Question

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

• A. $4$
• B. $2$
• C. $5$
• D. $3$

Answer 1

Answer: (D)

$3$

Answer 2

In $[FeF_{6}]^{3-}$ and $[Fe(CN)_{6}]^{30}$ both, Fe is present as $Fe^{3+}$

$F^{-}$ being a weak field ligand, is not capable to pair up these unpaired electrons but $CN^{-}$ does this
Hence, is case of
$[FeF_{6}]^{3-}=[Ar]$

Number of unpaired electrons $=5$
Magnetic moment, $\mu_{1}=\sqrt{5\left(5+2\right)}=\sqrt{35}$
In case of
$\left[Fe \left(CN\right)_{6}\right]^{3-} =\left[Ar\right]$

Number of unpaired electron $=1$
Magnetic moment, $\mu_{2}=\sqrt{1\left(1+2\right)}=\sqrt{3}$
Ratio of $\mu_{1}$ and $\mu_{2}\cdot \frac{\mu_{1}}{\mu_{2}}=\frac{\sqrt{35}}{\sqrt{3}}$
$=3.41=3$