## Hybridization Labeling

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

Christine Chen 1H
Posts: 64
Joined: Fri Sep 28, 2018 12:18 am

### Hybridization Labeling

Can somebody explain to me what the labeling of the hybridization (the stuff in parentheses) signifies?
For reference, this is HW question 4.95 from Edition 6; the structure is CH2=CHCHO
Attachments

daisyjimenezt
Posts: 30
Joined: Fri Sep 28, 2018 12:26 am

### Re: Hybridization Labeling

I think it shows what two atoms the sigma and pi bonds are between

angelagd3l
Posts: 32
Joined: Fri Sep 28, 2018 12:26 am

### Re: Hybridization Labeling

Should we be able to draw similar examples on Test 3?

Danny Elias Dis 1E
Posts: 60
Joined: Fri Sep 28, 2018 12:19 am

### Re: Hybridization Labeling

Can someone please explain the labeling on the lone pairs of electrons? How does that work for the oxygen for example?

StudentD2B
Posts: 43
Joined: Fri Sep 28, 2018 12:27 am

### Re: Hybridization Labeling

I believe the labeling is specifying the different types of hybridized orbitals that are participating in the bond. (sp, sp2, sp3, sp3d, or sp3d2). For example on the top right double bond between oxygen and carbon its saying that the sigma bond is between the hybridized sp2 orbital of the carbon and the hybridized sp2 orbital of the oxygen, and the pi bond is between the 2p orbital of the carbon and the 2p orbital of the oxygen. (sp2 hybridization being a combination of the one s-orbital and 2 p-orbitals)

StudentD2B
Posts: 43
Joined: Fri Sep 28, 2018 12:27 am

### Re: Hybridization Labeling

As for the lone pairs on the oxygen I believe that the "Osp2" denotation is saying that the lone pairs on the oxygen are in one of the hybridized sp2 orbital. There are three:
$h_{1}=s + 2^{\frac{1}{2}}p_{y}$
$h_{2}= s+(\frac{3}{2})^{\frac{1}{2}}p_{x}-(\frac{1}{2})^{\frac{1}{2}}p_{y}$
$h_{3}= s-(\frac{3}{2})^{\frac{1}{2}}p_{x}-(\frac{1}{2})^{\frac{1}{2}}p_{y}$

I seriously doubt you need to know the equations and I'm not sure how you tell which of the sp2 hybrid orbitals the electrons are in, but I believe that what it is saying