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I Hund's rules

  1. Aug 31, 2016 #1
    Can anyone explain the second rule, because the Wikipedia page is not very clear?

    Hund's zeroth rule - Ignore all inner shells and focus on the outermost shell.
    Hund's first rule - Put the electrons such that they maximize spin, ##s##.

    So far so good. Hund's second rule appears to be simply that the we should maximize ##l## (after rule 1). That is, if I have say ##3p^2##, then the two electrons should be in ##l=1## and ##l=0##. Is there anything more to it than that?

    https://en.wikipedia.org/wiki/Hund's_rules talks about singlets and triplets and that the second rule is never used until Ti. I don't understand either of these statements. What is the meaning of singlets and triplets in this context? Also, in my example above, we used Hund's second rule for a 3p orbital so why is Ti the lowest element where this rule is used? Thank you.
     
    Last edited: Aug 31, 2016
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  3. Aug 31, 2016 #2

    DrClaude

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    Staff: Mentor

    You are mixing up ##l## and ##m_l##. If you have 3p2, then you have two electrons with ##l=1##. When you add up the two orbital angular momenta with ##l=1##, you can get ##L=2, 1, 0##, so the second rule would tell you that the ground state would be ##L=2##. (This is not actually the case for 3p2 because of the first rule, see below.)

    A term with ##S=0## is called is singlet (because ##2S+1=1##) while a term with ##S=1## is called a triplet (because ##2S+1=3##). The first rule tells you that the triplet will be lowest in energy. In the case of 3p2, the possible terms are 3P, 1D, and 1S. Therefore, by the first rule, the ground state is 3P. Since there are no other triplet terms, you don't need the second rule. It is only when you have two d electrons that multiplicity alone will not tell you which is the ground state.
     
  4. Sep 1, 2016 #3
    Thank you for your reply. Can I check what exactly 3P, 1D, and 1S notation means? Sorry, if this is an obvious question but I'm guessing the 3 and the 1 refer to the triplet and the singlet but what are the P, D, and S?
     
  5. Sep 1, 2016 #4

    DrClaude

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    Spectroscopic notation: ##^{2S+1}L_J##, with ##S## the total spin, ##L## the total orbital angular momentum, and ##J## the total angular momentum. ##L## is expressed as a capital letter, similar to single-electron orbitals. In what I wrote, I omitted the possible values of ##J## (you need the third Hund rule to find which ##J## for the ground state).
     
  6. Sep 3, 2016 #5
    Thank you. Sorry to ask more questions but how did you know that ##^3P, ^1D##, and ##^1S## were the available options? In other words, why is it that the way spin adds up also determine ##L##? For example, why can I not have ##S=1## and ##L=0## which is the state ##^3S## in spectroscopic notation?
     
  7. Sep 4, 2016 #6

    DrClaude

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    There is no 3S because of the Pauli exclusion principle. The only way to figure this out is to write all possible microstates (combinations of ##m_l## and ##m_s##), and then see which terms they lead to.

    Have a look at this document for more information on how to do that,
     
  8. Sep 4, 2016 #7
    Thank you for the detailed document. I think I mostly get it except for how the ##L## value is known from a given ##M_L##. If I look at the microstates, for example, the fifth one in your table ##M_L = 0## but the ##L## value of this can be either ##0, 1## or ##2##. How is it that it is labelled as ##^3P## and not ##^3S##?
     
  9. Sep 4, 2016 #8

    DrClaude

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    ##^3##P is made up of ##M_L = 1, 0, -1## with ##M_S = 1, 0, -1##, you need all those microstates to form the term. Once you have used the microstate with ##M_L=0## and ##M_S=1## to build up ##^3##P, there is no such microstate left to make a ##^3##S. That's why have have to form the terms starting from the highest value of ##L##.
     
  10. Sep 4, 2016 #9
    Ah I see! Thank you a million DrClaude. You've really helped me finally get it!
     
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