Q.
Spin only magnetic moment of Fe(II) ion is equal to
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a
1.6×10-23 A.m2
b
2.6×10-23 A.m2
c
3.6×10-23 A.m2
d
4.5×10-23 A.m2
answer is D.
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Detailed Solution
Spin only magnetic moment, μs=nn+2 B.M ; n=number of unpaired electrons ;One B.M (Bohr Magneton) = 9.273×10-24 A.m2 ;Sc3+, Ti4+, Cu+, Zn2+…..n=0, μs=0 ; DiamagneticTi3+, Cu2+…..n=1, μs=3=1.732 B.M=1.732×9.273×10-24 A.m2 ;Ti2+, V3+, Ni2+…..n=2, μs=8=2.82 B.M=2.82×9.273×10-24 A.m2 ;Cr3+, V2+,Co2+…..n=3, μs=15=3.87 B.M=3.87×9.273×10-24 A.m2 ;Co3+, Fe2+, Mn3+, Cr2+…..n=4, μs=24=4.9 B.M=4.9×9.273×10-24 A.m2 ;Fe3+, Mn2+…..n=5, μs=35=5.9 B.M=5.9×9.273×10-24 A.m2 ;μobs>μcal ......Fe2+, Co2+, Ni2+ ; In these cases orbital contributon cannot be ignored.
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