Question

Explain hybridisation on Ni in [Ni(CN)\(_4\)]\(^{2-}\).

Hint:

Strong field ligands like CN\(^-\) cause pairing → dsp\(^2\) (square planar). Weak field ligands → sp\(^3\) (tetrahedral).

Answer 1

Step 1: Oxidation state of Ni.
In [Ni(CN)\(_4\)]\(^{2-}\), Ni is in +2 oxidation state. Electronic configuration of Ni atom = [Ar] 3d\(^8\)4s\(^2\). For Ni\(^{2+}\) = [Ar] 3d\(^8\).
Step 2: Effect of strong ligand (CN\(^-\)).
CN\(^-\) is a strong field ligand (according to spectrochemical series). It causes pairing of 3d electrons. Thus configuration becomes: 3d\(^{10}\).
Step 3: Hybridisation.
Now Ni\(^{2+}\) uses one 3d, one 4s, and two 4p orbitals → dsp\(^2\) hybridisation. This gives a square planar geometry.
Step 4: Conclusion.
Therefore, Ni in [Ni(CN)\(_4\)]\(^{2-}\) undergoes dsp\(^2\) hybridisation with square planar shape.