The oxidation state of sulphur in the anions SO32-, S2O42- and S2O2-6 follows the order

Question

The oxidation state of sulphur in the anions SO32-, S2O42- and S2O2-6 follows the order (A) S2O2-6 < S2O2-4 < SO2-3 (B) S2O2-4 < SO2-3 < S2O2-6 (C) SO2-3 < S2O2-4 < S2O2-6 (D) S2O2-4 < S2O2-6 < SO2-3

Tardigrade Admin · Accepted Answer

overset+3S2O2-4 < overset+4SO2-3 < overset+5S2O2-6

Q. The oxidation state of sulphur in the anions $S O_{3}^{2 -}$ , $S_{2} O_{4}^{2 -}$ and $S_{2} O_{6}^{2 -}$ follows the order

$S_{2} O_{6}^{2 -} < S_{2} O_{4}^{2 -} < S O_{3}^{2 -}$

$S_{2} O_{4}^{2 -} < S O_{3}^{2 -} < S_{2} O_{6}^{2 -}$

$S O_{3}^{2 -} < S_{2} O_{4}^{2 -} < S_{2} O_{6}^{2 -}$

$S_{2} O_{4}^{2 -} < S_{2} O_{6}^{2 -} < S O_{3}^{2 -}$

$S_{2} + 3 O_{4}^{2 -}$ $< S + 4 O_{3}^{2 -} <$ $S_{2} + 5 O_{6}^{2 -}$

Q. The oxidation state of sulphur in the anions SO32−​, S2​O42−​ and S2​O62−​ follows the order

S2​O62−​<S2​O42−​<SO32−​

S2​O42−​<SO32−​<S2​O62−​

SO32−​<S2​O42−​<S2​O62−​

S2​O42−​<S2​O62−​<SO32−​