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Trying to learn alone some aspects of quantum mechanics is, sometimes, a struggle. Reading the excellent paper by Piwowarska [1] I was hoping to, finally, understand what is the origin of the so-called “fictitious spin $1/2$” in a $\mathrm d^7$ cobalt ion.

Fig. 2 Fig. 2 Visualization of distinction between the electronic, efective, and fctitious spins and respective spin Hamiltonians in terms of the spin levels of $\ce{Co^2+}(\mathrm{3d^7})$ ions

If my cobalt is high spin $(3/2$ is the electronic spin $S),$ then the effective spin $(\widetilde{S})$ is also $3/2,$ but where is the fictitious spin $({S}')$ $1/2$ arising from if the molecule is, in fact, high spin?

Reference

  1. Piwowarska, D.; Gnutek, P.; Rudowicz, C. Origin of the Ground Kramers Doublets for $\ce{Co^2+}(\mathrm{3d^7})$ Ions with the Effective Spin 3/2 Versus the Fictitious ‘Spin’ ½. Appl Magn Reson 2019, 50 (6), 797–808. DOI: 10.1007/s00723-018-1080-4. (Open Access)
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Here's what I believe they're trying to say.

First, note that the discussion is mostly limited to high spin complexes (3 unpaired electrons out of the 7 d electrons), so the figure you reproduced is only representative of high spin states.

Second, recall that when D > 0 (ie left side of figure), the five d orbital energy levels split so that we have three t2g orbitals that are lower in energy than two e2 orbitals because of crystal field splitting. When D < 0 (right side of figure), the reverse is true, two e2 orbitals are lower in energy than three t2g orbitals.

In the absence of a crystal field, there is no difference in energy across the five orbitals, so for the case of 7 d electrons, we have a 4A state, which is the pre-split state in both cases in the figure.

In typical simple crystal field models, D>0 is associated with octahedral (i.e. OH) coordination and D<0 with tetrahedral (TH) or cubic coordination, but the authors give examples of TH with both D>0 and D<0, so the sign of D should not be assumed based on geometry here (nor should geometry be assumed based on sign).

Looking at the figure, then, we can start with the right side (because it's easier to understand) and see the transition from 4A to a state of D<0. That is the application of the crystal field. Since D<0, we know that the two e2 orbitals are lower in energy than the three t2g orbitals. Since we have high-spin d7, our electronic state is e24t2g3, with all three t2g electrons unpaired.

What we see in the figure is that there are two pairs of energy levels, with the states with spin $\pm \frac32$ at lower energy than those with spin $\pm \frac12$. The spin $\pm \frac32$ and $\pm \frac12$ values refer to the sum of the spins of the three unpaired electrons. The $\pm \frac32$ states are those in which all three t2g electrons have the same spin (either all $+\frac12$ or all $-\frac12$), while spin $\pm \frac12$ means that one of the three has a different spin than the other two. We know from Hund's rule that unpaired electrons in degenerate orbitals have the lowest energy when they all have the same spin direction, due to the exchange energy term. That is the source of the splitting in energy between the two possible total spin states.

We now get to the "fictitious spin" part. This is actually relatively simple. It has nothing to do with actual spin and simply indicates the number of experimentally relevant spin states, but is expressed in $\frac 12$ increments to have the familiar form of a spin value. Thus, fictitious spin of $\frac12$ just means two possible states, as one would observe for an electron with possible spins $\pm \frac12$. A fictitious spin of 1 would mean three possible states, as we would have for the case of two unpaired electrons.

In the figure, the fictitious spin is $\frac12$ because under the experimental conditions, only the two lower energy levels (ie actual spin of $\pm \frac32$) are observed. The two states with spin $\pm \frac12$ are higher enough in energy that they do not occur.

The left side can be interpreted similarly, except that here, when D>0, the three t2g orbitals are lower in energy than the two e2 orbitals, and the electron configuration is t2g5e22, with one of the t2g and both of the e2 electrons unpaired. Now, since the unpaired electrons are not in three degenerate orbitals, the lower energy states have the unpaired t2g opposite spin to the unpaired e2 electrons (which by Hund's rule again are the same spin).

The fictitious spin is interpreted in the same way on this side of the figure, simply meaning that only two states are accessible under the conditions of interest.

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  • $\begingroup$ Thank you so much for your detailed answer! $\endgroup$ – Henrique Junior Aug 19 '20 at 11:36

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