Sulfur Trioxide Hybridization

[Ray Guo 4C]

Why is SO3 sp2 hybridized? If two of the three p orbitals are used in hybridization, how are the three pi bonds formed with only one unhybridized p orbital left? Can d orbitals form pi bonds with p orbitals?

[CateJensen3K]

I would draw out the lewis structure in order to figure out the shape. The shape coordinates with the hybridization orbitals.
steric #2= linear=sp
steric #3= trigonal planar=sp2
steric #4= tetrahedral=sp3
steric #5= trigonal bipyramidal=sp3d
steric #6= octahedral=sp3d2

They also coordinate with the steric number, so if you are unsure about their shape I would also know that as well.
steric number= number of atoms around the central atom + number of lone pairs on the central atom

For SO3 the most stable lewis structure is one where S is double bonded to all three Os. Since we count multiple bonds as 1, the steric number = 3+0 (no lone pairs)
and the shape is trigonal planar. This makes SO3 have three sp2 hybridization.

[Artin Allahverdian 2H]

If you look at the Lewis structure of this molecule, a central S atom is bonded to 3 O atoms. Each S-O bond is a double bond, and S has no lone pairs. Thus, S has 3 regions of electron density (a double bond counts as a single region of electron density, and there are 3 of them, hence 3 regions of e- density). If there 3 regions of e- density, this corresponds to sp2 hybridization. Now since they are double bonds connecting S and O, each double bonds has 1 pi bond and 1 sigma bond, making a total of 3 pi bonds and 3 sigma bonds. When looking at the sigma bond the S has 2sp^2 hybridization. Whenever there is a double bond, the pi bond will have an unhybridized p orbital. Thus, S has 2p hybridization with respect to the pi bond.

[Ray Guo 4C]

If you look at the Lewis structure of this molecule, a central S atom is bonded to 3 O atoms. Each S-O bond is a double bond, and S has no lone pairs. Thus, S has 3 regions of electron density (a double bond counts as a single region of electron density, and there are 3 of them, hence 3 regions of e- density). If there 3 regions of e- density, this corresponds to sp2 hybridization. Now since they are double bonds connecting S and O, each double bonds has 1 pi bond and 1 sigma bond, making a total of 3 pi bonds and 3 sigma bonds. When looking at the sigma bond the S has 2sp^2 hybridization. Whenever there is a double bond, the pi bond will have an unhybridized p orbital. Thus, S has 2p hybridization with respect to the pi bond.

Can you elaborate on what a 2p hybridization is? And how does that hybridization explain the formation, or composition, of the three pi bonds?

[Ray Guo 4C]

I guess I wasn't clear enough about my question. I do understand the relationship between steric number and type of hybridization. What I am trying to ask is what (hybrid) orbitals is the sulfur atom using to form the three pi bonds. Normally pi bonds are formed by p orbitals, but in this case, only one p orbital on the sulfur atom is left unhybridized and ready to form one pi bond. Then how are the other two pi bonds formed?