# Why isn't just a fully occupied valence s orbital stable?

Why arent elements like calcium with an electron configuration of $$[\text{noble gas}]\ n\mathrm{s}^2$$ stable, although all the populated orbitals are fuully filled? Why is it necessary to obtain an octet configuration instead? (Ignoring the trivial cases of hydrogen and helium)

• Because it is an outer orbital that must be filled (although the value of "outer" is not intuitively obvious to the casual observer for transition metals). Sep 30 '20 at 4:11
• I actually think this is a good question as it’s something that’s rarely discussed in introductory courses.
– Jan
Oct 1 '20 at 9:46

Elements with just 1 or 2 electrons in the most outer $$\ce{s}$$ orbitals, like $$\ce{K}$$ or $$\ce{Ca}$$, have very well screened off the central charge of the kernel. As consequence, the effective charge of the kernel and related energy needed to ionize the atom are relatively low.
OTOH, elements with just few missing electrons in the most outer $$\ce{p}$$ orbital, like $$\ce{O}$$ or $$\ce{Cl}$$, do not have well screened off the central kernel charge, which is relatively high, what leads to considerable electron affinity to some extra electrons.