The spin-only magnetic moment of Fe^{2+} ion | vedclass

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(C) The electronic configuration of $Fe^{2+}$ is $[Ar] 3d^{6}$.It has $4$ unpaired electrons as shown in the $d$-orbital diagram:$d^{6} = [\uparrow\do

Vedclass · Accepted Answer

$5$

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(C) The electronic configuration of $Fe^{2+}$ is $[Ar] 3d^{6}$.It has $4$ unpaired electrons as shown in the $d$-orbital diagram:$d^{6} = [\uparrow\downarrow] [\uparrow] [\uparrow] [\uparrow] [\uparrow]$The spin-only magnetic moment is calculated using the formula:$\mu = \sqrt{n(n+2)} \ BM$Where $n$ is the number of unpaired electrons.$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.89 \ BM$Rounding to the nearest integer,the value is $5 \ BM$.

The spin-only magnetic moment of $Fe^{2+}$ ion (in $BM$) is approximately

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