Why is the oxidation number of oxygen in #O_2F_2# #+1#?

1 Answer
Apr 18, 2017

Answer:

Because oxidation number is the charge the atom in a molecule would have...........

Explanation:

........would have if the bonding electrons were distributed to the MOST electronegative atoms. Fluorine is MORE electronegative than oxygen (in fact fluorine is the most electronegative element on the Table, and the most reactive).

So when we do this for #"FOOF"# (so-named because of its EXTREME reactivity). We get a formal oxidation states of #""stackrel(-I)F-stackrel(+I)O-stackrel(+I)O-stackrel(-I)F#.

What is the oxygen oxidation state in #OF_2#? Is this usual?