Question

What is the geometry of $\mathrm{Ni}{\left(\mathrm{CO}\right)}_4$?

Answer 1

Oxidation state of Nickel.

• The oxidation state of Nickel in the ground state is $+2$ .
• In complex, $\mathrm{Ni}{\left(\mathrm{CO}\right)}_4$ Nickel exists in $0$ oxidation state.

Electronic configuration of Nickel:

• The electronic configuration of Nickel is $\left[\mathrm{Ar}\right]3{\mathrm{d}}^{8}4{\mathrm{s}}^2$.
• The electronic configuration of Nickel in $\mathrm{Ni}{\left(\mathrm{CO}\right)}_4$ is $\left[\mathrm{Ar}\right]3{\mathrm{d}}^{10}$

Hybridization and geometry of $\mathrm{Ni}{\left(\mathrm{CO}\right)}_4$:

• The geometry of $\mathrm{Ni}{\left(\mathrm{CO}\right)}_4$ is tetrahedral that is formed by ${\mathrm{sp}}^3$ hybridization.
• In $\mathrm{Ni}{\left(\mathrm{CO}\right)}_4$ complex , $\mathrm{Ni}$ exist in $0$ oxidation state.
• Carbonyl $\left(\mathrm{CO}\right)$ is a strong field ligand, it pushes all the electrons in the $3\mathrm{d}$ orbital.
• Thus, the hybridization is ${\mathrm{sp}}^3$, Since all electrons are paired so this complex is diamagnetic in nature.
• The formation of $\mathrm{Ni}{\left(\mathrm{CO}\right)}_4$ complex is given below:

• The geometry of this complex is tetrahedral as the hybridization is ${\mathrm{sp}}^3$.