Without looking at data in the text, sketch a qualitative graph of the third ionization energy versus atomic number for the elements Na through Ar, and explain your graph.

When it comes to ionization energy, there is a general trend where it increases across a period (from left to right) and decreases down a group (from top to bottom) in the periodic table. This occurs because as we move across a period, the effective nuclear charge increases, making it more difficult for electrons to be removed. On the other hand, as we go down a group, the atomic size increases and there is more electron shielding, making electrons easier to be removed. Since we are dealing with elements Na through Ar, which belong to the third period, we need to focus on the trend within this period.

The third ionization energy refers to the energy required to remove the third electron from an atom. As we move across the third period, the first ionization energy will initially have a significant increase due to the increasing nuclear charge. However, since we are focusing on the third ionization energy, some unique characteristics must be taken into consideration when sketching a graph. First, we need to understand that the second and third ionization energies are significantly higher than the first ionization energy. Thus, there will be a sharp increase in ionization energy between the third and second ionization energies for each element. This is because, after removing the first electron, an electron needs to be removed from a shell closer to the nucleus, which requires more energy.

Using the information above, we can now sketch the graph. The graph should show the increasing trend of the third ionization energy as we move across the period from Na to Ar. There will be some fluctuations due to the electron configuration of each element, but the general pattern remains. Some specific points to consider include: 1. A slight decrease can be observed at Mg and Al due to their lower electron shielding. 2. A noticeable decrease can be observed at Si due to its half-filled p-orbital. 3. A sharp increase between P and S due to the electron configuration of sulfur.

The graph can be explained as follows: 1. The overall increasing trend is due to the general ionization energy trend across periods, where the effective nuclear charge increases, making it harder for electrons to be removed. 2. Decreases observed at Mg, Al, and Si can be attributed to changes in electron shielding or orbital configuration. 3. A sharp increase between P and S is due to sulfur's full p-orbital, requiring more energy to remove an electron. 4. The magnitude of the third ionization energy is much higher compared to the first and second ionization energies due to the increased energy required for removing electrons closer to the nucleus. By following these steps, a qualitative graph of the third ionization energy versus atomic number for the elements Na through Ar can be sketched with an explanation that accounts for the general trend and specific fluctuations for each element.