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Orbital craziness

  1. Sep 14, 2004 #1

    ShawnD

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    When looking at the orbital names, why does the fourth period of elements go 4s, 3d, 4p? Why is it 3d? Why not 4d?
     
  2. jcsd
  3. Sep 14, 2004 #2
    It is because the 4s orbital happens to have a lower energy than the 3d orbital. If you notice on the periodic table, you do not start getting the transition metals until the 4th period due to this. If you want to know why the 3d orbital has a higher energy than the 4s, well you need three semesters of university level physics for that.

    Orbital filling gets even more complicated than than when you start filling up the d and f orbitals. In the lanthanide series, for instance, Ceriium with an atomic number of 58, has the orbital strucutre [Xe] 4f1 5d1 6s2, but Praseodymium, with an atomic number of 59, has the structure [Xe] 4f3 6s2! So there are general rules for the lower orbitals but things start getting crazy as you get the higher energy orbitals.
     
  4. Sep 15, 2004 #3
    To understand it concretly, you need to know Quantum Physics properly (as so-crates has mentioned). For now, you can remember that as a general rule, of two subshells, the subshell with a lower value of (n+l) where n = principal quantum number of the shell and l = azimuthal/angular momentum quantum number, is filled first. For 4s, n+l = 4 (as n = 4 and for the s-subshell, l = 0) and for 3d, n+l = 3 + 2 = 5. So 4s is filled before 3d.

    In case the value of n+l is equal for two subshells, the subshell having a lower value of the principal quantum number is filled first. This pretty much sums up what is called the Aufbau Principle. It seems hardly convincing at first and the actual reason can be explained convicingly only using quantum physics.

    Hope that helps...

    Cheers
    Vivek
     
  5. Sep 15, 2004 #4

    Gokul43201

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    And as far as the Qunatum Mechanics goes, if you solve for the radial part of the wavefunction, you find that the Enegry Eigenstates go like : (if this makes no sense, ignore it and simply look at the line below)

    E(n,l) = -E1/(n+l)^2

    This is what maverick was talking about - "The energy increases with increasing (n+l)".

    Additional Note : Remember, l = 0, 1, 2, and 3 are respectively designated the labels s, p, d, and f.
     
  6. Sep 15, 2004 #5

    chem_tr

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    By the way, do you happen to know that the designations s, p, d, and f come from sharp, principal, diffuse, and fundamental, respectively?
     
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