The ground state hass= 1 andℓ= 1 as we predicted. Other states labeled 2pare the ones that
Hund’s first two rules determined to be of higher energy. They are both spin singlets so its the
symmetry of the space wavefunction that is making the difference here.
26.6.3 Nitrogen Ground State
Now, withZ= 7 we have three valence 2P electrons and the shell is half full. Hund’sfirst rule ,
maximum totals, tells us to couple the three electron spins tos=^32. This is again the symmetric spin
state so we’ll need to make the space state antisymmetric. We now have the truly nasty problem
of figuring out which totalℓstates are totally antisymmetric. All I have to say is 3⊗ 3 ⊗3 =
(^7) S⊕ (^5) MS⊕ (^3) MS⊕ (^5) MA⊕ (^3) MA⊕ (^1) A⊕ (^3) MS. Here MS means mixed symmetric. That is; it is
symmetric under the interchange of two of the electrons but not with the third. Remember, adding
two P states together, we get totalℓ 12 = 0, 1 ,2. Adding another P state to each of these gives total
ℓ= 1 forℓ 12 = 0,ℓ= 0, 1 ,2 forℓ 12 = 1, andℓ= 1, 2 ,3 forℓ 12 = 2. Hund’s second rule, maximumℓ,