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Topic: Valence Bonding Theory - Clarification  (Read 1645 times)

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Offline airplanet56

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Valence Bonding Theory - Clarification
« on: July 01, 2013, 11:52:34 AM »
I'm learning about sp, sp2, sp2, sp3d, and sp3d2 hybrid orbitals. What is the need for these orbitals? Is it because electrons from s subshells from one atom have to fill up p subshells or d subshells from another atom? Does this action always require hybrid orbitals?

Offline Corribus

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Re: Valence Bonding Theory - Clarification
« Reply #1 on: July 01, 2013, 12:08:21 PM »
It helps to explain the shapes of molecules. 
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 magician4

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Re: Valence Bonding Theory - Clarification
« Reply #2 on: July 01, 2013, 12:20:52 PM »
a molecule always will try to become as energy-low as possible

electronic configuration on the other hand is only one aspect of the whole picture, with repulsive effects between the substituents on the other hand being another important factor.

 :rarrow:  sometimes geometry overrules , with respect to total energy balance

example:

methane does have 4 hydrogen substituents, and they're repulsive with respect to each other
arranging 3 of them under 90° (and the fourth one wheresoever), thereby forming 3 bonds of s-p-sigma type and one bond of s-s  sigma type (i.e. not hybridizating the carbon at all) will lead to high repulsive effects between the former hydrogens
making room between those hydrogens would give energetic relief, but would be costly as carbon has to form sp3-orbitals as a consequence
as a net effect this investment pays: arranging the substituents in a tetrahedral geometry will lessen repulsive effects, allow for better ("shorter") bondlength C-H , and will overcompensate the investment for hybridization by far

with PH3 on the other hand, the picture is different: phosphor is lager in diameter , hence the repulsive effects in between the hydrogens (and, additionally,  the free electron pair) won't be that hefty
in this case you'll see next to no hybridization at all: the hydrogens will be arranged under next to 90° (instead of 109,8 , as it would be the case with sp3 hybridization)  , in fact forming the s-p bonds mentioned at the methane (if only as a theoretical option there: here they become real), with the free electron pair remaining in the s-orbital at the phosphor


that's what hybrid orbitals are good for, and that's why they sometimes occur whenever the energy situation of the whole molecule will look better with their use

regards

Ingo
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