What is the hybridisation of metal ion in $[Ni(CO)_4]$ compound?

$sp^3$

The correct answer is Option (1) → $sp^3$

To find hybridisation in coordination compounds, check:

• Oxidation state of metal
• d-electron configuration
• Geometry of complex

CO is a strong field ligand and forms low-spin complexes.

Explanation

In $[\text{Ni}(\text{CO})_4]$:

• $\text{CO}$ is a neutral ligand $\rightarrow$ Oxidation state of $\text{Ni} = 0$

• Atomic configuration of Ni ($Z = 28$):

$Ni = [Ar] 3d^8 4s^2$

In $Ni^0$, electrons rearrange due to strong field $CO$ ligands:

$3d^8 4s^2 \rightarrow 3d^{10}$ (paired), $4s$ and $4p$ empty

Since four ligands are present, the geometry is tetrahedral, which requires $sp^3$ hybridisation (4 equivalent orbitals).

No d-orbitals are used in hybridisation here, so $dsp^2$ or $spd^2$ are not possible.