Question

What is the hybridisation of the metal in following complex?

$Ni(CO{)}_4$

• A. $ds{p}^2$
• B. $sp{d}^2$
• C. $s{p}^3$
• D. None of these

Answer 1

Answer: (C)

$s{p}^3$

Option (C) is correct.

$Ni(CO{)}_4$ shows $s{p}^3$ hybridisation and has tetrahedral geometry.

The $Ni$ atom in nickle carbonyl has zero oxidation state as the carbonyl ligands are neutral. The electronic configuration of $Ni$ in the gas phase is $\left[Ar\right]3d^{8}4s^2$. When it is combined with CO to form $Ni(CO{)}_4$ its electronic configuration is considered to be $\left[Ar\right]3d^{10}$, that is, the $3d$ shell is completely filled. Empty $4s$ and three $4p$ orbitals undergo $s{p}^3$ hybridization.
Each $CO$ donates two electrons to four $s{p}^3$ hybrid orbitals.