Draw as many resonance structures as you can for the following ni...

Question

Draw as many resonance structures as you can for the following nitrogen-containing compounds. Not all will obey the octet rule. Use curved arrows to depict the conversion of one structure into another.
(d) N2O31ONNO22

Accepted Answer

Well, everyone's in this video. We want to go ahead and give all possible resonance structures for this molecule right here and 204 and we of course use curve arrows represent the conversion of one structure to another resident structure. So first things first, we can go ahead and calculate the amount of valence electrons in this molecule. So we see here that for our nitrogen we have two atoms of nitrogen. Each nitrogen atom will contribute five valence electrons because it's in group 58, so two times five is equal to 10 and that's how many valence electrons nitrogen contributes. And for oxygen here we have four items of auction. Each will give us six valence electrons, so four times six will be equal to 24. We take the sum of these two products here to give us a total of 34 valence electrons that present in this molecule of n T 04. So now we can use this information here to draw the first structure and then following up with the other possible structures. So we know that from our compact um resemblance here, we have two auctions per ever nitrogen. So how can draw this is again, we have the nitrogen here And oxygen usually like to have two bonds. However, because there's two oxygen's per nitrogen, we can only give one double bond to the oxygen and the other one will have a single bond. So we can say that this one here will have the double bond which will need two lone pairs to complete its octet. And then if we have a single bond oxygen, then we need three bonds to give us its octet because it has three bonds or three lone pairs here, this oxygen will have a positive one formal charge. Same thing for the neighboring auction here we'll just give it a single bond and three lone pairs. So I have one In -1 from a charge and then a double bond auction to the left here. This will make our nitrogen is having a formal charge of plus one per atom here. And then we can see that our two positive one formal charges will cancel out with the two negative one formal charges because this molecule is neutral. So these charges must go ahead and cancel each other. Now using our curved arrows will go ahead and represent our our predict our second resident structured. So this pi bond here on the left side, we go and break that and form a another lone pair. And then whenever auction items here will give us well give up a lone pair to form a double bond. So we can go ahead and drop that curved arrow. So following our curved arrow, we're basically moving around our lone pairs and pi bonds. So all the right side here stays the same, nothing is touched there. So just go ahead and redraw that. Okay and now for the left side here we study the furthest to the left. This oxygen here we're breaking that pi bond so have a single bond. And then they'll have an oxygen with now three lone pairs instead of two. Now for the bottom oxygen. Right over here we're doing is using one of the lone pairs to create a double bond. So we have double bond to oxygen and they will have two lone pairs. Now instead of three Of course the auction to the left. Now with the three long has a formal charge of -1. And again these charges will cancel. All right, next one we're gonna go ahead and now manipulate the auction item on the right side on the bottom. Specifically, we'll go ahead and use one of them in pairs to create a double bond. And then the other auction item towards the right will go ahead and break the pi bond and create a lone pair onto our auction atom. So we follow those curved arrows again, the now left side of our atom or molecule will not change. So I'll just rewrite everything or everything as is Alright, so for the oxygen for this to the right here we're breaking our pi bond together as a single bond, oxygen. So of course we'll have three lone pairs giving a negative one formal charge. And for the bottom auction will have a double bond. So now we have a positive one chart here and then we have two negative one charges here. So again this or this molecule is neutral and last. We can go ahead and manipulate our auction item to the left here, forming a double bond. And then the oxygen atom here, we can go ahead and break the pi bond to create a lone pair. So you can see here that the right side of the molecule is not touched. So I'll just redraw everything as is. So we have a nitrogen with a single bond, oxygen three lone pairs giving a formal one charge negative one charge, double bond oxygen to lone pairs that's neutral is connected to another nitrogen here. So for the left oxygen right over here, what we're doing is making a double bond and it has two lone pairs now instead of three. So there's no charge there. But the oxygen on the bottom here is now having a single bond with three lone pairs. So I'll have a negative one charge. Of course, a nitrogen is still a plus one formal charge. So we see here that these four structures are going to be all the possible resonance structures of N +204. And this is going to be my final answer for this problem

Draw as many resonance structures as you can for the following nitrogen-containing compounds. Not all will obey the octet rule. Use curved arrows to depict the conversion of one structure into another. (d) N2O31ONNO22

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