Give the oxidation state of underlined. H2S2–––O8.
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Solution
Let the oxidation number of S in ${ H }_{ 2 }{ S }_{ 2 }{ O }_{ 8 }$ be x. The oxidation states of H and O are +1 and -2 respectively. Since the atom as whole is neutral, following equation holds true:
2×(+1)+2×(x)+8×(−2)=0⟹2+2x−16=0⟹2x=16−2⟹x=142=+7
But for sulphur the oxidation state can not be greater than +6. Therefore there is peroxy linkage in the compound. As we know that peroxy oxygen has an oxidation state of -1. Whenever, there is peroxy linkage, sulphur is always in its highest oxidation state +6. Let n be the number of peroxy linkage.