Consider all the MOs of some isolated molecule. (It could be a single atom too; I'll use MO to refer to AOs as well.) Number them in increasing order of the number of nodes (node = surface where the wave function has zero density). Orbitals with the same number of nodes can be numbered in any order. Now you have a sequence of orbitals $O_1, O_2, ...$. Let their respective energies be $E_1, E_2, ...$.
It seems to be "common knowledge" that $E_n \le E_{n+1}$ for any such system and any $n$. As Martin put it to me yesterday, "An orbital with 47 nodes can never be lower in energy than one with only 46." (Follow up on Counting Nodal Planes in cyclopropane.)
For various reasons, given below, I think this cannot be true in general, and I'd really like to know under what conditions it is known to be true. "Known" here can mean either a rigorous statement with a reference to a proof (a trivial example would be: It is true for a one-electron atom; we can calculate the energies exactly) or a precise statement with empirical justification (something like "no counterexamples are known for class X of molecules" - again with a reference).
Important: Please, I'm not looking for an explanation that re-states the rule in some equivalent or even looser fashion ("more nodes means the orbital is larger and less dense, therefore it must be higher in energy").
Why do I think the statement can't be always true? Well, a calcium atom has a filled 4s orbital and empty 3d orbitals. If this does not count as a counterexample, please explain what notion of orbital energy the statement applies to.
In general, I'm happy to believe that two MOs must satisfy the rule if they have "comparable" sets of nodes (say, constant $n_x$ and $n_y$ in the example I'm about to discuss), but I'd like to understand what is meant by "comparable" in general. In a molecule with no symmetry, are there comparable MOs at all? If so, how do we know if two given MOs are comparable?
The case of a much simpler system, the 3D rectangular box, may also be relevant. The energy levels for such a box are of course $$ \frac{\hbar^2\pi^2}{2m}\biggl( \frac{n_x^2}{L_x^2}+ \frac{n_y^2}{L_y^2}+ \frac{n_z^2}{L_z^2} \biggr), $$ where $n_x$, $n_y$ and $n_z$ are one more (or one less, if you count the walls) than the number of nodal planes in the respective direction. If we take $L_x = L_y = 1$ and $L_z = 0.1$ (say), the wavefunction with $n_x=5$, $n_y=1$, $n_z =1$ has energy $5^2+ 1^2 + (1/0.1)^2 = 126$ and 4 nodes (or 10, if you count the walls), while the wavefunction for $n_x=1$, $n_y=1$, $n_z=2$ has energy $402$ and 1 node (or 7). So clearly the rule is not true here.
Granted that molecules are not boxes, but this does show that arguments based simply on the number of sign changes are not rigorous, so they don't answer my question.
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