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For metal complexes with $A_{2}$ or $E$ ground state terms there is angular momentum contribution to the magnetic moment which is generally positive for more than half-filled subshells and negative otherwise, and related to the spin-only magnetic moment by $$m_\mathrm{eff} = m_\mathrm{SO}\left(1-\frac{\alpha\lambda}{\Delta}\right)$$

where $\Delta$ is the ligand field splitting parameter, $\alpha$ is the orbital angular momentum degeneracy of the state and $\lambda$ is a constant.

My questions:

  • When do exceptions to this behaviour occur ? For example I have found that $\ce{Co(bipy)_{3}(ClO4)}$ has a lower experimental magnetic moment than the spin only formula predicts despite being a $\ce{Co(I)}$ (d8) compound with a $^{3}\!A_{2g}$ ground state, but I do not understand why.
  • Is the deviation from the spin only formula always positive for $T$ states ? If not can we predict its sign similarly ? EDIT: a simple way to do this is via Hund's third rule. If the d-orbitals are half or more full, the highest J state will be lowest in energy, and then the deviation for the T state will be positive ! Exactly why is probably rather subtle.
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  • $\begingroup$ Just one little thing: your experimental data may or may not be good. It happens with published, cited data, too. There are several factors (impurities, anisotropy, mainly) that can influence measure values. Anisotropy is especially important in materials with large angular momentum... $\endgroup$
    – Greg
    May 14, 2015 at 11:39

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When do exceptions to this behaviour occur ? For example I have found that $\ce{Co(bipy)_{3}(ClO4)}$ has a lower experimental magnetic moment than the spin only formula predicts despite being a $\ce{Co(I)}$ (d8) compound with a $^{3}\!A_{2g}$ ground state, but I do not understand why.

According to ADVANCES IN INORGANIC CHEMISTRY AND RADIOCHEMISTRY, Volume 12, page 193 "The tris(bipyridal) cobalt(I) cation has a magnetic momment of 3.2-3.4 B.M. which is independent of temperature from 70K to 340K".

Since there are two unpaired electrons, the spin only value is $\sqrt{2(2+2)} =2.83$. So your experimental data are inconsistent with the literature, as 3.2-3.4 is greater than the spin only value, 2.83.

Is the deviation from the spin only formula always positive for $T$ states ? If not can we predict its sign similarly ?

No, the deviation from the spin only value is not always positive for complexes with T ground states. The main characteristic of complexes with T ground states is that the magnetic moment is temperature dependent, because the spin-orbital coupling splits the T state into states that differ in energy by an amount close to kT.

For example $\ce{Cs2VCl6}$ has a magnetic moment of 1.4 at 80K and 1.8 at 300K, with the spin only formula predicting 1.73. So deviation can be negative, and isn't necessarily positive or negative even for the same compound, depending upon temperature.

See section 4.3 of Dr. Mark D. Spicer's magnetism lecture for more information.

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  • $\begingroup$ My value for the magnetic moment is 2.05 BM (source unpublished, 2005) - is there a modern source corroborating your higher value ? $\endgroup$
    – J. LS
    Apr 7, 2015 at 16:41
  • $\begingroup$ @J.LS Not exactly, J. Chem. Soc., Dalton Trans., 1997, Pages 519–530 cites to 2 references that have values of 2.53 and 2.89, but these are not really more modern, even though the citing article is. $\endgroup$
    – DavePhD
    Apr 7, 2015 at 17:42
  • $\begingroup$ Why might you get a value lower than the spin only magnetic moment in this case ? Is this a TIP effect ? $\endgroup$
    – J. LS
    Apr 10, 2015 at 10:19
  • $\begingroup$ @J.LS I don't really have a good answer for you. The value 2.05 is significantly lower than the 3 indepedent published values (2.53, 2.89 and 3.2-3.4) as well as the value of 2.88 for the isoelectronic Ni(II) complex. There is discussion in the Dalton Trans. reference of whether the complex is really Co(II) with a bipy being reduced, but the magnetic moment values are said to rule this out. With a different magnetic moment value, perhaps a different conclusion would be reached. $\endgroup$
    – DavePhD
    Apr 10, 2015 at 12:59

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