Can someone please double check if I have oxidised the number of sulfur correctly for the following ions? The question is in the description box below.

Question

My answers are #S_20^2-(_4) and s=4+
and the other one s=3

1058 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-24T05:07:31

Your sulfur-based polyatomic anions are:

#"S"_2"O"_4^(2-)#

#"S"_2"O"_3^(2-)#

Dithionite has four oxygens, each of the typical oxidation state of #-2# (as this is not a peroxide), so by conservation of charge...

#overbrace(2xxO_S)^(S_2) + overbrace(4 xx -2)^(O_4) = overbrace(-2)^"charge"#

and each sulfur has on average, #color(blue)(O_S = +3)#.

Thiosulfate has three oxygens, each of the typical oxidation state of #-2# (as this is not a peroxide), so by conservation of charge...

#overbrace(2xxO_S)^(S_2) + overbrace(3 xx -2)^(O_3) = overbrace(-2)^"charge"#

and each sulfur has on average, #color(blue)(O_S = +2)#.

In reality this is not true; the second one is a sulfate analog, so we distinguish between the central sulfur and the outer sulfur...

#"S"_2"O"_3^(2-)# then has a central sulfur oxidation state of #+4# and outer sulfur oxidation state of #-2# (just like each oxygen atom).