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 \(I_{4}\).

The ionization energies are the energies required to remove consecutive electrons from the aluminum atom. The trend we observe in the given exercise is that the ionization energy values are increasing successively, i.e., \(I_1 < I_2 < I_3 < I_4\). This can be explained by the electronic configuration of aluminum and the effective nuclear charge acting on the electrons. The electronic configuration of aluminum is \(1s^2 2s^2 2p^6 3s^2 3p^1\). When electrons are successively removed from aluminum, the ionization energy increases due to two reasons: 1. The effective nuclear charge acting on the electrons increases, making the remaining electrons more firmly attached and harder to remove. 2. The ion becomes smaller, which in turn increases the electrostatic attraction between the nucleus and the electrons, requiring more energy to remove the electrons.

To explain the large increase in ionization energy between \(I_{3}\) and \(I_{4}\), let's consider the electronic configurations of each ion: - \( \mathrm{Al}^{3+} \): \(1s^2\, 2s^2\, 2p^6\) - \( \mathrm{Al}^{4+} \): \(1s^2\, 2s^2\, 2p^5\) When moving from \( \mathrm{Al}^{3+} \) to \( \mathrm{Al}^{4+} \), we are removing an electron from the 2p orbital, which is closer to the nucleus and already completely filled in the \( \mathrm{Al}^{3+} \) ion. Removing an electron from a filled inner shell requires much more energy compared to removing an electron from an outer shell. This is due to the increased electrostatic attraction between the electron and the nucleus. Moreover, this transition also represents the removal of an electron from a stable noble gas-like configuration, which is highly unfavorable. As a result, the ionization energy required in this step increases drastically, leading to a large increase between \(I_{3}\) and \(I_{4}\).