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How many unpaired electrons are present in Co^(3+), Fe^(2+)? Calculate their magnetic moment.

Answer»

Solution :`Co (Z = 27)Co^(3+)[Ar] 3d^(6)`
`Fe (Z = 26)Fe^(2+)[Ar]3d^(6)`
The number of unpaired ELECTRONS are 4 as FOLLOWS: Ar
Their magnetic moment is `mu= SQRT(4(4+2)) = sqrt(24) = 4.89 mu_(B)`


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