JEE Main 2023 — Coordination Compounds Question with Solution

For spin-only magnetic moment, we use the formula:

$$\mathrm{Magnetic}\mathrm{moment}=\sqrt{n(n+2)}\cdot \mathrm{BM}$$

where $n$ is the total number of unpaired electrons.

For $${\left[\mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{3+}$$, the electronic configuration of $\mathrm{F}{\mathrm{e}}^{3+}$ 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

$$=\sqrt{5(5+2)}\cdot \mathrm{BM}\approx 5.92\mathrm{BM}$$

For $${\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]}^{3-}$$, the $\mathrm{F}{\mathrm{e}}^{3+}$ 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

$$=\sqrt{1(1+2)}\cdot \mathrm{BM}\approx 1.732\mathrm{BM}$$

Therefore, the answer is 5.92 B.M. and 1.732 B.M.