The magnetic moment is measured in Bohr Magneton (BM).

Spin only magnetic moment of $\mathrm{Fe}$ in $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$ and $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$ complexes respectively is :

<p>For spin-only magnetic moment, we use the formula:<br/><br/> $\mathrm{Magnetic ~moment} = \sqrt{n(n+2)} \cdot \mathrm{BM}$<br/><br/> where $n$ is the total number of unpaired electrons.</p> <p>For $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$, the electronic configuration of $\mathrm{Fe^{3+}}$ is $\mathrm{d^5}$. Here, all the five electrons will be unpaired due to the high spin nature of Fe$^{3+}$. Hence, $n=5$ and magnetic moment<br/><br/> $= \sqrt{5(5+2)} \cdot \mathrm{BM} \approx 5.92 \mathrm{BM}$</p> <p>For $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$, the $\mathrm{Fe^{3+}}$ ion has electronic configuration $\mathrm{d^5}$. In this case, the strong field ligand cyanide ($\mathrm{CN}^{-}$) will pair up the electrons in the $\mathrm{d}$ orbital. So, there will be only one unpaired electron. Hence, $n=1$ and magnetic moment<br/><br/> $= \sqrt{1(1+2)} \cdot \mathrm{BM} \approx 1.732 \mathrm{BM}$</p> <p>Therefore, the answer is 5.92 B.M. and 1.732 B.M.</p>