Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers, and students in the field of chemistry. It only takes a minute to sign up.
Sign up to join this community
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.
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.
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.
Yep. Fluorine is -1, oxygen 0 and hydrogen +1. The sum of all oxidation states for a neutral molecule has to equal 0.
