JEE Chemistry 2023 Question with Solution

Given: Four statements about reduction potential, stabilization of oxidation states, reducing behaviour of $\mathrm{Cr}^{2+}$, and magnetic moment of $\mathrm{V}^{2+}$ are to be checked.

Find: Which statements are correct.

For statement (A), compare the standard reduction potentials:

Since the value for manganese is greater than that for iron, statement (A) is incorrect.

For statement (B), higher oxidation states of first-row d-block elements are stabilized by oxide ion because strong metal-oxygen bonds are formed. Hence statement (B) is correct.

For statement (C), $\mathrm{Cr}^{2+}$ acts as a reducing agent and can reduce $\mathrm{H}^+$ to hydrogen:

Also,

This confirms that aqueous $\mathrm{Cr}^{2+}$ can liberate hydrogen from dilute acid. So statement (C) is correct.

For statement (D), $\mathrm{V}^{2+}$ has three unpaired electrons, so its magnetic moment is approximately $3.87\,\mathrm{BM}$, which does not lie in the range $4.4\text{-}5.2\,\mathrm{BM}$. Therefore statement (D) is incorrect.

Thus, only statements (B) and (C) are correct. Therefore, the correct option is B.

(A) The claim says iron has greater $\mathrm{M}^{3+}/\mathrm{M}^{2+}$ reduction potential than manganese. But

and

Hence the claim is false.

(B) Oxide ion stabilizes higher oxidation states of transition metals by strong bonding, so this statement is true.

(C) Because $\mathrm{Cr}^{2+}$ is readily oxidized to $\mathrm{Cr}^{3+}$, it can reduce acid to hydrogen gas. Thus this statement is true.

(D) For $\mathrm{V}^{2+}$, the number of unpaired electrons is $3$, giving spin-only magnetic moment near $3.87\,\mathrm{BM}$, not $4.4\text{-}5.2\,\mathrm{BM}$. So this statement is false.

Therefore the correct set is (B), (C) only.