A) 5.91 BM
B) 4.89 BM
C) 3.87 BM
D) 6.92 BM
Correct Answer: A
Solution :
\[{{K}_{3}}[Fe{{F}_{6}}]\] \[F{{e}^{3+}}=[Ar]3{{d}^{5}}4{{s}^{0}}\] Number of unpaired electrons = 5 Magnetic moment \[=\sqrt{n(n+2)}=\sqrt{5(5+2)}\] \[=\sqrt{35}\]= 5.91 BM.You need to login to perform this action.
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