What is the correct order of oxidation states | vedclass

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(B) For $SO_3^{2-}$: Let the oxidation state of $S$ be $x$. Then $x + 3(-2) = -2$,so $x = +4$.For $S_2O_4^{2-}$: Let the oxidation state of $S$ be $x$

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$S_2O_4^{2-} < SO_3^{2-} < S_2O_6^{2-}$

Vedclass · Answer

(B) For $SO_3^{2-}$: Let the oxidation state of $S$ be $x$. Then $x + 3(-2) = -2$,so $x = +4$.For $S_2O_4^{2-}$: Let the oxidation state of $S$ be $x$. Then $2x + 4(-2) = -2$,so $2x = +6$,$x = +3$.For $S_2O_6^{2-}$: Let the oxidation state of $S$ be $x$. Then $2x + 6(-2) = -2$,so $2x = +10$,$x = +5$.The oxidation states are $+4$,$+3$,and $+5$ respectively.Thus,the correct order is $S_2O_4^{2-} (+3) < SO_3^{2-} (+4) < S_2O_6^{2-} (+5)$.

What is the correct order of oxidation states of sulfur in the following anions: $SO_3^{2-}$,$S_2O_4^{2-}$,and $S_2O_6^{2-}$?

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