1. Identify the hybridisation of each central atom/ion

We match each species in LIST-I with the correct hybridisation in LIST-II.

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2. For $\mathrm{PF}_5$

• Central atom: $\mathrm{P}$
• Number of bond pairs around P = 5
• Geometry: trigonal bipyramidal
• Hybridisation: $\mathrm{sp}^3d$

So, $A \to II$

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3. For $\mathrm{SF}_6$

• Central atom: $\mathrm{S}$
• Number of bond pairs around S = 6
• Geometry: octahedral
• Hybridisation: $\mathrm{sp}^3d^2$

So, $B \to III$

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4. For $\mathrm{Ni(CO)}_4$

• Nickel is in oxidation state $0$
• Electronic configuration of $\mathrm{Ni}$: $[Ar]3d^84s^2$
• In $\mathrm{Ni(CO)}_4$, due to strong field ligand CO, electrons pair up to give effective arrangement suitable for tetrahedral complex.
• Known hybridisation of $\mathrm{Ni(CO)}_4$ is $\mathrm{sp}^3$

So, $C \to IV$

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5. For $[\mathrm{PtCl}_4]^{2-}$

• Oxidation state of Pt: $x+4(-1)=-2 \Rightarrow x=+2$
• So Pt is $\mathrm{Pt}^{2+}$
• $\mathrm{Pt}^{2+}$ is a $d^8$ system
• For 5d metal $\mathrm{Pt}^{2+}$, $d^8$ complexes are typically square planar
• Square planar hybridisation is $\mathrm{dsp}^2$

So, $D \to I$

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6. Final matching

Thus we get: $A-II,\; B-III,\; C-IV,\; D-I$

This corresponds to Option C.

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7. Comparison with stored answer

Stored correct answer = C

Our derived answer = C

So they agree.