What is the oxidation number of oxygen in #H_2O_2#?

1 Answer
Sep 29, 2016

Answer:

The oxidation state of oxygen in peroxide is #-I#.

Explanation:

By definition, the oxidation number is the charge left on the central atom when all the bonds around it are broken, with the charge going to the most electronegative atom.

When we do this for water, #OH_2#, we get #O^(-II)# and #2xxH^(+I)#. But when we break the peroxo bond, the #O-O# bond, the charge, the two electrons in the bond, are presumed to be shared by EACH oxygen atom, in that each atom clearly has the same electronegativity:

#H-O-O-Hrarr2xxH-O*#. Now this peroxo radical clearly has a #-I# oxidation state, because the #H# is less electronegative than oxygen.

And so the take home message: treat the peroxide oxygen as a #-I# oxidation state.