For each of the following substances, identify the intermolecular force or forces that predominate. Using your knowledge of the relative strengths of the various forces, rank the substances in order of their normal boiling points: Al2O3, F2, H2O, Br2, ICl, NaCl.

Video transcript

Hello everyone today. We have been given the phone problem. The overall polarity of a molecule is the net sum of the individual bond polarities within a molecule. This compound here has three possible structures. Two of them are polar overall and one is not identified. The polar and non polar structures provide your reasoning. So the first thing we wanna do is We want to draw out our three possible structures. So the way this is written now is this methyl group here is connected to this essential carbon. So we have one methyl group here that's bonded to a carbon and then we have two chlorine is here that seem to be bonded to two different types of carbons. So what we're gonna do is we're gonna draw these two carbons linked together and then we are going to draw one chlorine to that carbon and one chlorine to that carbon. And then the last group is that metal group that's attached to that second carbon. And so now essentially what we have to do is we have to draw another bond line between these two carbons because carbon likes to bond four times and this completes its octet. So this is the one possible structure. The second possible structure is to have the two methyl groups on the same carbon and two chlorine on another carbon. So how is that going to look? That's going to be represented like this. We're gonna have a metal here bonded to a carbon and that carbon is going to be bonded by another metal that will be double bonded to a carbon and then the second carbon is bonded to to chlorine. Lastly, this can also be represented by the two chlorine groups being on opposite carbons as well as the chlorine is being on opposite carbons, but also on opposite sides. So how is that going to look? Well, what we can do is you can draw chlorine, they're bonded to a carbon and that carbon is double bonded to a second carbon. The first carbon is also bonded to a methyl group on the second carbon. The top right group is going to be a methyl group and the bottom right group is going to be a chlorine. We're just going to draw in the lone pairs around Each of these chlorine bleach chlorine gets six electrons to complete its octet. And so now we have the three possible structures. Now, according to our periodic trend that electro negativity increases as you go up and to the right on the periodic table, we can see that chlorine is more electro negative carbon. And so we can do that by drawing the dye pole arrows going from carbon to chlorine for each of these structures. There we go. So we note that in the first two structures. So, in this structure here and in this structure here, so in the first two structures, the disciple arrows point in the same direction, so die pole is going to be in the same direction and the same can be said for our second structure here, since they're drawn in the same direction, they are not canceled out. So the first structure stipe holes are not canceled out. And the second structures, die polls are not canceled out as well. And so this is going to indicate that they are both polar molecules in the third structure, However, the dyp holes point in opposite directions, right, there's one going to the top left and it's going to the bottom right, so it's the opposite direction. And since they're going in the opposite direction, they are going to cancel out. And since they cancel out, this is indicative of a non polar molecule. So in short, our first structure here is polar, as is our second structure. However, our third and final structure is going to be non polar. And with that we've answered the question overall, I hope this helped. And until next time.