The formation of the oxide ion `O_(g)^(2-)` requires first an exothermic and then an endothermic step as shown below:
`O_(g)+e^(-) rarr O_(g)^(-)`, `DeltaH=-142 kJ mol^(-1)`
`O(g)+e rarr O_(g)^(2-)`, `DeltaH=844kJ mol^(-1)`
This is because:

Options

A. `O^(-)` ion has comparatively larger size than oxygen atom
B. oxygen has high electron affinity
C. `O^(-)` ion will tend to resist the addition of another electron
D. oxygen is more electronegative

Answer

To understand why the formation of the oxide ion \( O^{2-} \) involves both an exothermic and an endothermic step, we can break down the process into two main steps:

### Step 1: Formation of the Oxide Ion \( O^{-} \)

1. **Reaction**: The first step involves the reaction of a neutral oxygen atom \( O(g) \) with an electron \( e^{-} \) to form the oxide ion \( O^{-} \).
\[ O(g) + e^{-} \rightarrow O^{-}(g) \]
2. **Enthalpy Change**: The enthalpy change for this reaction is given as \( \Delta H = -142 \, \text{kJ mol}^{-1} \). The negative sign indicates that this process is exothermic, meaning energy is released when the electron is added to the neutral oxygen atom.

### Step 2: Formation of the Dioxide Ion \( O^{2-} \)

3. **Reaction**: The second step involves adding another electron to the already negatively charged oxide ion \( O^{-} \) to form the dioxide ion \( O^{2-} \).
\[ O^{-}(g) + e^{-} \rightarrow O^{2-}(g) \]
4. **Enthalpy Change**: The enthalpy change for this reaction is given as \( \Delta H = +844 \, \text{kJ mol}^{-1} \). The positive sign indicates that this process is endothermic, meaning energy is absorbed when the second electron is added.

### Explanation of the Energy Changes

5. **Reason for Exothermic Step**: The first step is exothermic because adding an electron to a neutral atom typically releases energy. This is due to the attraction between the negatively charged electron and the positively charged nucleus of the oxygen atom.

6. **Reason for Endothermic Step**: The second step is endothermic because adding an electron to an already negatively charged ion \( O^{-} \) leads to electron-electron repulsion. The presence of the negative charge on \( O^{-} \) creates a repulsive force when another electron is introduced. To overcome this repulsion and successfully add the second electron, energy must be supplied to the system, resulting in a positive enthalpy change.

### Conclusion

The formation of the oxide ion \( O^{2-} \) requires first an exothermic step (formation of \( O^{-} \)) followed by an endothermic step (formation of \( O^{2-} \)) due to the repulsion experienced when adding an electron to an anion.

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