# How is the oxidation number of a monoatomic ion related to its charge?

Nov 14, 2016

How? Well, the oxidation number of a mono-atomic ion is EQUAL to its charge.

#### Explanation:

$\text{Oxidation number}$ is the charge left on an atom when all the bonding pairs are broken, with the charge assigned to the most electronegative atom. Note that this is an hypothetical exercise, but it can be useful in certain circumstances.

When oxygen undergoes reduction, we would write:

$\frac{1}{2} {O}_{2} + 2 {e}^{-} \rightarrow {O}^{2 -}$ $\left(i\right)$

The oxygen atom has undergone reduction by accepting 2 electrons to give the oxide anion. From where do the electrons come? Well typically a metal:

$N a \rightarrow N {a}^{+} + {e}^{-}$ $\left(i i\right)$

To write the overall redox equation, we sum the oxidation and reduction equations, weighting one of the half equations so that electrons are removed:

$\left(i\right) + 2 \times \left(i i\right) =$

$2 N a \left(s\right) + \frac{1}{2} {O}_{2} \left(g\right) \rightarrow N {a}_{2} O \left(s\right)$

Elements thus have a formal oxidation state of $0$, and monoatomic ions have an oxidation number, positive or negative, equal to the charge of the ion. Capisce?