oxidation no of S in Na2S4O6.
Oxidation number of sulphur in Na2S4O6 :
When we write the formula of Na2S4O6
Two S atoms namely S2 and S3 are joined together, and with two other S atoms named S1 and S4. Therefore their oxidation state will be zero. S atoms namely S1 and S4 have oxidation number +5 each.
Hence resultant oxidation number of sulphur in Na2S4O6 :
(0 + 0 + 5 + 5) / 4 = 2.5
So, oxidation state of S2 and S3 = 0
and oxidation state of S1 and S4 = +5 each
So, resultant O.N. of S = 2.5