Compatibility of spectroscopic term symbols with Pauli Exclusion Principle

Question

The ground electronic configuration of Carbon is $1s^2$$2s^2$$2p^2$

$l_1=1$ and $l_2=1$ $\implies$ $L=2,1,0$

$s_1 = \frac{1}{2}$ and $s_2=\frac{1}{2}$ $\implies S=1,0$

So the terms are $^{3}D,^{1}D,^{3}P,^{1}P,^{3}S,^{1}S$. However, only $^{1}D,^{3}P,^{1}S$ will survive because of Pauli exclusion principle.

My question is how does the Pauli Exclusion Principle play a role in retaining certain terms like $^{1}D,^{3}P,^{1}S$ while rejecting others?

EDITED:

I think I now understand that it is because the Pauli Exclusion principle demands the total wave function to be anti-symmetric. Hence, the L and S combinations must together be anti-symmetric. However, I still need to know how to classify L and S as symmetric and anti-symmetric depending on its value.

Answer 1

The two electrons in the $p^2$ shells must have at least one different quantum numbers. The available quantum numbers are $n$, $\ell$, $m_{\ell}$, $s$, and $m_s$ (the spin projection).

Because they are fermions, Pauli forces to differ in at least one quantum number.

$L=2$ term

This gives you a $D$ term. In order for the two $\ell = 1$ electrons to make a $L=2$ total angular momentum, they must be aligned. So $m_{\ell_1} = m_{\ell_2}$.

The only quantum number that can differ is the spin projection: $m_{s_1} = -m_{s_2}$.

Vectorially, this would give you a total spin of $\mathbf{s}_1 + \mathbf{s}_1 = \mathbf{0}$, corresponding to the total spin quantum number $S=0$.

Hence why you should only retain the $S=0$ terms: $^{1}D_{2}$.

$L=1$ term

This gives you the $P$ term.

This would require maths that you can find here but you can show that the orbintal angular momentum part of the state is a triplet and it is antisymmetric. Hence, you need the spin part to be symmetric, i.e. to be a spin triple and hence $S=1$.

Hence $^3P_{2}$.

$L=0$ term

This gives you the $S$ term. The reason is the same as for the $L=0$ term. Hence you need different spin orientations. So $S=0$ and $^{1}S_0$.

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The actual proof is (I think) quite mathematical with Slater determinants, but the general rule of thumb is that:

For odd numbers for the total angular momentum $(L= 1,3,5, ...)$ the spatial wavefunction is antisymmetric upon particle exchange.

You know that $S=0$ is always antisymmetric and $S=1$ always symmetric, so you could derive which $D, P, S$ also from that.