2015 AP Chemistry free response 6 | Studying for the AP Chemistry exam? | Chemistry | Khan Academy

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 clear because 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 has an electron cloud. I'm drawing a large anion here, a large, a large, a large anion right over here; that's the 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.

What they're saying is that you start to have significant covalent character. Remember, covalent bonds are when electrons are essentially shared between two atoms. When a cation has a polarizing effect on a large anion.

So one way to think about it is those electrons that are out here, well, they're going to want to get away from all their negative friends. Negative charges repel each other, and they might want to spend more time closer to this cation.

So they're going to start having covalent character; they’re going to be more and more shared between the two. There really is a spectrum between ionic bonds and covalent bonds. Over here, you might have a polarizing effect; these things spend more time on the left-hand side than on the right-hand side.

So you're going to have more of the electrons and more of the negative charge hang out over here than on the right. That's what they're talking about.

As a result, the student hypothesizes that salts composed of small cations and large anions should have relatively low melting points. All right, so it should take less energy to break them out of their kind of 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 you think about what the cations would be— the cations that make up these compounds— you have lithium, you have sodium, and you have potassium.

Then the anions that make up these compounds, you have iodide, you have iodide, and you have fluoride. Let me write this way: fluoride and iodide. So which of these are small and which of these are big?

Well, let's compare lithium, sodium, and potassium. We see lithium, sodium, and potassium; they are 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, so size increases.

Size increases as we go down like this. Size increases. 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 and potassium is the largest.

Now if you look at the anion, well, you have once again fluorine and iodine are in the same group. Iodine is below it; it has way more electrons, and you've added way more shells. Here is actually the important part.

Since iodine is larger than fluorine, well, iodide, which is when they gain an electron, is going to be larger than fluoride. So iodide is larger, and fluorine is smaller.

Let’s see, lithium iodide is a case of a small cation and a large anion, and so that's actually the direction— that's kind of the best example of what the student is talking about: small cation and large anion.

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 we hold on to sodium.

Well, sodium fluoride— yes, sodium fluoride is interesting because we have a larger cation. So this is a larger cation, and then we have a smaller anion. Notice that it has the highest melting point.

So these are two that are good to compare. So part A, I’m going to do it in that blue: compare lithium iodide, which has a small cation plus a large anion, to sodium fluoride, which has a larger cation plus a smaller anion.

And we see their melting points. Their melting points are consistent with the hypothesis. The small cation in lithium iodide has a lower melting point— I’ll just write melting point lower, and actually much lower than sodium fluoride’s melting point.

So at least if you compare those two, they seem to be good extremes. We took the lowest melting point, which completely typified what the student was saying about having a low melting point, and we looked at the highest melting point, which had a larger cation and a smaller anion.

So just looking at that, it seems to be consistent with the student hypothesis.

All right, 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.

All right, so let's think about how we can form a basic solution. Essentially, we would need to nab some hydrogens from the water molecules to have some hydroxide laying around.

And there are a couple of candidates here that could do that. You have all of these halides: you have the iodide, you have the fluoride. The important thing to recognize is that hydrogen iodide and, you don’t see or hydrochloric acid, since we’re dissolving it in water.

Now, hydro– not hydrochloric, hydriodic acid is a strong acid. So let’s see, HI— hydriodic acid, strong acid— while hydrochloric acid is a weaker acid.

So a strong acid is not going to be good at nabbing hydrogens. In fact, it's a strong acid; it wants to give away its hydrogens really, really badly. So this— 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. We could use either the lithium fluoride or the sodium fluoride.

Let’s just keep using sodium fluoride. So if you take sodium fluoride, 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 going to focus on sodium fluoride. Sodium fluoride is what we select. We select sodium chloride; that’s the first part we identified, a compound that can be dissolved in water to produce a basic solution.

Now let’s draw— let’s do first the ionic equation, and 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 anion dissolved in our aqueous solution.

And it is going to be in equilibrium— in equilibrium— with... actually, let me draw it this way: plus H₂O because that’s what it’s going to react with.

Obviously, I don’t have to say it’s an aqueous solution; it is the aqueous solution. So what'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— let me just copy and paste it here just so we don’t have to deal with it on two rows. So that is going to be in equilibrium.

Let me get the right tool out; that is going to be in equilibrium with this— this fluoride nabbing a hydrogen. So for instance, we could say HF in our aqueous solution plus OH⁻ in our aqueous solution, and then you still have the sodium cation in our aqueous solution.

As you can see, we’re going to form an equilibrium, and so you're going to 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 going strongly.

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. You put it in water; it becomes hydrofluoric.

It’s likely to donate it. When it’s in its gaseous form or by itself, we would call it hydrogen fluoride, but now when we put it in a solution, we can consider it to be hydrochloric acid.

So if you want your net ionic equation, well, we don’t have to worry too much about these sodiums; they're on both sides of this. So if you just focus on this right over here, this is your net ionic equation.