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The ratio of spin-only magnetic moment values μeff [Cr(CN)6]3−/μeff [Cr(H2O)6]3+\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+}μeff [Cr(CN)6]3−/μeff [Cr(H2O)6]3+ is _________.
Spin magnetic moment of [Cr(CN)6]3−(t2 g3 eg0)\left[\mathrm{Cr}(\mathrm{CN})_6\right]^{3-}\left(\mathrm{t}_{2 \mathrm{~g}}^3 \mathrm{e}_{\mathrm{g}}^0\right)[Cr(CN)6]3−(t2 g3 eg0)
μ1=3(3+2)=15 BM\mu_1=\sqrt{3(3+2)}=\sqrt{15} \mathrm{BM}μ1=3(3+2)=15 BM
Spin magnetic moment of
[Cr(H2 O)6]3+(t2 g3 eg0) {\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}\left(\mathrm{t}_{2 \mathrm{~g}}^3 \mathrm{e}_{\mathrm{g}}^0\right)} \\ [Cr(H2 O)6]3+(t2 g3 eg0)
μ2=3(3+2)=15 BM \mu_2=\sqrt{3(3+2)}=\sqrt{15} \mathrm{BM} \\ μ2=3(3+2)=15 BM
μ1μ2=1515=1 \frac{\mu_1}{\mu_2}=\frac{\sqrt{15}}{\sqrt{15}}=1 μ2μ1=1515=1
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