The spin only magnetic moment can be calculated as follows
μ=n(n+2)BM
n=Number of unpaired electrons
[Fe(H2O)6]3+=d5(Highspincomplex)=5.unpaired electrons
μ=5×7=35
=5.92B.M.
[Fe(CN)6]3+=d5(Lowspincomplex)=1
μ=1×3=3
=1.732B.M.
The magnetic moment is measured in Bohr Magneton (BM). Spin only magnetic moment of Fe in [Fe(H2O)6]3+ and [Fe(CN)6]3− complexes respectively is:
Held on 11 Apr 2023 · Verified 6 Jul 2026.
6.92B.M. in both
3.87B.M. and1.732B.M.
5.92B.M. and 1.732B.M.
4.89B.M. and 6.92B.M.
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