2
$\begingroup$

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).

$\endgroup$

1 Answer 1

2
$\begingroup$

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."

$\endgroup$
1
  • $\begingroup$ Thanks for the reference! My confusion was solved quite a while ago. Didn't know about this though! $\endgroup$
    – Sagnik
    Apr 3, 2018 at 13:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.