# 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.

• You may find useful these links for text formatting ( not to be applied to titles ): notation , chem/math formula/equation formatting and upright vs italic // Generally, only math variables and scientific quantities are in italic, most other things,like chemical formulas and orbital symbols are upright. Jan 5, 2022 at 14:34
• @Poutnik I will edit my Q in some time accordingly as it can be useful. Jan 5, 2022 at 14:41
• Sure, no hurry because of me, just for you to be aware for future cases. Jan 5, 2022 at 14:42

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 = $$n+1$$.
• $$\ce{Mn}$$ 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 $$(S) = 0+1=1$$.