# Grounds state of carbon - Hunds Rule

I have a question about Hund's second rule for the ground state of carbon. Why is it that when S=1, L has to be 1 instead of 2? I don't get the whole symmetric/ antisymmetric argument. Why is L=2 symmetric and L=1 antisymmetric?

The $1s$ and $2s$ levels are filled, giving $j=0$. There remains two electrons to place in the $2p$ states. By Hund's rule, they will have $S=1$, which implies the spin wave function for the pair is symmetric under exchange of the electrons. This means the spatial part of the wavefunction must be antisymmetric for the total spatial$\times$spin wavefunction to be antisymmetric, as required by the fermionic nature of the electrons.
When combining two $\ell=1$ states, the states with total angular momentum $L=2$ and $L=0$ are symmetric, while states of total angular momentum $L=1$ are antisymmetric.
You can check the symmetries of the spatial part from the properties of the CG coefficients $C^{LM}_{\ell_1m_1;\ell_2m_2}$ under exchange of $\ell_1=\ell_2=1$: $$C^{LM}_{1m_1;1m_2}=(-1)^L C^{LM}_{1m_2;1m_1}$$ indicating that the even $L$'s are symmetric and the odd $L$ antisymmetric.