It is relatively well known that the discrepancy between the observed and experimental magnetic moments in the first row transition metals is attributed to spin-orbit coupling when it comes to metal ions with $A_2$ and $E$ ground states. For example, Cu(II) has an $E_{2g}$ ground state term has the observed moment is slightly higher (1.96 BM for a phenanthroline complex) than the theoretically predicted "spin only" value (~1.73 BM, attributed to 1 unpaired electron). I also know that this is calculated by a formula which makes use of the spin-orbit coupling coefficient:

$μ_{eff}$ = $μ_{s.o.}$(1-$α λ$ /$Δ_o$)

where, $α$ takes up the values 2 (for E ground state) and 4 (for $A_2$ ground state), $\lambda$ is the spin orbit coupling coefficient, $Δ_o$ is the crystal field splitting energy.

My question relates to the possible effect of a Jahn Teller distortion on the magnetic moment of, say, Cu(II). Intuitively, a distortion should reduce the orbital contribution to angular momentum by the loss of orbital degeneracy. On the other hand, for Cu(II), for example, the $e_g$ set is magnetically non active. So does splitting basically have no effect on the orbital part of the angular momentum? The $t_{2g}$ set should be magnetically inactive as well because of being fully filled for Cu(II).


1 Answer 1


Planar complexes may be affected by this effect- not orthogonal complexes . ref-(2017) https://aip.scitation.org/doi/pdf/10.1063/1.4974805

"The most striking result is that the planar copper(II) (3d9 ) and nickel(II) (3d8 ) chelation afforded very strong ferromagnetic coupling often with 2J ≥ 300 K, owing to the orthogonal orbital arrangement between metal ion 3dx2-y2 and oxygen 2pz orbitals. Such a structure-magnetism relationship has recently been extended to coordination to rare earth metal ions."

  • $\begingroup$ Thanks for the reference! My confusion was solved quite a while ago. Didn't know about this though! $\endgroup$
    – Sagnik
    Commented Apr 3, 2018 at 13:33

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