Which of the following systems will have the larger ionization energy? Explain your reasoning. i) an electron at a distance of \(500 \mathrm{pm}\) from a nucleus with charge \(+2\) ii) an electron at a distance of \(700 \mathrm{pm}\) from a nucleus with charge \(+2\)

Diving into the subject of ionization energy requires a basic understanding of electrical interaction. It's rooted in the fundamental laws of physics that describe how charged particles, such as protons and electrons, interact. According to Coulomb's Law, the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

In the context of an atom, the positively charged nucleus exerts an attractive force on negatively charged electrons. This force is what keeps electrons in place around the nucleus. When we talk about ionization energy, we're essentially discussing the energy required to overcome this electrical interaction to liberate an electron from an atom's grasp.

Given two charged particles—or in our case, a nucleus and an electron—the strength of the electrical interaction will determine the ionization energy needed. The stronger the interaction (that is, the closer the electron is to the nucleus or the greater the charge of the nucleus), the higher the ionization energy required to remove the electron.

The charge of an atomic nucleus plays a crucial role in determining an atom's ionization energy. This charge is given by the number of protons in the nucleus, with each proton contributing a single positive charge. For instance, helium has an atomic nucleus charge of +2 because it has two protons.

This positive charge draws electrons towards the nucleus due to electrostatic attraction, and the more protons there are, the stronger this pull becomes. This is why elements with a higher nuclear charge generally have higher ionization energies. Electrons in such an environment feel a considerably stronger pull and require more energy to be stripped away from the atom.

In our exercise, both nuclei have a charge of +2, implying that they are equally effective at attracting their respective electrons. Consequently, in this comparison, the atomic nucleus charge is not the differentiating factor; both systems have the potential to exert the same magnitude of force on their electrons. However, as we will see, this is only part of the story.

The electron-nucleus distance, often measured in picometers (pm), is a measure of how far an electron is from the nucleus of an atom. It is a critical factor in determining the ionization energy because, as previously mentioned, the electrical interaction between the electron and the nucleus weakens as this distance increases.

In the context of our exercise, one electron is 500 pm away while the other is 700 pm away, both from nuclei with the same charge. The one closer at 500 pm is held more tightly by the nucleus than the one further away at 700 pm, due to the inverse square law governing electrostatic forces. Here, a shorter electron-nucleus distance translates to a significantly increased attraction, therefore a higher ionization energy would be required to remove the electron.

To conclude, the crucial insight from this concept is that with all other factors being equal, an electron that's closer to the nucleus will have a greater binding energy, demanding more ionization energy for its release. Hence, in the exercise presented, the electron at 500 pm distance has a larger ionization energy compared to the electron at 700 pm.