Question

The oxidation state of sulphur in the anions $SO_3^{2-}$, $S_2O_4^{2-}$ and $S_2O^{2-}_6$ follows the order

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

Answer 1

Answer: (B)

$\overset{+3}{S_2}O^{2-}_4$ $< \overset{+4}{S}O^{2-}_3 < $ $\overset{+5}{S_2}O^{2-}_6$