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Topic: Predicting Hybridization  (Read 4718 times)

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Offline Quietus

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Predicting Hybridization
« on: June 23, 2009, 12:17:09 AM »
Hi guys,

Sorry if this is a really elementary question, but I would like some clarification because I don't feel that I completely understand this.

Methylene (CH2) exists in one of two ways; spin paired, or spin unpaired.
I must predict which type of hybridization the carbon adopts.

For spin paired, the two non-bonding electrons must be in the same orbital to be spin paired, so I assumed correctly that the hybridization would be sp2. My question here is why would it form this way with an empty p orbital? do hybridized orbitals always fill before the non-hybridized orbitals do?

For spin unpaired, the answer is sp, leaving two sp orbitals each bonded to a hydrogen 180 degrees apart, and two p orbitals each with one of the two non-bonding electrons. Why wouldn't it be sp3? Do hybridized orbitals only form if there are enough electrons around to fill them?

Thanks!

Offline lancenti

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Re: Predicting Hybridization
« Reply #1 on: June 24, 2009, 03:50:55 AM »
I'm of the opinion that hybrid orbitals always fill first on the basis that they're of a lower energy. Also, if they weren't both in the hybrid orbital, they wouldn't be spin-paired anymore. This answers your first question.

As for spin unpaired, there are two bond-pairs and two electrons. Adopting an sp3 configuration would result in the two bond pairs being only 109.5 degrees apart, whereas the sp configuration would be 180 degrees apart. Sure, the two lone electrons are 90 degrees from each other and the bond pairs but that should be less of an issue than bond pairs being closer to each other. I read it somewhere, can't remember the exact explanation, that electrons spinning in pairs tend to be of a higher energy.

Offline FeLiXe

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Re: Predicting Hybridization
« Reply #2 on: June 24, 2009, 12:16:11 PM »
you could also think of it in the way that the two p-orbitals are higher in energy than hybrid orbitals. it is better to have them only singly occupied and fully occupied lower energy sp-hybrids. then degenerate sp3 hybrid orbitals partially filled

and of course it also has to be said that neither model is completely accurate: the triplet carbene is not completely linear and the singlet does not have exactly a 120° angle (http://en.wikipedia.org/wiki/Carbene#Structure_and_bonding)
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