s character

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

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s character

Question: What is s-character and why does the s character of a hybrid orbital increases the bond angle?

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Re: Why does increases in s character increase bond angle?

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.

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Re: s character

Can someone go over with me how to determine hybridization? I can kind of grasp the concept, but can someone help me solidify the idea. Please and thanks!

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Re: s character

The number of hybrid orbitals depends on how many regions of electron density there are, which is the sum of the number of bonded atoms and lone pair electrons. If there are 4 regions of electron density for say carbon, then you will need 4 hybrid orbitals made up of 3 p and 1 s, giving you 4 equal sp3 hybrid orbitals. If there were 3, then you would need 2 p and 1 s orbitals, giving you sp2 hybridization.