Re: Why aren't the full amount of electrons in certain shells complete?

It is not exactly true that a shell must be complete before 
the next shell starts to form in an atom. The reason is that the 
energy of an electron depends both on the shell and on a quantity 
called "angular momentum". Electrons with low angular momentum in 
a shell tend to be closer to the nucleus and have lower energies 
than electrons with high angular momentum in the same 
shell. This happens because an electron that tends to be far 
from the nucleus feels more strongly the repulsive force 
of the electrons that are closer to the nucleus and this 
"pumps up" its energy, so to speak. 

Like energy, angular momentum in an atom can take 
only discrete values. Angular momentum values can be 
labeled with a number l, which takes integer values 
starting with 0, and increases with increasing angular 
momentum. Electrons with l=0 are called "s" electrons, 
electrons with l=1 are called "p" electrons, with 
l=2, 3, 4, 5, ... are called "d", "f", "g", "h" ... 
electrons. 

As we add electrons to an atom, they take the shell 
and the angular momentum that gives them the 
lowest energy. According to the Pauli 
exclusion principle, in any given shell there can 
be up to 2*(2*l+1) (that's two times l+l+1) electrons 
in a shell with the same l value. Therefore we can have 
only 2 s electrons, 6 p electrons, 10 d electrons, 
and so forth until the shell is complete. Within the 
SAME shell s electrons have lower energy than p electrons, and 
p electrons have lower energy than d electrons and so on 
as we explained above. 
Thus s electrons start to fill the M shell and d electrons 
complete the shell, but how do energies of those d electrons 
compare to the energy of the s electrons of the NEXT shell? 

It so happens that the d electrons of the M shell can have 
slightly higher energies than the s electrons of the N shell.  
Actually their energies depend on how many electrons the 
atom has, therefore the available places in the M shell compete with 
the N shell for the lowest energy to offer to the electrons. 
An authoritative book on the subject, "The Theory of Atomic 
Structure and Spectra" by Robert D. Cowan, p. 115, says that 
this fact can only be demonstrated quantitatively with a complex 
calculation, and only in some cases. Calculations of many-electron 
structures are very difficult and its accuracy is very limited. 
What happens in reality can be deduced from experiment though. 
Table 4-3 of that book shows that the N shell starts to fill 
with s electrons in potassium when the M shell only has 2 s electrons 
and 6 p electrons. The missing d electrons of the M shell start to 
appear in scandium. As we add electrons, it can happen that the M 
shell can gain or loose an extra electron from the N shell. This 
happens for example when we go from vanadium to chromium and 
manganese or when we go from nickel to cooper and zinc. The same 
happens when the O shell starts to form before the N shell is 
complete in rubidium. This situation only happens 
with neutral and singly ionized (i.e. those with an electron 
missing) atoms. More ionized atoms (those missing two or more 
electrons) don't show this behavior 
because the energies of their electrons don't depend so strongly 
on angular momentum.

Vladimir Escalante