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Ch.9 - Molecular Geometry and Bonding Theories

Chapter 9, Problem 8b

The drawing below shows the overlap of two hybrid orbitals to form a bond in a hydrocarbon. (b) Which of the following could be the identity of the hydrocarbon: (i) CH4, (ii) C2H6, (iii) C2H4, or (iv) C2H2?

Diagram showing the overlap of two hybrid orbitals forming a bond in a hydrocarbon.

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Video transcript

Hello. Everyone in this video. We're going to go ahead take a look at this orbital overlap right here and we're trying to see what possible molecule can be from these choices right over here. So from this picture, what I can gather is that we have a set of two sp hybridized orbital's. So what that means is that there's a bond is formed and what that bond is going to be a triple bond. So when we know it's made of triple bonds, so triple bond is made up of one sigma bond And two Pi bonds. So because we have a triple bond, the ending for the nomenclature is going to be Y N. E. So taking a look at our choices here, then the only one with this Y. E ending is going to be Butin. So then our only option, Our only answer is going to be two, which is a few time. And this is going to be my final answer for this question. Thank you all so much for watching