Carbon Ground State

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

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 \bgroup\color{black}$(1s)^2(2s)^2(2p)^1$\egroup is assumed and the state of the sixth electron is given. Some states have other excited electrons and are indicated by a superscript. Different \bgroup\color{black}$j$\egroup states are not shown since the splitting is small. Electric dipole transitions are shown changing \bgroup\color{black}$\ell$\egroup by one unit.


The ground state has \bgroup\color{black}$s=1$\egroup and \bgroup\color{black}$\ell=1$\egroup as we predicted. Other states labeled \bgroup\color{black}$2p$\egroup 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.

Jim Branson 2013-04-22