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) \[{{S}_{2}}O_{4}^{2-}<{{S}_{2}}O_{6}^{2-}<SO_{3}^{2-}\]
Correct Answer: C
Solution :
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 \[{{S}_{2}}{{O}_{4}}^{-\,-}<S{{O}_{3}}^{-\,-}<{{S}_{2}}{{O}_{6}}^{-\,-}\]
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