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Predicting the molecular dipole moment based on the molecular geometry.  Created by Jay.
Video transcript
Now that we understand how to draw dot structures and we know how to predict the shapes of molecules, let's use those skills to analyze the polarity of molecules, using what's called the dipole moment. So to explain what a dipole moment is, let's look at this situation over here on the right, where we have a positively charged proton some distance away from a negatively charged electron. And let's say they're separated by a distance of d here. We know that a proton and an electron have the same magnitude of charge, so both have a magnitude of charge Q equal to 1.6 times 10 to the negative 19. So of course, a proton would have positively charged Q, so let's go ahead and make this a positively charged Q. And an electron would have a negatively charged Q, like that. If we were to calculate the dipole moment, the definition of a dipole moment, symbolized by the Greek letter mu, dipole moment is equal to the magnitude of that charge, Q, times the distance between those charges, d. So mu is equal to Q times d. And we're not really going to get into math in this video, but if you were to go ahead and do that calculation, you would end up with the units of Debyes. So you would get a number, and that number would be in Debyes here. So we're more concerned with analyzing a dipole moment in terms of the molecular structure, so let's go ahead and look at the dot structure for HCl. So if I look at this covalent bond between the hydrogen and the chlorine, I know that that covalent bond consists of two electrons. And chlorine is more electronegative than hydrogen, which means that those two electrons are going to be pulled closer to the chlorine. So I'm going to go ahead and show that here with this arrow. The arrows is pointing in the direction of movement of electrons, so those electrons in yellow are going to move closer towards the chlorine. So chlorine is going to get a little bit more electron density around it, and so we represent that with a partial negative charge. So we do a lowercase Greek delta here, and it's partially negative since it has increase in electron density, one way of thinking about it. And since hydrogen is losing a little bit of electron density, it's losing a little bit of negative charge, and so it is partially positive. So we go ahead and draw a partial positive sign here. And so we're setting up a situation where we are polarizing the molecule. So this part of the molecule over here on the right is increasing electron density, and so that is our partial negative side. That's one pole. And then this other side here is losing some electron density, and so it's partially positive, so we have it like that. So that's where the positive sign comes in. You can think about on this arrow here, this little positive sign giving you the distribution of charge in this molecule. And so you have these two poles, a positive pole and a negative pole. And if you think about those two poles as having a center of mass, you could have a distance between them, and you could calculate the dipole moment for this molecule. And so when you calculate the dipole moment for HCl, mu turns out to be equal to approximately 1.11 Debyes. And so we have a polarized bond, and we have a polarized molecule. And so therefore we can say that HCl is relatively polar. It has a dipole moment. So that's kind of how to think about analyzing these molecules. Let's do another one here. Let's do carbon dioxide. So I know that the CO2 molecule is linear, so after you draw the dot structure you're going to get a linear shape, which is going to be important when we're trying to predict the dipole moment. If I analyze the electrons in this carbon-oxygen bond-- so we have a double bond between carbon and oxygen-- oxygen is more electronegative than carbon. So oxygen's going to try to pull those electrons closer to itself. And so we go ahead and draw our arrow or vector pointing towards the right here. And so we have a bond dipole situation here. On the left, we have the exact same situation. Oxygen is more electronegative than carbon, and so these electrons are going to be pulled closer to this oxygen. So we draw another arrow or another vector in this case. So even though we have these individual bond dipoles, if you think about this molecule as being linear-- and you can see we have these two vectors that are equal in magnitude, but opposite in direction-- those two vectors are going to cancel out. And therefore we would not expect to have a dipole moment for the molecule. There's no molecular dipole here. So mu turns out to be equal to 0. A simplistic way of thinking about this would be like a tug of war. You have these really strong atoms, these oxygens, but they're equally strong. And if they're pulling with equal force in opposite directions, it's going to cancel out. So the individual bond dipoles cancel out, so there's no overall dipole moment for this molecule. And carbon dioxide is considered to be nonpolar. Let's go ahead and analyze a water molecule over here on the right. So the electrons in this covalent bond between the hydrogen and oxygen, oxygen is more electronegative than hydrogen, so those electrons are going to be pulled closer to the oxygen. Same thing for this bond over here. And we also have lone pairs of electrons on our central atom to think about. And that's of course going to increase the electron density going in this direction for that lone pair and in this direction for that one pair. And so even though we know the geometry of the water molecule is bent, and it's hard to represent that on this two-dimensional surface here. If you use a molymod set, you will kind of see that your net dipole moment would be directed upward in this case. And so the individual bond dipoles are going to add to give you a molecular dipole, in this case pointed up, and so therefore you're going to have a dipole moment associated with your water molecule. So mu turns out to be approximately 1.85, and we could consider water to be a polar molecule. Let's do two more examples. So on the left is CCl4, or carbon tetrachloride. And so you can see that we have a carbon bonded to chlorine here, and since this is a straight line, this means in the plane of the page. And so we know the geometry is tetrahedral around out this carbon, so let's go ahead and analyze that as well. So I have a wedge drawn here, which means this chlorine is coming out at you in space. And then I have a dash back here meaning this chlorine back here is going away from you in space. So that's how to think about it, but it's really much easier to go ahead and make this using a molymod set. And you can see that however you rotate this molecule, it's going to look the same in all directions. So a tetrahedral arrangements of four of the same atoms around a central atom, you can turn the molecule over. It's always going to look the same in three dimensions. And that's really important when you're analyzing the dipole moment for this molecule. So let's go ahead and do that. We'll start with our electrode negativity differences. So if I look at this top carbon-chlorine bond-- these two electrons in this top carbon-chlorine bond-- chlorine is more electronegative than carbon. And so we could think about those electrons being pulled closer to the chlorines. Let me go ahead and use green for that. So those two electrons are going in this direction. And it's the same thing for all of these chlorines. Chlorine is more electronegative than carbon, so we can draw these individual bond dipoles. We can draw four of them here. And in this case we have four dipoles, but they're going to cancel out in three dimensions. So again, this is a tough one to visualize on a two-dimensional surface. But if you have the molecule in front of you, it's a little bit easier to see that if you keep rotating the molecule, it looks the same. And so these individual bond dipoles cancel, there's no dipole moment for this molecule, and so mu is equal to 0. And we would expect the carbon tetrachloride molecule to be nonpolar. Let's look at the example on the right, where we have substituted in a hydrogen for one of the chlorines. And so now we have CHCl3, or chloroform. So now if we analyze the molecule-- so let's think about this bond in here-- carbon is actually a little bit more electronegative than hydrogen, so we can show the electrons in that bond in red moving towards the carbon this time. And once again, carbon versus chlorine, chlorine is more electronegative, so we're going to have a bond dipole in that direction, which we can do for all our chlorines here. And so hopefully it's a little bit easier to see in this case. In this case, the individual bond dipoles are going to combine to give you a net dipole located in the downward direction for this molecule. So I'm attempting to draw the molecular dipole, the dipole for the entire molecule, going a little bit down in terms of how I've drawn this molecule. And so since we have a hydrogen here, there's no upward pull in this case to balance out the downward pull. And so we would expect this molecule to have a dipole moment. And so mu turns out to be approximately 1.01 for chloroform, so it is certainly more polar than our carbon tetrachloride example.