Question

What is the geometry of ${\left[\mathrm{Ni}\left({\mathrm{CN}}_4\right)\right]}^{2-}$?

Answer 1

Calculation of oxidation state of $\mathrm{Ni}$

Consider the oxidation state of $\mathrm{Ni}$ is x.

$$\begin{gathered} x+\left(-1\times 4\right)=-2 \\ x=-2+4=+2 \end{gathered}$$

Electronic configuration of ${\mathrm{Ni}}^{2+}$

• The electronic configuration of $\mathrm{Ni}$ is $\left[\mathrm{Ar}\right]3{\mathrm{d}}^{8}4{\mathrm{s}}^2$.
• So, the electronic configuration of ${\mathrm{Ni}}^{2+}$ is $\left[\mathrm{Ar}\right]3{\mathrm{d}}^8$.

Effect of cyanide ligand

• $\mathrm{CN}$ is a strong field ligand. The electron will be paired.

Hybridization and geometry of ${\left[\mathrm{Ni}\left({\mathrm{CN}}_4\right)\right]}^{2-}$:

• ${\left[\mathrm{Ni}\left({\mathrm{CN}}_4\right)\right]}^{2-}$ has a square planar geometry that is formed by ${\mathrm{dsp}}^2$hybridization.
• This complex is diamagnetic as it has no unpaired electrons.
• The two unpaired $\mathrm{d}$ electrons participate in hybridization with the ${\mathrm{CN}}^{-}$ ligands forming ${\mathrm{dsp}}^2$ hybridization.

Formation of ${\left[\mathrm{Ni}\left({\mathrm{CN}}_4\right)\right]}^{2-}$ complex: