Boron Ground State

Boron, with $Z=5$ has the 1S and 2S levels filled. They add up to $j=0$ as do all closed shells. The valence electron is in the 2P state and hence has $\ell=1$ and $s={1\over 2}$. Since the shell is not half full we couple to the the lowest $j=\vert\ell-s\vert={1\over 2}$. So the ground state is $^2P_{1\over 2}$.

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