What hybridization do you expect for the atom indicated in red in each of the following species? (a) CH3CO2-

Video transcript

Hello everyone in this video. We want to go ahead and identify the orbiters that overlap to form bonds one here between the carbon and chlorine and bond to between the carbon and oxygen. And we can go ahead and ignore residents in this problem. So for bond one we have carbon, let's actually write this out. So we're here dealing with bond one and carbon here has three bonding groups. I'll just simplify that as B. G. So that's going to have a hybridization of sP two. And then for chlorine here we can see that we have one bond and three lone pairs. So one bonding group and three lone pairs are simplified long pairs as LP. So that again has a hybridization of S. P. Three because there's four total electron groups. Now we can see that the orbital's that are used to form this bond. Then between the chlorine and carbon is going to be from carbon sp two and chlorine sp three. Now we go ahead and look at bond too. So the bond between our carbon and auction difference here is that we have a double bond. So carbon has three bonding groups. So B. G. That has a hybridization then of S. P. To the auction has one bonding group as well as two lone pairs. So this is S. P. Two hybridized because there's a total of three bond or three electron groups. So then for our sigma bond. So carbon auctions sigma bond that deals with our carbon sp two interacting with auctions sP two orbital. And then for our carbon auction sigma bond this one is different. It's an interaction between RP orbital's, so we have carbons. P orbital reacting with oxygen is pure metal. So the highlight answers here are going to be my final answer for this problem.