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Main content
Current time:0:00Total duration:10:03
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)

Video transcript

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 goes up to 900 kilojoules per mole and then again and general we see this increase in ionization energies going over to neon so going across a period is an increase in the ionization energy that's because as we go across our period there's an increase in the effective nuclear charge so increase in the effective and remember the formula for that is the effective nuclear charge is equal to on 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 3 so 3 protons in the nucleus so positive 3 charge and lithium's electron configuration we know is 1s2 2s1 so two electrons in our 1s orbital and one electron in the 2's orbital beryllium has one more proton and one more electron so one more proton in the nucleus so a +4 charge and for beryllium the electron configuration is 1s2 2s2 so two electrons in the 1s orbital and then two electrons in the 2'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 +3 charge in the nucleus so the effective nuclear charge is equal to positive 3 and from that we subtract the average number of inner electrons shielding or outer electrons in this case we have these 2 and 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 repelled 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 +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 right so there's an attractive force between the outer electron and and our nucleus now the actual calculation for this on z2 s I should say 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 so it's a similar situation we have two inner electrons that are shielding this outer electron right they're repelling this outer electron shielding the outer electron 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 up 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 a 2 s orbital over here 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 to work so the outer electron for beryllium let's just choose this one again it's 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 the effective nuclear charge is only plus 1 for this outer electron and because of this the beryllium atom is smaller right the 2'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 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 a 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 we saw to the ionization what energy went from positive 520 kilojoules roll to 900 kilojoules per mole and we said that was because of the increased effective nuclear charge for beryllium but as you 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 there's a slight decrease in the ionization energy and let's look at the electron configuration of boron see if we can explain that boron has 5 electrons so the electron configuration is 1s2 2s2 and 2p 1 so that fifth electron of boron goes into a 2p orbital and the 2p orbital is higher in energy than a 2's orbital which means the electron the 2p orbital is on average further away from the nucleus and the two electrons in the u.s. orbital so we just get this out really quickly let's say that's my 2's orbital I have two electrons in there and this 1 electron in the 2p orbital is on average further away from the nucleus so those two electrons in the 2's orbital actually can repel this electron in the 2p orbital so there's a little bit extra shielding there of the 2p electron from the full attraction of the nucleus right so even though we have 5 protons a nucleus and a positive 5 charge for boron the fact that these 2'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 the 2p orbital away due to these 2'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 and 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 1,400 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 7 electrons to think about so its electron configuration is 1s2 2s2 and 2p threes that 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 2's orbital and the 2p orbitals so for nitrogen here's our 2's orbital we have two electrons in there so let's draw in our two electrons and for our 2p orbitals we have 3 electrons so here are the 2p orbitals and let's draw in our 3 electrons using orbital note let's do the same thing for oxygen so there's a 2 s orbital for oxygen which is full so we'll sketch in those two electrons we have 4 electrons in the 2p orbitals which are on the 2p orbitals there's one electron there's 2 there's 3 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 2p orbital its 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 2p orbital from there on we see our our general trend again the ionization energy for fluorine is up to 16 81 and then again for neon we see an increase in the ionization energy due to the increased effective nuclear charge