What is the hybridisation of metal ion in \( [Ni(CO)_4] \) compound?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(sp^3\)
According to the valence bond theory, the hybridisation of the metal in a complex depends on its oxidation state, its d-electron count and the strength of the ligands. In \( \ce{[Ni(CO)4]} \) carbon monoxide is a neutral ligand, so nickel is in the zero oxidation state.
Nickel (Z = 28) is \( \mathrm{[Ar]\,3d^{8}\,4s^{2}} \). CO is a strong field ligand; in its presence the two 4s electrons are pushed into the 3d orbitals and pair up, giving \( \mathrm{3d^{10}} \). The empty 4s orbital and the three 4p orbitals then mix to form four \( sp^3 \) hybrid orbitals, which accept the four lone pairs from the CO molecules.
So nickel in \( \ce{[Ni(CO)4]} \) is \( sp^3 \) hybridised; the complex is tetrahedral and diamagnetic (no unpaired electron).
Why the other options are incorrect
Option (2) \(dsp^2\) — Incorrect. This uses one inner 3d orbital and gives a square planar complex, as in \( \ce{[Ni(CN)4]^2-} \), where \( \ce{Ni^2+} \) (3d8) pairs its electrons to free one 3d orbital. In \( \ce{[Ni(CO)4]} \) all five 3d orbitals are full, so none is available.
Option (3) \(spd^2\) — Incorrect. This is not a standard hybridisation; a four-coordinate complex is either tetrahedral (sp3) or square planar (dsp2).
Option (4) \(sp^2d\) — Incorrect. This too is not a standard hybridisation for a four-coordinate complex.
NCERT (2026-27 edition):
Werner’s Theory of Coordination Compounds – “… He further postulated that octahedral, tetrahedral and square planar geometrical shapes are more common in coordination compounds of transition metals. Thus, [Co(NH3)6]3+, [CoCl(NH3)5]2+ and [CoCl2(NH3)4]+ are octahedral entities, while [Ni(CO)4] and [PtCl4]2– are tetrahedral and square planar, respectively.”