# How can an oxidation number be determined?

Oct 8, 2016

Oxidation state/number is the charge left on the central atom, when all the bonding pairs of electrons are broken with the charge assigned to the more electronegative atom.

#### Explanation:

Redox chemistry is conceived to occur on the basis of addition OR subtraction of electrons. Typically, an element in a compound or complex (or the element itself) is assigned an oxidation number. The sum of the oxidation numbers always equals the charge on the ion.

I will consider a few simple examples: $M n {O}_{4}^{-}$, and $C {r}_{2} {O}_{7}^{2 -}$.

TO begin, elemental oxygen, ${O}_{2}$, has a ZERO oxidation number. Typically it reacts to give the oxide, ${O}^{2 -}$. We conceive that the oxygen molecule has picked up 4 electrons to give $2 \times {O}^{2 -}$. From where does it get the electrons? Well, of course from the metal, the which formally LOSES electrons upon oxidation.

Thus for $M n {O}_{4}^{-}$, the oxidation numbers of $M n$ and of $O$ must sum to $- 1$, the charge on the ion. Thus $M {n}_{\text{ON}} + 4 \times \left(- 2\right)$ $=$ $- 1$. So $M {n}_{\text{ON}}$, the oxidation number of the metal is clearly $+ V I I$ (typically we use Roman numerals to specify the oxidation number).

And likewise for $C {r}_{2} {O}_{7}^{2 -}$, $2 \times C {r}_{\text{ON}} + 7 \times \left(- 2\right) = - 2$. It does not take too much arithmetic to work out that $C {r}_{\text{ON}} = + V I$.