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The difference in the oxidation numbers of the two types of sulphur atoms in Na2S4O6 is:
- 5
- 6
- 4
- 8
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The oxidation state of the first sulphur: Sulphur is attached to the same sulphur atoms hence there is no change in the electronic arrangement of this sulphur atom, therefore, its O.S. is zero.
The oxidation state of second sulphur: This atom of sulphur is attached to the three oxygen atoms two are doubly bonded and one is singly bonded
Therefore, for doubly bonded oxygen will exert the charge +2 and for singly bonded oxygen will +1. hence on the total, the O.S. of Second sulphur is +5.
S will have oxidation number =+5,0
The difference in oxidation number =5
The oxidation state of the first sulphur: Sulphur is attached to the same sulphur atoms hence there is no change in the electronic arrangement of this sulphur atom, therefore, its O.S. is zero.
The oxidation state of second sulphur: This atom of sulphur is attached to the three oxygen atoms two are doubly bonded and one is singly bonded
Therefore, for doubly bonded oxygen will exert the charge +2 and for singly bonded oxygen will +1. hence on the total, the O.S. of Second sulphur is +5.
The difference in oxidation number =5
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