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

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 !

Forgot to attach.

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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
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