Question

What is the hybridization of ${\left[{\mathrm{PtCl}}_4\right]}^{2-}$ and how?

Answer 1

Hybridization of ${\left[{\mathrm{PtCl}}_4\right]}^{2-}$:

• Platinum has an atomic number $78$.
• The electronic configuration of platinum is $\left[\mathrm{Xe}\right]4{\mathrm{f}}^{14}5{\mathrm{d}}^{9}6{\mathrm{s}}^1$.
• In ${\left[{\mathrm{PtCl}}_4\right]}^{2-}$ complex, platinum exists in $+2$ oxidation state.
• The electronic configuration of ${\mathrm{Pt}}^{2+}$ is $\left[\mathrm{Xe}\right]4{\mathrm{f}}^{14}5{\mathrm{d}}^8$.
• There is strong repulsion that takes between the electron of platinum and ligands which leads to a strong crystal field splitting.
• Thus, the splitting breaks the degeneracy of ${\mathrm{d}}_{{\mathrm{x}}^2-{\mathrm{y}}^2}$ and ${\mathrm{d}}_{{\mathrm{z}}^2}$ and stabilizes more to the square planar arrangement than the tetrahedral.
• Thus, the hybridization of ${\left[{\mathrm{PtCl}}_4\right]}^{2-}$complex is ${\mathrm{dsp}}^2$ and has square planar geometry.