What is the sulfur oxidation state in #SO_3^(2-),SO_4^(2-),S_2O_3^(2-)#?

1 Answer
Nov 3, 2017

Do you mean #SO_3^(2-)#, the #"sulfite ion"#?

Explanation:

I make it #S(+IV)# for sulfite.... How? Well the sum of the individual oxidation numbers equals the charge on the ion. And thus for #"sulfate"#, #SO_4^(2-)#, we got #4xx(-II)+S_"oxidation number"=-2#, and thus #S_"oxidation number"=+VI#. We note that oxygen generally assumes an oxidation number of #-II# in its compounds, and it does so here...

For #SO_3^(2-)#, the same procedure gives #S(+IV)#.

For #S_2O_3^(2-)#, #"thiosulfate"#, I like to think that sulfur has replaced ONE of the oxygens of #"SULFATE ion"#, and assumed its #-II# oxidation state, and thus we got #S(-II)# and #S(+VI)#, i.e an average oxidation number of #S(+II)#. Capisce?