Coupling constant (Jab) - Why more unpaired electrons means a smaller coupling?

Question

I hope this is not a "homework" question, but I'm having a bad time trying to figure this out. Available literature proposes 3 equations to calculate the coupling constant during a Broken-Symmetry approach:

\begin{align*} J(1) = \frac{-\left ( E_{HS}-E_{BS} \right )}{S_{max}^{2}} \\ J(2) = \frac{-\left ( E_{HS}-E_{BS} \right )}{S_{max}\left ( S_{max}+1 \right )} \\ J(3) = \frac{-\left ( E_{HS}-E_{BS} \right )}{\left \langle S_{HS}^{2} \right \rangle-\left \langle S_{BS}^{2} \right \rangle} \end{align*}

I'm intrigued by the fact that, from the equations, the more the system have unpaired electrons, the minor will be $J_{ab}$. Why does this happen? Doesn't more unpaired electrons increase magnetic momenta (and an increase in magnetic coupling))?

Answer 1

I hope I have understood your question correctly.

For illustrative purposes, I will restrict myself to the Hartree-Fock picture. In a broken spin system, e.g., UHF, the spin state is not a good quantum number. That is to say that the UHF wavefunction need not be an eigenfunction of $S^2$ or $S_z$. For example two complexes with a single unpaired electron each will combine three singlet states (two ionic states and one open-shell) and three triplet states ($m_s\in \{-1,0,1\}$).

The open-shell singlet ($|\alpha\phi_1 \beta\phi_2|-|\alpha\phi_2 \beta\phi_1|$) and the $m_s=0$ triplet ($|\alpha \phi_1 \beta \phi_2|+|\alpha\phi_2 \beta\phi_1|$) states are not single Slater determinants. The singlet states need not be of the same energy, whereas the triplet states have to be degenerate for a Hamiltonian without explicitly spin-dependent components such as spin-orbit coupling. Hence, you have to accommodate for the multiple counting.

For higher spins, the degeneracy grows with $S_{max}$ and must be corrected accordingly. This does not imply that coupling necessarily decreases with higher spins.