Start from an octahedral complex

In an octahedral field, the five $d$-orbitals split into:

• lower energy: $t_{2g} = (d_{xy}, d_{xz}, d_{yz})$
• higher energy: $e_g = (d_{x^2-y^2}, d_{z^2})$

This happens because the ligands approach along the Cartesian axes.

Now remove both axial ligands

The axial ligands are along the $z$-axis. Removing them means the metal now experiences ligand repulsion only from the four ligands in the $xy$-plane.

So we must see how each $d$-orbital is affected:

(i) $d_{x^2-y^2}$

• Its lobes point directly along the $x$ and $y$ axes.
• These directions still contain ligands.
• Hence it experiences maximum repulsion and becomes the highest energy orbital.

(ii) $d_{xy}$

• Its lobes lie in the $xy$-plane, between the axes.
• It still interacts with the planar ligands, but less strongly than $d_{x^2-y^2}$.
• So its energy is raised, but not as much.

(iii) $d_{z^2}$

• Two lobes lie along the $z$-axis, where ligands have been removed.
• Therefore the strong axial repulsion disappears.
• Its energy decreases significantly.
• The torus in the $xy$-plane gives some interaction, so it is not the lowest.

(iv) $d_{xz}$ and $d_{yz}$

• These orbitals have lobes in planes involving the $z$-axis.
• Since there are no ligands along $z$, they experience the least repulsion.
• Thus they become the lowest energy orbitals and remain degenerate.

Resulting ordering

Therefore, after removing the two axial ligands, the splitting pattern becomes:

$d_{x^2-y^2} > d_{xy} > d_{z^2} > d_{xz} = d_{yz}$

Interpretation

This is the splitting pattern for a square planar / tetragonally distorted limit obtained by axial ligand removal.

Matching with the options

The correct diagram must show:

• highest: $d_{x^2-y^2}$
• then: $d_{xy}$
• then: $d_{z^2}$
• lowest degenerate pair: $d_{xz}, d_{yz}$

This corresponds to Option A.

Comparison with stored answer

Stored correct answer = A.

My derived answer = A.

Hence, they agree.