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Calculated magnetic moment value for $\mathrm{Fe}^{2+}$ ion in $\mathrm{BM}$ is
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$4.90$
$\mathrm{Fe}^{2+}\left(\mathrm{d}^6\right):$ Unpaired electrons $(\mathrm{n})=4$
$\therefore \quad \mu_{\text {s.o. }}=\sqrt{4(4+2)}=2 \sqrt{6}=4.89 \simeq 4.90 \mathrm{BM}$
$\therefore \quad \mu_{\text {s.o. }}=\sqrt{4(4+2)}=2 \sqrt{6}=4.89 \simeq 4.90 \mathrm{BM}$
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