Term Symbol for Carbon He2s^22p^2

[FloridaGators]

Term Symbol for Carbon Total Angular Momentum

1. Write the term symnbols for carbon atom He2s²2p² which has two equivalent p electrons taking into account the Pauli exclusion principle.
2...
3. Since the carbon has two equivalent p-electrons, we know the l₁=1 and that l₂=1 as well. Also the spins for both the electrons are 1/2. So the sum of the spin is either 1 or 0. I'm not quit sure where to go from here. l = 2, 1 or 0, meaning we have S, P and D terms. I tabulated a table of 15 elements permuting the various l, s terms for the two electrons making no two have the same numbers. Is there an easier way to do this?

So, Far I've gotten ¹D₂, ³P_2,1,0, ¹P₁ and ³S₁. Are there all the terms?

 

[nickjer]

You need to use Hund's rules here. Unless you are supposed to list all possible choices.

 

[FloridaGators]

Hund's rules would just tell me which state is the lowest in energy of all the possible ones right? But the ones I listed are not Pauli forbidden states right and they are all possible?

 

[nickjer]

You are missing a term symbol and have some that don't belong. If you want to list all possible term symbols you will have to make tables. Read here for more details: http://en.wikipedia.org/wiki/Term_symbol#Term_symbols_for_an_electron_configuration

 

[FloridaGators]

Ya i actually just read that, but I still don't understand.

You have a state where
electron 1: l = 0 s = +1/2
electron 2: l = 1 s = -1/2

Here L = 1, and S = 0
So doesn't this correspond to a state of ¹P₁.. why is this term not included?

 

[nickjer]

Lets assume you have the 2 electrons in that configuration. Then you know:

m_l = +1, and m_s = 0

But you also know, L=1 or 2 can give you m_l = +1. So you don't know which L it is. Also, you know S=0 or 1 can give you m_s = 0. You usually check the higher L's and S's first before you check the next lowest L or S. And since I can make states with m_l = +2, +1, 0, -1, -2 using m_s=0 only (I am unable to do so with m_s=+/-1), then I know L=2 and S=0, and I take those possible electron configurations away.

Then you might ask about,
e1: l=0, s=-1/2
e2: l=1, s=+1/2

Well, since I already checked L=2 with S=1,0 and only L=2 S=0 worked out. Now I will check L=1 and S=1,0. Turns out, L=1 S=1 can exist, which does have a state m_l = 1, m_s = 0, and I can also make m_l=1 and m_s = +/-1 states that weren't used yet. So I pull out that possible electron configuration.

Now I am unable to make anymore m_l=1, and m_s=0 states. So poor L=1, S=0 can't exist.

Basically it is a combination game, where you try all possible combinations of L,S that can use up all the possible electron configurations exactly.

Also, once you include 3 electrons this game can become very difficult very fast. But using Hund's rules makes it very easy to find the ground state with any number of electrons.