Question

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.

Short answer

The graph of the third ionization energy versus atomic number for the elements Na through Ar shows an increasing trend as we move from left to right across the period, mainly due to increasing nuclear charge and decreasing atomic radius. However, there are exceptions for Aluminium and Sulfur, which exhibit slight drops in ionization energy due to their electron configurations. Removing the third electron from Aluminium breaks the 3rd energy shell, while in Sulfur, this results in a more stable, half-filled orbital. The ionization energy reaches its peak at Argon due to its full electron configuration, making it the most difficult element for electron removal.

Step-by-step solution

Identify the elements and their atomic numbers

We are asked to analyze the elements from Sodium (Na) to Argon (Ar). Their atomic numbers and electron configurations are as follows: - Na (Sodium): \(Z=11\), [2,8,1] - Mg (Magnesium): \(Z=12\), [2,8,2] - Al (Aluminium): \(Z=13\), [2,8,3] - Si (Silicon): \(Z=14\), [2,8,4] - P (Phosphorus): \(Z=15\), [2,8,5] - S (Sulfur): \(Z=16\), [2,8,6] - Cl (Chlorine): \(Z=17\), [2,8,7] - Ar (Argon): \(Z=18\), [2,8,8]

Determine the third ionization energy trends according to the factors.

In this step, we consider the factors affecting ionization energy and apply them to the elements within our range: 1. Nuclear charge increases as we move from left to right across a period. Thus, we expect ionization energy to increase for each successive element. 2. Electron shielding doesn't hugely affect this range since the electrons are all in the same energy level. Therefore, we will not consider it a significant factor for our graph. 3. Atomic radius decreases as we move from left to right across a period. This means the ionization energy should increase as atomic radius decreases.

Sketch the qualitative graph.

Based on the periodic trends mentioned before, the graph for the elements should show an increasing third ionization energy from Na to Ar. However, there will be an exception for Aluminium and Sulfur, as their electron configurations make them relatively more stable. Aluminium has three valence electrons, and removing the third electron from it involves breaking the 3rd energy shell. In the case of sulfur, the third electron's removal will make it half-filled, which is also more stable. Consequently, both Aluminium and Sulfur will have a slight drop in ionization energy compared to what is expected. This graph should look like: - Ionization energy increases from Na to Mg. - There is a small drop in ionization energy for Al. - Ionization energy rises again from Al to Si and continues increasing to P. - For S, there is another minor drop in ionization energy. - Ionization energy increases from S to Cl and reaches the highest value for Ar.

Explain the graph

The graph of the third ionization energy versus atomic number for the elements Na through Ar shows an increasing trend as we move from left to right across the period. This increase is mainly attributed to the increasing nuclear charge and decreasing atomic radius. The exceptions for Aluminium and Sulfur demonstrate that electron configurations play a significant role in determining ionization energy. Aluminium's small drop is due to breaking the 3rd energy level to remove the third electron. On the other hand, Sulfur's drop is a result of attaining a more stable electron configuration with half-filled orbitals. Finally, the ionization energy reaches the highest value at Argon, having a full electron configuration, making it the most stable and difficult to remove an electron from it.