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The geometry and magnetic behaviour of the complex $[Ni(CO)_4]$ are
Explanation
$Ni(0)$, $3d^{10}$, $sp^3$ → tetrahedral and diamagnetic.
Detailed Solution
$Ni(28) : [Ar]\,3d^8 4s^2$
$\because$ CO is a strong field ligand, the 4s electrons pair into 3d, giving $3d^{10}$.
For four CO ligands the hybridisation is $sp^3$, and thus the complex is diamagnetic with tetrahedral geometry.
$\because$ CO is a strong field ligand, the 4s electrons pair into 3d, giving $3d^{10}$.
For four CO ligands the hybridisation is $sp^3$, and thus the complex is diamagnetic with tetrahedral geometry.