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NEET Practice Questions (MCQs) with Answers & Solutions

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NEET 2025

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Match List-I with List-II:

List-I (Example) — A. Humidity, B. Alloys, C. Amalgams, D. Smoke

List-II (Type of Solution) — I. Solid in solid, II. Liquid in gas, III. Solid in gas, IV. Liquid in solid

Choose the correct answer:

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

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Explanation

Humidity = liquid in gas (II); Alloys = solid in solid (I); Amalgams = liquid (Hg) in solid (IV); Smoke = solid in gas (III).

NEET 2025

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The correct order of decreasing basic strength of the given amines is:

A. N-methylaniline > benzenamine > ethanamine > N-ethylethanamine B. N-ethylethanamine > ethanamine > benzenamine > N-methylaniline C. N-ethylethanamine > ethanamine > N-methylaniline > benzenamine D. benzenamine > ethanamine > N-methylaniline > N-ethylethanamine

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Explanation

Aliphatic amines are more basic than aromatic. Secondary (N-ethylethanamine) > primary aliphatic (ethanamine) > N-methylaniline > aniline (benzenamine).

NEET 2025

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Among the following, choose the ones with equal number of atoms. A. 212 g of $\text{Na}_2\text{CO}_3$ [molar mass 106 g], B. 248 g of $\text{Na}_2\text{O}$ [molar mass 62 g], C. 240 g of NaOH [molar mass 40 g], D. 12 g of $\text{H}_2$ [molar mass 2 g], E. 220 g of $\text{CO}_2$ [molar mass 44 g]. Choose the correct answer:

A. A, B, and C only B. A, B, and D only C. B, C, and D only D. B, D, and E only

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Explanation

A: 2 mol × 6 atoms = 12 $N_A$; B: 4 mol × 3 = 12 $N_A$; C: 6 mol × 3 = 18 $N_A$; D: 6 mol × 2 = 12 $N_A$; E: 5 mol × 3 = 15 $N_A$. Equal (12 $N_A$): A, B and D.

NEET 2025

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Match List I with List II:

List I (Name of Vitamin) — A. Vitamin B₁₂, B. Vitamin D, C. Vitamin B₂, D. Vitamin B₆

List II (Deficiency disease) — I. Cheilosis, II. Convulsions, III. Rickets, IV. Pernicious anaemia

Choose the correct answer:

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

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Explanation

B₁₂ → Pernicious anaemia (IV); D → Rickets (III); B₂ → Cheilosis (I); B₆ → Convulsions (II).

NEET 2025

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The correct order of decreasing acidity of the following aliphatic acids is:

A. $(\text{CH}_3)_3\text{CCOOH} > (\text{CH}_3)_2\text{CHCOOH} > \text{CH}_3\text{COOH} > \text{HCOOH}$ B. $\text{CH}_3\text{COOH} > (\text{CH}_3)_2\text{CHCOOH} > (\text{CH}_3)_3\text{CCOOH} > \text{HCOOH}$ C. $\text{HCOOH} > \text{CH}_3\text{COOH} > (\text{CH}_3)_2\text{CHCOOH} > (\text{CH}_3)_3\text{CCOOH}$ D. $\text{HCOOH} > (\text{CH}_3)_3\text{CCOOH} > (\text{CH}_3)_2\text{CHCOOH} > \text{CH}_3\text{COOH}$

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Explanation

More alkyl (+I) groups decrease acidity. HCOOH (no alkyl) is strongest; the bulky $(\text{CH}_3)_3\text{CCOOH}$ is weakest.

NEET 2025

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Given below are two statements:

Statement I: Ferromagnetism is considered as an extreme form of paramagnetism.

Statement II: The number of unpaired electrons in a $\text{Cr}^{2+}$ ion (Z = 24) is the same as that of a $\text{Nd}^{3+}$ ion (Z = 60).

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

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

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Explanation

Ferromagnetism is an extreme form of paramagnetism (I true). $\text{Cr}^{2+}$ ($3d^4$) has 4 unpaired electrons while $\text{Nd}^{3+}$ ($4f^3$) has 3 — not equal (II false).

NEET 2025

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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:

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

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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

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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:

A. 83.1 B. $2.077\times10^5$ C. 0.033 D. 0.021

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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

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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:

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

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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

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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)?

A. 2 B. 3 C. 5 D. 6

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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 $= 2+1+2+1 = 6$.

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