Now, let's look at this final one, d. We're going to say let's look at the easier one. We have H connected to N here. So it's going to be H bonding for its intermolecular force, which is a polar force. But now, we have to figure out what the heck is the force in IF₄^-. Here, it's not easy to see, so we have to draw it out. And we have to draw it because we know it's not ionic, we know it doesn't have H bonding. So it's not one of those 2. This compound is going to either be dipole-dipole if it's polar or London dispersion if it's non-polar. And the only way we can tell that is if we draw it out. So, I will go in the center because I is less electronegative. Iodine is in group 7a, so it has 7 valence electrons. Then we have 4 fluorines. Fluorines in group 7a, so it has 7 valence electrons. Fluorine only makes one bond, so there go the bonds. Now, also remember that minus 1 means what? Minus 1 means we gain another electron. So we're gonna add 1 more electron to the mix. Okay. So 1, 2, 3, 4, 5. There was one here also. We're going to say how many electrons do we have left? We have an unpaired electron here, here and these 2 are paired up. You see those 2 electrons that are separated? Bring them close together. We should pair up our lone pairs. So there they go. So we have 2 lone pairs. Now, we go over the rules. It has lone pairs around the central elements, so we're going to use the rules for rule 2. So first, the central element must be connected to the same elements. Iodine is only connected to fluorines. 2, the central element must be less electronegative. The central element is less electronegative, so it follows 2b. 2c, we use dipole arrows to point to the more electronegative element. So this dipole arrow points to fluorine. It gets cancelled out by this dipole arrow which heads in the opposite direction. This dipole arrow points to fluorine and it gets cancelled out by this arrow that points to the other fluorine. All our element dipole arrows cancel out. Then we have a lone pair dipole arrow that points this way and one that points the opposite way. So they also cancel each other out. Based on all the rules for rule 2, this compound is non-polar. And because this element is non-polar, it forces London dispersion. But here's the most important thing. 1 is non-polar and 1 is polar. Because there are differences in polarity, no solution is formed. So that's what we'd say for part d. Now, hopefully, you guys are working towards being able to draw these compounds faster and being able to identify the intermolecular force.

When it comes to solutions, for a solution to form, both compounds need to be polar or both need to be non-polar. Likes dissolve likes. Now, that we've seen this, I want you guys to attempt to do this practice question on your own. So in this one, we have to figure out which of the following statements is true. Meaning that one could be the correct answer or more than one answer could be the correct answers. So go over what we know about intermolecular forces, about polar and non-polar. And I'll give you guys a huge hint. If it ends with ane, e.g., methane, pentane, all that means is that compound is what we call an alkane. Alkanes are compounds with only carbon and hydrogen. And that should be a huge hint. If your compound has only carbons and hydrogens, what can you say about its polarity? Is it polar or non-polar?