= Solution
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 \b[$J^P=5/2^+$]. 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 \b[$J^P=1/2^-$]. <beryllium-10> is an <even-even nucleus>; pairing its four <protons> and six <neutrons> gives \b[$J^P=0^+$].
<boron-10> needs an additional qualification. The fifth <proton> and fifth <neutron> each occupy $1p_{3/2}$. <addition of angular momentum> permits
$$
J=0,1,2,3,\qquad P=(-1)^{1+1}=+1.
$$
The experimentally correct ground-state assignment is \b[$3^+$], 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.<Energy levels of boron-10 by TUNL (2004)>[Table 10.18, ground state and first excited state, physical page 1]{ref}
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