Oxidation state of sulphur in anions $\mathrm{SO}_{3}^{2-}, \mathrm{S}_{2} \mathrm{O}_{4}^{2-}$ and $\mathrm{S}_{2} \mathrm{O}_{6}^{2-}$ increases in the orders:

Question

Aicharya · Accepted Answer

In $\mathrm{SO}_{3}$
$x+3(-2)=-2 ; x=+4$
In $\mathrm{S}_{2} \mathrm{O}_{4}^{--}$
$2 x+4(-2)=-2$
$2 x-8=-2$
$2 x=6 ; \quad x=+3$
In $\mathrm{S}_{2} \mathrm{O}_{6}^{2-}$
$2 x+6(-2)=-2$
$2 x=10 ; \quad x=+5$
hence the correct order is $\mathrm{S}_{2} \mathrm{O}_{4}^{--} < \mathrm{SO}_{3}^{--} < \mathrm{S}_{2} \mathrm{O}_{6}^{-}$ *

Oxidation state of sulphur in anions $\mathrm{SO}_{3}^{2-}, \mathrm{S}_{2} \mathrm{O}_{4}^{2-}$ and…

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