Why Pt(II), Pd(II) and Au(III) always form square planar complex, irrespective of the strenght of ligand?
Dear student
Please find the solution to the asked query:
Square planar complexes are a higher energy structure.
The metals like Pt(II), Pd(II) have bigger d orbitals compared to other metals , due to which they can hold more electron density and this electron density can spread out in a larger d orbital, thus forming a square planar complex.
Also
Tetragonal distortion forms a square planar complex in which the ligands along the z-axis are completely removed.
This distortion to square planar complexes is prevalent for d8 configurations and elements in the 4th and 5th periods such as: Rh (I), Ir (I), Pt(II), Pd(III), and Au (III)
{you will learn more about tetragonal distortion in higher classes]
Hope this information will clear your doubts regarding the topic. If you have any other doubts please ask here on the forum and our experts will try to solve them as soon as possible.
Regards
Please find the solution to the asked query:
Square planar complexes are a higher energy structure.
The metals like Pt(II), Pd(II) have bigger d orbitals compared to other metals , due to which they can hold more electron density and this electron density can spread out in a larger d orbital, thus forming a square planar complex.
Also
Tetragonal distortion forms a square planar complex in which the ligands along the z-axis are completely removed.
This distortion to square planar complexes is prevalent for d8 configurations and elements in the 4th and 5th periods such as: Rh (I), Ir (I), Pt(II), Pd(III), and Au (III)
{you will learn more about tetragonal distortion in higher classes]
Hope this information will clear your doubts regarding the topic. If you have any other doubts please ask here on the forum and our experts will try to solve them as soon as possible.
Regards