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(C) The oxidation state of $Ni$ in $Ni(CO)_4$ is $0$.Thus,the electronic configuration of $Ni(0)$ is $[Ar] 3d^8 4s^2$.In the presence of the strong fi

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tetrahedral and diamagnetic

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(C) The oxidation state of $Ni$ in $Ni(CO)_4$ is $0$.Thus,the electronic configuration of $Ni(0)$ is $[Ar] 3d^8 4s^2$.In the presence of the strong field ligand $CO$,the $4s$ electrons pair up with the $3d$ electrons.This results in a fully filled $3d$ subshell and the use of one $4s$ and three $4p$ orbitals for $sp^3$ hybridisation.$Ni(CO)_4$ has tetrahedral geometry ($sp^3$ hybridisation) and is diamagnetic due to the absence of unpaired electrons.

$Ni(CO)_4$ is

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