0
$\begingroup$

This question already has an answer here:

Why only fluorine, oxygen and nitrogen can form hydrogen-bonds with the hydrogen of another molecule?

$\endgroup$

marked as duplicate by Mithoron, aventurin, DavePhD, Linear Christmas, Tyberius Apr 26 '18 at 19:20

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 4
    $\begingroup$ That's a common misconception, that is thoroughly incorrect. $\endgroup$ – Mithoron Apr 26 '18 at 17:35
3
$\begingroup$

No, that's not necessary. Amongst the strongest hydrogen bonds are formed by $\ce{ N, O, F}$ because of their high electronegativity.

There's evidence that the ions $\ce{Cl-, I-, Br- }$ form hydrogen bonds that are much stronger than those of the covalently bonded atoms

$\ce{Cl}$ can form weak hydrogen bonds, but $\ce{Br}$ and $\ce{I}$ form very weak bonds if at all.

A sulfur atom can also function as weak acceptor., but the $\ce{SH-}$ ion forms much stronger bonds.

Hydrogen bonding has been directly observed between a negatively charged carbon and an $\ce{-OH}$ group in the same molecule

Isocyanides' ($\ce{R- N+ #C- }$) carbon atom can also act as an acceptor (forming a rather strong Hydrogen bond).


Reference: March's Advanced Organic Chemistry, Chapter 3: Bonding weaker than Covalent

$\endgroup$
  • $\begingroup$ Just to avoid future confusion: the NOF hydrogen bonds are not necessarily stronger than others either. $\endgroup$ – Linear Christmas Apr 26 '18 at 19:02
  • 1
    $\begingroup$ I think they are, why not? @LinearChristmas . Other atoms form hydrogen bonding but the strongest is formed by N O F atoms... $\endgroup$ – santimirandarp Apr 27 '18 at 1:10
  • $\begingroup$ @santimirandarp While many strong and common hydrogen bonds are formed with NOF, it is not a sufficient criterion of assigning superiority when comparing strengths of two hydrogen bonds. In other words: while many strong H-bonds are indeed of the NOF variety, there are still cases where others are stronger. One must still carry out meticulous experiments, or perform accurate calculations to make such assignments. See my earlier comment on another question for a concrete example. $\endgroup$ – Linear Christmas Apr 27 '18 at 20:43

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