Question

Both $\left[Ni\right(CO{)}_4]$ and $\left[Ni\right(CN{)}_4{]}^{2-}$ are diamagnetic. The hybridization of nickel in these complexes, respectively, are :

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

Answer 1

Answer: (B)

$s{p}^3,ds{p}^2$

Option (B) is correct.

Hybridisation of $Ni$ atom in $Ni(CN{)}_4^{2-}$ and $Ni(CO{)}_4$ are $ds{p}^2$ and$s{p}^3$ respectively.

In $\left[Ni\right(CN{)}_4{]}^{2-}$, there is $N{i}^{2+}$ ion for which the electronic configuration in the valence shell is $3d^{8}4s^0$. In presence of strong field $C{N}^{-}$ ions, all the electrons are paired up. The empty orbitals (one $3d$, one $4s$ and two $4p$ orbitals) undergo $ds{p}^2$ hybridization to make bonds with four $C{N}^{-}$ ligands which result in square planar geometry. Since all the electrons are paired, $\left[Ni\right(CN{)}_4{]}^{2-}$ is diamagnetic.

In $\left[Ni\right(CO{)}_4]$, the valence shell electronic configuration of ground state $Ni$ atom is $3d^{8}4s^2$. All of these 10 electrons are pushed into $3d$ orbitals and get paired up when strong field $CO$ ligands approach $Ni$ atom. The empty $4s$ and three $4p$ orbitals undergo $s{p}^3$ hybridization and form bonds with four $CO$ ligands to give $Ni(CO{)}_4$ with tetrahedral geometry. Since all the electrons are paired, $Ni(CO{)}_4$ is diamagnetic.