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MGIMS WARDHA MGIMS WARDHA Solved Paper-2005

  • question_answer
    The magnetic moment of\[{{K}_{3}}[Fe{{(CN)}_{6}}]\]is found to be 1.7 BM. How many unpaired electron (s) is/are present per molecule?

    A) 1                             

    B) 2

    C) 3                             

    D) 4

    Correct Answer: A

    Solution :

    Magnetic moment of\[{{K}_{3}}[Fe{{(CN)}_{6}}]=1.7\text{ }BM\]Magnetic moment \[=\sqrt{n(n+2)}\]n= number of unpaired electrons present in molecule\[1.7=n\sqrt{(n+2)}\]n2 + 2n - 2.89 = 0 then n = 0.97 or 1


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