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

In the complex $[Ni(CN)_4]^{2-}$, the central metal ion is Ni in the +2 oxidation state. The ground state electronic configuration of Ni is $[Ar] 3d^8 4s^2$, so $Ni^{2+}$ has a $3d^8$ configuration. Because $CN^-$ is a strong field ligand, it causes the pairing of the two unpaired $3d$ electrons. This leaves one $3d$ orbital vacant. This empty inner $3d$ orbital, along with one $4s$ orbital and two $4p$ orbitals, undergoes $dsp^2$ hybridization to form four equivalent hybrid orbitals directed towards the corners of a square. These empty hybrid orbitals accept electron pairs from four cyanide ligands. Thus, the complex has a square planar geometry and $dsp^2$ hybridization.