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AP Chem: SAP‑1 (EU), SAP‑1.A (LO), SAP‑1.A.4 (EK), SAP‑2 (EU), SAP‑2.A (LO), SAP‑2.A.2 (EK), SAP‑2.A.3 (EK)

- [Instructor] 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, so 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 kilojoules per mole. Beryllium's goes up to
900 kilojoules per mole, and then again, in general,
we see this increase in ionization energies going over to neon. So, going across a period, there's 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, increase in Z effective. And remember, the formula for that is 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 atomic number three, so three protons in the nucleus, so positive three charge, and lithium's electron
configuration we know is one s two, two s one. So, two electrons in our one s orbital, and one electron in the two s orbital. Beryllium has one more proton and one more electron. So one more proton in the
nucleus, so a plus 4 charge, and for beryllium, the
electron configuration is one s two, two s two. So two electrons in the one s orbital, and then two electrons
in the two s orbital. Let's calculate the
effective nuclear charge for both of these, and first, we'll start with lithium. So for lithium, lithium has a plus three
charge in the nucleus, so the effective nuclear charge
is equal to positive three, 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 three charge. So we know that like charges repel, so this electron is going to repel this electron a little bit, and this electron repels this electron. And these two inner core
electrons of lithium have a shielding effect, they protect the outer electron from the full positive three charge. So there's two shielding electrons, so for a quick effective
nuclear charge calculation positive three minus two
gives us a value of plus one for the effective nuclear charge. So, it's like this outer
electron of lithium is feeling a nuclear charge of plus one, which pulls it toward the nucleus, right? So, there's an attractive
force between the outer electron and our nucleus. Now, the actual calculation for this um, Z is-- S I should say, does not
have to be an integer, and the actual value for
lithium is approximately one point three, but our
quick, crude calculation tells us positive one. 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 four, and from that, we subtract
the number of inner electrons that are shielding the outer electrons. So, it's a similar situation, we have two inner electrons
that are shielding this outer electron, they're repelling this outer electron, shielding the outer electron
from the full positive four charge of the nucleus. SO we say there are two inner electrons, so the effective nuclear charge is positive four minus two, giving us an effective nuclear
charge of positive two. In reality, the effective
nuclear charge is approximately one point nine, and that's because beryllium
has another electron in its two s orbital over here, which does effect this
electron a little bit. It repels it a little bit, and so it actually deceases
the effective nuclear charge to about, one point nine. But again, for a quick
calculation, positive two works. So, the outer electron for beryllium, let's just choose this one again, is feeling an effective nuclear charge of positive two, 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. The effective nuclear
charge is only plus one for this outer electron, and because of this, the
beryllium atom is smaller, right? The two s orbital gets
smaller, and the atom itself is smaller. Beryllium is smaller than lithium. So this outer electron here, let me switch colors again, this outer electron 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, and that's the reason for
the higher ionization energy. So beryllium has an ionization energy of positive 900 kilojoules per mole, compared to lithium's of
520 kilojoules 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 kilojoules per mole to 900 kilojoules 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 kilojoules per mole for beryllium to only 800 kilojoules 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 one s two, two s two, and two p one. So that fifth electron
goes into a two p orbital, and the two p orbital is higher in energy than a two s orbital, which means the electron in the two p orbital is on average, further
away from the nucleus that the two electron
in the two s orbital. So if we just sketch
this out really quickly, let's say that's my two s orbital, I have two electrons in there, and this one electron in the two p orbital is on average further
away from the nucleus. So, those two electrons
in the two s orbital actually can repel this
electron in the two p orbital. So, there's a little bit
extra shielding there of the two p electron
from the full attraction of the nucleus, right? So, even though we have
five protons in the nucleus, and a positive five charge for boron, the fact that these two s 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 electron in the two p orbital away due to these two s 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 kilojoules per mole, down to about 1300 kilojoules
per mole for oxygen. So, 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 it's electron configuration is one s two, two s two, and two p three. So that takes care of all seven electrons. For oxygen, we have another electron, so one s two, two s two, two p four is the electron configuration for oxygen. Let's just draw using orbital notation the two s orbital and the two p orbital. So for nitrogen, here's our two s orbital. We have two electrons in there, so let's draw in our two electrons. And for our two p 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 two s orbital
for oxygen, which is full, so we'll sketch in those two electrons, and we have four electrons
in the two p orbitals. So let me draw in the two p orbitals. There's one electron, there's two, there's three, and 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 two p orbital, it's repelled by the electron
that's already there, which means it's easier to
remove one of those electrons, so electrons have like charges, and like charges repel. And so that's the reason for this slight decrease in ionization energy. So, 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 two p 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.