Does oxygen in water have an sp3 orbital structure?

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The discussion centers on the molecular geometry of water (H2O) and its bond angle of 104.5°, which deviates from the expected angles of sp3 hybridization. Participants question why water's bond angle is described as a "greatly expanded" p-orbital angle rather than a "slightly contracted" sp3 angle. Dr. Richard F.W. Bader's work is referenced, suggesting that the bond angle's deviation can be attributed to reduced repulsion between hydrogen nuclei. The conversation critiques the limitations of valence bond theory in explaining water's structure, arguing that traditional hybridization models may not accurately represent chemical bonding. Ultimately, the energy difference between s and p orbitals in oxygen is highlighted as a factor that makes s-p hybridization energetically unfavorable in water.
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The angle of the water H2O molecule is 104.5°, the angle of ammonia H3N is 107°.

The angle between 2 p-orbitals is 90°, the angle between 2 sp3 hybrid orbitals is 109°28', the tetrahedral angle.

Why is it assumed that water is a "greatly" expanded p-orbital angle, rather then a "slightly" contracted sp3 orbital?
 
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Please explain to me what you mean by a 'greatly' expanded p-orbital angle for water and who told you or where did you see this?
 
chemisttree said:
...expanded p-orbital angle for water?

from "An Introduction to the Electronic Structure of Atoms and Molecules"
Dr. Richard F.W. Bader, Professor of Chemistry / McMaster University / Hamilton, Ontario

http://www.chemistry.mcmaster.ca/esam/Chapter_6/section_4.html

"The actual bond angle in the water molecule is 104.5°. The opening of the angle to a value greater than the predicted one of 90° can be accounted for in terms of a lessening of the repulsion between the hydrogen nuclei."
 
OK, let's look at his entire discussion relative to the point.
The actual bond angle in the water molecule is 104.5°. The opening of the angle to a value greater than the predicted one of 90° can be accounted for in terms of a lessening of the repulsion between the hydrogen nuclei. The assumption we have made is that the maximum amount of electron density will be transferred to the binding region and hence yield the strongest possible bond when the hydrogen and oxygen nuclei lie on the axis which is defined by the direction of the 2p orbital. For a given internuclear separation, this will result in the maximum overlap of the orbitals. Because an orbital with l ¹ 0 restricts the motion of the electron to certain preferred directions in space, bond angles and molecular geometry will be determined, to a first rough approximation, by the inter-orbital angles.
You note he uses the word 'assumption' and 'to a first rough approximation' in his discussion. This suggests to you that the valence bond theory doesn't adequately describe what is observed (although the professor never explicitly states that)
He then goes on to describe hybridization which better describes what is observed but with different examples. He is a poor teacher, that's all.
 
The point is that the energy difference between s and p orbitals in oxygen is fairly large, so that s-p hybridization is energetically unfavorable in water.
In fact, valence bond theory gives an exceedingly accurate description of H2O:
http://dx.doi.org/10.1016/0166-1280(88)80277-X
I don't think that Baader is a bad teacher. It is more the other way round: Most introductory chemistry text try to give a "one fits it all" description of chemical bonding in terms of hybrids which is often not physically correct.
 
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