Question

Quenching of orbital contribution to magnetic moment

Answer 1

When the orbital contribution to the magnetic moment is zero, we say that the orbital contribution to the magnetic moment is quenched. When the ground state of a complex is either A or E, the orbital contribution will be quenched. In other words, there is no orbital angular momentum for these states.

Answer 2

Answer:

Quenching of angular momentum means a spatialsymmetrythath must be stable in time. If you apply an external magnetic or electric field then this condition could change drastically.

Explanation:

• In particular, magnetic and spintronic characteristics depend heavily on orbital angular momentum. Therefore, regulating orbital angular momentum is crucial for both basic science and cutting-edge technology applications.
• Numerous attempts have been made to control the orbital angular momentum-induced quenching effects of ligand fields to change magnetic characteristics. In both molecular and solid-state magnetic materials, however, documented variations in the size of orbital angular momentum are currently negligible. Furthermore, there are currently no efficient ways to modify orbital angular momentum. Here, we discuss a dynamic bond method for achieving a significant shift in orbital angular momentum. We have created a Co(II) complex that switches between coordination numbers six and seven.
• Interactions (involving localized angular momentum states) tend to couple angular momentum states.
• Spin-orbit coupling for example (in absence of anything else) couples states with quantum numbers |S, mS, L, mL> into new ones with quantum numbers |J,mJ>, which are linear combinations of the former ones with well-defined coefficients (named Clebsch-Gordan coefficients). mS and mL then no longer are "good quantum numbers", because (most of) the new eigenstates are linear combinations of states with different mL and mS. However, the rules of angular momentum coupling have it that in each state the expectation (not eigen-) values <mL> and <mS> are still proportional to mJ, the new "good" quantum number (to be found in any decent QM or atomic physics textbook; the latter statement is also known as Wigner-Eckart theorem). under these circumstances, the orbital moments are thus different but not "quenched".
• About angular momentum states of atoms in a lattice. The surrounding atoms create an electrostatic potential of some symmetry. This (point group) symmetry is usually characterized by some axis with the "highest" rotational symmetry (this need not be unique [e.g. in octahedral or tetragonal local symmetry, encountered in many oxides], then pick one at random; In practice the choice can be tricky in sites of low symmetry).
• This potential (its angular dependence, to be precise) can then be expanded as a series in the spherical harmonics. The nonvanishing terms are not arbitrary then. If you have a q-fold rotation axis, the expansion will only have nonzero terms in those Ylm where m is an integer multiple of q. As a result, the potential only couples orbital momentum states with quantum numbers mL1, and mL2  such that mL2-mL1=nq (n being an integer) in so far as they exist. [The strength of coupling is given by the corresponding coefficient of  the potential expansion.] This may then lead to orbital momentum quenching.
• Example 1: one d-electron (l=2, ml=-2...2) in a four-fold symmetric site (q=4). The potential couples those orbital angular momentum states which differ in multiples of q=4 in their (orientational) orbital momentum quantum number. only one pair of such states exists, i.e. ml=-2 and ml=2, which are energetically degenerate. [Of course there are two such pairs, one for each mS.] Inevitably then, new eigenstates form and these are (|ml=-2> + ml=2>)/sqrt(2) and (|ml=-2> - ml=2>)/sqrt(2), also known as |xy> and |x2-y2 >.
• In cubic sites, their energetic splitting is known as the "t2g-eg"-splitting and is of the order of eV in oxides. In this case, you easily see that the expectation value of the orbital momentum in these states is zero because ml=-2 and ml=+2 contribute with the same weight in both states. The orbital moment is said to be quenched.
• Note: in reality, there will be spin-orbit coupling, too, and maybe an externally applied field as well, etc. As said above, soc 'prefers' a different linear combination of the ml states. In compounds with 3d elements, soc is usually much weaker than the electrostatic potential, so that orbital moment quenching remains dominant (soc then only 'unquenches' the orbital momentum a little bit. This is not unimportant, though, since it gives rise to magnetocrystalline anisotropy, consult good magnetism textbooks for more details on this). Typical experimental Zeeman energies are much smaller yet and therefore will not unquench the orbital moments.