# What is the oxidation number of each hydrogen atom in a compound usually?

Jul 18, 2016

$+ 1$

#### Explanation:

The thing to remember about oxidation numbers is that they are assigned with the electronegativity of the atoms in mind.

In order to assign an oxidation number to an atom bonded to another atom, you must assume that the more electronegative of the two atoms "takes" all the bonding electrons.

Hydrogen, as you know, has a single electron surrounding its nucleus, which consequently acts as its sole valence electron.

As a result, hydrogen can only form single bonds with other atoms because it can only share one electron. Therefore, any atom bonded to hydrogen that is more electronegative than hydrogen will "take" this valence electron for itself.

This will give hydrogen a $\textcolor{b l u e}{+ 1}$ oxidation number, since it only "lost" $1$ electron to the more electronegative atom.

Keep in mind that when hydrogen is bonded to less electronegative atoms, such as in metal hydrides, it "takes" the bonding electron shared by the other atom for itself.

This gives it a $\textcolor{b l u e}{- 1}$ oxidation number.

Take, for example, hydrogen in water, $\text{H"_2"O}$. The two hydrogen atoms are bonded to an oxygen atom, which is significantly more electronegative than hydrogen.

This means that each of the two hydrogen atoms will have a $\textcolor{b l u e}{+ 1}$ oxidation number, since they each "lose" their bonding electron. Consequently, oxygen will have a $\textcolor{b l u e}{- 2}$ oxidation number.

$\stackrel{\textcolor{b l u e}{+ 1}}{\text{H")_ 2 stackrel(color(blue)(-2))("O}}$

In sodium hydride, on the other hand, hydrogen is more electronegative than sodium, so it will have a $\textcolor{b l u e}{- 1}$ oxidation number. Consequently, sodium will have a $\textcolor{b l u e}{+ 1}$ oxidation number.

$\stackrel{\textcolor{b l u e}{+ 1}}{\text{Na") stackrel(color(blue)(-1))("H}}$

The most common oxidation number of hydrogen will be $\textcolor{b l u e}{+ 1}$, since that's the oxidation state it has in most of the compounds it forms.