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

Question

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.

Answer 1

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 $(S) = 0+1=1$.