We need to match each octahedral aqua complex with the Crystal Field Stabilization Energy (CFSE) values given.

1. Determine oxidation state and $d$-electron count

All ligands are $H_2O$, which is a neutral ligand. So the metal oxidation state equals the complex charge.

A. $[Ti(H_2O)_6]^{2+}$

Ti: $Z=22$

Neutral Ti: $[Ar]3d^2 4s^2$

$Ti^{2+} \Rightarrow 3d^2$

So, $d^2$ configuration.

B. $[V(H_2O)_6]^{2+}$

V: $Z=23$

Neutral V: $[Ar]3d^3 4s^2$

$V^{2+} \Rightarrow 3d^3$

So, $d^3$ configuration.

C. $[Mn(H_2O)_6]^{3+}$

Mn: $Z=25$

Neutral Mn: $[Ar]3d^5 4s^2$

$Mn^{3+} \Rightarrow 3d^4$

So, $d^4$ configuration.

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

Fe: $Z=26$

Neutral Fe: $[Ar]3d^6 4s^2$

$Fe^{3+} \Rightarrow 3d^5$

So, $d^5$ configuration.

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2. Nature of ligand

$H_2O$ is a weak field ligand, so these octahedral complexes are high spin.

In octahedral field:

• each electron in $t_{2g}$ contributes $-0.4\Delta_0$
• each electron in $e_g$ contributes $+0.6\Delta_0$

Thus,

$$\text{CFSE} = (-0.4\times n_{t_{2g}} + 0.6\times n_{e_g})\Delta_0$$

The table gives only the numerical coefficient.

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3. Calculate CFSE for each complex

A. $d^2$

High-spin octahedral filling: $t_{2g}^2 e_g^0$

$$\text{CFSE} = 2(-0.4\Delta_0) = -0.8\Delta_0$$

So A $\to$ IV.

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B. $d^3$

High-spin octahedral filling: $t_{2g}^3 e_g^0$

$$\text{CFSE} = 3(-0.4\Delta_0) = -1.2\Delta_0$$

So B $\to$ I.

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C. $d^4$ (high spin)

High-spin octahedral filling: $t_{2g}^3 e_g^1$

$$\text{CFSE} = 3(-0.4\Delta_0) + 1(+0.6\Delta_0)$$ $$= -1.2\Delta_0 + 0.6\Delta_0 = -0.6\Delta_0$$

So C $\to$ II.

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D. $d^5$ (high spin)

High-spin octahedral filling: $t_{2g}^3 e_g^2$

$$\text{CFSE} = 3(-0.4\Delta_0) + 2(+0.6\Delta_0)$$ $$= -1.2\Delta_0 + 1.2\Delta_0 = 0$$

So D $\to$ III.

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4. Final matching

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

This corresponds to Option A.

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5. Comparison with stored correct answer

Stored correct answer: A

My derived answer: A

So, the answer agrees with the stored correct answer.