According to the nuclear shell model, $^{19}F$ has one unpaired proton in the $6$-fold degenerate $1d_{5/2}$ state, which means the orbital angular momentum is $l = 2$ and the total angular momentum is $j = l+ \frac12 = 5/2$. If we follow the [rule highlighted by Lubos Motl][1] that nuclear states with higher $m_j$ are filled first, then the one unpaired proton goes in the $m_j = 5/2$ state predicting a nuclear spin of $5/2$. However, we know that $^{19}F$ has a nuclear spin of $1/2$.

Why is that? Is the selection rule claimed by Lubos Motl simply wrong? If so, what is the appropriate way to fill the states and calculate nuclear spin? For example, [why does $^{23}Na$ have a nuclear spin of $3/2$][2]?


  [1]: https://physics.stackexchange.com/a/90763/76347
  [2]: https://physics.stackexchange.com/q/90741/76347