I am confused about the energies of the orbitals after Jahn-Teller distortion of an octahedral complex. Take elongation for example.

To my understanding, since the ligands on the $z$ axis move away, there is less orbital overlap with the $d_{z^2}$ metal orbital, leading to weaker orbital interactions, leading to the antibonding orbital being lowered in energy.

There are two things I don't get: 1. The lowering of $d_{xz}$ and $d_{yz}$ and 2. the raising in energy of the $d_{x^2-y^2}$ and $d_{xy}$ orbitals.

1. The $d_{xz}$ and $d_{yz}$ don't seem to have the correct symmetry to sigma bond with any $p$ orbitals of the ligands. They are nonbonding in a perfect octahedron, so how can the be lowered in energy after distortion? Maybe they only appear lower relative to the $d_{xy}$?

2. Are they not actually raised in energy, only appearing higher relative to the $d_{z^2}$ and $d_{xz}$ and $d_{yz}$ orbitals? Or are the equatorial bonds shortened so there is a stronger interaction, raising the $d_{x^2-y^2}$? If so, then why is the $d_{xy}$ orbital raised?

Is there an explanation for the orbital energies using MO theory? If you could clear up my confusion that would be great. Thanks!

The d_xz and d_yz orbitals don't seem to be able to interact with a ligand on the z axis Elongation on the z axis

  • $\begingroup$ The diagram assumes a bonding pi interaction with xz and yz in addition to the sigma interaction with the z2. That's why xz and yz go down relative to xy. The "raising" of xy and z2 is an artifact of centering both diagrams on the average energy. The actual molecular anti-bonding orbitals that are mostly comprised of z2 and xy don't change energy unless the equatorial ligands get closer to the central atom (which can happen). $\endgroup$
    – Andrew
    Jan 24, 2021 at 14:20
  • 1
    $\begingroup$ Does this answer your question? I would like to point to my answer there which goes into some detail concerning the MO descriptions. If so, I can drope a duplicate close vote ;) $\endgroup$
    – Jan
    Jan 26, 2021 at 14:29
  • $\begingroup$ @Jan It doesn't fully answer my question. $\endgroup$
    – bud
    Jan 31, 2021 at 0:24

2 Answers 2


I disagree with your statement that

"The dxz and dyz don't seem to have the correct symmetry to sigma bond with any p orbitals of the ligands. They are nonbonding in a perfect octahedron"

Consider the case of a halide ligand like Cl-, which has filled p orbitals perpendicular to the axis of the M-Cl sigma bond. If these are aligned with the coordinate axes, they can each interact with proper symmetry with one of the off-axis t2g orbitals. In the axial position, these orbitals match up in symmetry with dxz and dyz.

In the perfect octahedron, all three t2g orbitals are affected equally (each interacts with the four ligands in its plane), so whether or not the ligand has p orbitals (and therefore pi bonding) does not qualtitatively change the splitting diagram. It just changes the magnitude of the splitting.

When the octahedron is distorted, however, the interactions with each of the t2g are no longer the same in magnitude. Just as with the sigma bond, as the axial ligand moves away from the metal center, the pi bonding interaction also weakens, so the d orbitals involved in pi bonding (which as you noted are anti-bonding orbitals) lower in energy.

The reason that the dxy appears to be raised in energy is likely just an artifact of centering the orbitals on the average energy in both cases. Alternatively, the diagram could be based on the assumption that as the axial ligands move away from the metal, the equatorial ligands move closer, increasing the strength of the sigma and pi bonding of those ligands. In the pi case, this would raise the energy of all three t2g orbitals because all of them can engage in pi bonding with equatorial ligands, but the dxy would go up the most since it can pi bond with all four equatorial ligands, whereas the other two can only pi bond with two of the four equatorial ligands.

UPDATE: Here's an image from some old lecture notes showing the difference in convention of alignment between the "MO" view and "d-orbital view" in which the "d-orbital" diagram recenters the energies around the average, even though only the dxz and dyz MO (or more properly the antibonding orbitals primarily composed of them) energies actually changes:

orbital diagrams


We are not trying to maximize overlap in this diagram, quite the opposite. You can imagine it rather as pushing negative charges, the negative ligands ions, unto the electrons on the central atom while looking at the orbital energies during this process. That is a repulsive interaction between two negative charges, that raises the energy of orbitals that closer to the negative charges.

The $d_{xy}$ orbital and the $d_{x^2-y^2}$ orbitals are in the xy-plane with 4 ligands close to them. The $d_{yz}$ and $d_{xz}$ orbitals are in the yz/xz -plane where 2 ligands are further away. This means they go down in energy, as the repulsive interaction with the approaching ligands is lower. The $d_{x^2-y^2}$ is actually the "worst" energetically, since it points directly towards the 4 closest ligands. This is why it is the highest in energy. The $d_{z^2}$ orbital goes down since we just increased the distance with the ligands.

"MO" explanation:

The explanation for the $d_{xz},d_{yz}$ and $d_{xy}$ remains the same. You are correct when you say that their symmetry doesn't fit any bonding ligand orbital symmetries. The lowering and raising of the energies is purely due to the new geometry. When we go from octahedral $O_h$ symmetry to $D_{4h}$ symmetry, we lose the energetic degeneracy due to symmetry. I.e. the degeneracy in the octahedral geoemetry is due to the fact that these orbitals form a basis for the the irreducible $T_{2g}$ representation. This degeneracy of the set $(d_{xz},d_{yz}, d_{xy})$ is lifted in $D_{4h}$. The orbitals belong in $D_{4h}$ to $E_g$ $(d_{xz}, d_{yz})$ and $B_{2g}$$(d_{xy})$. There is no symmetry reason for these two different sets to be degenerate in $D_{4h}$, i.e. the geometry with the elongated z-bonds. And the argument for there relative energy ordering is just the same as given before. These nonbonding orbitals are more stable if they are further apart from negative charge, i.e. the ligands. When we draw these plots we typically assume total orbital energy conservation, and to achieve this we must push one up if we lower two others. I.e. the ones in the z-plane go down and the other goes up such that the "center of energy" remains constant.

The $d_{z^2}$ orbital shown in your figure corresponds to an antibonding molecular orbital. By increasing the distance of the ligands by elongating the z-distance, the overlap becomes smaller and the energy splitting between bonding and antibonding becomes smaller, i.e. the bonding orbital(which is not shown in your figure as it lies way below the d-orbitals) goes up a bit and the antibonding goes down a bit. The $d_{x^2-y^2}$ was in the octahedral geoemetry degenerate with the $d_{z^2}$ orbital as both together formed a basis for the $E_g$ representation. This degeneracy is lifted in $D_{4h}$. In this case we conserve the center of energy of the $E_g$ set. Since we know that the $d_z^2$ orbital goes down, we must have the remaining $d_{x^2-y^2}$ orbital go up.

You should also rather look at actual molecular orbital diagrams instead of looking at crystal field diagrams. Its easy to get confused by mixing up ideas from crystal field theory, ligand field theory and molecular orbital theory.

I used the following character tables

Answer to question in comment:

My wording "more stable" was perhaps a bit vague. I mean that the non bonding orbitals are lowered in energy with respect to their energy in the octahedral geometry. Why is this ? Well, the energy does not change due to overlap with some orbitals. It changes due to the lower magnitude of the coulomb interaction of the ligand orbitals with the non bonding orbitals when the ligand orbitals are further away, which is the case in the elongated geometry. Even if orbitals can't mix, they still affect each others energy simply due to the fact that they represent negative charge distributions. Bringing non mixing orbitals together raises their energy even if their shape doesn't change at all. You could calculate the potential energy due this interaction by $$ E_{pot} \propto \int dr_1 dr_2 \frac{|\phi_{nonbonding}(r_1)|^2 \sum_{\text{Remaining orbitals}}|\phi_b(r_2)|^2}{|r_1-r_2|} $$ This positive energy contribution to the total orbital energy decreases with distance between the orbitals and thus the nonbonding orbital energy decreases as the ligands move away. A term like this is also part of MO theory.

  • $\begingroup$ Sorry, but I'm looking for an MO theory explanation. $\endgroup$
    – bud
    Jan 22, 2021 at 22:28
  • $\begingroup$ @user103515 I added a small MO discussion. $\endgroup$
    – Hans Wurst
    Jan 24, 2021 at 12:35
  • $\begingroup$ Thanks for the MO part. "These nonbonding orbitals are more stable if they are further apart from negative charge, i.e. the ligands." What is the MO explanation for this? How can they be lowered in energy when they are nonbonding? "When we draw these plots we typically assume total orbital energy conservation." Why is this? Do you have a source I can read up on? $\endgroup$
    – bud
    Jan 31, 2021 at 0:18
  • $\begingroup$ @user103515 I have added an explanation why non bonding orbital energies change, at the end of my answer. I am afraid that i have no source, my answer is based on lectures that i heard quite some time ago. $\endgroup$
    – Hans Wurst
    Jan 31, 2021 at 12:40

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