If `EA_1 and EA_2` for oxygen atom `-142kJmol"^(-1) and +844kJmol^(-1)` . The energy released form `2O+2e^(-)rarr2O^(-)` will be

Options

A. `-986kJmol^(-1)`
B. `-702kJmol^(-1)`
C. `-284kJmol^(-1)`
D. `-1688kJmol^(-1)`

Answer

To solve the problem, we need to calculate the total energy released when two oxygen atoms each gain one electron to form two O⁻ ions. We will use the given electron affinities (EA₁ and EA₂) for oxygen.

### Step-by-Step Solution:

1. **Understand Electron Affinity**:
- Electron affinity is the energy change when an electron is added to a neutral atom in the gas phase. A negative value indicates that energy is released when the electron is added.
- For oxygen, we have:
- EA₁ = -142 kJ/mol (energy released when one electron is added to O to form O⁻)
- EA₂ = +844 kJ/mol (energy required to add a second electron to O⁻ to form O²⁻)

2. **Calculate Energy Released for Two O Atoms**:
- When two oxygen atoms each gain one electron, we only consider EA₁ because we are not forming O²⁻ in this reaction.
- The total energy released when two electrons are added to two oxygen atoms is:
\[ \text{Total Energy} = 2 \times EA₁ = 2 \times (-142 \text{ kJ/mol}) = -284 \text{ kJ/mol} \]

3. **Final Answer**:
- The energy released from the reaction \(2O + 2e^- \rightarrow 2O^-\) is \(-284 \text{ kJ/mol}\).

### Final Result:
The energy released from the reaction is \(-284 \text{ kJ/mol}\).