Thursday, May 18, 2006

Lecture 005: Molecular Orbitals and Polarity

Lecture 005: Molecular Orbitals and Polarity
Speaker: Jean-Claude Bradley

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Speaker: We're going to resume with molecular orbitals, specifically we're going to be doing sp hybridization and then we can talk about the shapes.

We're going to use acetylene to show the hybridization, that's sp. This is very similar to the sp3 and sp2, it's just that we're going to take only one s orbital and only a p orbital and we're going to leave two p orbitals by themselves. Let's do the same thing, here's the atomic orbitals of carbon. We have 1s, 2s, 2px, 2py, 2pz, six electrons. Just like in the other two cases, a carbon atom by itself, not bonding inside of a molecule, has two lone electrons, two unpaired electrons. That's not enough for the bonding that we're going to require in acetylene, so what we're going to do is take the s orbital and one of the p orbitals. Same process, hybridization, and this is what we'll have left, the 1s is untouched. Now I have two orbitals that I'll call sp each and I have the two leftover p orbitals.

Now, when I fill this up, I put the two electrons in the 1s and the sp orbitals are used for bonding, so they'll contain only one unpaired electron, the other electron will come from the bond that it's making and then we have the last two electrons, each one in one leftover p orbital. That's the new hybridization and the hydrogen is the same. Hydrogen is not hybridized; it just has one electron in the one s orbital.

Let's draw a picture of this, this acetylene. Let's first draw the two sp orbitals. The sp orbitals are opposite to each other and we have bonding with a hydrogen on the left, so there's my s orbital overlapping with the sp orbital. Then I have the two sp orbitals from both carbons coming together and making a bond and the other hydrogen. I'm going to label this, there's an s orbital, there's an sp orbital, there's another sp orbital, we've got four sp orbitals in this molecule. If we look at the bonds, colored in blue here, we have an overlap between the s orbital and the sp orbital, that overlap is a bond. The two lobes are coming directly into each other, so that's a sigma bond. We have three sigma bonds in this molecule, of course there's the same thing happening on the other side between the other carbon and the other hydrogen. But, we also have those two sp orbitals of the carbons coming in to each other. The overlap of those two sp orbitals is head to head, so that's also a sigma bond. I'm drawing that third sigma bond here in blue, that overlap.

We've drawn now all the sp orbitals; we need to account for the p orbitals. The p orbitals are perpendicular to the sp orbitals, the first one will be easy to draw, it will be in the plane of the board. There's one of the p orbitals from one of the carbons, there's the other p orbital and just like in the case of ethylene, we have a side to side bonding between the two p orbitals and that's a pi bond. The pi bond, half of the pi bond will be on the upper portion of the board here, and half of it will be on the lower portion. Let's label this. Again, I'm going to show both sides here, make up one pi bond. So far we have one sigma bond and one pi bond. But, we have three bonds in acetylene; the third bond will come from the side by side overlap of the last p orbital. This one is a little trickier to draw because it's perpendicular to these orbitals right here, so what we have to do is draw another orbital coming in and out of the board. I'll do my best to represent that here, but it will be a little bit tricky. Let's make this one yellow. We have the two yellow orbitals coming directly towards you. In light blue, we have that second overlap. Here's the second pi bond. A little bit more labeling here, there's a p orbital, there's another p orbital and I think with that we have everything labeled. That's basically how a bond forms, so now, when you look at acetylene, there are three bonds, one of those bonds is a sigma, the other two are pi bonds. Questions?

We took a lot of time going through the molecular orbitals, trying to figure the shapes of all these orbitals. Let me just summarize the shapes by drawing them out. This is an approximation, if you take a look at your book, you'll see some nice computer generated graphics of the fine details of these, but the key things to remember are the simple shapes. An s orbital would just be a sphere and a p orbital would be like an infinity symbol. Now, sp orbitals, I will always draw all the sp orbitals because it's the relationship between the sp orbitals that counts. It actually looks very similar to a p orbital at first approximation. Again, if you look at the exact details of it, there are more details there, but the key thing is that we have two lobes that are pointing opposite of each other.

Sp2 is flat, and again, these are three sp2 orbitals. The sp3 are tetrahedral. This one's always trickier because it's a 3D structure that I'm attempting to draw. I'll show you some ways of representing a tetrahedral structure on a flat surface, but for now, just note that this is a 3D tetrahedral representation. The key thing that counts here is the angle between the lobes because that's what actually determines the shape of the molecule. Instead of having to go back and look though all of these notes to try to remember how all this connects, I'll summarize it for you in a table and give you a way of remembering it that's a little bit easier. Really, all you're going to have to worry about is sp, sp2 and sp3, so that's the hobs. The easiest way to figure out the hybridization is to count the groups of electrons around an atom, by a group of electrons; I'm referring to a group of bonds, bonds that are all together or a lone pair. I'll do several examples of that so you can see how you count them.

Whenever you have two groups of electrons, you'll have sp hybridization. Three groups will be sp2, four groups will be sp3. That's much easier than going through all the diagrams. What you need to know are the angles between. So, sp, the two lobes are opposite to each other and that corresponds to a 180 degree angle, sp2 is 120 and sp3 is 109.5. That's where you get the information about the shapes of the molecules. If we try to name these, the first sp will be linear, sp2 will be trigonal planar and the sp3 will be tetrahedral.

If I go back to some of the molecules that I gave you when we first did the Lewis structures, I told you not to worry about shape, just figure out how many electrons you have on the chart. Let's revisit these and see how you can predict the shape.

Let's look at H2 first, I'm not going to redo all of these, but I'll sample it. For H2, if you think about the hybridization, the H has a 1s orbital, we don't hybridization the 1s orbital. The only time you're going to do hybridization is when you mix s and p together. In this case, you just have two s orbitals and you can't form more than one bond to hydrogen, so it's always going to be in the same place, it's always going to be directly connected by one bond. I'm going to write here "no hybridization, just s orbitals."

Let's look at CH4. Now, when I drew CH4, I told you that it's not really the way it looks, well, now we now how it really looks. Count the number of groups of electrons around the carbon; we have four groups. In this case each bond corresponds to one group of electrons. If we go back to the little table, four groups correspond to sp3 and it will have 109.5 degrees and it will be tetrahedral. One way of drawing that is like this; I put two of the hydrogens in the plane of the board and I put another hydrogen coming out towards you and I show the fourth one in the back. That's one way to represent tetrahedral structure on a two dimensional surface.

The wedge that's filled in, that will always correspond to a bond that is coming towards you, it's going out of the page. A dotted bond generally shows an atom that is behind the page. If you look at this, this is one of the ways to represent a tetrahedron and I'm going to use this to make a certain point. Later on, I'm going to have to make another point and I'll have to use a different representation, but this one is convenient for what we are doing now.

The angle is the same, 109.5 degrees. Basically, that number comes from the fact that that's the farthest that you can put four groups around an atom and that's generally what's happening here. You'll notice that sp is 180, that's the farthest you can put two groups around one atom, 109.5 degrees is the largest angle you can have for four groups.

Let's look at another example that doesn't just have bonds; let's look at ammonia, NH3. Now, NH3 has four groups of electrons because I count each bond as one group and I count the lone pair as a fourth group. Four groups of electrons, again is sp3. Whenever I talk about hybridization, I'm always talking about an atom; I'm not talking about the molecule. The molecule has more than one atom, each atom will have it's own hybridization, but I'm only concerned about the nitrogen because the nitrogen is what gives shape to the molecule. What I'm trying to get here is what does this thing look like, so only the hybridization of nitrogen is really telling you that.

Now, sp3, again I've drawn this tetrahedral structure with the lone pair in the plane of the board, one of the hydrogen is in the plane, one hydrogen is coming out towards you and another one is pointing towards the back. The angle that we predict here, between all the bonds, between the two hydrogens and between the lone pair; in any of the hydrogens, we would predict 109.5. In reality, that's not exactly the angle because the lone pair takes up a little bit more room than the three sigma bonds, the angles might be a little off, by one or two degrees, but when I'm asking you the angles, I don't expect you to know that, so 109.5 will be the number that I'm looking for. Just keep in mind that unless you have four identical groups, it's not going to be exactly 109.5.

Let's look at BF3. Boron is a little confusing sometimes. Boron doesn't follow the octet rule, but that has nothing to do with the hybridization, the hybridization is going to be the same thing. When we look at boron, it's got three electrons; each one of those electrons will make a bond with fluorine. To complete the structure, I have to draw all the lone pairs on here.

If I'm going after the shape of a molecule, hopefully it's obvious that it's the hybridization of the boron that will give you the overall shape. There are no lone pairs on the boron compared to nitrogen, so how many groups of electrons around the boron? It's just three bonds. Three groups of electrons, go back to our little table here, that corresponds to sp2, corresponds to 120 degrees in trigonal planar. I know immediately, knowing what the hybridization is, I know that BF3 is flat and I know that the angle between the fluorines is 120 degrees.

Each atom here has it's own hybridization. What would the hybridization of the fluorine atom? Sp3. We have four groups of electrons; we have three lone pairs and one bond. The hybridization of the fluorine, sp2 would be the boron, but each fluorine would be sp3. The hybridization of the fluorine doesn't change the shape of the molecule; it just changes the relative angles between the lone pairs and the fluorine, so that is not going to change the way that you draw fluorine. I'm going to be asking you which specific atom that I want the hybridization for, if you look at the quiz, you'll see pretty much how that goes.

Let's see what else I can do here, I'll just pick one here. Azide, N3-. (N3 minus) If you remember, azide has negative charge on the end and positive charge in the middle. One of the things that you have to get straight about these formal charges is that they're not objects. When I put a negative symbol or a positive symbol, it's just a sign that that nitrogen has a plus charge, but it's not an object that you can do things with. One of the confusions is to count that as a group, it's not. You just ignore the charge, and the nitrogen in the middle has two groups of electrons, it has two double bonds. So, that would be two groups, which if we look at the table, corresponds to sp and 180 degrees. That's how you know that azide is straight; it's a linear molecule.

When we look at the hybridization of the terminal nitrogens, we have three groups, which will be sp2. For those nitrogens, the ends don't control the shape of the molecule, so you do need to know the hybridization, but it won't account for the shape.

I think I covered pretty much every case here, any questions about how to count a group? We'll go through another problem set; maybe something else will come up. That's pretty much all there is for that.

One of the things I'd like to do is revisit the example of ethylene. Ethylene double bonds to two carbons. I'm going to try to quickly sketch this. Each carbon has sp2 orbitals and each hydrogen has just an s orbital and what we have left over here are the p orbitals. I'm not going to label everything here; we already spent quite a bit of time doing that. The key thing I wanted to show is that the pi bond in this molecule acts like a rod that connects both sides of the double bond. You can think of it as each carbon atom is like a wheel on a car. If two wheels only have an axle between them, they can rotate relative to each other. But, if you put a rod through the top of a wheel, through the two front wheels, and another rod at the bottom, you effectively lock each wheel relative to the other. That means that if I turn one, I have to turn the other at the same time. That does something really important; it generates a new kind of isomer, that maybe you haven't looked at yet. Geometric or stereoisomer, I'll show an example. It means that I have two different groups, for example if I have two methyl groups, there are two ways in which those methyl groups can be located, either on the opposite sides of the double bond or on the same side. The reason that you have two different molecules is that you cannot rotate around a carbon-carbon double bond. You can rotate around a carbon-carbon single without much difficulty, but here it actually creates two different compounds that will have different properties. These are called geometric isomers. The ones that have the two groups that are the same on the opposite sides will be called trans. The others with the two groups that are the same, for example the methyl groups, on the same side will be called cis. Basically, that's a new kind of isomer that now you have to worry about every time you have a double bond and that's because of the pi bond.

The other kind of isomer would be a structural isomer. If I look at 1-butene, these, by the way, are both 2-butenes. If I look at 1-butene, it has the same number of carbons, the same number of hydrogens as 2-butene, but the connection between the atoms is different. In the case of 2-butene, I have CH3-CH=CH-CH3, but down here I have CH3-CH2=CH-CH2. Same number of atoms, but they are connected differently and that's what makes your dipole kind of isomer, the structural isomer. When you look a 2-butene, the trans and the cis have the same connectivity and so they are going to be geometric isomers, which is the difference. We're going to be doing some problems with that in the problem section.

Let's talk about the bond polarity. Let's start to predict some of the properties of these compounds based on doing the Lewis structures and getting the geometry right. There is a reason why we're spending a lot of time trying to figure out what these molecules look like because it will dictate how they behave and you need to know how to predict that.

The first thing I'll do is I'll draw methane, showing the correct tetrahedral geometry. We have four bonds and if you look at the electromagnetivity of hydrogen and the electromagnetivity of carbon, they're very similar. What happens is, if you're looking for what's the probability of finding an electron, it's going to be pretty much the same on the hydrogen or on the carbon. We say that those bonds are non polar, so anytime you see a carbon and a hydrogen, you can assume that those are going to be non polar. If I have four non polar bonds, then I can say that the entire molecule is non polar. Bonds are non polar and molecules are non polar.

If I replace one of the hydrogens with an electronegative element, like chlorine, this is what we end up with. We've got three non polar bonds, the carbon-hydrogen bonds, and we have one polar bond on the chlorine. What that means, if an atom is electronegative, it means it likes electrons, it will attract electrons towards it. We're going to put a delta symbol on the chlorine and a negative; we're going to say that the chlorine is slightly negative. This is not the same as a formal charge. A formal charge, if you put a plus there, that means plus one. This is just a slight charge, a slight increased probability of finding the electron on the chlorine, so it's not a formal charge.

Removing the electron density from the carbon, that makes it slightly positive. Whenever you have a separation of charge, even if it's a partial charge, you generate an electric field, just like you would in a capacitor on a macroscale. You can represent that electric field with an arrow, the same way that you do in physics and it is always from plus to minus. We would draw an arrow to represent that electric field, in the case of a molecule, we call that a dipole moment.

In chloromethane, we've got three non polar bonds and one polar bond. Because we don't have anything that's counterbalancing that one polar bond, the molecule itself will be polar. Let's keep adding chlorines and see where it takes us. Let's say I put two chlorines on that carbon and I still have two hydrogens left. Again, we don't look at the carbon-hydrogen, those are non polar. Each carbon chlorine bond will be polar, it will show slightly negative on the chlorine, slightly negative on the carbon, slightly negative on the chlorine and I'll use a different color here so it will be clearer. That means that we have one arrow going directly up and one arrow going to the right and down.

Just like in physics, whenever you have two vectors, you can add them up. You find that both of these arrows, if you imagine having a tree and tying ropes to it and pulling it, which direction is it going to fall, the way I drew the arrows, it will go up and to the right. The overall dipole moment for that molecule will be up and to the right, the way that I drew it here. I'll use yellow for that. That's the overall dipole moment for the molecule and the other two; we've got two polar bonds in the molecule. Two polar bonds, two non polar bonds and we say that the molecule is polar.

Let's add another chlorine to this. I'm going to have too much stuff on here, so I'm not going to put the plus and the minus; I'm just going to put the arrows. We've got three arrows and the reason I put the chlorine in the position I did, is because it's easier to see. You'll notice that they're not canceling out. Basically the three arrows, they're pointing in different directions, but they're all pointing down, that's the one thing they have in common. When we look at the overall dipole moment of this molecule, it will be positive and it will be pointing downward. We've got three polar bonds and the molecule is also polar.

Now, what happens if we have four chlorine atoms? Well, we have four vectors. Although it's a little difficult to see, drawing it like this, basically a tetrahedron represents a structure where the four groups are as far away from each other as they can be. It turns out that this is a completely symmetrical structure and if you are pulling evenly on those four bonds, they're all going to cancel out. That's the consequence of having a tetrahedral structure. We still have four polar bonds, but the molecule is non polar. Basically, that's the one thing you have to watch out for because I'll ask you, "Is this molecule polar or non polar?" and you'll have to go through the process of generating the Lewis structure, analyzing the shape, drawing the vectors and then figuring out if all these vectors cancel out or don't they. That's why you have to know the shapes to answer that question.

What happens in cases where we have formal charges? In the case of SO2, we have a negative charge on the oxygen and we have a positive charge on the sulfur. Whenever you have formal charges, that will dominate the electric field vectors in the molecule. Any slight electronegativity differences between the atoms, you can pretty much ignore. What we need to look at here is you have plus one and a negative one, you go from plus to minus, it will be a very strong dipole from the sulfur towards the oxygen. When you have formal charges, you just look at the charges, you put them down.

There's a complication with sulfur dioxide in that it has different resonance forms. So far, when I did CH4, CCL4, all of those, there was only one resonance form that you could draw, so you don't have to worry about any other complications, but that's not the case in SO2, there are two resonance forms. Whenever that happens, you need to draw each resonance form and figure out, for each one, where the vector lays. Let's do that. On the other resonance form the double bond will be on the left, single bond on the right, plus charge hasn't moved, it's still on the sulfur. Same thing, we draw an arrow from the sulfur to the oxygen.

If you think about what resonance actually means, resonance is a way of looking at the structure of a molecule using different representations. But, the molecule doesn't look like the hybrid on the left, doesn't look like the hybrid on the right and it's not going back and forth between the two. It actually looks like the average between the left and the right. What you need t do when you have more than one resonance form is superimpose the vectors from each. If I translate those vectors down, I have one vector going down and to the left, I have another one going down and to the right; clearly, they don't cancel out. What's left will be just a single arrow going down. What we predict for SO2 is that it will be a polar molecule and the direction of that dipole will be right between the two oxygens from the sulfur. Any questions?

Let's take a look at one more example. SO3 has three resonance forms. You're going to have two oxygens that have a negative charge; you have the sulfur that has a plus two charge. A plus two charge doesn't change in a three hybrids. Let's step through this and see what happens in terms of dipoles. The structure on the left, we have formal charges, so we're going to use that to draw the arrows. Always go from plus to minus, so we go from sulfur to each one of the negatively charged oxygens, that generates a vector pointing directly down for the left structure. The middle structure has vectors pointing down and to the left and straight up. That adds up to up and to the left. The third structure has a vector going straight up and another one going down and to the right. Those two add up to going up and to the right. Just like we did for SO2, we're going to translate those three vectors onto the same grid. This one going here, this one going up and to the left and this one going up and to the right. So, you're pulling with equal force in three different directions, hopefully it's obvious that those all cancel out. For SO3, there's no overall dipole moment, it's not polar. This one is a little bit more complicated than the other ones, it incorporates everything. It's got the resonance, it has the formal charges and it has an example of where all the vectors cancel out, which is important. Any questions on that?

Okay. I think we're probably going to be ready to start the problem set on Wednesday, so try to have it done, it will make more sense to you.


At 3:06 PM, Blogger Doug said...

When I hear about fluorine forming covalent bonds I usually hear that and sp3 orbital from some other atom shares an electron from a p orbital from fluorine. This always made me think that fluorine doesn't hybridize. But does it? I thought also that maybe it doesn't hybridize because of all of the torsional strain from having the lone pairs so close to each other in a sp3 tetrahydral geometric shape. That it was possible of lower energy for the fluorine atom to maintain a 2s orbital and 3 2p orbitals (with one of the 2p orbitals only holding to 1 electron).


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