Lecture 17. Review
Brindley, Amanda
UCI Chem 1A is the first quarter of General Chemistry and covers the following topics: Atomic structure; general properties of the elements; covalent, ionic, and metallic bonding; intermolecular forces; mass relationships. 
 Okay, so today. Okay, so we're going to finish up some hydrogen bonding things that I haven't quite finished up. Well, maybe we will. Okay. And then, so that won't take too much longer. And then we'll do a review lecture. I'm not sure if that whole packet that I gave you, if we'll get through all of it today or not. I actually, leaning forward, we probably won't. So whatever I don't finish in class today, I don't want us to feel rushed. I'm just going to do that Thursday night. So we'll have plenty of time over, spread over everything to at least finish it. So don't worry too much about that. Okay, so the rest of the hydrogen bonding that we need to talk about is why this is so important in biology. This is important not just in general, but also because of the demographics of this class tend to be pretty highly in the biological or variety of earth system science type things. Okay, so the first one that I want to talk about is in DNA. Now this is very, very important. Now you probably have a little bit of background either from high school or from your biology classes here on how DNA kind of looks. So you get this double helix structure. Okay, so we get this double helix structure and we have all of these bonded and those are bonded covalently or some other form. Covalently. Now the thing that holds these two together is what comes important in this chapter. What are all of these bonds? They're hydrogen bonds. So they're not actually covalent bonds that are holding them together. And that's really important because what do we need DNA to be able to do in order to live? It's got to be able to replicate in one other thing. Make mRNA, right? You've got to be able to replicate and make RNA. If you can't do those two things, we're going to have a little bit of trouble making new cells and making proteins and doing all the things that we need our DNA to tell our bodies to do. So in order to replicate it, the way that it works is you have to be able to split a word to make RNA. You have to be able to split the DNA strands. If they were covalently bonded, that would be a lot harder. You could still do it, but it would be a lot harder. But having these hydrogen bonds here strikes a good balance between they're fairly solid, they're strong, but they're not so strong that you can't undo them relatively easily. So we have proteins in our bodies that come through and can unzip them and rezip them back up as we go along. If it was covalently bonded, that would be tougher. If it was held together by some other force, that might be a lot harder too. The hydrogen bonds are kind of this balance. Now going a little bit more in depth with this, again, because of the demographics of this class, notice between the adenine and the thiamine, how many hydrogen bonds do we have? Two. What about between the guanine and the cytosine? Three. Three. Which ones are going to be stronger? Three. Yeah. So as you go along in your biology classes, you'll start at some point, probably your junior year, so you'll start calculating melting points for DNA strands and talking about how much, you know, when you start, especially in your molecular biology methods and lab courses, and you'll be calculating these, and you'll be calculating melting temperatures, how much you have to pull, how much energy or heat you have to put in to pull them apart and, you know, before you can let them come back together. And it's going to make a difference how much A's and T's you have and how many C's and G's you have. And that's why, because C's and G's bond, or they bond at three hydrogen bonds instead of two. So they stay together easier. So if you want to break them apart, it takes more energy. So that's, you know, when you're doing lab work and things like that in some of your higher level bioclasses, that'll come into play. Okay. The next place that I want to talk about, that it plays very important roles, is in proteins. So this is a picture of a protein, is one I just pulled off from this website down here. And all of these little yellow dotted lines are drawn in using hydra, or those are all your hydrogen bonds. Now, it's not all of them, of course, because there's a lot of them that are forming all your different structures, your secondary structures. But hydrogen bonds are largely responsible for your secondary structures. They play all of these important roles and tertiary structures that you can see. And so they're very, very important in protein formation. You need these proteins to stay in these very tight, there's not a lot of leeway between a lot of these structures on how much they can move. And hydrogen bonds are a large portion responsible to their structure. Without them, you wouldn't have as well of a defined structure as you do. Now in hydrogen bonding, just so you realize here, this is an amino acid. Okay? So this amino acid is what makes up all your proteins. So it's just chains and chains and chains of these, and big long formations. And so what happens is you can get your oxygens and your nitrogens to hydrogen bond together with your hydrogens that are on them. And that's what forms your structure is all these groups here. And we won't go into detail about the exact structures of that, I'll leave that for your higher level classes. But just sort of know how this works, know it's important. Be able to sort of explain to me why these sorts of things are important. The next one then to talk about real quickly is cellulose. So where do you know cellulose from? What does that make up? Your plants, right? That's sort of the structural component of your plants. And the way that cellulose works is it's these long chains of glucose. So you get these really, really long chains of glucose. Now on their own, a single long chain of glucose isn't going to be very sturdy. They're not going to be able to hold up a redwood with just individual chains of glucose. But then what happens is all of these chains, they're able to line up and they're able to form hydrogen bonds with each other. Look at all these OH groups here and these OH groups. Cellulose can all form hydrogen bonds in between chains and so you get these big long stacks of these chains and you form these fibrils and then that's what is able to make up all the structural components of your walls or your cell walls and things. So that is all I have for you on the rest of that chapter. So we're done with new midterm material. This was one thing I wanted. I meant to do this before but this is one thing I wanted to clarify from last time. There's a bit of a mistake. This is the correct ordering. So this is a problem that we did last class right kind of near the end. But there's a little bit of a mistake in the fact that I can't actually expect you to be able to rank all four of these because there's a big exception here that comes from the shape of the molecule. We didn't talk about that and we're not going to talk about that. So in a situation like this, this question, if it were to show up on an exam, would show up in this form. I could ask you to rank those three or I could ask you to rank those three. I would not ask you to rank all four. So I sent out an email to this extent but I know not everybody reads their emails. You should please read your emails. So just making sure that you guys realize that. So I would ask you to rank either this grouping or this grouping, not all four of them. So now under review material. All right. So this is, since you have done a large portion of your studying for the exam, it's not a bad idea as I'm lecturing here to try to beat me to the problems, right? You don't need to wait for me to write down the answer to write down the answer. Try to beat me to it. And then if I write down something different, you can say, wait, wait, I did something wrong. Figure out what you did wrong and clarify your mistake. Okay. So I have this drawing, I cl4 minus and I cl4 plus. And then I say for each one, name me the electron geometry, the molecular geometry, all of the bond angles, the hybridization of the central atoms. And then I want to know what orbitals is each atom used to bond with. And I want to know if it has a dipole. So quite a lot of information here, right? I wanted to include everything that I could possibly ask you about a molecule. Okay. So our first step is to pick one and draw it. So let's start with icl4 for no other reason than I listed it first. So we have icl4. First step to make sure that we don't do anything wrong later on is to count up all of our electrons. So we know we have seven from the i and seven from each of the chlorines. And then we have one extra one because of the electron, right? Can't forget about that. So we have 36 electrons. Now let's just go ahead and set up our system for the other one right here, too. So for icl4 plus, how many electrons are we going to have? 34. Good. 35 from the elements and then subtract one for the plus charge. Okay. Now let's draw these things out. So our central atom will be our i. And then we can try to draw out our chlorines. Now something I want to mention as, because this is a good place to mention it, you'll notice when I draw my Lewis structures, I tend to draw them in a way that looks like the shape that it would be in 3D. That's because in general when I'm drawing Lewis structures, I know what I'm drawing. I know ahead of time what I'm drawing. That's not to say that Lewis structures have to represent geometry. Lewis structures don't represent geometry. So when you're drawing these, you don't have to feel like you have to get the geometry right the first time. I mean if you get in the habit of drawing them a lot, you'll sort of just start to do it anyways and it is helpful at times. But notice when you did your sapling homework, sapling in purpose, draw it sort of in a way that didn't look anything like the actual molecule would look to make sure that you knew what you were doing as far as finding geometry. And that's fine. If I want to know the geometry, I'll ask you the geometry where you don't have to draw it that way. I say this especially so that you don't waste too much time on the trying to draw things if you need to. So we have this. Now how did I know to add in the two sets of lone pairs on the iodine? Well if you go through and you just add in four chlorines and you count up all of the electrons, we're missing four electrons, right? And so since we're missing four electrons, if we just count up the chlorines and the bonds, we know that those have to go on the iodine. I can't put more of them on the chlorine. OK, so that's our Lewis structure. So now we have to go through and answer all the questions based on it. So first of all, what's going to be the electron geometry? Electron geometry. So remember for electron geometry, you're counting in the lone pairs. You can see the lone pairs. Octahedral. Octahedral, right? How did you know that? What's the steric number? Six. So if you have a steric number of six, that's octahedral. What would you do if you didn't know that? Yeah, you needed to picture it. If you can't picture it, again, I'm fine with you bringing toothpicks. I'd rather not go to the full extent of candy and things of that sort in modeling kits just because it's too easy for the people around you to see. But if you want to bring some toothpicks and just kind of fiddle with them while you're doing this, I'm fine with that. So if you did that, the easiest way to get all of these to face each other or to fit into a shape would be to have them be like that, right? That gets you six. OK. So now comes the molecular geometry. And of course, I don't ever really mind you interrupting me with questions, but this is a place that's especially important if you have questions to interrupt me. So you can feel free to do it even more than normal. OK. So now we have our octahedral electron geometry, and we need to take out two of those and switch them out and put lone pairs there. So are we going to put two that are right next to each other as lone pairs or two that are across from each other? Across from each other. So we end up with this, right? So what shape is that? Square planar. Oh, I forgot one very important thing with my Lewis structure that would be a point off. What did I forget? Formal charge. There's a couple of different ways you can do formal charge. You could actually write it in where it belongs, like on the actual atom it belongs in. Or if I don't ask you to give me the formal charges of the atoms, you can just put it over top of the whole thing. That's fine. But you actually have to put it on the atom unless I ask you to do so. OK. So I'm just going to put it here. All right. Now our bond angles. So we've taken an octahedral geometry. We put some lone pairs on it. We have this geometry now. What is our bond angles? 90. Is it 90 or is it less than 90? Good. It's 90. Right, because even though you have the lone pairs on it, you have this lone pair pushing down this way and this lone pair pushing up this way. But they're pushing on each other equally. And so all the bond angles still end up 90. OK. Hybridization. So what was our steric number again? Six. So how many hybrid orbitals do we need? Six. So it's going to be sp3d2. OK. Yeah. So on this one, there's times where that's more OK than others. And on this one, it would be a little weird. I guess if you're talking from here to here, if you put 180, I wouldn't mark it wrong. On this one, you definitely don't have to put it though. OK. So now, next thing we have to do. Oh, bonding orbitals. OK. Oh, yeah. If you asked about hybridization before, would that be? If I asked you about the hybridization of Cl, there's two answers that I would be OK with. The one that I would prefer is to say unhybridized. The other one that you can use is you can use the fact that if you counted these lone pairs, you'd have one, two, three, and then the bond, so four, and you could base it off that. You could say sp3. But in general, we don't talk about the different ligands, the outside molecules, things that are only bonded to one atom. We just don't talk about those as being hybridized at all. In general, I just steer away from asking about it. What I will ask you though is this next thing. Yeah? Did you say that it bonds with the original atom? Yeah. That's what we're going to do right now, actually. So what I will ask you is exactly what she asked, is what is it bond with though? So the hybridization, I'll probably just ask you of central atoms, things that are bonded to more than one element. But what about what that bonds with? Yeah? If I asked you about hybridization, no. P is in a sort of hybridization. But for this next thing, yes. So what is a bond with? Would it bond with something? Well, yeah, let's do this next question now. So what does it bond with? So we need to break this down into two things. Iodine is going to bond with something and chlorine is going to bond with something. OK. So we have to figure out what these are now. So let's start with the chlorine. And I'm going to just do this on some scrap paper so I can write it out with a little bit more room. So let's start with the chlorine. What is the, if we were to draw the electron energy diagram of chlorine, what would it look like? Well, we have seven electrons, so it would look like that. So what orbital is empty, or not empty, but has a spot to bond with another something? P. So what is chlorine going to bond with? It'll bond with its P orbital, because that's where it has an empty spot for it. OK. So the chlorine is going to bond with its P orbital. Now let's talk about what the iodine is going to bond with. So why did we make hybrid orbitals? Why do we have this thing called hybridization that we talk about? Well, we have it because we need a way to spread out all of the bonds, or all of the atoms, I guess I should say, spread out all of the atoms and all of the lone pairs. So we picked our hybridization. We said it was sp3d2 based on the fact that we needed six things, six spots for all of these to bond with or put lone pairs in. So what is it going to use to bond with? First of all, do we really have any s or p orbitals left anymore? No, we've turned them into these hybrid orbitals. So what are we going to bond with? Our hybrid orbitals, right? We've made them to form those bonds and to put those lone pairs in. So iodine is going to go ahead and bond with its sp3d2 orbitals. I'm going to have to ditch the question there. So luckily, you all have the worksheet. OK, so we have one other thing that I could ask you about then here. Where are the lone pairs at? I should say the iodine lone pairs. Where are those? What type of orbitals are those in? Well, did that go into our sort of calculation about which hybrid orbitals we're going to make? Yeah, we want them spread out, right? We want them spread out just like the bonds were. So what orbitals would those be in? The hybrid orbitals, good. So those are also going to be in sp3d2 orbitals. So your lone pairs and each of your sigma bonds are formed by or are using your hybrid orbitals, in this case sp3d2. Yeah? So you have a separate plot value in combination of sp2, what do you use sp2 to use in the class p? So what he just asked, I want to say the question since a lot of you probably didn't hear it, is if you had something like an sp2 hybrid orbital, so you'd have the three sp2 orbitals and you would have a p orbital. When do you use the sp3 orbitals and when do you use the leftover pi orbitals? So we have two different types of bonds that we can make. We can make sigma bonds and we can make pi bonds. All of your sigma bonds are formed using your hybrid orbitals and then your pi bonds are formed using your leftover p orbitals. Okay, questions on this? Okay, so let's, oh yeah. Well, the thing is, is at that point you'd actually get into, if you had two leftover p orbitals, you would, both of those are going to form pi bonds. So it's either going to have two double bonds, like in the case of a carbon, it would have two double bonds or it would have a triple bond. Yes? Yeah, so he had, again, just repeating this, when do we use the hybrid orbitals and when do we use leftover p orbitals? We use hybrid orbitals in order to form our sigma bonds and to put our lone pairs in. And we use p orbitals to form our pi bonds. So either our double bonds, the second bonds and our double bonds, or the second and third bonds and our double, or triple bonds. Yes? So why does the central atom have to be hybrid? Say that a little bit louder. The central atom? You said that has to be hybrid orbitals. Those are hybrid orbitals because we need to get the right geometry so that we can spread everything out. And there's a couple of different ways of thinking about it. We can do hybrid orbitals or we can do MO theory, but MO theory for these would be quite complicated so we're sticking with hybrid orbitals. Okay, let's race through now and do ICL4 and it's all going to be very similar so try to beat me to it as we go along. Oh yeah, dipole, thank you. So does it have a dipole? So we have this shape. So are there any polar bonds? Very slightly, not by much, but a little bit. Does it have a dipole overall? No, they cancel, right? This one going this direction and this one going this direction cancel and this one this way and this one this one so they all cancel. So no dipole. Okay, next one. So we do something very similar. But now we've dropped two electrons. We've dropped a lone pair. So it's going to be pretty much the same, just minus one set of electrons. Okay so we have this. Now let's go through and do everything in pretty much the same order. In order to try to save some room I'm going to write up here a little bit though. So we have electron geometry and molecular geometry. So what do we think for the electron geometry? What's our steric number? Five, right? Remember you're counting atoms that are bonded to it and lone pairs, so five. So that gives you what shape? Trigonal bipyramidal, right? That's the one where you have sort of, I'll just make it real fast. This is the one where we have this shape. Now so we have sort of a triangle shape here. We can see that we have a pyramid on the top and a pyramid on the bottom, so trigonal bipyramidal. Now one of those is taken off and it's not an atom anymore, it's a lone pair, so it looks like this. So what shape is this molecular geometry? So remember this one is helpful to sort of tip it, ah, what did I do wrong? You took it off the wrong side, right? Get off that one. I went for the easy one for me to hold, not the correct one. There you go. Now what shape is it? Ah, right, tip it on its side, it's C-sa. So why did I take off this one? Why do you have to take this one off? Yeah, so what's the spawn angle? 120. 120 and what's the spawn angle? 90. So we want to give the lone pairs, these huge lone pairs taking up tons of room. Are we going to want to give them 120 to work with or 90 to work with? 120. So we have to take them off these. And what's that called? What's the space called? Equatorial, right? Think of it like a big globe. You have an equator going around the outside and then you have the axes going here. So this and this are axial. So then what shape is C-sa? Tip it on its side. Okay. So we have this. Okay, so now we have to work with bond angles which we sort of touched on. What were the bond angles again? So that's for the trigonal bipyramidal. What about for the C-sa? Less than 90 and less than 120. And again, 180 would work if you'd like. I don't necessarily care if you have it there or not, but think the 180 from the one all the way to the other. From the two axials. Yeah. Well, on the Lewis structure you can't see. You have to actually think about it in terms of a three-dimensional structure. So the equatorial ones are the ones that go along the 120 degree and then the axial are the ones that are the 90 degree. Drawing here. Oh, well axial and equatorial are just referring to individual atoms. So you can't really talk about it because the 90 is between an axial and an equatorial and then the 120 is between two equatorials. Yes. Do we have to put both? This? Yes. You have to put both. And putting 180 is a good idea. I'll add that there too. Okay. Hybridization. What is our steric number? Steric number is five. So what is our hybridization? SP3D. Okay. Next part. What do we use to bond with? So what is our CL going to bond with? Is it any different than over here? No, exact same situation. And what is it going to use to bond with? It's p orbital. It's exact same reason, right? Let's throw this back up for a minute. Exact same reason. That's what it has a spot to put it in. You don't need to hybridize those outside atoms. So we'll just leave them like that. Okay. Now, what is the iodine going to bond with? It only has sigma bonds. So we form sigma bonds with what kind of orbitals? Hybrid orbitals. So in this case, that is SP3D orbitals. And where are the lone pairs at? What orbital are those in? Good. In the same, right? I guess I should just say lone pair, not pairs. Just cancel out the s. Okay. So now, we have our C-sauce shape. Does it have a dipole? So first of all, first question we have to ask, are there polar bonds? Sure. Not super polar, but polar. Now are they going to cancel each other out? No. Some of it will, right? This one and this one will cancel a little bit, or most of the ways. Remember, they're actually all tilted a little bit. Most of this will cancel. And the side to side, or I guess in and out for you guys to compare a component of this will cancel. So if I hold this like this, which way did the dipole be? Yeah, to the side. It would be going in the direction like directly across the building. Okay. So let's do that. Let's write that down real quick. So yes, dipole. Oh, like a he- So on a Lewis structure, I wouldn't make you draw them like I have been. I could make you draw the delta positive and delta minus. The likelihood of me doing it on something like this is pretty small, just because you kind of have to draw it in between the CLs. And if you didn't draw your Lewis structure exactly right, the way, if I'm going to have you draw a dipole, I'll probably draw you the molecule. I'll draw you the molecule and have you draw the dipole in. So how do we decide which one has the negative charge? Which one's more electronegative? So how do we decide which one's more electronegative? Where it is on the periodic table. So which one's higher on the periodic table? iodine or chlorine? Chlorine. So which one's more electronegative? Chlorine. So the way I have this Lewis structure drawn, this is where the arrow designation might be a little bit easier to use. You draw like that. Yeah? I'm going to point on here, since I'm not sure which one's you're pointing to. But yes, this has a dipole, so yes, it's polar. If something has a dipole, it is polar. If something doesn't have a dipole, it isn't polar. Yes? Yeah? So the resonance doesn't really change much of anything when it comes to the charges. All it really does is it changes exactly how the charges are distributed. So you need to take, so think of like the sapling problems where you do something like NO3 and you had to draw all three, or NO3 I think it was, and you had to draw all three of the resonance structures. All that really changes there is that the charge was evenly distributed, right? Because it wasn't just belonging to one of the oxygens, it was belonging to all three equally, and so all of the dipoles can't, or all of the negative charges cancelled out. So what the resonance structures basically keep you from doing is having, it changes the way your formal charge functions into it. It's almost easier for like those ones that you've been doing on sapling and that I did in class, to just kind of think of, ignore where the charges actually are for that. Like ignore those formal charges and kind of think about it more, but other than that it doesn't actually change much. Okay. Questions? Yes? Yeah, you need more than 180. And you don't need to put that one there if you don't want to put it there. The 180 in this one is really sort of optional in my opinion. If I, I mean if I drew it, like if I drew you out of Lewis structure that actually showed the geometry and I drew this and I asked you to do it, then you'd have to be able to explain that it's more than 180, but if I just ask you for bond angles, I don't necessarily care about the 180 being there or not. Yeah? In the second part, if iodine has a double bond, would you say it's bond or has a double bond? Let's not talk about if this one, for this one just because if iodine has a double bond here, there's some other things going on, lots of other things going on, so we can't look at it with that molecule. But just know that you don't factor in that pi bond as a bond when you're counting hybridization. That's what you're really asking, right? Yeah, that you're counting, you don't count bonds to figure hybridization. If we can quiet the chatter just a little bit, I know you're talking about chem things, but it's just too loud. So if you're counting, you're counting numbers of atoms that are bonded to it and then lone pair, so you're not actually counting individual bonds when you figure hybridization. Yeah? Would you give us a problem where like say it would just be a high C on the board so that there would be like one unit process in a lone pair? Oh, you mean like a radical? Yeah, yeah, you mean a radical. You do. Where there's only, instead of a lone pair, you have one. Yeah, that's possible. And if you get something like that where you have, if you only have one electron and you don't have a pair, this shows up when you have an odd number of electrons. That's why this has a negative charge or a positive charge. You treat an electron that's by itself like that as a radical basically like a lone pair. And I say basically with one caveat that I'll talk about later on in the review packet. But for the sake of the hybridization and things of that, you count it just like a lone pair. Yeah, you count it exactly like a lone pair for the sake of naming the hybridization and me asking where it's at. There's just one little thing that changes that I'll talk about later on. Yes? Right, you can't. You wouldn't be able to because you'd have an odd number. So, okay. Let's move on to the next one. So I say draw the MO diagram for NO. Assume that the sigma is lower than the pi. Be sure to label all of your orbitals including your subscripts. Fill in any electrons that you need. Tell me what the bond order is and tell me if it's paramagnetic or diamagnetic. And then I added a question to it from when I sent it to you. So make sure you write that down. I also say would NO minus have a higher or a lower bond order? So let's draw this out. So this is one of those that I said I could ask. Yeah? For NO, is the sigma bond between the point of eye bond order and is one of the exceptions? No. So whenever you're dealing with a heteronuclear, it's what I tell you it is. So that's all you have to worry about for the heteronuclear ones. When you get into the heteronuclear things, there's a lot of weird things going on which is actually the sort of summary of the one slide that you probably none of you really liked all that much where I had the C1 psi 1 plus C2 psi 2. The summary of that slide in one sentence is when you have heteronuclear diatomics, a lot of things are going on that you don't really have to worry about too much. The way that they add, they add based on a weighted system that I can't expect you to know. So I'll tell you. This would be how I would word the question. And I would just tell you that this is lower than the pi. OK? All right. So let's start the drawing of this. So we know we're going to have some s orbitals. I'm fine with you just drawing the valence shell. I don't need you to draw in the 1s's. You should know they're there, know they exist, but I don't care if you draw them. That's kind of our starting point. So we can fill in for nitrogen. How many electrons do we have? Valence electrons? Five. And how many for oxygen? Six. OK. And now we can draw this. We can say, well, sigma is lower than pi. So we have sigma and then pi like this. We're pretending these are on exactly even levels because my straight line wasn't totally straight. OK. And now let's label them so we don't forget. So this is sigma 2s. Sigma 2p. Pi 2p. Pi 2p star. And sigma 2p star. Do we care for when you're drawing these that we know where the labels refer to? Make sure they're closer to where they belong. You can also put them off to the side. A lot of people write the labels off to the side if they run out of room. That's fine. I tend to forget to leave myself enough room when I'm drawing the outline so I write them below. But you're welcome to write them here and here and here and here and here and here. Don't forget your stars. Don't forget your subscripts. Now how many electrons total do we have here? We have 11 so we can go fill in. If we have that. OK. So we've labeled our atomic orbitals. We've labeled our molecular orbitals. We've included our subscripts. We've filled in all our electrons. Now I want to know what the bond order is. How do we know what the bond order is? We count up all our electrons. How many are in our bonding orbitals here? Two, four, six, eight. How many are in our anti-bonding? Two, three. So our bond order is 2.5. So that cover, oh no that doesn't cover everything. So is this paramagnetic or diamagnetic? Why? Careful how you say that. It has unpaired electrons, right? Has unpaired electrons. So paramagnetic is unpaired electrons. OK. Now I say does NO minus have a higher or lower bond order? So think about it a sec. Don't say it yet. Give people some time to think about it. Now there's a couple ways to do this. One is to redraw the entire MO diagram. I see that's the favorite way from all of your groans. Is that what you want to do? Is that what you want to do on a 50 minute exam? And I know my exams are a little on the long side, so a long 50 minute exam? No. Is that what I want you to do? No. What do I want you to do? OK, quiet a little bit. I want you to look and see where is that electron going to be either added to in this case, because it's a negative charge, or taken away from in a positive case. So where is this electron going to be added to if we add it here? Right here, right? So is that an anti-bonding or a bonding? It's an anti-bonding. So is that going to take away from the bond order or add to the bond order? It's going to take away from the bond order. So is it going to be smaller or bigger? Yeah. Now just for fun, let's say I did ask you to draw me the MO diagram of the NO minus or NO plus or something like that. Where would I put this negative charge as far as the nitrogen and the oxygen goes? Where should I put it? Good, the oxygen. Why? It's more electronegative. So you put the electron, you put the negative charge on the oxygen. And where would you take an electron away from? The nitrogen. Which remember that really doesn't play a role in the MO part of the diagram. It doesn't play much of a role in here, because you just know you're adding or taking one away. But it does matter over here. OK. Any questions, hand raised that I missed? Yeah. Oh, thank you. Don't forget the E on the side. That was on the checklist of things to make sure you draw. And that's like the easiest one to forget as you just saw. And the better you get at them, the easier that E is to forget. Yes. So stability has to do with a lot more than just bond order. Because you're also dealing with exact electron configurations and things of that sort. So for this one, we wouldn't really get into it too much. OK. Anything else? OK. Next one. And again, we're not feeling too rushed here, because whatever I don't cover here, I'll cover in my review session on Thursday. OK. How does constructive and destructive interference relate to molecular orbital diagrams? So this would be a sort of essay type question, right? I think on your last one, this is kind of similar to, I don't know, I asked you something similar to how you'd answer this one. OK. So how does this have anything to do with the MO diagrams? So let me give you some prompts here. What are bonding orbitals caused by? So when we take, when we take and we make MO molecular orbitals, we're taking two atomic orbitals, which are waves, right? Those are just waves, and we're overlapping them. And we can overlap them two different ways, right? We can overlap them constructively, and we can overlap them destructively. So how would you overlap them constructively? Would that form a bonding orbital or an anti-bonding orbital? Bonding. How would you overlap them destructively? What would you form? Destructive. So of bonding orbitals and anti-bonding orbitals, which one has the star? Anti-bonding. Which one do you fill first? Bonding. So that's all I would expect for that one. I'm not going to write it in essay form, but. OK. Questions on that one? No? OK. I thought it was worthwhile to do some more of these. We've just covered it, so it's new, but more examples of new things are, or I think, a very good thing. OK. Rank the following in order of boiling point. So when we rank by order of boiling point, what else are we really ranking by? Intermolecular forces, right? The higher the intermolecular force, what happens to the boiling point? It goes up. If you increase the intermolecular forces, you increase the boiling point. So if we look at this first one, we just have that. What's that called? That type of structure? Aligned structure, right? What's another way of drawing it? You can draw it like this. Make sure you know how to draw both ways. If you draw a Lewis structure, do you have to include the lone pairs? Yes. If you draw a line structure, do you include the lone pairs? No. If I draw you a line structure and ask you to tell me things about it, am I going to draw on the lone pairs? Where do you think a common mistake comes in there? People forget the lone pairs exist, right? Anything you're doing that has to do with lone pairs, which would be like what? What has to do with lone pairs? What do you figure out that you have to count numbers of lone pairs for? Steric number, which is your geometry, your hybridization, all those sorts of things. If you forget the lone pair, are you going to get those right? No. So on line structures, it's not a bad idea to go in and draw on your lone pairs. Now we have that. So which one has the, I just say rank and order boiling point. So let's say which one would have the higher boiling point, this one or that one? So what do we have to look at? We have to look at our intermolecular forces. What sort of intermolecular forces does this one have? Di-pole and LDFs. What about this one? All of them, right? Hydrogen bonding, well I shouldn't say all of them because there's some that we don't really include, but the three main ones that we check for. Hydrogen bonding, dipole and LDFs. Now is there any weird something going on with size here? Are they like, is one drastically bigger than the other? Say about the same. About the same. Sure, maybe not exactly, but close enough. The size thing, number of electron things are sort of an approximate number. So which one's going to be bigger? The one with hydrogen bonding or the one without? So this is going to be your largest or highest, I guess I should say. All right, now let's look at this one. Yeah? If I ask you to rank them, all you have to do is write the names. Like, no, if I ask you to rank them, if I say rank them, all you have to do, I'll tell you which way to rank them, I'll say low to high or high to low. You just have to write down the name of the compound and let it, that's it. If I ask you which one's higher, you just write down the name and you're done. You'd be a little, you might, and this one, you'd be a little hard for us to be able to do it without drawing it though. So is the hydrogen bond the resource to be the OH? The hydrogen bond on this one, it's made possible because of this OH. So I'm going to use some scrap paper here a minute. So remember, that's not where the hydrogen bond is forming. The hydrogen bond has to form between two different molecules. So you would make another molecule here. And this would be your hydrogen bond. So very good point. Remember your hydrogen bond, if I draw you one molecule and I say point to the hydrogen bond, you can't do it. You would have to draw another molecule and then point to that hydrogen bond between here and here. Okay. Oh, we are over time. Okay, now a big deal. We will finish this on Thursday. So come to my meeting on Thursday. Don't forget, as you're packing up, that Miranda has a review session tonight as well. I think hers is question and answer. So come with lots of questions to hers. If we run out of time or if we finish the worksheet in mind, we'll switch to question and answer as well.
 Okay, so today.  Okay, so we're going to finish up some hydrogen bonding things that I haven't quite finished  up.  Well, maybe we will.  Okay.  And then, so that won't take too much longer.  And then we'll do a review lecture.  I'm not sure if that whole packet that I gave you, if we'll get through all of it today  or not.  I actually, leaning forward, we probably won't.  So whatever I don't finish in class today, I don't want us to feel rushed.  I'm just going to do that Thursday night.  So we'll have plenty of time over, spread over everything to at least finish it.  So don't worry too much about that.  Okay, so the rest of the hydrogen bonding that we need to talk about is why this is  so important in biology.  This is important not just in general, but also because of the demographics of this class  tend to be pretty highly in the biological or variety of earth system science type things.  Okay, so the first one that I want to talk about is in DNA.  Now this is very, very important.  Now you probably have a little bit of background either from high school or from your biology  classes here on how DNA kind of looks.  So you get this double helix structure.  Okay, so we get this double helix structure and we have all of these bonded and those  are bonded covalently or some other form.  Covalently.  Now the thing that holds these two together is what comes important in this chapter.  What are all of these bonds?  They're hydrogen bonds.  So they're not actually covalent bonds that are holding them together.  And that's really important because what do we need DNA to be able to do in order to live?  It's got to be able to replicate in one other thing.  Make mRNA, right?  You've got to be able to replicate and make RNA.  If you can't do those two things, we're going to have a little bit of trouble making new  cells and making proteins and doing all the things that we need our DNA to tell our bodies  to do.  So in order to replicate it, the way that it works is you have to be able to split a  word to make RNA.  You have to be able to split the DNA strands.  If they were covalently bonded, that would be a lot harder.  You could still do it, but it would be a lot harder.  But having these hydrogen bonds here strikes a good balance between they're fairly solid,  they're strong, but they're not so strong that you can't undo them relatively easily.  So we have proteins in our bodies that come through and can unzip them and rezip them back  up as we go along.  If it was covalently bonded, that would be tougher.  If it was held together by some other force, that might be a lot harder too.  The hydrogen bonds are kind of this balance.  Now going a little bit more in depth with this, again, because of the demographics of  this class, notice between the adenine and the thiamine, how many hydrogen bonds do we  have?  Two.  What about between the guanine and the cytosine?  Three.  Three.  Which ones are going to be stronger?  Three.  Yeah.  So as you go along in your biology classes, you'll start at some point, probably your  junior year, so you'll start calculating melting points for DNA strands and talking about  how much, you know, when you start, especially in your molecular biology methods and lab  courses, and you'll be calculating these, and you'll be calculating melting temperatures,  how much you have to pull, how much energy or heat you have to put in to pull them apart  and, you know, before you can let them come back together.  And it's going to make a difference how much A's and T's you have and how many C's and  G's you have.  And that's why, because C's and G's bond, or they bond at three hydrogen bonds instead  of two.  So they stay together easier.  So if you want to break them apart, it takes more energy.  So that's, you know, when you're doing lab work and things like that in some of your  higher level bioclasses, that'll come into play.  Okay.  The next place that I want to talk about, that it plays very important roles, is in proteins.  So this is a picture of a protein, is one I just pulled off from this website down here.  And all of these little yellow dotted lines are drawn in using hydra, or those are all  your hydrogen bonds.  Now, it's not all of them, of course, because there's a lot of them that are forming all  your different structures, your secondary structures.  But hydrogen bonds are largely responsible for your secondary structures.  They play all of these important roles and tertiary structures that you can see.  And so they're very, very important in protein formation.  You need these proteins to stay in these very tight, there's not a lot of leeway between  a lot of these structures on how much they can move.  And hydrogen bonds are a large portion responsible to their structure.  Without them, you wouldn't have as well of a defined structure as you do.  Now in hydrogen bonding, just so you realize here, this is an amino acid.  Okay?  So this amino acid is what makes up all your proteins.  So it's just chains and chains and chains of these, and big long formations.  And so what happens is you can get your oxygens and your nitrogens to hydrogen bond together  with your hydrogens that are on them.  And that's what forms your structure is all these groups here.  And we won't go into detail about the exact structures of that, I'll leave that for your  higher level classes.  But just sort of know how this works, know it's important.  Be able to sort of explain to me why these sorts of things are important.  The next one then to talk about real quickly is cellulose.  So where do you know cellulose from?  What does that make up?  Your plants, right?  That's sort of the structural component of your plants.  And the way that cellulose works is it's these long chains of glucose.  So you get these really, really long chains of glucose.  Now on their own, a single long chain of glucose isn't going to be very sturdy.  They're not going to be able to hold up a redwood with just individual chains of glucose.  But then what happens is all of these chains, they're able to line up and they're able  to form hydrogen bonds with each other.  Look at all these OH groups here and these OH groups.  Cellulose can all form hydrogen bonds in between chains and so you get these big long stacks  of these chains and you form these fibrils and then that's what is able to make up all  the structural components of your walls or your cell walls and things.  So that is all I have for you on the rest of that chapter.  So we're done with new midterm material.  This was one thing I wanted.  I meant to do this before but this is one thing I wanted to clarify from last time.  There's a bit of a mistake.  This is the correct ordering.  So this is a problem that we did last class right kind of near the end.  But there's a little bit of a mistake in the fact that I can't actually expect you to be  able to rank all four of these because there's a big exception here that comes from the shape  of the molecule.  We didn't talk about that and we're not going to talk about that.  So in a situation like this, this question, if it were to show up on an exam, would show  up in this form.  I could ask you to rank those three or I could ask you to rank those three.  I would not ask you to rank all four.  So I sent out an email to this extent but I know not everybody reads their emails.  You should please read your emails.  So just making sure that you guys realize that.  So I would ask you to rank either this grouping or this grouping, not all four of them.  So now under review material.  All right.  So this is, since you have done a large portion of your studying for the exam, it's not a  bad idea as I'm lecturing here to try to beat me to the problems, right?  You don't need to wait for me to write down the answer to write down the answer.  Try to beat me to it.  And then if I write down something different, you can say, wait, wait, I did something wrong.  Figure out what you did wrong and clarify your mistake.  Okay.  So I have this drawing, I cl4 minus and I cl4 plus.  And then I say for each one, name me the electron geometry, the molecular geometry, all of the  bond angles, the hybridization of the central atoms.  And then I want to know what orbitals is each atom used to bond with.  And I want to know if it has a dipole.  So quite a lot of information here, right?  I wanted to include everything that I could possibly ask you about a molecule.  Okay.  So our first step is to pick one and draw it.  So let's start with icl4 for no other reason than I listed it first.  So we have icl4.  First step to make sure that we don't do anything wrong later on is to count up all  of our electrons.  So we know we have seven from the i and seven from each of the chlorines.  And then we have one extra one because of the electron, right?  Can't forget about that.  So we have 36 electrons.  Now let's just go ahead and set up our system for the other one right here, too.  So for icl4 plus, how many electrons are we going to have?  34.  Good.  35 from the elements and then subtract one for the plus charge.  Okay.  Now let's draw these things out.  So our central atom will be our i.  And then we can try to draw out our chlorines.  Now something I want to mention as, because this is a good place to mention it, you'll  notice when I draw my Lewis structures, I tend to draw them in a way that looks like  the shape that it would be in 3D.  That's because in general when I'm drawing Lewis structures, I know what I'm drawing.  I know ahead of time what I'm drawing.  That's not to say that Lewis structures have to represent geometry.  Lewis structures don't represent geometry.  So when you're drawing these, you don't have to feel like you have to get the geometry  right the first time.  I mean if you get in the habit of drawing them a lot, you'll sort of just start to do  it anyways and it is helpful at times.  But notice when you did your sapling homework, sapling in purpose, draw it sort of in a  way that didn't look anything like the actual molecule would look to make sure that you  knew what you were doing as far as finding geometry.  And that's fine.  If I want to know the geometry, I'll ask you the geometry where you don't have to draw  it that way.  I say this especially so that you don't waste too much time on the trying to draw things  if you need to.  So we have this.  Now how did I know to add in the two sets of lone pairs on the iodine?  Well if you go through and you just add in four chlorines and you count up all of the  electrons, we're missing four electrons, right?  And so since we're missing four electrons, if we just count up the chlorines and the  bonds, we know that those have to go on the iodine.  I can't put more of them on the chlorine.  OK, so that's our Lewis structure.  So now we have to go through and answer all the questions based on it.  So first of all, what's going to be the electron geometry?  Electron geometry.  So remember for electron geometry, you're counting in the lone pairs.  You can see the lone pairs.  Octahedral.  Octahedral, right?  How did you know that?  What's the steric number?  Six.  So if you have a steric number of six, that's octahedral.  What would you do if you didn't know that?  Yeah, you needed to picture it.  If you can't picture it, again, I'm fine with you bringing toothpicks.  I'd rather not go to the full extent of candy and things of that sort in modeling kits  just because it's too easy for the people around you to see.  But if you want to bring some toothpicks and just kind of fiddle with them while you're  doing this, I'm fine with that.  So if you did that, the easiest way to get all of these to face each other or to fit  into a shape would be to have them be like that, right?  That gets you six.  OK.  So now comes the molecular geometry.  And of course, I don't ever really mind you interrupting me with questions, but this is  a place that's especially important if you have questions to interrupt me.  So you can feel free to do it even more than normal.  OK.  So now we have our octahedral electron geometry, and we need to take out two of those and switch  them out and put lone pairs there.  So are we going to put two that are right next to each other as lone pairs or two that  are across from each other?  Across from each other.  So we end up with this, right?  So what shape is that?  Square planar.  Oh, I forgot one very important thing with my Lewis structure that would be a point  off.  What did I forget?  Formal charge.  There's a couple of different ways you can do formal charge.  You could actually write it in where it belongs, like on the actual atom it belongs in.  Or if I don't ask you to give me the formal charges of the atoms, you can just put it  over top of the whole thing.  That's fine.  But you actually have to put it on the atom unless I ask you to do so.  OK.  So I'm just going to put it here.  All right.  Now our bond angles.  So we've taken an octahedral geometry.  We put some lone pairs on it.  We have this geometry now.  What is our bond angles?  90.  Is it 90 or is it less than 90?  Good.  It's 90.  Right, because even though you have the lone pairs on it, you have this lone pair pushing  down this way and this lone pair pushing up this way.  But they're pushing on each other equally.  And so all the bond angles still end up 90.  OK.  Hybridization.  So what was our steric number again?  Six.  So how many hybrid orbitals do we need?  Six.  So it's going to be sp3d2.  OK.  Yeah.  So on this one, there's times where that's more OK than others.  And on this one, it would be a little weird.  I guess if you're talking from here to here, if you put 180, I wouldn't mark it wrong.  On this one, you definitely don't have to put it though.  OK.  So now, next thing we have to do.  Oh, bonding orbitals.  OK.  Oh, yeah.  If you asked about hybridization before, would that be?  If I asked you about the hybridization of Cl, there's two answers that I would be OK  with.  The one that I would prefer is to say unhybridized.  The other one that you can use is you can use the fact that if you counted these lone  pairs, you'd have one, two, three, and then the bond, so four, and you could base it off  that.  You could say sp3.  But in general, we don't talk about the different ligands, the outside molecules, things that  are only bonded to one atom.  We just don't talk about those as being hybridized at all.  In general, I just steer away from asking about it.  What I will ask you though is this next thing.  Yeah?  Did you say that it bonds with the original atom?  Yeah.  That's what we're going to do right now, actually.  So what I will ask you is exactly what she asked, is what is it bond with though?  So the hybridization, I'll probably just ask you of central atoms, things that are bonded  to more than one element.  But what about what that bonds with?  Yeah?  If I asked you about hybridization, no.  P is in a sort of hybridization.  But for this next thing, yes.  So what is a bond with?  Would it bond with something?  Well, yeah, let's do this next question now.  So what does it bond with?  So we need to break this down into two things.  Iodine is going to bond with something and chlorine is going to bond with something.  OK.  So we have to figure out what these are now.  So let's start with the chlorine.  And I'm going to just do this on some scrap paper so I can write it out with a little  bit more room.  So let's start with the chlorine.  What is the, if we were to draw the electron energy diagram of chlorine, what would it  look like?  Well, we have seven electrons, so it would look like that.  So what orbital is empty, or not empty, but has a spot to bond with another something?  P. So what is chlorine going to bond with?  It'll bond with its P orbital, because that's where it has an empty spot for it.  OK.  So the chlorine is going to bond with its P orbital.  Now let's talk about what the iodine is going to bond with.  So why did we make hybrid orbitals?  Why do we have this thing called hybridization that we talk about?  Well, we have it because we need a way to spread out all of the bonds, or all of the  atoms, I guess I should say, spread out all of the atoms and all of the lone pairs.  So we picked our hybridization.  We said it was sp3d2 based on the fact that we needed six things, six spots for all of  these to bond with or put lone pairs in.  So what is it going to use to bond with?  First of all, do we really have any s or p orbitals left anymore?  No, we've turned them into these hybrid orbitals.  So what are we going to bond with?  Our hybrid orbitals, right?  We've made them to form those bonds and to put those lone pairs in.  So iodine is going to go ahead and bond with its sp3d2 orbitals.  I'm going to have to ditch the question there.  So luckily, you all have the worksheet.  OK, so we have one other thing that I could ask you about then here.  Where are the lone pairs at?  I should say the iodine lone pairs.  Where are those?  What type of orbitals are those in?  Well, did that go into our sort of calculation about which hybrid orbitals we're going to  make?  Yeah, we want them spread out, right?  We want them spread out just like the bonds were.  So what orbitals would those be in?  The hybrid orbitals, good.  So those are also going to be in sp3d2 orbitals.  So your lone pairs and each of your sigma bonds are formed by or are using your hybrid  orbitals, in this case sp3d2.  Yeah?  So you have a separate plot value in combination of sp2, what do you use sp2 to use in the  class p?  So what he just asked, I want to say the question since a lot of you probably didn't  hear it, is if you had something like an sp2 hybrid orbital, so you'd have the three  sp2 orbitals and you would have a p orbital.  When do you use the sp3 orbitals and when do you use the leftover pi orbitals?  So we have two different types of bonds that we can make.  We can make sigma bonds and we can make pi bonds.  All of your sigma bonds are formed using your hybrid orbitals and then your pi bonds are  formed using your leftover p orbitals.  Okay, questions on this?  Okay, so let's, oh yeah.  Well, the thing is, is at that point you'd actually get into, if you had two leftover  p orbitals, you would, both of those are going to form pi bonds.  So it's either going to have two double bonds, like in the case of a carbon, it would have  two double bonds or it would have a triple bond.  Yes?  Yeah, so he had, again, just repeating this, when do we use the hybrid orbitals and when  do we use leftover p orbitals?  We use hybrid orbitals in order to form our sigma bonds and to put our lone pairs in.  And we use p orbitals to form our pi bonds.  So either our double bonds, the second bonds and our double bonds, or the second and third  bonds and our double, or triple bonds.  Yes?  So why does the central atom have to be hybrid?  Say that a little bit louder.  The central atom?  You said that has to be hybrid orbitals.  Those are hybrid orbitals because we need to get the right geometry so that we can spread  everything out.  And there's a couple of different ways of thinking about it.  We can do hybrid orbitals or we can do MO theory, but MO theory for these would be quite  complicated so we're sticking with hybrid orbitals.  Okay, let's race through now and do ICL4 and it's all going to be very similar so try to  beat me to it as we go along.  Oh yeah, dipole, thank you.  So does it have a dipole?  So we have this shape.  So are there any polar bonds?  Very slightly, not by much, but a little bit.  Does it have a dipole overall?  No, they cancel, right?  This one going this direction and this one going this direction cancel and this one this  way and this one this one so they all cancel.  So no dipole.  Okay, next one.  So we do something very similar.  But now we've dropped two electrons.  We've dropped a lone pair.  So it's going to be pretty much the same, just minus one set of electrons.  Okay so we have this.  Now let's go through and do everything in pretty much the same order.  In order to try to save some room I'm going to write up here a little bit though.  So we have electron geometry and molecular geometry.  So what do we think for the electron geometry?  What's our steric number?  Five, right?  Remember you're counting atoms that are bonded to it and lone pairs, so five.  So that gives you what shape?  Trigonal bipyramidal, right?  That's the one where you have sort of, I'll just make it real fast.  This is the one where we have this shape.  Now so we have sort of a triangle shape here.  We can see that we have a pyramid on the top and a pyramid on the bottom, so trigonal bipyramidal.  Now one of those is taken off and it's not an atom anymore, it's a lone pair, so it looks  like this.  So what shape is this molecular geometry?  So remember this one is helpful to sort of tip it, ah, what did I do wrong?  You took it off the wrong side, right?  Get off that one.  I went for the easy one for me to hold, not the correct one.  There you go.  Now what shape is it?  Ah, right, tip it on its side, it's C-sa.  So why did I take off this one?  Why do you have to take this one off?  Yeah, so what's the spawn angle?  120.  120 and what's the spawn angle?  90.  So we want to give the lone pairs, these huge lone pairs taking up tons of room.  Are we going to want to give them 120 to work with or 90 to work with?  120.  So we have to take them off these.  And what's that called?  What's the space called?  Equatorial, right?  Think of it like a big globe.  You have an equator going around the outside and then you have the axes going here.  So this and this are axial.  So then what shape is C-sa?  Tip it on its side.  Okay.  So we have this.  Okay, so now we have to work with bond angles which we sort of touched on.  What were the bond angles again?  So that's for the trigonal bipyramidal.  What about for the C-sa?  Less than 90 and less than 120.  And again, 180 would work if you'd like.  I don't necessarily care if you have it there or not, but think the 180 from the one all  the way to the other.  From the two axials.  Yeah.  Well, on the Lewis structure you can't see.  You have to actually think about it in terms of a three-dimensional structure.  So the equatorial ones are the ones that go along the 120 degree and then the axial are  the ones that are the 90 degree.  Drawing here.  Oh, well axial and equatorial are just referring to individual atoms.  So you can't really talk about it because the 90 is between an axial and an equatorial  and then the 120 is between two equatorials.  Yes.  Do we have to put both?  This?  Yes.  You have to put both.  And putting 180 is a good idea.  I'll add that there too.  Okay.  Hybridization.  What is our steric number?  Steric number is five.  So what is our hybridization?  SP3D.  Okay.  Next part.  What do we use to bond with?  So what is our CL going to bond with?  Is it any different than over here?  No, exact same situation.  And what is it going to use to bond with?  It's p orbital.  It's exact same reason, right?  Let's throw this back up for a minute.  Exact same reason.  That's what it has a spot to put it in.  You don't need to hybridize those outside atoms.  So we'll just leave them like that.  Okay.  Now, what is the iodine going to bond with?  It only has sigma bonds.  So we form sigma bonds with what kind of orbitals?  Hybrid orbitals.  So in this case, that is SP3D orbitals.  And where are the lone pairs at?  What orbital are those in?  Good.  In the same, right?  I guess I should just say lone pair, not pairs.  Just cancel out the s.  Okay.  So now, we have our C-sauce shape.  Does it have a dipole?  So first of all, first question we have to ask, are there polar bonds?  Sure.  Not super polar, but polar.  Now are they going to cancel each other out?  No.  Some of it will, right?  This one and this one will cancel a little bit, or most of the ways.  Remember, they're actually all tilted a little bit.  Most of this will cancel.  And the side to side, or I guess in and out for you guys to compare a component of this  will cancel.  So if I hold this like this, which way did the dipole be?  Yeah, to the side.  It would be going in the direction like directly across the building.  Okay.  So let's do that.  Let's write that down real quick.  So yes, dipole.  Oh, like a he-  So on a Lewis structure, I wouldn't make you draw them like I have been.  I could make you draw the delta positive and delta minus.  The likelihood of me doing it on something like this is pretty small, just because you  kind of have to draw it in between the CLs.  And if you didn't draw your Lewis structure exactly right, the way, if I'm going to have  you draw a dipole, I'll probably draw you the molecule.  I'll draw you the molecule and have you draw the dipole in.  So how do we decide which one has the negative charge?  Which one's more electronegative?  So how do we decide which one's more electronegative?  Where it is on the periodic table.  So which one's higher on the periodic table?  iodine or chlorine?  Chlorine.  So which one's more electronegative?  Chlorine.  So the way I have this Lewis structure drawn, this is where the arrow designation might  be a little bit easier to use.  You draw like that.  Yeah?  I'm going to point on here, since I'm not sure which one's you're pointing to.  But yes, this has a dipole, so yes, it's polar.  If something has a dipole, it is polar.  If something doesn't have a dipole, it isn't polar.  Yes?  Yeah?  So the resonance doesn't really change much of anything when it comes to the charges.  All it really does is it changes exactly how the charges are distributed.  So you need to take, so think of like the sapling problems where you do something like  NO3 and you had to draw all three, or NO3 I think it was, and you had to draw all three  of the resonance structures.  All that really changes there is that the charge was evenly distributed, right?  Because it wasn't just belonging to one of the oxygens, it was belonging to all three  equally, and so all of the dipoles can't, or all of the negative charges cancelled out.  So what the resonance structures basically keep you from doing is having, it changes  the way your formal charge functions into it.  It's almost easier for like those ones that you've been doing on sapling and that I did  in class, to just kind of think of, ignore where the charges actually are for that.  Like ignore those formal charges and kind of think about it more, but other than that  it doesn't actually change much.  Okay.  Questions?  Yes?  Yeah, you need more than 180.  And you don't need to put that one there if you don't want to put it there.  The 180 in this one is really sort of optional in my opinion.  If I, I mean if I drew it, like if I drew you out of Lewis structure that actually showed  the geometry and I drew this and I asked you to do it, then you'd have to be able to explain  that it's more than 180, but if I just ask you for bond angles, I don't necessarily care  about the 180 being there or not.  Yeah?  In the second part, if iodine has a double bond, would you say it's bond or has a double  bond?  Let's not talk about if this one, for this one just because if iodine has a double bond  here, there's some other things going on, lots of other things going on, so we can't  look at it with that molecule.  But just know that you don't factor in that pi bond as a bond when you're counting hybridization.  That's what you're really asking, right?  Yeah, that you're counting, you don't count bonds to figure hybridization.  If we can quiet the chatter just a little bit, I know you're talking about chem things,  but it's just too loud.  So if you're counting, you're counting numbers of atoms that are bonded to it and then lone  pair, so you're not actually counting individual bonds when you figure hybridization.  Yeah?  Would you give us a problem where like say it would just be a high C on the board so  that there would be like one unit process in a lone pair?  Oh, you mean like a radical?  Yeah, yeah, you mean a radical.  You do.  Where there's only, instead of a lone pair, you have one.  Yeah, that's possible.  And if you get something like that where you have, if you only have one electron and you  don't have a pair, this shows up when you have an odd number of electrons.  That's why this has a negative charge or a positive charge.  You treat an electron that's by itself like that as a radical basically like a lone pair.  And I say basically with one caveat that I'll talk about later on in the review packet.  But for the sake of the hybridization and things of that, you count it just like a lone  pair.  Yeah, you count it exactly like a lone pair for the sake of naming the hybridization  and me asking where it's at.  There's just one little thing that changes that I'll talk about later on.  Yes?  Right, you can't.  You wouldn't be able to because you'd have an odd number.  So, okay.  Let's move on to the next one.  So I say draw the MO diagram for NO.  Assume that the sigma is lower than the pi.  Be sure to label all of your orbitals including your subscripts.  Fill in any electrons that you need.  Tell me what the bond order is and tell me if it's paramagnetic or diamagnetic.  And then I added a question to it from when I sent it to you.  So make sure you write that down.  I also say would NO minus have a higher or a lower bond order?  So let's draw this out.  So this is one of those that I said I could ask.  Yeah?  For NO, is the sigma bond between the point of eye bond order and is one of the exceptions?  No.  So whenever you're dealing with a heteronuclear, it's what I tell you it is.  So that's all you have to worry about for the heteronuclear ones.  When you get into the heteronuclear things, there's a lot of weird things going on which  is actually the sort of summary of the one slide that you probably none of you really  liked all that much where I had the C1 psi 1 plus C2 psi 2.  The summary of that slide in one sentence is when you have heteronuclear diatomics,  a lot of things are going on that you don't really have to worry about too much.  The way that they add, they add based on a weighted system that I can't expect you to  know.  So I'll tell you.  This would be how I would word the question.  And I would just tell you that this is lower than the pi.  OK?  All right.  So let's start the drawing of this.  So we know we're going to have some s orbitals.  I'm fine with you just drawing the valence shell.  I don't need you to draw in the 1s's.  You should know they're there, know they exist, but I don't care if you draw them.  That's kind of our starting point.  So we can fill in for nitrogen.  How many electrons do we have?  Valence electrons?  Five.  And how many for oxygen?  Six.  OK.  And now we can draw this.  We can say, well, sigma is lower than pi.  So we have sigma and then pi like this.  We're pretending these are on exactly even levels because my straight line wasn't totally  straight.  OK.  And now let's label them so we don't forget.  So this is sigma 2s.  Sigma 2p.  Pi 2p.  Pi 2p star.  And sigma 2p star.  Do we care for when you're drawing these that we know where the labels refer to?  Make sure they're closer to where they belong.  You can also put them off to the side.  A lot of people write the labels off to the side if they run out of room.  That's fine.  I tend to forget to leave myself enough room when I'm drawing the outline so I write them  below.  But you're welcome to write them here and here and here and here and here and here.  Don't forget your stars.  Don't forget your subscripts.  Now how many electrons total do we have here?  We have 11 so we can go fill in.  If we have that.  OK.  So we've labeled our atomic orbitals.  We've labeled our molecular orbitals.  We've included our subscripts.  We've filled in all our electrons.  Now I want to know what the bond order is.  How do we know what the bond order is?  We count up all our electrons.  How many are in our bonding orbitals here?  Two, four, six, eight.  How many are in our anti-bonding?  Two, three.  So our bond order is 2.5.  So that cover, oh no that doesn't cover everything.  So is this paramagnetic or diamagnetic?  Why?  Careful how you say that.  It has unpaired electrons, right?  Has unpaired electrons.  So paramagnetic is unpaired electrons.  OK.  Now I say does NO minus have a higher or lower bond order?  So think about it a sec.  Don't say it yet.  Give people some time to think about it.  Now there's a couple ways to do this.  One is to redraw the entire MO diagram.  I see that's the favorite way from all of your groans.  Is that what you want to do?  Is that what you want to do on a 50 minute exam?  And I know my exams are a little on the long side, so a long 50 minute exam?  No.  Is that what I want you to do?  No.  What do I want you to do?  OK, quiet a little bit.  I want you to look and see where is that electron going to be either added to in this case,  because it's a negative charge, or taken away from in a positive case.  So where is this electron going to be added to if we add it here?  Right here, right?  So is that an anti-bonding or a bonding?  It's an anti-bonding.  So is that going to take away from the bond order or add to the bond order?  It's going to take away from the bond order.  So is it going to be smaller or bigger?  Yeah.  Now just for fun, let's say I did ask you to draw me the MO diagram of the NO minus  or NO plus or something like that.  Where would I put this negative charge as far as the nitrogen and the oxygen goes?  Where should I put it?  Good, the oxygen.  Why?  It's more electronegative.  So you put the electron, you put the negative charge on the oxygen.  And where would you take an electron away from?  The nitrogen.  Which remember that really doesn't play a role in the MO part of the diagram.  It doesn't play much of a role in here, because you just know you're adding or taking one  away.  But it does matter over here.  OK.  Any questions, hand raised that I missed?  Yeah.  Oh, thank you.  Don't forget the E on the side.  That was on the checklist of things to make sure you draw.  And that's like the easiest one to forget as you just saw.  And the better you get at them, the easier that E is to forget.  Yes.  So stability has to do with a lot more than just bond order.  Because you're also dealing with exact electron configurations and things of that sort.  So for this one, we wouldn't really get into it too much.  OK.  Anything else?  OK.  Next one.  And again, we're not feeling too rushed here, because whatever I don't cover here, I'll  cover in my review session on Thursday.  OK.  How does constructive and destructive interference relate to molecular orbital diagrams?  So this would be a sort of essay type question, right?  I think on your last one, this is kind of similar to, I don't know, I asked you something  similar to how you'd answer this one.  OK.  So how does this have anything to do with the MO diagrams?  So let me give you some prompts here.  What are bonding orbitals caused by?  So when we take, when we take and we make MO molecular orbitals, we're taking two atomic  orbitals, which are waves, right?  Those are just waves, and we're overlapping them.  And we can overlap them two different ways, right?  We can overlap them constructively, and we can overlap them destructively.  So how would you overlap them constructively?  Would that form a bonding orbital or an anti-bonding orbital?  Bonding.  How would you overlap them destructively?  What would you form?  Destructive.  So of bonding orbitals and anti-bonding orbitals, which one has the star?  Anti-bonding.  Which one do you fill first?  Bonding.  So that's all I would expect for that one.  I'm not going to write it in essay form, but.  OK.  Questions on that one?  No?  OK.  I thought it was worthwhile to do some more of these.  We've just covered it, so it's new, but more examples of new things are, or I think, a  very good thing.  OK.  Rank the following in order of boiling point.  So when we rank by order of boiling point, what else are we really ranking by?  Intermolecular forces, right?  The higher the intermolecular force, what happens to the boiling point?  It goes up.  If you increase the intermolecular forces, you increase the boiling point.  So if we look at this first one, we just have that.  What's that called?  That type of structure?  Aligned structure, right?  What's another way of drawing it?  You can draw it like this.  Make sure you know how to draw both ways.  If you draw a Lewis structure, do you have to include the lone pairs?  Yes.  If you draw a line structure, do you include the lone pairs?  No.  If I draw you a line structure and ask you to tell me things about it, am I going to  draw on the lone pairs?  Where do you think a common mistake comes in there?  People forget the lone pairs exist, right?  Anything you're doing that has to do with lone pairs, which would be like what?  What has to do with lone pairs?  What do you figure out that you have to count numbers of lone pairs for?  Steric number, which is your geometry, your hybridization, all those sorts of things.  If you forget the lone pair, are you going to get those right?  No.  So on line structures, it's not a bad idea to go in and draw on your lone pairs.  Now we have that.  So which one has the, I just say rank and order boiling point.  So let's say which one would have the higher boiling point, this one or that one?  So what do we have to look at?  We have to look at our intermolecular forces.  What sort of intermolecular forces does this one have?  Di-pole and LDFs.  What about this one?  All of them, right?  Hydrogen bonding, well I shouldn't say all of them because there's some that we don't  really include, but the three main ones that we check for.  Hydrogen bonding, dipole and LDFs.  Now is there any weird something going on with size here?  Are they like, is one drastically bigger than the other?  Say about the same.  About the same.  Sure, maybe not exactly, but close enough.  The size thing, number of electron things are sort of an approximate number.  So which one's going to be bigger?  The one with hydrogen bonding or the one without?  So this is going to be your largest or highest, I guess I should say.  All right, now let's look at this one.  Yeah?  If I ask you to rank them, all you have to do is write the names.  Like, no, if I ask you to rank them, if I say rank them, all you have to do, I'll tell  you which way to rank them, I'll say low to high or high to low.  You just have to write down the name of the compound and let it, that's it.  If I ask you which one's higher, you just write down the name and you're done.  You'd be a little, you might, and this one, you'd be a little hard for us to be able to  do it without drawing it though.  So is the hydrogen bond the resource to be the OH?  The hydrogen bond on this one, it's made possible because of this OH.  So I'm going to use some scrap paper here a minute.  So remember, that's not where the hydrogen bond is forming.  The hydrogen bond has to form between two different molecules.  So you would make another molecule here.  And this would be your hydrogen bond.  So very good point.  Remember your hydrogen bond, if I draw you one molecule and I say point to the hydrogen  bond, you can't do it.  You would have to draw another molecule and then point to that hydrogen bond between here  and here.  Okay.  Oh, we are over time.  Okay, now a big deal.  We will finish this on Thursday.  So come to my meeting on Thursday.  Don't forget, as you're packing up, that Miranda has a review session tonight as well.  I think hers is question and answer.  So come with lots of questions to hers.  If we run out of time or if we finish the worksheet in mind, we'll switch to question  and answer as well.
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Okay, so today. Okay, so we're going to finish up some hydrogen bonding things that I haven't quite finished up. Well, maybe we will. Okay. And then, so that won't take too much longer. And then we'll do a review lecture. I'm not sure if that whole packet that I gave you, if we'll get through all of it today or not. I actually, leaning forward, we probably won't. So whatever I don't finish in class today, I don't want us to feel rushed. I'm just going to do that Thursday night. So we'll have plenty of time over, spread over everything to at least finish it. So don't worry too much about that. Okay, so the rest of the hydrogen bonding that we need to talk about is why this is so important in biology. This is important not just in general, but also because of the demographics of this class tend to be pretty highly in the biological or variety of earth system science type things. Okay, so the first one that I want to talk about is in DNA. Now this is very, very important. Now you probably have a little bit of background either from high school or from your biology classes here on how DNA kind of looks. So you get this double helix structure. Okay, so we get this double helix structure and we have all of these bonded and those are bonded covalently or some other form. Covalently. Now the thing that holds these two together is what comes important in this chapter. What are all of these bonds? They're hydrogen bonds. So they're not actually covalent bonds that are holding them together. And that's really important because what do we need DNA to be able to do in order to live? It's got to be able to replicate in one other thing. Make mRNA, right? You've got to be able to replicate and make RNA. If you can't do those two things, we're going to have a little bit of trouble making new cells and making proteins and doing all the things that we need our DNA to tell our bodies to do. So in order to replicate it, the way that it works is you have to be able to split a word to make RNA. You have to be able to split the DNA strands. If they were covalently bonded, that would be a lot harder. You could still do it, but it would be a lot harder. But having these hydrogen bonds here strikes a good balance between they're fairly solid, they're strong, but they're not so strong that you can't undo them relatively easily. So we have proteins in our bodies that come through and can unzip them and rezip them back up as we go along. If it was covalently bonded, that would be tougher. If it was held together by some other force, that might be a lot harder too. The hydrogen bonds are kind of this balance. Now going a little bit more in depth with this, again, because of the demographics of this class, notice between the adenine and the thiamine, how many hydrogen bonds do we have? Two. What about between the guanine and the cytosine? Three. Three. Which ones are going to be stronger? Three. Yeah. So as you go along in your biology classes, you'll start at some point, probably your junior year, so you'll start calculating melting points for DNA strands and talking about how much, you know, when you start, especially in your molecular biology methods and lab courses, and you'll be calculating these, and you'll be calculating melting temperatures, how much you have to pull, how much energy or heat you have to put in to pull them apart and, you know, before you can let them come back together. And it's going to make a difference how much A's and T's you have and how many C's and G's you have. And that's why, because C's and G's bond, or they bond at three hydrogen bonds instead of two. So they stay together easier. So if you want to break them apart, it takes more energy. So that's, you know, when you're doing lab work and things like that in some of your higher level bioclasses, that'll come into play. Okay. The next place that I want to talk about, that it plays very important roles, is in proteins. So this is a picture of a protein, is one I just pulled off from this website down here. And all of these little yellow dotted lines are drawn in using hydra, or those are all your hydrogen bonds. Now, it's not all of them, of course, because there's a lot of them that are forming all your different structures, your secondary structures. But hydrogen bonds are largely responsible for your secondary structures. They play all of these important roles and tertiary structures that you can see. And so they're very, very important in protein formation. You need these proteins to stay in these very tight, there's not a lot of leeway between a lot of these structures on how much they can move. And hydrogen bonds are a large portion responsible to their structure. Without them, you wouldn't have as well of a defined structure as you do. Now in hydrogen bonding, just so you realize here, this is an amino acid. Okay? So this amino acid is what makes up all your proteins. So it's just chains and chains and chains of these, and big long formations. And so what happens is you can get your oxygens and your nitrogens to hydrogen bond together with your hydrogens that are on them. And that's what forms your structure is all these groups here. And we won't go into detail about the exact structures of that, I'll leave that for your higher level classes. But just sort of know how this works, know it's important. Be able to sort of explain to me why these sorts of things are important. The next one then to talk about real quickly is cellulose. So where do you know cellulose from? What does that make up? Your plants, right? That's sort of the structural component of your plants. And the way that cellulose works is it's these long chains of glucose. So you get these really, really long chains of glucose. Now on their own, a single long chain of glucose isn't going to be very sturdy. They're not going to be able to hold up a redwood with just individual chains of glucose. But then what happens is all of these chains, they're able to line up and they're able to form hydrogen bonds with each other. Look at all these OH groups here and these OH groups. Cellulose can all form hydrogen bonds in between chains and so you get these big long stacks of these chains and you form these fibrils and then that's what is able to make up all the structural components of your walls or your cell walls and things. So that is all I have for you on the rest of that chapter. So we're done with new midterm material. This was one thing I wanted. I meant to do this before but this is one thing I wanted to clarify from last time. There's a bit of a mistake. This is the correct ordering. So this is a problem that we did last class right kind of near the end. But there's a little bit of a mistake in the fact that I can't actually expect you to be able to rank all four of these because there's a big exception here that comes from the shape of the molecule. We didn't talk about that and we're not going to talk about that. So in a situation like this, this question, if it were to show up on an exam, would show up in this form. I could ask you to rank those three or I could ask you to rank those three. I would not ask you to rank all four. So I sent out an email to this extent but I know not everybody reads their emails. You should please read your emails. So just making sure that you guys realize that. So I would ask you to rank either this grouping or this grouping, not all four of them. So now under review material. All right. So this is, since you have done a large portion of your studying for the exam, it's not a bad idea as I'm lecturing here to try to beat me to the problems, right? You don't need to wait for me to write down the answer to write down the answer. Try to beat me to it. And then if I write down something different, you can say, wait, wait, I did something wrong. Figure out what you did wrong and clarify your mistake. Okay. So I have this drawing, I cl4 minus and I cl4 plus. And then I say for each one, name me the electron geometry, the molecular geometry, all of the bond angles, the hybridization of the central atoms. And then I want to know what orbitals is each atom used to bond with. And I want to know if it has a dipole. So quite a lot of information here, right? I wanted to include everything that I could possibly ask you about a molecule. Okay. So our first step is to pick one and draw it. So let's start with icl4 for no other reason than I listed it first. So we have icl4. First step to make sure that we don't do anything wrong later on is to count up all of our electrons. So we know we have seven from the i and seven from each of the chlorines. And then we have one extra one because of the electron, right? Can't forget about that. So we have 36 electrons. Now let's just go ahead and set up our system for the other one right here, too. So for icl4 plus, how many electrons are we going to have? 34. Good. 35 from the elements and then subtract one for the plus charge. Okay. Now let's draw these things out. So our central atom will be our i. And then we can try to draw out our chlorines. Now something I want to mention as, because this is a good place to mention it, you'll notice when I draw my Lewis structures, I tend to draw them in a way that looks like the shape that it would be in 3D. That's because in general when I'm drawing Lewis structures, I know what I'm drawing. I know ahead of time what I'm drawing. That's not to say that Lewis structures have to represent geometry. Lewis structures don't represent geometry. So when you're drawing these, you don't have to feel like you have to get the geometry right the first time. I mean if you get in the habit of drawing them a lot, you'll sort of just start to do it anyways and it is helpful at times. But notice when you did your sapling homework, sapling in purpose, draw it sort of in a way that didn't look anything like the actual molecule would look to make sure that you knew what you were doing as far as finding geometry. And that's fine. If I want to know the geometry, I'll ask you the geometry where you don't have to draw it that way. I say this especially so that you don't waste too much time on the trying to draw things if you need to. So we have this. Now how did I know to add in the two sets of lone pairs on the iodine? Well if you go through and you just add in four chlorines and you count up all of the electrons, we're missing four electrons, right? And so since we're missing four electrons, if we just count up the chlorines and the bonds, we know that those have to go on the iodine. I can't put more of them on the chlorine. OK, so that's our Lewis structure. So now we have to go through and answer all the questions based on it. So first of all, what's going to be the electron geometry? Electron geometry. So remember for electron geometry, you're counting in the lone pairs. You can see the lone pairs. Octahedral. Octahedral, right? How did you know that? What's the steric number? Six. So if you have a steric number of six, that's octahedral. What would you do if you didn't know that? Yeah, you needed to picture it. If you can't picture it, again, I'm fine with you bringing toothpicks. I'd rather not go to the full extent of candy and things of that sort in modeling kits just because it's too easy for the people around you to see. But if you want to bring some toothpicks and just kind of fiddle with them while you're doing this, I'm fine with that. So if you did that, the easiest way to get all of these to face each other or to fit into a shape would be to have them be like that, right? That gets you six. OK. So now comes the molecular geometry. And of course, I don't ever really mind you interrupting me with questions, but this is a place that's especially important if you have questions to interrupt me. So you can feel free to do it even more than normal. OK. So now we have our octahedral electron geometry, and we need to take out two of those and switch them out and put lone pairs there. So are we going to put two that are right next to each other as lone pairs or two that are across from each other? Across from each other. So we end up with this, right? So what shape is that? Square planar. Oh, I forgot one very important thing with my Lewis structure that would be a point off. What did I forget? Formal charge. There's a couple of different ways you can do formal charge. You could actually write it in where it belongs, like on the actual atom it belongs in. Or if I don't ask you to give me the formal charges of the atoms, you can just put it over top of the whole thing. That's fine. But you actually have to put it on the atom unless I ask you to do so. OK. So I'm just going to put it here. All right. Now our bond angles. So we've taken an octahedral geometry. We put some lone pairs on it. We have this geometry now. What is our bond angles? 90. Is it 90 or is it less than 90? Good. It's 90. Right, because even though you have the lone pairs on it, you have this lone pair pushing down this way and this lone pair pushing up this way. But they're pushing on each other equally. And so all the bond angles still end up 90. OK. Hybridization. So what was our steric number again? Six. So how many hybrid orbitals do we need? Six. So it's going to be sp3d2. OK. Yeah. So on this one, there's times where that's more OK than others. And on this one, it would be a little weird. I guess if you're talking from here to here, if you put 180, I wouldn't mark it wrong. On this one, you definitely don't have to put it though. OK. So now, next thing we have to do. Oh, bonding orbitals. OK. Oh, yeah. If you asked about hybridization before, would that be? If I asked you about the hybridization of Cl, there's two answers that I would be OK with. The one that I would prefer is to say unhybridized. The other one that you can use is you can use the fact that if you counted these lone pairs, you'd have one, two, three, and then the bond, so four, and you could base it off that. You could say sp3. But in general, we don't talk about the different ligands, the outside molecules, things that are only bonded to one atom. We just don't talk about those as being hybridized at all. In general, I just steer away from asking about it. What I will ask you though is this next thing. Yeah? Did you say that it bonds with the original atom? Yeah. That's what we're going to do right now, actually. So what I will ask you is exactly what she asked, is what is it bond with though? So the hybridization, I'll probably just ask you of central atoms, things that are bonded to more than one element. But what about what that bonds with? Yeah? If I asked you about hybridization, no. P is in a sort of hybridization. But for this next thing, yes. So what is a bond with? Would it bond with something? Well, yeah, let's do this next question now. So what does it bond with? So we need to break this down into two things. Iodine is going to bond with something and chlorine is going to bond with something. OK. So we have to figure out what these are now. So let's start with the chlorine. And I'm going to just do this on some scrap paper so I can write it out with a little bit more room. So let's start with the chlorine. What is the, if we were to draw the electron energy diagram of chlorine, what would it look like? Well, we have seven electrons, so it would look like that. So what orbital is empty, or not empty, but has a spot to bond with another something? P. So what is chlorine going to bond with? It'll bond with its P orbital, because that's where it has an empty spot for it. OK. So the chlorine is going to bond with its P orbital. Now let's talk about what the iodine is going to bond with. So why did we make hybrid orbitals? Why do we have this thing called hybridization that we talk about? Well, we have it because we need a way to spread out all of the bonds, or all of the atoms, I guess I should say, spread out all of the atoms and all of the lone pairs. So we picked our hybridization. We said it was sp3d2 based on the fact that we needed six things, six spots for all of these to bond with or put lone pairs in. So what is it going to use to bond with? First of all, do we really have any s or p orbitals left anymore? No, we've turned them into these hybrid orbitals. So what are we going to bond with? Our hybrid orbitals, right? We've made them to form those bonds and to put those lone pairs in. So iodine is going to go ahead and bond with its sp3d2 orbitals. I'm going to have to ditch the question there. So luckily, you all have the worksheet. OK, so we have one other thing that I could ask you about then here. Where are the lone pairs at? I should say the iodine lone pairs. Where are those? What type of orbitals are those in? Well, did that go into our sort of calculation about which hybrid orbitals we're going to make? Yeah, we want them spread out, right? We want them spread out just like the bonds were. So what orbitals would those be in? The hybrid orbitals, good. So those are also going to be in sp3d2 orbitals. So your lone pairs and each of your sigma bonds are formed by or are using your hybrid orbitals, in this case sp3d2. Yeah? So you have a separate plot value in combination of sp2, what do you use sp2 to use in the class p? So what he just asked, I want to say the question since a lot of you probably didn't hear it, is if you had something like an sp2 hybrid orbital, so you'd have the three sp2 orbitals and you would have a p orbital. When do you use the sp3 orbitals and when do you use the leftover pi orbitals? So we have two different types of bonds that we can make. We can make sigma bonds and we can make pi bonds. All of your sigma bonds are formed using your hybrid orbitals and then your pi bonds are formed using your leftover p orbitals. Okay, questions on this? Okay, so let's, oh yeah. Well, the thing is, is at that point you'd actually get into, if you had two leftover p orbitals, you would, both of those are going to form pi bonds. So it's either going to have two double bonds, like in the case of a carbon, it would have two double bonds or it would have a triple bond. Yes? Yeah, so he had, again, just repeating this, when do we use the hybrid orbitals and when do we use leftover p orbitals? We use hybrid orbitals in order to form our sigma bonds and to put our lone pairs in. And we use p orbitals to form our pi bonds. So either our double bonds, the second bonds and our double bonds, or the second and third bonds and our double, or triple bonds. Yes? So why does the central atom have to be hybrid? Say that a little bit louder. The central atom? You said that has to be hybrid orbitals. Those are hybrid orbitals because we need to get the right geometry so that we can spread everything out. And there's a couple of different ways of thinking about it. We can do hybrid orbitals or we can do MO theory, but MO theory for these would be quite complicated so we're sticking with hybrid orbitals. Okay, let's race through now and do ICL4 and it's all going to be very similar so try to beat me to it as we go along. Oh yeah, dipole, thank you. So does it have a dipole? So we have this shape. So are there any polar bonds? Very slightly, not by much, but a little bit. Does it have a dipole overall? No, they cancel, right? This one going this direction and this one going this direction cancel and this one this way and this one this one so they all cancel. So no dipole. Okay, next one. So we do something very similar. But now we've dropped two electrons. We've dropped a lone pair. So it's going to be pretty much the same, just minus one set of electrons. Okay so we have this. Now let's go through and do everything in pretty much the same order. In order to try to save some room I'm going to write up here a little bit though. So we have electron geometry and molecular geometry. So what do we think for the electron geometry? What's our steric number? Five, right? Remember you're counting atoms that are bonded to it and lone pairs, so five. So that gives you what shape? Trigonal bipyramidal, right? That's the one where you have sort of, I'll just make it real fast. This is the one where we have this shape. Now so we have sort of a triangle shape here. We can see that we have a pyramid on the top and a pyramid on the bottom, so trigonal bipyramidal. Now one of those is taken off and it's not an atom anymore, it's a lone pair, so it looks like this. So what shape is this molecular geometry? So remember this one is helpful to sort of tip it, ah, what did I do wrong? You took it off the wrong side, right? Get off that one. I went for the easy one for me to hold, not the correct one. There you go. Now what shape is it? Ah, right, tip it on its side, it's C-sa. So why did I take off this one? Why do you have to take this one off? Yeah, so what's the spawn angle? 120. 120 and what's the spawn angle? 90. So we want to give the lone pairs, these huge lone pairs taking up tons of room. Are we going to want to give them 120 to work with or 90 to work with? 120. So we have to take them off these. And what's that called? What's the space called? Equatorial, right? Think of it like a big globe. You have an equator going around the outside and then you have the axes going here. So this and this are axial. So then what shape is C-sa? Tip it on its side. Okay. So we have this. Okay, so now we have to work with bond angles which we sort of touched on. What were the bond angles again? So that's for the trigonal bipyramidal. What about for the C-sa? Less than 90 and less than 120. And again, 180 would work if you'd like. I don't necessarily care if you have it there or not, but think the 180 from the one all the way to the other. From the two axials. Yeah. Well, on the Lewis structure you can't see. You have to actually think about it in terms of a three-dimensional structure. So the equatorial ones are the ones that go along the 120 degree and then the axial are the ones that are the 90 degree. Drawing here. Oh, well axial and equatorial are just referring to individual atoms. So you can't really talk about it because the 90 is between an axial and an equatorial and then the 120 is between two equatorials. Yes. Do we have to put both? This? Yes. You have to put both. And putting 180 is a good idea. I'll add that there too. Okay. Hybridization. What is our steric number? Steric number is five. So what is our hybridization? SP3D. Okay. Next part. What do we use to bond with? So what is our CL going to bond with? Is it any different than over here? No, exact same situation. And what is it going to use to bond with? It's p orbital. It's exact same reason, right? Let's throw this back up for a minute. Exact same reason. That's what it has a spot to put it in. You don't need to hybridize those outside atoms. So we'll just leave them like that. Okay. Now, what is the iodine going to bond with? It only has sigma bonds. So we form sigma bonds with what kind of orbitals? Hybrid orbitals. So in this case, that is SP3D orbitals. And where are the lone pairs at? What orbital are those in? Good. In the same, right? I guess I should just say lone pair, not pairs. Just cancel out the s. Okay. So now, we have our C-sauce shape. Does it have a dipole? So first of all, first question we have to ask, are there polar bonds? Sure. Not super polar, but polar. Now are they going to cancel each other out? No. Some of it will, right? This one and this one will cancel a little bit, or most of the ways. Remember, they're actually all tilted a little bit. Most of this will cancel. And the side to side, or I guess in and out for you guys to compare a component of this will cancel. So if I hold this like this, which way did the dipole be? Yeah, to the side. It would be going in the direction like directly across the building. Okay. So let's do that. Let's write that down real quick. So yes, dipole. Oh, like a he- So on a Lewis structure, I wouldn't make you draw them like I have been. I could make you draw the delta positive and delta minus. The likelihood of me doing it on something like this is pretty small, just because you kind of have to draw it in between the CLs. And if you didn't draw your Lewis structure exactly right, the way, if I'm going to have you draw a dipole, I'll probably draw you the molecule. I'll draw you the molecule and have you draw the dipole in. So how do we decide which one has the negative charge? Which one's more electronegative? So how do we decide which one's more electronegative? Where it is on the periodic table. So which one's higher on the periodic table? iodine or chlorine? Chlorine. So which one's more electronegative? Chlorine. So the way I have this Lewis structure drawn, this is where the arrow designation might be a little bit easier to use. You draw like that. Yeah? I'm going to point on here, since I'm not sure which one's you're pointing to. But yes, this has a dipole, so yes, it's polar. If something has a dipole, it is polar. If something doesn't have a dipole, it isn't polar. Yes? Yeah? So the resonance doesn't really change much of anything when it comes to the charges. All it really does is it changes exactly how the charges are distributed. So you need to take, so think of like the sapling problems where you do something like NO3 and you had to draw all three, or NO3 I think it was, and you had to draw all three of the resonance structures. All that really changes there is that the charge was evenly distributed, right? Because it wasn't just belonging to one of the oxygens, it was belonging to all three equally, and so all of the dipoles can't, or all of the negative charges cancelled out. So what the resonance structures basically keep you from doing is having, it changes the way your formal charge functions into it. It's almost easier for like those ones that you've been doing on sapling and that I did in class, to just kind of think of, ignore where the charges actually are for that. Like ignore those formal charges and kind of think about it more, but other than that it doesn't actually change much. Okay. Questions? Yes? Yeah, you need more than 180. And you don't need to put that one there if you don't want to put it there. The 180 in this one is really sort of optional in my opinion. If I, I mean if I drew it, like if I drew you out of Lewis structure that actually showed the geometry and I drew this and I asked you to do it, then you'd have to be able to explain that it's more than 180, but if I just ask you for bond angles, I don't necessarily care about the 180 being there or not. Yeah? In the second part, if iodine has a double bond, would you say it's bond or has a double bond? Let's not talk about if this one, for this one just because if iodine has a double bond here, there's some other things going on, lots of other things going on, so we can't look at it with that molecule. But just know that you don't factor in that pi bond as a bond when you're counting hybridization. That's what you're really asking, right? Yeah, that you're counting, you don't count bonds to figure hybridization. If we can quiet the chatter just a little bit, I know you're talking about chem things, but it's just too loud. So if you're counting, you're counting numbers of atoms that are bonded to it and then lone pair, so you're not actually counting individual bonds when you figure hybridization. Yeah? Would you give us a problem where like say it would just be a high C on the board so that there would be like one unit process in a lone pair? Oh, you mean like a radical? Yeah, yeah, you mean a radical. You do. Where there's only, instead of a lone pair, you have one. Yeah, that's possible. And if you get something like that where you have, if you only have one electron and you don't have a pair, this shows up when you have an odd number of electrons. That's why this has a negative charge or a positive charge. You treat an electron that's by itself like that as a radical basically like a lone pair. And I say basically with one caveat that I'll talk about later on in the review packet. But for the sake of the hybridization and things of that, you count it just like a lone pair. Yeah, you count it exactly like a lone pair for the sake of naming the hybridization and me asking where it's at. There's just one little thing that changes that I'll talk about later on. Yes? Right, you can't. You wouldn't be able to because you'd have an odd number. So, okay. Let's move on to the next one. So I say draw the MO diagram for NO. Assume that the sigma is lower than the pi. Be sure to label all of your orbitals including your subscripts. Fill in any electrons that you need. Tell me what the bond order is and tell me if it's paramagnetic or diamagnetic. And then I added a question to it from when I sent it to you. So make sure you write that down. I also say would NO minus have a higher or a lower bond order? So let's draw this out. So this is one of those that I said I could ask. Yeah? For NO, is the sigma bond between the point of eye bond order and is one of the exceptions? No. So whenever you're dealing with a heteronuclear, it's what I tell you it is. So that's all you have to worry about for the heteronuclear ones. When you get into the heteronuclear things, there's a lot of weird things going on which is actually the sort of summary of the one slide that you probably none of you really liked all that much where I had the C1 psi 1 plus C2 psi 2. The summary of that slide in one sentence is when you have heteronuclear diatomics, a lot of things are going on that you don't really have to worry about too much. The way that they add, they add based on a weighted system that I can't expect you to know. So I'll tell you. This would be how I would word the question. And I would just tell you that this is lower than the pi. OK? All right. So let's start the drawing of this. So we know we're going to have some s orbitals. I'm fine with you just drawing the valence shell. I don't need you to draw in the 1s's. You should know they're there, know they exist, but I don't care if you draw them. That's kind of our starting point. So we can fill in for nitrogen. How many electrons do we have? Valence electrons? Five. And how many for oxygen? Six. OK. And now we can draw this. We can say, well, sigma is lower than pi. So we have sigma and then pi like this. We're pretending these are on exactly even levels because my straight line wasn't totally straight. OK. And now let's label them so we don't forget. So this is sigma 2s. Sigma 2p. Pi 2p. Pi 2p star. And sigma 2p star. Do we care for when you're drawing these that we know where the labels refer to? Make sure they're closer to where they belong. You can also put them off to the side. A lot of people write the labels off to the side if they run out of room. That's fine. I tend to forget to leave myself enough room when I'm drawing the outline so I write them below. But you're welcome to write them here and here and here and here and here and here. Don't forget your stars. Don't forget your subscripts. Now how many electrons total do we have here? We have 11 so we can go fill in. If we have that. OK. So we've labeled our atomic orbitals. We've labeled our molecular orbitals. We've included our subscripts. We've filled in all our electrons. Now I want to know what the bond order is. How do we know what the bond order is? We count up all our electrons. How many are in our bonding orbitals here? Two, four, six, eight. How many are in our anti-bonding? Two, three. So our bond order is 2.5. So that cover, oh no that doesn't cover everything. So is this paramagnetic or diamagnetic? Why? Careful how you say that. It has unpaired electrons, right? Has unpaired electrons. So paramagnetic is unpaired electrons. OK. Now I say does NO minus have a higher or lower bond order? So think about it a sec. Don't say it yet. Give people some time to think about it. Now there's a couple ways to do this. One is to redraw the entire MO diagram. I see that's the favorite way from all of your groans. Is that what you want to do? Is that what you want to do on a 50 minute exam? And I know my exams are a little on the long side, so a long 50 minute exam? No. Is that what I want you to do? No. What do I want you to do? OK, quiet a little bit. I want you to look and see where is that electron going to be either added to in this case, because it's a negative charge, or taken away from in a positive case. So where is this electron going to be added to if we add it here? Right here, right? So is that an anti-bonding or a bonding? It's an anti-bonding. So is that going to take away from the bond order or add to the bond order? It's going to take away from the bond order. So is it going to be smaller or bigger? Yeah. Now just for fun, let's say I did ask you to draw me the MO diagram of the NO minus or NO plus or something like that. Where would I put this negative charge as far as the nitrogen and the oxygen goes? Where should I put it? Good, the oxygen. Why? It's more electronegative. So you put the electron, you put the negative charge on the oxygen. And where would you take an electron away from? The nitrogen. Which remember that really doesn't play a role in the MO part of the diagram. It doesn't play much of a role in here, because you just know you're adding or taking one away. But it does matter over here. OK. Any questions, hand raised that I missed? Yeah. Oh, thank you. Don't forget the E on the side. That was on the checklist of things to make sure you draw. And that's like the easiest one to forget as you just saw. And the better you get at them, the easier that E is to forget. Yes. So stability has to do with a lot more than just bond order. Because you're also dealing with exact electron configurations and things of that sort. So for this one, we wouldn't really get into it too much. OK. Anything else? OK. Next one. And again, we're not feeling too rushed here, because whatever I don't cover here, I'll cover in my review session on Thursday. OK. How does constructive and destructive interference relate to molecular orbital diagrams? So this would be a sort of essay type question, right? I think on your last one, this is kind of similar to, I don't know, I asked you something similar to how you'd answer this one. OK. So how does this have anything to do with the MO diagrams? So let me give you some prompts here. What are bonding orbitals caused by? So when we take, when we take and we make MO molecular orbitals, we're taking two atomic orbitals, which are waves, right? Those are just waves, and we're overlapping them. And we can overlap them two different ways, right? We can overlap them constructively, and we can overlap them destructively. So how would you overlap them constructively? Would that form a bonding orbital or an anti-bonding orbital? Bonding. How would you overlap them destructively? What would you form? Destructive. So of bonding orbitals and anti-bonding orbitals, which one has the star? Anti-bonding. Which one do you fill first? Bonding. So that's all I would expect for that one. I'm not going to write it in essay form, but. OK. Questions on that one? No? OK. I thought it was worthwhile to do some more of these. We've just covered it, so it's new, but more examples of new things are, or I think, a very good thing. OK. Rank the following in order of boiling point. So when we rank by order of boiling point, what else are we really ranking by? Intermolecular forces, right? The higher the intermolecular force, what happens to the boiling point? It goes up. If you increase the intermolecular forces, you increase the boiling point. So if we look at this first one, we just have that. What's that called? That type of structure? Aligned structure, right? What's another way of drawing it? You can draw it like this. Make sure you know how to draw both ways. If you draw a Lewis structure, do you have to include the lone pairs? Yes. If you draw a line structure, do you include the lone pairs? No. If I draw you a line structure and ask you to tell me things about it, am I going to draw on the lone pairs? Where do you think a common mistake comes in there? People forget the lone pairs exist, right? Anything you're doing that has to do with lone pairs, which would be like what? What has to do with lone pairs? What do you figure out that you have to count numbers of lone pairs for? Steric number, which is your geometry, your hybridization, all those sorts of things. If you forget the lone pair, are you going to get those right? No. So on line structures, it's not a bad idea to go in and draw on your lone pairs. Now we have that. So which one has the, I just say rank and order boiling point. So let's say which one would have the higher boiling point, this one or that one? So what do we have to look at? We have to look at our intermolecular forces. What sort of intermolecular forces does this one have? Di-pole and LDFs. What about this one? All of them, right? Hydrogen bonding, well I shouldn't say all of them because there's some that we don't really include, but the three main ones that we check for. Hydrogen bonding, dipole and LDFs. Now is there any weird something going on with size here? Are they like, is one drastically bigger than the other? Say about the same. About the same. Sure, maybe not exactly, but close enough. The size thing, number of electron things are sort of an approximate number. So which one's going to be bigger? The one with hydrogen bonding or the one without? So this is going to be your largest or highest, I guess I should say. All right, now let's look at this one. Yeah? If I ask you to rank them, all you have to do is write the names. Like, no, if I ask you to rank them, if I say rank them, all you have to do, I'll tell you which way to rank them, I'll say low to high or high to low. You just have to write down the name of the compound and let it, that's it. If I ask you which one's higher, you just write down the name and you're done. You'd be a little, you might, and this one, you'd be a little hard for us to be able to do it without drawing it though. So is the hydrogen bond the resource to be the OH? The hydrogen bond on this one, it's made possible because of this OH. So I'm going to use some scrap paper here a minute. So remember, that's not where the hydrogen bond is forming. The hydrogen bond has to form between two different molecules. So you would make another molecule here. And this would be your hydrogen bond. So very good point. Remember your hydrogen bond, if I draw you one molecule and I say point to the hydrogen bond, you can't do it. You would have to draw another molecule and then point to that hydrogen bond between here and here. Okay. Oh, we are over time. Okay, now a big deal. We will finish this on Thursday. So come to my meeting on Thursday. Don't forget, as you're packing up, that Miranda has a review session tonight as well. I think hers is question and answer. So come with lots of questions to hers. If we run out of time or if we finish the worksheet in mind, we'll switch to question and answer as well.
structure as you do. Now in hydrogen bonding, like you have to get the geometry right the first point of eye bond order and is one of the very good thing. OK. Rank the following in order intermolecular forces, you increase the boiling