1. Determine the oxidation state, $d$-electron count, geometry, hybridisation, and magnetic nature of each complex.

(A) $[\mathrm{MnBr}_4]^{2-}$

• Let oxidation state of Mn be $x$.

• Since $\mathrm{Br}^- = -1$: $x + 4(-1) = -2$ $x - 4 = -2 \Rightarrow x = +2$

• Mn: $Z=25$ $\mathrm{Mn}^{2+} = 3d^5$

• Coordination number $=4$, so geometry is tetrahedral with weak ligand $\mathrm{Br}^-$.

• Tetrahedral complexes use $sp^3$ hybridisation.

• For tetrahedral $d^5$, all five electrons remain unpaired.

So, $(A)$ is $sp^3$ and paramagnetic $\Rightarrow$ (IV).

(B) $[\mathrm{FeF}_6]^{3-}$

• Let oxidation state of Fe be $x$. $x + 6(-1) = -3$ $x - 6 = -3 \Rightarrow x = +3$

• Fe: $Z=26$ $\mathrm{Fe}^{3+} = 3d^5$

• $\mathrm{F}^-$ is a weak field ligand, so no pairing occurs.

• Coordination number $=6$, hence octahedral.

• Weak-field octahedral complex uses outer orbital hybridisation $sp^3d^2$.

• High-spin $d^5$ has unpaired electrons, so it is paramagnetic.

So, $(B)$ is $sp^3d^2$ and paramagnetic $\Rightarrow$ (II).

(C) $[\mathrm{Co}(\mathrm{C}_2\mathrm{O}_4)_3]^{3-}$

• Oxalate $\left(\mathrm{C}_2\mathrm{O}_4\right)^{2-}$ is $-2$.

• Let oxidation state of Co be $x$: $x + 3(-2) = -3$ $x - 6 = -3 \Rightarrow x = +3$

• Co: $Z=27$ $\mathrm{Co}^{3+} = 3d^6$

• Coordination number $=6$ (three bidentate oxalate ligands), so octahedral.

• $\mathrm{Co}^{3+}$ has high charge; with oxalate, electrons pair up to give low-spin octahedral complex.

• Thus inner orbital complex with hybridisation $d^2sp^3$.

• Low-spin $d^6$ octahedral has all electrons paired, hence diamagnetic.

So, $(C)$ is $d^2sp^3$ and diamagnetic $\Rightarrow$ (I).

(D) $[\mathrm{Ni}(\mathrm{CO})_4]$

• CO is neutral, so oxidation state of Ni is $0$.

• Ni: $Z=28$ $\mathrm{Ni}^0 = 3d^8 4s^2$

• In $[\mathrm{Ni}(\mathrm{CO})_4]$, due to strong-field ligand CO, electrons pair to give effective configuration suitable for tetrahedral $sp^3$ hybridisation.

• This complex is well known to be tetrahedral and diamagnetic.

So, $(D)$ is $sp^3$ and diamagnetic $\Rightarrow$ (III).

2. Final matching

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

This corresponds to Option C.

3. Comparison with stored correct answer

Stored correct answer = C

Our derived answer = C

Hence, the answer agrees with the stored correct answer.