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Name: Margaret H.
Status: educator
Age: 30s
Location: N/A
Country: N/A
Date: 12/17/2004

1. I understand that ozone must be hybridized because the bond lengths between the oxygen atoms are the same. For the Lewis dot resonance structure, the central oxygen atoms has a double bond with another oxygen atom. I can understand how each of these atoms are sp2 hybridized. However, how is the other oxygen atom sp2 hybridized? Does a dative covalent bond somehow allow for/create a p-orbital?

2. Does the negative charge on an atom also allow for/create a p-orbital? In the acetate ion, does the oxygen with the negative charge become sp2 hybridized because the lone pair with the negative charge moves to a p-orbital? If so, why?

Dear Margaret,

Of course, these are excellent questions. Here's my own take on this. Basically, if the molecule is going to be symmetrized by resonance, all of the contributing orbitals must be symmetrized as well - including their hybridizations.

(1) In order for the electrons to be delocalized, they must occupy a delocalized state, which is generally referred to as a pi molecular orbital (same as in benzene). This molecular orbital is best described as the completely in-phase combination of p orbitals localized on each atom. This in term is most easily interpreted by assuming that each atom is sp2 hybridized, leaving an empty p orbital on each atom which can overlap with the other two.

(2) Acetate actually has a lot in common with ozone, except that instead of an O at the middle we have CH3-C. The need to invoke sp2 hybridization here for both O atoms is the same as in the case of ozone; we have two equivalent bonds which are in resonance, which means that there is a delocalized pi molecular orbital formed by the overlap of a p orbital on C with p orbitals on the two oxygen atoms. It doesn't have anything to do directly with the presence or absence of formal charge on the atoms.

Fundamentally, whenever you have resonance between equivalent structures, all of the atoms involved in the resonance are going to have to have equivalent contributions to the molecular orbitals, so they will all have to use the same hybridization model.

Hope this helps!

Dr. Topper

Question 1. You have to be careful about "explaining" chemical bonding in terms of approximate models whether it is Lewis dot octet rules, VSEPR, hybridization, or molecular orbitals. Each has its place and utility; however, each also has examples where it fails, especially the simple models. Ozone is one example where the "simple" models just do not work very well. The tip off is the fact that ozone is blue. That means it has low energy excited electronic states (like oxygen). Another flag is the fact that the bond length of O2 and O3 is 0.1207 and 0.1278 nm, respectively. Almost the same! There has been quite a bit of theoretical work done on the electronic structure of ozone. The sites below can provide you with some of that work, at varying levels of complexity. As you might guess I have an interest in the bonding in O3!!

Question 2. Again you have to use the model "that works" in the case of acetate anion, CH3--CO2(-1) the charge is considered to be delocalized equally over the two oxygen atoms. Lewis octet rules do not handle that concept "naturally". The two oxygen atoms are electronically identical. So in order to retain the Lewis model one has to talk about partial electronic charges, the electron jumping back and forth between the two oxygen atoms. At best that stretches the model trying to explain a concept (delocalization) that it was not intended to handle in its original conception.

Vince Calder


2- I do not think charge has much to do with determining the hybridization state.

I think (ground-state) electrons are just magic little quantum computers that feel out the molecular landscape, invisibly think about what distribution has lowest energy, and then manifest that distribution.

What influences their decision is: the number of electrons present vs. the set of quantum states (orbitals) available to be filled (much like the Lewis picture), and the skeletal pattern of the nuclei to which they conform, which cannot be quickly moved by the electrons.

The electron we think of as adding a negative charge, is from the hybridization viewpoint just another electron to be accommodated like the others. Either accommodated, or jettisoned into free space if the energy penalty of the molecule's net charge is too high. We know from experience that the charge is not jettisoned.

Acetate (CH3CO2-) has one stretched pi-orbital from O to C to the other O, making it perfectly symmetrical, just because that is the lowest-energy method of accommodating the next electron, the one we consider our ion's charge. This has lowest energy because the negative charge is distributed among 2 atoms and because the orbital-sets of all 3 atoms are filled. Any other solution loses one of those. 3-atom pi-bonds do not fit into Lewis diagram sketches easily. Neither does the 6-carbon pi-bond of the benzene ring.

Yes, you could say that adding a negative charge to one oxygen results in a lone pair with negative charge. But then, the electrostatic energy penalty of uncompensated charge can be greatly reduced by splitting the charge into two separated pieces. A "pi" orbital from 0-C-O accomplishes this, and also retains most of the bonding energy advantage of the C=O double bond that "existed" when the lone pair existed. So it happens instantly, allowing no time that the "other state" ever exists.

Of course, this charge-distribution is roughly equal to a linear sum of two lone-pair/double-bond electron-density arrangements: one with charge on Oxygen A and double-bond on Oxygen B, other with charge on B and double-bond on A. So this molecule can react with things much as if it had a point-charge on Oxygen A, a up to half of the times you apply a test. If a positive charge is brought close to one oxygen, the distribution can easily shift so the negative charge is on the nearer oxygen a larger amount of the time, say 75%.

It is only when a real nucleus is immediately adjacent to one of the oxygens, waiting to be bonded, that the electrons need to definitively choose a one-sided distribution.

1- Ozone, I do not understand yet. Oxygen too, for that matter. Why would it be energetically favorable for O2 to be a di-radical rather than a double-bond? (di-radical: two radicals in one molecule, a single bond, with one half-filled lone pair on each oxygen atom.) And yet, due to it's paramagnetism, we know it has a large percentage of the di-radical electron-arrangement. and the bond strength is too low to be a double bond. If O2 does not have a clean Lewis structure model, I will not expect O3 to do it either.

If both bond lengths are the same, how can you choose two O atoms to be the central ones? I can rationalize, saying you are imagining a certain stage in a reaction where one O atom approaches an O2 molecule. Then your question becomes, why is an arrangement with two good single-bonds and two radical electrons at the ends have lower energy than one double bond and one coordination-bond? (is that your "dative" bond?) Two things: oxygen is small and tight with it's electrons. Another oxygen will not get much from trying to attach itself with a coordination bond. Those bonds are more likely to work when a small highly electronegative divalent atom (only oxygen, really) takes advantage of a larger, less electronegative atom with a large number of lone-pair electrons (sulfur, chlorine, iodine...) Also, two single-bonds usually have higher bond-strength (more energy-advantage) than one double-bond. It almost makes sense, really, that O3 should be single-bonded.

The O atom in the center position of O3 has two single bonds. Its Lewis octet is filled. It has too low an atomic number to have "d" orbitals, which are usually what allows for expansion beyond the Lewis octet. So the radical electrons at the ends are mostly "stranded" by themselves, they cannot use the central atom as a bridge to form a shared orbital, another bond. They could use an empty higher orbital of the oxygen, but I think these are energetically similar to being out in free space. So the radical electrons must either jump directly to each other, eventually bending the molecule to make an equilateral O3 ring, or they stay by themselves. 3-member rings are extremely strained, and it seems that is not what O3 prefers to be. Bear in mind that either of these states (bridging over central, or ring-closed) might actually be present in some small percentage. Only a first-principles molecular-orbital calculator would know for sure. Is that my old classmate, or the leading-edge software he sells, or neither at this time? Regrettably I do not know the exact status.

Why O2 is not more double-bonded is what bothers me.

Lewis octets just are not a very good summary of the full picture in the case of oxygen-oxygen bonding.

Jim Swenson

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