InterviewSolution
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InterviewSolution
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A compound of vanadium has a magnetic moment of `1.73BM`. Work out the electronic configuration of vanadium in the compound |
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Answer» Step I. Calculation of the number of unpaired electrons. Magnetic moment `(mu)=sqrt(n(n+2)) " " ` (Here n= number of unpaired electrons) As `" " mu=1.73` BM, therefore, `1.73=sqrt(n(n+2))` `(1.73)^(2)=n(n+2) or 3=n^(2)+2n`. `n^(2)+2n-3=0 " "or (n-1)(n+3)=0`. Therefore n=1 Step II. Configuration of vanadium ion. The electronic configuration of vanadium (Z=23) is : `1s^(2) 2s^(2) 2p^(6)3s^(2)3p^(6)3d^(3)4s^(2)` Since the ion has only one unpaired electron, therefore , its configuration is : `1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(1)` Thus, vanadium is present as `V^(4+)` ion or V (IV). |
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