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)

  • 1
    $\begingroup$ 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). $\endgroup$ Sep 30, 2020 at 4:11
  • 1
    $\begingroup$ I actually think this is a good question as it’s something that’s rarely discussed in introductory courses. $\endgroup$
    – Jan
    Oct 1, 2020 at 9:46

1 Answer 1


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.

Combining these 2 kinds of elements leads to energy release, as energy spent by ionization of the former is compensated by energy release by capturing electrons by the latter and by a covalent contribution to the element bond.

In case of solid ionic compounds, there is additional large energy release du the ionic lattice energy.

  • $\begingroup$ No, it's not compensated. There's still covalent component needed for such reactions to happen. $\endgroup$
    – Mithoron
    Sep 30, 2020 at 13:56
  • $\begingroup$ Hmm, you may be right. $\endgroup$
    – Poutnik
    Sep 30, 2020 at 13:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.