Spectroscopic term of ground state electron configuration of Carbon atom

[boyu]

The ground state electron configuration of Carbon atom is $$1s^{2}2s^{2}2p^{2}$$

For the electrons, $$1s^{2}2s^{2}$$, $$L=0$$, $$S=0$$

So only consider electrons of $$2p^{2}$$, and

$$s_{1}=s_{2}=1/2 ---> S=0,1$$
$$l_{1}=l_{2}=1 ---> L=0,1,2$$

For S=0, L=0; J=0, so we have $$^{1}S_{0}$$
For S=0, L=1; J=1, so $$^{1}P_{1}$$
For S=0, L=2; J=2, so $$^{1}D_{2}$$

For S=1, L=0; J=1, so $$^{3}S_{1}$$
For S=1, L=1; J=0,1,2, so $$^{3}P_{0}, ^{3}P_{1}, ^{3}P_{2}$$
For S=1, L=2; J=1,2,3, so $$^{3}D_{1}, ^{3}D_{2}, ^{3}D_{3}$$

The above is based on my derivation. However, the correct answer is actually:
$$^{1}S_{0}, ^{1}D_{2}, ^{3}P_{0}, ^{3}P_{1}, ^{3}P_{2}$$

My question is: where are all the other possible terms? Where is wrong in my derivation:

 

[DrDu]

Not all the states you are constructing are compartible with the Pauli principle.
Namely you have six p-type spin orbitals and there are 15 (2 out of 6) possibilities.
Counting the multiplicities of the correct states you should also get 15.
The total number of states of your approach is 36 which would result from filling up the p orbitals with hypothetically distinguishable electrons.

 

[boyu]

How to get 15 from counting the multiplicities from correct states? The spin multiplicities are 1, 1, 3, 3 & 3.

 

[DrDu]

You have to count the multiplicities of J not of S or L. The multiplicities of S and L are dispersed over states with different J.

 

[boyu]

Many thanks! ^_^