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 $\ce{Cu^2+}$:

Cu2+ electronic configuration

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.

SO42- MO diagram

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.