Nitrogen Ground State

Now, with $Z=7$ we have three valence 2P electrons and the shell is half full. Hund's first rule , maximum total $s$, tells us to couple the three electron spins to $s={3\over 2}$. This is again the symmetric spin state so we'll need to make the space state antisymmetric. We now have the truly nasty problem of figuring out which total $\ell$ states are totally antisymmetric. All I have to say is $3\otimes 3\otimes 3 = 7_S \oplus 5_{MS} \oplus 3_{MS} \oplus 5_{MA} \oplus 3_{MA} \oplus 1_A \oplus 3_{MS}$. Here MS means mixed symmetric. That is; it is symmetric under the interchange of two of the electrons but not with the third. Remember, adding two P states together, we get total $\ell_{12}=0,1,2$. Adding another P state to each of these gives total $\ell=1$ for $\ell_{12}=0$, $\ell=0,1,2$ for $\ell_{12}=1$, and $\ell=1,2,3$ for $\ell_{12}=2$. Hund's second rule, maximum $\ell$, doesn't play a role, again, because only the $\ell=0$ state is totally antisymmetric. Since the shell is just half full we couple to the the lowest $j=\vert\ell-s\vert={3\over 2}$. So the ground state is $^4S_{3\over 2}$.

$m_\ell$	e
1	$\uparrow$
0	$\uparrow$
-1	$\uparrow$
$s=\sum m_s={3\over 2}$
$\ell=\sum m_\ell=0$

The chart of nitrogen states is similar to the chart in the last section. Our prediction of the ground state is again correct and a few space symmetric states end up a few eV higher than the ground state.