[FeF_6] $ [Orissa JEE 2005]
A) 5.91 BM
B) 4.89 BM
C) 3.87 BM
D) 6.92 BM
Correct Answer: A
$ K_3[FeF_6] $
$ F{e^{3+}}=[Ar]3d^{5}4s^{0} $ Number of unpaired electrons = 5 Magnetic moment $ =\sqrt{n(n+2)}=\sqrt{5(5+2)} $
$ =\sqrt{35} $ = 5.91 BM.
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