Question

Based on their valence-shell electron configurations which of the following species would you expect to have the lowest ionization energy: (a) \(\mathrm{C}_{2}{ }^{+}\); (b) \(\mathrm{C}_{2}\); (c) \(\mathrm{C}_{2}{ }^{-}\)?

Short answer

\(\text{C}_2{}^{-}\text{ has the lowest ionization energy.}Ionization energy decreases with increased electron-electron repulsion, which is greatest in an anion.}

Step-by-step solution

- Understanding Ionization Energy

Ionization energy is the energy required to remove an electron from an atom or molecule in its gaseous state. Generally, ionization energy increases across a period from left to right and decreases down a group in the periodic table. Additionally, the more electrons there are in the outer shell, the lower the ionization energy due to increased electron-electron repulsion.

- Analyze the Valence Electron Configuration

To determine the ionization energy, analyze the valence electron configuration of each species. The species with the highest electron count in their valence shell will likely have the lowest ionization energy because it's easier to remove an electron due to increased electron-electron repulsion. (a) \(\mathrm{C}_{2}{ }^{+}\) has one fewer electron than \(\mathrm{C}_{2}\), making it harder to remove an electron. (b) \(\mathrm{C}_{2}\) has a complete set of valence electrons, which generally leads to a higher ionization energy than if it had one more or one fewer electron. (c) \(\mathrm{C}_{2}{ }^{-}\) has one more electron than \(\mathrm{C}_{2}\), increasing electron-electron repulsion and therefore should have the lowest ionization energy.

- Comparing the Species

Compare the species based on their valence electron configuration. The neutral molecule \(\mathrm{C}_{2}\) will have a higher ionization energy compared to its ions because adding or removing an electron destabilizes the electron configuration, making it easier to remove another electron. Between \(\mathrm{C}_{2}{ }^{+}\) and \(\mathrm{C}_{2}{ }^{-}\), the latter's additional electron means more electron-electron repulsion, making it easier to remove an electron and thus, it has the lower ionization energy.

Key concepts

Valence-Shell Electron Configuration

The valence-shell electron configuration refers to the arrangement of electrons in the outermost shell of an atom or molecule. This shell greatly influences an atom's chemical properties, including its ionization energy. Atoms strive to reach a stable configuration, often resembling the noble gases with full outer shells, which is known as the octet rule.

For instance, in the context of the exercise, \texttt{C\(_2^+\)} has one less electron in its valence shell compared to \texttt{C\(_2\)}, this loss of an electron makes its ionization energy higher because the remaining electrons are held more tightly by the nucleus. Conversely, \texttt{C\(_2^-\)}, with an additional electron, experiences an increase in electron-electron repulsion, which can destabilize the valence shell, making it easier to remove an electron. Thus, understanding the effect of adding or removing electrons in the valence shell is crucial in predicting the ionization energy for different species.

Electron-Electron Repulsion

Electron-electron repulsion is a concept that emerges from the fact that electrons are negatively charged, and like charges repel each other. This repulsion is significant in the outermost shell of an atom because it affects how tightly electrons are held.

In a species with more electrons, such as \texttt{C\(_2^-\)}, the added electron increases the repulsion among all electrons in the valence shell. As this repulsion goes up, it becomes easier for an electron to be removed, subsequently lowering the ionization energy. In our exercise, we see that \texttt{C\(_2^-\)} is predicted to have the lowest ionization energy, primarily due to this increased electron-electron repulsion.

Periodic Table Trends

The periodic table is more than just a chart of elements; it's a map of trends that helps us predict various properties of atoms, including ionization energy. One of the key trends is that ionization energy generally increases across a period from left to right and decreases down a group. This has to do with atomic size and effective nuclear charge.

As we move from left to right across a period, protons are added to the nucleus, which pulls electrons closer, making them harder to remove. Moving down a group, the electrons are farther from the nucleus and more shielded by inner electron shells, which makes them easier to remove. These trends underscore the reasoning in our exercise, as we consider not just the charge of the species (its ionic state), but also where its atom would lie on the periodic table to intuit its ionization energy.