Are there stronger oxidizing agents than fluorine gas, so it could oxidize fluoride to fluorine?

Also, in case of oxygen, fluorine gas can oxidize oxygen gas to the exotic dioxygenyl ion. Can anything oxidize fluorine gas to $\ce{F+}$ ion?

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    $\begingroup$ Fluorine gas can't oxidize oxygen gas to dioxygenyl ion; $\ce{PtF6}$ can. $\endgroup$ Commented Jun 24, 2016 at 21:00
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    $\begingroup$ Fluorine gas by itself can't. Throw BF3 into the mix and it does. $\endgroup$
    – Ian Bush
    Commented Sep 1, 2017 at 7:01
  • $\begingroup$ Where do alpha particles fit in this scheme? I think He++ ions would be pretty potent oxidizers too. $\endgroup$
    – Curt F.
    Commented May 6, 2021 at 17:02

3 Answers 3


There are stronger oxidization agents than fluorine gas. I can name two in particular. But I don't think either would actually oxidize fluorine gas in it's pure state, for obvious reasons that you'll see shortly.

Dioxygen Difluoride. Affectionately named 'FOOF'. This is an extremely, extremely potent oxidizer. It also lights ice on fire. In fact, it lights essentially everything on fire or blows up on contact with said things.

Chlorine Trifluoride. Arguably, worse. This gas burns through practically everything, except for a few specific structural metals, but only because these form a very thin passivization layer. If this scratches, the tank burns. A notable incident with this gas, a 900kg spill of it burnt through 30cm of concrete, then 90cm of gravel underneath that concrete. It burnt over a meter into the ground.

As a foreword: I have no idea why you'd want to get within three miles of these things. Please don't light yourself on fire. It's not fun. (Trust me).

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    $\begingroup$ How are you quantifying stronger? $\endgroup$
    – Ian Bush
    Commented Jun 24, 2016 at 19:49
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    $\begingroup$ More readily performs a redox reaction. $\endgroup$ Commented Jun 24, 2016 at 21:00
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    $\begingroup$ Does FOOF or ClF3 has a higher standard electrode potential than fluorine gas? If yes, how much? $\endgroup$
    – Nissa
    Commented Jun 25, 2016 at 14:50
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    $\begingroup$ @Nissa They would dissolve your electrodes, so the question is sort of moot. $\endgroup$
    – Mithoron
    Commented Jul 30, 2016 at 12:39
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    $\begingroup$ Note that many standard redox potentials are just formal, not really measured due kinetic or other reasons, but derived from the reaction $\Delta G_\mathrm{r} = nFE^{\circ}$ $\endgroup$
    – Poutnik
    Commented Oct 9, 2020 at 11:35

I'm not sure, but I think krypton difluoride is the strongest oxidizing agent I know of that's fairly stable at normal temperatures and pressures. (It decomposes at ~10% per hour at ~25°C, but is stable at dry ice temperature or below. Wikipedia claims it is "the most powerful known oxidizing agent known" and "more powerful even than elemental fluorine due to the even lower bond energy of Kr–F compared to F–F, with a redox potential of +3.5 V for the KrF2/Kr couple".) It is a colorless crystalline solid at room temperature that reacts with xenon to directly form XeF6 without any extra input of energy, and reacts with gold to form the hexafluoroaurate salt KrFAuF6, with gold in the +5 oxidation state (both releasing krypton gas). It can also be used to oxidize ClF5 to the ClF6+ cation. I don't know how it reacts with N2, Cl2, O2, or O3, but I feel like it must oxidize them, since things that oxidize xenon at all (such as platinum hexafluoride) can generally oxidize O2 to O2+ (dioxygenium). (I don't think O3+ is stable, but I imagine ozone could just be broken down into O2+ ions. In any case, it's probably easier to oxidize than XeF4, in which case KrF2 should be able to oxidize it.) Obviously, atomic fluorine would probably be stronger, but at normal pressures and temperatures it would spontaneously react with itself to form F2. (Atomic F and O can exist in the interstellar medium, though, with O being common even in the Earth's thermosphere.) I think that putting such an inert gas between the two fluorine atoms acts as kind of an approximation of atomic fluorine.

On the subject of ridiculously unstable things, positively charged things naturally attract electrons, so perhaps they are the real strongest oxidizing agents. (The only problem is that the compounds they form might break down as soon as the charge is balanced.) One might suppose that the F9+ ion (a bare fluorine nucleus) would be the best oxidizer, and I imagine it would be much better than F2, but if you're just interested in how good it is at stealing electrons from things in the short term, then charge matters much more than the electronegativity of the neutral atom, so a bigger nucleus would probably be a better "oxidizing agent". Thus, I might propose bare califonium or mendelevium nuclei as the best fairly stable particles for stealing electrons. However, these elements are, in fact, metals, so a lot of this work would be undone chemically whenever the charge went back to normal, which would happen quickly as it would basically just steal electrons from a as large an area as possible, so that positive charge would be spread out over a lot of only slightly positive, not-very electronegative, things, even if the electrons taken away to charge the ions were shot into deep space or something.

That being said, these considerations do shed some light on why electricity is often used in making strong oxidizing agents like krypton difluoride, as well as ozone, perchlorates, xenon fluorides, and even O2 by electrolysis of water. (It also makes strong reducing agents like alkali metals.) An F2+ ion is a much better oxidizing agent than neutral F2, and a Kr- ion is probably a much better reducing agent than caesium (it's certainly better than neutral rubidium, which has the same electron configuration). When you use electricity or UV light or extreme heat on a mixture of F2 and Kr you can get these ions and even more extreme ones, and that is how you can make KrF2. (Actually, the hot-wire synthesis method produces short-lived atomic F, which is also a better oxidizing agent than F2, as I mentioned.) The F2 doesn't have to oxidize the krypton: The F2+ or atomic F (or F+ or F3+ or F4+ or whatever*) does, and the Kr-F bonds just have to not immediately fall apart once the charge rebalances to make neutral molecules.

*My guess is that F2+ ions would sometimes bond to neutral F2 molecules to form ions like F3+ or F4+ (F3+ being isoelctronic with OF2), which is why I included them. So that might be an answer for what can oxidize fluorine.

  • $\begingroup$ Note: Kr- is probably irrelevant. I'm not sure if it can even exist, since noble gasses are thought to negative electron affinity, which means that it releases energy for Kr- to decay into neutral Kr and a free electron (e-). In any case, F2 or F oxidizing it would probably just result in Kr and F2(-) or F- and no chemical bond. $\endgroup$
    – Mr. Nichan
    Commented Oct 9, 2020 at 19:48
  • $\begingroup$ :/ You probably had better kept to redox potential - there's is estimated one for KrF2 when I've never seen it for ClF3 or FOOF. $\endgroup$
    – Mithoron
    Commented Feb 2, 2021 at 13:48

Fluoride was inadvertently oxidized to fluorine in an attempt to oxidize oxygen to the +4 oxidation state.

Crawford and Klapötke[1] showed through quantum mechanical calculations that the $\ce{OF3^+}$ ion is vibrationally stable and potentially accessible. They therefore attempted the reaction

$\ce{OF2 +F2 +AsF5 -> OF3^+ +AsF6^-}$

but it did not work; instead they observed

$\ce{2OF2 +AsF5 -> O2^+ +AsF6^- +(3/2)F2}.$

Oxygen difluoride would not ordinarily oxidize fluoride to fluorine. Presumably it combines with the Lewis acid $\ce{AsF5}$ to generate a stronger oxidant.


  1. Margaret-Jane Crawford, Thomas M. Klapötke (1999), "The trifluorooxonium cation, OF3+", Journal of Fluorine Chemistry, 99(2), 151-156, ISSN 0022-1139, https://doi.org/10.1016/S0022-1139(99)00139-6.

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