Hey, guys in this question were given three compounds and wrasse defying residents structures for each of them, given the curved lines using and using them as guides. So, in our first problem, in part A, we have a benzene ring with two metal attachments, and we have curved arrows more moving in the thes directions. Remember, when we have curved arrows like this with the full arrowhead, that means a full pair of electrons is moving from that bond to the next. So let me go ahead and draw my ah benzene ring benzene ring first, and then we have our two metal attachments. So when we move this pair of electrons over to this bond right here, this is going to form a double bond here. And when we do that, this carbon can't support five bond. Since we'll have a double bond here, one bond here and a double one here. So that's why this double bond on my right side is going to move to this bottom right part of the hex or the benzene ring. Now, when we get to this carbon, this carbon is going to have five bonds on it. Because of this double bond and this double bond. So this Oh, that. Yeah, this double one, this double bond. And there's going to be a hydrogen here as well. So that's five bonds. Carbon can't hold five bonds. So that's why this double one then moves to this bond right here. And when we look at this, we don't have to put any formal charges or anything because none of the, um, octet it's of the, um, carbons making up the benzene ring were changed in any way because they're basically the bonds are basically in the same places. So this is our complete residents structure for this part a moving on to part B. We have this, uh, Alka off keen and with ah, look what looks to be what was an alcohol attachment. And this Ah, oxygen is donating a pair of electrons to this bond right here. And this is going to go ahead and push this pair of electrons from this bond over to this bond which will complete this carbons octet. So let's go ahead and draw this out first. So when one pair of electron moves over to this carbon, we have a bond here. This carbon would have five bonds because of the double bond here and the making of a double bond here, as well as the hydrogen here. That's why this double bond must move over to this carbon. Now that we have that looking at all of our carbons and all of our other Adams, none of the formal charges have changed or the formal charges are all now zero. So when we have resident structures, we want to make sure that our net formal charge is the same as it was in the beginning. In the beginning, initially, we have a net formal charge of positive one here and negative one here. So that's a net of zero. And now for a residence structure. Here we have a net of zero because all of our four more charters are now zero. So this is there now complete resident structure. And now, for part C, we have this pen maintain ring and with some attachments to that, So we have a lot of electrons moving, so lets go ahead first, draw out our structure. We have the paint in ring carbon here, hydrogen here, oxygen on and oxygen. So starting off with this oxygen right here. This oxygen is donating a pair of electrons to this bond right here. So let's go ahead and do that. This carbon would have five Barnes, if this double one doesn't move, so it has to move to this pond right here in the painting ring. And now this carbon right here would have to support five bonds, which it can't to it. It moves the double wand electrons to the oxygen since its very electra negative, and it can handle the negative charge. So this now has a negative formal charge, and we have a negative negative one formal charge to start off with. So this structure is now are a complete residents structure.