For a trigonal planar complex (e.g. $\ce{[AgCl_3]^2-}$):
According to this image (source):
The $x^2 - y^2$ and $xy$ orbitals are on the same energy level. However, assuming that one of the $\ce{Cl}$ ligands lies on the $y$ axis (or the $x$ axis), wouldn't the $x^2 - y^2$ orbital be higher in energy? Thanks in advance.
Say you put a ligand in the trigonal complex (or an equatorial ligand in a trigonal bipyramidal complex) on the $x$-axis, where $z$ is the "axial" axis. Then that ligand is directly aligned with an $x^2-y^2$ lobe and falls on a "seam" of the $xy$ orbital. But the other two ligands in the plane are closer to an $xy$ lobe ($15°$ off) than to an $x^2-y^2$ lobe ($30°$ off). These effects balance so that both in-plane $d$ orbital see the same total metal-ligand intetaction.
The most basic crystal field argument includes point-symmetric charges approaching the central metal in a way as the ligands would. Then, any orbitals that are symmetry-equivalent will end up at the same energy, and depending on how much these point towards the point-symmetric approaching charges they will be raised or lowered. Therefore, we should be looking at:
which orbitals are symmetry-equivalent, i.e. transform the same way in the point group (here: $D_\mathrm{3h}$. Warning: Link leads to a MathJax-heavy page!)
where the ligands are approaching from
Thankfully, the character tables already include the symmetry of the orbitals: look at the final columns of the table where the orbital designators are written (s orbitals are always in the totally symmetric irreducible representation, i.e. $\mathrm a_1'$ here). We see that $z^2$ is in the row $\mathrm a_1'$, $x^2-y^2$ and $xy$ are both in the row $\mathrm e'$ and $xz$ and $yz$ are both found in $\mathrm e''$. Thus, the orbitals must end up in three different groups of distinct energy:
The first two should be immediately obvious, the third maybe not. However, it is easily explained if you remember how $\mathrm{d}_{xy}$ and $\mathrm{d}_{x^2-y^2}$ look with respect to the principal axis (the $z$ axis): they are both perpendicular to $z$ and differ only by their alignment, $\mathrm{d}_{x^2-y^2}$ having lobes pointing along the coordinate axes while $\mathrm{d}_{xy}$’s lobes point in-between the coordinate axes. A simple $C_8$ rotation ($45^\circ$) around the principal axis will transform one onto the other.
More importantly, however, one of the main elements of symmetry is $C_3$ — the rotation around the $z$ axis by $120^\circ$. Applying this will transform either of these two orbitals into something that is in-between a pure $\mathrm{d}_{xy}$ and a pure $\mathrm{d}_{x^2-y^2}$ orbital: it can be understood as a linear combination of the two. However, any operation of symmetry must transform everything onto something symmetry-equivalent and thus the two must be symmetry-equivalent and therefore degenerate.
It remains to be shown why your diagram shows the orbitals at the energy levels it does. Simply speaking, there are no ligands in $z$ direction so anything containing $z$ contribution should be stabilised. This obviously applies to $\mathrm{d}_{z^2}$ and also to $\mathrm{d}_{xz}$ and $\mathrm{d}_{yz}$. Why is the former slightly higher in energy than the latter? The crystal field model’s answer is that $\mathrm{d}_{z^2}$ has a central ‘hoop’ that is still pointing towards the ligands in the $xy$ plane while the other two have a nodal plane in the $xy$ plane. Thus, the latter two should have a lower energy.
The highest energy obviously needs to be given to all orbitals in the $xy$ plane as that is the plane in which the ligands are. This applies to both $\mathrm{d}_{xy}$ and $\mathrm{d}_{x^2-y^2}$ as I have shown above that they are symmetry-equivalent. Therefore, these two are found at the highest energy.
