oxidation no of S in Na2S4O6.

Oxidation number of sulphur in  Na2S4O6 :

When we write the formula of Na2S4O

 

 

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

  • 93
What are you looking for?