What are the oxidation numbers of S atoms in | vedclass

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(B) The structure of the tetrathionate ion $(S_4O_6^{2-})$ consists of a chain of four sulphur atoms. In this structure,the two central sulphur atoms

Vedclass · Accepted Answer

$5, 0, 0, 5$

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(B) The structure of the tetrathionate ion $(S_4O_6^{2-})$ consists of a chain of four sulphur atoms. In this structure,the two central sulphur atoms are bonded to each other and to other sulphur atoms,so their oxidation state is $0$. The two terminal sulphur atoms are each bonded to three oxygen atoms (two via double bonds and one via a single bond) and one sulphur atom. For each terminal sulphur atom: $x + 2(-2) + 1(-1) = -1$ (since the terminal group is $-SO_3^-$),which simplifies to $x - 5 = -1$,so $x = +5$. Thus,the oxidation numbers of the four $S$ atoms are $+5, 0, 0, +5$.

Column-$I$	Column-$II$
$A$. $\underline{N}H_4 Cl$	$P$. $+1$
$B$. $H\underline{N}O_3$	$Q$. $+5$
$C$. $\underline{N}_2 O$	$R$. $-3$
$D$. $\underline{N}H_2 OH$	$S$. $-1$

What are the oxidation numbers of $S$ atoms in $S_4O_6^{2-}$?

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Column-$I$ Column-$II$
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