Q. The magnetic moment of $K_{3}\lbrack Fe(CN)_{6}\rbrack$ is found to be 1.7 BM. How many unpaired electron (s) is/are present per molecule?
A) 1
B) 2
C) 3
D) 4
Explanation
(a)
Magnetic moment of
$K_{3}\lbrack Fe(CN)_{6}\rbrack$=1.7 BM
Magnetic moment=$\sqrt{n(n + 2)}$
$n =$number of unpaired electrons present in molecule
1.7=$\sqrt{n(n + 2)}$
$- n^{2} + 2n - 2.89 = 0\ then\ n = 0.97\ or\ 1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.