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I tried to draw the frontier molecular orbitals of the following biradical structure:

enter image description here

(A) At the top I've drawn the HOMO and LUMO, because I know that the reaction coordinate belongs to the $\ce{C_{$2v$}}$ point group. These two orbitals of the HOMO (green) lies in the O-C-O plane. In each $p$ orbital I would locate one electron (triplett state is favoured because of Hunds rule). Oxygen has six valence electrons, so I must bring four electrons in orbitals of each oxygen. I can place two electrons in a $p$ orbital perpendicular to the greenish depicted orbital. Then I would place the other two electrons in a $sp$ hybrid orbital.

(B) In the second possibility (at the right at the bottom) I also used a $sp$ hybridization for the oxygen atom. The two radicals are placed in the $p$ orbital perpendicular to the O-C-O plane. I guess the structure is less favoured than the first possibility (with the two radical in the O-C-O plane) because there are unfavoured filled-filled interactions between pink-pink and green-green while above there is onyl a pink-pink destabilizing interaction.

(C) As a third possibility I've used a $sp^2$ hybridization of the oxygen. The radicals are now perpendicular to the plane. Without exact calculation of the hybrid orbitals I would predict that 4 electrons in $sp^2$ hybrid orbitals plus one electron in a $p$ orbital is lower in energy as case (A) (3 electrons in $p$ orbital and two electrons in a $sp$ hybrid orbital). However the green lobes (both two electrons and hence repulsive interaction) here in the case (C) in the $sp^2$ hybrid orbitals seems to come closer (120°) than the two electrons in the orthogonal $p$ orbitals (also filled-filled) in case (A).

Are these reasons why the HOMO is best described by case (A)>(C)>(B)?

Probably this question has similiarties to the hybridization in a previous question (please note I'm a novice concerning hybridization) Why allene cannot be described with an allyl system?

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I think you are asking what does the HOMO and LUMO of the 1,3-dioxabiradical look like.

Let's start by using cyclopropane as a model compound. In cyclopropane the $\ce{C-H}$ bonds are approximately $\ce{sp^{2.46}}$ hybridized and the $\ce{C-C}$ bond is approximately $\ce{sp^{3.74}}$ hybridized (see here). That's a lot of p-character in the $\ce{C-C}$ sigma bond. Dioxirane is likely similarly hybridized. As soon as the 1,3-biradical forms (as soon as we break the $\ce{O-O}$ bond), the HOMO and LUMO will look like what is shown in the "initial" drawing below. If the biradical lives long enough to undergo rotation about the $\ce{C-O}$ bond, then the HOMO and LUMO will look like what is shown in the "rotated" drawing.

enter image description here

These drawings just point out the conformational dependence of the HOMO and LUMO. Note that the two lone pairs on each oxygen have not entered the discussion. They are orthogonal to, and have no effect upon, the HOMO and LUMO orbitals.

Imagine bringing two oxygen atoms closer to each other. If we push them closer and the orbitals containing the odd electrons are pointing towards each other as in the "initial" geometry, we will form a sigma bond and we know what the HOMO and LUMO look like for a sigma bond (just like what is pictured in the "initial" geometry, only the orbitals are closer together). If we push them closer and the odd electrons are pointing parallel to each other as in the "rotated" geometry, we will form a pi bond and we know what the HOMO and LUMO look like for a pi bond (just like what is pictured in the "rotated" geometry, only the orbitals are closer together). In the 1,3-biradical the carbon atom holds the 2 oxygen atoms together, they are closer than they would be at infinity, but further apart then they would be if there was a bond between them. This tells us that the HOMO-LUMO energy separation is greater than it would be at infinity (0), but less than if they were closer together.

We could refine our model by noting that, according to Bent's rule, we would expect the orbitals containing the oxygen lone pairs to have a bit more s-character that the carbon $\ce{C-H}$ orbital in cyclopropane. Therefore, in dioxirane and the corresponding 1,3-dioxabiradical, we would expect both the $\ce{O-O}$ bond and the orbital containing the single electron in the biradical to have even more p-character than $\ce{sp^{3.74}}$.

Again, however, to re-emphasize an earlier point, the lone pairs on oxygen have no bearing on our discussion. That said, if we would like to speculate on the hybridization of the orbitals containing the lone pairs, we may do so. Most texts tell us that the oxygen lone pairs in water, ethers and alcohols are in (roughly) $\ce{sp^3}$ orbitals. However data from photoelectron spectroscopy experiments seems to suggest that one lone pair is in (roughly) a p-orbital while the other lone pair is in an $\ce{sp}$ orbital.

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  • $\begingroup$ IR data shows also that the C-H bond in cyclopropane has $sp^2$ character. I tried to figure out the C-C hybridization without your formula (just with the "valence orbitals"). 3 C gives totally 3 s + 9 p orbitals. Now I subtract 3x(s+2p) for the 9 $sp^2$ orbitals and I obtain 3 p orbitals. In the lecture the cyclopropane has been described with Walsh orbitals and three perpendicular p orbitals (Möbius set with smaller $\beta$ than the Hückel set of the Walsh orbitals pointing to the center of the cycle). $\endgroup$
    – laminin
    Jun 15, 2015 at 20:57
  • $\begingroup$ Frankly I think this is not the truth, because I expect a part of the $sp^2$ hybridization coming from 3 H s orbitals therefore instead of 3 perpendicular p orbitals I get 3 sp orbitals for the C-C bond. Different webpages treat the Möbius banana bonds as p orbitals. Möbius sp orbitals would propably give more overlap than just p orbitals, yes? Are these sp orbitals also perpendicular to the Walsh orbitals as the p orbitals would do? $\endgroup$
    – laminin
    Jun 15, 2015 at 20:58
  • $\begingroup$ A few comments: 1) both Coulson's theorem and pKa data suggest that the cyclopropane $\ce{C-H}$ bond is neither $\ce{sp^2}$ nor $\ce{sp^3}$, but rather something in between, 2) you're analysis is off because there are only 6 $\ce{C-H}$ bonds in cyclopropane, not 9, 3) "I expect a part of the sp2 hybridization coming from 3 H s orbitals" , the substituent (H in this case) electronegativity will affect the carbon hybridization. The $\ce{C-C}$ bonds have much more p character than sp; p (90° bond angle) will provide for much better overlap in cyclopropane than sp (180° bond angle) $\endgroup$
    – ron
    Jun 15, 2015 at 21:52
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    $\begingroup$ Google "Coulson's theorem" and read about it if you have time. $\endgroup$
    – ron
    Jun 15, 2015 at 21:52
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    $\begingroup$ 1) google cyclopropane bond angle, most values are 114-116° (example); 2) OK, now I see what you're doing, but again, the C-H bonds are more like sp^2.5; 3) if the C-C bonds are sp hybridized (they are not) you would need 2 sp orbitals on each carbon and they would point 180° from each other, impossible in cyclopropane $\endgroup$
    – ron
    Jun 15, 2015 at 22:41

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