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What should be the oxidation state of $\ce{F}$ in $\ce{HOF}$.

As fluorine is the most electronegative element in the periodic table, it should be $-1$. But when I googled it, I found that many sources say it is $+1$ based on the fact that it is an oxidising agent.

Is there any theoretical way in which we can predict its oxidation no. or at least explain this anomaly.

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    $\begingroup$ Any compound involving both oxygen and fluorine is likely to be an exception to the normal rules of thumb — for everything, not just oxidation states. $\endgroup$ – zwol Aug 30 '16 at 17:54
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    $\begingroup$ Also, keep in mind that "oxidation state" of anything in a covalent molecule is merely an approximation to what is really going on. I'd expect the electron distribution in $\ce{HOF}$ to be basically that the O and the F have ganged up on the H, stolen its electron, and are sharing it between them with some, but not a lot, of polarization toward the F — on the order of $\ce{H+}$, $\ce{O^{-0.4}}$, $\ce{F^{-0.6}}$. $\endgroup$ – zwol Aug 30 '16 at 17:58
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    $\begingroup$ Related: What should be the oxidation state of oxygen in HOF (hypofluorous acid)? $\endgroup$ – zwol Aug 30 '16 at 19:55
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While I agree with the outcome of Jan's assignment of oxidation numbers, it is very necessary to point out that this is not the only possible assignment, (probably still) not even the most common one.

I also agree that sources on the Internet can be wrong, but at the time Jan was writing his answer, the most authoritative source, the IUPAC themselves, was wrong in their definition. In 2016 the official rules changed, which I have discussed in an answer to Electronegativity Considerations in Assigning Oxidation States.

Prior to 2016 the official rules of the IUPAC indeed assign $\mathbf{+1}$ to fluorine. This is counter-intuitive, as you stated, but that is the way the strict application of the rules give it to us. For me this is one of the most disturbing flaws of these rules. What you called an anomaly is indeed due to an incomplete set of rules.
I have laid out the different possible assignments in my answer to What should be the oxidation state of oxygen in HOF (hypofluorous acid)?.

Luckily this flaw has (recently) been addressed in the new more comprehensive definition. In brief the it is: The oxidation state of an atom is the charge of this atom after ionic approximation of its heteronuclear bonds.
With this you would indeed assign $-1$ to fluorine, $+1$ to hydrogen, and $0$ to oxygen.

Nevertheless, one should be aware, that there are two systems out there, which produce contradictory results. I expect the old rules to be present in popular text books for quite a while, so this answer may serve as a reference for the issue.

A general flaw of all systems assigning oxidation numbers is that there is not really a correct way of doing it. This is especially true if you try to consider fractional oxidation numbers, which are still not really covered in the new definition. Oxidation states/numbers are nothing you can measure, hence there is also no unique way of correctly predicting them. They were introduced as a tool to better understand and keep track of the changes in the charge density. They usually do not come close to measured or calculated partial charges and they often oversimplify bonding situations of molecules.

Does that mean we should abandon the concept? No. It is a very helpful tool, especially when considering redox reactions. It gives you a formal framework to track a reaction and to make simple predictions. With the newer, much more comprehensive framework, the assignment became less ambiguous and more in line with chemical intuition.

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  • $\begingroup$ Hey wait, can you give a reference where fluorine is clearly rendered in the +1 oxidation state? The one you have does NOT seem to say that! $\endgroup$ – Oscar Lanzi May 31 at 11:16
  • $\begingroup$ @OscarLanzi Is this kind of tone your comment tries to convey really necessary? I am sorry that my posts don't automatically update as soon as a rule changes, I was quoting those still in place in 2014 (see web.archive.org/web/20141102154648/http://goldbook.iupac.org/…), and considering those rules, as I have reported, the assignment is +1 for fluorine. The newer definition doesn't make this ridiculous assignment anymore, so I guess I should update all those posts, but I actually won't have time for that. $\endgroup$ – Martin - マーチン May 31 at 12:01
  • $\begingroup$ Sounds to me as if this answer is obsolete. Maybe explicitly state that the rules are changed from the original answer? $\endgroup$ – Oscar Lanzi May 31 at 13:02
  • $\begingroup$ @OscarLanzi I don't think this answer is anywhere close to obsolete, as with many outdated concepts, they will linger in textbooks for probably more than a decade to come. Additionally, I have already stated that I should update all posts that are referencing the old rules, but that will take time. Time that I am currently neither able nor willing to invest. So, it seems like you still have all the other usual tools at your disposal to make your objection heard... $\endgroup$ – Martin - マーチン May 31 at 13:38
  • $\begingroup$ @oscar hopefully you are now satisfied with this version of the answer. $\endgroup$ – Martin - マーチン yesterday
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There is no anomaly in here. Fluorine’s oxidation state in $\ce{HOF}$ is $\mathrm{-I}$ as the theory says. Hydrogen’s is $\mathrm{+I}$. This leaves oxygen with an oxidation state of $\mathrm{\pm 0}$.

That final fact is where the oxidative power comes from. Oxygen, the second-most electronegative element in the periodic table, has the same oxidation state as in molecular form. It is also bonded to an even more electronegative fluorine. Even in hydrogen peroxide oxygen would me more reduced. Thus, any reaction of $\ce{HOF}$ involves oxygen gaining electrons, not fluorine.

Remember that many sources on the internet can be wrong.

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  • $\begingroup$ thanks for the explanation. Can you show any reaction of HOF where it acts as an oxidising agent so that the oxidation states might be justified? $\endgroup$ – Amritansh Singhal Aug 30 '16 at 16:55
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    $\begingroup$ Well... the official instructions of the all-mighty IUPAC do indeed dictate an oxidation state of +1 for fluorine. Only the extended version (see here) will come to the more logical conclusion that you are reaching, which is technically not the common convention. $\endgroup$ – Martin - マーチン Aug 31 '16 at 7:13
  • $\begingroup$ @Martin-マーチン There are instances when I don’t give a damn about what IUPAC says — e.g, they also propose double bonds in sulphones. $\endgroup$ – Jan Aug 31 '16 at 8:01
  • $\begingroup$ I think I made it clear what I think about these rules, but saying that sources on the internet may be wrong is absolutely misleading here. Especially because the common convention is the one proposed by IUPAC and found in almost all textbooks. And in that set of rules, fluorine's oxidation number is +1. $\endgroup$ – Martin - マーチン Aug 31 '16 at 8:11
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    $\begingroup$ You might be happy/interested to see that the IUPAC has updated their definition... but then again, you might be completely ambivalent, because you hate them. $\endgroup$ – Martin - マーチン yesterday
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Yep. Fluorine is -1, oxygen 0 and hydrogen +1. The sum of all oxidation states for a neutral molecule has to equal 0.

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protected by andselisk May 31 at 11:12

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