Does fluorine ever form a double or triple bond? I wonder if seeming lack of such higher order bonds is related to the electronegativity of fluorine.

  • 4
    $\begingroup$ Who said it doesn't? $\endgroup$
    – andselisk
    Dec 4 '17 at 10:15
  • $\begingroup$ Oxtoby-principle of modern chemistry $\endgroup$
    – TRIPLE
    Dec 4 '17 at 10:21
  • 3
    $\begingroup$ It was a rhetorical clickable question. There is $\ce{F=N}$ bond known to exist. $\endgroup$
    – andselisk
    Dec 4 '17 at 11:11
  • 1
    $\begingroup$ For all intents and purposes, it doesn't... $\endgroup$
    – orthocresol
    Dec 4 '17 at 12:36

The compound $\ce{NF}$, which is isoelectronic to $\ce{O2}$ is known, has been isolated in matrices and characterised and subjected to calculations. Like for dioxygen, three different states of this molecule are known: one triplet and two singlet states. Without performing any sophisticated analysis of its orbitals, we can expect a bond order of 2, and thus a double bond.

Harbison performed calculations on this compound. He came to the conclusion that the most stable triplet state is best described using only a single $\ce{N-F}$ bond. The two singlet states which require full electron pairing, however, display a much shorter $\ce{N-F}$ bond distance and are thus better described by an $\ce{N=F}$ double bond. Adding formal charges would lead to: $$\ce{\overset{-}{N}=\overset{+}{F}}$$


G. S. Harbison, J. Am. Chem. Soc., 2002, 124, 366–367. DOI: 10.1021/ja0159261.

  • $\begingroup$ Going down this line, one wonders what would happen as the atomic number of $\ce{X}$ in $\ce{XF}$ decreases. Apparently boron monofluoride $\ce{BF}$ still has only partial double bond character. Beryllium monofluoride $\ce{BeF}$ would also be an interesting albeit more complicated case... $\endgroup$ Oct 3 '20 at 23:54

For most purposes, i.e. writing Lewis Structures and such things, fluorine always forms single bonds. In higher level chemistry, it may be possible (like in the comments). Fluorine, like you said, is very electronegative, and therefore it doesn't like to "share" its electrons, leading to almost always making single bonds.


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.