4.95 Composition of Bonds/Hybridization of Lone Pairs


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Pauline Tze 3B
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4.95 Composition of Bonds/Hybridization of Lone Pairs

Postby Pauline Tze 3B » Sun Nov 06, 2016 9:30 pm

Consider the bonding in CH2=CHCHO. (a) Draw the most important Lewis structure. Include all nonzero formal charges. (b) Identify the composition of the bonds and the hybridization of each lone pair—for example, by writing (H1s,C2sp2).

Hi, I'm confused by part b. Can someone explain how you determine what energy level the hybridization has and what this question specifically means by "hybridization of each lone pair"?

I'm especially confused by why the lone pairs of oxygen are hybridized Osp2 while in the sigma and pi bonds it's hybridized O2sp2 and O2p respectively in the solutions manual.

Thank you!

Rochelle Ellison 2H
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Joined: Wed Sep 21, 2016 3:00 pm

Re: 4.95 Composition of Bonds/Hybridization of Lone Pairs

Postby Rochelle Ellison 2H » Mon Nov 07, 2016 8:51 pm

To describe the composition of the bonds you first determine if its a sigma or a pi bond. Single bonds are sigma bonds, double bonds have both a sigma and a pi bond, and triple bonds have 2 pi and 1 sigma. Next you determine the hybridization of each atom and apply that to each of the bonds. The energy level is just the level that the valence electrons are in as those are the electrons that are hybridized to form the bond. The hybridization of the lone pairs is just asking what hybrid orbitals the lone pairs occupy so for oxygen which has 2 lone pairs and 1 carbon-oxygen double bond, the lone pairs occupy sp^2 hybrid orbitals (you could also call them 2sp^2 hybrid orbitals because they are using the valence electrons from the 2nd energy level. Finally, sigma bonds will often involve the hybridized orbitals while pi bonds will involve the unhybridized orbitals because the unhybridized p orbitals can only overlap side by side after the hybridized orbitals overlap end to end when 2 atoms form multiple bonds. So that's why the sigma bond involves oxygen's sp^2 orbital while the pi bond involves the only orbital left which is the unhybridized 2p orbital.


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