Question

Consider the following ionization energies for aluminum: $$ \begin{aligned} \mathrm{Al}(g) \longrightarrow \mathrm{Al}^{+}(g)+\mathrm{e}^{-} & I_{1}=580 \mathrm{kJ} / \mathrm{mol} \\ \mathrm{Al}^{+}(g) \longrightarrow \mathrm{Al}^{2+}(g)+\mathrm{e}^{-} & I_{2}=1815 \mathrm{kJ} / \mathrm{mol} \\ \mathrm{Al}^{2+}(g) \longrightarrow \mathrm{Al}^{3+}(g)+\mathrm{e}^{-} & I_{3}=2740 \mathrm{kJ} / \mathrm{mol} \\ \mathrm{Al}^{3+}(g) \longrightarrow \mathrm{Al}^{4+}(g)+\mathrm{e}^{-} & I_{4}=11,600 \mathrm{kJ} / \mathrm{mol} \end{aligned} $$ 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 ionization energies of aluminum show an increasing trend as electrons are removed from orbitals further from the nucleus, making it more difficult to remove subsequent electrons. The large increase between I3 and I4 can be attributed to the transition from removing valence electrons (3s and 3p orbitals) to an inner shell electron (2p orbital), and the increased stability of the electron configuration of the Al³⁺ ion, which has a noble gas configuration (1s² 2s² 2p⁶). Removing an electron from this stable configuration requires significantly more energy.

Step-by-step solution

Write down the electron configuration of aluminum

First, we need to write down the electron configuration of aluminum. Aluminum has an atomic number of 13. Therefore, its electron configuration is 1s² 2s² 2p⁶ 3s² 3p¹.

Analyze the trend in ionization energies

The ionization energies of aluminum show an increasing trend. For the first ionization energy (I1), an electron is removed from the 3p orbital. For the second ionization energy (I2), an electron is removed from the 3s orbital. For the third ionization energy (I3), the last valence electron is removed from the 3s orbital. For the fourth ionization energy (I4), an electron is removed from the inner shell (2p orbital). As we remove electrons from the orbitals further from the nucleus, we notice that the ionization energies increase. This is because electrons closer to the nucleus are more tightly bound, and it takes more energy to remove them. Moreover, as we remove electrons, the effective nuclear charge experienced by the remaining electrons increases, making it harder to remove the next electron.

Explain the large increase between I3 and I4

The large increase between I3 and I4 can be explained by considering the electron configurations of Al³⁺ and Al⁴⁺ ions. When all three valence electrons are removed (in the case of I3), the electron configuration of Al³⁺ becomes 1s² 2s² 2p⁶, which corresponds to the electron configuration of a noble gas (Neon). Noble gas configurations are extremely stable due to completely filled energy levels. Therefore, removing an electron from a noble gas configuration requires a significantly higher amount of energy than removing an electron from a partially filled orbital, as in the case of the first three ionizations of aluminum. In the case of I4, an electron is removed from the 2p orbital, which is an inner shell. The ionization energy increases drastically because inner shell electrons are more tightly bound to the nucleus and require more energy to be removed. Additionally, by removing one of the core electrons from Al³⁺, the Al⁴⁺ ion would have a less stable electron configuration. So, the large increase in ionization energy between I3 and I4 can be attributed to the transition from removing valence electrons to inner shell electrons and the increased stability of the electron configuration of the Al³⁺ ion (having a noble gas configuration).

Key concepts

Electron Configuration

The electron configuration of an atom is like its blueprint, detailing how electrons are arranged around its nucleus. For aluminum, which has an atomic number of 13, its electron configuration is written as 1s² 2s² 2p⁶ 3s² 3p¹. This configuration follows the Aufbau principle, meaning electrons fill subshells in order of increasing energy levels.

Aluminum's valence electrons, which are the outermost electrons involved in chemical reactions, are found in the 3p and 3s orbitals. Understanding the electron configuration is key to explaining many of the atom's chemical properties, such as its ionization energy. When electrons are removed during ionization, they typically come from the highest energy level that contains electrons, or in simple terms, the outermost shell. This configuration will change as we remove electrons, explaining why successive ionization energies increase as we move from valence to core electrons.

Effective Nuclear Charge

Effective nuclear charge is a crucial concept that helps explain why the ionization energies of aluminum increase significantly with each removed electron. The term refers to the net positive charge experienced by valence electrons, taking into account both the actual nuclear charge and the shielding effect provided by inner shell electrons.

In aluminum, when an electron is removed, the remaining electrons are slightly more attracted to the nucleus due to reduced electron-electron repulsion. This increased attraction happens because the inner electrons continue to shield the outer electrons, but now there are fewer outer electrons for the same number of protons in the nucleus.

As a result, the effective nuclear charge increases for the remaining electrons, making each subsequent electron harder to remove and increasing the ionization energy. The concept of effective nuclear charge can be summed up in the equation: \ \(Z_{eff} = Z - S\) where \(Z\) is the atomic number and \(S\) is the number of shielding electrons.

Noble Gas Configuration

Noble gas configuration is an important idea when discussing ionization energies, especially when considering large jumps in these energies, like the increase between \(I_3\) and \(I_4\) for aluminum. When the electrons are removed to form \(Al^{3+}\), its configuration mirrors that of the noble gas neon, which is 1s² 2s² 2p⁶.

Noble gases are characterized by their full energy levels, which make them extremely stable and less likely to react chemically. Attaining a noble gas configuration is the drive behind many chemical reactions because it requires the least amount of energy to maintain.

This stability is precisely why removing an electron from \(Al^{3+}\), and breaking its noble gas configuration, requires much more energy. To disrupt the filled 2p shell and promote the electron to a higher energy state significantly raises the ionization energy needed, demonstrating the inherent stability and resistance to change in noble gases. This is why the transition from \(Al^{3+}\) to \(Al^{4+}\) involves such a significant increase in energy demands.