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 (integer)
1
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.
<br/><br/>Despite the difference in ligands (CN<sup>-</sup> and H<sub>2</sub>O), both complexes have a low-spin configuration because CN<sup>-</sup> is a strong field ligand and H<sub>2</sub>O 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<sub>2g</sub> orbital and no unpaired electrons in the e<sub>g</sub> orbital.
<br/><br/>Using the spin-only magnetic moment formula, $\mu = \sqrt{n(n+2)}$, where n is the number of unpaired electrons, we find that :
<br/><br/>For $[\mathrm{Cr}(\mathrm{CN})_{6}]^{3-}$, the spin-only magnetic moment is $\mu_{1}=\sqrt{3(3+2)} = \sqrt{15}$ BM.
<br/><br/>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.
<br/><br/>Therefore, the ratio of the spin-only magnetic moments is $\frac{\mu_{1}}{\mu_{2}} = \frac{\sqrt{15}}{\sqrt{15}} = 1$.
<br/><br/>So, the spin-only magnetic moment of $[\mathrm{Cr}(\mathrm{CN})_{6}]^{3-}$ is equal to the spin-only magnetic moment of $[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}]^{3+}$.
About this question
Subject: Chemistry · Chapter: Coordination Compounds · Topic: Ligands and Coordination Number
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