Coordination Geometries of Temperature Dependent Nickel Complex

To determine the coordination geometries around $\text{Ni}^{2+}$ in the given complex $\left[\text{NiCl}_2\{\text{P}(\text{C}_2\text{H}_5)_2(\text{C}_6\text{H}_5)\}_2\right]$, let us analyze its electronic configuration and crystal field splitting behavior:

1. Oxidation State and Electronic Configuration:

   • The central metal ion is nickel in the $+2$ oxidation state: $\text{Ni}^{2+}$.
   • The electronic configuration of $\text{Ni}^{2+}$ is $[\text{Ar}]\,3d^8$.
   • The coordination number of $\text{Ni}^{2+}$ is $4$ (surrounded by two chloride ions and two phosphine ligands).

2. Four-coordinate $d^8$ Geometries: Four-coordinate complexes of $d^8$ transition metal ions can exist in two distinct geometries:

   • Tetrahedral Geometry ($sp^3$ hybridization): In a tetrahedral crystal field, the $d$-orbitals split into a lower energy $e$ set ($d_{z^2}, d_{x^2-y^2}$) and a higher energy $t_2$ set ($d_{xy}, d_{yz}, d_{xz}$). The electronic configuration is: $e^4 \, t_2^4$ This results in $2$ unpaired electrons, making the complex paramagnetic.

   • Square Planar Geometry ($dsp^2$ hybridization): In a square planar crystal field, the strong ligand field splits the $d$-orbitals such that the $d_{x^2-y^2}$ orbital is significantly higher in energy compared to the others ($d_{xz}, d_{yz}, d_{z^2}, d_{xy}$). All 8 electrons are paired in the four lower-energy orbitals: $(d_{xz}, d_{yz})^4 \, (d_{z^2})^2 \, (d_{xy})^2 \, (d_{x^2-y^2})^0$ This results in $0$ unpaired electrons, making the complex diamagnetic.

3. Conclusion: The complex exists in a temperature-dependent equilibrium between the paramagnetic tetrahedral form and the diamagnetic square planar form. Thus, the geometries in the paramagnetic and diamagnetic states are tetrahedral and square planar, respectively.

Therefore, the correct option is C.