For example, if a single electron fills the 2p orbitals. Since all orbitals are degenerate, there is no reason why there would be any preference for px, py, pz. Is the electron hence in a superposition of these 3 orbitals?

And what if we consider hydrogen for n=2, where 2s, 2px, 2py and 2pz are degenerate?

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    $\begingroup$ It is worse than that. While the orbitals are degenerate, you have no way of knowing which of them should be called $p_x$, etc. $\endgroup$ Sep 21, 2018 at 12:55
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    $\begingroup$ @Ivan, yeah that makes sense, but is it then true that the electron probability is symmetric around the core, i.e. it exists in a superposition of all possible degenerate orbitals? $\endgroup$
    – Stikke
    Sep 21, 2018 at 13:28
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    $\begingroup$ Yes, it is symmetric all right. $\endgroup$ Sep 21, 2018 at 13:45
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    $\begingroup$ If one would like to describe this mathematically, choosing a configuration like $2s^2 p_x^1$ will give a wrong (to high) energy. To fix this you have to mix in the other 2 configurations ($2s^2 p_y^1$ and $2s^2 p_z^1$). See chemistry.stackexchange.com/questions/82656/… for details. $\endgroup$
    – Feodoran
    Sep 21, 2018 at 16:12
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    $\begingroup$ Thanks! And another follow-up question: what happens if a second electron fills the singly occupied p-shell? Even if they cannot be placed in the same orbital there would still be degeneracy right? How does that work then? $\endgroup$
    – Stikke
    Sep 22, 2018 at 12:14


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