Question

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

• A. $s{p}^3$ , $s{p}^3$ , $ds{p}^2$
• B. $ds{p}^2$ , $s{p}^3$ , $s{p}^3$
• C. $s{p}^3$ , $ds{p}^2$ , $ds{p}^2$
• D. $s{p}^3$ , $ds{p}^2$ , $s{p}^3$

Answer 1

Answer: (D)

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 $C{N}^{-}$ is a strong ligand and as it approaches the metal ion, the electrons must pair up.

In $[NiC{l}_4{]}^{2-},C{l}^{-}$ provides a weak field ligand. It is therefore, unable to pair up the unpaired electrons of the $3d$ orbitals.