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

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

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 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##