JEE Main 2023 — Coordination Compounds Question with Solution

In both given complexes, namely, $$[\mathrm{Cr}(\mathrm{CN}{)}_6{]}^{3-}$$ and $$[\mathrm{Cr}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6{]}^{3+}$$, the chromium atom is in the +3 oxidation state. This results in a $$d^3$$ configuration for the chromium ion.

Despite the difference in ligands (CN^- and H₂O), both complexes have a low-spin configuration because CN^- is a strong field ligand and H₂O is a weak field ligand. This configuration is denoted as $${\mathrm{t}}_{2g}^3{\mathrm{e}}_{\mathrm{g}}^0$$, indicating that there are three unpaired electrons in the t_2g orbital and no unpaired electrons in the e_g orbital.

Using the spin-only magnetic moment formula, $$\mu =\sqrt{n(n+2)}$$, where n is the number of unpaired electrons, we find that :

For $$[\mathrm{Cr}(\mathrm{CN}{)}_6{]}^{3-}$$, the spin-only magnetic moment is $${\mu }_1=\sqrt{3(3+2)}=\sqrt{15}$$ BM.

And for $$[\mathrm{Cr}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6{]}^{3+}$$, the spin-only magnetic moment is also $${\mu }_2=\sqrt{3(3+2)}=\sqrt{15}$$ BM.

Therefore, the ratio of the spin-only magnetic moments is $$\frac{{\mu }_1}{{\mu }_2}=\frac{\sqrt{15}}{\sqrt{15}}=1$$.

So, the spin-only magnetic moment of $$[\mathrm{Cr}(\mathrm{CN}{)}_6{]}^{3-}$$ is equal to the spin-only magnetic moment of $$[\mathrm{Cr}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6{]}^{3+}$$.