Let's determine the hybridisation of each complex one by one.

1. $\mathrm{K}_2[\mathrm{Ni}(\mathrm{CN})_4]$

The complex ion is $[\mathrm{Ni}(\mathrm{CN})_4]^{2-}$.

• Oxidation state of Ni: $x+4(-1)=-2 \Rightarrow x=+2$
• So, $\mathrm{Ni}^{2+}$ has configuration: $[\mathrm{Ar}]\,3d^8$
• $\mathrm{CN}^-$ is a strong field ligand, so pairing occurs.
• Coordination number $=4$.
• For $d^8$ metal ion with strong field ligands, geometry is square planar with hybridisation: $dsp^2$

So, $A \to \text{III}$

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

• Here Ni is in oxidation state $0$.
• Configuration of $\mathrm{Ni}$: $[\mathrm{Ar}]\,3d^8 4s^2$
• In $\mathrm{Ni}(0)$ carbonyl complex, electrons rearrange to give filled $3d$ orbitals and bonding through outer orbitals.
• Coordination number $=4$.
• $[\mathrm{Ni}(\mathrm{CO})_4]$ is tetrahedral.
• Tetrahedral hybridisation: $sp^3$

So, $B \to \text{I}$

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3. $[\mathrm{Co}(\mathrm{NH}_3)_6]\mathrm{Cl}_3$

The complex cation is $[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$.

• Oxidation state of Co: $x+6(0)=+3 \Rightarrow x=+3$
• So, $\mathrm{Co}^{3+}$ has configuration: $[\mathrm{Ar}]\,3d^6$
• $\mathrm{NH}_3$ is a ligand that causes pairing for $\mathrm{Co}^{3+}$.
• Coordination number $=6$, so octahedral complex.
• Since pairing occurs in $3d$, it is an inner orbital complex with hybridisation: $d^2sp^3$

So, $C \to \text{IV}$

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4. $\mathrm{Na}_3[\mathrm{CoF}_6]$

The complex ion is $[\mathrm{CoF}_6]^{3-}$.

• Oxidation state of Co: $x+6(-1)=-3 \Rightarrow x=+3$
• So, $\mathrm{Co}^{3+}$ is: $[\mathrm{Ar}]\,3d^6$
• $\mathrm{F}^-$ is a weak field ligand, so no pairing occurs.
• Coordination number $=6$, octahedral.
• Hence it forms an outer orbital complex with hybridisation: $sp^3d^2$

So, $D \to \text{II}$

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Final matching

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

This corresponds to Option D.

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Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So, the answer agrees with the stored answer.