NEET Chemistry NEET PYQ-The d-and f-Block Elements

  • question_answer
    The reaction of aqueous \[KMn{{O}_{4}}\] with \[{{H}_{2}}{{O}_{2}}\] in acidic conditions gives             [AIPMT 2014]

    A) \[M{{n}^{4+}}\]and \[{{O}_{2}}\]

    B) \[M{{n}^{2+}}\] and \[{{O}_{2}}\]

    C) \[M{{n}^{2+}}\]and \[{{O}_{3}}\] 

    D) \[M{{n}^{4+}}\]and \[Mn{{O}_{2}}\]

    Correct Answer: B

    Solution :

    [b] The reaction of aqueous \[KMn{{O}_{4}}\] with \[{{H}_{2}}{{O}_{2}}\] in acidic medium is
    \[3{{H}_{2}}S{{O}_{4}}+2KMn{{O}_{4}}+5{{H}_{2}}{{O}_{2}}\to \]
    \[5{{O}_{2}}+2MnS{{O}_{4}}+8{{H}_{2}}O+{{K}_{2}}S{{O}_{4}}\]
    In the above reaction, \[KMn{{O}_{4}}\] oxidises \[{{H}_{2}}{{O}_{2}}\] to \[{{O}_{2}}\] and itself \[[MnO_{4}^{-}]\] gets reduced to \[M{{n}^{2+}}\] ion as \[MnS{{O}_{4}}\]. Hence, aqueous solution of \[KMn{{O}_{4}}\] with \[{{H}_{2}}{{O}_{2}}\] yields \[M{{n}^{2+}}\] and \[{{O}_{2}}\] in acidic conditions.


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