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

Question

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

Toppr Admin · Accepted Answer

O-S- of S in SO2-x2212-3-x-x2212-6-x2212-2x-4O-S- of S in S2O2-x2212-4-2x-x2212-8-x2212-2x-3O-S- of S in S2O2-x2212-6-2x-x2212-12-2x-5-xA0-The correct order of increasing O-S- is-S2O2-x2212-4-lt-SO2-x2212-3-lt-S2O2-x2212-6

The oxidation state of sulphur in the anions SO2−3,S2O2−4 and S2O2−6 follows the order :SO2−3<S2O2−4<S2O2−6S2O2−6<S2O2−4<SO2−3S2O2−4<SO2−3<S2O2−6S2O2−4<S2O2−6<SO2−3

O.S. of S in SO2−3,x−6=−2x=+4O.S. of S in S2O2−4,2x−8=−2x=3O.S. of S in S2O2−6,2x−12=2x=+5 The correct order of increasing O.S. is;S2O2−4<SO2−3<S2O2−6

O.S. of S in $S O_{3}^{2 -},$
$x - 6 = - 2$
$x = + 4$

O.S. of S in $S_{2} O_{4}^{2 -},$
$2 x - 8 = - 2$
$x = 3$

O.S. of S in $S_{2} O_{6}^{2 -},$
$2 x - 12 = 2$
$x = + 5$

The correct order of increasing O.S. is;

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