Question

Give the oxidation state of underlined.
$H_2\underset{–}{S_2}{O}_8$.

Answer 1

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:

$2\times (+1)+2\times \left(x\right)+8\times (-2)=0⟹2+2x-16=0⟹2x=16-2⟹x=\frac{14}{2}=+7$

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.

$2\times (+1)+2\times (+6)+n\times (-1)-(n-8)\times (-2)=0⟹2+12-n+2n-16=0⟹n=16-12-2⟹n=+2$

Hence, there is one peroxy linkage between 2 oxygen atoms. Therefore, the oxidation state of sulphur is +6