Concept:The first electron affinity is usually negative (exothermic) because the nucleus attracts the electron. The second electron affinity is always positive (endothermic) due to electrostatic repulsion.
Formula:$$ \text{O}^-_{(g)} + e^- \longrightarrow \text{O}^{2-}_{(g)} \quad \Delta H = +\text{ve (Endothermic)} $$
Solution:- When adding an electron to a neutral Oxygen or Chlorine atom (Opts A & C), the nucleus pulls it in, releasing energy (negative EA).
- When attempting to add an electron to an already negative ion like \( \text{O}^{-1} \), the incoming negative electron is strongly repelled by the negative charge of the ion.
- Energy must be forcibly put into the system to overcome this repulsion and attach the second electron, resulting in a positive electron affinity.
Why other options are incorrect:- Option A & Option C: First electron affinities are exothermic (negative).
- Option B: While theoretically endothermic, Chlorine rarely forms a \( 2^- \) ion. Oxygen strictly forms the \( \text{O}^{2-} \) oxide ion, making it the classic textbook example for positive 2nd EA.
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