From the spin-only formula we can predict that for tetrahedral cobalt (II) complexes $$m_{eff} = 3.87 \mu_B $$

This ignores orbital angular momentum effects, which result in higher magnetic moments for the tetrahedral halide complexes (Hund's 3rd rule results in spin-orbit coupling "together"). Why do the Co(II) cyanide and Co (III) alkyl complexes have lower magnetic moments than the spin only formula predict ? I have quoted values of 2.15 $\mu_B$ and ~3 $\mu_B$ for these cases.


1 Answer 1


The question isn't really clear exactly what complexes you are referring to. However, cyanide is quite a high field ligand, so perhaps the Co(II) cyanide compound is low spin while the 3.87 value is for high spin (3/2). I don't see how it could be low spin if it is really tetrahedral, but isn't $\ce {Co(CN)_4^2-}$ square planar?

For Co(III) alkyl, alkyl ligands can also be relatively high field, and it is at least theoretically possible to have low spin (s=1) Co(III) tetrahedral complex. However, it is rare for tetrahedral complexes to be low spin because the ligand field splitting energy is lower than for octahedral.

If those are not the reasons, see Iron, Cobalt, and Nickel Complexes having Anomalous Magnetic Moments Quarterly Review Chemical Society volume 22, pp 457-498.

  • $\begingroup$ The cyanide complex is indeed square planar, which explains the magnetic data - my textbook didn't make that clear. As for Co(III) alkyl, assuming an e(3) t2(3) configuration in a tetrahedral ligand field you would have a magnetic moment of 4.89 - but the observed values are much lower. I assume square planar geometry would be out of the question with a ligand such as norbornyl (tetra Co(III) complex magnetic moment 3.15). $\endgroup$
    – J. LS
    Dec 11, 2014 at 21:30
  • $\begingroup$ norbornyl complexes were the first ever discovered low-spin tetrahedral complex of first-row transition metals. pubs.rsc.org/en/content/articlehtml/1986/c3/c39860001491 $\endgroup$
    – DavePhD
    Dec 11, 2014 at 21:43
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    $\begingroup$ but it is Co(IV) norbornyl complex that is low spin tetrahedral $\endgroup$
    – DavePhD
    Dec 11, 2014 at 22:01
  • $\begingroup$ @J.LS magnetic moment of 3.15 would mean cobalt (III) norbornyl complex is low spin (s=1) tetrahedral, so it looks like both Co(III) and Co(IV) norbornyl complex are both extremely rare low spin tetrahedral complexes $\endgroup$
    – DavePhD
    Dec 12, 2014 at 11:49
  • $\begingroup$ @DavePhD we may also go the other way: metal(V) complexes "should be" low spin ... unless they not only are tetrahedral but also have an exceptionally pi-donating ligand. Like this case, with cobalt at that. $\endgroup$ Oct 27, 2023 at 15:13

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