# Value of L for the ground state of Vanadium (II)

## Main Question or Discussion Point

Hi,

I've been working through my lecture notes from last year and I encountered this example of Russel-Saunders term symbols.

I'm confused by the value of L being given as 1 + 2 = 3.

What I recalled was that L is defined for two electrons by the Clebsch-Gordan series L = l(1)+l(2), l(1) + l(2) - 1,...,Mod (l(1) - l(2)). As the two outer electrons of V (III) are in the d subshell (l = 2), surely L = 4, 3, 2, 1, 0 and hence for the ground state we maximize l per Hundt's second law so L = 4.

Any help greatly appreciated !

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DrDu
You have to take into account that the two electrons are supposed to have the same spin due to the first rule of Hund. Hence they cannot occupy the same orbital due to Pauli's principle, i.e. to maximize L one will be in the orbital with l=2 but the other electron can then only occupy the orbital with l=1.

Why do they have to occupy orbitals with different values of l rather than just ml ? If they are d2 electrons, the value of l must be the same ? l = 1, 2 would have one electron in a p orbital, surely.

DrDu
Yes, you are obviously right. They have to occupy orbitals with the different ml.
The argument goes like this:
Write down all possible combinations of the ml. Then look for the maximal value of ML, this must coincide with the maximal value of L.
If you want to obtain all possible values of L you have to remove the ladder of values ML =L_max, L_max-1, ... -L_max+1, -L_max and look for the next largest remaining value of ML. Eventually you have to repeat the process several times.

• 1 person
Ah, thanks, that makes it clear - ML can take values of 3, 2, 1, 0, -1, -2, -3, so L = 3 for ground state.

V(ii) - v+

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