The magnetic moment of Mn^{4+} ion is approxi | vedclass

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(C) The electronic configuration of $Mn$ $(Z=25)$ is $[Ar] \ 3d^5 \ 4s^2$.
For $Mn^{4+}$,the configuration is $[Ar] \ 3d^3$.
Here,the number of unpa

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$4$

Vedclass · Answer

(C) The electronic configuration of $Mn$ $(Z=25)$ is $[Ar] \ 3d^5 \ 4s^2$.
For $Mn^{4+}$,the configuration is $[Ar] \ 3d^3$.
Here,the number of unpaired electrons $(n)$ is $3$.
The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ B.M.$.
$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ B.M.$.
Rounding to the nearest integer,we get $4 \ B.M.$.

The magnetic moment of $Mn^{4+}$ ion is approximately ...... $B.M.$

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