Question

The ratio of magnetic moment (spin only value) between $[Fe{F}_6{]}^{3-}$ and $[Fe(CN{)}_6{]}^{3-}$ is approximately

• A. 4
• B. 2
• C. 5
• D. 3

Answer 1

Answer: (D)

In $[Fe{F}_6{]}^{3-}$ and $[Fe(CN{)}_6{]}^{30}$ both, Fe is present as $F{e}^{3+}$

$F^{-}$ being a weak field ligand, is not capable to pair up these unpaired electrons but $C{N}^{-}$ does this
Hence, is case of
$[Fe{F}_6{]}^{3-}=[Ar]$

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 ${\mu }_2\cdot \frac{{\mu }_1}{{\mu }_2}=\frac{\sqrt{35}}{\sqrt{3}}$
$=3.41=3$