As fluorine is the most electronegative element in the routine table, it need to be $-1$. But when ns googled it, I found that many sources to speak it is $+1$ based on the truth that that is one oxidising agent.

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Is there any theoretical method in which we can predict its oxidation no. Or in ~ least explain this anomaly.



There is no anomaly in here. Fluorine’s oxidation state in $\ceHOF$ is $\mathrm-I$ together the concept says. Hydrogen’s is $\mathrm+I$. This pipeline oxygen v an oxidation state the $\mathrm\pm 0$.

That final truth is wherein the oxidative power comes from. Oxygen, the second-most electronegative facet in the periodic table, has the exact same oxidation state as in molecular form. That is also bonded come an even much more electronegative fluorine. Also in hydrogen peroxide oxygen would certainly me more reduced. Thus, any type of reaction that $\ceHOF$ entails oxygen acquiring electrons, not fluorine.

Remember that numerous sources on the internet can be wrong.

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answer Aug 30 "16 in ~ 9:40

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While ns agree with the result of Jan"s assignment the oxidation numbers, it is really necessary to point out the this is no the only feasible assignment, (probably still) not also the most typical one.

I additionally agree that sources on the Internet can be wrong, however at the time Jan was creating his answer, the many authoritative source, the IUPAC themselves, to be wrong in your definition. In 2016 the main rules changed, which ns have disputed in response to Electronegativity Considerations in Assigning Oxidation States.

Prior to 2016 the main rules the the IUPAC without doubt assign $\mathbf+1$ to fluorine. This is counter-intuitive, together you stated, yet that is the means the strict application of the rules offer it to us. Because that me this is just one of the many disturbing flaws of this rules. What you dubbed an anomaly is indeed as result of an incomplete collection of rules.I have laid the end the different possible assignments in mine answer to What must be the oxidation state of oxygen in HOF (hypofluorous acid)?.

Luckily this flaw has (recently) been addressed in the new much more comprehensive definition. In quick the it is: The oxidation state of one atom is the fee of this atom ~ ionic approximation of its heteronuclear bonds.With this you would certainly assign $-1$ to fluorine, $+1$ to hydrogen, and $0$ come oxygen.

Nevertheless, one must be aware, that there room two equipment out there, i m sorry produce inconsistent results. I mean the old rule to be present in renowned text books for rather a while, for this reason this answer may serve together a referral for the issue.

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A basic flaw of all solution assigning oxidation numbers is that there is no really a correct way of doing it. This is especially true if you try to think about fractional oxidation numbers, which space still not really covered in the brand-new definition. Oxidation states/numbers are nothing you deserve to measure, thus there is also no unique method of effectively predicting them. Lock were introduced as a device to better understand and keep track of the transforms in the fee density. Lock usually do not come close to measured or calculation partial charges and they regularly oversimplify bonding cases of molecules.

Does that mean we must abandon the concept? No. The is a really helpful tool, specifically when considering oxidization reactions. It offers you a formal structure to monitor a reaction and also to make simple predictions. Through the newer, much an ext comprehensive framework, the assignment came to be less pass out and more in line through chemical intuition.