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. Note that the chart method is clearly easier to use in this case. 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.