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What should be the oxidation state of oxygen in $\ce{HOF}$ (hypofluorous acid)?

Sources on the internet have confused me. Most state its oxidation state to be 0, while the others state that it is -2 (considering the oxidation state of fluorine to be +1, which is a bit surprising for me as it's the most electronegative element in the periodic table)


Source stating the oxidation state to be -2:

J. Chem. Educ. 1972, 49 (4), 299:
The two possibilities brought forth in the Note are (1) assign hydrogen +1, oxygen -2, and fluorine +1 or (2) assign H +1, oxy- gen 0, and fluorine -1. On the grounds that $\ce{HOF}$ is a powerful oxidizing agent, the conclusion reached in the Note is that (1) should be favored.

Other sources stating the oxidation state to be 0:

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Until the (recent) redefinition of the IUPAC, the concept of oxidation states was not as well defined as one would expect. I have discussed the issues of the old version and outlined the new definition in more detail in an answer to Electronegativity Considerations in Assigning Oxidation States.

When you apply the official pre-2016 definition (via the Internet Archive) from the IUPAC gold book, then you have to assign $\text{OS}(\ce{H}) = +1$ as it is not a compound with a metal, $\text{OS}(\ce{O}) = -2$ as it is no peroxide compound, and leaving the disturbing $\text{OS}(\ce{F})= \color\red{+1}$.

Going with the on electronegativity based alternative description, you will still assign $\text{OS}(\ce{H}) = +1$ as it has the lowest electronegativity. Then you assign $\text{OS}(\ce{F})= -1$, because of the highest electronegativity. This leaves a oxidation state of $\text{OS}(\ce{O}) = 0$.

This assignment is matched by the 2016 definition, the summary version of which is: The oxidation state of an atom is the charge of this atom after ionic approximation of its heteronuclear bonds. The definition refers to the use of Allen electronegativities (see Pure Appl. Chem. 2016, 88 (8), 831–839 for more detail). Consequently the heteronuclear approximation yields:

heteronuclear approximation of HOF with oxidation states

The true charges are of course something completely different. Based on a DF-BP86/def2-SVP calculation I ran some population analyses with Multiwfn 3.4.1 (a newer version is available). The NBO charges are taken from a prior version of this answer, but I since have lost access to the program and the files used.

\begin{array}{lrrr}\hline \text{Method} & \ce{H} & \ce{O} & \ce{F} \\\hline \text{Hirshfeld} & +0.18 & -0.10 & -0.09 \\ \text{ADCH} & +0.38 & -0.29 & -0.09 \\ \text{VDD} & +0.18 & -0.10 & -0.09 \\ \text{Mulliken} & +0.20 & -0.06 & -0.15 \\ \text{Löwdin} & +0.09 & +0.01 & -0.10 \\ \text{Becke} & +0.38 & -0.21 & -0.17 \\ \text{ADC Becke} & +0.38 & -0.29 & -0.09 \\ \text{CM5} & +0.35 & -0.26 & -0.08 \\\hline \text{QTAIM} & +0.61 & -0.44 & -0.17 \\\hline \text{NBO} & +0.45 & -0.32 & -0.12 \\\hline \end{array}

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  • $\begingroup$ So what conclusion can be made from this? Do they show fractional OS? $\endgroup$ – Saharsh Aug 25 '14 at 13:19
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    $\begingroup$ @Alpha The lesson to be learned here is, that oxidation states are hardly telling the truth, but they can be a helpful concept in understanding reactions. Unfortunately there is a very thin line between using and overusing a concept. Personally I am not a big fan of fractional oxidation states and would always go for the electronegativity argument. $\endgroup$ – Martin - マーチン Aug 25 '14 at 15:50
  • $\begingroup$ So can we deduce anything about oxidation states based on the NBO partial charges? $\endgroup$ – Jan Aug 30 '16 at 21:35
  • $\begingroup$ @Jan No. I find it hard to deduce an oxidation state from any charge, but then again I do not put much stock into oxidation states in the first place. $\endgroup$ – Martin - マーチン Aug 31 '16 at 4:51
  • $\begingroup$ @Martin Your answer is helpful. However, are those partial charges a representation of the actual distribution of charge within the molecule? If so, how come oxygen has more negative charge than fluorine? $\endgroup$ – ghosts_in_the_code Jan 4 '17 at 16:03
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$\ce{F}$ is the most electronegative element, so it is always -1 oxidation state, except in $\ce{F2}$, where it is zero. So, in $\ce{HOF}$, $\ce{H}$ has +1, $\ce{O}$ has zero and $\ce{F}$ has -1 oxidation states. There is no way $\ce{F}$ will ever have oxidation state +1. In $\ce{OF2}$ (oxygen fluoride, and not fluorine oxide) $\ce{O}$ has +2 oxidation state!

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  • $\begingroup$ Hi, welcome to our site! "There is no way F will ever have oxidation state +1. " I understand your sentiment, but did you have a look at the accepted answer? It clearly shows how F can have a OS of +1. Can you please explain by what reasoning have you contradicted it? $\endgroup$ – Gaurang Tandon Jun 7 '18 at 4:01
  • $\begingroup$ Please see my answer for a more detailed analysis. While I agree with your process, it is unfortunately not the one that is guided by the common definitions. I find this contradiction quite spurious and I am dissatisfied with the current status quo. It is true that fluorine should never have a positive oxidation state, and for $\ce{OF2}$ this is even more ridiculous. Unfortunately any attempts to change the definition have been unfruitful so far. $\endgroup$ – Martin - マーチン Dec 10 '18 at 10:08
  • $\begingroup$ IUPAC is right for normal compounds. It simply fail for the case when O and F have a direct bond. $\endgroup$ – Ionel Mugurel Ciobîcă Dec 11 '18 at 19:16
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The oxidation state in my opinion will be zero.Because as flourine is the most electronegative element it will never show positive ooxidation state.It's maximum oxidation state is zero.

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    $\begingroup$ Could you elaborate on this a bit more? At this point, there is no source for your answer and it doesn't give much more information than is already presented in Martin's answer. @MohammedSafwanA $\endgroup$ – Tyberius May 16 '18 at 14:29
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$\ce{H}$ is less electronegative than $\ce{O}$, so you pretend the electrons in the $\ce{H-O}$ bond "belong" to $\pu{O}$; $\ce{H}$ is $+1$. By similar reasoning, $\ce{F}$ is $-1$.
You can work out the oxidation number of $\ce{O}$, either by counting electrons, or by using the fact that adding up all the oxidation numbers of any chemical species has to give you the charge of that species, which in this case of course is $0$.

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    $\begingroup$ Please see my answer for a more detailed analysis. While I agree with your process, it is unfortunately not the one that is guided by the common definitions. $\endgroup$ – Martin - マーチン Dec 10 '18 at 10:05

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