Wednesday, April 26, 2006

Lecture 004: More Lewis Structures and Molecular Orbitals

Lecture 004: More Lewis Structures and Molecular Orbitals
Speaker: Jean-Claude Bradley

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Speaker: We're going to resume, I'm going to do a few more Lewis structures for you. One thing we didn't cover yet, I think, were some charges, so we can see how that turns out.

I'm going to start with SO2, we're doing a Lewis structure for SO2. I'm going to go through these fairly quickly because we spent enough time on this last Wednesday. Sulfur and oxygen both have six valence electrons. We're going to do six times three. We're going to need eight. We have 24 that we need, we're going to share 24 minus 18, we have six that we're going to share. We have six divided by two, for three bonds. We have a situation where we have a sulfur, one type of atom, and we have oxygens, a whole bunch of other oxygens so I gave you a heuristic when that happens. First thing is you don't make a triangle out of this, it's going to be in a straight line and you're going to put the sulfur in the middle. We're going to put the sulfur here and then we have three bonds to connect to the rest. I put an oxygen on this side, I put an oxygen on this side, I require two bonds only to link that, but I have to use up all three, so I'm going to put the other one here.

As we continue, we're going to complete the octets. Sulfur already has six electrons, so it only needs one lone pair. The oxygen at the end requires three lone pairs. Now, we go and look at charges. The oxygen on the left has six electrons for charges, oxygen normally has six, so it's uncharged. Sulfur, five electrons for charges, it normally has six, so it's missing an electron, which means that it's going to be positive. The oxygen on the end, on the right side, has seven electrons for charges, it has one more than it normally has, so it's going to be minus one. The plus and the minus add up to zero, that has to match up with the original problem, which does not have a charge. That's SO2.

Now, this is actually not the correct way to draw SO2 because I haven't taken into account the shape of the molecule. When we do this problem again, you'll see that that's not the way it looks, it has a different shape, but I'm going to deal with that seperately because we have to deal with more theory before we can deal with that.

Next we have SO3. Available, again, there are all six valence electrons. We have 24 available electrons. We need eight times four to complete the octets, that's 32. That means that we have to share 32 minus 24, eight. The number of bonds we have will be eight divided by two, for four. I have four atoms. You don't make three member rings, you don't make four member rings either if you can do something different. It doesn't mean that three and four member ring compounds don't exist, it's just that they're higher energy, until we come across an example, that's not something you would do.

If we have a sulfur and then three oxygens, we'll put the sulfur in the middle again and the three oxygens around it. We're going to use up three bonds to minimally connect that structure and we have to use up one more bond. Again, it would be tempting to connect two oxygens together, but that would make a triangle, so we're not going to do that. Instead, we're going to put that next bond between any of the three oxygens and the sulfur, it doesn't matter which one, and we'll see why that is a bit later on.

I've used up the four bonds, I now complete the octets. We find that the oxygens on the bottom each have one negative charge, the oxygen on the top is uncharged and the sulfur has four electrons for charges, sulfur normally has six, so it has a double positive charge. I have two plus and I have a minus and a minus that adds up to zero, which is what we had in the original problem, so that all works out.

Let's look at something that has an added element. Azide, which is N3-, this is the first example where we have a charge and we're going to have to account for that in the available electrons. Nitrogen is five, so five times three would be 15, but then we have to add the one negative charge that's in the original molecule, that would be plus one. That adds up to 16.

Now for the octets, the octet doesn't care about charge at all, you're just counting up to eight electrons, so the charge has no effect on the requirement for the electrons per atom, that would just be eight times three. We have to share 24 minus 16, or eight electrons. That means that we have eight divided by two, for four bonds. Once again, we have three atoms, we're going to put them in a straight line. There's no central atom, so there's only one way to do that, we're going to put the three nitrogens in a row. I'll use up two of the four bonds to connect the structure. I have to put the other two bonds, so I'm just going to do what I've been doing so far, is just add them to the existing nitrogen-nitrogen single bonds to make double bonds. I have used all the bonds up, I complete the octets, the two nitrogens on the end require two lone pairs. Nitrogen in the middle already has eight electrons for the octet rule. We have six electrons on the nitrogens on the end. Nitrogen normally has five, so that would be negative one. I have a negative one on the left, I have a negative one on the right and the central nitrogen has, for charges, four electrons, so that corresponds to the charge of plus one. Overall, I have two minuses and one plus that adds up to an overall charge of minus one, which is consistent with the original problem, so that works out.

That's all you do with charges, so if you have a positive charge or a negative charge, you simply add or delete from the available. If you forget about that, you'll notice that here you'll get an odd number and that'll give you fractional bonds, so you'll never have fractional bonds, you'll always have a real number of bonds at the end.

We'll do a couple more here. Sulfuric acid, here we have a whole bunch of atoms and we'll see how we're going to deal with that. Available we have two hydrogens, that's two times one plus six for the sulfur, plus six times four for the oxygens. That gives us a total of 32. How many do we need? Here, you got to be careful, for some reason it's easy to make a mistake with the hydrogen, the hydrogen needs two electrons each. That's two times two and then all the other atoms require eight, so we have five atoms and all of that adds up to 44. That means we have to share 44 minus 32, which is 12, the total number of bonds in this molecule will be six. Now, you have a lot of atoms here, so the first thing you have to look at is simplifying this problem. H2 in the SO4, hydrogen can only bond to one atom, so that's not going to be going anywhere in the middle, what I'm going to put in the middle then will be the sulfur.

If you notice, there's a bit of a pattern here, we went from SO to SO2 to SO3, so now I have SO4. You just continue that pattern, you just keep putting oxygens around the sulfur. I'm going to first worry about linking all of my oxygens and now I'm going to link up my hydrogens. I can't put a hydrogen on the sulfur because it'd be too many. You can only have eight electrons for the octet rule, the only place I can possibly put the hydrogens is on one of the oxygens, that's what will dictate where these will go. The way that I've drawn this, it looks like there are different ways I can position the hydrogens, I could put the hydrogen on the top, on the right, but as we will learn shortly, all those positions are the same. The molecule doesn't look like the way I drew it, it doesn't really matter where you put it, so I'll just put it at the bottom here.

I have used up all of the six bonds, that's important. You don't add anything else, you just complete the octets. Each of the terminal oxygens will end up with six electrons. The oxygens on the OHs are 4, then when we look at charges, we'll get minus ones for the oxygens terminating. The sulfur, we count four electrons for charges, sulfur normally has six, so that would give me plus two. We have a plus two and we have two negative charges. That adds up to zero, which was the original problem.

The next one I'd like to so is nitrous oxide, N2O. Each nitrogen will be five, plus six for the oxygen, 16, we will need eight times three. We're going to share 24 minus 16 for eight. Divide eight by two, so we have four bonds. Here we have a situation where we have an oxygen, one type of atom and another type of atom, the nitrogen, we have more nitrogen than oxygen. If you follow the heuristic I gave you, you'd be tempted to put the oxygen in the middle and the nitrogens around it. The problem with that is that you're going to end up with an oxygen that has four bonds on it, that's not something that you're ever going to see, so you will know that that can't be the right structure. As you do more and more of these, you'll see a certain number of bonds are just not possible on certain atoms. That would be a case of not putting the oxygen in the middle. Instead, we're going to put the oxygen on the end, because once again, we're not going to make a three membered ring, so the only other option is to go N-N-O.

If I use up the two bonds to put the minimum in the structure, now I have another two bonds to put and if I put one between the two nitrogens. I have a choice now, I can put another bond, make a triple bond between the nitrogens or I could make a double bond between the nitrogen and the oxygen. It doesn't really matter which one you do because those are both what we call reasonance forms for that structure, that's a good place to talk about reasonance. Let's say I put the double bond between the nitrogen and the oxygen. Let's complete the structure. If I complete the octets, I need four electrons on the left, nitrogen in the middle is fine and I need two electrons on the right. That gives me a nitrogen on the left that is negative one, and I have six electrons for charges. Nitrogen in the middle, that is plus, and an oxygen that is neutral.

We're talking about reasonance forms. I can draw this in a different way by moving the electrons around. Using this red marker, I'm going to draw an arrow from one of the lone pairs from the nitrogen. I'm using an arrow that has a full head, we'll see half-arrows later on in this course that are like the arrows we drew for the electron spin, for the up and down. This one means that two electrons move, whenever you see an arrow like that, it has to start from the source of electrons, usually a lone pair or a bond, and that's telling you to take those two electrons and put them between the two nitrogens. If I were to do that and just leave the structure like that, I have a problem because my central nitrogen would have five bonds and we can never have that happen. What happens at the same time is two electrons, between the nitrogen and the oxygen, move over and get dumped on the oxygen. Doing that at the same time, I never have a situation where I have five bonds on the nitrogen, it happens simultaneously. If we follow what this is telling us, it means put a triple bond between the two nitrogens and lead a single bond between the nitrogen and the terminal oxygen. What I've done by moving electrons around is I've shifted around the distribution of charge because the nitrogen on the left, if you count for charges, five, is now neutral. The nitrogen in the middle, nothing has changed. It had four bonds, still has four bonds, so that's still positive. But, the oxygen now has a negative charge, which it didn't before.

I'm going to draw an arrow like this double-sided arrow, to show that these two are reasonance forms of the same molecule. If you remember, reasonance forms don't inter-change, they're basically two different ways of looking at the same molecule. It's not that N2O splits between these two views, it's going to be an average of the two views and there's going to be a different contribution from one or the other, depending on the stability of each reasonace form. That means that the nitrogen in the middle, what's the charge on there? On one reasonance form it's plus one, the other reasonance form it's plus one, regardless of what contribution from each, the charge is going to be exactly plus one, that's a formal charge. The oxygen, however, goes from zero to minus one. Depending on which one of those reasonance forms has a lower energy, we know it's going to be a charge somewhere between zero and minus one. If you look at what the reasonance form really tells you, it's what's the probability of finding that extra charge on the oxygen. It's going to be a probability that it's greater than zero, but less than 100 percent because we have two reasonance forms here.

We're still using Lewis structures because we need Lewis structures to do reactions. These arrows here are what we are going to use to do reactions and you cannot use the arrow unless you have one of the reasonance forms. That's why we keep coming back to that, that's why they're useful. Do we have any questions on this?

Student: The formal charge? What is a formal charge?

Speaker: Yes, the formal charge, the plusses and the minuses that I've been drawing, those are all the formal charges.

Student: How did you do that [inaudible] but I know that we get the same charge.

Speaker: Let's pick a charge like the nitrogen on the left. You're asking me how I determine the charge on it?

Student: How you do the other [inaudible]

Speaker: I'm not sure for the other formula. The formula that I'm giving you, you count the electrons and the lone pairs fully, four, and then you count one electron per bond, six, and then you compare that with what nitrogen normally has.

Student: [inaudible]

Speaker: That's another way of stating the same thing. The way that I explain it is a little more intuitive because I'm not drawing a formula. I'm basically saying, "How many electrons effectively do I have there for charges? I have six. How many do I normally have? I normally have five." It makes sense if I have an extra electron. The electrons are negatively charged, therefore that should be minus one. Makes sense? Anything else?

We covered resonance forms a little bit, let me give you another example of a resonance form. When we looked at SO3, and I was putting down the last of the four bonds, I put it between one of the oxygens, Now why did I pick that specific oxygen, why only did I have reason to pick up one oxygen over the other. The reason for that is that basically I am drawing one of the three resonance forms of SO3. That double bond has no reason to select one of those oxygens. If I used the same kind of curved arrow formalism, I am going to do pretty much the same thing I did in the last example, I am going to start from one of the lone pairs, one of the negatively charged oxygen, dump those two electrons between the S and the O and at the same time take two electrons from the double bond and dump that on oxygen that is now neutral. Following what this tells me, I have removed two electrons from the double bond, each bond is two electrons. I have only one bond left, but I have three lone pairs and that's now negative. The oxygen that was negative is now neutral and has a double bond and everywhere else in the molecule that I haven't drawn an arrow to or from, you can't touch that. That's one of the fundamental things about organic chemistry. The way to not get confused is to really understand how the arrows work and to really follow what they tell you to do. You don't touch anything that you haven't drawn an arrow from or to, thats why this O minus stays like that on the left.

There is one more form that we can draw here, the oxygen on the left, two electrons can go between the S and the O making a double bond and I can move that double bond over like that. Now the double bond is on the left hand side and I have drawn the third resonance form.

Based on this, this is a little different than the last problem. The last problem we had a triple bond between two nitrogens versus a double bond between a nitrogen and oxygen. Those are two clearly different structures, they have two different energies, they have to have different energies. One will be more stable than the other and so we know that they will not be 50-50. But in this one, there's absolutely no difference between those three oxygens. We know for the fact that each one of these resonance forms represents exactly a third of what SO3 looks like. Now we can be accurate, we can say that the charge on the sulfur is plus two, it's the average of the three numbers. Each one of these forms has plus two, therefore the charge on sulfur is plus two. I'm going to say average charge is plus two.

Now, what about the oxygen on the lower left? What is the average charge? One third, you've got three structures. Follow what happens here, I haven't moved the atoms around. I have minus one, minus one and zero. That gives me minus two divided by three structures, so.
it will be minus two thirds. Again, the only reason we can be very confident about that is that we know that they are all the same energy.

If we go back to the previous problem, between these two forms we can probably take a guess as to which one will be a better representation. As you do more and more of these Lewis structures and start to see and draw atoms, you'll see there are certain things that you'll see all the time. One of the things that you see is a nitrogen with four bonds and a positive charge, that you will see all the time. The other thing you will see is an oxygen with the single bond and one negative charge on it, that's something we've seen many, many times over. Oxygen likes to have an electron on it, that's a fairly stable configuration. If you look at the periodic table, oxygen is more electronegative than nitrogen, so you can probably project that the negative charge will prefer to sit on the oxygen than sit on the nitrogen on the left. Seeing a nitrogen with negative charge is not something that you will never be forced to draw, but if there is a choice, the electrons would be rather on the oxygen. We can predict from this - this is a better representation, these are not equal, so we would predict that the oxygen would probably have a negative charge above minus 0.5, it would be something between 0.5 and one.

The other thing I want to talk about sulfur, that can be a little confusing depending on what book you use, you will sometimes see sulfur will have more than four bonds. For example, you might sometimes see sulfur drawn like this with double bonds on each one. That's something after you master how to do the Lewis structures that satisfy the octet rule, then that's fine, we can draw it like that. When you see it in books, it will make sense, but I will not accept this. When I ask you questions, I'm asking you for the Lewis structures that satisfy the octet rule. If you look at the quiz, I'll be asking you things like how many bonds are on an atom or how many lone pairs are there or what's the charge. I'm referring to the Lewis structures that satisfy the octet rule just the way we've done them, but you might be confused if you look in books and see more than four bonds on sulfur. It's a different convenience of writing it.

Any questions on resonance, hybrids, or curved arrow formalism? That's what I just introduced. Again, if you do the problems and look at the quiz, you can practice this material.

Next thing I want to look at is just a couple of definitions of Lewis skeletal and condensed structural formulas. What we've been doing are Lewis structures. For example, if you look at propane, when I draw a Lewis structure, I have no choice but to draw every single bond in the molecule. That's the only acceptable way to draw propane according to a Lewis dot structure. If I have formal charges, if I have lone pairs, if you miss any element of that, it's a wrong structure, Lewis is the most detailed structure. I'm going to use that when it's necessary, but sometimes it's not necessary. If I'm not doing anything with, say, the CH3 at the end when we're doing a reaction, I may not always draw it because I don't need to, but if I'm doing something to one of the hydrogens on that end methyl group, then I have to draw it out otherwise it won't make sense, what I'm drawing. I'll use it depending on the application. If I want to draw things a little more quickly, what I can do is use a skeletal formula or a skeletal structure and it's a lot quicker to draw. That's propane, it's just two lines. In a skeletal formula, each corner and each termination is a carbon. You're expected to add as many hydrogens as is necessary for each of those carbons to have four bonds. That's always going to be the case for normal molecules. It means I can draw structures that are very complicated really quickly.

Another example of a skeletal formula that you might be more familiar with is benzene. When you draw benzene like that, you don't put in the hydrogens, that doesn't mean they don't exist. You're expected to know that each carbon has three bonds, so there's one bond is left over and that will be a hydrogen. That's very convenient if you just want to draw a structure quickly.

Next, we have condensed. A condensed formula is what you might see, for example in the quizzes, you might see in the problem sets that are typed and it would look like this for propane. I'm just drawing CH3-CH2-CH3. The advantage of that is you can do that with a simple text editor and you also have some assumptions here. For example, if you write CH3-CHO, again you're expected to know that each carbon has four bonds. The only way you can draw this structure, if I translate that into a Lewis structure, it would look like this. CH3 is okay, then I have a carbon. The only way I can have only one hydrogen and only one oxygen on that carbon is if I have a double bond for the oxygen. Drawing CH3-CHO, when you draw this out to do a problem with it, you would draw it like that, C double bond arrow H.

There are other ways of representing molecules. For example, I can use a molecular formula. A molecular formula, of let's say, benzene, C6H6. The problem with a molecular formula is that it doesn't tell you how to connect up the atoms. For the very smallest ones, like we've been doing the Lewis structures of, there's probably only one really stable answer. When you start to add a whole bunch of carbons and hydrogens, there are many, many solutions to it and you can see that from the problem set. There are a couple problems like that, where you have to come up with three structures that match the molecular formula. Generally, you're going to want at least a skeletal formula to know what compound specifically you're dealing with.

The other kind of formula is an empirical formula. An empirical formula is the result you would get if you analyzed the compound, let's say for how much carbon and how much hydrogen it has. If you have an unknown compound, you can combust it and from the results, you'll know that for every carbon, you'll have one hydrogen. For benzene, the empirical formula would be CH because it's a one to one relationship and in an empirical formula you always give the lowest possible ratio, so C6H6 has to simplify to CH. That gives you even less information because it could be C2H2, C3H3, you just don't know. Basically, you use all of these different pieces of information depending on the problem that you are doing and the information that you are given.

Another thing I want to talk about in terms of definitions, is acids and bases. Probably the definition of acids and bases that you've been introduced to is the Lowry-Bronsted representation. A Lowry-Bronsted acid donates a proton. It donates an H-plus or a proton and a base would receive it. For example, if we have sodium bicarbonate, NaHCO3 and we throw in some HCl, what will happen is the H from the HCl will end up over on the HCO3 and now we'll have NaCl and H2CO3. Carbonic acid, and sodium chloride. Because the proton moves from the HCl to the bicarbonate, that defines the HCl as being the acid and the bicarbonate being the base.

A molecule by itself is not an acid or a base. It's only an acid or a base in relationship with another molecule. If I take sodium bicarbonate, again NaHCO3, and react it with NaOH, the proton from the bicarbonate, will be the one that's moving. That will give me Na2CO3, and H2O. In this case, the bicarbonate is actually the acid and the hydroxide is the base. That's fine, but in organic chemistry, we need a little bit more broader definition of acid/base for the kinds of things we're going to look at, that would be the Lewis acid/base. I think the easiest way to remember the Lewis acid/base, is the Lewis base donates a lone pair and the Lewis acid accepts the lone pair.

So again, we have a situation where we don't have a transfer of protons, but we have an acid/base reaction. An example would be ammonia, NH3, reacting with boron trifluoride, BF3. Again, using the curved arrow formalism, if I start the arrow at the lone pair on the nitrogen and I dump it on the boron, then I make a bond or a complex between the ammonia and the boron, the boron now becoming negative and the nitrogen becoming positive. In this case, the ammonia is the Lewis base because it donates a lone pair, BF3 is the Lewis acid because it accepts the lone pair. That's basically all there is to Lewis acid/base. It's something that will come up again and again throughout this course.

Let's move on from doing Lewis structures. Let's try to understand how to predict the shapes of these things. Like I said, the only thing I was focusing on in the beginning was you knowing how many bonds and electrons and the charges. Now let's see if you can predict these shapes. In order to do that, we're going to have to look at molecular orbitals. When we look at electronic configuration, the very first thing that we looked at in this course, I did the first ten atoms in the periodic table and showed how the electrons show up in the shells, in the orbitals. When you make molecules out of collections of atoms, you also get orbitals, but they're molecular orbitals. Something happens when you go from an isolated atom to an atom that is participating in a covalent bond. Let's take a look at the simplest example, two hydrogen atoms coming together to make H2. If I go back to the electronic configuration, hydrogen is very simple, I just have one electron and the one s. I have one electron in the one s from each hydrogen, they come together and now we end up making H2.

Graphically, you'll remember the s orbitals are a sphere, so I've got two spheres coming together. When they overlap, I form a bond. There are my two s orbitals. The bond that we're making in the case of H2 is a sigma bond because the two orbitals are coming in head to head. We'll see an example of where that's not the case and that'll be a different type of bond, but this is a sigma bond. In this case, we didn't have to do anything to the s orbitals because it's pretty obvious, I have one electron in each, they come together, two electrons make one bond and everything works out. The problem is that doesn't work if you try to do the same thing with carbon.

Let's say I tried to show what CH4 would look like. With CH4, if I go back to the electronic configuration of carbon, we have 1s, 2s, 2px, 2py and 2pz. I have six electrons total in carbon. We want to make four bonds with four different hydrogens. Hydrogen has one electron and the one s. The problem with this is I have only two unpaired electrons in carbon, so the way that I drew this, it doesn't look like there's anyway for me to make more than two bonds to the carbon, and that's true, if the carbon stayed in that form, you could never make CH4. What happens, the valence electrons, remember what those are, the 2s and 2p, they'll actually mix or what's called hybridize. The 1s does not changes at all, it's never involved in the bonding of carbon and the 2s and the 3p orbitals end up giving you four orbitals of identical energy that are called sp3. So, I have four sp3 orbitals. With those sp3 orbitals, I now have four electrons, each one of these single electrons can form a bond with the hydrogen's single electron.

This process, again we'll do this a couple times, is called hybridization. We say that the one s and the three p orbitals hybridize to become transform into four seperate sp3 orbitals. The reason they're called sp3 is that I use up one s orbital and three p orbitals, that's pretty straightforward. The thing is, what do these look like? You know what a p orbital looks like, you know what an s orbital looks like, but what does an sp3 orbital look like? It turns out if you calculate this, you find that it looks like a tetrahedron. The four sp3 orbitals arrange themselves around the carbon like this. I'll give you some better ways to represent a 3D structure, but this'll do. We have four lobes and what I'm trying to show here is a tetrahedron. We have like a little pyramid at the bottom and then you have one of the lobes coming up. Those four orbitals overlap with the one s orbital of the hydrogen, each one of them. Basically, the overlap between the sp3 orbital and the s orbital is a sigma bond, each one of those is a sigma bond. Continuing the labeling here, the red part, each one is an sp3 orbital and the circles are s orbitals from the hydrogens. So this is a very rough drawing, I'm trying to emphasize the fact that the overlap - if you look in any organic chemistry book, it looks a little bit better - It merges with the s orbitals, it's not exactly like this, but you can think of it that way, that they overlap, and that overlap creates the bonds.

This is tetrahedral. We've used one s orbitals, and three p orbitals, is there another way that orbitals can hybridize? Yes, if we look at ethylene, we have a double bond. Again, let's draw the electronic configuration for carbon, 1s, 2s, 2px, 2py, 2pz. Fill it in with six electrons and then we have the hydrogen that just has one in the one s. In this case, carbon needs to form bonds to only three atoms, it doesn't have to form bonds, like in CH4, it has four bonds to make, in ethylene it only has to make three bonds. It's not going to use an sp3 hybridization, it only needs two of the p orbitals, and one of the s orbital, so it's going to be an sp2. What that means is it will use up the one s and two of the p orbitals. Let's keep track of this, the one s does not change, now I have instead of a different s orbital and two p orbitals, I'll have three identical sp2 orbitals. I have not used up one of the p orbitals, that doesn't disappear, that remains as a p orbital. We just drag that p orbital down, that's going to make a difference, as you'll see shortly. What that means is each one of these sp2 orbitals will make a bond with either a carbon or a hydrogen, so we're going to put one electron in each of these sp2 orbitals. That leaves one of the electrons left and that electron has to go into the p orbital. Again, when we label, this is the hybridization.

This one is actually a little bit easier to draw than the sp3 because it's flat. At least it's initially easier to draw. If I put the carbons like this, I will put the hydrogens at 120 degree angles. The sp2 orbitals are flat, they're coplanar. I'm drawing each one on each carbon, the hydrogen still is just a sphere, it's just the one s. The overlap between the carbon and the hydrogen is the bond. Again, we have one lobe ramming into another lobe, that is the definition of a sigma bond, head to head interaction. If you look at the two sp2 orbitals in the middle, they're also getting into each other head to head, so that's also a sigma bond between the two carbons. The only thing I haven't labeled here are the orbitals on the H and the C, there's the s orbital. Each one of the lobes around the carbon is an sp2 orbital. That's basically how it's laid out. What you're looking at right here is flat, it's in the plane of the screen.

One thing I have not accounted for is the leftover p orbital. That actually is not in the plane of the screen, that will be perpendicular to the screen. It's a little hard to draw and again, if you look in the book, you'll get nice computer generated pictures, but I'm showing you how to construct this. This is flat, I'm going to try to draw something with two lobes, like this, which is the leftover p orbital. You want to picture these with one lobe coming out towards you, one lobe behind the screen and they're each one p orbital. What happens with these p orbitals, they each have one electron and they can overlap onto each other, but they don't overlap head to head, they overlap side to side like this. That's not a sigma bond, that's a pi bond.

The pi bond is very different from the sigma bond in that it is distributed evenly between the two sides of the molecule. I drew a yellow area on the top, I drew a yellow area on the bottom, that's not two pi bonds. That's the same pi bond, it's just distributed across the space like that. I'm going to try to show that by putting the two arrows like that.

What I'm really drawing here is the probability of finding an electron in space, there's 50 percent chance of finding it on one side and 50 percent chance on the other. The only thing I haven't drawn here is the p, there's one p orbital and there's the other p. Like I said, this is not a great picture, it shows you how to construct it from scratch, but you'll probably want to take a look at a computer generated picture to see really how the distribution works out.

What I think I'll do is I'll pause here, when we resume we're going to look at sp hybridization, and after we go over a couple things, we'll do some problems. That's it, we'll see you next week.


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