Question

Why is ${\left[{\mathrm{PtCl}}_4\right]}^{2-}$ square planar?

Answer 1

Square Planar Complexes

• Square planar complexes have the highest crystal field splitting energy compared to the other complexes.
• As a result, the majority of square planar complexes have low spin, and strong field ligands..
• Examples: $\left[{\mathrm{PtCl}}_2{\left({\mathrm{NH}}_3\right)}_2\right]$ and ${\left[{\mathrm{PtCl}}_4\right]}^{2-}$.

${\left[{\mathrm{PtCl}}_4\right]}^{\mathbf{2}\mathbf{-}}$ square planar

• Platinum contains six valence electrons.
• Chlorine is the more electronegative element having a $-1$ charge and it is a weak ligand.
• In ${\left[{\mathrm{PtCl}}_4\right]}^{2-}$, the oxidation state of the central atom platinum is $+2$.
• The electronic configuration of ${\mathrm{Pt}}^{2+}$ is as follows:

${\mathrm{Pt}}^{2+}=\left[\mathrm{Xe}\right]5d^{8}6s^0$

• The orbitals are now distorted and hybridized as the extremely electronegative chlorine ion attacks the platinum ion with an additional electron.
• The four chlorine atoms undergo $ds{p}^2$ hybridization as a result of the $d$ orbital having four pairs of electrons.
• The $ds{p}^2$ hybridization complex should be tetrahedral when compared to the general structure of the molecules or ions, but there is one abnormality that is the size of the platinum is too large for it to make bonds with the ligands.
• A significant repulsion between the platinum and the electron and ligand occurs, resulting in a strong crystal field splitting.
• As a result, the splitting eliminates the degeneracy of $d_{x^2-y^2}$ and $d_{z^2}$.
• This degeneracy favors the square planar over the tetrahedral.

Structure of ${\left[{\mathrm{PtCl}}_4\right]}^{\mathbf{2}\mathbf{-}}$