Chemistry - Coordination Chemistry Question with Solution | TestHub

The spin only magnetic moment can be calculated as follows

$\mathrm{\mu }=\sqrt{\mathrm{n}\left(\mathrm{n}+2\right)}\mathrm{BM}$

$\mathrm{n}=$Number of unpaired electrons

${\left[\mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{3+}={\mathrm{d}}^5(\mathrm{High} \mathrm{spin} \mathrm{complex})=5$.unpaired electrons
$\mathrm{\mu }=\sqrt{5\times 7}=\sqrt{35}$
$=5.92 \mathrm{B}.\mathrm{M}.$
${\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]}^{3+}={\mathrm{d}}^5(\mathrm{Low} \mathrm{spin} \mathrm{complex})=1$
$\mathrm{\mu }=\sqrt{1\times 3}=\sqrt{3}$
$=1.732 \mathrm{B}.\mathrm{M}.$