Using Hund's rule, find the fundamental term of an atom

[fluidistic]

Homework Statement

Using Hund's rule, find the fundamental term of an atom whose last incomplete subshell contains 3 electrons d. Do the same with 4 electrons p.

Homework Equations

The 3 Hund's rules.

The Attempt at a Solution

By fundamental term I'm guessing they mean the ground state term.
Despite knowing Hund's rules, I don't know how I can start the problem without having the term symbols of the atom(s).
I somehow made an attempt, for the 3 electrons d.
Since the subshell is d, it means each electrons will have the quantum numbers [itex]l=2[/itex]. So that [itex]L=6[/itex]. Also, since there are 3 electrons, [itex]S=3/2[/itex]. Thus J goes from [itex]L-S=9/2[/itex] to [itex]L+S=15/2[/itex].
That d shell is half filled so that J will equal to [itex]9/2[/itex] for the ground state.
L could in principle for 6 (but in practice no?!), this would make my answer as [itex]^{16}H_{9/2}[/itex]. Which is -I'm 100% sure-, totally wrong.
What I don't understand is that L seems to go from 0 to 6 as if one adds the [itex]m_l[/itex] 's of the 3 electrons instead of the [itex]3 l[/itex]'s.
Any help is appreciated.

 

[Jolb]

By Hund's zeroth rule, all the other shells are filled, so they contribute nothing to spin or angular momentum.

There are 5 d orbitals and 3 p orbitals. Hund's first rule is that they have the maximum spin allowed for these shells following the exclusion principle. Hence, for d³, s=3/2, and for p⁴, the spin is 1/2+1/2+1/2-1/2 = 1.

Hund's second rule is that L is the maximum for given S in each shell. d's separate orbitals have L = 2, 1, 0, -1, -2, while p's are 1, 0, -1. Using what we found in the last step, d³ has L = 2+1+0 = 3, and P⁴ has 1+0+-1+1 = 1.

Hund's third rule is that J = L+S for more-than-half filled shells, and J=|L-S| for less-than-half filled shells. (notice that both equations give the same answer for exactly half-filled shells). Hence for d³, J = |3/2-3|=3/2, and for p⁴, J=1+1=2.

In the "Hieroglyphic" notation, these would be:

d³ = ⁴F_3/2
p⁴ = ³P₂Edit: in retrospect, it seems like the source of your confusion is that there are actually 4 Hund's rules! The zeroth rule is crucial and allows you to neglect everything besides the outermost shells. Also, remember the exclusion principle: there can only be two electrons per orbital, and if there are two, their spins must be antiparallel.

Edit 2: corrections are in bold. Now it should make sense.

 

[fluidistic]

By Hund's zeroth rule, all the other shells are filled, so they contribute nothing to spin or angular momentum.

There are 5 d orbitals and 3 p orbitals. Hund's first rule is that they have the maximum spin allowed for these shells following the exclusion principle. Hence, for d³, s=3/2, and for p⁴, the spin is 1/2+1/2+1/2-1/2 = 1.

Hund's second rule is that L is the maximum for given S in each shell. d's separate orbitals have L = 2, 1, 0, -1, -2, while p's are 1, 0, -1. Using what we found in the last step, d³ has L = 2+1+0 = 3, and P⁴ has 1+0+-1+1 = 1.

Hund's third rule is that J = L+S for less-than-half filled shells, and J=|L-S| for more-than-half filled shells. Hence for d³, J = |3/2-3|=3/2, and for p⁴, J=1+1=2.

In the "Hieroglyphic" notation, these would be:

d³ = ⁴F_3/2
p⁴ = ³P₂


Edit: in retrospect, it seems like the source of your confusion is that there are actually 4 Hund's rules! The zeroth rule is crucial and allows you to neglect everything besides the outermost shells. Also, remember the exclusion principle: there can only be two electrons per orbital, and if there are two, their spins must be antiparallel.

Thank you, I was not expecting a full and detailed answer (in fact the rules of the forum forbids this!) but I have nothing else to say than a big thank you, that helped me to understand a lot. I'm lacking books on this topic.
I've carefully been through your sentences and I've a small doubt.
Why did you consider that for the d³ electrons the shell was more than half filled and for the p⁴ electrons it was less than half filled? Except this point, I understood everything you made, thanks to your comments.

 

[Jolb]

I think trying to answer your question without giving the full answer would be much more trouble--the best way I could possibly explain this would be by working an example for you, and this is actually a much better example problem than I could think of easily. Plus, a good example would include both a more-than-half-filled example and a less-than-half-filled example, like this one. Hund's rules are one of the most heuristic things you learn in QM--I personally have never heard a satisfactory theoretical discussion of them. So please forgive me, Gods of PF.Oops, I think I made a typo above. I had "more" and "less" reversed.
Anyway, p shells have 3 orbitals. Two electrons can inhabit each orbital. Thus there can be at most 6 electrons in a P orbital. 4>6/2. So p⁴ is MORE than half filled, so you use J=L+S

I'll go back and find my typo so the post makes sense.

 

[fluidistic]

I think trying to answer your question without giving the full answer would be much more trouble--the best way I could possibly explain this would be by working an example for you, and this is actually a much better example problem than I could think of easily. Plus, a good example would include both a more-than-half-filled example and a less-than-half-filled example, like this one. Hund's rules are one of the most heuristic things you learn in QM--I personally have never heard a satisfactory theoretical discussion of them. So please forgive me, Gods of PF.


Oops, I think I made a typo above. I had "more" and "less" reversed.
Anyway, p shells have 3 orbitals. Two electrons can inhabit each orbital. Thus there can be at most 6 electrons in a P orbital. 4>6/2. So p⁴ is MORE than half filled, so you use J=L+S

I'll go back and find my typo so the post makes sense.

Thank you for everything. You left everything as clear as possible to me.