October 19, 2019, 09:06:09 PM
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Topic: Hybridization of Oxygen in Water  (Read 367 times)

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Offline Varlam

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Hybridization of Oxygen in Water
« on: June 26, 2019, 04:38:12 PM »
Electon confuguration of Oxygen is 1s2 2s2 2Px2 2Py1 2Pz1

Why does Oxygen form sp3 hybridization, when it simply can share two 1s1 electrons of hydrogens with its 2Py1 and 2Pz1 to make its each orbital have 2 electons.

Offline Corribus

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Re: Hybridization of Oxygen in Water
« Reply #1 on: June 26, 2019, 10:22:26 PM »
I don't really love the hybridization model except as a tool to understand molecular geometry, but neither here nor there. I think the first thing to point out is that it's clear what you suggest doesn't happen, because the true geometry of water isn't consistent with the kind of bonding you propose. As to the why, well - there are a lot of ways you could answer that, some due to energy, some due to symmetry. I think most of those explanations would be, based on your questions, a little bit beyond your level. So let's just consider what a water molecule would look like without the hybridization: you'd have two hydrogens in perpendicular orientation, then you'd have a lone pair perpendicular in the third direction, and another lone pair (from the 2S orbital of oxygen) spread spherically around the oxygen nucleus, overlapping with the other lone pair and the electrons in the two O-H bonds. Do you see any problems with this, from an energy standpoint?
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline sgojja

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Re: Hybridization of Oxygen in Water
« Reply #2 on: August 18, 2019, 05:28:39 PM »
The concept of hybridisation is completely derived from experiments. you note the angle of a molecule, you try to define it from theoretically. that is why a particular hybridisation cant be justified. on basis of vsepr, attraction and repulsion, molecules and shapes can be justified (position of lone pairs, in specific)
but, there is an explaination. we say, hybridisation lowers down the energy. what do we mean by that? in this case, if bond with p orbital is made, then, it will be weaker because p are located far from nucleus. Have you ever noticed that every hybridisation requires a s orbital? inclusion of s orbital is a necessity for every single bond, closer the electron cloud to the nucleus, more will be attraction. hence, stability will increase.
more s character leads to more electronegativity, why? as s character increases, electron cloud becomes nearer to the nucleus, hence strong attractive forces are in action.

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