Aspirin has the following connections among atoms. Complete the electron-dot structure for aspirin, tell how many s bonds and how many p bonds the molecule contains, and tell the hybridization of each carbon atom.

Video transcript

So everyone in this video, we're given this incomplete structure here. So we have to first go ahead and add any double bonds necessary as well as lone pairs to fulfill all the atoms octet and then we have to determine the hybridization of carbonation and the oxygen atoms present in our molecule. And then we have to also give the number of sigma and pi bonds that are seen in this molecule. So first we go ahead, evaluate our flooring atoms. So flooring likes to have one bond and three lone pairs. So you see in the structure here that we have three flooring atoms each have one single bond. So let's go ahead and add the three lone pairs per atom. I'll do this in the color red. Alright, so we have added the three lone pairs and now flooring has its octet. Now oxygen likes to have two bonds and two lone pairs. So you can see here in our molecule that we have these two here. So this auction here already has two bonds. So it just needs two more lone pairs. So we can go ahead and add that again. I'll do this in red. As for this auction here, we only have one single bonds but we can also add another bond to fulfill its requirements of having two bonds. So we'll do some purple. So now that we have to bond on this auction, we can go ahead and add the necessary two lone pairs to fulfill the auctions of tits. All right. So now all the oxygen atoms has fulfilled its octet rule. We'll see we're also going to deal with hydrogen and nitrogen and carbons. So for higher regions it does not need a trans like other atoms do. It fulfills or it requires two electrons. Because of the duet rule and because of the two electrons, it only needs one bond. So that's why they're only at the end of the molecules is always at the end. It's never a central item. So that's already something that we don't need to consider. So, carbon likes to have four bonds. So you see here that the carbons are usually in the ring here except for this carbon this carbon this carbon already has the four bonds um one bond here here and then we have a double bond. So that's four bonds. And then for carbon we have four different groups. We have the three car Florence here and then one other carbon bonds there. But for these rings here, we need to go ahead and add alternating double bonds. So we can put one here here and here and for the right six member ring, you can add one here here and here. Next adam will evaluate is nitrogen. So the nitrogen likes to have three bonds and two lone pairs. So you see here for the central nitrogen here, there's already three bonds, one to the hydrogen and two to the carbon. So we can go ahead and just add one more long pair. As for our nitrogen to the right six member ring here, we see that since we already had this double bond, this does have 33 bonds. So we just need to add one more long pair to fulfill nitrogen octet. So now we have added all the necessary double bonds and lone pairs to fulfill the atoms octet. So now we can go ahead and count the number of sigma and pi bonds. So we have double bonds and single bonds in our structure here. And all the violent bonds have a sigma bond, regardless if there's one bond to bond and three bonds, that's just the minimum amount that it has. And in any other remaining bonds are going to be our pi bonds. So in our case we can go ahead and just count single and double bonds as having a presence of one sigma bond. So this does have a lot of bonds. I'll try to do this a little bit slower. So we have 123456789, 10 11 12 13 14 16 17 18 1920 22 23 24 25 27 28 29 30. So we have counted 30 total bonds. So there's going to be 30 sigma bonds in this molecule. And like I said, any other bonds, if there's a double bond, triple bond, the other remaining bonds were going to be pi bonds. So here we're gonna just count the double bonds as having a presence of that pi bond. So here we have 123456 and seven. So there's seven pi bonds present in this molecule. Now we go ahead and determine the hybridization of the carbon nitrogen oxygen atoms. So you can see here that there's going to be carbon, nitrogen oxygen atoms with single bonds as well as double bonds. And they all have similar connectivity ease. So for carbon nitrogen oxygen with the single bonds only they all have three groups or four groups. So let's just take into account this carbon for example, with the three floors and one carbon. So here we have 1234. We have four different groups. That's going to be a sp sp three hybridization. And we see here for our nitrogen here we have three bonds and one lone pair. Again that's S. P. Three. And for auction, let's take into this one to account because it only has single bonds. So we have two bonds here and two lone pairs. That's four different groups. That sp three. So then we can go ahead and conclude that our carbon nitrogen auction with specifically are single bonds Is all going to be sp three hybrid ice. So that's one answer. That's another one. And now we can go ahead and look at the carbon nitrogen and oxygen with our double bonds. So again they have similar connectivity ease. So let's consider carbon with a double bond. So this right here, it's going to have three groups connected to it. So there's 12 and three. And again, we're looking at this carbon as a central atom. So there's 12 and three different groups. That's Sp two hybrid ice. Now for high nitrogen with a double bond, which is this one right here. So there's two bonds here and then one lone pair. So that's S. P. Two and oxygen with a double bond. Is this carbon neutral group here? So there's one bond and two lone pairs. Again, that's S P two. So all carbon nitrogen oxygen atoms with double bonds are all SP two hybridized. So this highlighted answers are all going to be my final answers for this problem.