1. Find oxidation state and d-configuration of the central metal ion

• (A) $[\mathrm{CoF_6}]^{3-}$

  Let oxidation state of Co be $x$: $x+6(-1)=-3 \Rightarrow x=+3$ So, $\mathrm{Co^{3+}} : 3d^6$

• (B) $[\mathrm{NiCl_4}]^{2-}$

  Let oxidation state of Ni be $x$: $x+4(-1)=-2 \Rightarrow x=+2$ So, $\mathrm{Ni^{2+}} : 3d^8$

• (C) $[\mathrm{Co(NH_3)_6}]^{3+}$

  Since $\mathrm{NH_3}$ is neutral: $x=+3$ So, $\mathrm{Co^{3+}} : 3d^6$

• (D) $[\mathrm{Ni(CN)_4}]^{2-}$

  Let oxidation state of Ni be $x$: $x+4(-1)=-2 \Rightarrow x=+2$ So, $\mathrm{Ni^{2+}} : 3d^8$

2. Decide strong field / weak field and geometry

(A) $[\mathrm{CoF_6}]^{3-}$

• $\mathrm{F^-}$ is a weak field ligand.
• For octahedral $\mathrm{Co^{3+}}$ ($d^6$), weak field gives high-spin octahedral complex.
• High-spin octahedral complexes use outer orbital hybridisation: $sp^3d^2$
• So, $\boxed{(A) \to (III)}$

(B) $[\mathrm{NiCl_4}]^{2-}$

• $\mathrm{Cl^-}$ is a weak field ligand.
• $\mathrm{Ni^{2+}}$ is $d^8$.
• With weak field tetrahedral preference, $[\mathrm{NiCl_4}]^{2-}$ is tetrahedral.
• Tetrahedral hybridisation is: $sp^3$
• So, $\boxed{(B) \to (II)}$

(C) $[\mathrm{Co(NH_3)_6}]^{3+}$

• $\mathrm{NH_3}$ is a stronger ligand than $\mathrm{F^-}$.
• For $\mathrm{Co^{3+}}$ ($d^6$), pairing occurs, giving low-spin octahedral complex.
• Low-spin octahedral inner orbital complex uses: $d^2sp^3$
• So, $\boxed{(C) \to (I)}$

(D) $[\mathrm{Ni(CN)_4}]^{2-}$

• $\mathrm{CN^-}$ is a strong field ligand.
• $\mathrm{Ni^{2+}}$ is $d^8$.
• Strong field causes pairing and forms square planar complex.
• Square planar hybridisation is: $dsp^2$
• So, $\boxed{(D) \to (IV)}$

3. Final matching

Thus,

This corresponds to Option D.

4. Comparison with stored answer

Stored correct answer = D

Our derived answer = D

So, the answer agrees with the stored answer.