Carbon Ground State

Carbon, with $Z=6$ has the 1S and 2S levels filled giving $j=0$ as a base. It has two valence 2P electrons. Hund's first rule , maximum total $s$, tells us to couple the two electron spins to $s=1$. This is the symmetric spin state so we'll need to make the space state antisymmetric. Hund's second rule, maximum $\ell$, doesn't play a role because only the $\ell=1$ state is antisymmetric. Remember, adding two P states together, we get total $\ell=0,1,2$. The maximum state is symmetric, the next antisymmetric, and the $\ell=0$ state is again symmetric under interchange. This means $\ell=1$ is the only option. Since the shell is not half full we couple to the the lowest $j=\vert\ell-s\vert=0$. So the ground state is $^3P_{0}$. The simpler way works with a table.

$m_\ell$	e
1	$\uparrow$
0	$\uparrow$
-1
$s=\sum m_s=1$
$\ell=\sum m_\ell=1$

We can take a look at the excited states of carbon to get an appreciation of Hund's rules. The following chart shows the states of a carbon atom. For most states, a basis of $(1s)^2(2s)^2(2p)^1$ is assumed and the state of the sixth electron is given. Some states have other excited electrons and are indicated by a superscript. Different $j$ states are not shown since the splitting is small. Electric dipole transitions are shown changing $\ell$ by one unit.

The ground state has $s=1$ and $\ell=1$ as we predicted. Other states labeled $2p$ are the ones that Hund's first two rules determined to be of higher energy. They are both spin singlets so its the symmetry of the space wavefunction that is making the difference here.