Solve the following

In both $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$ and $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}, \mathrm{Fe}$ exists in the $+3$ oxidation state i.e., in $d^{5}$ configuration.

Since CN− is a strong field ligand, it causes the pairing of unpaired electrons. Therefore, there is only one unpaired electron left in the d-orbital.

Therefore,

$\mu=\sqrt{n(n+2)}$

$=\sqrt{1(1+2)}$

$=\sqrt{3}$

$=1.732 \mathrm{BM}$

On the other hand, H2O is a weak field ligand. Therefore, it cannot cause the pairing of electrons. This means that the number of unpaired electrons is 5.

Therefore,

$\mu=\sqrt{n(n+2)}$

$=\sqrt{5(5+2)}$

$=\sqrt{35}$

$\simeq 6 \mathrm{BM}$

Thus, it is evident that $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$ is strongly paramagnetic, while $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$ is weakly paramagnetic.