Arrange the following molecules in the increa | vedclass

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Question

(A) $1$. For $SO_3^{2-}$: Let the oxidation number of $S$ be $x$. Then $x + 3(-2) = -2$,which gives $x - 6 = -2$,so $x = +4$. $2$. For $S_2O_4^{2-}$:

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) $1$. For $SO_3^{2-}$: Let the oxidation number of $S$ be $x$. Then $x + 3(-2) = -2$,which gives $x - 6 = -2$,so $x = +4$. $2$. For $S_2O_4^{2-}$: Let the oxidation number of $S$ be $y$. Then $2y + 4(-2) = -2$,which gives $2y - 8 = -2$,so $2y = +6$,$y = +3$. $3$. For $S_2O_6^{2-}$: Let the oxidation number of $S$ be $z$. Then $2z + 6(-2) = -2$,which gives $2z - 12 = -2$,so $2z = +10$,$z = +5$. $4$. Comparing the values: $+3 $5$. Therefore,the increasing order is $S_2O_4^{2-} < SO_3^{2-} < S_2O_6^{2-}$.

Arrange the following molecules in the increasing order of the oxidation number of $S$: $SO_3^{2-}$,$S_2O_4^{2-}$,and $S_2O_6^{2-}$.

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