Find oxidation state and electronic configuration of Ni in each complex

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

• CO is a neutral ligand.
• So oxidation state of Ni is $0$.
• Atomic number of Ni $= 28$.
• Ground-state configuration of Ni: $\mathrm{Ni}: [\mathrm{Ar}]\,3d^8 4s^2$
• In bonding description for $\mathrm{Ni}(0)$ with strong-field ligand CO, electrons pair to give effective configuration allowing tetrahedral $sp^3$ hybridisation.
• $[\mathrm{Ni(CO)_4}]$ is known to be tetrahedral and diamagnetic.

Therefore, for $[\mathrm{Ni(CO)_4}]$, hybridisation is: $sp^3$

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

• CN$^-$ has charge $-1$.
• Let oxidation state of Ni be $x$: $x + 4(-1) = -2$ $x = +2$
• So Ni is $\mathrm{Ni}^{2+}$.
• Electronic configuration of $\mathrm{Ni}^{2+}$: $\mathrm{Ni}: [\mathrm{Ar}]\,3d^8 4s^2$ $\mathrm{Ni}^{2+}: [\mathrm{Ar}]\,3d^8$

Effect of ligand CN$^-$

• CN$^-$ is a strong-field ligand.
• It causes pairing of electrons in $3d$ orbitals.
• For $d^8$ metal ion with strong ligand in coordination number 4, the complex becomes square planar using inner-orbital hybridisation.
• Hence hybridisation is: $dsp^2$
• All electrons become paired, so the complex is diamagnetic.

Match with options

• $[\mathrm{Ni(CO)_4}] \rightarrow sp^3$
• $[\mathrm{Ni(CN)_4}]^{2-} \rightarrow dsp^2$

So the correct option is: $\boxed{\text{B: } sp^3,\ dsp^2}$

Comparison with stored answer

• Derived answer: B
• Stored correct answer: B
• They agree.