What is the value of spin only magnetic momen | vedclass

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(D) The atomic number of $Mn$ is $25$. The electronic configuration of $Mn$ is $[Ar] 3d^5 4s^2$. For $Mn^{2+}$,the configuration is $[Ar] 3d^5$. This

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$5.92$

Vedclass · Answer

(D) The atomic number of $Mn$ is $25$. The electronic configuration of $Mn$ is $[Ar] 3d^5 4s^2$. For $Mn^{2+}$,the configuration is $[Ar] 3d^5$. This means there are $5$ unpaired electrons $(n = 5)$. The spin only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$. Substituting $n = 5$,we get $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.

What is the value of spin only magnetic moment for $Mn^{2+}$ in $BM$?

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