Question

The hybridization of A and B are:

A. $Ni(CN{)}_4^{2-}$ B. $Ni(CO{)}_4$

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

Answer 1

The correct option is A $ds{p}^2,s{p}^3$

Hybridisation of $Ni(CN{)}_4^{2-}$ 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 $4d$, $3s$ and two $4p$ orbitals undergo $ds{p}^2$ hybridization to make bonds with $C{N}^{-}$ ligands in square planar geometry. Thus, $\left[Ni\right(CN{)}_4{]}^{2-}$ 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 $CO$ ligands to give $Ni(CO{)}_4$. Thus, $Ni(CO{)}_4$ is diamagnetic.