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Periodic trends and Coulomb's law

Periodic trends (such as electronegativity, electron affinity, atomic and ionic radii, and ionization energy) can be understood in terms of Coulomb's law, which is Fₑ = (qq₂)/r². For example, consider first ionization energy: Coulomb's law tells us that the greater the nuclear charge (q₁) and the shorter the distance between the nucleus and the outermost electron (r), the stronger the attraction between the nucleus and the electron. As a result, the electron will require more energy to remove. Created by Sal Khan.

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  • leafers ultimate style avatar for user Charles Liu
    At , Mr.Khan said that Florine and Chlorine wants to lose an electron. However, they want to gain an electron I think. Am I right?
    (5 votes)
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  • stelly green style avatar for user Taimas
    Why radius is decreasing from left to right?
    (9 votes)
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    • leaf red style avatar for user Richard
      As you go from left to right across a period the effective nuclear charge of the atoms increases because the number of protons is increasing. A greater effective nuclear charge means the positive charge of the protons from the nucleus is felt more strongly by the valence electrons resulting in a stronger force of attraction. A stronger force of attraction between the nucleus and the valence electrons means that the atomic radius will decrease as the valence electrons are pulled in closer to the nucleus. This kind of force of attraction is electrical and is described by Coulomb's Law which states that the force between unlike charges is inversely proportional the distance between them.

      Hope that helps.
      (19 votes)
  • aqualine ultimate style avatar for user Aditya Kumbhare
    at , the video said that noble gases Neon and Argon have an effective charge of 8. but if the outer shell is full, shouldn't the electrons in the outer shell be counted as part of the core electrons, leaving an effective charge of 0?
    (9 votes)
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    • mr pants purple style avatar for user Ryan W
      They’re talking about the effective nuclear charge that the outermost electrons feel, a rough calculation for this is number of protons - number of core electrons

      Ne has 10 protons and 2 core electrons, 10 - 2 = 8
      (12 votes)
  • male robot hal style avatar for user Joshua Newburn
    At , Sal says that when you release energy, that is conventionally recognized as a "negative affinity." This contradicts what he said just before about Flourine releasing energy and having "high affinity." Could I get some clarification about the meaning of "affinity?"
    (11 votes)
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  • blobby green style avatar for user Peter Patterson
    At , it says that the radius of an atom can be decreased by the Coulomb's force. But at , we can see that the decreasing radius will make the Coulomb's force even stronger. So why do electrons stay on the shells instead of crashing into the nucleus?
    (2 votes)
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    • leaf red style avatar for user Richard
      So the issue with thinking electrons could crash into the nucleus is that it assumes electrons are solid particles orbiting the nucleus according to Coulomb's Law much like how planets orbit the sun because of gravity. This is using classical mechanics to describe electrons when in reality they behave much more differently. To accurately describe the nature of electrons, we must resort to quantum mechanics. Under quantum mechanics the electron is a quantized wave function which occupies certain probable regions around the nucleus called orbitals. And using this understanding of the electron, electrons in the atom do enter the nucleus. In fact, electrons in the s orbitals tend to peak at the nucleus. An electron in an atom spreads out according to its energy. The states with more energy are more spread out. All electron states overlap with the nucleus, so the concept of an electron "crashing into" the nucleus does not really make sense. Electrons are always partially in the nucleus. Hope that starts to clarify things a little more.
      (14 votes)
  • duskpin ultimate style avatar for user Oblivion
    I don't quite understand how the radius of an atom decreases / increases in the Periodic Table. At , Sal says that the radius of an atom decreases as you go from left to right in the Table--why exactly is that? Shouldn't the radius get larger, since there are more shells?
    (2 votes)
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    • leaf red style avatar for user Richard
      If you move left to right along the same period (row), then you’re increasing the atomic number of the elements and therefore the number of protons in the nucleus. More protons mean a greater effective nuclear charge, or a greater attractive force felt by the atom’s electrons to the nucleus. If this attractive force is greater, then the electrons are brought in closer to the nucleus resulting in a smaller atom.

      If you move up to down along the same group (column), then you’re increasing the number of electron shells for the atom. Higher electron shells make electrons orbit the nucleus at greater distance from the nucleus resulting in a larger atom.

      Putting these two trends together, then the smallest atoms are in the upper right of the periodic table. They have the most protons for their periods, and the least number of electron shells for their groups.

      Hope that helps.
      (12 votes)
  • blobby green style avatar for user Y K
    A little confused about radius.

    How does radius get smaller as we go to the right? If we're going to the right, we have more electrons meaning we're filling more shells which increases radius and repulsion.

    For example, Li only fills 1s^2 2s^1, but Ne fills 1s^2 2s^2 2p^6. It's going out further (increasing radius) because of having more electrons, isn't it?
    (4 votes)
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    • leaf red style avatar for user Richard
      As we move from left to right along the same period (row) on the periodic table, both the atomic number (number of protons) and number of valence electrons are increasing. The protons provide an attractive force to the valence electrons while the other valence electrons repel each other. These are two conflicting force at work, but the attractive force from the protons is greater in magnitude so the valence electrons feel a greater net attractive force as you move to the right. A greater attractive force means the valence electrons are pulled in closer to the nucleus and the atom has a smaller atomic radius as a result.

      Hope that helps.
      (5 votes)
  • blobby green style avatar for user Tyrus Reidt
    So the noble gasses's core electrons are the atomic number of the previous noble gas and not their atomic number (i.e. Ar has 10 core electrons and not 18)? In the valence electron video in lesson five, he said that Ar has 18 core electrons...
    (2 votes)
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    • leaf red style avatar for user Richard
      The core electrons for any atom are all the electrons of the atom which aren't valence electrons. For noble gases the valence electrons are the s and p electrons of the highest electron shell. All other electrons would be considered core electrons.

      For helium it only has two 1s electrons which constitute the valence shell. There isn't a shell below the first electron shell so it has no core electrons.

      Neon's valence electrons are the 2s and 2p electrons (in total eight) and the two 1s electrons in the lower shell are the core electrons.

      Argon also has eight valence electrons, but this time the 3s and 3p electrons meaning the remaining ten in the lower first and seconds shells are core electrons. So your rule does work, but only for noble gases up to argon.

      Krypton's valence electrons are the 4s and 4p electrons (again in total eight) which means the remaining 28 electrons are core electrons. The reason your rule begins to break down at krypton is because the filling of the d subshell adds an extra 10 electrons to the core electrons. So it's no longer simple the previous noble gas's (Argon) atomic number because of the d electrons. It breaks down even more when you get to the heavier noble gases and you have to worry about the f electrons too.

      The previous video talked about using noble gas notation to write the electron configurations of higher atomic number elements. The idea is that noble gases have completed shells so elements can build off the electron configurations of previous noble gases to express their own electrons configuration.

      So writing calcium's electron configuration would look like:
      1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)4s^(2), normally.
      But with noble gas notation we can use the noble gas prior to calcium, argon, as a foundation. Argon's electron configuration is: 1s^(2)2s^(2)2p^(6)3s^(2)3p^(6), which we can condense by saying [Ar].
      So calcium's electron configuration can also be written as:
      [Ar]4s^(2)

      Noble gas notation is not saying that a noble gas like argon has 18 core electrons, rather that it has a complete valence shell.

      Hope that helps.
      (8 votes)
  • piceratops tree style avatar for user rexchidera01
    Thanks to Sal Khan and all the other engineers who have helped me succeed in Chemistry.
    (4 votes)
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  • boggle purple style avatar for user lily J
    What's the difference between electron affinity and electronegativity? Aren't they both mean the likelihood of gaining an electron? And please, please answer in COMMON words, do not use words that are so professional that I can't understand a word of it...
    Thanks! :)
    (2 votes)
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    • blobby green style avatar for user ezelisondo
      Electron Affinity: How much energy is released when an atom gains an electron. (How much an atom "likes" gaining an electron) Electronegativity: How strongly an atom attracts electrons when it is in a compound. (How good an atom is at "pulling" shared electrons towards itself)
      (1 vote)

Video transcript

- [Instructor] In this video we're going to look at trends for the periodic table of elements for dimensions like ionization energy, atomic and ionic radii, electron affinity, and electro negativity. And to do so, we're going to start with a very fundamental idea in chemistry or physics, and that's Coulomb's Law. And for our point of view, we can view Coulomb's Law as saying that the magnitude of the force between two charged particles is going to be proportional, that just means proportional right there, is going to be proportional to the charge on the first particle times the charge on the second particle, divided by the distance between those two particles, squared. When we're thinking about it in context of the periodic table of elements and various atoms, you can view q1 as the effective positive charge from the protons in the nucleus of an atom. You can view q2 as the charge of an electron. Now any given electron is going to have the same negative charge, but as we try to understand trends in the period table of elements, it's really the outer most shell electrons, the valence electrons, that are most interesting. Those are the ones that describe the reactivity. And so when we think about the distance between the two charges we're mainly going to be thinking about the distance between the nucleus and those outer most valence electrons. Now we can view this effective charge, I'll call it z-effective, as being equal to the difference between the charge in the nucleus, so you can just view this as the atomic number, atomic number or the number of protons that a given element or an atom of that element has, and the difference between that and what is often known as S, or how much shielding there is. Now there is complicated models for that, but for an introductory chemistry class, this is often approximated by the number of core electrons. Remember, we really want to think about what's going on with the valence electrons. And so if you imagine a nucleus here, do that orange color, that has protons in it. And so you have core electrons. Let's say these are the core electrons in the first shell, and then you have some core electrons in the second shell. And let's say the valence electrons are in the third shell. So let's say these are some valence electrons here, they're blurred around, they're in these orbitals. Those valence electrons, which have a negative charge, are going to be attracted to the positive charge of the nucleus but they're also going to be repulsed by all these core electrons that are in between them. And so that's why an approximation of the effective charge that these valence electrons might experience is going to be the charge of the nucleus minus, and this is an approximation, the number of core electrons that you have. So if we use that roughly as a way to think about z-effective, what do you think are going to be the trends in the periodic table of elements? What would be the effective charge for the Group I elements over here? Well, Hydrogen has no core electrons and it has an atomic number of 1. So 1 minus 0 is going to have an effective charge of roughly 1. Lithium atomic number of 3, minus 2 core electrons that are in 1-S, so once again you're going to have 3 minus 2, effective charge of 1. So roughly speaking, all of these Group I elements have an effective charge of 1. What if you were to go to the halogens? What's the effective charge there? Well if you look at Flourine, atomic number of 9, has 2 core electrons in the first shell, so has an effective charge of 7. Chlorine actually has an effective charge of 7 for the same reason. Atomic number of 17, but 10 core electrons. If you go even further to the right, to the noble gases, you see that Helium is going to have an effective charge of 2, atomic number of 2 minus 0 core electrons. But then when you get to Neon, you have an atomic number of 10, and then minus only 2 core electrons. And you'll see as you go down these noble gases, other than Helium, they have an effective charge of 8. And so the general trend is, your effective charge is low at the left, effective charge low for Group I, and then when you go to the right of the periodic table, you have a z-effective, is going to be high. So within a given period, or within a given row in the periodic table of elements, your outer electrons, your valence electrons, are in the same shell. But the effective charge is increasing as you go from left to right. So this q1 right over here is going to be increasing. So what is that going to do to the radius of the atom? Well, Coulomb's Law will say that the magnitude of the attractive force between those opposite charges is going to be stronger. And so even though you're adding electrons as you go from left to right within a row, within a period, the atoms in general are actually going to get smaller. Let me write it this way. So as you go from left to right, generally speaking, radius decreases. Now what's the trend within a column? Well one way to think about it is, as you go down a column, as you go down a Group, you're filling shells that are further out. And so you'd expect radius to increase as you go down a column, or down a Group. Or you could say radius decreases as you go up a group. So radius decreases. So overall what's the trend in the periodic table of elements? Well radius is going to decrease as you go up and to the right. And so you could draw an arrow something like this. And it is indeed the case that by most measures, Helium is considered to be the smallest atom, a neutral Helium atom. And Francium is considered to be the largest atom. So could we use this to think about other trends in the periodic table of elements? What about, for example, ionization energy? Just as a reminder, the first ionization energy is the minimum energy required to remove that first electron from a neutral version of that element. And since it's the minimum energy, it's going to be one of those outer most electrons. It's going to be one of the valence electrons. And so what's going to drive that? Well you can imagine the ionization energy is going to be high in cases where the Coulomb forces are high. And what are the situations where the Coulomb forces are high? Well this is going to be a situation where you have a high effective charge and where you have a low radius. Low radius makes the Coulomb forces high. And effective charge makes the Coulomb forces high. So where is that true? So you have the lowest radii at the top right and you have the highest effective charge at the right. So you would expect the highest ionization energies to occur in the top right. So high ionization energy. And that actually makes intuitive sense. These noble gases are very stable. They don't want to release an electron. So it's going to take a lot of energy to take one of those electrons away. Fluorine or Chlorine, they're so close to completing a shell, the last thing they want to do is lose an electron. So once again, it takes a lot of energy to take that first electron away. On the other hand, if you go to something like Francium, it has one valence electron. And that valence electron is pretty far from the nucleus. And there's a low effective charge despite all the protons because there's so much shielding from all those core electrons. So it's not surprising that it doesn't take a ton of energy to remove that first electron from Francium. Now another trend that we can think about, which is in some ways the opposite, is electron affinity. Ionization energy is talking about the energy it takes to remove an electron. Electron affinity thinks about how much energy is released if we add an electron to a neutral version of a given element. So high electron affinity elements, these are the ones that really want electrons. So they should have a high Coulomb force between their nucleus and the outer most electrons. And so that means they should have a high effective z, and that also means that they should have a low r. So one way to think about it, you're going to have a similar trend with the one difference that the noble gases don't like gaining or losing electrons. But we do know that the Flourines and the Chlorines of the world can be become more stable if the gain an electron. They can actually release energy. So you actually have high electron affinities for the top right, especially the Halogens. And you have low electron affinities at the bottom left. Now there's one little quirk in chemistry conventions, people will generally say that Fluorine and Chlorine and the things in the top right that aren't noble gases, have a high electron affinity. And it is the case that energy is released when you add an electron to a neutral version of them. It just happens to be that the convention, and this can get a little confusing, is that when you release energy you have a negative electron affinity. But generally speaking, when they say a high electron affinity, this thing's going to release more energy when it's able to grab an electron. Now a notion that is related to electron affinity is electro negativity. And the difference between the two can sometimes be a little bit confusing. Electro negativity is all about when an atom shares a pair of electrons with another atom, how likely is it to attract that pair to itself versus for the pair to be attracted away from it to the other one? And so you can imagine it correlates very strongly with electron affinity. Things that release energy when they're able to be ionized to grab an electron, if they form a bond and they're sharing a pair of electrons, they are more likely to hog those electrons. Electron affinity is easier to measure. You can actually see when this element's in a gaseous state if you add electrons how much energy is released, it's normally measured in kilojoules per mole of the atom in question. While electro negativity isn't as clear cut on how to measure it, but it can be a useful concept in future videos as we think about different atoms sharing pairs of electrons and where do the electrons spend most of their time. So I'll leave you there. We started with Coulomb forces and we were able to intuit a whole bunch of trends just thinking about Coulomb's Law and the periodic table of elements.