I found this statement when browsing the Wikipedia article for atomic orbitals:

"Orbitals of multi-electron atoms are qualitatively similar to those of hydrogen."

Is this true? Googling around I could only found this article where in page 50 it seems to address how to obtain the wave function of atoms with multiple electrons, but I don't have the necessary background to understand if it proves the statement or not.

Please include academic sources or a brief proof if possible.

I find it surprising that adding electrons wouldn't change the shape of the orbitals substantially, but that's what is implied when I've studied chemistry.

  • $\begingroup$ Minor comment to the post (v2): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. $\endgroup$ – Qmechanic May 7 at 19:24
  • $\begingroup$ What shapes? The spherical harmonics are quite spherical. $\endgroup$ – Pieter May 7 at 19:29
  • $\begingroup$ @Pieter Are they? en.wikipedia.org/wiki/Spherical_harmonics#/media/… $\endgroup$ – safesphere May 7 at 20:38
  • $\begingroup$ @safesphere Those are the standing waves with stationary nodes and antinodes, where the expectation of angular momentum is zero. Look at commons.wikimedia.org/wiki/… $\endgroup$ – Pieter May 7 at 20:44
  • $\begingroup$ @Pieter So which shape defines the probability of finding the electon, spherical or, say, figure-8? $\endgroup$ – safesphere May 7 at 20:51

An approximation that seems to work well for the multi-electron case is the Hartree-Fock method.

In Hartree-Fock, we assume the mean-field approximation. Each electron feels the repulsion from other electrons based on their average, not instantaneous, positions. (This assumption prevents Hartree-Fock from predicting van der Waals forces.)

We thus modify the hydrogen Hamiltonian by introducing two new operators. One is the average Coulombic repulsion between electrons, and the other is the exchange interaction. However, because we're using the average position of the electrons, then for our spherical atom these operators don't have an angular dependence. Thus the spherical harmonics are still separable as in the hydrogen case, so roughly the shape of the orbitals must remain the same. The only part that can change is the radial part of the wavefunction. Doing the calculations, you'll see that the radial part of the wavefunctions are squeezed or stretched a little bit due to Coulombic repulsion and the exchange interaction between electrons, and the increased Coulombic attraction to the nucleus. But as Wikipedia says, qualitatively they don't change much until you introduce multiple atoms.

Without the mean-field approximation, I suppose even the angular shape would change, but that's beyond me.

  • $\begingroup$ Thanks. I don't have enough background knowledge to fully understand or back your argument, but it does make some intuitive sense and it offers a mathematical explanation. I am marking it as the answer, but then again (for future readers) I can't verify if it is 100% accurate. It makes me doubt that @ggcg mentioned ignoring the interactions in the hamiltonian on his answer. Does that mean interactions between the electrons would change the hamiltonian, and therefore the general shape of the orbitals? $\endgroup$ – J Physics FM May 8 at 18:27
  • $\begingroup$ @J Physics FM Yes, the electrons add new terms to the Hamiltonian. I added a little more to my answer. $\endgroup$ – HiddenBabel May 8 at 21:21
  • $\begingroup$ @JPhysicsFM it may help to note that use of H orbitals is an approximate method to describe psi function for many electrons as a simplistic combination of such simple orbitals. So "high-Z atom orbitals" are similar to H orbitals because this was done by hand, to simplify the analysis (mutual repulsion of electrons was simplified into radial central field and this allows use of H orbitals). If we want to stay exact, the situation for high-Z atoms is much more complicated than in H, since there is strong repulsion between multiple electrons and it is not as simple as one central radial field. $\endgroup$ – Ján Lalinský May 9 at 22:51

What is meant is that the quantum numbers of the hydrogen solutions are still relevant for multi-electron orbitals. You still have shells 1s, 2sp, 3spd, etc. There is not a "proof". What is there is excellent agreement between quantum chemical calculation and experiment.

  • $\begingroup$ It makes sense. Could you please include in your answer if we could expect the shapes of orbitals to change substantially in multiple electrons atoms compared to the shape of orbitals on a hydrogen atoms? I know the shape of the orbitals does change substantially when another nucleus (atom) is introduced, but I am not so sure for multiple electrons on a single atom. Thanks for answering. $\endgroup$ – J Physics FM May 7 at 18:39
  • $\begingroup$ see this for hydrogen hyperphysics.phy-astr.gsu.edu/hbase/Chemical/eleorb.html .found this youtube.com/watch?v=Ewf7RlVNBSA $\endgroup$ – anna v May 7 at 19:28

(1) if you ignore the interaction between electrons in the Hamiltonian then yes, it would be true.

(2) for the outer most electron this is probably a good approximation.

Are you referring to orbits as predicted by the Bohr model or Schreodinger's equation? The statement may be only true relative to a particular modeling approach.

Also, it may be that the effects of the other electrons average out over time leaving a dominant contribution from the nucleus. Lastly, don't just look at Wikipedia, go to a physical chemistry text.

  • $\begingroup$ Excuse my ignorance, but how would the effect of the electrons average out over time? It's not like they are going anywhere... Concerning the model, I am talking about the orbitals predicted by Bohrs model but, from the little I know, I understand that Bohrs predictions are closely related to Shrodingers ecuation. Thanks for answering. $\endgroup$ – J Physics FM May 7 at 18:41
  • $\begingroup$ No, all the e- are buzzing around the nucleus. The "orbitals" represent stationary states, or eigenstates of the energy operator. The Bohr model fails quite miserably for complex atoms so I would not say it is closely related to Schrodinger's equation. It is well known that we get corrections to the model using the full machinery of QM. Also, in the context of the Bohr model the prediction would hold for light ionized atoms like Li, etc. Here you have a nucleus with Ze charge and only 1 e flying around it. $\endgroup$ – ggcg May 7 at 18:49
  • $\begingroup$ Had I known you were specifically referring to Bohr I would have made the point about Li. I may edit my answer. $\endgroup$ – ggcg May 7 at 18:56
  • 1
    $\begingroup$ Why the heck is this down voted? $\endgroup$ – ggcg May 7 at 21:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.