How do we write spin multiplicity for Mn(2+), Mn(7+)?

Mn has atomic number = 25


Since it is an exception to electronic configuration , unlike having = $\mathrm{3d^7}$ , it has electronic configuration = $\mathrm{3d^5,4s^2}$.


Formula for spin multiplicity = 2|S| + 1.


Spin multiplicity of Mn = 2(0) + 1 = 1 since there are no unpaired electrons.


I’m having difficulty with the further ions multiplicity.


For $\ce{Mn2+}$ , there are 5 unpaired electrons in d sub-shell. The 2e- of $\mathrm{4s^2}$ are lost.


M = 2(5/2) + 1 = 6. Now , according to my textbook. Mn = Mn2+ in terms multiplicity which I do not see as correct.


M for $\ce{Mn7+}$ ,


Electron configuration =$\mathrm{2s^2}$, $\mathrm{3d^0}$


Therefore , here I think it should be 2(2/2) + 1 = 3.




*Spin multiplicity* = $2S+1$, where $S$ is the total spin angular momentum. Now $S = \frac{n}{2}$ where $n$ represents total number of unpaired electrons. So now we can write **spin multiplicity** = $n+1$.


Now coming to your question,


* $\ce{Mn}$ has 5 unpaired electrons in it. Therefore, its spin multiplicity $(S) = 5+1 = 6$.
* Similarly, $\ce{Mn^{2+}}$ has 5 unpaired elections in it. Therefore, its spin multiplicity $(S) = 5+1=6$.
* $\ce{Mn^{7+}}$ has zero number of unpaired electrons. Therefore, its spin multiplicity $(S) = 0+1=1$.


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