Question

The correct order of energies of d-orbitals of metal ions in a square planar complex is:
A. \[{d_{xy}} = {d_{yz}} = {d_{zx}} > {d_{{x^2} - {y^2}}} = {d_{{x^2}}}\]
B. \[{d_{{x^2} - {y^2}}} = {d_{{z^2}}} > {d_{xy}} = {d_{yz}} = {d_{zx}}\]
C. \[{d_{{x^2} - {y^2}}} > {d_{{z^2}}} > {d_{xy}} = {d_{yz}} = {d_{zx}}\]
D. \[{d_{{x^2} - {y^2}}} > {d_{xy}} > {d_{{z^2}}} > {d_{zx}} = {d_{yz}}\]

Answer 1

Hint: The bonding interaction between transition metals and the ligands attached to them with a coordinate bond is given by crystal field theory. Crystal field theory explains the breaking of degeneracies of electron orbital states. These orbitals are usually d or f orbitals.

Complete step-by-step answer:
In a square planar, there are four ligands attached to the transition metal. But the difference is that the electrons of the ligands are only and only attracted to xy plane. Due to this reason, any orbital in the xy plane has a higher energy level. Square planar complex has four different energy levels. The highest energy is of \[{d_{{x^2} - {y^2}}}\] followed by \[{d_{xy}}\] and then \[{d_{{z^2}}}\]. The energies of \[{d_{zx}}\] and \[{d_{yz}}\] are equal to each other and low as compared to other d orbitals.
The splitting energy from highest orbital to lowest orbital is greater for square planar as compared to octahedral. For a square planar complex, splitting energy is larger than pairing energy. So, square planar complexes are usually low spin complexes.

Therefore, the correct answer is D, that is, \[{d_{{x^2} - {y^2}}} > {d_{xy}} > {d_{{z^2}}} > {d_{zx}} = {d_{yz}}\].

Note: Crystal field stabilization energy is the stability which arises from placing a transition metal ion in a crystal field generated by ligands. Crystal field theory helps us to understand some colours and some magnetic properties. As soon as a ligand approaches metal ions, electrons from the ligand will be nearer to some of the d orbitals and farther away from others, this leads to loss of degeneracy.