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Is there a relation (or formula) between the sign of the magnetic susceptibility and the gyromagnetic ratio of a nucleus in NMR?

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I should preface this with the caveat that I'm a chemist and not a physicist, so it's possible that I haven't understood your question correctly.

Magnetic susceptibility, $\chi$, is an electronic property, not a nuclear property, so one could simply write the question off as not making sense. That being said, let's look a little closer as I believe we can demonstrate an answer to this question.

First, magnetic susceptibilities are typically divided into diamagnetic susceptibilities (which are negative and arise from paired electrons), paramagnetic susceptibilities (which are positive and arise from unpaired electrons), and then a few others (ferromagnetic, ferrimagnetic, etc. which arise from the cooperative behavior of unpaired electrons).

Second, the shell theory of the nucleus provides a nice picture when discussing the NMR properties of different isotopes as we can picture protons and neutrons existing in discrete nuclear energy levels and being paired or unpaired, much like chemists picture electrons in atomic or molecular orbitals with their discrete energy levels. One of the neat results from this approach is that any isotope with an even number of protons and and even number of neutrons will have all of the protons paired up and all of the neutrons paired up, resulting in a total nuclear spin = 0. Thus $\ce{^4He}$, $\ce{^12C}$, and $\ce{^16O}$ are all expected to be spin 0. On the other hand, an isotope with an odd number of protons will have an unpaired proton; and an odd number of neutrons gives rise to an unpaired neutron. Having a single unpaired nucleon gives rise to a half-integer nuclear spin (1/2, 3/2, 5/2, etc.) where the unpaired nucleon provides a unit of 1/2 to the angular momentum and any remaining integer spin comes from the nuclear equivalent of orbital angular momentum ($l=0,1,2,$ etc.) Having both an unpaired proton and neutron gives you an integer-spin nucleus (1, 2, 3, 4, etc.) where 1 unit of angular momentum comes from the two unpaired nucleons and the rest comes from the shell theory equivalent of orbital angular momentum.

OK, so why I am boring you with all of this detail? Well, nuclei with unpaired nucleons are our nuclear analogues to atoms or molecules with unpaired electrons. Thus, all NMR active isotopes (spin > 0) can be considered to exhibit nuclear paramagnetism. It's the presence of the unpaired nucleon that gives rise to a non-zero spin angular momentum. On the other hand, all NMR inactive isotopes (spin = 0) would be expected to exhibit nuclear diamagnetism.

Now, at this point I'm venturing beyond what I actually know, as I have never seen any values reported for nuclear magnetic susceptibility. However, one would expect that, analogous to the electronic case, nuclear paramagnetism should have a positive susceptibility and nuclear diamagnetism should have a negative susceptibility. In other words, the only nuclei that should have negative nuclear magnetic susceptibilities have a spin of 0 and no gyromagnetic ratio. Any nucleus that is NMR active should have a positive nuclear magnetic susceptibility; but, some NMR active isotopes have positive gyromagnetic ratios and others have negative gyromagnetic ratios.

So the answer to your question (if I've understood it correctly) would be NO, there is no formula relating the sign of the gyromagnetic ratio to the magnetic susceptibility.

In the case of normal (electronic) magnetic susceptibility, the nucleus plays no role (or at least such a small role that it's completely negligible) and in the case of the extremely weak nuclear magnetic susceptibility, all NMR active nuclei have positive susceptibilities regardless of the sign of their gyromagnetic ratios and all NMR inactive nuclei have negative susceptibilities and no gyromagnetic ratios.

As a footnote that you may or may not find interesting. Picturing integer-spin nuclei as the nuclear equivalent of di-radicals gives a nice rationalization for why they are so rare (only ~8 stable isotopes with integer spin vs. ~112 stable isotopes with half-integer spin vs. I don't know how many isotopes with spin = 0). Just like molecules, the di-radical isotopes are quite unstable, with some exceptions (e.g., $\ce{^2H}$ is spin=1 and has an unpaired proton and neutron, but is a stable isotope; similarly, molecular oxygen is probably the most abundant di-radical (note: you need M.O. theory to see why it's a di-radical, but you can easily show that liquid $\ce{O2}$ is paramagentic - go do a youtube search) and is relatively stable, at least compared to many other di-radicals).

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    $\begingroup$ Your last paragraph makes me want to shout ‘Oh nitrogen, why do you have to be special! D=’ $\endgroup$
    – Jan
    Commented Oct 21, 2015 at 17:28
  • $\begingroup$ I also indented to point out that we can not yet predict gyromagnetic ratios from the known nuclear structure (neither the sign nor the magnitude of the gyromagnetic ratio). I believe that we are able to predict the values of the nuclear spin (I vaguely recall that the shell theory is able to predict the spin angular momentum for excited nuclear states, which is something neat to think about). However, my understanding of nuclear structure pretty much ends with the shell theory (and a rather superficial understanding at that). $\endgroup$
    – S. Burt
    Commented Oct 21, 2015 at 18:14

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