So when I took inorganic chemistry, we had a section on molecular symmetry and group theory (actually, I’m pretty sure that’s the name of the book we used). Anyway, we learned that - IIRC - if two orbitals have the same irreducible representation, you can hybridize them in the general sense of the term - i.e. combine and re-normalize their wavefunctions.
One of the “higher” symmetry groups that corresponds to tetrahedral molecules has p-orbitals corresponding to one irreducible representation and s-orbitals corresponding to another (I don’t feel like looking at a character table, so I can’t tell you which ones right now) which means, among other things, that you can’t hybridize p and s orbitals in tetrahedral molecules, right? So how does something tetrahedral become sp3 hybridized, like they teach in organic chemistry?
I’m sure I’m operating on any number of flawed assumptions, and since I took inorganic a while ago I may not be remembering correctly. But how do you reconcile the two?
This is, admittedly, a WAG, as I am a physics grad student and have not dealt with anything resembling this kind of chemistry in years:
This may be your flawed assumption. What you state as known is that a shared irreducible representation allows hybridization. This in and of itself does not imply that a shared irreducible representation is required for hybridization. In other words, it may be a sufficient but not necessary condition.
An sp3 orbital is not tetrahedral. There are four separate sp3 orbitals in a tetrahedral molecule that you describe. You can, in fact, add more s or p character to one of these orbitals, but then the molecule would only nominally be tetrahedral. In cyclopropane, for example, the carbon-carbon bonds have a higher portion of p character leaving the carbon-hydrogen bonds with more s character.
I’ll probably have to look through the books tomorrow to figure out what rule you are thinking of. As far as I know, s, p, d … orbitals can all be linearly combined to make hybrids. The hybrids, themselves being linear combinations of s, p, d … orbitals, can be linearly combined with s, p, d … orbitals. There are selection rules for bonding that sound similar to what you are saying.
:smack: Yeah, I know. Imprecise language and all that.
The selection rules are what I was thinking of, I believe. Take a look at this powerpoint I found from a quick google:
I forgot that the A1 representation always forms a basis for an s orbital, and therefore you can combine it with the p orbitals (which also are based on the A1 representation). Sorry for this incredibly esoteric concern.
Yes we are thinking about the same selection rules. I mostly think about these things in terms of bonding rather than hybridization. Since you know that an sp3 orbital is not tetrahedral, I’m not really sure that I understand your question. It’s true that you can’t hybridize two orbitals of different symmetry. You can’t hybridize a pz with a px. There are no p2 orbitals. You can hybridize a px, pz and an s though. That’s an sp2, obviously.