3

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?

Stikke
  • 471
  • 2
  • 8
  • 4
    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. – Ivan Neretin Sep 21 '18 at 12:55
  • 2
    @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? – Stikke Sep 21 '18 at 13:28
  • 3
    Yes, it is symmetric all right. – Ivan Neretin Sep 21 '18 at 13:45
  • 3
    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 https://chemistry.stackexchange.com/questions/82656/what-constraints-are-imposed-on-a-wavefunction-by-the-symmetry-of-the-system/82682#82682 for details. – Feodoran Sep 21 '18 at 16:12
  • 1
    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? – Stikke Sep 22 '18 at 12:14
  • 1
    You would mix three different states as well ($\ce{2s^2p^1_xp^1_y}$, $\ce{2s^2p^1_yp^1_z}$ and $\ce{2s^2p^1_zp^1_x}$). – schneiderfelipe Sep 24 '18 at 11:59

0 Answers0