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The magnetic moment of a metal ion of fi...

The magnetic moment of a metal ion of first transition series is 2.83 BM. Therefore it will have how many unpaired electrons?

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The correct Answer is:
2
` mu = sqrt( n(n+2))" " 2 :83 = sqrt(n(n+2))`
` n(n+2)=8" " n^2+2n -8=0 " " therefore n=2`
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Presence of unpaired electrons makes a species paramagnetic. Each unpaired electron has a magnetic moment associated with its spin angular momentum and orbital angular momentum. For the compounds of the first series of transition metals, the contribution of the orbital angular momentum is effectively quenched and hence is considerably important. For these, the magnetic moment is determined by the number of unpaired electrons and is calculated by using the spin-only formula, i.e., mu = sqrt(n(n+2)) where n=number of unpaired electrons and mu =magnetic momentin (in Bohr magneton, B.M.). If n=1, then mu = 1.73 B.M . The magnetic moment of a transition metal ion is found to be 3.87 B.M. The number of unpaired electrons present in it is:

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