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

1 Answer
Oct 23, 2015

Answer:

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#