Formation of `F^(-)` is exothermic but formation of `O2^(-)` is endothermic. Explain in two or three sentences.

Explanation

To understand why the formation of \( F^- \) is exothermic while the formation of \( O^{2-} \) is endothermic, we can analyze the electronic configurations and the energy changes involved in each process.

1. **Formation of \( F^- \)**: A neutral fluorine atom has the electronic configuration \( 1s^2 2s^2 2p^5 \). When it gains one electron, it achieves the noble gas configuration of neon, \( 1s^2 2s^2 2p^6 \). This transition releases a significant amount of energy because the atom becomes more stable by reaching a full outer shell. Thus, the formation of \( F^- \) is an exothermic process, meaning energy is released.

2. **Formation of \( O^{2-} \)**: A neutral oxygen atom has the electronic configuration \( 1s^2 2s^2 2p^4 \). When it gains its first electron, it forms \( O^- \), which is an exothermic process as it moves towards a more stable configuration. However, gaining a second electron to form \( O^{2-} \) requires overcoming increased electron-electron repulsion in the already negatively charged \( O^- \) ion. This repulsion means that energy must be supplied to add the second electron, making the formation of \( O^{2-} \) an endothermic process.

In summary, the formation of \( F^- \) is exothermic due to the stability gained from achieving a noble gas configuration, while the formation of \( O^{2-} \) is endothermic because of the repulsion between electrons that requires energy input.