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Caslculate the magnetic moments of Fe^(2+) and Fe^(3+) |
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Answer» Solution :`FE^(2+)`, there are 4 unpaired electrons `mu= SQRT(4(4 + 2)) = sqrt(4xx6) =sqrt24 = 4.89 B.M` In `Fe^(3+)` there are 5 unpaired electrons. `mu = sqrt(5(5+2)) = sqrt(5xx7)=sqrt135 =5.91 B.M` |
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