What is the #"oxidation number"#?

1 Answer
Apr 10, 2017

Answer:

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.