Why does the 4s sub-shell fill before the 3d sub-shell in a potassium atom?
Hi marz; This is a difficult problem that is not adequately dealt with in the usual texts on Inorganic Chemistry. Briefly the answer is as follows: The 3d AOs are ALWAYS at a lower energy than the 4s AOs in atoms (but see below) although the difference in energy is comparatively small. So why are the 4s AOs filled first? The total energy of the atom depends not only the Coulombic attraction between the positive nucleus and the negatively charged electrons, but also on the repulsive electron-electron interactions. Sophisticated calculations show that when the last electron of K is placed in the 4s AO rather than the 3d AO there is less e-e repulsive energy and the [Ar] 4s1 configuration actually has a lower energy than the [Ar]3d1. (The 4s e is slightly further from the nucleus on average than the 3d electron.) It is the energy of the K atom as a whole that dictates the electronic configuration not the individual energies of the 3d and 4s AOs. Of course this applies to all third row (Fo
