Past exam of the advanced quantum physics course of the University of Cambridge / 2019 / Section A / 1 / Solution

The closed $1s^2$ and $2s^2$ subshells contribute zero total orbital angular momentum and spin, so only the two $2p$ electrons matter. The first of Hund's rules maximizes their total spin, giving $S=1$. The Pauli exclusion principle then requires an antisymmetric spatial wavefunction. Two parallel-spin electrons cannot occupy the same $m_l$ orbital: the greatest possible orbital projection is $1+0=1$, so the maximal allowed $L$ within this triplet is $L=1$.

The ground term is therefore ${}^{3}P$, in term symbol notation ${}^{2S+1}{L}_J$. Since the $p^2$ subshell contains fewer than half of its six possible electrons, the third of Hund's rules places the smallest $J$ lowest. Addition of angular momentum gives $J=0,1,2$, in that energy order within the ground term. ThusThese are the fine structure partners of the same electronic term, with magnetic multiplicities $1,3,5$, rather than different principal-shell excitations. The ordering agrees with the tabulated lowest neutral-carbon levels.^[ref]