and I thought that only p orbitals can form pi bonds
Not exactly, its just in general chemistry you always see p orbitals forming pi bonds. A pi bond has electron density above and below the bonding axis and there is a nodal plane (an area where there is no electron density) through the bonding axis. A pi bond can form between two p orbitals or between a p and a d orbital. Try drawing a d orbital (not dz^2, but any of the others) next to a p orbital and you'll see that there will be some overlap. It's just like being able to form a sigma bond between two s orbitals or an s and p orbital (or an s and d orbital or p and d orbital or two p orbitals for that matter). Where the electron density is defines what type of bond.
If P is sp3 hybridized, then how do we explain that there are 10 electrons around P? There will be two electron that aren't accounted for? Or am I being stupid? The oxygen atom involved in the double bond is sp2 hybridized. One electron of one of the hybridized 2p orbital forms the pi bond with one of the hybridized 3p orbital of P.
Can someone please explain the whole hybridization process? Perhaps this is above a general chemistry course, but as I'm studying chemistry I want to know
Hybridization is a way to explain the geometries of molecules, which come from experimental measurements. For example, we know that CH
4 is tetrahedral in shape from spectroscopy. If you look at the orientation of C's atomic orbitals as they are, it doesn't work, all of the p orbitals are perpendicular. But, if you take the 2s and 3 2p orbitals of carbon, combine them (using some advanced "magic" math), out comes 4 sp
3 hybrid orbitals that give bond angles that are 109.5 degrees apart, which match the experimental bond angles of methane.
So, if I look at phosphate, I see a molecule with 4 bonds (never mind one is a double bond). From VSEPR, I know that if an atom is bonded to 4 things and has no lone pairs, the bonds have a tetrahedral geometry, like CH
4. To explain this using hybridization, I can say that the 3s and 3p orbitals form 4 sp
3 hybrid orbitals that would contain 1 electron each. All of the sigma bonds are formed between these and a p orbital from oxygen. So, 8 pairs of electrons around P so far.
Now, to get the double bond. We still have one valence electron left from P that can be "promoted" to a d orbital. The d orbital can then form a pi bond with a p orbital from one of the oxygen atoms. So now P has 10 electrons around it. This is a way to explain how P "expands its octet" using its empty d orbitals.
BTW, things like phosphate are really tricky to explain with hybridization, which is why most gen. chem professors steer clear of it. But it can be done, and I hope this helps some more.
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Topic: Phosphate and hybridization (Read 32508 times)