Spin of Nucleus

In the nuclear shell model, $^{14}$N is seen as an inert $^{12}$C core, with 2 extra nucleons. The ground state for the strong nuclear force is going to put these nucleons in the same spatial state (a symmetric S state) and the same spin state (which for spin 1/2 particles is the spin-1 triplet state). Anti-symmetrization of the wave function is accomplished in the isospin sector via the singlet isospin wave function:

$$ |I=0, I_3=0rangle = frac 1 {sqrt 2}[|prangle|nrangle - |nrangle|prangle]$$

It's important to note that individual nucleons do not have a definite "proton" or "neutron" identity with isospin. It is exactly the same as a spin 1/2 particle not having a definite spin.

So we can treat each $^{14}$N as indistinguishable vector bosons.

You can work out the Clebsch-Gordan coefficients for combining two vectors, but the general structure is:

$$ {bf 3} otimes {bf 3} = {bf 5}_S oplus {bf 3}_A oplus {bf 1}_S $$

which means the spin-2 and spin-0 multiplets are symmetric under interchange and the spin-1 combination is antisymmetric. For example:

$$|J=1,J_z=0rangle = frac 1 {sqrt 2}[|1,+1rangle_1|1,-1rangle_2 - |1,-1rangle_1|1,+1rangle_2]$$

while:

$$|J=0,J_z=0rangle = frac 1 {sqrt 3}[|1,+1rangle_1|1,-1rangle_2 + |1,-1rangle_1|1,+1rangle_2 - |1,0rangle_1|1,0rangle_2 ]$$

The molecular wave function then needs to have the same symmetry as the spin wave function to make the overall wave function symmetric.