What is the usual oxidation number of oxygen? Of hydrogen?

1 Answer
Dec 2, 2016

Answer:

Well, in their compounds it is usually #-II(O)# and #+I(H)#. See here for other examples.

Explanation:

The oxidation number is the charge left on the atom of interest when we conceptually break the #"element-element"# bonds with the charge (the electrons!) assigned to the most electronegative atom.

Do this for water #H-O-H#, and we get #H^+ + HO^-#; do this again for hydroxide, we get #H^+ + O^(2-)#. And thus the oxidation numbers for hydrogen and oxygen in water are #+I# and #-II#. I reiterate that this is an entirely conceptual exercise. The rigmarole can sometimes help to balance redox equations.

In metal hydrides, #MH#, #MH_2#, the oxidation of hydrogen is #-I# (because here it is the more electronegative element and gets the electron).

In peroxides, i.e. #HO-OH#, because the electrons are conceived to be shared in the peroxo, #O-O#, linkage, the oxidation number of #O# is #-I#.

Of course for #"elemental hydrogen"#, #H_2#, and for #"elemental oxygen"#, #O_2#, neither element has accepted nor donated electrons. The oxidation state of an element is a big fat #0#.

When these 2 elements react together, when the oxygen is #"reduced"# and the hydrogen is #"oxidized"#, electron transfer is conceived to have occurred, and the two elements are now assigned their normal oxidation numbers, which are?

#H_2(g) + 1/2O_2(g) rarr H_2O(l)#