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I was studying some coordination chemistry and somewhere around the text it was stated that one could know the magnetic moment of a complex from its metal because since all ligands were often diamagnetic, closed shell molecules, the unpaired electrons were centered on the metal anyway. It is implied that the spin of all of those unpaired electrons is the source of the paramagnetism.

So my question is this: From the common isotopes of C, H, N, S, O and P only oxygen-16 and carbon-12 have zero spin. Then why don't the spin of the other nuclei do not contribute to the paramagnetism of the sample the way electrons do? Particularly hydrogen, which is often quite abundant in organic ligands and has a spin of 1/2, same as the electron.

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    $\begingroup$ The paragraph was obvisouly talking about electron spin, not nuclear spin. Nuclear spin has very little contribution to magnetic properties, as nuclear magneton is much smaller than the corresponding electronic magnetic moment (Bohr magneton), due to the much higher charge/mass ratio of the electron $\endgroup$
    – Greg
    Nov 23 at 6:19
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    $\begingroup$ Do you refer to compounds like TEMPO, subject to electron paramagnetic resonance (EPR)/electron spin resonance (ESR)? $\endgroup$
    – Buttonwood
    Nov 23 at 6:53
  • $\begingroup$ See also Nuclear_magneton $\endgroup$
    – Poutnik
    Nov 23 at 8:47
  • $\begingroup$ @Greg I believe you have nailed the question. Would you mind to write an answer covering those bases so I can close the question? $\endgroup$
    – urquiza
    Nov 23 at 11:14
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As requested, this is my answer: The question is essentially about why nuclear contributions to magnetic behavior is generally neglected.

It is true that both nuclei and the electrons can carry unpaired spin in a material. The magnetic moment associated with these spin is proportional to two constants: the nuclear magneton in the case of nucleus and the Bohr magneton in case of electrons. These are the constants essentially telling you how strong the given spin interact with the magnetic field, and they are inversely proportionate with the mass of the particle. In other words, they depend on how big is the charge of the particle compared to its mass. Since electrons are about 1800 times lighter than a nucleon, the magnetic properties of a material will primarily dependent on the electronic spin and the contribution of nuclear spin is negligible in most cases.

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    $\begingroup$ So if I understand it well, nonzero nucleus spin leads to very slight paramagnetism contribution, which in total effect just very slightly decreases the diamagnetism of the material, but cannot cause paramagnetism without electron contribution. $\endgroup$
    – Poutnik
    Nov 23 at 15:27
  • $\begingroup$ @Poutnik yes, essentially that is what happens $\endgroup$
    – Greg
    Nov 23 at 16:50
  • $\begingroup$ @Poutnik and about the electrons, there is the spin and also the orbital magnetic moments to consider. It's just that, for light elements, the orbital component is often negligible in comparison to the spin component. $\endgroup$
    – urquiza
    Nov 24 at 13:44
  • $\begingroup$ @urquiza Yes, I am aware about both electron angular momenta. $\endgroup$
    – Poutnik
    Nov 24 at 13:48
  • $\begingroup$ @urquiza Orbital momentum generally breaks down due to low symmetry, its contribution is low even in lighter transition metal complexes (except cobalt, but that is it). $\endgroup$
    – Greg
    Nov 25 at 2:28

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