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Electron Configuration from Term Symbols in Lanthanides

  1. Jun 9, 2015 #1
    Hi, I'm new to the forum, and I am very sorry if this has been asked before.
    I am Japanese and therefore my English might be a little bit awkward.

    I want to ask about term symbols and if they can be used to derive electron configuration in lanthanides.

    First of all, I do not major in Physics. I major in Chemistry. Therefore I am not very fond of physical or quantum physical aspect of lanthanides. The reason I am here is none of the people in my lab (not even the teachers) really know about term symbols. They know how they can use it, but they do not have the technical background that leads to the use of term symbols.

    1) First of all, can term symbol be used as an effective way of deriving electron configuration? (If a term symbol is given for an atom, does it specify the electron configuration?).

    2) In lanthanides, which should be considered: L-S coupling or j-j coupling (or something else)?

    3) Does quantum number J (total angular quantum number) determine anything about electron configuration of an atom? I have found this true for some atoms (especially in lighter atoms or smaller number of electrons), but not for some other atoms.

    I was really confused about trivalent Europium (and Terbium) where there are 6 (or 8 in Terbium) electrons in 7 different orbitals. I can understand ground state term symbol 7F. The 4f orbitals should look like:
    +3 +2 +1 0 -1 -2 -3
    Parentheses are 4f orbitals, "up" means electron in its +1/2 state, and the numbers below the parentheses are azimuthal quantum numbers.
    However, does the total angular momentum quantum number "J" mean anything in this case? As long as total spin quantum number satisfies 7 = 2S + 1, all of the electron spins are +1/2, and due to Pauli principle, two electron cannot be in a same orbital unless the spin is changed. In that case, 7F1, 7F2, ..., 7F6 makes no sense at all. There are no possible combination of quantum number "l" and "s" that can satisfy "j" that makes up total quantum number "J".

    On the other hand, it made sense for 1D2 excited state of Praseodymium, because there are multiple possible state for spin because S = 0 (s1 = +1/2, s2 = -1/2 OR s1 = -1/2, s2 = +1/2), as well as the azimuthal quantum numbers, and the value of total angular quantum number can determine what kind of electron configuration it is:
    Pr(III) 1D2: (up)()()()(down)()()

    I am most likely having some terrible misunderstanding here and there, but because I really have no one to ask, I can't figure it out on my own. Please help me.

    Thank you very much.
  2. jcsd
  3. Jun 9, 2015 #2


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    add 1) It is easier the other way round: To calculate the possible term symbols from the known configuration. The ground state term symbol can be inferred using Hund's rules.
    2) For Lanthanides, L-S coupling is appropriate. This may be insufficient for actinides.
    3) J doesn't tell much about configuration.
    In calculating J, take in mind that it is the absolute value of a vector (like S and L). Hence J=S+L is only one possibility and in fact ##|L-S|\le J \le L+S##.
    You also have to be careful in your statement about the configuration in Pr. It is possible to obtain L=2 e.g. from l_1=+3 and l_2=+2 als ## l_1-l_2 \le L=2 \le l_1+l_2##.
  4. Jun 9, 2015 #3

    Thank you for the reply.

    I think now I'm starting to understand. You said that S is also the absolute value of a vector. Does that mean
    |s1-s2| ≤ S ≤ s1+s2
    as well?

    I might have learned this in quantum chemistry but I forgot.
    Last edited: Jun 9, 2015
  5. Jun 9, 2015 #4


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    Yes, exactly. E.g. from two spins, you can form a singlet and a triplet (corresponding to S=0=s1-s2 and S=1=s1+s2, respectively).
  6. Jun 9, 2015 #5


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    don't forget the absolute sign |s1-s2|.
  7. Jun 9, 2015 #6
    Oh yes, that's it. Now the stuff about term symbol and about the singlet and triplets are adding up in my mind.


    thank you for point that out. I forgot about it.
  8. Jun 9, 2015 #7


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    I just noted that I made an error here. Both l_2 and l_2 equal 3. What you wrote down are the z components, i.e. ##l_{1z}=+3##, ##l_{2z}=-1##. The z components add up like scalars, i.e.: ##L_z=l_{1z}+l_{2z}=2 ##. Now L_z can range from -L to +L, so the configuration you wrote down can contribute to both L=2 and L=3.
  9. Jun 9, 2015 #8
    Hmmm...I'm confused.

    If I'm understanding you right, then when we consider an electron configuration with information about which electron is in which specific orbitals (azimuthal), because they give no information about the angular momentum vector but only the size of certain component (x, y, z) of the vector, we cannot figure out term symbols, which are represented by the size of angular momentum vector.
  10. Jun 9, 2015 #9


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    Not quite. The problem is that term symbols are usually combinations of configurations. In the case of your example, a value of L_z=2 can result from several configurations (l_1z,l_2z): (+3,-1), (+1,+1) and (+2,0). Some superposition will correspond to the state with L=3 while another one will correspond to L=2.
  11. Jun 9, 2015 #10
    Actually, I understood until that part.

    What I didn't understand was whether or not the reason behind the inability to determine electron configuration from term symbols comes from the fact that term symbols are considered by the size of the vector itself while electron configuration (e.g. -3, -2, -1, 0, 1, 2, 3 for f-orbitals) considers only z-component of the size of the vector (meaning that the two are looking at different things). If so, then I can understand what you said above.
  12. Jun 9, 2015 #11


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    The point is that the f-orbitals are all energetically degenerate, and also all the possible configurations, as long as spin orbit coupling is not taken into account. So you have to form appropriate linear combinations of the configurartions. In principle, this is not very different from what you do in forming molecular orbitals from atomic orbitals. As soon as two atoms begin to interact when you bring them close to each other, you have to form appropriate bonding and anti-bonding linear combinations of the atomic orbitals to describe bonding. As soon as you include electron electron interaction, you also have to use superposistions of different configurations. This is called configuration interacion (CI).
    Hence, there is no reason to expect that an atom is well described in terms of a single configuration when spin orbit interaction is taken into account.

    Let's hammer home the case of Praseodymium III you mentioned:
    First we have to write down all possible combinations of l_1z and l_2z and the corresponding values for L_z (assuming singlet spin):
    L_z=6: (3,3)
    L_z=5: (3,2)
    L_z=4: (3,1), (2,2)
    L_z=3: (3,0), (2,1)
    L_z=2: (3, -1), (2,0),(1,1)
    L_z=1: (3, -2), (2,-1),(1,0)
    L_z=0: (3,-3), (2,-2), (1,-1), (0,0)
    and corresponding values for negative L_z.
    The state (3,3) can only belong to L=6. But then we have 2L+1=13 states belonging to L=6, with L_z=6,5,4...-5,-6.
    Hence if we remove the corresponding number of states, the next state with maximal L_z is L_z=4 which must belong to L=4. Again, we remove 9 states, one for every L_z.
    The next value is L=2 and the last one L=0.
    Hence you can see that the configuration (3,-1) you considered may contribute to singlet I, G, D.
    For triplet spin, the analysis is completely analogous, only that the configurations with equal l_z are missing, as no two electrons can occupy the same orbital.
    So you get L=5,3,1 instead or triplet H,F,P.
    For more than two electrons the situation becomes a little bit more involved. However, there exists a nice method using so called Weyl diagrams.
    I urge you to read:
    A modern approach to ls coupling in the theory of atomic spectra. Journal of chemical education, 1998, 75. Jg., Nr. 1, S. 110.
  13. Jun 13, 2015 #12
    So informative!

    Thank you very much for the help. I'll check out the reference you talked about later.
    I think this cleared up pretty much most of my problems!
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