 OK, so let's do this problem. It says, using condensed electron configurations, write the reactions for the formation of the common ions of lead. So the first thing you need to do is remember what the symbol for lead is, and that is db. And then, of course, well, write the condensed electron configuration. So how do you do that? You find the noble gas prior to the atom that you're looking for. In lead's case, since lead's nuclear charge is 82, OK? So the noble gas before that is xenon. So the condensed electron configuration is going to start with xenon. And then, well, where do we go? 6s, 6s, there's two electrons in it. And then, this is the weird one, right? So remember, look at the extended version of the periodic table of this one, it freaks you out. But the lanthanides are here at 57. So this is put in between them. So when we do that, we've got the 4f, deep in our f orbitals, and there are 14 of them, because they're all filled up. And then after that, you've got the d-shell. But that's the 5d. So I guess this problem is good in some way, so you can know how to label these things. And of course, if you count those, there's 10 of those. And then, if you look, well, it's in the p-block. So it's the 6p, because it's on period 6. And there's two of them. So remember what we said. When we have filled d's and filled f orbitals, when we have those things filled, it gives it some increased stability. And when those are filled, they become, even though they're in the valence shell, they become core electrons once the entire sub-shell is filled. So we have a filled f and a filled d here. So do you guys see that? So they are now essentially inner or core electrons. So you would expect lead to form two different ions. The first one should be obvious. Well, maybe nothing is obvious in chemistry. I shouldn't say that. But the first one should be they should both kind of jump out at you, I guess, is what I should say. And remember we said that, well, once you've become full, those guys are stable. So those ones are stable. So you wouldn't expect to lose those electrons. And in fact, you don't. So lead will form two ions. The first one is Pb2 plus, as you might expect. Why? Because the very outer electrons are these 6p2 electrons. So it'll lose those 6p. And then it asks us, well, write the condensed electron configuration for that. So if you guys want to help me, you can. But if you don't want to, I can just read it from you. You guys are such a help. So is everybody cool with that? I guess they're saying if you want, you can say plus 2 electrons. It should be obvious that that occurred. I wouldn't mind if you just wrote that. I'm going to get that out of here. The other one that it should form is the after the loss of the s electron. And this is the typical behavior of these P block that is to help form two regularly occurring ions after the P's leave and then after the s's leave. Or are removed, I guess I should say. So that's the condensed electron configurations for the atom and the two common ions, that you would expect. Are there any questions about this though? Again, we have the indium problem for example 2. So check that one out too. Good job, guys.