Chemistry - Coordination Chemistry Question with Solution | TestHub

$\left[\mathrm{Fe}\right({\mathrm{H}}_2\mathrm{O}{)}_6{]}^{3+ }$ has magnetic moment value of $5.92 \mathrm{BM}$ because ${\mathrm{H}}_2\mathrm{O}$  is a weak ligand and pairing will not occur. It involves ${\mathrm{sp}}^3{\mathrm{d}}^2$ hybridisation with five unpaired electrons (because magnetic moment is equal to $5.92 \mathrm{BM}$) and outer orbital complex is formed. Magnetic moment can be calculated by the formula ${\mathrm{\mu }}_{\mathrm{m}}=\sqrt{\mathrm{n}(\mathrm{n}+2)}$ $\mathrm{BM}$ where, ${\mathrm{\mu }}_{\mathrm{m}}=$magnetic moment, $\mathrm{n}=$number of unpaired electron(s).

$$\begin{gathered} {\mathrm{\mu }}_{\mathrm{m}}=1.74 \mathrm{BM}, \mathrm{n}=1 \\ {\mathrm{\mu }}_{\mathrm{m}}=5.92 \mathrm{BM}, \mathrm{n}=5 \end{gathered}$$

In $\left[\mathrm{Fe}\right(\mathrm{CN}{)}_6{]}^{3–}$complex ${\mathrm{CN}}^{-}$ is stronger ligand than ${\mathrm{H}}_2\mathrm{O}$ according to spectrochemical series.  Hence, four electrons will pair itself. It involves ${\mathrm{d}}^2{\mathrm{sp}}^{3 }$hybridisation with one unpaired electron and inner orbital complex is formed and has a value of only $1.74 \mathrm{BM}$.

In Bothe the complexes the oxidation state of iron is +3.