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Question
Calculate magnetic moment of `Fe_((aq))^(2+) ion (Z=26).`
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Solution
Data: Fe2+
Z=26
Electronic configuration of
`Fe^(2+)=1s^2 2s^2 2p^6 3s^2 3p^6 4s^@3d^6`
No. of unpaired electrons = 4
To find: Magnetic moment = u
Formula: `u=sqrt(n(n+2))B.M.`
Solution: `u=sqrt(4(4+2))`
`=sqrt24`
u = 4.90 B.M.
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