Write the ground-state configurations of the neutral atoms

• $\mathrm{Cr}$ : $[Ar] \, 3d^5 4s^1$
• $\mathrm{Mn}$ : $[Ar] \, 3d^5 4s^2$
• $\mathrm{Ni}$ : $[Ar] \, 3d^8 4s^2$
• $\mathrm{V}$ : $[Ar] \, 3d^3 4s^2$

Rule for forming cations of transition elements

Electrons are removed first from the $4s$ orbital and then from the $3d$ orbital.

Find the electronic distribution of each species

(A) $\mathrm{Cr^{2+}}$

Neutral $\mathrm{Cr}$ is $3d^5 4s^1$. Remove two electrons:

• first from $4s$: $3d^5$
• then one from $3d$: $3d^4$

So, $\mathrm{Cr^{2+}} \to 3d^4$. Hence, (A) $\to$ (III).

(B) $\mathrm{Mn^{+}}$

Neutral $\mathrm{Mn}$ is $3d^5 4s^2$. Remove one electron from $4s$:

$\mathrm{Mn^+} : 3d^5 4s^1$

Hence, (B) $\to$ (IV).

(C) $\mathrm{Ni^{2+}}$

Neutral $\mathrm{Ni}$ is $3d^8 4s^2$. Remove two $4s$ electrons:

$\mathrm{Ni^{2+}} : 3d^8$

Hence, (C) $\to$ (I).

(D) $\mathrm{V^{+}}$

Neutral $\mathrm{V}$ is $3d^3 4s^2$. Remove one electron from $4s$:

$\mathrm{V^+} : 3d^3 4s^1$

Hence, (D) $\to$ (II).

Final matching

Match with given options

This corresponds to Option A.