I'm looking at the transition state for an SN2 reaction mechanism.

The reaction is

 

2-bromooctane + OH- ----> 2-octanol + Br-

 

So, how exactly is the transition state sp2?

I thought sp2 only had three bonds at most?

What's going on?

the point is that it's planar, and sp2 hybridisation would explain that. I guess you'd have to assume that both the leaving group and the attacking group are not (entirely) bound at that stage.

I see. Fair explanation, hermanntrude.

Thank you.

I can see your point, though, and i'm only really trying to simplify... the transition state essentially has 3 normal bonds and two half-bonds, and would have a trigonal bipyramidal shape, but that requires an sp3d hybridisation, but that's not possible in carbon, since it doesn't have the unused d orbitals available. I suspect the answer to your question would become an essay quite easily, given half a chance and a lot more knowledge than I posess

You should be careful of saying that carbon doesn't have the avaliable d-orbitals avaliable to form a hybrid orbital. Carbon does have d-orbitals they are just empty and are far too high in energy to be used for any type of useful bonding.

 

With regards to the orginal question, i expect they are asking for you too know that there are three full bonds and two partially broken/formed bonds.

You should be careful of saying that carbon doesn't have the avaliable d-orbitals avaliable to form a hybrid orbital. Carbon does have d-orbitals they are just empty and are far too high in energy to be used for any type of useful bonding.

 

With regards to the orginal question, i expect they are asking for you too know that there are three full bonds and two partially broken/formed bonds.

 

in all hybridisation examples ive ever seen, the atomic orbitals which were hybridised all had the same value for n. If that were the case here, we'd need some 2d orbitals. These don't exist so my statement remains valid.

in all hybridisation examples ive ever seen, the atomic orbitals which were hybridised all had the same value for n. If that were the case here, we'd need some 2d orbitals. These don't exist so my statement remains valid.

 

While that is true for the first p-block elements, bonding of the d- and f-block does involve hybridisation between different principal quantum numbers does happen. The 3d and 4s orbitals are very close to be equal in energy allowing them to be hybridised in a model. Granted its a prety poor model compared to other bonding models but thats ok.

While that is true for the first p-block elements, bonding of the d- and f-block does involve hybridisation between different principal quantum numbers does happen. The 3d and 4s orbitals are very close to be equal in energy allowing them to be hybridised in a model. Granted its a prety poor model compared to other bonding models but thats ok.

 

I've learned something new. I had no idea there were actual examples of hybridisation between orbitals of different principle quantum numbers. Can you give me some links or diagrams? i'd love to see what the hybrid orbitals look like.

I shall have a look through my inorganic books and hopefully be able to you a reference for it. The main account for hybdrisation is that the orbtials are close enough in energy to overlap.

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http://en.wikipedia.org/wiki/Orbital_hybridisation

 

That doesn't have some pictures but shows where sp3d hybridisation occurs

penta coordinate carbon is possible.

penta coordinate carbon is possible.

 

adianadiadi, this is a science forum. You can't just make a statement without backing it up. Give us some examples... what hybridisation is used? how and when is a pentacoordinate carbon possible?

Look up a 3-center, 4 electron bond. I don't see why the mechanism couldn't use that as a high-energy intermediate with sp2 hybridization of the central carbon.