0
$\begingroup$

I understand that the reason when you plot ionization energy against atomic number, you get peaks at the noble gases, is because they have full valence shells.

When I was trying to think this through, I wasn't sure why the outer electron is more tightly bound if the outer shell was full, as the more electrons in the outer shell, the more the nuclear charge is screened off and so the lower the potential energy of the electron being added, making it less bound? I know that my argument is flawed somewhere but I really only have a qualitative understanding of why full valence shells are most stable.

Can someone provide a more detailed explanation of this process?

Improve this question
$\endgroup$
4
  • $\begingroup$ The higher stability is related to the exchange energy $\endgroup$
    – user238497
    Commented Nov 26, 2019 at 12:49
  • $\begingroup$ @Rama Not particularly - this is old-fashioned screening, and the bulk of the physics and its intuition comes from the direct interaction without invoking anything as fancy as exchange. Or can you provide relevant literature that shows that the higher stability is indeed impossible to model unless exchange is accounted for? $\endgroup$
    – Emilio Pisanty
    Commented Nov 26, 2019 at 17:52
  • $\begingroup$ In this chemistry link vias.org/genchem/atomstruct_12433_05.html a Bohr type model is discussed, but continues several pages later to invoke quantization to explain the figure. $\endgroup$
    – anna v
    Commented Nov 26, 2019 at 18:07
  • $\begingroup$ @EmilioPisanty My highschool chemistry textbook states that the stability of half/fully filled orbital arises due to the concept of exchange energy (though it does not go deep into the topic, also I have very basic knowledge of QM so can't say much about it). If you wanna have a look then check this link on page no. 61. Note that the link contains a PDF. $\endgroup$
    – user238497
    Commented Nov 26, 2019 at 18:43

1 Answer 1

Reset to default
0
$\begingroup$

the more electrons in the outer shell, the more the nuclear charge is screened off and so the lower the potential energy of the electron being added, making it less bound?

Yes, but, all things considered, you'd rather have that last electron on an orbital even farther out, where the rest of the electrons can be that much more effective at screening the central charge.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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