Oxidation state of sulphur in anions SO2-3. S2O2-4 and S2O2-6 increases in the orders:

Question

Oxidation state of sulphur in anions SO2-3. S2O2-4 and S2O2-6 increases in the orders: (A) S2O2-6< S2O2-4< SO2-3 (B) SO2-6< S2O2-4< SO2-6 (C) S2O2-4< SO2-3< S2O2-6 (D) S2O2-4< S2O2-6< SO2-3

Tardigrade Admin · Accepted Answer

In SO3-- x+3(-2)=-2; x=+4 In S2O4 2x+4(-2) =-2 2x-8 = -2 2x=6; x=+3 In S2O2-6 2x+6(-2)-2 2x= 10; x=+5 hence the correct order is S2O4-- < SO3-- < S2O6--

Q. Oxidation state of sulphur in anions $S O_{3}^{2 -} . S_{2} O_{4}^{2 -}$ and $S_{2} O_{6}^{2 -}$ increases in the orders:

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

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

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

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

In $S O_{3}^{--}$
$x + 3 (- 2) = - 2; x = + 4$
In $S_{2} O_{4}$
$2 x + 4 (- 2) = - 2$
$2 x - 8 = - 2$
$2 x = 6; x = + 3$
In $S_{2} O_{6}^{2 -}$
$2 x + 6 (- 2) - 2$
$2 x = 10; x = + 5$
hence the correct order is
$S_{2} O_{4}^{--} < S O_{3}^{--} < S_{2} O_{6}^{--}$

Q. Oxidation state of sulphur in anions SO32−​.S2​O42−​ and S2​O62−​ increases in the orders:

S2​O62−​<S2​O42−​<SO32−​

SO62−​<S2​O42−​<SO62−​

S2​O42−​<SO32−​<S2​O62−​

S2​O42−​<S2​O62−​<SO32−​