In hydrogen, all orbitals with the same principal quantum number 'n' (1,2,3...) are degenerate, regardless of the orbital angular momentum quantum number'l' (0,1...n-1 or s,p,d..). However, in atoms with more than one electron, orbitals with different values of l for a given value for n are not degenerate. Why is this? Surely the radial distribution functions are similar for hydrogen (in that there's still penetration of orbitals and so on). Or is it that orbitals with different values of l are degenerate for a value of n greater than would be occupied in that particular atom's ground state?
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5$\begingroup$ This excellent answer by user Philipp seems to answer your question precisely. $\endgroup$– Nicolau Saker NetoOct 29, 2014 at 11:36
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2 Answers
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The answer is right there in your question. It's only the interaction of multiple electrons in an atom like He, Li, Be, etc. that makes the different angular momenta wave functions differ in energy.
Consider this.. For the one electron system, why should a $p$ or $d$ orbital differ in energy from an $s$. What makes them differ?
In the multi-electron case, the $p$ orbitals have different spatial extent, different angular components, so the electron density caused by an electron in that orbital will interact differently with the other electrons. In other words, you need to have more than one electron for the "shape" of the $p$ and $d$ and $f$ orbitals to matter to the other electrons.
In the H atom, there's only one electron, so there's no electron-electron repulsion to differentiate the $s$, $p$, and $d$ orbitals.
answered Oct 29, 2014 at 13:21
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$\begingroup$ I'm sorry to say that I dont find this answer at all satisfactory. For example why is degeneracy lost in the case of the He atom. If we are to discuss it in terms of orbitals that are occupied then we are only dealing with the 1s orbital. (not the p and d that the above answer appeals to). $\endgroup$ Apr 18, 2019 at 21:30
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$\begingroup$ @Dr.EricScerri - no problem - you can of course write your own answer if you wish. (I expect that you've thought about how to explain this far more than I have.) That's how StackExchange works. $\endgroup$ Apr 20, 2019 at 0:10
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$\begingroup$ Actually I have always had great difficulty in connecting removal of degeneracy with the mere presence of more than one electron. This is why I have been searching for answers. Let me raise a further question in the hope of getting clear on this issue. He of course has 2 electrons. Will the energies of transitions from 1s to 2s be the same as between 1s and 2p? I think the answer is YES. But this is a case in which the relative penetration of electrons in orbitals fails to give an answer as I see it. www.ericscerri.com $\endgroup$ Apr 21, 2019 at 1:06
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2$\begingroup$ @Dr.EricScerri - No, for Helium, the transitions from 1s to 2s and 1s to 2p have different energies. (e.g., NIST tables. That's probably worth a separate question. $\endgroup$ Apr 22, 2019 at 0:30
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This was asked a long time ago, but I think I'll add an answer.
The answer is penetration and shielding. For a hydrogen atom, if its electron is promoted to 2s or 2p, the only thing that determines its energy is its radial distance, determined by quantum number n. There is no electron-electron repulsion, obviously.
For Helium, however, if one of its electrons is promoted to 2s or 2p, there's still an electron remaining in the core 1s shell, so the other electron in the 2s or 2p orbitals will be shielded because of electron-electron repulsion. Since 2s is shielded less by the core than 2p, it experiences a higher Zeff and is able to penetrate the nucleus more and hence has a lower potential than 2p, which is shielded more by the core and has a lower Zeff than 2s. Therefore, 2s is lower in energy than 2p because of penetration and shielding.
