Ionization energy trends | Atomic models and periodicity | High school chemistry | Khan Academy

We're now going to think about ionization energy trends.

What's ionization energy? It's the energy required to remove the highest energy electron from an atom. To think about this, let's look at some data. So right over here is ionization energy plotted against atomic number for elements in the periodic table of elements.

First, we can look group by group. For example, if we look at group one, that's the first column, the leftmost column in the periodic table of elements, starting with hydrogen. Here you can see that as we go down that column, it looks like, generally speaking, our ionization energy is decreasing. For group two, we see a similar pattern; as we go down that second column, our ionization energy, generally speaking, looks like there's a few slight anomalies here, but it looks like it's decreasing.

So as you go down a group, the data seems to imply that ionization energy decreases. It takes less and less energy to remove that highest energy electron. Now let's look at the periods. Remember, the periods are the rows in the periodic table. So if we look at that second row in the periodic table, as we go from left to right, it looks like, generally speaking, ionization energy is increasing. It becomes harder and harder to remove that highest energy electron.

Now, there are some anomalies here; it looks like it briefly decreases, then it increases, briefly decreases again. But the general trend is that, as you go from left to right along that period, energy is increasing. We see the same thing in period 3. Once again, there are some anomalies here, but the trend seems to generally hold. We could also look at period 4 and so on and so forth.

So if we think about ionization energy, what we saw is that as we go down a group right over here, it becomes easier and easier to remove that highest energy electron. In another way to think about it, ionization energy decreases if you go from left to right along a period. Right over here, we saw from the data that ionization energy increases.

Now let's think about why this is. Well, if you look at this trend along, or get an intuition for why this is. If you look at this trend along a group, over here, we already said that you're going to have the same number of valence electrons, but those valence electrons are further and further out. They're at higher and higher energy shells, and so you have a lot more shielding from all of those core electrons.

So that causes some interference. We've already seen that as you go down a group, your atomic radius increases, so you're also going to have less of that effective nuclear force, which is essentially how much the force is between when you consider the nucleus, when you consider the shielding from the electrons in between, and when you think about the distance of those outer electrons.

It makes sense that it's easier and easier to remove the highest energy electron from, say, cesium because it's further out. There are more electrons shielding it from that nucleus, even though there's a lot of protons in that nucleus than, say, in the case of hydrogen.

Now, as you go from left to right along a period, we already talked about the fact that you're adding electrons, but you're either backfilling into a lower energy shell or you're adding at that same outer valence shell. But as you're doing so, as you're going from left to right, you have more and more protons.

So those protons, as you add more and more, you're going to have a stronger positive charge. It's going to pull more and more on those outer electrons. Remember, you have the same number of core electrons because you're just adding to the outer shells right over here.

So it's going to pull harder and harder on them and bring them closer and closer. We saw that atomic radius decreases as you go from left to right, and so it makes sense that ionization energy increases. It's going to be harder and harder to pull off that highest energy electron from, say, bromine than it would be from, say, potassium.