Question

Expalin why $\left[Fe\right(H_{2}O{)}_6{]}^{3+}$ has magnetic moment value of 5.92 BM whereas $\left[Fe\right(CN{)}_6{]}^{3-}$ has a value of only 1.74 BM?

Answer 1

As we know, ${\mu }_m=\sqrt{n(n+2)}$ BM
where, ${\mu }_m$=magnetic moment
${\mu }_n$=number of unpaired electrons
${\mu }_m$=1.74 i.e., n=1
and ${\mu }_m$=5.92 i.e., n=5
$\left[Fe\right(CN{)}_6{]}^{3-}$ involves $d^{2}s{p}^3$ hybridiastion with one unpaired electron (as shown by its magnetic moment 1.74BM) and $\left[Fe\right(H_{2}O{)}_6{]}^{3+}$ involves $s{p}^3{d}^2$ hybridisation with five unpaired electrons (because magnetic moment equal to 5.92BM).
$C{N}^{-}$ is stronger ligand than $H_{2}O$ according to spectrochemical series. ${△}_0>P$ for $C{N}^{-}$ hence, fourth electron will pair itself. Whereas for water pairing will not happen for $\left[Fe\right(CN{)}_6{]}^{3-}$ the electronic configuration of $F{e}^{3+}$ is



One unpaired electron
For $\left[Fe\right(H_{2}O{)}_6{]}^{3+}$ the electronic configuration of $F{e}^{3+}$ is



Five unpaired electron
Hence, $\left[Fe\right(CN{)}_6{]}^{3-}$ and $\left[Fe\right(H_{2}O{)}_6{]}^{3+}$ are inner orbital and outer orbital complex respectively.