Question

$Ni(CO{)}_4$ is

• A. tetrahedral and paramagnetic
• B. square planar and diamagnetic
• C. tetrahedral and diamagnetic
• D. square planar and paramagnetic

Answer 1

Answer: (C)

The oxidation state of $Ni$ in $Ni(CO{)}_4$ is $0$.
Thus, the electronic
configuration of $Ni(O)$ is $[Ar]3d^{8}4s^0$

As $CO$ is a strong ligand pairing of electrons occur.

$Ni(CO{)}_4$ has tetrahedral geometry ($s{p}^3$ hybridisation) and is diamagnetic due to absence of unpaired electrons.