Skip to main content
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?

Verified Solution
Video duration:
1m
This video solution was recommended by our tutors as helpful for the problem above.
763
views
Was this helpful?

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