Energy diagram for hybridized oxygen in CO

AI Thread Summary
The discussion centers on the confusion regarding the energy diagram for hybridized oxygen in carbon monoxide (CO). Participants express concerns about the apparent discrepancy in electron count, noting that oxygen has six electrons but the diagram shows only five. The conversation highlights the importance of understanding bonding and anti-bonding orbitals, suggesting that the diagram may be intended to illustrate hybridization effects rather than a literal electron count. There is skepticism about the necessity of hybridization for oxygen in a diatomic molecule like CO. Overall, the discussion emphasizes the complexities of molecular orbital theory and the potential misinterpretations of hybridization in this context.
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Homework Statement


Please look at the screenshot attached. The answer is "Yes", but I am not sure why.

Homework Equations


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The Attempt at a Solution


Oxygen have six electrons, but the energy diagram shows only five. That was why I thought it was wrong, but the diagram shown is correct. Can anyone explain why?
 

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There have to be six electrons. Oxygen isn’t charged in CO. The diagram must be wrong.
 
TeethWhitener said:
The diagram must be wrong.
..., unless, it's for bonding electrons from both carbon and oxygen? Seem to recall some sort of quasi-quantum mechanical argument for "bonding/anti-bonding orbitals."
 
The MO diagram for CO is isoelectronic to N2 and will have either 6 or 10 electrons, depending on whether you choose to include the σ-σ* bonding-antibonding pair generated from the 2s electrons. The point of the exercise is doubtless to highlight the energy splitting due to the hybridization of s and p orbitals on oxygen, but in no world is 5 electrons correct.
 
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How the heck do you hybridize a diagram?
Furthermore, there is no reason why O in CO should be hybridized at all.
 
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I agree, it doesn’t make much sense for a diatomic. I imagine the point was to show the effects of hybridization on s and p orbital energies, but that’s probably the least of the problems with this example.
 
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