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Is this interpretation correct? (diagrams of Carbon Monoxide and Carbon Dioxide)

  1. Jul 10, 2018 #1

    Is this correct for C(triple bond)O ie Carbon Monoxide upload_2018-7-10_19-10-45.png
    This is for CO2. Please check both.
    ALso check everything ie hybridized orbitals, bondings etc. I'm trying to clear concepts.
    (Note: red= unhybridized, black= hybridized)

    Thank you.
    Last edited: Jul 10, 2018
  2. jcsd
  3. Jul 10, 2018 #2


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    First, the orbitals are fine. Some details can be improved though.

    I'm going to explain CO2 first. It is very close to being right. You forgot to draw bonding (black scribble between orbitals) for the other side (negative phase) of the Pz and Py orbitals. I'm sure this is a careless mistake because you did it for CO. Otherwise, they are good.

    As for the CO, the orbitals are a mixture of a resonance structure ⊕:C≡O:⊖ ↔ :C=O:: (sorry the four electrons on the oxygen is wrong, two electrons belong on sp and the other two on Pz, but I cannot write that here). In that sense, the orbitals are fine, but the bond order cannot be discretely written in the diagram.

    EDIT: You forgot to color code CO for the unhybridized or hybridized.
    Last edited: Jul 10, 2018
  4. Jul 11, 2018 #3
    Okay, the silly mistakes apart. I wanted to ask you something, will side atoms (or ligands. here O) show hybridization too? People on my other posts have said they won't..
  5. Jul 11, 2018 #4


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    I think I need to clear the misunderstanding first. DrDu meant to say that hybridization won't fully explain what happens in PF3. Any elements beyond 2nd period is going to be quite complicated. As an example, as you may already know, compounds containing phosphorous can sometimes show hypervalency (such as PF5), which you won't see for 1st and 2nd period elements. Larger atoms tend to be quite complicated where conventional understanding of bonding may or may not apply.

    Carbon and oxygen in CO2 and CO can surely be explained well by hybridized orbitals.
  6. Jul 11, 2018 #5


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    CO2 is an example where you can see that different assumptions about hybridization lead to comparable bonding schemes. In the case of CO2 you have several choices:
    a) take the O's to be sp2 hybridized, which is what you apparently did. Note that you can draw two equivalent structures depending on whether the sp2 hybrids on the left oxygen atom lie in the xy plane or xz plane. You can even get resonance stabilization by considering both structures at once.
    b) Take the O's to be sp hybridized. Then you get again at least two resonance structures depending on whether the pi bond between the o on the left and the central c lies in the xy or xz plane. There will be a bit more repulsion between the lone pairs on O and the pi bonds as in case a.
    c) Consider the oxygens as completely unhybridized. This will lead to somewhat weaker sigma bonds as in case b.
    d) Consider all atoms to be sp3 hybridized. This will lead to so-called "banana bonds" between the atoms (https://en.wikipedia.org/wiki/Bent_bond). Banana bonds are energetically favourable over sigma-pi type bonds as the "bananas" are at greater distance from each other which reduces electrostatic repulsion between the bonds. There is some confusion about this even on wikipedia. The reason is that in molecular orbital theory banana-bonds and sigma-pi bonds are equivalent while in valence bond theory, they are not.
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