The formation of oxide ion `O^(2-)(g)` from oxygen atom requires first an exothermic and then an endothermic step as shown below
`O(g)+e^(-) rarr O^(-)(g), DeltaH^(-) = - 141 kj mol^(-1)`
`O^(-)(g) +e^(-) rarr O^(2-) (g), DeltaH^(-) =+ 780 kj mol^(-1)`
Thus, process of formation of `O^(2-)` in gas phase is unfavourable even through `O^(2-)` is isoelectronic with neon. It is due to the fact that A) oxygen is more electronegative B) addition of electron in oxygen results in larget size of the ion C) electron repulsion outweights the stability gained by achieving noble gas configuration D) `O^(-)` ion has comparatively smaller size than oxygen atom

Options

A. Oxygen is more electronegative
B. Addition of electron in oxygen results in larger size of the ion
C. Electron repulsion outweights the stability gained by achieving noble gas configuration
D. `O^(-)`ion has comparatively smaller size than oxygen atom.

Explanation

To solve the question regarding the formation of the oxide ion \( O^{2-} \) from an oxygen atom, we need to analyze the steps involved in the process and understand the thermodynamic implications.

### Step-by-Step Solution:

1. **Understanding the Process**:
The formation of the oxide ion \( O^{2-} \) involves two steps:
- The first step is the addition of one electron to the oxygen atom, forming the \( O^{-} \) ion. This process is exothermic, meaning it releases energy:
\[ O(g) + e^{-} \rightarrow O^{-}(g), \quad \Delta H = -141 \, \text{kJ mol}^{-1} \]
- The second step is the addition of another electron to the \( O^{-} \) ion, forming \( O^{2-} \). This process is endothermic, meaning it requires energy:
\[ O^{-}(g) + e^{-} \rightarrow O^{2-}(g), \quad \Delta H = +780 \, \text{kJ mol}^{-1} \]

2. **Analyzing the Energetics**:
The first step is favorable because it releases energy. However, the second step is unfavorable because it requires a significant amount of energy to overcome the repulsion between the two negatively charged electrons being added to the already negatively charged \( O^{-} \) ion.

3. **Isoelectronic Configuration**:
The question mentions that \( O^{2-} \) is isoelectronic with neon, meaning both have the same number of electrons (10 electrons). However, having a noble gas configuration does not guarantee stability if there are significant repulsive forces present.

4. **Identifying the Main Issue**:
The unfavorable formation of \( O^{2-} \) can be attributed to the strong electron-electron repulsion that occurs when the second electron is added to the \( O^{-} \) ion. This repulsion outweighs the stability gained from achieving a noble gas configuration.

5. **Conclusion**:
Based on the analysis, the correct answer is:
**C) Electron repulsion outweighs the stability gained by achieving noble gas configuration.**