• question_answer Oxidation state of sulphur in anions $\operatorname{S}\operatorname{O}_{3}^{2-},{{\operatorname{S}}_{2}}\operatorname{O}_{4}^{2-},$ and ${{\operatorname{S}}_{2}}\operatorname{O}_{6}^{2-}$ increases  in the orders:     JEE Main  Online Paper (Held On 22 April 2013) A)  ${{\operatorname{S}}_{2}}\operatorname{O}_{6}^{2-}<{{\operatorname{S}}_{2}}\operatorname{O}_{4}^{2-}<\operatorname{S}\operatorname{O}_{3}^{2-}$B)  $\operatorname{S}\operatorname{O}_{3}^{2-}<{{\operatorname{S}}_{2}}\operatorname{O}_{4}^{2-}<{{\operatorname{S}}_{2}}\operatorname{O}_{6}^{2-}$C)  ${{\operatorname{S}}_{2}}\operatorname{O}_{4}^{2-}<\operatorname{S}\operatorname{O}_{3}^{2-}<{{\operatorname{S}}_{2}}\operatorname{O}_{6}^{2-}$D)  ${{S}_{2}}O_{4}^{2-}<{{S}_{2}}O_{6}^{2-}<SO_{3}^{2-}$

In $SO_{3}^{-\,-}$ $x+3(-2)=-2;x=+\,4$ In ${{S}_{2}}O_{4,}^{-\,-}$ $2x+4(-2)=-2$ $2x=-8=-2$ $2x=6;$              $x=+3$ In ${{S}_{2}}O_{6}^{2-}$ $2x+6(-2)=-2$ $2x=10;$           $x=+5$ Hence the correct order is ${{S}_{2}}{{O}_{4}}^{-\,-}<S{{O}_{3}}^{-\,-}<{{S}_{2}}{{O}_{6}}^{-\,-}$