The spin only magnetic moment of Mn^{4+} ion | vedclass

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(C) The electronic configuration of $Mn$ $(Z=25)$ is $[Ar] \ 3d^{5} \ 4s^{2}$.For $Mn^{4+}$ ion,we remove four electrons,resulting in the configuratio

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$4$

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(C) The electronic configuration of $Mn$ $(Z=25)$ is $[Ar] \ 3d^{5} \ 4s^{2}$.For $Mn^{4+}$ ion,we remove four electrons,resulting in the configuration $[Ar] \ 3d^{3}$.This ion contains $n=3$ unpaired electrons.The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$.Substituting $n=3$,we get $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$.Rounding to the nearest integer,the value is $4 \ BM$.

The spin only magnetic moment of $Mn^{4+}$ ion is nearly (in $BM$)

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