# How do you calculate the oxidation number of an ion?

##### 1 Answer
Aug 19, 2016

For simple ions, i.e. $N {a}^{+}$, $C {l}^{-}$, $F {e}^{2 +}$, $F {e}^{3 +}$, the oxidation number of the element is simply the charge on the ion.

#### Explanation:

For more complex ions, the SUM of the oxidation numbers equals the charge on the ion. The sum of the oxidation numbers also equals the charge on the molecule in a NEUTRAL molecule.

Let's consider $C {l}^{-}$, $C l {O}_{2}$, $C l {O}_{4}^{-}$. Typically the oxidation number of oxygen in its oxides is $- I I$, and it is here in these examples.

Given what I have said the oxidation number of $C l$ in chloride ion, $C {l}^{-}$, is $- I$. The oxidation number of $C l$ in $C l {O}_{2}$ is $+ I V$, and the oxidation number of $C l$ in perchlorate, $C l {O}_{4}^{-}$, is $+ V I I$.

I don't what level you are at. If you are at A level, you simply have to know what I wrote in the opening statement. If you are an undergrad you do have to know how to assign oxidation states in more complicated molecules and ions. Good luck.