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Hund's rule

  1. Mar 17, 2012 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    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.



    2. Relevant equations
    The 3 Hund's rules.


    3. 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.
     
  2. jcsd
  3. Mar 19, 2012 #2
    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 = 4F3/2
    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.

    Edit 2: corrections are in bold. Now it should make sense.
     
    Last edited: Mar 19, 2012
  4. Mar 19, 2012 #3

    fluidistic

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    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.
     
  5. Mar 19, 2012 #4
    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.
     
    Last edited: Mar 19, 2012
  6. Mar 19, 2012 #5

    fluidistic

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