Molecular covalent bonds across energy levels

AI Thread Summary
The discussion centers on clarifying concepts related to molecular orbitals and bonding, particularly focusing on valence electrons and their roles in chemical bonding. It highlights the simplification in educational resources that often exclude inner electrons to emphasize outermost electrons, which are crucial for bonding. The conversation explains that including inner electrons would lead to equal numbers of bonding and antibonding orbitals, negating their contribution to overall bonding. Participants note that using elements with only s and p orbitals simplifies the understanding of bonding, while acknowledging the complexity introduced by d orbitals. The importance of electronegativity in determining bond type, as introduced by Pauling, is also discussed, alongside Koopman's theorem, which relates ionization energy and electron affinity to molecular orbital energies. The exchange of resources, including a recommended book on atomic and molecular structure, further enriches the discussion.
nomadreid
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In examples for molecular covalent bonds on the Internet (e.g., the site given in the full text), all components, with the exception of hydrogen, have the same energy level n. Yet there are bonds formed from orbitals of different energy levels, even though the bonding is more likely for identical n, no? So isn't it just whether the orbitals are s's, p's, an sp hybrid, or whatever? Why the n?
I am sure this is an elementary question; I'm just trying to clarify some points that were poorly explained to me years ago in secondary school. I know that a full answer would involve solving Schrödinger's equation etc., but keeping this on the level of valence electrons,...) I was confused by the sites, e.g. , https://chem.libretexts.org/Bookshe...1.7:_Molecular_Orbitals_and__Covalent_Bonding, using only components , besides H, that had the same n for all components. After all, isn't the whole idea of a group the similarities across peiods?
 
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I think it is a case of giving you the simplest possible examples, using only the outermost electrons, where n, the principle quantum number, is the main contribution to the energy of the outer electrons, and l the azimuthal quantum number making a smaller contribution to the energy of the orbitals. They are deliberately choosing elements with only s and p orbitals to keep it simple but still show that predictions can be made.

If the full inner electrons were included, they would show that the number of bonding and antibonding orbitals would be the same, all would be full, and so make no contribution to the overall bonds as the bonding orbitals are canceled by the antibonding orbitals. It's only the outermost orbitals, when they are not full, that are likely in their examples to show any overall net bonding.

You could draw them all out for the sodium example and see this for yourself. Then repeat for potassium, and as the diagram becomes bigger you will see why they ignore the ones which don't contribute overall to the bonding.

And when it gets to elements with d orbitals, it gets even harder to draw.
 
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Thanks, @DrJohn. The reason for not including the inner electrons is clear, but I thought it would be even simpler if, instead of saying for example, ns1 and ns1 give nσ, say that, among the valence electrons, ns1 and ms1 give σ, or something similar. But your explanation that that they are simply using the simplest examples makes sense, though. It just worried me to think that I was missing something...
 
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caz, thanks. Looks good. downloaded!
 
If the orbitals are of vastly different energy, you'd rather get an ionic bond.
 
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Thanks, DrDu. I never noticed that. Very helpful!
 
You are welcome!
This reasoning can be formalized. Pauling already introduced the electronegativity, which determines the ionicity of the bonds, in terms of the electron affinity and ionization energy.
According to Koopman's theorem, both are given basically in terms of the LUMO and HOMO orbital energies. https://en.wikipedia.org/wiki/Koopmans'_theorem
 
DrDu, thanks again; once again, your comments are helpful and the Wiki article is as well.
 
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