We need to match each complex with its crystal field stabilization energy (CFSE) in units of $\Delta_0$.

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

A. $\mathrm{[Cu(NH_3)_6]^{2+}}$

• $\mathrm{NH_3}$ is neutral
• So Cu is in $+2$ oxidation state
• Cu: $[Ar]3d^{10}4s^1$
• $\mathrm{Cu^{2+}}: 3d^9$

So, A is a $d^9$ octahedral complex.

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B. $\mathrm{[Ti(H_2O)_6]^{3+}}$

• $\mathrm{H_2O}$ is neutral
• So Ti is in $+3$ oxidation state
• Ti: $[Ar]3d^24s^2$
• $\mathrm{Ti^{3+}}: 3d^1$

So, B is a $d^1$ octahedral complex.

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C. $\mathrm{[Fe(CN)_6]^{3-}}$

Let oxidation state of Fe be $x$: $x+6(-1)=-3$ $x=+3$ Thus Fe is $\mathrm{Fe^{3+}}$.

• Fe: $[Ar]3d^64s^2$
• $\mathrm{Fe^{3+}}: 3d^5$

$\mathrm{CN^-}$ is a strong field ligand, so this is low-spin octahedral $d^5$.

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D. $\mathrm{[NiF_6]^{4-}}$

Let oxidation state of Ni be $x$: $x+6(-1)=-4$ $x=+2$ Thus Ni is $\mathrm{Ni^{2+}}$.

• Ni: $[Ar]3d^84s^2$
• $\mathrm{Ni^{2+}}: 3d^8$

$\mathrm{F^-}$ is a weak field ligand, but for octahedral $d^8$, the configuration is still $t_{2g}^6e_g^2$.

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2. Calculate CFSE for each

For octahedral complexes:

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

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

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A. $d^9$ : $t_{2g}^6 e_g^3$

$\text{CFSE} = 6(-0.4)+3(0.6)$ $= -2.4+1.8=-0.6\Delta_0$ So, A $\to$ I

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B. $d^1$ : $t_{2g}^1 e_g^0$

$\text{CFSE} = 1(-0.4) = -0.4\Delta_0$ So, B $\to$ IV

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C. low-spin $d^5$ : $t_{2g}^5 e_g^0$

$\text{CFSE} = 5(-0.4) = -2.0\Delta_0$ So, C $\to$ II

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D. $d^8$ : $t_{2g}^6 e_g^2$

$\text{CFSE} = 6(-0.4)+2(0.6)$ $= -2.4+1.2=-1.2\Delta_0$ So, D $\to$ III

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3. Final matching

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

This corresponds to Option A.

4. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

They agree.