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Question
Calculate the oxidation number of sulphur atom in the following compounds:
\[\ce{Na2S4O6}\]
Solution
Let x= oxidation number of sulphur, and +1 is oxidation number of \[\ce{Na}\], –2 is oxidation number of \[\ce{O}\], also we can assume total charge on compound = 0 then solving we get.
+2 + 4x – 12 = 0
x = +2.5
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