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Anti-Bonding/Bonding Molecular Orbitals

  1. May 8, 2010 #1
    Can someone explain to me, what are the factors/reasons that when certain atoms bond together, they form anti-Bonding or Bonding Molecular Orbitals? I am confused, and this is not in my syllabus(equivalent to Grade 12).

    P.S. It would be helpful if you included wavefunctions as well.
     
  2. jcsd
  3. May 8, 2010 #2

    Borek

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    Staff: Mentor

    Both bonding and antibonding orbitals are formed whenever two atoms get close enough.

    What will happen then depends on how many valence electrons are there to fill these orbitals. Bonding orbitals are filled first, as they have lower energy. If - once bonding orbitals are filled - there are no valence electrons left - you have a stable molecule, with lower energy that initial two atoms. However, if after filling bonding orbitals, there are still valence electrons left, they have no choice but to fill antibonding orbitals. Antibonding orbitals have much higher energy, so when tehy are filled molecule has high energy - and is not energetically favorable (compared to separated atoms). Thus electrons filling antibonding orbitals effectively destroy the molecule.

    Final effect may depend on exact number of electrons, for example if there are 2 electrons on bonding orbital and 1 electron on antibonding orbital, molecule has lower energy than separated atoms, so it is still stable.
     
  4. May 8, 2010 #3

    alxm

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    A simplified answer (MO theory) is that the orbitals of the atoms come together to form molecular orbitals.

    Much like classical waves, the atomic orbitals (wave functions) can then interfere constructively or destructively.

    Considering, say, H2, your two atomic s-orbitals come together constructively to form a σ orbital,
    where you have increased density between the two atoms (s1 + s2 -> σ) - thus it's called a bonding orbital.
    But they can also come together 'destructively' to form a σ*, where you have a decreased density between the two atoms (s1 - s2 -> σ*), hence anti-bonding.

    The σ orbital is greater than zero everywhere (because the s orbitals are), whereas the σ* is zero (= has a node)
    in the plane that's equidistant from both atoms, where the two s-orbitals cancel out completely.
    What you can infer from this, is that going from one atom to the other, the anti-bonding orbital has a sharper
    curvature (passes through a greater range of values) than the bonding orbital, and with quantum-mechanical wave functions
    (as well with classical waves), a sharper curvature means higher energy.
    So this tells us that anti-bonding orbitals are usually higher in energy than the corresponding bonding orbitals. Which is true. (and good news for chemistry)
     
  5. May 8, 2010 #4
    If you want the "why" answered and not just memorize textbook stuff, then check out Cohen-Tannoudji "Quantum Mechanics" (I believe Vol 1) the chapter about the electron wavefunction of the hydrogen molecule. I might do it myself later!
    Basically taking superpositions of orbitals is a QM approximation method. You have to think about all the Maths in order to understand what bonding and antibonding means. Determining the coefficients is a two parameter problem. And there are probably more reasons that explain bonding and anti-bonding more specifically.

    If you don't dig through the Maths, then you would be just citing other people and not have a clue what you are talking about.
     
  6. May 9, 2010 #5
    To Borek:

    Are MOs linked to sigma and pi orbitals? And if i wanted to depict all these MOs in a diagram, what would it look like?


    Is the definition of a node "Region where no electrons can be found"? Cos i learnt that orbitals are just regions where there is a high chance of finding electrons, if i wanted to be certain of the possible locations of an electron, the orbital would be the size of the universe. This conflicts with the node's definition, cos there should be no area where an electron can never be found?

    And why is that good news for chemistry?

    Is the Math tough? Im not a fan of it, but i'll do my best. Thanks so much for sharing that title, and yes, i want the why, and i don't want to learn by memorizing. Im sort of an 'oddball' in my school :)
     
  7. May 9, 2010 #6
    Ah no. The Maths is just as easy as a standard QM undergrad question about the Schrödinger equation. The Maths is simple, but you have to think about how exactly everything follows, what it means and so on. E=mc^2 is simple, but there is much behind it.

    It isn't easy to extract the meaning from Maths, but the point is *all* of the information is contained in the equations, so the meaning also must be somewhere.
     
  8. May 9, 2010 #7

    alxm

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    Yes, an orbital node is an area where there is zero electron density. That definition of orbital isn't quite correct; An orbital gives a complete description of the electron probability-density that corresponds to a certain energy level or "pattern of motion" of the electron. And yes, they're infinitely big - but since they drop off exponentially, this isn't of much concern. The way you usually see them drawn is by choosing a cut-off, say the volume within which the electron has a 90% probability of being.

    It's good news for chemistry that antibonding orbitals are higher in energy, or we wouldn't have nearly as many stable compounds.


    Well, Gerenuk is correct in that I (intentionally, btw) didn't give an exact description; but rather the MO/LCAO-model picture (which is really sufficient theory for most chemists). Obviously if you want to get a "deeper" understanding you have to learn the maths. I couldn't really recommend Cohen-Tannoudji for a 12-grader though (unless you're really good at math). You'd need to study calculus first.
     
  9. May 9, 2010 #8
    Thanks so much guys, I had a hard time reading university reference books (Chemistry : The Central Science, Pearson) and I didn't understand what the MOs were all about.

    By the way, can MOs be depicted collectively in a diagram? And are they related to sigma and pi bonds?

    To alxm : Yea, I learn calculus in school. Im from Singapore btw, so i'm not sure what the syllabii in the other countries are. I used grade 12 as an approximation, cos many internet users are from the US and UK. :)
     
  10. May 9, 2010 #9

    Borek

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    You can have both σ bonding/antibonding and π bonding/antibonding. You may think about σ/π (plus δ) or bonding/antibonding as of two separate properties of MO.
     
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