(c) Is the p bond in NO2- localized or delocalized?

Video transcript

Hello everyone in this video we're trying to find if the pie pond in this N. 03 minus ion exhibits the localization. So first let's calculate how many valence electrons there are in RN. and then we'll draw the lower structure for that. So we have the compound and three minus. We can see that we have one Atom of nitrogen and three atoms of oxygen. So of course we need to keep that in mind. We were calculating the electrons. So from nitrogen there's going to be one Adam. So one times and then each nitrogen atom has five valence electrons because instant group 58. And then for oxygen we have three of those atoms To apply by there's going to be six electrons per oxygen Adam because it's in group 68. So one times five is equal to 5, 3 times six equal to 18. And then adding the five and 18. We got a total of 23 electrons. And let's not forget this negative charge here. This negative charge here means that there's going to be one more electron. So go ahead add the one electron Giving us a total of 24 electrons to work with for our lower structure. So our lower structure, we have a central adam of nitrogen Of course connected to three oxygen atoms. So 1, 2 and three. We're going to give any one of the oxygen atoms a double bond that's going to have to long pairs and the rest is going to have three lone pairs giving each a -1 formal charge and then the nitrogen will have a positive one formal charge. If you add the charter's up, it should lead goal to just negative one. So we have negative one That's -1 is -2. The plus one is negative one. This is going to be our correct LewiS structure and we can see that there is going to be resonance. So let's throw out the resident structure. So again, we'll draw the Lewis structure again the center as nitrogen and then we have this oxygen with a double bond. Then we have two options with single bond And three lone pairs. All right. And we can see here, what we can do for residents is we can take one of the lone pairs from the this one of these oxygen's with a negative one charge to form a double bond and have one of the pi bonds go ahead and break off to create a lone pair. So, drawing the arrow pushing will have something like this. And this. So, the resident structure is going to look like this again. Center Adam as nitrogen. Now, I'll have a double bond here, still a single bond here. Get up on church and then another three lone pairs and the nitrogen will still have that plus one formal charge. So this is going to be our resident structures. You see that the pipeline between the nitrogen and oxygen that's going to go ahead and occur and participate in our resonance structure. And because it does participate between any of the oxygen atoms, So, like I said, either one of the oxygen atoms can go ahead and use one of their long parents to create that pi bond and then that pi bond will break, which means that we will have correspondence in resonance structure. Because these pI bonds do participate in residents, it means that we do exhibit D localization. So the answer is put this on the side here, answer is yes. It does participate in de localization, and that is going to be our final answer for this problem.