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Ionization energy: period trend

An element's first ionization energy is the energy required to remove the outermost, or least bound, electron from a neutral atom of the element. On the periodic table, first ionization energy generally increases as you move left to right across a period. This is due to increasing nuclear charge, which results in the outermost electron being more strongly bound to the nucleus. Created by Jay.

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  • blobby green style avatar for user Goh Zheng Feng
    I still dont really get it about the decreasing of IE from N to O. Can give me a simpler explanation?
    (29 votes)
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    • blobby green style avatar for user bibhorbista
      You know the maximum capacity of p orbital is 6 electrons. so N has three 2p electrons and O has four 2p electrons. This case is kind of exception. Since Nitrogen is half filled coz it has 3 electrons out of 6 electrons that can be occupied in p orbital it has higher ionization energy compared to that of oxygen.
      Half filled and full filled orbtials usually tend to show higher ionization energy . Thus the noble gas elements have high ionization energy. (FYI Fluorine has the highest ionization energy)
      (14 votes)
  • duskpin sapling style avatar for user Melody Tan
    If I'm not mistaken, the electrons (electron clouds or whatever it is called) are in constant motion around an atom. If this is the case how can the inner shell electrons "shield" them constantly? Wouldn't the ionization energy change?
    (10 votes)
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    • piceratops seed style avatar for user RogerP
      Although we can describe electrons in terms of mathematics, actually knowing what they look like is beyond us. You have probably heard of wave-particle duality which describes the behaviour of sub-atomic particles, such as electrons - they have the properties of being both waves and particles at the same time. If you wish to think of electrons as particles, then you have to accept they are everywhere at once, which is down to their wave function properties - they are not moving around from one place to another. If they are everywhere at once, then they are providing constant shielding.

      This is quantum mechanics which is extremely weird and can be very counter-intuitive.
      (14 votes)
  • piceratops ultimate style avatar for user Jeremy Duke
    Do electrons move between shells? If the orbitals are simply probability zones, would those zones ever overlap? Are the numbers of electrons per orbital also just be an average or a probability?
    (5 votes)
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    • starky ultimate style avatar for user ♪♫  Viola  ♫♪
      1. Electrons jump up to higher shells when excited and fall back down and release that energy as photons.
      2. Yes, orbitals can overlap.
      3. Not quite sure what you mean. Orbitals are probability locations of where a certain number of electrons are. We aren't uncertain about the umber of electrons, but their location and momentum. So I guess yes, if we are uncertain about whether there are 2 electrons in an s orbital, that could mean that there are fewer or more in there at a point in time.
      (5 votes)
  • hopper cool style avatar for user archi31
    What exactly is Effective Nuclear Charge?
    (4 votes)
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  • starky seedling style avatar for user Jenny Kim
    What does it mean by shielding?
    (3 votes)
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  • male robot hal style avatar for user RobinZhangTheGreat
    So, is ionization energy the energy needed to remove an electron from an atom? I'm confused. I thought ionization energy was to remove all the outer electrons.
    (3 votes)
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  • blobby green style avatar for user T
    Does this mean that Zeff remains constant going down a group?
    (2 votes)
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    • leaf red style avatar for user Richard
      Using the simplified definition of effective nuclear charge they use in the video, yes. Because if effective nuclear charge is: Zeff = Z – S, and we consider only the shielding of core electrons then the effective nuclear charge should be the same as the number of valence electrons which is constant in a group.

      In reality, no. The shielding effect of lower electrons does not cancel the attraction effect of protons in a 1:1 ratio. So the actual effective nuclear charge numbers are often higher then what we would calculate using the simplified definition in the video. They actually hint at this at . A more accurate way of calculating effective nuclear charge includes the use of Slater’s rules which better calculates the shielding.

      Hope that helps.
      (4 votes)
  • leafers seed style avatar for user bipul.soti
    I understand why the p orbital pairing means O has a lower ionization energy than C, but why does this not apply to F (i.e. why does ionization energy increase for F)? Two pairs are made, so doesn't that mean there should be even more repulsion between the electrons?
    (2 votes)
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    • leaf red style avatar for user Richard
      Well first, oxygen has a higher first ionization energy compared to carbon. Oxygen’s first ionization energy is 1313.9 kJ/mol, while carbon is 1086.5 kJ/mol.

      Fluorine has a high first ionization energy because it has such a high effective nuclear charge. Sure it has another paired p electron, but it also has more protons than the previous elements in the same period. So there is an increase in the repulsive force due to the extra electron, but there is also an attractive force due to the extra protons too. Of these two conflicting forces, the attractive force is higher in magnitude resulting in a net attractive force. A high effective nuclear charge means it’s difficult to remove electrons resulting in a high ionization energy.

      Hope that helps.
      (4 votes)
  • stelly orange style avatar for user royaleshu16
    so does the shielding effect remain same or increase accross the period?
    (2 votes)
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    • leaf red style avatar for user Richard
      Yes, going from left to right across a period the electron shielding increases. Jay hints at this in the video, but core electrons are not the only electrons which shield valance electrons, the other valence electrons in the same shell also provide shielding. So as we move from left to right along a period, the number of valence electrons increases and so too does the shielding as a result.

      Keep in mind the effective nuclear charge also increases as we move from left to right because the number of protons increases. If both the nuclear charge and the shielding increase, conflicting forces, as we move from left to right and the effective nuclear charge, that must mean the increase in protons is able to overcome the increase in shielding from the extra electrons.

      Hope that helps.
      (4 votes)
  • old spice man blue style avatar for user Owen
    How can average number of shielding electrons be anything other than an integer? Is it possible to have a fraction of an electron, or am I misunderstanding something?
    (2 votes)
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    • leafers sapling style avatar for user Isaac
      Electrons are quantum phenomena (probability clouds!), so it is entirely possible to have a fraction of an electron charge in the right general location to perform shielding. Some electrons, as a result, are less useful for shielding than others, hence the decimal output. For a more precise calculation of the decimal portion, Slater's rules can be used, although those calculations are a little more involved.
      (3 votes)

Video transcript

- [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.