2
$\begingroup$

In a paper on chemical periodicities of elements, Cao et al. show the number of valence electrons in a periodic table (https://doi.org/10.1515/pac-2019-0901). Apart from Boron and Aluminum shown as group 3 and 13, the table matches what is taught about main group elements in textbooks except for the entry for fluorine.

enter image description here

Fluorine is listed as 5 "active" valence electrons, implying perhaps that the 2s electrons do not participate in bonding. Why is fluorine treated differently than oxygen (or does oxygen make compounds where the 2s electrons are more involved in bonding than those of fluorine)?

If I look at the effective nuclear charges experienced by the 2s and 2p electrons in nitrogen, oxygen and fluorine, there is no jump in the energy difference that would place fluorine in a special category (values found on Wikipedia from Clementi et al 1963 and 1967):

enter image description here

$\endgroup$
6
  • 2
    $\begingroup$ Basic beryllium acetate, maybe? $\endgroup$ Commented Nov 2, 2021 at 15:43
  • 2
    $\begingroup$ This "periodic table" looks so bad I wouldn't put it in a post here, much less any paper. Not to mention it's wildly inaccurate. For starters all noble gases can be protonated - somehow their electrons are "active" enough. $\endgroup$
    – Mithoron
    Commented Nov 2, 2021 at 16:14
  • $\begingroup$ @Mithoron The only way possible to discuss a bad figure is to post it. It is what I am asking about, not an illustration where I have a choice whether to use a high-quality figure or not. $\endgroup$
    – Karsten
    Commented Nov 2, 2021 at 16:32
  • $\begingroup$ I was taught many years ago that the F 2s orbital is so low in energy that it does not contribute meaningfully to bonding. Eg in SF6 or BF3, the ligand orbitals are essentially completely 2p. But I don't have any actual calculations to back that up. $\endgroup$
    – Andrew
    Commented Nov 2, 2021 at 19:25
  • 1
    $\begingroup$ And why 11 in column 11 ? $\endgroup$
    – Maurice
    Commented Nov 5, 2021 at 19:57

1 Answer 1

2
$\begingroup$

Here is an example where the fluorine 2s orbitals are described as too tightly bound to the nucleus to participate in covalent bonds:

enter image description here

This is for the $\ce{SF6}$ molecule, discussing why the 2s orbitals are non-bonding. The full text with figures is e.g. here and the open access MO diagram is e.g. here.

Here is another, simpler example (image credit: Nick Greeves' Chemtube3D, also used in Libretext document), the MO diagram of hydrogen fluoride:

enter image description here

The OPs question shows a table with effective nuclear charges for the first and second period. If instead you look at ionization energies and include the third period elements, you do find that the s electrons in oxygen and fluorine hold the record when comparing outer shell electrons, and it might make sense to set them aside when making molecular orbital diagrams:

enter image description here

Source: https://www.colby.edu/chemistry/PChem/notes/AOIE.pdf

$\endgroup$
1
  • $\begingroup$ I really hate this classification of bonding and non-bonding and anti-bonding the image implies of using. In my point of view is stupid to base something on a reference that cannot be true any more. And it totally interferes when you later try to understand back-bonding. In MO theory there is no such thing as non-bonding, it can't be, it is using canonical (akin) orbitals, which are delocalised, therefore you cannot unambiguously define bonds. Anyway, in that regard, orbitals are the worst tool to try to capture bonds. $\endgroup$ Commented Sep 2, 2022 at 20:46

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.