Chemistry NEET Questions — Free MCQ Practice

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Chemistry MCQs for NEET — Practice Questions with Answers

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NEET 2024

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Match List I with List II.

List I (Reactions)	List II (Reagents/Condition)
A. Bicyclohexyl → 2 cyclohexanone	I. PhCOCl / Anhyd. AlCl₃
B. Benzene → diphenyl ketone	II. CrO₃
C. Cyclohexanol → cyclohexanone	III. KMnO₄ / KOH, Δ
D. Ar–CH₂CH₃ (with –COOK) → ozonolysis product	IV. (i) O₃ (ii) Zn–H₂O

Choose the correct answer from the options given below:

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

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Explanation

A-III (KMnO₄/KOH), B-I (Friedel–Crafts acylation), C-II (CrO₃), D-IV (ozonolysis).

NEET 2024

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Activation energy of any chemical reaction can be calculated if one knows the value of:

A. rate constant at two different temperatures B. rate constant at standard temperature C. probability of collision D. orientation of reactant molecules during collision

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Explanation

Arrhenius: $\ln(k_2/k_1) = (E_a/R)(1/T_1 - 1/T_2)$ — need $k$ at two $T$.

NEET 2024

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The most stable carbocation among the following is:

A. 1-methylcyclohexyl cation (tertiary, on cyclohexane ring) B. H₃C–CH₂–C⁺(H)–CH(CH₃)–CH₃ (secondary) C. CH₃–C⁺(H)–CH₂–CH(CH₃)–CH₃ (secondary) D. cyclopentyl-CH₂⁺ (primary)

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Explanation

(A) is a 3° carbocation with maximum hyperconjugation and inductive stabilisation.

NEET 2024

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Given below are two statements:

Statement I: The boiling point of three isomeric pentanes follows the order n-pentane > isopentane > neopentane

Statement II: When branching increases, the molecule attains a shape of sphere. This results in smaller surface area for contact, due to which the intermolecular forces between the spherical molecules are weak, thereby lowering the boiling point.

In the light of the above statements, choose the most appropriate answer from the options given below:

A. Statement I is incorrect but Statement II is correct. B. Both Statement I and Statement II are correct. C. Both Statement I and Statement II are incorrect. D. Statement I is correct but Statement II is incorrect.

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Explanation

Both standard NCERT facts — II is the correct explanation of I.

NEET 2024

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Which one of the following alcohols reacts instantaneously with Lucas reagent?

A. $(CH_3)_3C\text{-OH}$ (2-methyl-2-propanol, tertiary) B. $CH_3\text{-}CH_2\text{-}CH_2\text{-}CH_2OH$ (n-butanol, primary) C. $CH_3\text{-}CH_2\text{-}CH(\text{OH})\text{-}CH_3$ (2-butanol, secondary) D. $CH_3\text{-}CH(CH_3)\text{-}CH_2OH$ (isobutyl alcohol, primary)

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Explanation

Lucas test: 3° alcohols react instantaneously (immediate turbidity).

NEET 2024

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The Henry's law constant ($K_H$) values of three gases (A, B, C) in water are $145$, $2\times 10^{-5}$ and $35$ kbar, respectively. The solubility of these gases in water follow the order:

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

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Explanation

Solubility $\propto 1/K_H$; B has the smallest $K_H$, A the largest.

NEET 2024

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Which plot of $\ln k$ vs $\dfrac{1}{T}$ is consistent with Arrhenius equation? (Options are sketches of straight lines through the first quadrant: A — decreasing line with negative slope; B–D — increasing lines.)

A. A — straight line with negative slope (decreasing) B. B — increasing straight line (positive slope) ending on the y-axis C. C — increasing straight line through origin D. D — increasing straight line with positive slope

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Explanation

$\ln k = \ln A - E_a/(RT)$ → slope $= -E_a/R < 0$.

NEET 2024

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On heating, some solid substances change from solid to vapour state without passing through liquid state. The technique used for the purification of such solid substances based on the above principle is known as:

A. Chromatography B. Crystallization C. Sublimation D. Distillation

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Explanation

Direct solid → vapour transition is sublimation.

NEET 2024

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Intramolecular hydrogen bonding is present in

A. HF B. o-nitrophenol (ortho) C. p-nitrophenol (para) D. m-nitrophenol (meta)

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Explanation

Only in the ortho isomer are –OH and –NO₂ close enough to form an intramolecular H-bond.

NEET 2024

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1 gram of sodium hydroxide was treated with 25 mL of $0.75$ M HCl solution, the mass of sodium hydroxide left unreacted is equal to:

A. 200 mg B. 750 mg C. 250 mg D. Zero mg

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Explanation

25 mmol NaOH, 18.75 mmol HCl. Remaining NaOH = 6.25 mmol $\times$ 40 $=$ 250 mg.

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Frequently Asked Questions

Yes. You can attempt every Chemistry question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.