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JCECE Engineering JCECE Engineering Solved Paper-2012

  • question_answer
    The magnetic moment of a transition metal ion is\[\sqrt{15}BM\]. Therefore, the number of unpaired electrons present in it is

    A) \[4\]                                     

    B) \[1\]

    C) \[2\]                                     

    D)  \[3\]

    Correct Answer: D

    Solution :

    Magnetic moment\[=\sqrt{n(n+2)}\] where, \[n=\]number of unpaired electrons Given\[\mu =\sqrt{15}\]                 \[\sqrt{15}=\sqrt{n(n+2)}\]                 \[15=n(n+2)\]                 \[{{n}^{2}}+2n-15=0\] \[(n+5)(n-37)=0\]                 \[n=3\]                 \[(\because \,\,n\ne -5)\]


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