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  4. The ratio of magnetic moment (spin only value) between [FeF6]3- and [Fe(CN)6]3- is approximately

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Q. The ratio of magnetic moment (spin only value) between $ [FeF_{6}]^{3-} $ and $ [Fe(CN)_{6}]^{3-} $ is approximately

AMUAMU 2014Coordination Compounds

Solution:

In $[FeF_{6}]^{3-}$ and $[Fe(CN)_{6}]^{30}$ both, Fe is present as $Fe^{3+}$
image
$F^{-}$ being a weak field ligand, is not capable to pair up these unpaired electrons but $CN^{-}$ does this
Hence, is case of
$[FeF_{6}]^{3-}=[Ar]$
image
Number of unpaired electrons $=5$
Magnetic moment, $\mu_{1}=\sqrt{5\left(5+2\right)}=\sqrt{35}$
In case of
$\left[Fe \left(CN\right)_{6}\right]^{3-} =\left[Ar\right]$
image
Number of unpaired electron $=1$
Magnetic moment, $\mu_{2}=\sqrt{1\left(1+2\right)}=\sqrt{3}$
Ratio of $\mu_{1}$ and $\mu_{2}\cdot \frac{\mu_{1}}{\mu_{2}}=\frac{\sqrt{35}}{\sqrt{3}}$
$=3.41=3$