# Total orbital angular momentum in a filled p subshell

From "Physical Chemistry", P. Atkins, J. De Paula, 9th edition, page 358

"A closed shell has zero orbital angular momentum because all the individual orbital angular momenta sum to zero. Therefore, when working out term symbols, we need consider only the electrons of the unfilled shell."

So, a configuration like $$[\text{He}]1s^2$$ or $$[\text{He}]1s^2 2p^6$$ should have $$L = 0$$, right? The first presents $$L = 0$$ for sure, since $$l = 0$$ for $$s$$ orbitals and applying the Clebsch-Gordan series we obtain $$0$$. Doing the same for the second configuration, on the other hand, I obtain values other than $$0$$. Could someone explain to me better?

One way of looking at this is to consider that if $$L = 1$$ (for example), then there must be a state with $$M_L = -1$$ (because $$M_L$$ ranges from $$-L$$ to $$L$$ in integer steps).
However, $$M_L$$ is not obtained by a Clebsch–Gordan series: instead, it's just a simple addition of the individual $$m_l$$'s of each electron. That is:
$$M_L = m_{l,1} + m_{l,2} + m_{l,3} + m_{l,4} + m_{l,5} + m_{l,6}$$
and for a p subshell, the allowed values of $$m_l$$ are $$-1$$, $$0$$, $$+1$$ (there are two of each). So $$M_L$$ for a full p subshell must be $$0$$, and it's then a contradiction to have $$L \neq 0$$ as that would imply that $$M_L$$ can have a nonzero value.
This is basically a manifestation of the Pauli exclusion principle, because it is precisely that which forces the sum of $$m_l$$'s to be zero. If you read around in Atkins, it will mention how certain term symbols are forbidden due to the Pauli exclusion principle. This is a similar case.