um, example would be a question that says, find the Louis structure for the compound end to age four. I don't know what that looks like, so let's just go ahead and jump into it. The first thing I do is I take the four atoms, and I say, which one has the highest bonding preference? Obviously one of the nitrogen. Okay, So what I do is I take my biggest 1/1 of the highest bonding preference, and they put it in the middle, and then I propose a sigma pawn framework. Okay, Notice that even put random here because I'm trying to emphasize that you're not supposed to just know this from the first shot. Ah, lot of students feel confused because they're like, How am I supposed to know this? This is trial and error. So what I did here was I literally just made, like, the dumbest version that I could. I just put a bunch of atoms around the end. Okay, so I said, Okay, one of the H is could go this to this end. The other h could go here. The other h could go here, and then one of the ends can go there. Okay. What I'm basically doing is I'm just trying to get the first the first Adam in the middle toe Phillips octet, and then any other atoms, I'm attaching them somewhere else. So basically attached four atoms that end because the most city can handle. I had one hydrogen left over. Then I was like, Hey, you know, I'll just put it on that end and see what happens. Okay, so now if I go to my second rule, it says fill in the octet with lone pairs so that first nitrogen that's in the middle fills its octet, has four bonds. But the nitrogen that I put on the side does not. This one would need to Lone pairs. That's what I did in this drawing. I went ahead and added those two lone pairs. And now what I did is I go ahead and calculate formal charge. This is the part that is different from Gen. Kim and Jim. Kim usually wouldn't do this, But in Oracle, I always want you to calculate formal charge. Now that we do that, what we find is that I'm gonna have a positive charge on the end because the end wants to have five electrons, but it only has four. And then I have a negative charge on that end because that end has six. Is that cool? So far, this is not very good. I can probably do better than this. So what I'm gonna do is if I get Teoh a structure that has lots of charges or lots of adjacent charges. I have two options. Either I can rearrange the signal. Bon framework. Meaning I can rearrange where these where these single bonds are. Or I can add pi bonds to remove excess lone pairs. Okay, so in this case, I have two different options. I could try to make a double bond, but I would run into a big problem if I made a double bond right? The way it is drawn right now. The problem is that let's say that I tried to put a dull bond here to get rid of these electrons. Then what would happen is that that nitrogen would wind up breaking its octet. This one right here would now have 10 electrons. So there's actually nowhere that I could make a double bond. That makes sense. Okay, there's actually nowhere that I could make double bond That wouldn't violate someone's octet. So that means that I have to change the way change around the signal on framework. So what I'm thinking is this This nitrogen wants more electrons. This nitrogen wants less electrons and wants less electrons the way that I could make that happen, possibly by moving a bond. If I could move one of these h is over here, Then what that would do is I would take away one lone pair and replace it with a stick. And remember that a stick only counts as one, whereas a lone pair accounts is to So let me redraw this looking like this and and each each each h. Okay, Now, I still have to make sure that everything fulfills its octet. So I'm gonna have to add lone pairs. So I'm gonna need a lone pair here and a lone pair there now, like, go ahead and a re calculate formal charges and see if this is any better. So the formal charge of this nitrogen would now be zero because of the fact that it's got three bonds on a lone pair, which is the way nitrogen wants to be. And this one is also zero. So that's it. We found a good Louis structure. Okay, so obviously this one is kind of easy, but I'm just showing you that is the right Lewis structure for this molecule. And we got there through trial and error. We didn't just immediately know what the answer Waas, alright.