Question

The hybridisation states of $[Ni(CO)_4]$ , $[Ni(CN)_4]^{2-}$ and $[NiCl_4]^{2-}$ species are respectively

• A. $sp^3$ , $sp^3$ , $dsp^2$
• B. $dsp^2$ , $sp^3$ , $sp^3$
• C. $sp^3$ , $dsp^2$ , $dsp^2$
• D. $sp^3$ , $dsp^2$ , $sp^3$

Answer 1

Answer: (D)

$sp^3$ , $dsp^2$ , $sp^3$

Answer 2

In the formation of $[Ni(CO)_4]$, oxidation state of $Ni$ is $O$ and its electronic
configuration is $[Ar] \,3d^8\,4s^2$.
Hence, we have

In $[Ni(CN)_4]^{2-}$, $Ni$ is in $+2$ oxidation state and $CN^-$ is a strong ligand and as it approaches the metal ion, the electrons must pair up.

In $[NiCl_4]^{2-}, Cl^-$ provides a weak field ligand. It is therefore, unable to pair up the unpaired electrons of the $3d$ orbitals.