The oxidation states of sulphur in the anions | vedclass

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(B) To find the oxidation state of sulphur $(S)$ in each anion,we use the rule that the sum of oxidation states equals the charge of the ion.$1$. For

Vedclass · Accepted Answer

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

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(B) To find the oxidation state of sulphur $(S)$ in each anion,we use the rule that the sum of oxidation states equals the charge of the ion.$1$. For $S_2O_4^{2-}$: $2x + 4(-2) = -2 \implies 2x - 8 = -2 \implies 2x = +6 \implies x = +3$.$2$. For $SO_3^{2-}$: $x + 3(-2) = -2 \implies x - 6 = -2 \implies x = +4$.$3$. For $S_2O_6^{2-}$: $2x + 6(-2) = -2 \implies 2x - 12 = -2 \implies 2x = +10 \implies x = +5$.Comparing the values: $+3 Therefore,the order is $S_2O_4^{2-} < SO_3^{2-} < S_2O_6^{2-}$.

The oxidation states of sulphur in the anions $SO_3^{2-}$,$S_2O_4^{2-}$ and $S_2O_6^{2-}$ follow the order:

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