The correct option is C Both are tetrahedral
[Ni(CO)4] and [NiCl4]2−
are both tetrahedral.
Using VBT, in the case of [Ni(CO)4],
28Ni→[Ar]3d84s2
Since, CO (carbonyl) is a strong and neutral ligand
∴ it pairs up `3d` orbitals to give [Ar]3d104so configuration. 4s and 4p orbitals participate in hybridisation shows sp3 hybridisation.
Using VBT, in the case of [NiCl4]2−,
Ni2+→[Ar]3d84s0
Cl− is a weak ligand.
∴ no pairing occurs and it will show sp3 hybridisation and thus the geometry will be tetrahedral.