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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.

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

Reference:

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

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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.

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