Question

Consider the following ionization energies for aluminum: $$\begin{array}{c}{\operatorname{Al}(g) \longrightarrow \mathrm{Al}^{+}(g)+\mathrm{e}^{-} \quad I_{1}=580 \mathrm{kJ} / \mathrm{mol}} \\\ {\mathrm{Al}^{+}(g) \longrightarrow \mathrm{Al}^{2+}(g)+\mathrm{e}^{-} \quad I_{2}=1815 \mathrm{kJ} / \mathrm{mol}} \\ {\mathrm{Al}^{2+}(g) \longrightarrow \mathrm{Al}^{3+}(g)+\mathrm{e}^{-} \quad I_{3}=2740 \mathrm{kJ} / \mathrm{mol}} \\ {\mathrm{Al}^{3+}(g) \longrightarrow \mathrm{Al}^{4+}(g)+\mathrm{e}^{-} \quad I_{4}=11,600 \mathrm{kJ} / \mathrm{mol}}\end{array}$$ a. Account for the trend in the values of the ionization energies. b. Explain the large increase between \(I_{3}\) and \(I_{4}\)

Short answer

The trend of increasing ionization energies for aluminum ions (Al(g) to Al^3+(g)) is due to the increase in the positive charge of the ion, which makes it harder to remove an electron. The large increase between \(I_3\) and \(I_4\) is because the fourth electron removal breaks the stable noble gas configuration of Al^3+(g), requiring significantly more energy.

Step-by-step solution

Part a: Trend in Ionization Energies

The ionization energy is the energy required to remove an electron from an atom or an ion in the gas phase. Generally, ionization energies increase across a period from left to right (with some exceptions) and decrease down a group in the periodic table. This is because as you move across a period, the number of protons in the nucleus increases, creating a stronger positive charge that attracts the electrons, making it harder to remove an electron. On the other hand, as you move down a group, the atomic size increases because electrons are being added to higher energy levels, which are further from the nucleus, reducing the attraction between the nucleus and the electrons, making it easier to remove an electron. In the case of aluminum: Al(g) has the electron configuration \([Ne]3s^23p^1\). When it loses its first electron, it becomes Al^(+(g)) with the electron configuration \([Ne]3s^2\). The first ionization energy, \(I_1\), is 580 kJ/mol. Al^(+(g)) has the electron configuration \([Ne]3s^2\). When it loses its second electron, it becomes Al^2+(g) with the electron configuration \([Ne]3s^1\). The second ionization energy, \(I_2\), is 1815 kJ/mol. Al^2+(g) has the electron configuration \([Ne]3s^1\). When it loses its third electron, it becomes Al^3+(g) with the electron configuration \([Ne]\). The third ionization energy, \(I_3\), is 2740 kJ/mol. So far, the ionization energies increased from \(I_1\) to \(I_3\), which is consistent with the periodic trend. This is because removing an electron from a more positively charged ion takes more energy as the remaining electrons experience more attraction towards the positively charged nucleus. Now, let's look at the fourth ionization energy.

Part b: Large Increase between \(I_{3}\) and \(I_{4}\)

The large increase between \(I_3\) and \(I_4\) is due to the breaking of an electron shell. In this case, we are removing an electron from a noble gas configuration, which is more stable and requires a significantly higher amount of energy. Al^3+(g) has the electron configuration \([Ne]\). To remove the fourth electron, we have to break the full noble gas configuration and move to the inner 2p subshell of Al^4+(g). This electron is much closer to the nucleus and experiences a much stronger attraction, resulting in the high ionization energy, \(I_4\), of 11,600 kJ/mol. In conclusion, the trend of increasing ionization energies in aluminum can be explained by the general periodic trend and the large increase between \(I_3\) and \(I_4\) is due to breaking a more stable noble gas electron configuration, which requires a significantly greater amount of energy.

Key concepts

Electron Configuration

In an atom, electrons are arranged in energy levels or shells, which are like layers surrounding the nucleus. Each level can hold a certain number of electrons, and the way these electrons are distributed is known as the electron configuration. Let's consider aluminum, which has an initial electron configuration of \([Ne]3s^23p^1\). This notation indicates that aluminum starts with the electron configuration of the noble gas neon, followed by one electron in the 3p orbital and two in the 3s orbital.
As electrons are removed, such as during ionization, the configuration changes. Notice how aluminum changes configuration as electrons are sequentially removed:

• Neutral Al: \([Ne]3s^23p^1\)
• Al⁺: \([Ne]3s^2\)
• Al²⁺: \([Ne]3s^1\)
• Al³⁺: \([Ne]\)

With each electron removed, the subsequent ionization energy increases because the resulting ion has a stronger positive charge, making it increasingly difficult to detach more electrons.
This fundamental understanding of electron configurations is crucial for topics such as ionization energies and chemical bonding.

Noble Gas Configuration

Noble gas configuration refers to the electron arrangement resembling that of the nearest noble gas. Noble gases are known for being incredibly stable due to their full valence electron shells. This stability is why elements tend to attain a noble gas configuration during ionization or chemical reactions. In the case of aluminum, when it forms the Al³⁺ ion, it achieves the electron configuration \([Ne]\), mirroring the configuration of neon, a noble gas.
When discussing ionization, reaching a noble gas configuration can significantly impact the required energy. For instance, the huge leap from the third to the fourth ionization energy in aluminum (from 2740 kJ/mol to 11600 kJ/mol) is because removing an electron beyond the noble gas configuration involves breaking this stable setup. Such an ionization requires a lot more energy, emphasizing the secure nature of the noble gas arrangement.

Periodic Table Trends

The periodic table is structured in such a way that it reveals trends in various properties of elements, including ionization energies. Generally, ionization energy increases as you move across a period (left to right) and decreases as you move down a group (top to bottom).
Moving across a period, additional protons in the nucleus increase the positive charge, thereby strengthening the pull on electrons and resulting in higher ionization energies. Conversely, moving down a group, new electron shells are added, which places electrons further from the nucleus. The increased distance and added inner-shell electrons that shield the outer electrons weaken the nucleus-electron interaction, reducing the ionization energy.
This fundamental understanding aids in predicting the behavior of elements like aluminum as seen in the ionization energy values provided. Notably, the trend continues with increased values from \(I_1\) to \(I_3\), aligning with expected outcomes based on periodic trends, while \(I_4\) highlights the influence of breaking into a stable electron configuration.