What is the half-life of a first-order reaction with rate constant k = 0.693 s⁻¹?

The correct answer is B: 1.0 s. For first-order reactions: t₁/₂ = ln(2)/k = 0.693/0.693 = 1.0 s. The half-life of a first-order reaction is independent of initial concentration.

The IUPAC name of the compound with structure CH3-CH(OH)-CH2-CHO is:

The correct answer is B: 3-hydroxybutanal. The longest carbon chain contains 4 carbons with the aldehyde group (CHO) at position 1. The hydroxyl group is at position 3, giving 3-hydroxybutanal.

For a reaction with rate law Rate = k[A]⁰[B]¹, if [B] is increased 4 times while [A] remains constant, the rate increases by:

The correct answer is B: 4 times. Since the reaction is first-order in B and zero-order in A, Rate ∝ [B]. Increasing [B] by 4 times increases Rate by 4 times

In the complex reaction: A + B → I (fast equilibrium), I + B → C (slow), the rate law derived from this mechanism is:

The correct answer is B: Rate = k[A][B]². Rate = k₂[I][B]. From equilibrium: [I] = K₁[A][B]. Therefore: Rate = k₂K₁[A][B]², which is second-order in B

In the extraction of iron from Fe₂O₃ using CO as reducing agent, the rate-determining step involves:

The correct answer is B: Diffusion of CO through the solid layer of unreacted ore. In solid-state reduction, diffusion through the ash layer is often the rate-determining step, creating a diffusion barrier.

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Inorganic Chemistry

Chemistry questions for JEE Main — Physical, Organic, Inorganic Chemistry.

49 Q 5 Topics Take Mock Test

Difficulty: All Easy Medium Hard 21–30 of 49

Topics in JEE Chemistry

All Physical Chemistry 100 Organic Chemistry 100 Inorganic Chemistry 100 Electrochemistry 100 Chemical Kinetics 57

Q.21 Medium Inorganic Chemistry

The crystal field splitting energy (Δ) for [Co(H₂O)₆]²⁺ is approximately 9,300 cm⁻¹. The complex will be:

A High spin with 4 unpaired electrons

B Low spin with 2 unpaired electrons

C High spin with 3 unpaired electrons

D Cannot be determined

Correct Answer: A. High spin with 4 unpaired electrons

EXPLANATION

Co²⁺ is d⁷. Water is a weak field ligand with small Δ, resulting in high spin configuration (t₂g⁵ eg²) with 4 unpaired electrons.

Test

Q.22 Medium Inorganic Chemistry

The spin-only magnetic moment of Fe²⁺ in its ground state is approximately:

A 2.82 BM

B 4.90 BM

C 5.92 BM

D 3.87 BM

Correct Answer: B. 4.90 BM

EXPLANATION

Fe²⁺ has 4 unpaired electrons (d⁶ configuration). Magnetic moment = √[n(n+2)] = √[4×6] = 4.90 BM

Test

Q.23 Medium Inorganic Chemistry

The reduction potential of Fe³⁺/Fe²⁺ is +0.77 V. This indicates:

A Fe³⁺ is easily reduced

B Fe²⁺ is easily oxidized

C Both A and B

D Neither A nor B

Correct Answer: C. Both A and B

EXPLANATION

Positive E° means Fe³⁺ is easily reduced and Fe²⁺ is easily oxidized. Fe³⁺ is a strong oxidizing agent relative to Fe²⁺.

Test

Q.24 Medium Inorganic Chemistry

Which statement about borax is correct?

A It is a simple tetrahedral borate

B Its formula is Na₂B₄O₅(OH)₄·8H₂O

C It contains only BO₃ units

D It is insoluble in water

Correct Answer: B. Its formula is Na₂B₄O₅(OH)₄·8H₂O

EXPLANATION

Borax is a complex borate with formula Na₂B₄O₅(OH)₄·8H₂O, containing both triangular BO₃ and tetrahedral BO₄ units.

Test

Q.25 Medium Inorganic Chemistry

Which of the following is a non-stoichiometric compound?

A NaCl

B Fe₀.₉₅O

C CaCO₃

D H₂O

Correct Answer: B. Fe₀.₉₅O

EXPLANATION

Fe₀.₉₅O is a non-stoichiometric compound with variable composition. It contains both Fe²⁺ and Fe³⁺ ions due to defects in crystal structure.

Test

Q.26 Medium Inorganic Chemistry

In the extraction of copper using electrorefining, what is the anode material?

A Pure copper

B Blister copper

C Graphite

D Iron

Correct Answer: B. Blister copper

EXPLANATION

Blister copper (impure copper from smelting) acts as the anode. Pure copper is the cathode. Impurities fall as anode mud.

Test

Q.27 Medium Inorganic Chemistry

The magnetic moment of [Fe(CN)₆]⁴⁻ is approximately:

A 0 B.M.

B 2.82 B.M.

C 4.90 B.M.

D 5.92 B.M.

Correct Answer: A. 0 B.M.

EXPLANATION

CN⁻ is a strong field ligand causing pairing in d-orbitals. Fe²⁺ becomes diamagnetic with μ = 0 B.M. (t₂g⁶eg⁰ configuration).

Test

Q.28 Medium Inorganic Chemistry

Which compound shows the phenomenon of disproportionation?

A H₂O₂

B NaClO

C Both A and B

D Neither A nor B

Correct Answer: C. Both A and B

EXPLANATION

H₂O₂ disproportionates: H₂O₂ → H₂O + ½O₂. NaClO: 3ClO⁻ → 2Cl⁻ + ClO₃⁻. Both show disproportionation.

Test

Q.29 Medium Inorganic Chemistry

The hybridization of sulfur in SF₆ is:

A sp³

B sp³d

C sp³d²

D d²sp³

Correct Answer: C. sp³d²

EXPLANATION

SF₆ has 6 bonding pairs and 0 lone pairs on S. Coordination number = 6, requiring sp³d² hybridization with octahedral geometry.

Test

Q.30 Medium Inorganic Chemistry

Which halide of boron exists as a dimer in the gaseous state?

A BF₃

B BCl₃

C BBr₃

D BI₃

Correct Answer: B. BCl₃

EXPLANATION

BCl₃ forms a dimer (B₂Cl₆) in gaseous state due to bridging chlorine atoms. BF₃ remains monomeric due to strong B-F bonds and p-p repulsion.

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