In this problem, we're being asked to write the electron configuration of several different ions. And if it doesn't say so, assume that you're doing it for ground states, meaning all of the electrons in their lowest possible energy states. When you're doing if for an ion, you always want to begin by doing it for the atom itself. So let's do oxygen, first. If we look at a periodic table and locate oxygen here, and we find the noble gas that comes before it, which is helium, and then we work our way across until we get to the oxygen. So this is the 2s block or, that isn't a two, 2s. And then we have two, then we come into this area and that's the 2p, and we count over and we get 4. So that is oxygen, and we'll write that down here. Helium, 2s2 2p4. Now we're ready to do it for the ion, which is O2 minus. This means we have two more additional electrons to place in. They go into the next available energy states. Well, we've only put four of the electrons in the P and we know that a P is three orbitals. Therefore, it holds a total of six electrons. So we could put those two additional into the P that will give me helium 2s2 2p6. Now, let's look again at the periodic table. If we're putting those two additional electrons in at 2p6, we now have the configuration of neon. So it'd be equally correct to call this Ne in square brackets. So it's got the electron configuration of that noble gas. Let's move to the next one. We're going to do phosphorous first. So we find phosphorous on the periodic table, which is right here. We find the noble gas that comes before it, and that's neon. And we work our way to phosphorus, so we're going to cross until we hit here. This is 3s, we have 2 in the P block, that's a green area. And then we have 3p and we count over 3, 3p3. So it's neon 3s2 3p3. Okay. So now we're ready to do P with a 3 minus charge. We're going to put the three additional electrons in the next available location. So that would be neon 3s2 and 3p, now we can go up to 6. Again, if we look at the periodic table and those three additional electrons going on gives us an argon electron configuration. So we can also equally call this Ar in square brackets. Okay. Let's do a cation. In a cation, sorry, this is c. In a cation, again, we want to do magnesium first, and then in a cation, you're removing electrons. So let's find magnesium on the periodic table. I'm going to actually clear out this page, so that we cannot have things getting in the way. And we're finding magnesium, it's right here. So the noble gas that comes before magnesium is neon. One, two, three, this is a 3s and magnesium is two boxes over. So it's 3s2. So neon 3s2. So magnesium with a 2 plus charge, you're now going to be removing two electrons. And they come from the outermost shell, highest n available, so it's these two electrons we're going to take out, and we would put neon. And so that would be the electron configuration magnesium with a 2 plus charge. Now, very often, students will think, well, magnesium's got 12, let's just move back two and let's do the electron configuration that's two before it. And that works for some elements, but that doesn't work for all elements, as we're going to see here. Let's do chromium. Chromium is d. All right. So we're doing d. Chromium's electron configuration, as we locate it here. The noble gas that comes before chromium is argon, so we have argon. We come over to this block and we count down, one, two, three, four. So this is a 4s. And we have two, and then we enter the d block, and d's start with the number three, so this is 3d, and we count over, one, two, three, four. Now, we learned that when you do electron configurations of the transition elements, if we have four or nine here, we're going to promote one up. So chromium is actually argon 4s1 3d5, so let's write that down. Argon 4s1 3d5. So now, we're ready to do chromium 3 plus, so we have to remove 3 electrons. We do not remove the last three we put in. We look at the valence shell and we take out the highest end first. So we're going to take one of these away, and then we go to the d and remove a couple of those. So that's going to give me argon 3d3. So if we look at our periodic table, it would not be correct to say, okay, it's got three less electrons, so let's count back three and let's do the electron configuration of that element. So we did that, we would have argon 4s2 3d1. And that is not the electron configuration of chromium with a 3 plus charge. Last one is silver, plus, so let's do silver, first. Find silver on the periodic table, it's down here. So find the noble gas that comes before it, it's krypton. So we write, well, let's go ahead and write it down here. Krypton, work our way over to silver. So this is one, two, three, four, five, 5s2. And then we enter this area, that's 3d, this is 4d, and we count over and get this 9, so 4d9. Again, if we're one away from filling it up, which is 10, we would promote up. So instead of being that, it's krypton 5s1 4d10. So that's electron configuration of silver. Krypton 5s1 4d10. Now we're going to have silver plus, and this is the only electron stable ion of silver. And that's because it's going to remove that one electron from here. So this is going to be krypton, and we're going to completely fill up that d, oops, 4d subshell, and that's a very stable configuration. So don't forget, it's important that you do the electron configuration of the atom first and then remove electrons. It's not so important for the anions, negative charged ions. because you're just going to put in the next available ones no matter what. But if you get in the habit of always doing it for the atom first, then you'll remember. Okay, I look at that and then I remove them from the highest n before I start removing them from n's that are smaller.
