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

##### 1 Answer
Dec 2, 2016

Well, in their compounds it is usually $- I I \left(O\right)$ and $+ I \left(H\right)$. See here for other examples.

#### Explanation:

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

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

In metal hydrides, $M H$, $M {H}_{2}$, the oxidation of hydrogen is $- I$ (because here it is the more electronegative element and gets the electron).

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

Of course for $\text{elemental hydrogen}$, ${H}_{2}$, and for $\text{elemental oxygen}$, ${O}_{2}$, 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 $\text{reduced}$ and the hydrogen is $\text{oxidized}$, electron transfer is conceived to have occurred, and the two elements are now assigned their normal oxidation numbers, which are?

${H}_{2} \left(g\right) + \frac{1}{2} {O}_{2} \left(g\right) \rightarrow {H}_{2} O \left(l\right)$