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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.

Hybridisation	Geometry	Complex structure
$(A) \ sp^3$	$(i) \ \text{Square planar}$	$(p) \ [Fe(CN)_6]^{3-}$
$(B) \ d^2sp^3$	$(ii) \ \text{Tetrahedral}$	$(q) \ [ZnCl_4]^{2-}$
$(C) \ dsp^2$	$(iii) \ \text{Octahedral}$	$(r) \ [Ni(NH_3)_4]^{2+}$
-	$(iv) \ \text{Linear}$	$(s) \ [Ag(CN)_2]^-$

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

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