The oxidation states of S in Peroxodisulphuri | vedclass

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Question

(A) $1$. In Peroxodisulphuric acid $(H_2S_2O_8)$,there is a peroxide linkage $(-O-O-)$. Let the oxidation state of $S$ be $x$. The structure is $HO_3S

Vedclass · Accepted Answer

$+6, +2.5$

Vedclass · Answer

(A) $1$. In Peroxodisulphuric acid $(H_2S_2O_8)$,there is a peroxide linkage $(-O-O-)$. Let the oxidation state of $S$ be $x$. The structure is $HO_3S-O-O-SO_3H$. The oxidation state of $S$ is calculated as: $2(1) + 2(x) + 6(-2) + 2(-1) = 0$,which gives $2 + 2x - 12 - 2 = 0$,so $2x = 12$,$x = +6$.$2$. In Sodium tetrathionate $(Na_2S_4O_6)$,the structure is $Na^+O_3S-S-S-SO_3^-Na^+$. The two terminal $S$ atoms are bonded to three oxygen atoms and one sulfur atom,having an oxidation state of $+5$. The two central $S$ atoms are bonded to two other sulfur atoms,having an oxidation state of $0$. The average oxidation state is $(2 	imes 5 + 2 	imes 0) / 4 = +2.5$. However,the question asks for the oxidation states present. The values are $+5$ and $0$ for the atoms,or $+2.5$ as an average. Given the options,the most appropriate representation for the set is $+6$ and $+2.5$.

The oxidation states of $S$ in Peroxodisulphuric acid $(H_2S_2O_8)$ and Sodium tetrathionate $(Na_2S_4O_6)$ are respectively:

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