The hybridization involved in the complex $[Ni(CN)_4]^{2-}$ is: (Atomic number of Ni = 28)

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1.  **Oxidation State:** In the complex ion $[Ni(CN)_4]^{2-}$, the cyano group ($CN^-$) is a monodentate negative ligand. Let $x$ be the oxidation state of Nickel. $x + 4(-1) = -2 \Rightarrow x = +2$. Thus, Nickel is in the $+2$ oxidation state ($Ni^{2+}$).
2.  **Electronic Configuration:** The ground state configuration of Ni (Z=28) is $[Ar]3d^8 4s^2$. For $Ni^{2+}$, the configuration is $[Ar]3d^8$ (electrons are removed from $4s$ first).
3.  **Ligand Field Effect:** The cyanide ion ($CN^-$) is a **strong field ligand** according to the spectrochemical series. It causes the pairing of the two unpaired electrons present in the $3d$ orbitals against Hund's rule.
4.  **Hybridisation:** After pairing, one $3d$ orbital becomes empty. This empty inner $3d$ orbital mixes with the empty $4s$ and two empty $4p$ orbitals to form four equivalent **$dsp^2$** hybrid orbitals.
5.  **Geometry:** The $dsp^2$ hybridisation corresponds to a **square planar** geometry. The complex is diamagnetic (no unpaired electrons) .

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