What is the correct order of the oxidation st | vedclass

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(B) The oxidation state of $S$ is calculated as follows:For $SO_3^{2-}$: $x + 3(-2) = -2 \implies x = +4$For ${S_2}O_4^{2-}$: $2x + 4(-2) = -2 \implie

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${S_2}O_4^{2-} < SO_3^{2-} < {S_2}O_6^{2-}$

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(B) The oxidation state of $S$ is calculated as follows:For $SO_3^{2-}$: $x + 3(-2) = -2 \implies x = +4$For ${S_2}O_4^{2-}$: $2x + 4(-2) = -2 \implies 2x = +6 \implies x = +3$For ${S_2}O_6^{2-}$: $2x + 6(-2) = -2 \implies 2x = +10 \implies x = +5$Comparing the oxidation states: $+3 Thus,the order is ${S_2}O_4^{2-} < SO_3^{2-} < {S_2}O_6^{2-}$.

What is the correct order of the oxidation state of sulfur in the anions $SO_3^{2-}$,${S_2}O_4^{2-}$,and ${S_2}O_6^{2-}$?

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