# Bond Angles, Bond Lengths and Hybridization

1. Nov 2, 2014

Consider the hypothetical molecule 1,3,5-cyclohexatriene (not benzene, I am referring to a molecule where no resonance takes place - a molecule with three C=C bonds and 3 C-C bonds). Each carbon atom in this molecule would be participating in one double bond, and so each carbon atom would be sp2 hybridized. This would suggest bond angles of 120° throughout the cyclic structure, which is only true of a regular hexagon. However, C=C bonds are shorter than C-C bonds. Three C=C bonds and three C-C bonds would certainly result in a somewhat distorted hexagon. Those are clearly two conflicting results. What are your thoughts on this?

2. Nov 3, 2014

### Staff: Mentor

Assumption that each carbon would be exactly sp2 is where you gone wrong. This is a strained ring, so it would be distorted.

3. Nov 3, 2014

Why and how wouldn't the C atoms be sp2 hybridized?

4. Nov 3, 2014

### Staff: Mentor

They can't be exactly sp2 for teh reason you have listed - it is impossible to have all bond angles equal to 120° and a flat molecule.

Think about cyclopropene - it has a double bond. Draw it. What are angles? It is possible these are sp2 atoms?

5. Nov 4, 2014

### DrDu

As I repeatedly pointed out in this forum, hybridization is not a property of the molecule but to some degree up to your personal choice.
Instead of sp$^2$, you could use three different hybrids, so that e.g. the shorter bond has more s character and the longer bond more p character.
Another possibility would be to use sp3 hybrids so that the double bonds are two banana bonds.
In the strained compounds Borek mentioned, this also has experimental consequences, to get the smaller bond angles, it is adavantageous for the orbitals forming the bonds in the ring to have more p character so that the orbital on C forming the CH bond has more s character. This increases the acidity of the CH bond.
http://en.wikipedia.org/wiki/Bent_bond

6. Nov 6, 2014

### PhaseShifter

I have to disagree with everyone who has posted here.

It's entirely possible to have an irregular hexagon that still has 120º angles at all six corners. What ever gave you the idea that it wasn't?

Try this:

Draw an equilateral triangle, and make marks at the 1/4 and 3/4 points along each edge.
Now draw straight lines connecting each mark to the nearest mark on another side.
You now have a planar figure with six 120º angles, where three sides are twice as long as the other three sides.

7. Nov 6, 2014

### Staff: Mentor

Sigh. That it is possible was my first idea, and I did some quick calculation to check it. And I got the correct result which supports the idea it possible, I just misunderstood what I got. It happens to still lie on my desk, making faces at me :(