2015 AP Chemistry free response questions
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2015 AP Chemistry free response 6
- [Voiceover] A student learns that ionic compounds have significant covalent character when a cation has a polarizing effect on a large anion. So what are they talking about? So if I have a cation, so this is my cation, and then this is my large anion, my large anion right over here. And actually, let me do it this way, let me make it a little bit clearer 'cause a cation still has electrons, so you have your nucleus here, of your cation, you have your electron cloud, you have your electron cloud of your cation right over here, but you have fewer electrons than you have protons, so you're going to have a positive charge. And that's why we call it a cation. And the anion, it still has a positive nucleus, and it has an electron cloud. I'm drawing a large anion here, a large, a large, a large anion right over here. That's an electron cloud, but it has more electrons in the cloud than it has protons in the nucleus, so it has a negative charge. That is the anion. Now what they're saying is you start to have significant covalent character, remember covelent bonds or electrons are essentially shared between two atoms when a cation has a polarizing effect on a large anion. So one way to think about is these electrons that are out here, well they're gonna want to get away from all their negative friends here. Negative charges repel each other and they might want to spend more time closer to this cation. And so their gonna start having covalent character, they're gonna be more and more shared between the two, and there really is a spectrum between ionic bonds and covalent bonds. And over here, you might have a polarizing effect. These things spend more time on the left hand side, the way I've drawn it, than on the right hand side. So you're gonna have more of the electrons and more of the negative charge hanging out over here than on the right, that's what they're talking about. And as a result, the student hypothesizes that salts composed of small cations and large anions should have relatively low melting points. Alright, so it should be easy to take less energy to break them out of their, kind of the salt lattice structure when they're in solid form. Select two compounds from the table, and explain how the data supports the student's hypothesis. Let's see, these are all salts here, and if these were to disassociate, or if you think about what the cations would be, the cations, the cations that make up, that make up these compounds, you have lithium, lithium, you have sodium, and you have potassium. And then the anions, the anions that make up these compounds, you have iodide, you have iodide and you have fluoride. Let me write it this way, flouride and iodide. So which of these are small and which of these are big? Well, let's compare lithium, sodium, and potassium. And we see lithium, sodium, and potassium, they're all group one elements. And in general, as we go down a group, as we go down a column in the periodic table, we're adding shells, and so size increases. So size increases as we go down like this. Size, size increases. And so of these, lithium is the smallest and potassium is the largest of these three. So lithium, let me write that down, lithium is the smallest, smallest, and potassium is the largest of these, of these three. Now if you look at the anions, well you have once again, fluorine and iodine are in the same, are in the same group, and iodine is below it, it has way more electrons and you've added way more shells here is actually the important part. And so you can see that since iodine is larger than fluorine, well iodide, which is when they gain an electron, it's gonna be larger than fluoride. So iodide, this one is larger, and the fluorine is smaller. So let's see, lithium-iodide is a case of small cation, small cation and large, and large anion. And so that's actually the direction, that's kind of the best example of small, of what the student is talking about, small cation and large anion. Lithium iodide, lithium iodide is a great example of that. And as we can see, it has the lowest melting point of everyone on the table. Now let's take the other extreme. What happens if we take a large cation and a small anion. Well, let's see, they don't have that combination here, but let's see, if let's see, sodium, well sodium fluoride is interesting because we have a larger, we have a larger cation, so this is a larger cation, and then we have a smaller anion, smaller anion. And notice, that has the highest, that has the highest melting point. And so these are two that are good to compare. So part a, we'll do that in blue, compare lithium iodide, which has small, small cation, plus large anion, to sodium fluoride which has larger cation plus smaller anion, smaller, smaller anion. And we see, and we see, that their melting point's, their melting point's, melting point's consistent with hypothesis, consistent with hypothesis. The small cation, or lithium iodide, I'll just write melting point, melting point lower, lower, and actually much lower than sodium fluoride's melting point. So at least if you compare those two and those seem to be good extremes, if we took the lowest melting point, which completely typified what the student was saying as having a low melting point, as we looked at the highest melting point which had a larger cation and a smaller anion. And so just looking at that, it seems to be consistent with the student's hypothesis. Alright, part B. Identify a compound from the table that can be dissolved in water to produce a basic solution. Write the net ionic equation for the reaction that occurs to cause the solution to be basic. Alright, so let's think about how we can form a basic solution. So essentially, we would need to nab some hydrogens from the water molecules in order to have some hydroxide laying around. And so there's a couple of candidates here that could do that. You have all of these halides, you have the iodide, you have the fluoride, and the important thing to recognize is that hydrogen iodide or hydrochloric acid, since we're dissolving it in water now, hydroiotic acid is a strong acid. So let's see, HI, hydroiotic acid, strong acid, strong acid, while hydrofluoric acid is a weaker acid, is a weaker acid. So a strong acid is not going to be good at nabbing hydrogens. In fact, it wants to give away its hyrdorogens really, really, really badly. So the iodine does not seem like a good candidate. The fluoride does seem like an interesting candidate. so let's take one of these, let's take one of these candidates out here, you can use either the lithium fluoride or the sodium fluoride, let's just take, I know we've already dealt with the sodium fluoride, so let's keep using that. So if you take sodium fluoride, sodium fluoride, and so let me, since they say write the net ionic equation for the reaction that occurs to cause the solution to be basic. So we're gonna focus on sodium fluoride, sodium, sodium fluoride is what we select. We select sodium fluoride, that's the first part. We identified a compound that can be dissolved in water to produce a basic solution. We select that, and now let's draw, let's do first the ionic equation, then we'll do the net ionic equation. So if you dissolve this in water, you're going to have sodium cations dissolved in our aqueous solution, plus fluoride anions dissolved in our, dissolved in our aqueous solution, and it is going to be in equilibrium, in equilibrium with, actually let me draw it, let me draw it this way, let me draw it this way, plus H2O, because that's what it's going to react with, and obviously, I don't have to say it's an aqueous solution, it is the aqueous solution, so it's making the solution aqueous. This is going to be in equilibrium with, and I'll go to the next row here. If I was actually taking the AP test, well actually, let me just, let me just copy and paste it here, just so we don't have to deal with it on two rows. So put it right over there. So that is going to be in, that is going to be in equilibrium. Let me get the right tool out. That is going to be in equilibrium with this fluoride nabbing a hydrogen. So hydrofluoric acid, we could say HF in our aqueous solution, plus OH minus in our aqueous solution, and then you still have the sodium cation in our aqueous solution. And so this, as you can see will perform at equilibrium, and so you're gonna have more hydroxide around, so you're forming your basic, you're forming your basic solution. If this was an iodide right over here, then this reaction would go strongly in that direction. If this was hydrogen iodide right over here, this wouldn't be some kind of nice equilibrium. This would be a strong acid, it would definitely want to get rid of its hydrogens. So that's why we don't want to use hydrogen iodide. We want to use hydrogen fluoride. When you put it in water, it becomes hydrofluoric, it's likely to donate, when it's in its gassiest form or it's by itself, we would call it a hydrogen fluoride, but now when we put it in a solution, we can consider it to be hydrofluoric acid. And so if you want your net ionic equation, well we don't have to worry too much about these sodiums that are on both sides of this. And so if you just focus on this right over here, this is your net ionic, net ionic equation.