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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 ...