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JEE Main & Advanced JEE Main Online Paper (Held On 08 April 2018)

  • question_answer
    The oxidation states of \[\text{Cr}\] in \[\text{ }\!\![\!\!\text{ Cr(}{{\text{H}}_{\text{2}}}\text{O}{{\text{)}}_{\text{6}}}\text{ }\!\!]\!\!\text{ C}{{\text{l}}_{\text{3}}}\], \[\text{ }\!\![\!\!\text{ Cr(}{{\text{C}}_{6}}{{\text{H}}_{6}}{{\text{)}}_{2}}\text{ }\!\!]\!\!\text{ }\], and \[{{\text{K}}_{\text{2}}}\text{ }\!\![\!\!\text{ Cr(CN}{{\text{)}}_{\text{2}}}{{\text{(O)}}_{\text{2}}}\text{(}{{\text{O}}_{\text{2}}}\text{)(N}{{\text{H}}_{\text{3}}}\text{) }\!\!]\!\!\text{ }\] Respectively are:                         [JEE Main Online 08-04-2018]

    A)  \[\text{+3, 0, and +6}\]        

    B)  \[\text{+3, 0, and +4}\]

    C)  \[\text{+3, +4 and +6}\]     

    D)  \[\text{+3, +2 and +4}\]

    Correct Answer: A

    Solution :

    \[\Rightarrow [Cr{{({{H}_{2}}O)}_{6}}]C{{l}_{3}}\to \] be oxidation state of chromium \[\text{x+6(0)=+3}\] \[\text{x = +3}\] \[\Rightarrow \]\[[Cr{{({{C}_{6}}{{H}_{6}})}_{2}}]\] \[x+2(0)=0\] \[x=0\] \[\Rightarrow {{\text{k}}_{\text{2}}}\text{ }\!\![\!\!\text{ Cr(CN}{{\text{)}}_{\text{2}}}{{\text{(O)}}_{\text{2}}}\text{(}{{\text{O}}_{\text{2}}}\text{)(N}{{\text{H}}_{\text{3}}}\text{) }\!\!]\!\!\text{ }\] \[x-2+2(-2)+1(-2)+0=-2\] \[x-2-4-2=-2\] \[x-6=0\] \[x=+6\]


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