# Nuclear spin multiplicity of radicals

How do you calculate the nuclear spin multiplicity of radicals such as $\ce{N2H+}$? How does the net charge on the molecular species effect its overall nuclear spin?

The short answer is that the nuclear spin and electronic spin are separate effects. So an electronic radical ($\ce{N2H+}$) does not affect the nuclear spin of the protons or the nitrogen ($\ce{^14N}$ has nuclear spin 1).
If there's are unpaired electrons, paramagnetic NMR requires understanding the coupling of the electron spin and the nuclear spin. It's possible, but not as straight-forward as typical diamagnetic NMR (for example $\ce{^1H}$ peaks may spread over 200-250 ppm, unlike ~8-12 ppm for normal protons).
The effect of the contact term arises from transfer of spin polarization to the observed nucleus. Spin polarization is a consequence of the very strong electron-nuclear (NMR detected) interaction. This coupling, also known by EPR spectroscopists as hyperfine coupling, is on the order of MHz, vs the usual internuclear (J) coupling observed in conventional NMR spectra, which are on the order of a few Hz. This difference reflects the large magnetic moment of an electron (−1.00 μB), which is much greater than any nuclear magnetic moment (e.g. for $\ce{^1H}$: 1.52×10−3 μB).