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Q.
$Ni(CO)_4$ is
KVPYKVPY 2014
Solution:
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 ($sp^3$ hybridisation) and is diamagnetic due to absence of unpaired electrons.
