Complete removal of both the axial ligands (along the z-axis) from an octahedral complex leads to which of the following splitting patterns? (relative orbital energies not on scale).

Square Planar Splitting

In a standard octahedral complex, the five $$d$$-orbitals split into two groups: the higher-energy $$e_g$$ pair ($$d_{x^2-y^2}$$, $$d_{z^2}$$) and the lower-energy $$t_{2g}$$ triplet ($$d_{xy}$$, $$d_{xz}$$, $$d_{yz}$$).

When the two ligands along the z-axis are completely pulled away:

• The electrostatic repulsion experienced by all $$d$$-orbitals containing a "z" component ($$d_{z^2}$$, $$d_{xz}$$, and $$d_{yz}$$) drops significantly, causing their energy levels to plummet.
• Conversely, the four remaining ligands in the $$xy$$-plane exert relatively stronger repulsion on the orbitals lying within that plane ($$d_{x^2-y^2}$$ and $$d_{xy}$$), causing their relative energies to rise.

Final Energy Level Ordering:

The resulting square planar crystal field splitting pattern from highest energy to lowest energy is:

$$\mathbf{d_{x^2-y^2} > d_{xy} > d_{z^2} > (d_{xz} = d_{yz})}$$

• $$d_{x^2-y^2}$$: Highest energy because its lobes point directly at the remaining four equatorial ligands.
• $$d_{xy}$$: Lies in the equatorial plane between the ligands, making it the second highest.
• $$d_{z^2}$$: Drops drastically below $$d_{xy}$$ due to the total absence of axial ligands.
• $$d_{xz}, d_{yz}$$: Remain a degenerate pair at the lowest energy level since they interact least with the equatorial plane.

Conclusion:

The correct diagram must show a 4-level splitting pattern ordered as $$d_{x^2-y^2}$$ at the top, followed by $$d_{xy}$$, then $$d_{z^2}$$, and the degenerate pair $$d_{xz}, d_{yz}$$ at the bottom.

Answer: Option D