Question

$\ce{[Fe(H2O)6]2 [Fe(CN)6]}$
Cationic and anionic species spin only magnetic moment in Bohr magneton is :

• A. 4.89 and 0
• B. 5.9 and 0
• C. 0 and 4.89
• D. 2.82 and 2.82

Answer 1

Answer: (A)

$\ce{[Fe^{+2} (H2O)6]^{2+}} \ce{[Fe^{+2} (CN)6]^{4-} } $
$\ce{ Fe^{2+} \Rightarrow [Ar] 3d^6 } \ce{ Fe^{2+} \Rightarrow [Ar] 3d^6 } $
$\ce{ H_2 O \Rightarrow WFL } \ce{CN^{-} \Rightarrow SFL}$
Number of unpaired electrons = 4 $ $ Number of unpaired electrons = 0
$\mu = \sqrt{ n(n +2)} = 4.89 BM \mu = 0 $