JEE MAIN - Chemistry (2023 - 11th April Morning Shift - No. 18)
The ratio of spin-only magnetic moment values $$\mu_{\text {eff }}\left[\mathrm{Cr}(\mathrm{CN})_{6}\right]^{3-} / \mu_{\text {eff }}\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$$ is _________.
Answer
1
Explanation
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 t2g orbital and no unpaired electrons in the eg 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 $$[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}]^{3+}$$.
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 t2g orbital and no unpaired electrons in the eg 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 $$[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}]^{3+}$$.
Comments (0)
