This is apparently a true statement:
"In a hydrogen atom, the 2s and 2p subshells have the same energy."
Why is this the case? Why would they be of the same energy? Wouldn't the 2p be slightly higher in energy than the 2s?
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5$\begingroup$ While addressing a different question, this should answer your question pretty precisely. $\endgroup$ – Philipp Mar 9 '16 at 8:38
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3$\begingroup$ Incidentally, the answer to your title is "no" for the general case. The hydrogen atom is a very specific case as mentioned by @Philipp. $\endgroup$ – Geoff Hutchison Mar 9 '16 at 15:06
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My chemistry teacher Mr Ong Chiau Jin answers:
Why are all the orbitals that have the same principal number in Hydrogen degenerate?
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?
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.
This may not be the perfect answer but it does capture the essence of the perfect answer, which is the idea of multi-electron interactions. Hope this helps.
answered Aug 20 '17 at 1:51
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1$\begingroup$ Do you think you could possibly find an authoritative source for this?
My chemistry teacher Mr Ong Chiau Jin answersdoes not a reference make, even though the information is/may be factually accurate. $\endgroup$ – NotEvans. Aug 20 '17 at 13:29 -
$\begingroup$ @NotEvans. No, I don't think I have come across this piece of information in the literature I have read before and hence, I do not have any authorative sources to cite. Do let me know if you have come across any such source. $\endgroup$ – Tan Yong Boon Aug 20 '17 at 14:54
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If you go through the trouble of setting up the Hamiltonian for a single electron in the vicinity of a point-sized, positively charged particle (our approximation of the proton) — i.e. that for the hydrogen atom — and then solve the Schrödinger equation, you will realise that the energy of said electron depends only on the principal quantum number (typically labelled $n$).
This is true for all hydrogen-like atoms, i.e. also for $\ce{He+, Li^2+},\dots$ (Note, however, that the energies of a helium-cation’s orbitals are naturally different from those of a hydrogen atom’s.)
It is not until you add additional electrons to the species that their potential induces a difference between the energies of s and p subshells of a given shell.
