Question

which one of the following atoms (designated by complete electron configuration) would have the most negative value for electron affinity?
1s² 2s² 2p⁶
1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹
1s² 2s² 2p⁶ 3s¹
1s² 2s² 2p⁶ 3s² 3p²
1s² 2s² 2p⁶ 3s²
1s² 2s² 2p⁵

Explanation:

Electron affinity is the energy change when an electron is added to a neutral atom. A more negative electron affinity means the atom has a greater tendency to accept an electron. Atoms with nearly - filled sub - shells have a high tendency to accept electrons.
The electron configuration \(1s^{2}2s^{2}2p^{5}\) is of fluorine. Fluorine has a very high electron affinity because it is one electron short of a stable \(2p^{6}\) (noble - gas - like) configuration in its valence shell (\(n = 2\)).

• For \(1s^{2}2s^{2}2p^{6}\), it is a noble - gas (Neon) configuration. Noble gases have a very low tendency to accept electrons as their shells are completely filled.
• For \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}\) (Potassium), it has a single electron in the \(4s\) orbital. It has a relatively low electron affinity as it is more likely to lose the \(4s\) electron to achieve a noble - gas (\(Ar\)) - like configuration (\(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}\)).
• For \(1s^{2}2s^{2}2p^{6}3s^{1}\) (Sodium), it has a single electron in the \(3s\) orbital. It is more likely to lose the \(3s\) electron to achieve a \(Ne\) - like configuration (\(1s^{2}2s^{2}2p^{6}\)).
• For \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{2}\) (Silicon), it has two electrons in the \(3p\) orbital. It can accept electrons, but not as readily as an atom that is one electron short of a filled sub - shell.
• For \(1s^{2}2s^{2}2p^{6}3s^{2}\) (Magnesium), it has a filled \(3s\) orbital. It has a relatively low electron affinity as it is more stable with its filled \(3s\) orbital.

Answer:

\(1s^{2}2s^{2}2p^{5}\)