Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Tetrahedrally adapted d-orbitals?

  1. Feb 4, 2005 #1
    Does anyone have a link to a picture of the shapes of tetrahedrally adapted atomic d-orbitals?
    Aren't the familiar d(z2), d(x2-y2), d(xz), d(xy) and d(yz) orbitals just one of several possible sets of atomic d-orbital shapes? Or have I misunderstood this point?
    I would like to rationalise why in a tetrahedral ligand field a triply degenerate set will be able to interact with the ligands and a doubly degenerate set will not. For octahedral symmetry this (well, rather the inverse: two orbitals interact, three won't) is easy, because usually we depict the d-orbitals as above in cartesian coordinates. But shouldn't it be possible to make linear combinations of the d-orbitals to create what is to become one doubly and one triply degenerate set if a tetrahedral field is applied?
    I tried the question in the chemistry section, but without much success, when I realised it might be better placed here since it concerns atomic orbitals and quantum physics (although it applies to chemical complex forming and MO theory).
    Last edited: Feb 4, 2005
  2. jcsd
  3. Feb 11, 2005 #2
    Your talking hybridization when you modify the d orbitals to match the XRD data. You might find some ideas if you visit: www.orbitals.com .
    The guys here prefer to look at wave functions and would probably refer you to check a theory called Density Functional Theory (DFT) which might provide orbital-like images, but I'm not sure. You might also check for a software called "Schrodinger". Have not used that soft for a while, so I might have to revise the name. Will check a web search.
  4. Feb 11, 2005 #3
    Just to be correct : QM delivers all knowledge on atomic orbitals (the squared spherical harmonics) as well as hybridization. I am sure you have heard of these harmonics and their use in molecular orbital theory. The idea is the write the wavefunction of a molecule as a linear combination of the constituent atomic wave functions, provided to us by solving the eigenequation for both L² and L_z-operators.

    DFT is used to solve the Schrodinger equation (SE) in an approximative manner. The idea is to look at many interacting electrons as if they are all free and moving in a new socalled effective potential. This is basically the same as Hartree Fock states but in DFT you start from an arbitrary expression for the electron density and you determin the groundstate energy of the many particle system. In hartree Fock theory, you start from an arbitrary guess for the electron-wavefunction. By using the variational principle you can determin an upper limit for the system's ground state energy. Keep in mind that in DFT every observable (energy and so on) is written as a functional of the electron density because you can proove there is a 1-1-relation between the wavefunction of the system and the electron density. This is the Hohenbergh-Kohn-Theorema...

  5. Feb 11, 2005 #4


    User Avatar
    Science Advisor
    Homework Helper

    Hm...Ain't it a bit weird:what_are_electrons mentions DFT,yet he's pretty unclear with the Bohr model...
    C'est la vie...

  6. Feb 13, 2005 #5
    Yes, but are not these just the d_xy, d_xz, d_yz etcetera?

    If you want something that looks more tetrahedral, you need to mix in odd wavefunctions: 3p-orbitals. "Atom in a Box" at http://daugerresearch.com/orbitals/ can show you the orbitals. Actually, there is a tetrahedral example on the page: a superposition of <3,2,1> and <3,1,-1>.

    Hälsningar & groeten,

  7. Feb 14, 2005 #6
    depending on the cyrstal field the d orbitals are going to point in different directions. Changing the energy relative to the 2p orbitals. Aso you get a difference in the bonding of the [itex] d\gamma [/itex] and [itex] d\varepsilon [/itex] with the p orbitals.
  8. Feb 14, 2005 #7
    Hey.... I read that!
    BTW, I'm not the one making unsubstantiated correlations. Just 'cause I ask a question ....
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook