(A) The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.For $Mn^{2+}$

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(A) The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.For $Mn^{2+}$ $(Z=25)$,the electronic configuration is $[Ar] \ 3d^5$. Here,$n = 5$.$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.Thus,$Mn^{2+}$ has a magnetic moment of approximately $5.93 \ BM$.