1. Determine geometry and hybridization of each complex

We use:

• Coordination number $4$:
  • tetrahedral $\to sp^3$
  • square planar $\to dsp^2$
• Coordination number $6$ (octahedral):
  • inner orbital complex $\to d^2sp^3$
  • outer orbital complex $\to sp^3d^2$

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2. (A) $Ni(CO)_4$

• Oxidation state of Ni = $0$
• Electronic configuration of $Ni$: $[Ar]3d^8 4s^2$
• In $Ni(0)$ with strong-field ligand CO, electrons pair to give effectively filled $3d$ set.
• $Ni(CO)_4$ is a tetrahedral complex.

Hence hybridization is: $sp^3$ So, $A \to I$

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

• Let oxidation state of Ni be $x$: $x + 4(-1) = -2 \Rightarrow x=+2$
• So $Ni^{2+}$ is $3d^8$
• $CN^-$ is a strong-field ligand, causing pairing.
• For $d^8$ nickel with strong ligand in coordination number $4$, the complex becomes square planar.

Hence hybridization is: $dsp^2$ So, $B \to IV$

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4. (C) $[Co(CN)_6]^{3-}$

• Oxidation state of Co: $x + 6(-1) = -3 \Rightarrow x=+3$
• So $Co^{3+}$ is $3d^6$
• $CN^-$ is a strong-field ligand, so pairing occurs.
• This gives a low-spin octahedral complex using inner $3d$ orbitals.

Hence hybridization is: $d^2sp^3$ So, $C \to III$

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5. (D) $[CoF_6]^{3-}$

• Oxidation state of Co: $x + 6(-1) = -3 \Rightarrow x=+3$
• So $Co^{3+}$ is $3d^6$
• $F^-$ is a weak-field ligand, so no pairing occurs.
• Therefore it forms a high-spin octahedral complex using outer orbitals.

Hence hybridization is: $sp^3d^2$ So, $D \to II$

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6. Final matching

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

This corresponds to Option B.

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7. Comparison with stored answer

Stored correct answer = B

Our derived answer = B

So they agree.