# Does fluorine ever form a double or triple bond?

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.

• – andselisk Dec 4 '17 at 10:15
• Oxtoby-principle of modern chemistry – Spectrum Dec 4 '17 at 10:21
• It was a rhetorical clickable question. There is $\ce{F=N}$ bond known to exist. – andselisk Dec 4 '17 at 11:11
• For all intents and purposes, it doesn't... – 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: $$\ce{\overset{-}{N}=\overset{+}{F}}$$