Deduce the structures of [NiCl₄]^2– and [Ni(CN)₄]^2– considering the hybridization of the metal ion. Calculate the magnetic moment (spin only) of the species.  

 NiCl₄^2-, there is Ni²⁺ ion, However, in presence of weak field Cl^- ligands, NO pairing of d-electrons occurs. Therefore, Ni²⁺ undergoes sp³ hybridization to make bonds with Cl^- ligands in tetrahedral geometry. As there are unpaired electrons in the d-orbitals, NiCl₄^2- is paramagnetic. 
Since it have two unpaired electron electron therefore the magnetic moment :
$$\begin{gathered} \mathrm{Magnetic} \mathrm{moment} = \sqrt{\mathrm{n}(\mathrm{n}+2)} \\ = \sqrt{2(2+2)} \\ =\sqrt{8} \\ =2.82 \end{gathered}$$

In [Ni(CN)₄]^2-, there is Ni²⁺ ion for which the electronic configuration in the valence shell is 3d⁸ 4s⁰.

 In presence of strong field CN^- ions, all the electrons are paired up. The empty 4d, 3s and two 4p orbitals undergo dsp² hybridization to make bonds with CN^- ligands in square planar geometry. Thus [Ni(CN)₄]^2- is diamagnetic. 
Since there is no any unpaired electron therefore its magnetic moment is zero.