Question

What is the `"oxidation number"`?

Answer 1

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_2O`, metal oxides, `MO`, and `M_(y)O_z`, the oxygen is typically assigned a formal oxidation number of `-II`. For `OF_2`, a real gas, oxygen is formally assigned an oxidation number of `+II`. Why? Because of course, fluorine is MORE electronegative than oxygen, and so is assigned the electrons, and the negative charge.

In peroxides, e.g. `HO-OH`, 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_2O_2rarr2xxHO*`, i.e. `O^-I`. For `FO-OF`, we get `O^(+I)`. This is `"dioxygen difluoride"`, and its name is not a misnomer; it is as reactive as buggery; things treated with it go `"foof"` with alacrity. `OF_2` is known (and a little bit more stable than the former) and here we have `O^(+II)`. 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.