 Okay, so let's do this molecule here, ICL3. So what we say about this molecule already, ICL3, is that normal valence, ICL3? No, it's what? Expanded valence. Right? And which one of those atoms is the central atom, the iodine, and how do we know? This is the biggest one, okay? So we should expect it to be something like, let's just draw it like that, okay? So we've got something like that, but we've got ICL3, right? So CL in this case does not expand its valence, hopefully everybody notices that, okay? So I'm going to draw those other two CLs in a weird place. So this one, hopefully you guys would expect to have formed that bond like that, okay? So that one makes sense, but because iodine has expanded its valence, and remember, what do we need in order to form a bond? We need a half-filled orbital, right? So what happens is this orbital that's filled, if you will, kind of splits up and becomes two half-filled orbitals, okay? So when that happens, we get something like that, okay? Does everybody see that? So now, hopefully you see, we've got one, two, three bonds being formed. One, two lone pairs still around, okay? Does everybody see that? Any questions about that? Okay, so let's draw it. These are wedged and hashed bonds, okay? Well, not bonds, but in the lone pairs, okay? So if we look over here, I know we have this diagram, which you won't have during the exam or anything, so you have to remember. So how many, oh, sorry, I'm erasing that. How many electron groups does this iodine have around it? Five, right? So we're going to write number of electron groups. So this side wants you guys to do it, okay? So you've got to do these baby steps, okay? So five electron groups. So what is the electronic geometry then? Trigonal bipyramidal, why? Because that has the five electron groups around it, okay? Okay, so we only have three atoms around the central atom, right? So there's not five atoms, so it's not trigonal bipyramidal molecular geometry, but instead, it's T-shape. So again, you've got to go with, well, five electron groups and it must be trigonal. But since we've got only number of atoms, it's three, it must be T-shape. Is everybody okay with this analysis? So this is what you have to do for everyone, okay? Are there any questions about this one? This one's much harder, I think, than the ones that do the octet rule, you know? But honestly, I don't think it's something that's out of your realm of possibilities, okay? Yeah? But that should be the Lewis structure or the Lewis? So in this case, again, the molecular and the Lewis structure are the same. Because you've got to remember that this lone pair here is coming out towards us like this. And this lone pair is actually kind of going back away from us if you want to think about it that way, okay? So I think it's easier to see in this kind of diagram that the computer drew. But you're going to have to write things like this. You're going to have to write things like this, you know, on the page. But when you are drawing your lone pairs, you don't usually put a wedge or a dash or anything like that. So in actuality, drawing something like this and drawing something like this would be equivalent structures, if that makes sense, okay? And this, yes, would be the same as the Lewis structure as well, okay? This is a good question. Any other questions on this? So what are the angles? What would you expect them to be? So these two are on the equator there, okay? So this angle, what would you expect that to be? About 90 degrees. So look here, right, 90, 90. And if we had atoms here, we would expect around that equator to be 120, 120, and 120. So those are the bond angles for something that's a trigonal bipyramidal, it's 90 and 120, okay? So that's another good question, but this one doesn't have the 120 bond angles. Okay, oh, I guess it does have this one though, which is 180. Any other questions about this? Okay, like I said, I have, I think at least a dozen other expanded valence videos. Please watch them at your leisure to understand any ones that you don't understand, okay?