electronegativity - What should be the oxidation state of oxygen in HOF (hypofluorous acid)?

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)

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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:

Answer

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:

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}