States of $^{18}$O using nuclear shell model Using the nuclear shell model it can be proved that in the external shell $1d_{5/2}$ of $^{18}$O there are two neutrons. Thus, by composition of angular momentum, the possible states of the nucleus are:
$I^P$ = $0^+$ , $1^+$, $2^+$, $3^+$, $4^+$, $5^+$.
During the lecture my professor said that experimentally we can find only the states $0^+$ , $2^+$, $4^+$ and this lack can be explained through a reasoning based on isospin.
Why the states $1^+$, $3^+$, $5^+$ are prohibited?
 A: You (your professor) have selected a case, where $^{16}O$ is magic nucleus and it works as a core for the games of orbitals above and it is much easier to put strong statements like it can be proved that in the external shell $1d_{5/2}$ there are two nucleons in $^{18}O$. We could start a discussion here...
Neverthless:  you cannot find $5^+$, it would be against Pauli principle (same projections m=+5/2). Wouldnt it?
More generally, for identical nucleons, the isospin projection $T_z = t_z(1) + t_z(2) = 1$ and hence total isospin is $T=1$. This is a symetric wavefunction (look at the deuteron and pp and nn cases). This implies (from Pauli principle), that the space-spin part must be antisymetric .... 
$\Psi(j^2JM) = N \sum_{m,m'}{ (jmjm'|jjJM) [\phi_1(m)\phi_2(m')- \phi_1(m')\phi_2(m)]}$
can be rewritten as
$\Psi(j^2JM) = N \sum_{m,m'}{ (jmjm'|jjJM) - (jm'jm|jjJM) )\ \phi_1(m)\phi_2(m')}$
and for these two Clebsh -Gordan coefficients there is a rule
$\Psi(j^2JM) = N [1-(-1)^{2j-J}]  \sum_{m,m'}{ (jmjm'|jjJM) \  \phi_1(m)\phi_2(m')}$
In your case $2j=5$ is odd, so $J$ must be even, else the right side vanishes. And here you have the total angular momentum possible from $J=0,...(2j-1)$. With pure isospin $T=1$ as a gift. The $p-n$ interaction can have both isospins, which ends up with some mixing when $j_p \ne j_n$.
Check the Casten's book (I forgot the title, but it is quite famous).
