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(D) The oxidation state of $Ni$ in the complex $[Ni(CO)_4]$ is calculated as follows:$x + 4(0) = 0 \implies x = 0$The ground state electronic configur

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$sp^3$,diamagnetic

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(D) The oxidation state of $Ni$ in the complex $[Ni(CO)_4]$ is calculated as follows:$x + 4(0) = 0 \implies x = 0$The ground state electronic configuration of $Ni$ $(Z=28)$ is $[Ar] 3d^8 4s^2$.In the presence of the strong field ligand $CO$,the two $4s$ electrons pair up with the $3d$ electrons,resulting in a fully filled $3d$ subshell $(3d^{10})$.The $4s$ and $4p$ orbitals hybridize to form four $sp^3$ hybrid orbitals,which accept electron pairs from four $CO$ ligands.Since all electrons are paired,the complex is diamagnetic and exhibits $sp^3$ hybridisation.

The hybridisation and magnetic behavior of the complex $[Ni(CO)_4]$ are

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