# What are the oxidation numbers of S and O in the ion S2O3(2-)?

Oct 23, 2015

Oxygen would have an oxidation state of $- 2$, therefore sulfur would have an oxidation state of $+ 2$.

#### Explanation:

Let me explain:

So you have the whole compound that has a total charge of $\left(2 -\right)$. This means everything in the compound will have to 'add' up to $- 2$.

Break down the elements in the compound:

Oxygen's normal oxidation number is $- 2$. Because you have three oxygen atoms, the oxidation number is now

$- 2 \times \text{3 oxygen atoms} = - 6$

Remember, the whole compound is $- 2$, so we have to get the charge from $- 6$ up to $- 2$.

Sulfur's normal oxidation number in this case would be $+ 2$. There are two sulfur atoms so the number is now

$+ 2 \times \text{2 sulfur atoms} = + 4$

It's perfect.

${\overbrace{- 6}}^{\textcolor{b l u e}{\text{the oxygen atoms")) + underbrace((+4))_(color(red)("the sulfur atoms}}} = - 2$