Consider the first ionization energy of neon and the electron affinity of fluorine. (a) Write equations, including electron configurations, for each process. (b) These two quantities have opposite signs. Which will be positive, and which will be negative? (c) Would you expect the magnitudes of these two quantities to be equal? If not, which one would you expect to be larger?

For the ionization energy of neon (Ne), we have to remove one electron from a neutral Ne atom, creating a positive ion. The electron configuration of a neutral Ne atom is \([He]2s^22p^6\). Its first ionization energy equation and electron configurations will be: Ne(\([He]2s^22p^6\)) -> Ne+(\([He]2s^22p^5\)) + e^- For the electron affinity of fluorine (F), we need to add an electron to a neutral F atom, creating a negative ion. The electron configuration of a neutral F atom is \([He]2s^22p^5\). Its electron affinity equation and electron configurations will be: F(\([He]2s^22p^5\)) + e^- -> F-(\([He]2s^22p^6\))

Ionization energy is the energy needed to remove an electron, and it's always a positive value since the process requires energy input. So, the first ionization energy of neon will be positive. Electron affinity is the energy released when an electron is added to an atom, and it's usually negative since energy is released during the process. The electron affinity value of fluorine will be negative.

In general, the magnitudes of ionization energy and electron affinity can be different due to atomic structure and electron-electron repulsion forces. In this case, we're comparing the first ionization energy of neon and electron affinity of fluorine. Neon has a completely filled 2p orbital, making it very stable, and its ionization energy is expected to be high. Fluorine, on the other hand, has almost a completely filled 2p orbital and tends to gain an electron to achieve a stable electron configuration (similar to Ne). Therefore, fluorine has a high electron affinity. However, the first ionization energy of neon is expected to be larger than the electron affinity of fluorine. This is because removing an electron from the completely filled and stable 2p orbital of neon is more difficult compared to adding an electron to the almost-filled 2p orbital of fluorine.