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I am told that the sulfur trioxide molecule exhibits charge separation because of poor p-orbital overlap. Sulfur's 3p orbitals are much bigger than oxygen's 3p orbitals, and thus the $\ce{S=O}$ bond is best described as $\ce{S+ - O-}$.

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So is the author's argument suggesting that sulfur trioxide is unhybridized, and that it uses pure s orbitals to form sigma bonds and pure p orbitals to form pi bonds? Because that's the only way I can see sulfur as having three empty p orbitals to utilize. If we were to assign hybridization through counting the number of sigma pairs of electrons (3) we would arrive at sp2 hybridization, and there aren't enough p-orbitals to go around.

Also while we're on this topic, Wikipedia says that sulfur trioxide is best represented as having one $\ce{S=O}$ bond and two $\ce{S+ - O-}$ bonds. Now, does bond length differ between an $\ce{S+ - O-}$ and an $\ce{S=O}$ bond? Wikipedia suggests not, as only one bond length is given. If bond length is equivalent between the two, why?

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Sulfur trioxide is planar with 120 degree O-S-O bond angles. If the molecule were unhybridized and we used the 3 orthogonal p-orbitals for our sigma bonds, then we would expect the O-S-O angle to be 90 degrees. Further, in the unhybridized scheme, there are not enough s orbitals to form 3 sigma bonds that are pure s. Just considering s and p orbital involvement, we could say that the sulfur is $\ce{sp^2}$ hybridized. This would produce 3 $\ce{sp^2}$ orbitals for sigma bonds from sulfur to oxygen and leave one p orbital for one pi bond. We can draw 3 resonance structures for this arrangement, so we would expect all bond lengths to be equivalent.

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  • $\begingroup$ Right, but the author almost explicitly states that three p-orbitals are used. So where can the s-orbitals for the sigma bonds come from? The author's argument struck me as off. $\endgroup$
    – Dissenter
    Aug 19, 2014 at 19:06
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    $\begingroup$ I don't read it that way. Perhaps the author is saying that there are 3 S=O double bonds, but they are not such great double bonds - in fact each one is really about 1/3 of a double bond (as the resonance structures suggest). There is no way to create a planar molecule with 120 degree angles using 3 p orbitals for each of 3 S=O pi bonds. $\endgroup$
    – ron
    Aug 19, 2014 at 19:20
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    $\begingroup$ @Dissenter, ron: For a full orbital scheme of sulfur trioxide, see this answer of me. The bond order of about 1.4 is consistent with the resonance picture. If you do not want to employ the concept of hybridisation here, you could describe the MO with symmetry adapted ligand group orbitals. This of course would not change a thing, since these descriptions are mathematically equal. $\endgroup$
    – Martin - マーチン ♦
    Aug 20, 2014 at 3:57

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