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  3. Chemistry
  4. The magnetic moment of K3[Fe(CN)6] is found to be 1.7 BM . How many unpaired electron (s) is/are present per molecule ?

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Q. The magnetic moment of $ K_3[Fe(CN)_6] $ is found to be $ 1.7\, BM $ . How many unpaired electron $ (s) $ is/are present per molecule ?

AMUAMU 2019

Solution:

Given, magnetic moment of
$K_{3}[Fe(CN)_{6}]=1.7\,BM$
Magnetic moment $=\sqrt{n\left(n+2\right)}$
$n=$ number of unpaired electrons present in molecule $1.7=\sqrt{n(n+2)}$
$-n^{2}+2n-2.89=0$
Then $n=0.97$ or $1$