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Chemistry MCQs for NEET — Practice Questions with Answers

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NEET 2024

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Identify the major product C formed in the following reaction sequence:

$CH_3\text{-}CH_2\text{-}CH_2\text{-}I \xrightarrow{NaCN} A \xrightarrow[\text{Partial hydrolysis}]{OH^-} B \xrightarrow{NaOH/Br_2} C\ (\text{major})$

A. $\alpha$-bromobutanoic acid B. propylamine C. butylamine D. butanamide

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Explanation

NaCN substitution → butanenitrile (A); partial hydrolysis → butanamide (B); Hofmann bromamide → 1° amine with one less C → propylamine (C).

NEET 2024

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During the preparation of Mohr's salt solution (Ferrous ammonium sulphate), which of the following acid is added to prevent hydrolysis of $\text{Fe}^{2+}$ ion?

A. dilute sulphuric acid B. dilute hydrochloric acid C. concentrated sulphuric acid D. dilute nitric acid

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Explanation

Dilute H₂SO₄ suppresses hydrolysis of Fe²⁺ without introducing foreign cations/anions or oxidising Fe²⁺.

NEET 2024

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The rate of a reaction quadruples when temperature changes from $27^\circ C$ to $57^\circ C$. Calculate the energy of activation. Given $R = 8.314\ J\,K^{-1}\,mol^{-1}$, $\log 4 = 0.6021$.

A. 3804 kJ/mol B. 38.04 kJ/mol C. 380.4 kJ/mol D. 3.80 kJ/mol

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Explanation

$\log(k_2/k_1) = (E_a/2.303R)(1/T_1 - 1/T_2)$ → $E_a \approx 38.05$ kJ/mol.

NEET 2024

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Mass in grams of copper deposited by passing $9.6487$ A current through a voltmeter containing copper sulphate solution for 100 seconds is:

(Given: Molar mass of $Cu : 63\ g\,mol^{-1}$, $1F = 96487\,C$)

A. 0.0315 g B. 3.15 g C. 0.315 g D. 31.5 g

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Explanation

$m = \dfrac{M\,I\,t}{nF} = \dfrac{63 \times 9.6487 \times 100}{2 \times 96487} = 0.315$ g.

NEET 2025

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The ratio of the wavelengths of the light absorbed by a Hydrogen atom when it undergoes $n = 2 \to n = 3$ and $n = 4 \to n = 6$ transitions, respectively, is:

A. $\dfrac{1}{36}$ B. $\dfrac{1}{16}$ C. $\dfrac{1}{9}$ D. $\dfrac{1}{4}$

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Explanation

$\dfrac{1}{\lambda}\propto\left(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right)$. For $2\to3$: $\tfrac{5}{36}$; for $4\to6$: $\tfrac{5}{144}$. $\dfrac{\lambda_{2\to3}}{\lambda_{4\to6}} = \dfrac{5/144}{5/36} = \dfrac{1}{4}$.

NEET 2025

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Which of the following statements are true?

A. Unlike Ga that has a very high melting point, Cs has a very low melting point. B. On Pauling scale, the electronegativity values of N and Cl are not the same. C. Ar, K⁺, Cl⁻, Ca²⁺, and S²⁻ are all isoelectronic species. D. The correct order of the first ionization enthalpies of Na, Mg, Al, and Si is Si > Al > Mg > Na. E. The atomic radius of Cs is greater than that of Li and Rb.

Choose the correct answer from the options given below:

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

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Explanation

A is false (Ga has a low melting point). B is false (N and Cl have nearly equal Pauling electronegativity, ≈3.0). C is true (all have 18 electrons). D is false (correct order is Si > Mg > Al > Na). E is true. Hence C and E only.

NEET 2025

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Match List I with List II:

List I (Ion) — A. Co²⁺, B. Mg²⁺, C. Pb²⁺, D. Al³⁺

List II (Group Number in Cation Analysis) — I. Group-I, II. Group-III, III. Group-IV, IV. Group-VI

Choose the correct answer from the options given below:

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

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Explanation

Co²⁺ → Group IV (III); Mg²⁺ → Group VI (IV); Pb²⁺ → Group I (I); Al³⁺ → Group III (II). So A-III, B-IV, C-I, D-II.

NEET 2025

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Predict the major product "P" in the following sequence of reactions: 1-methylcyclopentene $\xrightarrow{\text{(i) HBr, benzoyl peroxide}}\xrightarrow{\text{(ii) KCN}}\xrightarrow{\text{(iii) Na(Hg)/C}_2\text{H}_5\text{OH}}$ P (Major)

A. 2-methylcyclopentyl)methanamine — cyclopentane bearing CH₃ and –CH₂NH₂ on adjacent carbons B. cyclopentane with CH₃ and –CH₂NH₂ on the same carbon C. cyclopentane bearing CH₃ and –NC (isocyanide) on adjacent carbons D. cyclopentane with CH₃ and –NC on the same carbon

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Explanation

(i) HBr with peroxide adds Br anti-Markovnikov to the less-substituted carbon (C2). (ii) KCN does SN2 → nitrile at C2. (iii) Na(Hg)/EtOH reduces –CN to –CH₂NH₂. Product: CH₃ and –CH₂NH₂ on adjacent ring carbons.

NEET 2025

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Energy and radius of first Bohr orbit of He⁺ and Li²⁺ are [Given $R_H = 2.18\times10^{-18}$ J, $a_0 = 52.9$ pm]:

A. $E_n(\text{Li}^{2+}) = -19.62\times10^{-18}$ J; $r_n(\text{Li}^{2+}) = 17.6$ pm; $E_n(\text{He}^+) = -8.72\times10^{-18}$ J; $r_n(\text{He}^+) = 26.4$ pm B. $E_n(\text{Li}^{2+}) = -8.72\times10^{-18}$ J; $r_n(\text{Li}^{2+}) = 26.4$ pm; $E_n(\text{He}^+) = -19.62\times10^{-18}$ J; $r_n(\text{He}^+) = 17.6$ pm C. $E_n(\text{Li}^{2+}) = -19.62\times10^{-16}$ J; $r_n(\text{Li}^{2+}) = 17.6$ pm; $E_n(\text{He}^+) = -8.72\times10^{-16}$ J; $r_n(\text{He}^+) = 26.4$ pm D. $E_n(\text{Li}^{2+}) = -8.72\times10^{-16}$ J; $r_n(\text{Li}^{2+}) = 17.6$ pm; $E_n(\text{He}^+) = -19.62\times10^{-16}$ J; $r_n(\text{He}^+) = 17.6$ pm

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Explanation

$E_n = -R_H\dfrac{Z^2}{n^2}$, $r_n = a_0\dfrac{n^2}{Z}$. Li²⁺ ($Z=3$): $E = -2.18\times9 = -19.62\times10^{-18}$ J, $r = 52.9/3 = 17.6$ pm. He⁺ ($Z=2$): $E = -8.72\times10^{-18}$ J, $r = 26.4$ pm.

NEET 2025

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Which of the following are paramagnetic? A. $[\text{NiCl}_4]^{2-}$, B. $\text{Ni(CO)}_4$, C. $[\text{Ni(CN)}_4]^{2-}$, D. $[\text{Ni(H}_2\text{O)}_6]^{2+}$, E. $\text{Ni(PPh}_3)_4$. Choose the correct answer from the options given below:

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

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Explanation

$[\text{NiCl}_4]^{2-}$ (tetrahedral) and $[\text{Ni(H}_2\text{O)}_6]^{2+}$ (octahedral) have 2 unpaired electrons → paramagnetic. $\text{Ni(CO)}_4$, $[\text{Ni(CN)}_4]^{2-}$ (square planar) and $\text{Ni(PPh}_3)_4$ are diamagnetic.

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