Ionization energy: period trend | Atomic structure and properties | AP Chemistry | Khan Academy

In this video, let's look at the periodic trends for ionization energy. So for this period, as we go across from lithium all the way over to neon. As we go this way across our periodic table, we can see in general there's an increase in the ionization energy.

So lithium is positive 520 KJ per mole; beryllium goes up to 900 KJ per mole, and then again in general we see this increase in ionization energies going over to neon. Going across a period is an increase in the ionization energy, and that's because as we go across our period, there's an increase in the effective nuclear charge, so an increase in Z effective.

Remember the formula for that: the effective nuclear charge is equal to the actual number of protons, which is Z, and from that we subtract S, which is the average number of inner electrons shielding our outer electrons.

So let's examine this in more detail looking at lithium and beryllium. Lithium has an atomic number of three, so three protons in the nucleus, so a positive 3 charge. Lithium's electron configuration we know is 1s² 2s¹, so two electrons in our 1s orbital and one electron in the 2s orbital.

Beryllium has one more proton and one more electron, so one more proton in the nucleus, so a positive 4 charge. For beryllium, the electron configuration is 1s² 2s², so two electrons in the 1s orbital and then two electrons in the 2s orbital.

Let's calculate the effective nuclear charge for both of these. First, we'll start with lithium. For lithium, it has a positive 3 charge in the nucleus. The effective nuclear charge is equal to positive 3, and from that, we subtract the average number of inner electrons shielding our outer electrons.

In this case, we have these two inner or core electrons that are shielding our outer electron, our valence electron, from this full positive 3 charge. We know that like charges repel, so this electron is going to repel this electron a little bit, and these two inner core electrons of lithium have a shielding effect. They protect the outer electron from the full positive 3 charge.

For a quick effective nuclear charge calculation, positive 3 minus 2 gives us a value of +1 for the effective nuclear charge. So it's like this outer electron of lithium is feeling a nuclear charge of +1, which pulls it toward the nucleus. There's an attractive force between the outer electron and our nucleus.

Now, the actual calculation for this, Z does not have to be an integer; the actual value for lithium is approximately 1.3, but our quick crude calculation tells us positive 1. Let's do the same calculation for beryllium.

So the effective nuclear charge for beryllium is equal to the number of protons, right, which for beryllium is positive 4. From that, we subtract the number of inner electrons that are shielding the outer electrons. It's a similar situation; we have two inner electrons that are shielding this outer electron. They repel this outer electron, shielding it from the full positive 4 charge of the nucleus.

So we say there are two inner electrons, so the effective nuclear charge is positive 4 minus 2, giving us an effective nuclear charge of positive 2. In reality, the effective nuclear charge is approximately 1.9, and that's because beryllium has another electron in its 2s orbital, which does affect this electron a little bit, repels it a little bit, and so it actually decreases the effective nuclear charge to about 1.9. But again, for a quick calculation, positive 2 works.

So the outer electron for beryllium, let's just choose this one again, is feeling an effective nuclear charge of positive 2, which means that it's going to be pulled closer to the nucleus. There's a greater attractive force on this outer electron for beryllium as compared to this outer electron for lithium, where the effective nuclear charge is only +1 for this outer electron.

Because of this, the beryllium atom is smaller. The 2s orbital gets smaller, and the atom itself is smaller. Beryllium is smaller than lithium. This outer electron, let me switch colors again, for beryllium is closer to the nucleus than the outer electron for lithium; it feels a greater attractive force, and therefore it takes more energy to pull this electron away from the neutral beryllium atom.

That's the reason for the higher ionization energy. Beryllium has an ionization energy of positive 900 KJ per mole compared to lithium's of 520 KJ per mole. So it has to do with the effective nuclear charge.

So far, we've compared lithium and beryllium, and we saw that the ionization energy went from positive 520 KJ per mole to 900 KJ per mole, and we said that was because of the increased effective nuclear charge for beryllium. But as we go from beryllium to boron, there's still an increased effective nuclear charge, but notice our ionization energy goes from 900 KJ per mole for beryllium to only 800 KJ per mole for boron.

So there's a slight decrease in the ionization energy, and let's look at the electron configuration of boron to see if we can explain that. Boron has five electrons, so the electron configuration is 1s² 2s² 2p¹. So that fifth electron of boron goes into a 2p orbital, and the 2p orbital is higher in energy than a 2s orbital, which means the electron in the 2p orbital is on average further away from the nucleus than the two electrons in the 2s orbital.

If we just sketch this out really quickly, let's say that's my 2s orbital; I have two electrons in there, and this one electron in the 2p orbital is on average further away from the nucleus. So those two electrons in the 2s orbital actually can repel this electron in the 2p orbital, creating a little extra shielding for the 2p electron from the full attraction of the nucleus.

Even though we have five protons in the nucleus and a positive five charge for boron, the fact that these 2s electrons add a little bit of extra shielding means it's easier to pull this electron away. So it turns out to be a little bit easier to pull this 2p orbital electron away due to these 2s electrons, and that's the reason for this slight decrease in ionization energy.

As we go from boron to carbon, we see an increase in ionization energy; from carbon to nitrogen, an increase in ionization energy again we attribute that to increased effective nuclear charge. But when we go from nitrogen to oxygen, we see a slight decrease again from about 1400 KJ per mole down to about 1300 KJ per mole for oxygen.

Let's see if we can explain that by writing out some electron configurations for nitrogen and oxygen. Nitrogen has seven electrons to think about, so its electron configuration is 1s² 2s² 2p³. This takes care of all seven electrons.

For oxygen, we have another electron, so 1s² 2s² 2p⁴ is the electron configuration for oxygen. Let's just draw using orbital notation the 2s orbital and the 2p orbital. So for nitrogen, here's our 2s orbital, we have two electrons in there. For our 2p orbitals, we have three electrons, so here are the two p orbitals, and let's draw in our three electrons using orbital notation.

Let's do the same thing for oxygen. So there's the 2s orbital for oxygen, which is full, so we'll sketch in those two electrons, and we have four electrons in the 2p orbitals. Let me draw in the two p orbitals—there's one electron, there's two, there's three. Notice what happens when we add the fourth electron; we're adding it to an orbital that already has an electron in it.

So when I add that fourth electron to the 2p orbital, it's repelled by the electron that's already there, which means it's easier to remove one of those electrons. Like charges repel, and so that's the reason for this slight decrease in ionization energy.

It turns out to be a little bit easier to remove an electron from an oxygen atom than nitrogen due to this repulsion in this 2p orbital. From there on, we see our general trend again—the ionization energy for fluorine is up to 1681, and then again for neon, we see an increase in the ionization energy due to the increased effective nuclear charge.