JEE Main & Advanced JEE Main Paper (Held On 22 April 2013)

  • 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)  \[{{S}_{2}}O_{4}^{2-}<{{S}_{2}}O_{6}^{2-}<SO_{3}^{2-}\]

    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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