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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.

## Want to join the conversation?

• 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?
• Listen to the full sentence, he says "the last thing they want to do is lose an electron", ie. they do not want to lose an electron
• 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?
• 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
• 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?"
• Why radius is decreasing from left to right?
• 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.
• 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?
(1 vote)
• 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.
• 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?
(1 vote)
• 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.

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?
• 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.
• 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...
(1 vote)
• 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.
• It is said 'by most measures Helium is considered to be the smallest atom', but those are all theoretical measurements; and on the list of the actual test of those hypotheses, the empirical measurements, hydrogen is less than half the predicted one, being thus much smaller than helium.

25 to 120 is almost five times smaller to the point it lists more half a dozen elements smaller than helium; although in single covalent bonds the different is less stark being 28 for hydrogen and 32 for helium.

That seems like a major things, a true plothole, with huge effects on the entire understand of the periodic table trends and thus atomic theory in general; the predictions mean nothing if they are so wrong that when actually measured we get wildly different results.

Yet I could not find a single article tackling the issue; there are several on the lanthanide and d-block contradictions, but naught on the hydrogen one.

The best hint I saw for it being ignored was due to discrepancy in muon measurements, as the Proton radius puzzle, and that it hints to these missing pieces of physics on the understanding of charges, but found absolutely no direct tackle on the hydrogen atom radius topic.

Is there anywhere with active research and hypotheses covering that for me to look it up?
Or is there any reason I did not find for that elephant in the room to be so ignored?
• The difficulty in measuring the size of an atom is that it requires you measuring the positions of electrons from the nucleus. This is difficult because electrons are not solid particles in the classical physics sense, but rather more as a wave using quantum mechanics. Electron's orbits around nuclei are defined as wave functions where there are basically regions of high probability of finding an electron at any instant. So there isn't a solid boundary that defines the edge of an atom.

However, on average the distance of helium's electrons to its nucleus is smaller than hydrogen and its electron.

Hope that helps.