Question

Explain, in terms of ionization energies, why the following ions do not form under normal conditions. (a) \(\mathrm{Ba}^{3+}\) (b) \(\mathrm{O}^{2+}\) (c) \(\mathrm{Cl}^{+}\)

Short answer

The ions \(\mathrm{Ba}^{3+}\), \(\mathrm{O}^{2+}\), and \(\mathrm{Cl}^{+}\) do not form because the required ionization energy to achieve these states is too high. Forming these ions would involve disrupting stable electron configurations, which requires more energy than is typically available under normal conditions.

Step-by-step solution

Explanation for \(\mathrm{Ba}^{3+}\)

The barium atom, \(Ba\), loses 2 electrons easily to get a stable configuration similar to Xenon (\(Xe\)) and forms \(Ba^{2+}\) because its first two ionization energies are relatively low. However, removing a third electron from \(Ba^{2+}\) to form \(Ba^{3+}\) means breaking into a stable electron configuration, and it would require a higher amount of energy (the third ionization energy) which is typically not available under normal conditions.

Explanation for \(\mathrm{O}^{2+}\)

Oxygen atom (\(O\)) readily gains two electrons to achieve a stable electronic configuration similar to Neon (\(Ne\)), forming \(O^{2-}\). However, to form \(O^{2+}\), it would mean removing two electrons from oxygen. The second ionization energy for oxygen is significantly high because it's removing an electron from a stable 2p orbital, hence it does not readily form under normal conditions.

Explanation for \(\mathrm{Cl}^{+}\)

Chlorine atom (\(Cl\)) readily accepts an electron to achieve a stable electronic configuration, forming \(Cl^{-}\). However, to form \(Cl^{+}\), it means removing an electron from chlorine. The ionization energy for chlorine is relatively high because it involves breaking a stable 3p orbital. \(Cl^{+}\) does not form readily under normal conditions.

Key concepts

Barium Ions

Barium ions are typically known to form a stable positive charge with a 2+ state, denoted as \( \mathrm{Ba}^{2+} \). This occurs because barium, being an alkaline earth metal, easily loses its two outermost electrons. This loss makes barium electronically analogous to the noble gas Xenon, which is a very stable arrangement due to its filled electron shell configuration.

• The first two electrons are removed with relatively low energy requirements, making the formation of \( \mathrm{Ba}^{2+} \) common under normal circumstances.
• Attempting to remove a third electron to form \( \mathrm{Ba}^{3+} \) requires disrupting this stable configuration.

When a stable, low-energy state is disturbed, the third ionization energy is considerably high. The energy needed to remove a third electron is not typically accessible under normal conditions, thereby preventing the formation of \( \mathrm{Ba}^{3+} \) ions in typical environments.

Oxygen Ions

Oxygen ions most commonly form a 2- charge, creating \( \mathrm{O}^{2-} \), because oxygen naturally tends to gain two electrons. This action completes its valence shell, aligning with the stable noble gas configuration of Neon.

• Gaining electrons is energetically favorable for oxygen, which is why \( \mathrm{O}^{2-} \) forms readily.
• However, forming \( \mathrm{O}^{2+} \) involves removing two electrons from oxygen.

The process includes removing electrons from the stable 2p orbital, and the energy required for such a removal (specifically the second ionization energy) is very high. This high energy makes the formation of \( \mathrm{O}^{2+} \) under normal conditions unfeasible, as it disrupts the stable, lower energy states of the electron configuration.

Chlorine Ions

Chlorine ions often form a negative charge, \( \mathrm{Cl}^{-} \), because chlorine gains an electron to fulfill its valence shell, achieving a noble gas-like electron configuration akin to Argon. This configuration is highly stable.

• The process of gaining an electron is easy for chlorine due to its high electron affinity, contributing to the predominant formation of \( \mathrm{Cl}^{-} \).
• In contrast, forming \( \mathrm{Cl}^{+} \) demands the removal of an electron from an already stable 3p orbital.

The ionization energy required to remove this electron is considerably high, indicating why \( \mathrm{Cl}^{+} \) does not naturally form under typical circumstances. Disrupting an electron arrangement that is already low-energy and stable is not favored unless under conditions where energy input is significantly high.