Second Ionisation Energy Confusion

Question

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?

Answer 1

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.

Answer 2

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.