$\mathrm{NiCl}_{2}\left\{\mathrm{P}\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2}\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)\right\}_{2}$ exhibits temperature depend-ent magnetic behaviour (paramagnetic/diamagnetic). The coordination geometries of $\mathrm{Ni}^{2+}$ in the paramagnetic and diamagnetic states are respectively

• A

  tetrahedral and tetrahedral
• B

  square planar and square planar
• Correct answer
  C

  tetrahedral and square planar
• D

  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.

Hence, the correct answer is option: 'C'.