# Finding the ground state $L$ and $S$ quantum numbers of an atom

In an example in class we were asked to determine the ground state total orbital angular moment and total spin angular momentum quantum numbers $\textbf{L}$ and $\textbf{S}$ of Nitrogen with electron configuration

$$N:[\mathrm{He}]\,2s^22p^3$$

We are told to used Hund's rules which were given as the following.

1. Find the maximum $\space M_S$ consistent with the Pauli Exclusion Principle. Set $S=M_s$

2. For that$\space M_S$, find the maximum $M_l$. Set $L=M_l$

It was then presented that the result is as follows.

$$\max(M_s)=\frac{3}{2}\implies S = \frac{3}{2}$$

$$\max(M_l)=0\implies L = 0$$

$$\therefore S = \frac{3}{2}\space \text{and} \space L = 0$$

I am trying to do homework problems similar to this and can not figure out how they reasoned $L = 0$ and was hoping to gain some clarification if at all possible.When moving to atoms with partially filled $\space f$ and $\space d$ orbitals I don't know where to start and I think it is because I am not sure how they approached this problem. Any help clarifying this process would be much appreciated.

• the maximum M l should be equal to 1(for the case of p orbital). Why has it been taken to be 0? – Rishabh Jain Dec 11 '17 at 8:38

The given spin state requires all the spins to be parallel, which means that that state (and the whole $S=3/2$ manifold by extension) is symmetric under exchange. However, the global state needs to be antisymmetric, which means that the orbital component also needs to be antisymmetric. Within a p shell, this can only be achieved by putting one electron each on the $m_L=1$, $0$ and -$1$ states (since any repetitions would vanish under antisymmetrization), and that then gives you $M_L=0$.