How to prove Nitrogen atom with total angular momentum $L=2$ and $L=1$ are not anti-symmetric? I'm working on Problem 5.13(d) in "Griffiths 《Introduction of Quantum Mechanics》 2nd Edition".
It asked to determine the nitrogen electron configuration by Hund's rule. And here is the answer:

It didn't prove mathmetically that L=2 and L=1 wavefunction are not anti-symmetric. I am trying to prove L=2 wavefunction is not anti-symmetric using Clebsch-Gordan coefficients by myself but got stuck.
Evidently, I don't know how to decompose the L=2 state into direct product of three electron state. The Clebsch-Gordan coefficients table seems only for two electrons.
For example the top ladder of the L=2 state $|LM\rangle=|22\rangle$, how to decomposed into three electron state $|l_1m_1\rangle|l_2m_2\rangle|l_3m_3\rangle$? I try to consider first decompose into $|l_{12}m_{12}\rangle|l_3m_3\rangle$ and then $|l_{12}m_{12}\rangle=|l_1m_1\rangle|l_2m_2\rangle$, but it still a mess. Here is what I've done:
According to Clebsch-Gordan coefficients table: $|LM\rangle=|22\rangle=\sqrt{\frac{2}{3}}|22\rangle|10\rangle-\sqrt{\frac{1}{3}}|21\rangle|11\rangle$, thus $|l_{12}m_{12}\rangle$ can be $|22\rangle or \quad |21\rangle$, thus then I decompose $|l_{12}m_{12}\rangle=|22\rangle$ or $|21\rangle$ into $\quad|l_1m_1\rangle|l_2m_2\rangle$ by Clebsch-Gordan coefficient again, then plug them in the above formula $|LM\rangle$. I finally got: $$|LM\rangle=|22\rangle=\sqrt{\frac{2}{3}}|11\rangle|11\rangle|10\rangle-\sqrt{\frac{1}{3}}(\sqrt{\frac{1}{2}}|11\rangle|10\rangle|11\rangle+\sqrt{\frac{1}{2}}|10\rangle|11\rangle|11\rangle).$$
But it's neither symmetric nor anti-symmetric. I don't know what's wrong.
 A: When dealing with $3$ (or more) particles, there exists states of mixed symmetry that are neither fully symmetric nor fully antisymmetric.  Than can occur when a value of total $L$ occurs more than once in the list of possible total angular momenta.
Thus, this will also happen with three spin-$1/2$ particles.  You can verify that in this case you get $S=3/2$ once and $S=1/2$ twice.  The two sets of states with $S=1/2$, i.e. $\vert 1/2,m_s\rangle_1$ and $\vert 1/2,m_s\rangle_2$, with $m_s=\pm 1/2$, are neither symmetric nor antisymmetric.  You can construct them by hand (there are $4$ in total) by first coupling two spins to get states of $s=1$ and $s=0$, and then construct a $S=1/2$ set from the $s=1$ states and the last spin-1/2 particle, and a second set from the $s=0$ state and the last spin-1/2 particles.
For instance, this last set is
\begin{align}
\frac{1}{\sqrt{2}}\left(\vert +\rangle\vert -\rangle-\vert -\rangle\vert +\rangle\right)\vert +\rangle\, ,\qquad 
\frac{1}{\sqrt{2}}\left(\vert +\rangle \vert -\rangle -\vert -\rangle \vert +\rangle\right)\vert -\rangle\, ,
\end{align}
which is antisymmetric under permutation of particles 1 and 2, but neither symmetric nor antisymmetric under permutation of 1 and 3, or permutation of $2$ and $3$.
It seems in fact that what you have done for your $\vert 22\rangle$ state is akin to this, i.e. you have found that $L=2$ states have mixed symmetry.
Constructing states of with a given permutation symmetry (symmetric, antisymmetric, or mixed symmetry) is usually non-trivial and requires quite a bit of patience but you are basically on the right track: you would construct a state of type $\vert L,L\rangle$ and examine its symmetry property under permutation.  All other states obtained by lowering with $L_-$ will have the same permutation symmetry.  What’s a bit tricky is that you might get sets of states of specific $L$, some of which might have a given permutation symmetry, some of which may have another type of permutation symmetry.  For instance if you place $3$ particles in the $n=1$ states of a 3D-harmonic oscillator, you can find some resulting $L=1$ states that are fully symmetric and some that have mixed symmetry.
Normally to keep track of all this one must use the machinery that comes with the representation theory of the permutation (or symemetric) group.
