## HW Problem 2F.15

$sp, sp^{2}, sp^{3}, dsp^{3}, d^{2}sp^{3}$

Gisela F Ramirez 2H
Posts: 61
Joined: Fri Sep 28, 2018 12:27 am

### HW Problem 2F.15

How does the bond angle and s-character relate? For this problem, the answer is that the bond angles increase as the s-character of the hybrids increase, why is this so?

Summer de Vera 2C
Posts: 65
Joined: Fri Sep 28, 2018 12:16 am

### Re: HW Problem 2F.15

Hey! I was also really confused on this question. I found an answer to this question from 2011, so hopefully this clarifies things a bit.
"Answer: s-character is the contribution of sigma type bond in a hybridization: sp3 = 25% s-character, 75% p-character sp2 = 33% s-character, 66% p-character sp = 50% s-character, 50% p-character The more s-character a bond has, the stronger and shorter the bond is. so an sp-sp bond is strongest, and sp3-sp3 bond is weakest.

The bond angle of sp3 is 109.5, sp2 is 120 and sp is 180. An sp orbital is half s character, sp2 is 1/3 s character and sp3 is 1/4 s character, so increasing the s character corresponds to increasing the bond angle.
Another way to think about it is that you want to keep all of the orbitals of the same shape as far apart as possible (typically we would actually say that we want them to overlap as little as possible). Recall that when you hybridize one s and one p orbital, you get two sp orbitals, similarly you can mix two p and one s to get three equivalent sp2 orbitals. It turns out that the best way to keep two orbitals from overlapping much is to put them on opposite sides of the atom (sp - 180 degrees) keeping three of the same orbital apart results in a trigonal planar type structure (sp2 - 120 degrees). This is a little bit of why the angles increase as you increase s character."

I think an important thing to keep in mind is the shape and angles of the molecule when thinking of the relationship between bond angle and s– character.