Question

The formation of oxide ion $\mathrm{O}^{2-}(g)$ requires first an exothermic and then an endothermic step as shown below: $\mathrm{O}(g) + e^- \longrightarrow \mathrm{O}^-(g); \Delta H^\circ = -142\mathrm{kJ}\mathrm{mol}^{-1}$ and $\mathrm{O}^-(g) + e^- \longrightarrow \mathrm{O}^{2-}(g); \Delta H^\circ = 844\mathrm{kJ}\mathrm{mol}^{-1}$. This is because

• A. Oxygen is more electronegative
• B. $\mathrm{O}^-$ ion has comparatively larger size than oxygen atom
• C. $\mathrm{O}^-$ ion will tend to resist the addition of another electron
• D. Oxygen has high electron affinity

Hint:

Second electron affinity is always endothermic due to repulsion.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Second electron affinity is always endothermic.
Step 2: Detailed Explanation:
After gaining one electron, $\mathrm{O}^-$ has a stable configuration and resists adding another electron due to electron-electron repulsion.
Step 3: Final Answer:
The $\mathrm{O}^-$ ion resists addition of another electron.