What is the hybridization of the compound \([Ni(CO)_4]\)?

sp³

The correct answer is option 2. \(sp^3\).

The valence shell electronic configuration of ground state \(Ni\) atom is \(3d^8 4s^2\). All of these 10 electrons are pushed into \(3d\) orbitals and get paired up when strong field \(CO\) ligands approach \(Ni\) atom. The empty \(4s\) and three \(4p\) orbitals undergo \(sp^3\) hybridization and form bonds with \(CO\) ligands to give \(Ni(CO)_4\). Thus \(Ni(CO)_4\) is diamagnetic.