What is the correct order of the oxidation st | vedclass

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(B) $1$. For $S_2O_4^{2-}$: Let the oxidation state of $S$ be $x$. $2x + 4(-2) = -2 \implies 2x - 8 = -2 \implies 2x = 6 \implies x = +3$.$2$. For $SO

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

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(B) $1$. For $S_2O_4^{2-}$: Let the oxidation state of $S$ be $x$. $2x + 4(-2) = -2 \implies 2x - 8 = -2 \implies 2x = 6 \implies x = +3$.$2$. For $SO_3^{2-}$: Let the oxidation state of $S$ be $x$. $x + 3(-2) = -2 \implies x - 6 = -2 \implies x = +4$.$3$. For $S_2O_6^{2-}$: Let the oxidation state of $S$ be $x$. $2x + 6(-2) = -2 \implies 2x - 12 = -2 \implies 2x = 10 \implies x = +5$.$4$. Comparing the oxidation states: $+3 $5$. Therefore,the correct order is $S_2O_4^{2-} < SO_3^{2-} < S_2O_6^{2-}$.

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

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