Q. What is the magnetic moment of${\ K}_{3}\left\lbrack FeF_{6} \right\rbrack$?
A) 3.87 BM
B) 4.89 BM
C) 5.91 BM
D) 6.92 BM
Explanation
(c)
$K_{3}\lbrack FeF_{6}\rbrack$
$Fe^{3 +} = \lbrack Ar\rbrack\ 3d^{5}\ 4s^{0}$
[image:9_image212.emf]
Number of unpaired electrons=5
Magnetic moment =$\sqrt{n(n + 2)} = \sqrt{s(s + 2)}$
$= \sqrt{35} = 5.91\ BM$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.