QJEEM23Ch8633: JEE Main 2023 Chemistry - Class 12 JEEM23Ch8633 Solution: The ratio of spin-only magnetic moment values is _________.

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 $[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}]^{3+}$ .