Question

Consider the following ionization energies for aluminum: $$ \begin{aligned} \mathrm{Al}(g) \longrightarrow \mathrm{Al}^{+}(g)+\mathrm{e}^{-} & I_{1}=580 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{Al}^{+}(g) \longrightarrow \mathrm{Al}^{2+}(g)+\mathrm{e}^{-} & I_{2}=1815 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ $$ \begin{array}{ll} \mathrm{Al}^{2+}(g) \longrightarrow \mathrm{Al}^{3+}(g)+\mathrm{e}^{-} & I_{3}=2740 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{Al}^{3+}(g) \longrightarrow \mathrm{Al}^{4+}(g)+\mathrm{e}^{-} & I_{4}=11,600 \mathrm{~kJ} / \mathrm{mol} \end{array} $$ a. Account for the trend in the values of the ionization energies. b. Explain the large increase between \(I_{3}\) and \(I_{4}\). c. Which one of the four ions has the greatest electron affinity? Explain. d. List the four aluminum ions given in order of increasing size, and explain your ordering. (Hint: Remember that most of the size of an atom or ion is due to its electrons.)

Short answer

In summary: a. The trend in ionization energies of aluminum is caused by an increase in effective nuclear charge (Z) as more electrons are removed. The more tightly an electron is held, the higher the energy required to remove it. b. The large increase between \(I_{3}\) and \(I_{4}\) is due to the need to remove an inner 2p electron, which is well-shielded and near the nucleus, requiring a considerable amount of energy. c. Al\(^{2+}\) has the greatest electron affinity due to its relatively less positive charge and the vacant 3s subshell. d. The order of increasing size for the aluminum ions is: Al\(^{4+}\) < Al\(^{3+}\) < Al\(^{2+}\) < Al\(^+\) < Al. This order is based on the increased effective nuclear charge as electrons are removed, leading to a decrease in ion size.

Step-by-step solution

a. Analyzing the trend of ionization energies

To account for the trend in the values of the ionization energies, we need to examine the electronic configurations of the aluminum ions. The electron configuration for neutral aluminum (Al) is \(1s^2 2s^2 2p^6 3s^2 3p^1\). As aluminum loses electrons and forms positively charged ions, the electronic configuration changes. For each ion, we'll write the electronic configuration and analyze why the ionization energy is increasing: 1. Al: \(1s^2 2s^2 2p^6 3s^2 3p^1\) - The first electron is removed from the outermost electron shell (3p). 2. Al\(^+\): \(1s^2 2s^2 2p^6 3s^2\) - The second electron is removed from the 3s subshell. 3. Al\(^{2+}\): \(1s^2 2s^2 2p^6 3s^1\) - The third electron is removed from the 3s subshell. With each ionization step, the effective nuclear charge (Z) increases, and the remaining electrons are held more tightly. Consequently, more energy is required to remove them, leading to an increase in ionization energy.

b. Explaining the large increase between \(I_{3}\) and \(I_{4}\)

The electron configuration for Al\(^{3+}\) is \(1s^2 2s^2 2p^6\). In this case, Al\(^{3+}\) has completely filled lower energy shells (1s, 2s, and 2p). To remove the fourth electron and form Al\(^{4+}\), we need to remove one of the inner 2p electrons, which are well-shielded and near the nucleus. This process requires a significant amount of energy, resulting in a large increase between \(I_{3}\) and \(I_{4}\).

c. Identifying the ion with the greatest electron affinity

Electron affinity is the energy change that occurs when a neutral atom gains an electron. A larger electron affinity means that the atom is more likely to attract and retain an additional electron. Since Al\(^{3+}\) ion has a fully filled lower energy shell with higher stability, it does not desire to gain an electron as much. On the other hand, Al, Al\(^+\), and Al\(^{2+}\) have vacancies in their valence shells, which means they are more likely to accept an electron. Thus, out of the four aluminum ions, Al\(^{2+}\) has the greatest electron affinity due to its relatively less positive charge and the vacant 3s subshell.

d. Listing the four aluminum ions in order of increasing size and explaining the order

The order of increasing size for the four aluminum ions is: Al\(^{4+}\) < Al\(^{3+}\) < Al\(^{2+}\) < Al\(^+\) < Al As the positive charge increases, the effective nuclear charge also increases, causing the remaining electrons to be held more tightly. With each removal of an electron, the ion size decreases because of the increasing effective nuclear charge. Consequently, Al\(^{4+}\) is the smallest ion, and Al\(^{3+}\), Al\(^{2+}\), and Al\(^+\) follow sequentially in increasing size, with neutral Al being the largest.

Key concepts

Aluminum Ions

Aluminum ions are simply forms of aluminum atoms that have lost some of their electrons, becoming positively charged. When aluminum (Al) loses electrons, it can form different ions, such as Al⁺, Al²⁺, Al³⁺, and Al⁴⁺. These ions each have a different number of electrons but share the same nucleus. As electrons are removed, the atom now becomes an "ion", and it carries a positive charge. This is because electrons, which are negatively charged, no longer balance the positive charge of the protons in the nucleus.
The more electrons are lost, the more positively charged the ion becomes. In the case of aluminum, it primarily forms the Al³⁺ ion in many of its compounds, as removing three electrons tends to result in a particularly stable electron arrangement. Understanding how and why these ions form requires exploring related atomic concepts.

Electron Configuration

Electron configuration is a way to express how electrons are arranged around the nucleus of an atom. For aluminum, the neutral atom has the electron configuration of \(1s^2 2s^2 2p^6 3s^2 3p^1\). This notation tells us that aluminum's electrons fill the lower energy shells first, starting with the 1s orbital and proceeding to higher orbitals like 3s and 3p.
When aluminum forms ions, its electron configuration changes:

• Al⁺: \(1s^2 2s^2 2p^6 3s^2\)
• Al²⁺: \(1s^2 2s^2 2p^6 3s^1\)
• Al³⁺: \(1s^2 2s^2 2p^6\)

The removal of electrons from subshells closest to the nucleus is why higher energy is required to take away additional electrons. This is because, with fewer electrons, the remaining electrons are attracted even more strongly to the positive charge of the nucleus, increasing the ionization energy necessary for further removals.

Electron Affinity

Electron affinity describes how eager an atom or ion is to gain an electron. Atoms with higher electron affinities are more likely to attract additional electrons, releasing energy in the process. For aluminum ions, as electrons are lost and positive charges increase, the affinity to accept an electron also changes.
For instance, neutral aluminum, Al, might loosely want to grab an electron to fill its partially occupied 3p subshell. In contrast, {Al³⁺} is much less inclined to gain an electron, as adding a new electron would disrupt its stable electron shell configuration. The ion Al²⁺, with a vacancy in its 3s shell, is most capable of accepting an additional electron, making it likely to have the greatest electron affinity among the ions discussed.

Atomic Size

Atomic size refers to the dimensions of an atom, often visualized as the "cloud" of electrons surrounding the nucleus. For ions, size changes as the charge changes due to loss or gain of electrons. When aluminum ions are formed by removing electrons, this naturally reduces the size of the cloud because there are fewer electrons, leading to less electron-electron repulsion.
Each removal causes the electron cloud to shrink slightly due to the increased pull from the nucleus, known as the effective nuclear charge. This results in different sizes for aluminum ions:

• Al⁴⁺ being the smallest since it has the least electron cloud
• Followed by Al³⁺, Al²⁺, and then Al⁺
• Finally, neutral Al is largest since it retains all its electrons

This understanding of atomic size is crucial to studying the behaviors and properties of elements in different chemical environments.