The geometry and magnetic behaviour of the complex $[\mathrm{Ni}(\mathrm{CO}){}_4]$ are:

Solution & Explanation

\textbf{Answer:} (B) $[\mathrm{Ni}(\mathrm{CO}){}_4]$ has tetrahedral geometry and is diamagnetic. \textbf{Solution:} CO is a neutral ligand, so the oxidation state of Ni is $0$: Ni(0). Ni is $[\mathrm{Ar}]3\mathrm{d}{}^{8}4\mathrm{s}{}^2$; in the zero oxidation state the configuration rearranges to $3d^{10}4s^0$ (the $4s$ electrons move into $3d$ under the strong-field CO). With a filled $3d^{10}$ set, all $d$ electrons are paired, so the complex is \textbf{diamagnetic}. The empty $4s$ and three $4p$ orbitals hybridise as $s{p}^3$, giving \textbf{tetrahedral} geometry. Hence option (B).