# How do you find an oxidation number of an element in a compound?

Oct 15, 2016

$\text{The oxidation number of an element in a compound.........}$

#### Explanation:

The oxidation state of an element in a compound is the charge left on the atom of interest when all the bonding pairs of electrons are formally broken, and the charged distributed to the most electronegative atom.

This definition is, I grant, a mouthful, but let's see how it is applied. We start with the water molecule, $O {H}_{2}$, which of course is neutral. The $O - H$ bond is of course composed of 2 electrons (this goes back to the idea of covalent bonds as the result of the sharing of 2 electrons between adjacent atoms), and when we break this bond, we are left with ${O}^{2 -}$ and $2 \times {H}^{+}$ (because oxygen is more electronegative than hydrogen, the 2 electrons in the bond devolve to oxygen). Thus given the definition, the oxidation state of $O$ $=$ $- I I$, and that of $H$ $=$ $+ I$.

Elements, of course, are considered to have $0$ oxidation states, because in their elemental state they neither accepted not donated electrons: cf oxidation of carbon:

$C \left(s\right) + {O}_{2} \left(g\right) \rightarrow C {O}_{2} \left(g\right)$

Elemental carbon is conceived to have been oxidized, i.e. it has lost 4 electrons to give $C \left(I V +\right)$, and dioxygen has accepted those same electrons to give $2 \times \left(- I I\right)$.

Now of course, these ideas of electron loss and electron gain are formalisms in that they have no actual reality, but often they allow us to balance redox reactions. I can give you a few ions with the oxidation numbers assigned: $C {O}_{3}^{2 -} , C \left(+ I V\right)$; $S {O}_{3}^{2 -} , S \left(+ I V\right)$; $S {O}_{4}^{2 -} , S \left(+ V I\right)$; $C l {O}_{4}^{-} , C l \left(+ V I I\right)$; $C l {O}_{3}^{-} , C l \left(+ V\right)$.

Anyway, I hope this spray hasn't managed to confuse you totally. Remember, these are simple ideas, and they use very simple arithmetic.