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MO diagram of CO

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    Hey I was wondering if anyone could explain/help me understand the molecular orbital diagram of CO, this the energy level diagram I have copied out of my textbook.





    I get the pi interactions ok it is the sigma interactions I am having trouble fully understanding
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    I can see why you're confused, I've checked the MO diagrams I have in textbooks to the one you've got there and...they're different :p:

    The oxygen 2s is not mixing with the carbon 2pz to produce the 4 sigma orbital - the gap in energy is too big to be realistic

    Anyway, if we start with the carbon 2s mixing with the oxygen and carbon pz orbitals - we can see that they're all fairly close in energy and so will mix to produce 3 new MOs (call them lower sigma, middle sigma, upper sigma). The 2px and 2py will simply interact as normal.

    The oxygen 2s will interact with the carbon 2s forming two new MOs (1 sigma and 2 sigma)

    if you're having problems the best thing to do would be to draw it from scratch and not just look at the finished article
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    Ok thanks, I am using Shriver and Atkins 5th edition Inorganic Chemistry textbook and they have a line from O2s to the 4 sigma orbital. One of the problems I have had searching the web is that none of the diagrams seem to be consistent with each other.

    I am still a bit confused though as to which are the bonding and anti bonding molecular orbitals, I get the MO diagrams of things like O2 and F2 and but can't get my head around this one. I think I am getting confused with the interactions between the s and pz orbitals and where the lines for the mixing of these orbitals should direct to.
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    I have the third edition of the same book and it's not the same :p:

    It can be difficult to determine which is bonding, anti-bonding and non-bonding when you have a lot of mixing take place. Just judging from an accurate MO diagram is sufficient, here I'd say the bonding orbitals are 1 sigma and 2 x 1 pi whilst the remaining occupied sigma orbitals are non-bonding giving us the familiar bond order of 3.
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    The diagram I have in my lecture notes has the two 2s orbitals mixing to give an occupied bonding and antibonding orbital, and then the 6 3p orbitals mixing to give a sigma bonding/antibonding and a doubly degenerate pi bonding/antibonding level, of which the lowest 5 orbitals are filled. As 4 of these are bonding and one is antibonding you end up with an overall bond order of 3.
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    There are 4 atomic orbitals that will overlap to give 4 molecular orbitals with sigma symmetry.

    You will always get one orbital which is more bonding than the atomic orbitals, (labelled one above) and one which is more anti-bonding (labelled four). The other two will be somewhere in between, and without doing detailed calculations you can't really say their relative energies.

    You can argue their relative energies by other means however. For example, if the  3\sigma was higher in energy than the  2\pi then there would be two unpaired electrons and CO would show paramagnetism (which it doesn't).

    Finally, a word on the lines connecting the orbitals. Technically all of the orbitals with the correct symmetry will contribute to the molecular orbitals, so it is perfectly correct to draw lines from all of them. However, as mentioned above the more significant contributions arise when the atomic orbitals are similar in energy, so lines are often drawn from these orbitals only. You'll see both done.
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    (Original post by Bradshaw)
    There are 4 atomic orbitals that will overlap to give 4 molecular orbitals with sigma symmetry.

    You will always get one orbital which is more bonding than the atomic orbitals, (labelled one above) and one which is more anti-bonding (labelled four). The other two will be somewhere in between, and without doing detailed calculations you can't really say their relative energies.

    You can argue their relative energies by other means however. For example, if the  3\sigma was higher in energy than the  2\pi then there would be two unpaired electrons and CO would show paramagnetism (which it doesn't).

    Finally, a word on the lines connecting the orbitals. Technically all of the orbitals with the correct symmetry will contribute to the molecular orbitals, so it is perfectly correct to draw lines from all of them. However, as mentioned above the more significant contributions arise when the atomic orbitals are similar in energy, so lines are often drawn from these orbitals only. You'll see both done.
    Ok thanks that clears it up slightly, does that mean for the 3 sigma MO I could draw connecting lines from the O2s and C2pz AO's and the diagram would still be correct. The C2pz seems pretty similar in energy so is there any reason why that one hasn't been connected on my diagram?
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    (Original post by illusionz)
    The diagram I have in my lecture notes has the two 2s orbitals mixing to give an occupied bonding and antibonding orbital, and then the 6 3p orbitals mixing to give a sigma bonding/antibonding and a doubly degenerate pi bonding/antibonding level, of which the lowest 5 orbitals are filled. As 4 of these are bonding and one is antibonding you end up with an overall bond order of 3.
    Yea that would be my first instinct tbh and seems easier to understand, I think that doesn't show that the pz orbitals can now mix with the s orbitals though.
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    (Original post by CoventryCity)
    Yea that would be my first instinct tbh and seems easier to understand, I think that doesn't show that the pz orbitals can now mix with the s orbitals though.
    No it doesn't. As Bradshaw said, it's common to not show it due to the poor energy match meaning the effect of mixing is pretty smal. It depends on the level your studying at to be honest, it's an approximation, but a pretty good one.
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    Qualitative molecular orbital theory = an exercise in hand waving :gah:
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    (Original post by EierVonSatan)
    Qualitative molecular orbital theory = an exercise in hand waving :gah:
    At least its easy though!

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