1. All given complexes are octahedral aqua complexes

Since each complex is of the form $[M(H_2O)_6]^{n+}$, the geometry is octahedral.

Also, $H_2O$ is a weak field ligand, so for $3d$ metals we generally get high-spin configurations.

In an octahedral field, the $d$-orbitals split as: $t_{2g} < e_g$

---

2. Find oxidation state and $d$-electron count for each metal

(A) $[Cr(H_2O)_6]^{3+}$

• $H_2O$ is neutral, so oxidation state of Cr is $+3$.
• Cr: $[Ar]3d^5 4s^1$
• $Cr^{3+}$: remove 3 electrons $\Rightarrow 3d^3$

Thus configuration in octahedral field: $t_{2g}^3 e_g^0$ So, $A \to (II)$

---

(B) $[Fe(H_2O)_6]^{3+}$

• Fe oxidation state $=+3$
• Fe: $[Ar]3d^6 4s^2$
• $Fe^{3+}$: remove 3 electrons $\Rightarrow 3d^5$

With weak field ligand $H_2O$, this is high spin: $t_{2g}^3 e_g^2$ So, $B \to (III)$

---

(C) $[Ni(H_2O)_6]^{2+}$

• Ni oxidation state $=+2$
• Ni: $[Ar]3d^8 4s^2$
• $Ni^{2+}$: remove 2 electrons $\Rightarrow 3d^8$

Octahedral arrangement: $t_{2g}^6 e_g^2$ So, $C \to (IV)$

---

(D) $[V(H_2O)_6]^{3+}$

• V oxidation state $=+3$
• V: $[Ar]3d^3 4s^2$
• $V^{3+}$: remove 3 electrons $\Rightarrow 3d^2$

Thus: $t_{2g}^2 e_g^0$ So, $D \to (I)$

---

3. Final matching

$A\to II, \quad B\to III, \quad C\to IV, \quad D\to I$

This corresponds to Option C.

---

4. Comparison with stored answer

Stored correct answer: C

My derived answer: C

They agree.