# What is the "oxidation number"?

Apr 10, 2017

Oxidation number is the charge left on the central atom, when all the bonding pairs of electrons are removed, with the charge assigned to the most electronegative atom.........

#### Explanation:

And you have forgotten the oxidation state of the oxygen we are breathing in now, ${O}_{2}$, which of course has an oxidation number of $0$.

Oxygen is in fact more electronegative than most elements, and thus for many compounds, e.g. ${H}_{2} O$, metal oxides, $M O$, and ${M}_{y} {O}_{z}$, the oxygen is typically assigned a formal oxidation number of $- I I$. For $O {F}_{2}$, a real gas, oxygen is formally assigned an oxidation number of $+ I I$. Why? Because of course, fluorine is MORE electronegative than oxygen, and so is assigned the electrons, and the negative charge.

In peroxides, e.g. $H O - O H$, the electrons in the $O - O$ bond are conceived to be shared, i.e when we break the bond and of course this is a conceptual exercise, we get: ${H}_{2} {O}_{2} \rightarrow 2 \times H O \cdot$, i.e. ${O}^{-} I$. For $F O - O F$, we get ${O}^{+ I}$. This is $\text{dioxygen difluoride}$, and its name is not a misnomer; it is as reactive as buggery; things treated with it go $\text{foof}$ with alacrity. $O {F}_{2}$ is known (and a little bit more stable than the former) and here we have ${O}^{+ I I}$. I would not touch either reagent on my current paygrade.

So why? Well, remember that our ideas of oxidation numbers are a conceptual label of convenience. They are a useful way of understanding how elements combine, especially in regard to redox reactions, but they do not have fundamental significance.