# Spin Multiplicities of Ions

The multiplicity is fundamentally defined as $$2S + 1$$ where $$S$$ is the total spin.

From what I understand, the multiplicity corresponds with the number of unpaired/paired electrons. For example, in the case of $$\ce{Cu^2+}$$:

The single unpaired electron in $$\ce{Cu^2+}$$ means that $$S=\frac{1}{2} \implies M=2$$.

However, my professor has mentioned that the electrons that give the ion the final charge (positive or negative) can be considered as unpaired, thus in the case of $$\ce{Cu^2+}$$, there are 2 unpaired electrons giving it a triplet $$M=3$$. I am having trouble understand how this arises, both physically and mathematically.

In addition, in the case of polyatomic ions, how do electrons pair? For example, in the case of $$\ce{SO4^2-}$$, looking at the MO digram, all the electrons are paired, including the 2 electrons that give it the final charge. However, I was told that the multiplicity should be 3 as it should be treated as having 2 unpaired electrons.

Source: Wikimedia Commons

I am more confused with regard to positive ions, such as $$\ce{Al^3+}$$, as I was told that the positive +3 charge can be regarded as having 3 unpaired electrons and thus $$s=\frac{3}{2}$$ and multiplicity of 4.* Since the electrons have been removed, I am having difficulty understanding where unpaired electrons arise from.

From what I have read, coupling does not affect this, and I cannot seem to find any explanation or equations that might support this. May I know if anyone has an explanation for this? Thank you!

*more accurately, maximum spin where all electrons have parallel spin

• Al3+ does not have any unpaired electrons, thus this cation exists only in the singlet spin state. Jun 8 '21 at 15:31