Question

Oxidation state of sulphur in anions $SO^{2-}_{3}. S_{2}O^{2-}_{4}$ and $S_{2}O^{2-}_{6}$ increases in the orders:

• A. $S_{2}O^{2-}_{6}< S_{2}O^{2-}_{4}< SO^{2-}_{3}$
• B. $SO^{2-}_{6}< S_{2}O^{2-}_{4}< SO^{2-}_{6}$
• C. $S_{2}O^{2-}_{4}< SO^{2-}_{3}< S_{2}O^{2-}_{6}$
• D. $S_{2}O^{2-}_{4}< S_{2}O^{2-}_{6}< SO^{2-}_{3}$

Answer 1

Answer: (C)

In $SO_{3}^{--}$
$x+3\left(-2\right)=-2; x=+4$
In $S_{2}O_{4}$
$2x+4\left(-2\right) =-2$
$2x-8 = -2$
$2x=6; x=+3$
In $S_{2}O^{2-}_{6}$
$2x+6\left(-2\right)-2$
$2x= 10; x=+5$
hence the correct order is
$S_{2}O_{4}^{--} < SO_{3}^{--} < S_{2}O_{6}^{--}$