I am currently working through the physics governing how electrons arrange themselves within an atom into orbitals, and I have two related questions. Note that all the atoms I talk about are concerning its ground states.
The first is: how is total orbital angular momentum, $L$ (capital $L$) related to $M_L$? I know that for two electrons (say Carbon), $L=l_1+l_2$, ..., $|l_1-l_2|$; so $L$ can be $2$, $1$, $0$. Is this because if $l=1$ (say for valence of Carbon), then $m_l$ can be $1$, $0$, $-1$? Then why can $L$ not be $-2$ or $-1$? And, what happens for Nitrogen where there are three valence electrons? Is $L=l_1+l_2+l_3$, ..., $|l_1-l_2-l_3|$, giving $3$, $2$, or $1$? I know that $L=0$ is a possibility, so I am confused there as well.
My next question is concerning the symmetry arguments of identical particles. I know that the total wavefunction must be anti-symmetric because electrons are fermions. Going back to Carbon, when determining $L$ values, we first look at $L=2$ (because of Hund's second rule). However, the spin wavefunction is symmetric according to Hund's first rule, so we have to find an anti-symmetric spatial wavefunction. I read that we can rule out $L=2$ because $L=2$ is symmetric, and $L=1$ is anti-symmetric so that is the correct $L$. I do not understand how we can prove that $L=2$ is symmetric. I've read somewhere that $L=2$ is symmetric because the "top of the ladder" i.e. the $|2,2\rangle$ state is symmetric so the rest must be symmetric. Please explain why that is the case. Following that, why must we have 6 symmetric states and 3 antisymmetric states?