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It is intuitive that the second ionisation will require more energy than the first. However, I'm having trouble explaining exactly why. I came across this other question on a similar topic: Why second ionisation energy greater than the first?

The answer made sense until the answer said:

Technically, there would be an increased attractive force ($\propto kq_1q_2/r^2$) from the nucleus, because the electrons on average will now be closer to the nucleus.

Considering, say, chlorine, when a valence electron is removed, it makes sense that the electrostatic repulsion between valence electrons would decrease. However, by removing 1 electron, why does this increase the attractive force? I don't see how would increase the coulombic force between the nucleus and a valence electron, since the magnitude of both charges are constant.

If it was the radius changing, why should the electrons be closer to the nucleus just because 1 electron is removed?

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  • $\begingroup$ Just to get a sense: Think in term of net charge. There is no charge for the first ionisation. After your test charge, ie the second electron to be removed, sense a net one $\endgroup$ – Alchimista Oct 28 '19 at 8:25
  • $\begingroup$ @Alchimista but why does having a net charge matter? Why would that affect the energy to remove 1 electron? The electron still has the same force of attraction to the nucleus essentially right? $\endgroup$ – John Hon Oct 28 '19 at 8:43
  • $\begingroup$ Beside that quantum mechanics applies, just think of a distribution of n positive and n negative charges. Upon removal of the first one you get a net charge. Obviously the field is now different and the force experienced by each charge is now greater. Don't look at protons and the electron to be removed only, look at the entire distribution. It can be an exercise in the electrostatic chapter. $\endgroup$ – Alchimista Oct 28 '19 at 9:13
  • $\begingroup$ @Alchimista No I don't see why the force experienced by each charge would be greater. Say we have 10 protons and 10 electrons in your arrangement. Each electron will experience 10 times the coloumbic force of 1 proton attracting it and 9 times the coloumbic force of 1 electron repelling it. If we remove 9 electrons, such that there's only one, then the attractive force felt must be the same. All that has decreased is the repulsion force. Is this what you mean by an "increase in force felt"? $\endgroup$ – John Hon Oct 28 '19 at 22:40
  • $\begingroup$ Well yes. Force is the resulting vector. That is why I invite you to look at a bunch of charges. Two plus and two minus suffice, they don't have to be specifically arranged. Just do it. $\endgroup$ – Alchimista Oct 29 '19 at 8:32
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Beside that the proportional relationship proposed could be discussed further, as r rearrange, all this statement are related to the sums of forces. They stated that there would be a greater attractive force because they discuss ionization and this is the case. But the attractive force is the result of all attractions and repulsions. Any distribution of electric charges would lead to this situation and is generally an exercise treated after Coulomb law. Any reader can do that providing the dynamics is not involved.

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In classical models of the atom, the radius of an electron subshell changes with the number of electrons in it (and with nuclear charge). This is essentially because the radius at which a set of electrons "settle" is a result of a balance involving the electrical forces, the magnetic forces and centrifugal forces. For example, the radius of $\ce{He}$ is 0.567 (in Bohr units). When it ionizes to $\ce{He+}$, one electron moves to infinity and the other settles at a radius of 0.500 because the force balance has changed with the departure of the first electron.

Having said that, for the larger elements the main driver for increasing ionization energy (within a given subshell) is the increase of effective nuclear charge as you peel away electrons. In the case of Chlorine, the first electron to ionize effectively feels a +1 nuclear charge as it speeds away from its orbit to infinity (17 protons screened by the remaining 16 electrons). If the second electron departs, it will feel a +2 nuclear charge since the nucleus is now only screened by the remaining 15 electrons, leading to an ionization energy that is roughly twice as high, and so on.

The 3p subshell radius only changes by a few percent as a result of these successive departures, so the contribution of the change in radius to the ionization energy is small. It's only when you start peeling away the 7th electron (from the 2p shell) that you see a significant increase in ionization energies due to the much smaller radius, followed by another step-change when you reach the 1s electrons.

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