0
$\begingroup$

Oxygen forms two hydrogen bonds as it has two lone pairs. So, why does fluorine only forms one hydrogen bond having three lone pairs?

Both oxygen and fluorine have nearly the same size, so steric hindrance should not be an issue.

I'm aware of the following facts, so please do not use them as an explanation:

  • hydrogen forms only one hydrogen bond;
  • fluorine forms one hydrogen bond in hydrogen fluoride $\ce{HF}$;
  • oxygen forms two hydrogen bonds in water $\ce{H2O}.$

Are there any compounds in which fluorine forms three hydrogen bonds?

Improve this question
$\endgroup$
3
  • $\begingroup$ The number of lone pairs has no importance in the calculation of the bonds of an atom. Only the number of electrons non included in lone pairs is important. Oxygen has $2$ such electrons. It can make $2$ bonds with $\ce{H}$ atoms. Fluor has only one such electron. So it can make only one bond with $\ce{H}$ atoms. $\endgroup$
    – Maurice
    Commented Dec 18, 2023 at 11:59
  • 1
    $\begingroup$ Be aware of possible confusion of "fluorine H-bond" as 1/ an ordinary polar covalent bond H-F as in hydrogen fluoride. 2/ Hydrogen bond as e.g. between $\ce{HF}$ molecules as $\ce{H-F\bond{...}H-F}$. $\endgroup$
    – Poutnik
    Commented Dec 18, 2023 at 12:17
  • 2
    $\begingroup$ Who said F only forms one H-bond, and in which contetxt? $\endgroup$
    – Ivan Neretin
    Commented Dec 18, 2023 at 14:51

1 Answer 1

Reset to default
6
$\begingroup$

Fluorine in hydrogen fluoride can form only a limited amount of hydrogen bonds because there is only one (protic) hydrogen atom per fluorine. Ammonium fluoride has enough protic hydrogens to form hydrogen bonds with all four electron pairs on each fluorine — and so they do, in a wurtzite-type arrangement of the ions.

Improve this answer
$\endgroup$
0

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.