Are all degenerate d-orbitals identical?

Question

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)

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Remarks:

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).

Answer 1

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.

EDIT:

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.

Answer 2

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).