1. The problem statement, all variables and given/known data Find the ground state of nitrogen in the form (2s+1)(L)(J), where s represents the total spin, L represents the total orbital angular momentum (s,p,d,f), and J represents the total angular momentum. Attempt at solution Well, the electron configuration is [He](2s)^2 (2p)^3. Now, we know that everything until the (2p)^2 contributes 0 total angular momentum. We have three electrons of orbital angular momentum 1 (p = 1). These can form orbitals or s, p, d, f (0, 1, 2, 3). Now, the spins of the three electrons are 1/2, so they can either form spins 1/2 or 3/2. By Hund's first, they will take spin 3/2 (if possible). Now, from this, we have the possible electron configurations to be 4S 3/2 4P 1/2, 4P 3/2, 4P 5/2 4D 1/2, 4D 3/2, 4D 5/2, 4D 7/2 4F 3/2, 4F 5/2, 4F 7/2, 4F 9/2 By Hund's third rule, since the atom is less than half filled, we know that the only possible configurations left are 4S 3/2 4P 1/2 4D 1/2 4F 3/2 Now, for Hund's second rule, 4F 3/2 should be the correct one; but this isn't the case; the total spin in this case is 3/2, and by the clebsch gordan table, we have: |3/2 3/2> = |1 1> |1/2 1/2> = (|1/2 1/2> |1/2 1/2>) |1/2 1/2>, which is symmetric. So we need the orbital angular momentum part of our configuration to be antisymmetric since electrons are fermions. Now, we look at F: we have 3 particles whose orbital angular momenta are 1 and we need to try to form 3. We can do this as follows: |3 3> = |2 2> |1 1> = |1 1> |1 1> |1 1> which is symmetric. So, 4F 3/2 doesn't work because the entire thing is symmetric. Now, we go on to the D levels: I need to have 3 particles of orbital angular momentum 1 have total orbital angular momentum 2. The answer says that this is also symmetric; how is this so?