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Destabilizing Molecules Through Electron Rearrangement?

  1. Jan 23, 2013 #1

    jaketodd

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    By changing the arrangement of electrons in a molecule's energy levels (without adding or removing any electrons), can the molecule become destabilized or come apart? An example or two would really help.

    Thanks,

    Jake
     
  2. jcsd
  3. Jan 24, 2013 #2
    sure.

    in a previous post in the chemistry forum i repeated a derivation of the hydrogen molecule ion, H2+. it took a while and can't find it anymore.

    long story short: it turns out that from the 2 atomic 1s orbitals on the 2 hydrogens, in the molecule you get 2 molecular orbitals, one of which is symmetric, and the other is antisymmetric.

    the symmetric orbital has lower energy than the antisymmetric orbital. thus, with a H2+ ion, the single electron goes into the symmetric orbital, and the entire molecule is stable. This also explains the neutral hydrogen molecule, since the symmetric orbital can hold 2 electrons of different spin states.

    but if one of the electrons was excited, say by a laser, the next higher molecular orbital up is the antisymmetric one with a node for the probability distribution between the 2 protons, and the molecule gets destabilized.
     
  4. Jan 24, 2013 #3

    sophiecentaur

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    Sounds to me like you're talking Allotropes, here.
     
  5. Jan 24, 2013 #4
    Allotropes means there is at least one metastable phase that is not the most stable thermodynamic phase, but is kept stable at room temperature by kinetics.

    for example, diamond is thermodynamically less stable than graphite, but since the relaxation time is unimaginably long, it is essentially stable, but the constituent molecules are the same.

    this has little to do with molecular electronic structure, but the OP asked about electronic effects.
     
  6. Jan 27, 2013 #5

    jaketodd

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    So what are the requirements for destabilization when promoting electrons up energy levels please? I am not understanding the symmetric, antisymmetric, the nodes, etc. Please help.

    Thanks,

    Jake
     
  7. Jan 27, 2013 #6
    sure.

    lets say you have 2 protons and an electron. this is a hydrogen molecule ion: H2+.

    let the distance between the protons be R and apply the Born-Oppenheimer approximation such that the protons do not move. The chemical dynamics of the molecule is determined by the electron motion alone. We write this as the electronic Hamiltonian

    He = -h^2/2m * ∇^2 - e^2/rA - e^2/rB + e^2/R

    if you let R approach infinity and look at one of the protons (lets call this one proton A) the potential looks very much like a hydrogen atom. Likewise, if you look at the other proton called B, the potential also looks like a hydrogen atom.

    ψA = β*e^rA/a0
    ψB = β*e^rB/a0

    where β is a constant of normalization. so for the case that R = infinity, the assertion that the molecular wavefunction ψM is the normalized sum of the 2 hydrogenic wavefunction is correct. if R is reduced to some finite number, here's the thing: while ψM = C1*ψA ± C2*ψB is no longer exactly true, it is a good approximation. Indeed it turns out to capture the correct qualitative behavior, although energy calculations require much more sophistication.

    ψM = C1*ψA ± C2*ψB

    we know that the nuclei are identical, so the electron distribution must be symmetric. The only way for this to happen is the absolute values of C1 and C2 are the same, otherwise the electron would spend more time at one of the protons despite both protons being identical. So |C1| = |C2|.

    So the molecular wavefunction is actually 2 wavefunctions:

    ψM+ = C1(ψA + ψB) and ψM- = C1(ψA - ψB)

    For ψM+ the wavefunction is very low far away from both protons and hit a peak at the nuclei. But it is nowhere zero and the same applies for its absolute square. That's the bonding orbital.

    For ψM- note that the wavefunction has a - sign. That means it crosses zero somewhere. When you graph its absolute square, it turns out that it crosses zero between the protons. Therefore, when you square it, there is a point in the graph that is a ZERO (node) and has NO ELECTRONS there. That is the anti-bonding orbital.

    When you promote an electron up from a bonding orbital to the anti-bonding orbital, you're giving existence to a system state that includes a node between the 2 protons, which separates them.

    Here's a wiki article that explains this in a better way and adds the energy calculations to boot.

    http://en.wikipedia.org/wiki/Holstein–Herring_method
     
    Last edited: Jan 27, 2013
  8. Jan 27, 2013 #7

    jaketodd

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    I really appreciate all you've laid out there, but I don't really follow. Maybe I could reduce my question to this: Let's start with an atom that has three electrons in its valence shell. Let's say these three electrons are bonded with three additional atoms. Now, if you were to excite an inner-shell electron to an energy level above the three bonded electrons level, it seems to me that there would be a new valence level defined, and the three bonds below it would break apart because they are no longer in the valence shell. Is this correct? Thank you!!
     
    Last edited: Jan 28, 2013
  9. Jan 28, 2013 #8
    OK, I think I understand your question now. This depends.

    The LUMO (lowest unoccupied molecular orbital) in a complicated molecule is not necessarily anti-bonding! I used hydrogen ions as an example, but almost all molecules aren't hydrogen ions. For some molecules, like oxygen, there are unoccupied bonding orbitals right above the occupied bonding orbitals.
     
  10. Jan 28, 2013 #9

    jaketodd

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    How about for main group elements? It says here: "For a main group element, a valence electron can only be in the outermost electron shell."
     
  11. Jan 28, 2013 #10
    From the wiki:

    "In chemistry, a valence electron is an electron that is associated with an atom, and that can participate in the formation of a chemical bond; in a single covalent bond, both atoms in the bond contribute one valence electron in order to form a shared pair. The presence of valence electrons can determine the element's chemical properties and whether it may bond with other elements..."

    Note that it says atoms and elements. You can only talk about valence electrons for elements and atoms. It is not a useful concept for molecules and solids. For molecules we talk about molecular (as opposed to atomic) orbitals and for solids we talk about bands.

    http://en.wikipedia.org/wiki/Molecular_orbital
     
  12. Jan 28, 2013 #11

    jaketodd

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    I see. Is the bonding for a particular element, within a molecule, specific to the bonds that are connected to that particular element, or do you have to take every element in the molecule (even ones that aren't bonded to the particular atom) into account?

    Thanks,

    Jake
     
  13. Jan 28, 2013 #12

    jaketodd

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    Let me ask a better question: Take an element such as Phosphorus, with its outer electrons all bonded in a big complex molecule. Now, let's leave Phosphorus' bonds as they are, but excite some of its more inner electrons (who are not participating in bonding) to the highest possible energy level (but not ionizing).

    Would that destabilize the Phosphorus' bonds to the molecule, anyone?

    Thanks!
     
  14. Jan 29, 2013 #13
    it depends on the specific molecule. it might, it might not, but get this: exciting its inner electrons would require huge energy. that energy can't be directed so specifically towards ONLY the inner electrons. it would definitely start ionizing stuff, and the bonds would break anyways. That's why x-rays are used to sterilize stuff; even though they come from core electrons they can still knock out valence electrons.
     
  15. Jan 29, 2013 #14

    jaketodd

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    Oh I see, thank you. You sure know a lot.
     
  16. Jan 29, 2013 #15
    But, if you very monochromatic light, you can target inner electrons, since there's a specific transition energy to go from an inner electron to some excited state. It works better if the molecules are cold to reduce Doppler broadening.
     
  17. Jan 29, 2013 #16

    ZapperZ

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    Please note that this is done all the time in many techniques, such as Auger spectroscopy. In x-ray photoemission, one is also exciting core-level electrons.

    None of these do any "harm" to the either the molecules, or the solid.

    Zz.
     
  18. Jan 29, 2013 #17

    jaketodd

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    None of this stuff destabilizes molecules? Now I'm confused.
     
  19. Jan 29, 2013 #18

    chemisttree

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  20. Jan 30, 2013 #19

    ZapperZ

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  21. Jan 30, 2013 #20

    ZapperZ

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    Do a search on those two techniques. If they destroy the original material, then we can't do repeated measurement on that same material. Yet, we do.

    Zz.
     
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