Transcription
In this lesson, we're going to talk about Van der Waals forces and London dispersion forces. Now, these forces, they exist in all molecules, but they are the predominant forces in nonpolar molecules. So, let me describe what they are with a visual illustration.
So, let's say we have two atoms with a positive charge at the center. And we know that in the electron cloud, the atoms are surrounded by a negative charge. The nucleus at the center has a positive charge. Now, dependent on the number of protons, it can be a positive eight charge, positive twelve charge. We're not going to focus on that particular number. So, when the electrons are evenly distributed in an atom, you're gonna have a non-polar atom.
Now, because the electrons are constantly moving, there's gonna be times where one side of the atom will have more electrons than the other side. In this case, the electron cloud is now distorted. So, we have an excess of negative charge on the left and a deficiency of negative charge on the right side. So, basically, we can say that this particular picture that looks like this. So, there's a net negative charge on the left. And because the right side is electron deficient, it's more positively charged down. Now, granted, there's still negative charge on the right side, but this positive charge means that it is less negative than the left side. So, now this atom is now polarized. It's now a dipole.
Now, this dipole doesn't last long. So, it may last for a very, very, very short time. So, it's a temporary dipole. What happens if we take this dipole and place it next to a polar atom where the electron cloud is evenly distributed? The electrons will feel a force that will accelerate them towards the positive side of that atom. And so, this electron cloud becomes distorted. And so, this atom becomes polarized. And so, what happens is now you have two polarized atoms.
Now, you need to be familiar with the term called polarizability. And it basically describes the probability of an atom's electron cloud being distorted due to the random motion of electrons. So, let's compare fluorine and iodine. Fluorine has nine electrons. Iodine has 53. So, because iodine has a lot more electrons than fluorine, it's more polarizable. It can create a temporary dipole more likely than a fluorine atom. So, a temporary dipole will be more likely to form in an iodine atom than a fluorine atom.
Now, notice that the positive side of this atom will be attracted to the negative side of the other one. So, these two, they will feel a force of attraction that accelerates them. And so, these forces that attract these atoms together, they're known collectively as the Van der Waals forces, also known as the London dispersion forces, LDF.
Now, this first example, it was a temporary dipole. A dipole is basically a polarized object or polarized molecule where one side is negative and the other side is positive. Now, the reason why it's a temporary dipole is because it can go back to its original state. The electrons on the left side can diffuse towards the right side because this negative charge is attracted to the positive charge. And so, it could be turned into its nonpolar state where the electron cloud is no longer distorted, the electrons are distributed evenly. So, keep that in mind. This temporary dipole doesn't really last long. It's dependent on the size of the atoms.
Now, the dipole on the right is an induced dipole. And the reason why it's called an induced dipole is because it was induced or created by this first dipole. When this dipole became, when the electron cloud of that dipole became distorted, it caused this, the electron cloud of this atom to become distorted, thus creating an induced dipole. So, make sure you understand the word induced dipole. It's just the dipole that was created by something else. So, it doesn't last long. It can revert back to its nonpolar state. And so, these forces that hold or that attract these dipole-induced dipole molecules together, and those are the Van der Waals forces or the London dispersion forces. They're very weak and they're found in every molecule. However, in nonpolar molecules, they are the dominant forces in nonpolar molecules.
So, these Van der Waals forces explain why some nonpolar molecules have higher boiling points than other nonpolar molecules. Let's consider the halogens, for example. Fluorine, chlorine, bromine, and iodine. Fluorine is a gas. Chlorine is a gas. Bromine is a red liquid. And iodine is basically a purple solid at room temperature, of course. Now, the boiling point of fluorine is -188 degrees Celsius. The boiling point of chlorine, it's -34 degrees Celsius. And for bromine, it's 59 degrees Celsius. And for iodine, it's even higher, 114 degrees Celsius.
Now, if we write down a molecular weight of these molecules, let me get my book out. The molar mass or molecular weight for fluorine is 38. Fluorine has nine electrons, as we said. The molar mass for chlorine is approximately 70.9. And chlorine as an atom has 17 electrons. Bromine's molar mass, this almost is like 159.8. And it has 35 electrons per. And for iodine, we said had 53 electrons. And its molar mass is about 253.8. So, as we travel down the periodic table in group 7A, that is the group where all the halogens are, notice that the number of electrons increases. And so, the molecules are becoming more polarizable due to the increased number of electrons. And so, the molecular weight and the number of electrons, they're proportional. So, as a molecule's molecular weight increases, the number of electrons will increase with it too.
Now, with molecules that have a high number of electrons or a higher molecular weight, those electrons will, I mean, those molecules will have more London dispersion forces or more LDF forces, or you could say Van der Waals forces, they're the same. And so, as that goes up, you could see the trend with the boiling point. The boiling point increases. So, iodine has the highest. So, make sure you understand this relationship. So, molecules with more electrons or higher molar mass values, they will have more Van der Waals forces and thus they will have a higher boiling point. So, boiling point is directly related to the amount of LDF forces that a molecule has.
So, here's a question for you. Consider the molecules methane, ethane, propane, and butane. So, which of the following four molecules will have the highest boiling point? So, looking at the chemical formula, this is C1H4. Ethane has two carbons, six hydrogens. Propane has three carbons, eight hydrogens. And butane has four carbons and 10 hydrogens. So, if we analyze the molar mass of each of these molecules, carbon is about 12. Hydrogen is 1. So, methane has a molar mass of 16. Methane is going to be 2 times 12, that's 24, plus 6, so approximately 30. Three times 12 is 36, plus 8, 44. Four times 12 is 48, plus 10, so 58. So, the molecule with the highest molar mass is gonna have the highest boiling point. So, it's gonna be butane. Methane has the lowest boiling point because it has the lowest molar mass. All of these molecules are nonpolar. Anytime you have a hydrocarbon that contains only carbon and hydrogen bonds, it's a nonpolar molecule. Therefore, the predominant intermolecular force acting on it will be London dispersion forces or Van der Waals forces.