Paragraph: The coordination number of Ni^ 2+ is 4 . NiCl_2+KCN...

In it unpaired orbital is not present, so it is diamagnetic in character (square planar shape) In chloro complex $\mathrm{K}_2 \left[\mathrm{Ni}(\mathrm{Cl})_4\right], complex ion is \left[\mathrm{Ni}^2\left(\mathrm{Cl}_4\right)\right]^{2-} and \mathrm{Ni}$ is present $\mathrm{Ni}^{2+}$ or $\mathrm{Ni}$($\mathrm{II}$),so $\mathrm{Ni}^2+=1$ s^2, 2 s^2 2 p^6, 3 s^2 3 p^6 3 d^8, 4 s^0 In \left.\left[\mathrm{Ni}_{\left(\mathrm{Cl}_4\right)}\right)\right]^2 ion,\mathrm{Ni}^{2+} is present as follows due to weaker ligand character is $\mathrm{Cl}^{-}$ ion. \left(\mathrm{Cl}^{-}\right.is$weak field ligand). So, it is unable to pair up the electron and$\mathrm{Ni}^{2+}$needs four empty orbitals to accommodate four$\mathrm{Cl}^{-}$ligand. Thus, \left(\mathrm{NiCl}_4\right)^{2-}$ shows $s p³$-hybridisation (Tetrahedral shapes)
Hence, due to presence of unpaired orbitals, it is paramagnetic in character.