Chemistry MCQs for NEET — Practice Questions with Answers

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NEET 2025

Match List I with List II:

List I (Mixture) — A. CHCl₃ + C₆H₅NH₂, B. Crude oil in petroleum industry, C. Glycerol from spent-lye, D. Aniline - water

List II (Method of Separation) — I. Distillation under reduced pressure, II. Steam distillation, III. Fractional distillation, IV. Simple distillation

Choose the correct answer:

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Explanation

CHCl₃ + aniline (large bp difference) → simple distillation (IV); crude oil → fractional (III); glycerol from spent-lye → distillation under reduced pressure (I); aniline–water → steam distillation (II).

NEET 2025

For the reaction $\text{A(g)} \rightleftharpoons 2\text{B(g)}$, the backward reaction rate constant is higher than the forward reaction rate constant by a factor of 2500, at 1000 K. [Given: $R = 0.0831\ \text{L atm mol}^{-1}\text{K}^{-1}$] $K_p$ for the reaction at 1000 K is:

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Explanation

$K_c = \dfrac{k_f}{k_b} = \dfrac{1}{2500}$. $\Delta n = 1$, so $K_p = K_c(RT) = \dfrac{0.0831\times1000}{2500} = 0.033$.

NEET 2025

Given below are two statements:

Statement I: Benzenediazonium salt is prepared by the reaction of aniline with nitrous acid at 273–278 K. It decomposes easily in the dry state.

Statement II: Insertion of iodine into the benzene ring is difficult and hence iodobenzene is prepared through the reaction of benzenediazonium salt with KI.

In the light of the above statements, choose the most appropriate answer:

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Explanation

Both statements are correct: benzenediazonium chloride is made at 273–278 K and decomposes when dry; iodobenzene is conveniently made via the diazonium salt + KI.

NEET 2025

How many products (including stereoisomers) are expected from monochlorination of the following compound: $(\text{CH}_3)_2\text{CH–CH}_2\text{–CH}_3$ (2-methylbutane)?

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Explanation

Four constitutional monochloro products; two of them (1-chloro-2-methylbutane and 2-chloro-3-methylbutane) contain a stereocentre giving a pair of enantiomers each. Total $= 2+1+2+1 = 6$.

NEET 2025

Among the given compounds I–III, the correct order of bond dissociation energy of the C–H bond marked with * is:

*H (I)*H (II)*H (III)
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Explanation

C–H bond strength rises with s-character: $sp$ (II, alkyne) > $sp^2$ (I, benzene) > $sp^3$ (III, cyclopropane). Hence II > I > III.

NEET 2025

Which one of the following compounds does NOT decolourize bromine water?

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Explanation

Cyclohexane is a saturated hydrocarbon and does not react with bromine water. Phenol and aniline brominate the ring; styrene adds across the C=C.

NEET 2025

The major product of the following reaction is: $\text{C}_6\text{H}_5\text{COCH}_2\text{CH}_2\text{CN} \xrightarrow{\text{(i) CH}_3\text{MgBr (excess)}}\xrightarrow{\text{(ii) H}_3\text{O}^+}$

OC≡N(i) CH₃MgBr (excess) (ii) H₃O⁺
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Explanation

Excess CH₃MgBr adds to the ketone (→ tertiary alcohol) and to the nitrile (→ imine, then ketone on hydrolysis). Product: a tertiary alcohol at the original carbonyl and a methyl ketone where the –CN was.

NEET 2025

Which of the following aqueous solution will exhibit highest boiling point?

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Explanation

Boiling-point elevation $\propto i\times m$. Effective concentrations: urea 0.01, KNO₃ 0.02, Na₂SO₄ 0.03, glucose 0.015. Na₂SO₄ ($i=3$) is highest.

NEET 2025

Match List-I with List-II:

List-I — A. Haber process, B. Wacker oxidation, C. Wilkinson catalyst, D. Ziegler catalyst

List-II — I. Fe catalyst, II. PdCl₂, III. [(PPh₃)₃RhCl], IV. TiCl₄ with Al(CH₃)₃

Choose the correct answer:

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Explanation

Haber → Fe (I); Wacker → PdCl₂ (II); Wilkinson → [(PPh₃)₃RhCl] (III); Ziegler → TiCl₄ with Al(CH₃)₃ (IV).

NEET 2025

5 moles of liquid X and 10 moles of liquid Y make a solution having a vapour pressure of 70 torr. The vapour pressures of pure X and Y are 63 torr and 78 torr respectively. Which of the following is true regarding the described solution?

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Explanation

Ideal (Raoult) pressure $= \tfrac13(63)+\tfrac23(78) = 73$ torr. Observed 70 torr < 73 torr → negative deviation.

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