I'm referring specifically to this graph, which appeared in a chemistry lecture but which the lecturer could not explain. We know that 2s is lower energy than 2p. But surely those electrons in 2p orbitals, which apparently experience greater effective nuclear charge for Z > 10, would have lower energy! Can someone reconcile my understanding of orbital energy levels and this diagram?
One big issue here is that we generally don't talk about different ionization energies of anything but the outermost orbitals. Once electrons are in "core," we usually don't care since they generally do not participate in any kind of everyday chemistry.
We generally consider the $2s$ orbital to be lower in energy than the $2p$ because the $2s$ has some density inside the radial node that we think of as "nuclear penetration," i.e., the $2s$ electrons are able to get closer to the nucleus through the $1s$ core electrons. Now, once you add a $3s$ electron, I wonder if the innermost radial probability regions of the $3s$ actually repel the $2s$ electrons more than the $2p$ electrons which are generally farther away. The differential repulsion would have to happen in such a way so that it more than compensates for any nuclear penetration effects of the $2s$ electrons. This would certainly align with the observation that the $2p$ electrons are more stabilized only when we add a $3s$ electron.
Note that the x-axis is atomic number. The graph does not allow you to compare the energy of Fluorine's 2s and 2p orbitals (Z=9) but rather shows the energy of Lithium's and Beryllium's 2s orbitals (Z=3 and 4) and the energy of Boron's through Fluorine's 2p orbitals (Z = 5 through 9). The energies of the orbitals are not comparable on the graph because the atomic numbers are not the same. To compare the energies of Fluorine's 2s orbitals to its 2p orbitals, you would need to calculate the effective nuclear charge felt by the 2s electrons in Fluorine and the effective nuclear charge felt by the 2p orbitals in Fluorine, both calculations using Z=9. You can do this using Slater's rules for assigning shielding constants felt by the electrons in these orbitals.
