Question

Both $\left[Ni\right(CO{)}_4]$ and $\left[Ni\right(CN{)}_4{]}^{2-}$ are diamagnetic. The hybridisation 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

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

In $\left[Ni\right(CO{)}_4]$, $Ni$ is in 0 oxidation state. So electronic configuration of $Ni=\left[Ar\right]4s^{2}3d^8$. Since $CO$ is a strong field ligand, it forces pairing of electrons.

In $\left[Ni\right(CN{)}_4{]}^{2-}$ $Ni$ is in +2 oxidation state. So the electronic configuration of Ni is $\left[Ar\right]4s^{0}3d^8$. Since $C{N}^{-}$ is also a strong ligand it will pair up the electrons and will get hybridised in the following manner