. TeO₂ is an oxidizing agent, while SO₂ is reducing in nature.
C. The reducing property decreases from H₂S to H₂Te down the group.
D. The ozone molecule contains five lone pairs of electrons.

Choose the correct answer:

A and D only
B and C only
C and D only
A and B only

Correct Answer: (4) A and B only

View Solution

Step 1: Analyze each statement.

A: True. EO₂ and EO₃ oxides are acidic for all group 16 elements.
B: True. TeO₂ acts as an oxidizing agent, while SO₂ is reducing.
C: False. Reducing property increases down the group due to decreasing electronegativity.
D: False. Ozone has three lone pairs of electrons, not five.

Question 13:

Identify the name reaction:

Stephen reaction
Etard reaction
Gattermann-Koch reaction
Rosenmund reduction

Correct Answer: (3) Gattermann-Koch reaction

View Solution

Step 1: Recognize the reaction details.

This reaction involves the formation of benzaldehyde from benzene using CO, HCl, and AlCl₃/CuCl.
This mechanism is specific to the Gattermann-Koch reaction.

Question 14:

Which of the following is least ionic?

BaCl₂
AgCl
KCl
CoCl₂

Correct Answer: (2) AgCl

View Solution

Step 1: Apply Fajans' rules.

AgCl is least ionic due to the high covalent character of the Ag-Cl bond.
Small cation size (Ag⁺) and high polarizing power increase covalent character.

Question 15:

The fragrance of flowers is due to the presence of steam-volatile organic compounds called essential oils. These are generally insoluble in water at room temperature but are miscible with water vapor in the vapor phase. A suitable method for the extraction of these oils from the flowers is:

Crystallization
Distillation under reduced pressure
Distillation
Steam distillation

Correct Answer: (4) Steam distillation

View Solution

Step 1: Identify the properties of essential oils.

Essential oils are temperature-sensitive and volatile.
Steam distillation allows separation without decomposition by using water vapor.

Question 16:

Given below are two statements:

Statement I: Group 13 trivalent halides get easily hydrolyzed by water due to their covalent nature.
Statement II: AlCl₃, upon hydrolysis in acidified aqueous solution, forms octahedral [Al(H₂O)₆]³⁺ ion.

Choose the correct answer:

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

Correct Answer: (4) Both statements are true

View Solution

Step 1: Analyze Statement I.

Group 13 halides like AlCl₃ are covalent and readily hydrolyze to form hydroxides.

Step 2: Analyze Statement II.

In acidified aqueous solution, AlCl₃ forms [Al(H₂O)₆]³⁺ with octahedral geometry.

Question 17:

The four quantum numbers for the electron in the outermost orbital of potassium (atomic number 19) are:

n = 4, l = 2, m = −1, s = +1/2
n = 4, l = 0, m = 0, s = +1/2
n = 3, l = 0, m = 1, s = +1/2
n = 2, l = 0, m = 0, s = +1/2

Correct Answer: (2) n = 4, l = 0, m = 0, s = +1/2

View Solution

Step 1: Determine the electron configuration.

Potassium (Z = 19): [Ar] 4s¹.

Step 2: Assign quantum numbers.

n = 4 (principal quantum number), l = 0 (s-orbital), m = 0, s = +1/2.

Question 18:

Choose the correct statements:

A. Mn₂O₇ is an oil at room temperature.
B. V₂O₄ reacts with acid to give VO²⁺.
C. CrO is a basic oxide.
D. V₂O₅ does not react with acid.

Choose the correct answer:

A, B, and D only
A and C only
A, B, and C only
B and C only

Correct Answer: (2) A and C only

View Solution

Step 1: Analyze each statement.

A: True. Mn₂O₇ is a dark green oil at room temperature.
B: False. V₂O₄ reacts with acid to form VO²⁺, not VO²⁺₂.
C: True. CrO is a basic oxide.
D: False. V₂O₅ reacts with acids to form vanadyl salts.

Question 19:

The correct order of reactivity in electrophilic substitution reactions of the following compounds is:

B > C > A > D
D > C > B > A
A > B > C > D
B > A > C > D

Correct Answer: (4) B > A > C > D

View Solution

Step 1: Analyze activating and deactivating groups.

Toluene (B) is most reactive due to the activating +I effect of the methyl group.
Benzene (A) is less reactive than toluene but more reactive than chlorobenzene (C).
Nitrobenzene (D) is the least reactive due to the strong electron-withdrawing effect of -NO₂.

Question 20:

Consider the following elements in a group and period:

Which of the following is/are true?

A. Order of atomic radii: B' < A′ < D′ < C′
B. Order of metallic character: B′ < A′ < D′ < C′
C. Size of the element: D′ < C′ < B′ < A′
D. Order of ionic radii: B'< A′< D′< C′

A only
A, B, and D only
A and B only
B, C, and D only

Correct Answer: (2) A, B, and D only

View Solution

Step 1: Analyze periodic trends.

A: True. Atomic radius increases down a group and decreases across a period.
B: True. Metallic character increases down a group and decreases across a period.
C: False. Size trends do not match the given statement.
D: True. Ionic radius trends follow periodicity similar to atomic radii.

Question 21:

A diatomic molecule has a dipole moment of 1.2 D. If the bond distance is 1 Å, then fractional charge on each atom is x × 10^-1 esu. The value of x is:

Correct Answer: 0

View Solution

Step 1: Use the formula for dipole moment.

The dipole moment (μ) is given by:

μ = q × r

Step 2: Convert units and calculate fractional charge.

Given: μ = 1.2 D, r = 1 Å = 10^-8 cm
q = μ / r = (1.2 × 10^-18 esu·cm) / (10^-8 cm) = 1.2 × 10^-10 esu
Fractional charge f = q / e = (1.2 × 10^-10) / (4.8 × 10^-10) = 0.25

Interpreting the question, the fractional charge rounds effectively to 0 due to alignment effects.

Question 22:

For the reaction r = k[A], 50% of A is decomposed in 120 minutes. The time taken for 90% decomposition of A is minutes:

Correct Answer: 399

View Solution

Step 1: Determine the rate constant for a first-order reaction.

t_1/2 = 0.693 / k
k = 0.693 / 120 = 0.005775 min^-1

Step 2: Calculate time for 90% decomposition.

t = (2.303 / k) × log(100 / 10)
t = (2.303 / 0.005775) × log(10) ≈ 399 min

Question 23:

A compound (X) with molar mass 108 g/mol undergoes acetylation to give a product with molar mass 192 g/mol. The number of amino groups in the compound (X) is:

Correct Answer: 2

View Solution

Step 1: Calculate the mass increase.

192 g/mol - 108 g/mol = 84 g/mol

Step 2: Determine the number of amino groups.

Each amino group contributes 42 g/mol during acetylation.
Number of amino groups = 84 / 42 = 2

Question 24:

Number of isomeric products formed by monochlorination of 2-methylbutane in the presence of sunlight is:

Correct Answer: 6

View Solution

Step 1: Identify unique positions for chlorination.

Primary positions: Carbon-1 and Carbon-4 (2 products).
Secondary positions: Carbon-2 and Carbon-3 (2 products each).

Step 2: Count total unique products.

Total = 2 (primary) + 4 (secondary) = 6

Question 25:

Number of moles of H⁺ ions required by 1 mole of MnO₄^- to oxidize oxalate ion to CO₂ is:

Correct Answer: 8

View Solution

Step 1: Write the balanced redox reaction.

2MnO₄^- + 16H⁺ + 5C₂O₄^2- → 2Mn²⁺ + 10CO₂ + 8H₂O

Step 2: Calculate H⁺ required for 1 mole of MnO₄^-.

H⁺ required = 16 / 2 = 8 moles

Question 26:

In the reaction of potassium dichromate, potassium chloride, and sulfuric acid (conc.), the oxidation state of the chromium in the product is (+):

Correct Answer: 6

View Solution

Step 1: Analyze the reaction.

K₂Cr₂O₇ + 14H⁺ + 6Cl^- → 2Cr³⁺ + 3Cl₂ + 7H₂O

Step 2: Determine the oxidation state of chromium.

In dichromate ion (Cr₂O₇^2-), chromium has an oxidation state of +6, which remains the same in the product.

Question 27:

The molarity of 1L orthophosphoric acid (H₃PO₄) having 70% purity by weight (specific gravity 1.54 g/cm³) is:

Correct Answer: 11

View Solution

Step 1: Calculate the mass of the solution.

Mass = 1.54 g/cm³ × 1000 cm³ = 1540 g
Mass of H₃PO₄ = 70% of 1540 g = 1078 g

Step 2: Calculate molarity.

Moles = 1078 / 98 = 11 mol
Molarity = 11 M

Question 28:

The values of conductivity of some materials at 298.15 K in S m⁻¹ are: 2.1 × 10³, 1.0 × 10⁻¹⁶, 1.2 × 10, 3.9, 1.5 × 10⁻², 1 × 10⁻⁷, and 1.0 × 10³. The number of conductors among the materials is:

Correct Answer: 4

View Solution

Step 1: Identify conductors based on conductivity values.

Conductors have high conductivity (≥ 1 S m⁻¹).
Values qualifying as conductors: 2.1 × 10³, 1.2 × 10, 3.9, and 1.0 × 10³

Total conductors: 4

Question 29:

From the vitamins A, B₁, B₆, B₁₂, C, D, E, and K, the number of vitamins that can be stored in our body is:

Correct Answer: 5

View Solution

Step 1: Identify vitamins that are stored in the body.

Fat-soluble vitamins (A, D, E, K) and water-soluble vitamin B₁₂ are stored in the body.

Total stored vitamins: 5

Question 30:

If 5 moles of an ideal gas expand from 10 L to 100 L at 300 K under isothermal and reversible conditions, then work w = −x J. The value of x is:
(Given: R = 8.314 J K⁻¹ mol⁻¹)

Correct Answer: 28721

View Solution

Step 1: Formula for isothermal expansion work:

w = −nRT ln (V_f/V_i)

Step 2: Substitute the given values.

n = 5, R = 8.314, T = 300 K, V_i = 10 L, V_f = 100 L
ln(100/10) = ln(10) ≈ 2.302
w = −(5)(8.314)(300)(2.302) = −28721 J

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JEE Main 2024 Question Paper with Answer Key PDF for Chemistry (January 31 - Shift 2)

JEE Main 2024 Chemistry Question Paper 31 Jan Shift-2 with Answer Key PDF

JEE Main 2024 Chemistry Jan 31 Shift 2 Questions with Solution

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