What is the order of the oxidation number of | vedclass

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Question

(A) To find the oxidation number of $S$ in each ion:$1$. In $SO_3^{2-}$: Let the oxidation number of $S$ be $x$. Then $x + 3(-2) = -2$,which gives $x

Vedclass · Accepted Answer

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

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

Column-$I$	Column-$II$
$A$. $Cr_2O_7^{2-}$	$1$. $+3$
$B$. $MnO_4^{-}$	$2$. $+4$
$C$. $VO_3^{-}$	$3$. $+5$
$D$. $FeF_6^{3-}$	$4$. $+6$
	$5$. $+7$

What is the order of the oxidation number of $S$ in $SO_3^{2-}$,$S_2O_4^{2-}$,and $S_2O_6^{2-}$?

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Column-$I$ Column-$II$

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