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Electron Configuration/Hund's Rule/LS Coupling

  1. Mar 24, 2016 #1
    1. The problem statement, all variables and given/known data
    1. Consider the electronic configurations [Ar]3d94s1 and [Ar]3d10 for an atom.
    (a) Identify this element.

    (b) Use Hund’s rules to explain which is the ground state. Show your work for full credit.

    2. Relevant equations
    J=L+S
    n2S+1LJ
    L= l1+l2
    S= s1 + s2 or S = (1\2)* number of electrons
    J = range from |L-S| to |L+S|

    3. The attempt at a solution
    I'm pretty sure that the element in question is Ni, because it doesn't say anything about being an ion. But I'm having some trouble understanding how to apply Hund's rules.

    So I'm trying to use the LS coupling scheme to get the total angular momentum J to make an energy level diagram but I'm not sure I've done it right because every example I've seen has used two electrons in the same shell (so same value n). So what do you do when you have two electrons in two different shells (different n's)?

    First I looked at only the unpaired electrons and used the above equations to get:
    3d9--> n=3, l=2, ml=-2,-1,0,1,2
    4s1--> n=4, l=0, ml=0
    L = 2
    S = 1
    J = 1, 2, 3​
    But then, when I go to put it all together, I come to the two different principle quantum numbers problem. So how would I use the n2S+1LJ notation in this case? Would there be two different notations for each n? And the ground state should be something along the lines of n3D1 right? Then for the 3d10 configuration, how would you apply Hund's rule if all the shells are full? I know that unfilled shells have lower energy, so I'm fairly certain that the 3d94s1 configuration would be the ground state, but how could I quantitatively describe the 3d10 state?

    Or am I just doing this all wrong?
     
  2. jcsd
  3. Mar 24, 2016 #2

    DrClaude

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    Staff: Mentor

    You can't do that. You would need to consider all electrons in a partially filled subshell, so all 9 3d electrons. However, you are save here by the fact that "holes" (empty orbitals) behave the same way as electrons. Hence, 3d9 gives the same term symbols as 3d1. So what you have done next actually applies.

    That's not the only possibility.

    You should list the electronic configuration: 3d94s1 2S+1LJ and 3d10 2S+1LJ


    I don't see how Hund's rule applies here. First, Hund's rule is for equivalent electrons. It won't tell you which term symbol from 3d94s1 is lower in energy (it would work for something like 4p2). Second, I don't see how Hund's rule can be used to distinguish between different electronic configurations.
     
  4. Mar 24, 2016 #3
    How could there be other possibilities?

    I thought that as an extension of Hund's rules, the term symbols do show the lowest energy because I thought that the terms with the maximum S and L values with the minimized J correspond to the state with the lowest energy?

    That's why I'm having such a difficult time with this!! My only other argument would be to just say that Hund's rule states that subshells are filled one electron at a time and then show the electrons in an energy level diagram and just say that atoms with un paired electrons have a lower energy than those with filled electrons. I just feel like that's not much of an answer.
     
  5. Mar 25, 2016 #4

    DrClaude

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    Staff: Mentor

    By the rules of addition of angular momenta. You have two electrons (or one hole and one electron) in different orbitals, what are the possible values of the total spin?

    But Hund's rule can only apply to equivalent electrons. Take Si for example. The ground electronic configuration is 3p2, giving the term symbols 1D, 3P, and 1S. Applying Hund's rule, we take the triplet to have the lowest energy, so the ground state is 3P, which is correct.

    Take now the excited configuration 3p4p. The possible term symbols are now 3D, 1D, 3P, 1P, 3S, 1S. Applying Hund's rule would give 3D as the ground state. It is actually 1P (yes, the singlet is lower in energy, which goes against Hund's first rule). Conclusion: Hund's rule only apply for equivalent electrons (and then again, only to find the ground state, they say nothing about the order of the other term symbols).

    As I said, I don't see how Hund's rule can apply to electronic configurations. How could it explain, for instance, why Fe is 3d64s2 while Ru is 4d75s.

    On top of that, I have checked and the actual ground configuration of Ni is 3d84s2, which is not even considered in the problem. It is a shame that you are stuck with such a bad exercise.
     
  6. Mar 26, 2016 #5
    That was one thing that was confusing me too. I know that you can have two possible values of S, either 0 or 1. But one place I read that you can just calculate it as S=(1\2)*the number of electrons, I just wasn't sure which electrons it was referring to. Like here for example, would it be all of the 9 electrons in 3d and the one in 4s? Because then that would give S=6 and thats not possible right?

    Yeah I'm still not entirely sure how the prof is wanting us to answer this. That's why I've been trying to figure out this whole term symbol business because I thought I'd be able to go through and find the term symbols of the ground state of each configuration and be able to compare them but its proving to be more difficult than I originally thought.

    I know but I have found a few sources who contend that the 3d94s1 configuration should actually be considered the ground state. Otherwise I'm not sure what the element in question would be. My only other guess was Cu+, but the problem doesn't say anything about an ion. But thank you for validating my frustration. I've been working on this problem for a forever and have been seriously questioning my sanity because it seems like it should be fairly straightforward.
     
  7. Mar 27, 2016 #6

    DrClaude

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    Staff: Mentor

    No, because the Pauli exclusion principle has to be followed. For all 10 electrons, you would need to list all possible microstates, and then see the possible values of S. As you can consider a 3d hole instead, that gives the sum of two spin 1/2 particles, so S = 0 or 1.
     
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