Question

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

Answer 1

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 `(2-)`. 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 xx "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 xx "2 sulfur atoms" = +4`

It's perfect.

`overbrace(-6)^(color(blue)("the oxygen atoms")) + underbrace((+4))_(color(red)("the sulfur atoms"))= -2`