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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.

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:

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

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:

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

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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?

A. Law of conservation of mass B. Law of constant proportion C. Law of multiple proportion D. Law of gaseous volume

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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:

A. +1, −1, and +6 B. +2, −2, and +6 C. +1, −2, and +4 D. +4, −4, and +6

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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.

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