Let's start with a question. Which of these two molecules has a greater dipole moment? Is it methyl chloride, also known as chloromethane, or is it carbon tetrachloride?
What would you say? Well, let's analyze the CH bond and the CCL bond before we talk about the dipole moment of the entire molecule. Carbon has an electronegativity value of 2.5.
and for hydrogen it's 2.1. So because carbon is more electronegative than hydrogen, it's going to pull the electrons toward itself, and so the bond becomes slightly polarized. Carbon will bear a partial negative charge and hydrogen will bear a partial positive charge.
However, because the electronegativity difference is less than 0.5, the C-H bond is defined to be relatively nonpolar. even though there's some small Eon difference. Now for the CCL bond, chlorine has an electronegativity value of 3.0 and so because it's more electronegative than carbon, it's going to pull the electrons toward itself.
So this time, chlorine has the partial negative charge and carbon has the partial positive charge. Now the carbon-chlorine bond is considered to be a polar bond because the electronegativity difference between carbon and chlorine is 0.5. If it's 0.5 or more it's defined to be polar.
Now notice the arrow is bigger for the CCL bond than for the CH bond because the electronegativity difference is bigger. And also chlorine is a lot bigger than hydrogen. The dipole moment depends on Not just the charges, but also the distance between the charges. And chlorine is a lot bigger than hydrogen.
And so the carbon-chlorine bond is going to have a much larger dipole moment than the carbon-hydrogen bond. The dipole moment is the product of the charge and the separation distance, or the distance between the opposite charges. So charge and the size of the atoms matter.
So for bigger atoms, the dipole moment will increase. Keep that in mind. Now let's analyze the molecular dipole moment of methyl chloride. So we have a chlorine atom, and we have three hydrogen atoms. Now as we said before, chlorine is more electronegative than carbon.
So to draw the dipole moment, the arrow will point towards chlorine. Now carbon is more electronegative than hydrogen. So we're going to draw a smaller dipole moment because those bonds are less polar.
But notice that all of the arrows, they point in one general direction. And that is they point upward. If you take the vector sum of those arrows, you'll see that the net dipole moment, it points towards chlorine.
And so overall, this is a polar molecule. The dipole moment is greater than zero. In fact, the experimental dipole moment of methyl chloride is 1.87, where D is the unit debi. Now what about carbon tetrachloride? What do we know about its molecular dipole moment?
Well, let's draw the structure for CCL4. So this time we have four chlorine atoms and no hydrogen atoms. So drawing the dipole moment of each carbon-chlorine bond is going to look like this. All of the arrows will point towards chlorine. And notice that all of the arrows emanate away from the carbon atom.
And so it happens in such a way that all of these arrows, they completely cancel. Keep in mind, this is not a 2D structure. It's a 3D tetrahedral structure. So the net dipole moment for CCL4 is zero. It completely cancels.
Even though the molecule has polar bonds, overall, this is a nonpolar molecule. Now, let's look at another example. Which of these two molecules have a greater dipole moment?
Is it the trans isomer on the left? or the cis isomer on the right. Feel free to pause the video and work on that problem. So let's start with the isomer on the left. So we know that the arrow is going to point towards the more electronegative chlorine atom.
And for the CH bond, we're going to draw a small arrow, but it's going to point towards the carbon atom, since carbon is more electronegative than hydrogen. Now for the other molecule, let's do the same thing. So which molecule is polar and which one is not? So if we look at these two arrows, notice that they're completely opposite to each other. They cancel out.
And if you look at these two arrows... they also cancel out. So therefore we could say that the molecule on the left has a dipole moment of zero.
So it's a nonpolar molecule. Now what about the molecule on the right side? Notice that these two arrows they don't completely cancel. They partially cancel but they do add up and produce a net dipole moment. And the same is true with these two arrows.
The net dipole moment is in the upward direction. And let me show you why. So we had an arrow going this way, and another arrow going that way.
So if we resolve those two vectors into its components, the one on the right has an x component going towards the right, and a y component going up. The one on the left has an x component going towards the left, and a y component going up. So notice that these two, they cancel because they're opposite to each other. However, the y components, they do not cancel as the x components, because the y components, they're pointing towards the same direction.
And so those two will add, producing a larger y component. Now let's talk about the other two arrows, which were smaller, but they were pointing towards the carbon atoms. Now notice that the X components cancel because they're opposite to each other, but the Y components, they don't cancel. They're pointing in the upward direction, and so they're going to add up, producing a larger Y component.
So if we add these two, we can see that the net dipole moment is in the upward positive Y direction. And so because it has a net molecular dipole moment, We could say that the cis isomer is polar, whereas the trans isomer is relatively nonpolar. Let's work on this problem. The CO bond has a length of 140 picometers and an experimental dipole moment of 0.7 Debye.
Calculate the percent ionic character of this bond. So how can we do that? Well, the first thing we need to do is start with an equation.
The percent ionic character is going to equal to the experimental dipole moment divided by the dipole moment if the atoms were 100% ionic. So we already have the experimental dipole moment. So that's given to us as 0.7 Debye.
What we need to do in this problem is we need to calculate the ionic dipole moment. And how can we do that? Well let's say if the carbon was attached to three R groups and it has a positive charge. And the oxygen is attached to one R group and let's say it has a negative charge.
And let's say that these ions are separated by a distance of 140 picometers and they have four ionic charges. What would be the dipole moment between those two ions? To calculate the dipole moment, it's equal to the charge times the separation distance. Now, a positive charge is equal to the charge of a proton.
A negative charge is equal to the charge of an electron. These two have the same magnitude of charge, but the sign is different. One is positive and the other is negative.
The charge of a proton is 1.6 times 10. to the negative 19 coulombs. The charge of an electron is the same number but it's negative 1.6. times 10 to the negative 19 coulombs.
And since we're trying to calculate a percentage, the negative sign is not important to us. Now the separation distance is 140 picometers. One picometer is 10 to the minus 12 meters. So let's go ahead and multiply those two numbers. And so you should get 2.24 times 10 to the negative 29 Coulomb meters.
Now we need to convert that to Debye. It's important to know that 1 Debye, spelled as D-E-B-Y-E, or Debbie, however you want to say it, that's equal to 3. 3 4 times 10 to the negative 30 kilometers so we need to convert kilometers into units of the body so we can write 1d over this number so these units will cancel So 2.24 times 10 to the negative 29 divided by 3.34 times 10 to negative 30, this is equal to 6.707. And so that's the dipole moment if the atoms were pure ions, or if they have full ionic charges. Now let's calculate the percent ionic character.
So it's going to be the experimentally observed dipole moment, which we said was 0.7 Debye, divided by the dipole moment if they were purely ionic. And that's 6.707 Debye times 100%. So I got 10.4% as my answer.
So we can say that the CO bond has a 10% iconic character, or a 10.4% iconic character. And that's it for this problem. Go ahead and try this problem for the sake of practice. So take a minute, pause the video, and work it out.
So we have an OH bond, and we're given the length of the bond, and it's 96 picometers. And we're also given the observe or the experimental. dipole moment and it's 1.5 dB.
Calculate the percent ionic character of this bond. So let's begin by calculating the dipole moment if the atoms in the OH bond were 100% ionic. So that would mean oxygen have a full negative charge and hydrogen has a full positive charge.
So this would be O plus. And this would be, I mean, O would have a negative charge, because it's more electronegative. So that would be O minus, and this would be H plus. So oxygen would have a full negative charge, H would have a full positive charge. So in that case, using this formula, Q would simply be the charge of a proton, which is 1.6 times 10 to negative 19. And this is not going to change, that number is going to just stay the same if you have positive and negative charges.
r is based on this number that's 96 picometers and we need to convert that to meters so it's 96 times 10 to the negative 10 meters I take that back that's supposed to be 10 to negative 12 meters so let's go ahead and multiply those two numbers And so you should get 1.536 times 10 to the negative 29 Coulomb meters. And let's convert that to units of Dubai. So we're going to divide it by 3.34 times 10 to the negative 30 Coulomb meters.
So always do it in such a way that the unit Coulombs and meters cancels. And so you should get 4.5988, or you could just round that to 4.6. So now these are the two numbers that we need at this point. So the percent ionic character is going to equal to the experimental or the observed dipole moment divided by the dipole moment if there were pure ions times 100%. And so that's going to be 1.5 Debye divided by 4.6 Debye times 100%.
And so I got 32.6%. So that's the percent ionic character of the OH bond based on the information we were given.