## Lack of Spin Pairing in Hybridized Orbitals

$sp, sp^{2}, sp^{3}, dsp^{3}, d^{2}sp^{3}$

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### Lack of Spin Pairing in Hybridized Orbitals

In today's lecture, Dr. Lavelle said that in a sp2 hybridization (in the molecule C2H2 for example) the fourth electron of the carbon would have to be in the unhybridized 2p orbital because the energy difference between the sp2 orbitals and unhybridized p orbital is very small. I know it has something to do with the repulsion between the two electrons if they were spin paired, but I am not exactly sure what the energy difference has to do with it or why the repulsion is too great in the hybridized orbital, even though electrons are usually spin paired in normal, unhybridized orbitals. I would appreciate if someone could clarify this concept.

Megan Sparrow 1A
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### Re: Lack of Spin Pairing in Hybridized Orbitals

During hybridization, the C-C sigma bond is formed when one sp orbital overlaps from each of the carbons, and two C-H bonds are created when the second sp orbital on each carbon overlaps with 1s orbital of hydrogen. In this, the carbon atom will have two half-filled 2p orbitals. These two pairs of p orbitals do not participate in the hybridization and instead form two pi bonds resulting in the creation of a triple bond.

An important thing to note is that in the formation of C2H2, the carbon atom needs extra electrons to form 4 bonds with hydrogen and other carbon atoms. As a result, one 2s2 pair is moved to the empty 2pz orbital. This is plausible since the energy difference is very low between the hybridized orbitals and non hybridized ones.
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Faith Lee 2L
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### Re: Lack of Spin Pairing in Hybridized Orbitals

Adding onto what Megan said, if we allowed spin pairing in the sp^2 orbitals, we would be left with 1 paired e- and 2 unpaired e- in the sp^2-orbital and no e- in the unhybridized p-orbital. This would only allow carbon to bond with 2 other atoms. By "moving" 1 of the e- in the paired sp^2-orbital to the empty p-orbital we now have 4 orbitals that can accomodate an extra electron (and therefore an extra bond) each. The small energy difference between the sp^2-orbital and the p-orbtital helps illustrate how "moving" the paired e- to the p-orbital is feasible since it does not require an enormous amount of energy to do so :)