The oxidation states of sulphur in the anions SO 3 2-, S 2 O 4 2- and S 2 O 6 2- follow the order -

Question

The oxidation states of sulphur in the anions SO 3 2-, S 2 O 4 2- and S 2 O 6 2- follow the order - (A) S2O4 2 - < S2O6 2 - < SO3 2 -  (B) S2O6 2 - < S2O42 - < SO3 2 -  (C) S2O4 2 - < SO3 2 - < S2O6 2 -  (D) SO3 2 - < S2O4 2 - < S2O6 2 -

Tardigrade Admin · Accepted Answer

Oxidation state of S in SO3 2 - x + ( - 2 × 3 ) = - 2 x = + 6 - 2 = + 4 Oxidation state of S in S2 O4 2 - 2 x 4- (- 2 × 4) = - 2 2x = + 8 - 2 = + 6 x = ( + 6 / 2 ) = + 3 Oxidation state of S in S2 O6 2 - 2 x 4 - (- 2 x× 6) = - 2 2x = 4 - 12 - 2 = 10 x = (10/ 2) = - 5 Hence, increasing order of oxidation states of S is S2O4 2 - < SO3 2 - < S2O6 2 -

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

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

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

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

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

Q. The oxidation states of sulphur in the anions SO3​2−,S2​O4​2− and S2​O6​2− follow the order -

S2​O42−​<S2​O62−​<SO32−​

S2​O62−​<S2​O42−​<SO32−​

S2​O42−​<SO32−​<S2​O62−​

SO32−​<S2​O42−​<S2​O62−​

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

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

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

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

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