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An atomic nucleus with odd proton number and odd neutron number. Both unpaired species can contribute to the total nuclear spin; addition of angular momentum gives possible couplings, while the nuclear residual interaction determines their ordering.

In the nuclear shell model, a level with total angular momentum $j$ has $2j+1$ substates for each nucleon species. The first three levels accommodate $2+4+2=8$ protons or eight neutrons. nuclear pairing makes a filled shell or a pair of like nucleons contribute $J=0$, and an orbital has spatial parity $(-1{)}^{\ell }$.

For oxygen-17, eight protons and eight neutrons form a closed core; the remaining neutron occupies $1d_{5/2}$, so $J^P=5/2^{+}$. For nitrogen-15, eight neutrons close their shell, while the seventh proton occupies $1p_{1/2}$. Equivalently it is a proton hole relative to the closed eight-proton shell, giving $J^P=1/2^{-}$. beryllium-10 is an even-even nucleus; pairing its four protons and six neutrons gives $J^P=0^{+}$.

boron-10 needs an additional qualification. The fifth proton and fifth neutron each occupy $1p_{3/2}$. addition of angular momentum permitsThe experimentally correct ground-state assignment is $3^{+}$, corresponding to aligned unpaired $j=3/2$ contributions in this simple description. However, one-particle level ordering alone does not select it over the other couplings: their splitting requires the nuclear residual interaction. Assuming a preference for the aligned proton-neutron coupling gives the intended prediction; without that assumption, the strictly deducible result is the allowed positive-parity multiplet. The evaluated ground state is $3^{+},T=0$, with a $1^{+},T=0$ level above it.^[ref]