Question

The correct order of magnetic moments (spin only values in B.M.) is:

• A. $[Fe(CN)_6 ]^{4-} > [MnCl_4 ]^{2-} > [CoCl_4]^{2-} $
• B. $ [MnCl_4 ]^{2-} > [Fe(CN)_6 ]^{4-} > [CoCl_4]^{2-} $
• C. $ [MnCl_4 ]^{2-} > [CoCl_4]^{2-} > [Fe(CN)_6 ]^{4-} $
• D. $[Fe(CN)_6 ]^{4-}> [CoCl_4]^{2-} > [MnCl_4 ]^{2-} $

Answer 1

The Correct Option is C

Magnetic moment of $\left[ MnCl _{4}\right]^{2-}$ is $5.9\, BM$.
$Mn$ is in $+2$ oxidation state having configuration of $[ Ar ] 3 d ^{5}$.
It has $5$ unpaired electrons.
$\mu= \sqrt{ n ( n +2)}$
$=\sqrt{(5(5+2)} $
$=5.9 \,BM$
Magnetic moment of $\left[ CoC 1_{4}\right]^{2-}$ is $3.9 \,BM$.
$Co$ is in $+2$ oxidation state having configuration of [Ar] 3d $^{7}$.
It has $3$ unpaired electrons.
$\mu= \sqrt{ n ( n +2)} $
$=\sqrt{(3(3+2)} $
$=3.9 \,BM$
Magnetic moment of $\left[ FeCN _{4}\right]^{2-}$ is $0.0 BM$.
$Fe$ is in $+2$ oxidation state having configuration of $[ Ar ] 3 d ^{6}$. In this complex $CN ^{-}$ion is a strong ligand and all electrons are paired and there are no unpaired electrons.
$\mu= \sqrt{ n ( n +2)} $
$=\sqrt{(0(0+2)}$
$=0 \,BM$
The decreasing order of the magnetic moments is:
$\left.\left[ MnCl _{4}\right]^{2-}>\left[ CoCl _{4}\right]^{2-}> Fe ( CN )_{6}\right]^{4-}$