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Pentacoordination of Zinc?

  1. Sep 23, 2004 #1

    chem_tr

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    Hello,

    I've heard rumors (frankly, not rumors, evidenced by some scientific papers), that zinc(II) ion has coordinated five ligands, four at a plane, and the fifth from top, to form a tetragonal monopyramidal environment.

    I have some doubts about this, because zinc (II) ion has d10 electronic configuration in normal circumstances, which means, according to crystal field theory, that octahedral stability energy is zero.

    What are your comments?

    chem_tr
     
  2. jcsd
  3. Sep 23, 2004 #2
    Hi chem_tr,

    Five ligands? Mmm.... strange. It would be necessary to calculate the stability energy. Are the five ligands identical? I´ve found the degeneracy split pattern for a square piramidal geomety:
    table

    Can you calculate the stability energy from this data? I´m not sure about it.
     
  4. Sep 23, 2004 #3

    chem_tr

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    Hello altered-gravity,

    Let me correct something first. There are four donors in a planar macrocyle (e.g., phthalocyanine), and the fifth one approaches from the perpendicular axis to the plane. There are also some examples in which zinc coordinates a pentadentate ligand, two thiolates, one pyridine nitrogen, and two azomethine nitrogens.

    I will look through your table in detail, thank you for your interest.
     
  5. Sep 24, 2004 #4
    Biochemical complexes are too much for my POOR "ab initio power" :biggrin:

    Anyway the case you are talkig about (phtalocyanine ring) is a square based piramidal geometry (C4v point group). In the table I post before, that geometry is supposed for five identical and simple ligands, and says that the degeneracy splits into four different energy levels, three monodegenerate and one doubly degenerate.

    In your case, although much more complicated, the point group is the same. So we may think that the energy scheme is the same (four levels). The task is to calculate the system energy provided that there are 10 electrons and the spin is 0. I think that the energy would be like this:
    [tex]E=\sum_{i} e_i -\sum_i \sum_{j>i} J_{ij} [/tex]

    i: each electron
    ei: energy of the level where "i" electron is
    Jij: Coulomb integral of electrons "i" and "j"

    I´m just speculating so don´t pay too much attention to me if you´re working seriously on this.
     
    Last edited: Sep 24, 2004
  6. Sep 25, 2004 #5

    chem_tr

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    Hello,

    C4v symmetry is valid for phthalocyanines, you are right. I have no idea about how to use these data on the table I downloaded from your post. Can you say that, by using the values on the table, it is possible/known for Zn to bind another ligand to reach pentacoordinated state?

    My mathematics has left me a long time ago, so I cannot make any comments about your double-sigma-containing huge formula :smile:

    Thank you for your interest
     
  7. Sep 25, 2004 #6
    Not exactly. That table gives you energies of the "d" orbitals for different geometries, but with ONLY ONE ligand type, for example Zn(OH)4 2-. From here you can calculate the square piramid geometry system energy, and compare it with others as planar square geom, in other words, the relative stability of the pentacoordinated. Then you can "imagine" or qualitatively aproach the perturbation that the fifth ligand creates when it aproaches the square planar system.

    The table gives energyes in terms of [tex]\Delta_0[/tex], this parameter is the energy perturbation of an octahedral geometry with the same ligand type, look at this:
    link
    this parameter must be calculated for each ligand type.

    For the phtalocyanine system you can only know that the energy pattern wold be similar due to the simetry, but not equal. If you want to calculate it exactly you must classically construct the whole field that all ligands create and start with parturbation theory calculus. This task colud be large with such complex ligands.

    I´m sorry if this doesn´t help you. Good luck.
     
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