Knowing the electron gain enthalpy values for `O rarr O^(Ө)` and `O rarr O^(2-)` as `- 141 kJ mol^-1` and `+ 702 kJ mol^-1` respectively, how can you account for the formation of a large number of oxides having `O^(2-)` species and not `O^(Ө)` ?

Explanation

To account for the formation of a large number of oxides containing the `O^(2-)` species instead of `O^(−)`, we can analyze the electron gain enthalpy values and the stability of these species. Here’s a step-by-step breakdown of the reasoning:

### Step 1: Understanding Electron Gain Enthalpy
- The electron gain enthalpy for the formation of `O^(−)` from `O` is `-141 kJ mol^-1`. This negative value indicates that the process is exothermic and favorable, meaning that it releases energy when an electron is added to an oxygen atom to form `O^(−)`.
- Conversely, the electron gain enthalpy for the formation of `O^(2-)` from `O^(−)` is `+702 kJ mol^-1`. This positive value indicates that the process is endothermic and not favorable, meaning that it requires energy input to add another electron to `O^(−)` to form `O^(2-)`.

### Step 2: Stability of the Species
- Despite the unfavorable formation of `O^(2-)`, this ion achieves a noble gas configuration (similar to neon) when it gains two electrons. This configuration is highly stable and energetically favorable in terms of overall stability.
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