1. Determine oxidation state and electronic configuration of the metal in each complex

For all nickel complexes here, nickel is in oxidation state $0$ or $+2$.

• $Ni: [Ar]3d^8 4s^2$
• $Ni^{2+}: [Ar]3d^8$
• $Mn^{2+}: [Ar]3d^5$

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2. Analyze each species

A. $[Ni(CO)_4]$

• $CO$ is a neutral ligand, so oxidation state of Ni is $0$.
• Thus $Ni^0$ has configuration $3d^{10}$.
• A $d^{10}$ tetra-coordinate carbonyl complex is tetrahedral.
• Since all electrons are paired, magnetic moment is $0$ BM.

So, A $\to$ III.

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B. $[Ni(CN)_4]^{2-}$

• Let oxidation state of Ni be $x$: $x + 4(-1) = -2 \Rightarrow x=+2$
• So metal is $Ni^{2+}$, i.e. $3d^8$.
• $CN^-$ is a strong field ligand, causing pairing.
• For $d^8$ Ni(II) with strong field ligands, the complex is square planar and diamagnetic.
• Hence magnetic moment = $0$ BM.

So, B $\to$ II.

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C. $[NiCl_4]^{2-}$

• Oxidation state of Ni: $x + 4(-1) = -2 \Rightarrow x=+2$
• So $Ni^{2+}$ is $3d^8$.
• $Cl^-$ is a weak field ligand.
• Therefore the complex is tetrahedral.
• Tetrahedral $Ni^{2+}(d^8)$ has 2 unpaired electrons.
• Magnetic moment: $\mu = \sqrt{n(n+2)} = \sqrt{2(2+2)} = \sqrt{8} \approx 2.8\,BM$

So, C $\to$ I.

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D. $[MnBr_4]^{2-}$

• Oxidation state of Mn: $x + 4(-1) = -2 \Rightarrow x=+2$
• So $Mn^{2+}$ is $3d^5$.
• $Br^-$ is a weak field ligand, so tetrahedral complex will be high spin.
• Thus there are 5 unpaired electrons.
• Magnetic moment: $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.9\,BM$

So, D $\to$ IV.

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3. Final matching

• A $\to$ III
• B $\to$ II
• C $\to$ I
• D $\to$ IV

This corresponds to Option C.

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4. Comparison with stored correct answer

Stored correct answer = C.

Our derived answer = C.

So they agree.