Term symbol of a molecule in an excited state

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Aleolomorfo
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Homework Statement


Considering the molecule of ##C_2^+## in an excited state with valence configuration ##2\sigma_g^2## ##2\sigma_u^2## ##1\pi_u^2## ##3\sigma_g^1##. Finding all the possible term symbol.

Homework Equations


Term symbol ##^{2S+1}\Lambda##
##L=0 \rightarrow \Sigma##
##L=1 \rightarrow \Pi##
##L=2 \rightarrow \Phi##

The Attempt at a Solution


The problem is solved but I have some doubts about the solution. First of all, the term symbol are: ##^2\Delta_g##, ##^2\Sigma_g##, ##^2\Sigma_g##, ##^4\Sigma_g##. This is ok.

First, If I want to find all the possible configuration with the above valence configuration I have to calculate:
$$\binom{\text{maximum number of electrons which can be "host" in the shell}}{\text{electrons in the shell}}$$
Not considering the full shell (##\pi## state can host 4 electrons, instead ##\sigma## states only 2):
$$\binom{4}{2}+\binom{2}{1}=8$$
But in the solution there is written that the possible states are 12, and I do not understand why.

Second, I do not understand how to calculate the degeneracies of each term. From my perspective should be ##(2S+1)(2\Lambda+1)##, so ##^2\Sigma_g## should be ##2\times (2\times 2+1)= 10##, but in the solution there are only 4 states.

I need some help to sort out my ideas, thank you in advance.
 
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Aleolomorfo said:
Not considering the full shell (##\pi## state can host 4 electrons, instead ##\sigma## states only 2):
$$\binom{4}{2}+\binom{2}{1}=8$$
But in the solution there is written that the possible states are 12, and I do not understand why.
Is adding the two binomial coefficients the right thing to do?

Aleolomorfo said:
Second, I do not understand how to calculate the degeneracies of each term. From my perspective should be ##(2S+1)(2\Lambda+1)##, so ##^2\Sigma_g## should be ##2\times (2\times 2+1)= 10##, but in the solution there are only 4 states.
I don't understand what you are doing here. And what is ##\Lambda## for a ##\Sigma## state?
 
DrClaude said:
Is adding the two binomial coefficients the right thing to do?
No, I have to multiply, thank you.

DrClaude said:
I don't understand what you are doing here. And what is ΛΛ\Lambda for a ΣΣ\Sigma state?
I see it is a bit confusing, but now I have understood reading again my book, thank you anyway