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Chemistry MCQs for NEET — Practice Questions with Answers

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NEET 2025
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Given below are two statements:

Statement I: Like nitrogen that can form ammonia, arsenic can form arsine.

Statement II: Antimony cannot form antimony pentoxide.

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

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Explanation

Arsenic forms arsine $\text{AsH}_3$ (Statement I correct). Antimony does form $\text{Sb}_2\text{O}_5$ (antimony pentoxide), so Statement II is incorrect.

NEET 2025
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Which among the following electronic configurations belong to main group elements? A. $[\text{Ne}]3s^1$, B. $[\text{Ar}]3d^3 4s^2$, C. $[\text{Kr}]4d^{10}5s^2 5p^5$, D. $[\text{Ar}]3d^{10}4s^1$, E. $[\text{Rn}]5f^0 6d^2 7s^2$. Choose the correct answer from the option given below:

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Explanation

A ($[\text{Ne}]3s^1$ = Na, s-block) and C ($[\text{Kr}]4d^{10}5s^2 5p^5$ = I, p-block) are main group. B, D are transition (d-block) and E is f-block.

NEET 2025
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Dalton's Atomic theory could not explain which of the following?

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Explanation

Dalton's theory explained the laws of conservation of mass, constant and multiple proportions, but could not explain Gay-Lussac's law of gaseous volumes.

NEET 2025
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Consider the following compounds: $\underline{\text{K}}\text{O}_2$, $\text{H}_2\underline{\text{O}}_2$ and $\text{H}_2\underline{\text{S}}\text{O}_4$. The oxidation states of the underlined elements in them are, respectively:

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Explanation

In $\text{KO}_2$, K is +1; in $\text{H}_2\text{O}_2$, O is −1; in $\text{H}_2\text{SO}_4$, S is +6.

NEET 2025
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If the half-life $(t_{1/2})$ for a first order reaction is 1 minute, then the time required for 99.9% completion of the reaction is closest to:

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Explanation

99.9% completion leaves $1/1000$ remaining. $\left(\tfrac12\right)^n = \tfrac{1}{1000}\Rightarrow n\approx10$ half-lives $= 10$ minutes.

NEET 2025
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The correct order of the wavelength of light absorbed by the following complexes is: A. $[\text{Co(NH}_3)_6]^{3+}$, B. $[\text{Co(CN)}_6]^{3-}$, C. $[\text{Cu(H}_2\text{O)}_4]^{2+}$, D. $[\text{Ti(H}_2\text{O)}_6]^{3+}$. Choose the correct answer:

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Explanation

Larger crystal-field splitting $\Delta$ means shorter absorbed wavelength. $\Delta$: B (CN⁻) > A (NH₃) > D (Ti³⁺/H₂O) > C (Cu²⁺/H₂O). So wavelength increases as B < A < D < C.

NEET 2025
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Which one of the following compounds can exist as cis-trans isomers?

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Explanation

1,2-Dimethylcyclohexane has two different groups on each of two ring carbons → cis and trans isomers exist. The others lack the required geometric restriction.

NEET 2025
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Phosphoric acid ionizes in three steps with their ionization constant values $K_{a_1}$, $K_{a_2}$ and $K_{a_3}$, respectively, while K is the overall ionization constant. Which of the following statements are true? A. $\log K = \log K_{a_1} + \log K_{a_2} + \log K_{a_3}$. B. $\text{H}_3\text{PO}_4$ is a stronger acid than $\text{H}_2\text{PO}_4^-$ and $\text{HPO}_4^{2-}$. C. $K_{a_1} > K_{a_2} > K_{a_3}$. D. $K_{a_1} = \dfrac{K_{a_3}+K_{a_2}}{2}$. Choose the correct answer:

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Explanation

$K = K_{a_1}K_{a_2}K_{a_3}\Rightarrow\log K = \sum\log K_{a_i}$ (A true). Successive ionization weakens: $K_{a_1} > K_{a_2} > K_{a_3}$ (C true) and $\text{H}_3\text{PO}_4$ is the strongest (B true). D is false.

NEET 2025
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Which one of the following reactions does NOT give benzene as the product?

COONa(1) PhCOONa + sodalime, Δ(2) n-hexane / Mo₂O₃, 773 K(3) HC≡CH / red-hot Fe, 873 KN₂⁺Cl⁻(4) PhN₂⁺Cl⁻ + H₂O, warm
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Explanation

Benzene diazonium chloride with warm water gives phenol (and N₂), not benzene. The other three (decarboxylation, aromatization of hexane, cyclic trimerization of ethyne) all give benzene.

NEET 2025
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If the molar conductivity $(\Lambda_m)$ of a $0.050\ \text{mol L}^{-1}$ solution of a monobasic weak acid is $90\ \text{S cm}^2\text{mol}^{-1}$, its extent (degree) of dissociation will be [Assume $\Lambda^\circ_+ = 349.6\ \text{S cm}^2\text{mol}^{-1}$ and $\Lambda^\circ_- = 50.4\ \text{S cm}^2\text{mol}^{-1}$]:

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Explanation

$\Lambda^\circ_m = 349.6 + 50.4 = 400$. $\alpha = \dfrac{\Lambda_m}{\Lambda^\circ_m} = \dfrac{90}{400} = 0.225$.

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