$\mathrm{NiCl}_{2}\left\{\mathrm{P}\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2}\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)\right\}_{2}$ exhibits temperature depend-ent magnetic behaviour (paramagnetic/diamagnetic). The coordination geometries of $\mathrm{Ni}^{2+}$ in the paramagnetic and diamagnetic states are respectively
tetrahedral and tetrahedral
square planar and square planar
tetrahedral and square planar
square planar and tetrahedral
Solution
In both states (paramagnetic and diamagnetic) of the given complex, $\mathrm{Ni}$ exists as $\mathrm{Ni}^{2+}$ whose electronic configuration is $[\mathrm{Ar}] 3 d^{8} 4 s^{0}$.
In the above paramagnetic state the geometry of the complex is $s p^{3}$ giving tetrahedral geometry.
The diamagnetic state is achieved by pairing of electrons in $3 d$ orbital.
Thus, the geometry of the complex will be $d s p^{2}$ giving square planar geometry.
