This discussion spun off from the comments to another question. The basic idea of that question was that electrons don't have a preferred order of filling in $p$ orbitals, i.e. the first electron will fill in the $p_x$ orbital just as well as it will fill in the $p_y$ or $p_z$ oribtal. This is because in the absence of electromagnetic fields, all p-orbitals are degenerate, and the coordinate axes are things we arbitrarily assign, so, the degenerate $p$ orbitals are all identical.

I wonder if the same can be said about $d$-orbitals. Four of them ($d_{xy}$, $d_{xz}$, $d_{yz}$, $d_{x^2-y^2}$) feel identical to me, as they all have four perpendicular and coplanar lobes (but I could be wrong). The fifth one - $d_{z^2}$ - doesn't look identical to me. Its positioning of the lobes is wildly different from the other four.

This makes me wonder,

Are all degenerate $d$-orbitals identical?

By identical, I mean that the electron filling order has no preference. (but this might not be the best definition of identical orbitals please mention if you have a better one)


  1. I didn't mention $f$-orbitals because they have even ridiculous orbital structures (4 types have six lobes, 2 have 8 lobes, and 1 seems to be a advanced version of $d_{z^2}$ with two rings. Crazy!), but in case their answer is remotely similar, please consider mentioning them in your answer for completeness.
  2. I have talked about atomic orbitals and not molecular orbitals (just in case that made the answer different).

2 Answers 2


Yes, they are identical.

One thing that we don't really teach well with orbitals is thinking about the symmetry of the orbital with respect to the name of the orbital.

$p_{x}$ has the same symmetry as the function $f(x,y,z)=x$.

Likewise, $d_{x^{2}-y^{2}}$ has the same symmetry as $f(x,y,z)=x^{2}-y^{2}$.

What about $d_{z^{2}}$? You should note that $d_{z^{2}}$ is really $d_{2z^{2}-x^{2}-y^{2}}$.

Imagine taking $d_{z^{2}-x^{2}}$ and $d_{z^{2}-y^{2}}$, summing them up, and renormalizing. So the symmetry is like: $z^{2}-x^{2} + z^{2} - y^{2}=2z^{2}-x^{2}-y^{2}$.

What you get is a big lobe of the same phase along the $z$-axis. And around it in the $x$-$y$-plane (but smaller), you see a smear of the opposite phase that's shaped kind of like a torus. That's why it looks different. It's really two of the other looking orbitals put together.

Why don't we just use $d_{z^{2}-x^{2}}$ and $d_{z^{2}-y^{2}}$ instead? Because they're not linearly independent with the 4 other "normal"-looking orbitals. $d_{x^{2}-y^{2}}$ summed with $d_{z^{2}-x^{2}}$ is $d_{z^{2}-y^{2}}$. The 5 conventional orbitals are just the nice linear combinations of the 5 spherical harmonic solutions to the angular part of the Schrodinger equation. See here.

You don't seem to have a problem with $d_{xy}$, $d_{zx}$, $d_{yz}$, and $d_{x^{2}-y^{2}}$ all being degenerate. So hopefully, you can see that $d_{z^{2}}$ is not really different and therefore also of the same energy as the other 4.


Based on comments, it's also important to point out that the pictures of orbitals with smooth surfaces are not real pictures of orbitals. These are boundary surface diagrams where we've drawn the surface that represents a constant probability of finding the electron for that orbital. This basically means that electron densities for $d_{z^{2}-x^{2}}$ and $d_{z^{2}-y^{2}}$ will smear together in the $x,y$-plane (the orbitals are the same phase) to create something that is symmetrical and shaped like a torus.

This analysis extends fully to $f$ orbitals. Just identify the full functional name for each orbital and apply the same analysis here.

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    $\begingroup$ Thanks for your answer! That's quite a lot of things you touched upon. I tried to interpret them, please let me know if I am wrong. 1. $f(x,y,z)=x$ represents $p_x$ orbital and it is symmetric about $y$ and $z$ axes. 2. Renormalization on Wikipeda refers to QFT, QED and perturbations makes no sense :( Do you have a simpler alternate word/meaning for what is its importance in this context? 3. "It's really two of the other looking orbitals put together." Are they joined head-to-tail? Why don't they have their tail at the origin (like all others)? $\endgroup$ Commented Feb 13, 2018 at 4:01
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    $\begingroup$ This doesn't appear to actually answer the question "does electron filling order exhibit a preference for particular d-orbitals?", or at least not explicitly. $\endgroup$
    – OrangeDog
    Commented Feb 13, 2018 at 10:57
  • $\begingroup$ $f(x,y,z)=x$ only has the same symmetry. That is not the $p$-orbital. Normalization simply means that the sum of the square of the wave function over all space should be 1. If I add two orbitals together, I need to normalize again because the sum of the squares of the sum over all space is probably not still 1. I don't understand what you mean by "tail" or "head-to-tail"... @GaurangTandon $\endgroup$
    – Zhe
    Commented Feb 13, 2018 at 13:56
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    $\begingroup$ @OrangeDog That's true, but the OP seems to think $d_{z^{2}}$ is different but doesn't have an issue with the other 4 being degenerate. I was hoping that by showing that $d_{z^{2}}$ is exactly like the other 4, that disconnection would go away. $\endgroup$
    – Zhe
    Commented Feb 13, 2018 at 13:57
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    $\begingroup$ @Zhe then your answer would be greatly improved by starting it with Yes. $\endgroup$
    – OrangeDog
    Commented Feb 13, 2018 at 13:58

By identical i guess you mean to be able to be transformed to one another by a symmetry operation. So:

I think it is the same reason why the orbitals 2s, 2px, 2py, and 2pz are all degenerate in the H atom. You can clearly see the symmetry between the 3 p orbitals but how can the spherical 2s be symmetric with the rest? The answer is that the Coulomb potential has a hidden symmetry that shows up in spaces of dimension higher than 3. Hence, in the same way that a 2 dimentional being would not be able to see the symmetry between the 3 p orbitals (the px and py will look like two disks that touch when projected to a plane while the pz will look like a simple disk) we cant see the symmetry between the s and p orbitals in the H atom (or dz2 and the rest ds) since we are only 3 dimentional beings. Ref: Molecular quantum mechanics by Atkins, p.95. For this reason you need to look at the math behind this. Unfortunaltely, being no physical chemist i wouldnt be able to show the math behind this (or it would take me a while figuring it out).

  • $\begingroup$ Nice! But, I got confused at this line: "The answer is that the Coulomb potential has a hidden symmetry that shows up in spaces of dimension higher than 3." Do you mean to say that all these orbitals are not 3d but actually 4d orbitals??? $\endgroup$ Commented Feb 13, 2018 at 9:18
  • $\begingroup$ No, they are not 4d. When i say dimention i mean it in a mathematical way and not changing the principal quantum number which is what you have done. For example when you plot an xy chart you are ploting in two dimentions. When you plot an xyz 3D graph (it is a surface now) you are ploting 3 dimentions. However you can plot in more than 3D, like xyzv, but now you wont be able to visualize the result bacuse this is how far our perception goes (in the same way you cant see in IR or UV). Have a look at the ref above as well for more details. It is quite a read though! $\endgroup$
    – Outlander
    Commented Feb 13, 2018 at 9:57
  • $\begingroup$ Sorry I didn't make my comment clearer I actually meant "...are not 3 dimensional but actually 4 dimensional orbitals???" $\endgroup$ Commented Feb 13, 2018 at 10:25
  • $\begingroup$ Sorry, i see what you mean and it is a good point. I think that the equations that describe the orbitals must be multidimentional as you say. In the answer by Zhe you can see that the function that describes the dz2 orbital is of the type f(x,y,z) which is already 4 dimentions. However i dont know the math behind it being a synthetic chemist. Maybe a physical chemist would like to jump in at this stage? $\endgroup$
    – Outlander
    Commented Feb 13, 2018 at 10:57

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