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

In summary, the example has a problem with its definition of L due to the presence of two electrons with the same spin and the need to take into account Pauli's principle.
  • #1
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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  • #2
Forgot to attach.
 

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  • #3
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.
 
  • #4
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.
 
  • #5
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.
 
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  • #6
Ah, thanks, that makes it clear - ML can take values of 3, 2, 1, 0, -1, -2, -3, so L = 3 for ground state.
 
  • #7
V(ii) - v+
 

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1. What is the ground state of Vanadium (II)?

The ground state of Vanadium (II) is the lowest energy state that an electron can occupy in the Vanadium (II) atom. It is also known as the "zero point energy" state.

2. How is the ground state of Vanadium (II) determined?

The ground state of Vanadium (II) is determined by the electron configuration of the atom, which is based on the number of electrons in each energy level. In the ground state, the electrons occupy the lowest energy levels first, following the Aufbau principle.

3. What is the significance of the ground state of Vanadium (II)?

The ground state of Vanadium (II) is important because it represents the most stable and lowest energy state of the atom. It also determines the chemical and physical properties of Vanadium (II) and how it interacts with other elements.

4. How does the ground state of Vanadium (II) affect its reactivity?

The ground state of Vanadium (II) affects its reactivity by determining the number of valence electrons available for bonding. In the ground state, Vanadium (II) has two valence electrons, making it highly reactive and capable of forming bonds with other elements.

5. Can the ground state of Vanadium (II) change?

Yes, the ground state of Vanadium (II) can change if the atom gains or loses electrons, which can occur through chemical reactions or the absorption or emission of energy. However, the ground state is the most stable state for the Vanadium (II) atom.

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