The odor of cinnamon oil is due to cinnamaldehyde, C9H8O. What is the hybridization of each carbon atom in cinnamaldehyde? How many s and how many p bonds does cinnamaldehyde have?

Video transcript

Hello. Everyone in this video, we're dealing with this structure here. This is a representation of our molecule with a formula of C 10 H 12 0. So we have to go ahead and determine the hybridization of each carbon atom in this molecule as well as providing the number of sigma bonds and pi bonds present in our lewis dot structure. So no matter what the number of bonds is, there's always one sigma bond. So if we have a double bond or triple bond, there's always going to be at least just one sigma bond and the rest are going to be pi bonds. Let's go ahead and count the number of single bonds that we have. And then of course, because we have a double bond, there's still going to be a sigma bond present. So we'll just go ahead and count the double bond as being a single bond as well. So we can go ahead and do the sigma bonds first. Let's go ahead and start maybe from the right. So we have 123456789, 10 11 12 13 14 16 17 18 21 22 23. So we just counted that we have a total of sigma bonds. Now going ahead to count for our pi bonds. And like I said, let's say, we have a double bond or triple bond, one of them will be a sigma bond and rest will be pi bonds here in the structure. We only have double bonds. So we'll go ahead and just count those double bonds as having the other bonds. As not as a pi bond. So for example, for this will just be one pi bond because one of them is sigma, so then we'll have 1234. We have a total of four pi bonds in the structure. So that's going to be one part of our answer ago and highlight that meaning that this is going to be our final answer. Another part of the question is asking for the hybridization of each carbon atom. So we see a lot of common carbons here in this structure, one of them being a carbon with a single bond, then carbon with a double bond each will have different hybridization. So we'll go ahead and start with the carbon with double bonds. And this seems more prominent here. Alright, so we'll go ahead and look at one of our carbons with a double bond. So maybe this one right here because they're all the same. So we have one electron group from the hydrogen, one electron group from the single bond carbon and one electron group from this double bond carbon. So even though this is a double bond is still connected to just one carbon atom and that's one electron group. So then we have 12 and three carbon or electron groups. So because we have three electronic groups, the hybridization for this is going to be S P two, so one from the s orbital and two from the pure orbital. So total will be three and this matches the three electron groups. Now, that's going to be one of our final answers. Another one then is going to be our carbon with just single bonds. So for example, then you take a look at this metal group here which is hanging off this auction atom. So this carbon here, we have one electron group from our oxygen and then three from our high regions. So we have a total of four electron groups, electron groups. All right, because we have four electron groups will add one more to RP. So B S. P three. Again, we have one orbital from the S orbital and three from the p orbital. So one plus three is four matching this four electric group rule. So this will be our final answer for this problem.