Why does the bond angle increase as the s-character of a hybrid orbital increases?

[Christine Lin 1H]

I was working on the chemistry textbook problems (specifically 2F.15) and I had trouble understanding this problem conceptually. What is the connections of increasing s-character and bond angle?

[Allison Peng 1D]

I think that it's because when you look at the hybridization of an orbital in a bond, as the bond angle decreases (more bonds on the central atom), more p-orbitals are involved in hybridization and bonding. For example, a linear bond is sp hybridized and has a bond angle of 180 degrees, and based on the hybridization you'd expect similar s- and p- character. However, a tetrahedral bond is sp3 hybridized and has a bond angle of 109.5 degrees, and you'd expect more p-character than s-character, since there are more p orbitals involved in hybridization. This is just how I think of it, so let me know what you think.

[Joanna Zhao 1J]

As the number of regions of electron density increases, p-character increases because there are more p orbitals contributing to hybridization.
2 regions -> sp hybridized (50% s-character) -> 180 degrees
3 regions -> sp2 hybridized (33% s-character) -> 120 degrees
4 regions -> sp3 hybridized (25% s-character) -> 109.5 degrees
As you can see, the bond angle is decreasing.

I guess one way to think about it conceptually is that 3 unhybridized p orbitals occupy axes x, y, and z and each orbital is 90 degrees apart from the others (image: http://d1j63owfs0b5j3.cloudfront.net/te ... 001860.jpg). As these p orbitals become increasingly hybridized with s orbitals, think of it like the severely small bond angles becoming increasingly "diluted" (so to speak) and so the bond angles increase.

[Christine Lin 1H]

As the number of regions of electron density increases, p-character increases because there are more p orbitals contributing to hybridization.
2 regions -> sp hybridized (50% s-character) -> 180 degrees
3 regions -> sp2 hybridized (33% s-character) -> 120 degrees
4 regions -> sp3 hybridized (25% s-character) -> 109.5 degrees
As you can see, the bond angle is decreasing.

I guess one way to think about it conceptually is that 3 unhybridized p orbitals occupy axes x, y, and z and each orbital is 90 degrees apart from the others (image: http://d1j63owfs0b5j3.cloudfront.net/te ... 001860.jpg). As these p orbitals become increasingly hybridized with s orbitals, think of it like the severely small bond angles becoming increasingly "diluted" (so to speak) and so the bond angles increase.

Thank you for your detailed response! I think I was confused by the "s-character" wording, but your explanation of increasing the regions/orbitals --> smaller bond angles. I think searching us the hybridized orbitals was helpful too. Thank you again Joanna :)

[Christine Lin 1H]

I think that it's because when you look at the hybridization of an orbital in a bond, as the bond angle decreases (more bonds on the central atom), more p-orbitals are involved in hybridization and bonding. For example, a linear bond is sp hybridized and has a bond angle of 180 degrees, and based on the hybridization you'd expect similar s- and p- character. However, a tetrahedral bond is sp3 hybridized and has a bond angle of 109.5 degrees, and you'd expect more p-character than s-character, since there are more p orbitals involved in hybridization. This is just how I think of it, so let me know what you think.

I agree that as the bond angle decreases and more p-orbitals are involved in hybridization, then the bond angle decreases. I think the problem (2F.15) was hinting that, since they included that the sp3 hybridized atom is 109.5° and the sp2 hybridized atom is 120°.

I think looking at the hybridized atoms was pretty helpful in understanding bond angles and s-character conceptually:

(Image from TA's slides).