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 understand the process of electron affinity (EA) for oxygen and how it relates to the energy released when two oxygen atoms gain two electrons.

1. **Understanding Electron Affinity**:
- The first electron affinity (EA1) of oxygen is given as -142 kJ/mol. This means that when one mole of oxygen atoms gains one mole of electrons, 142 kJ of energy is released.
- The second electron affinity (EA2) of oxygen is given as +844 kJ/mol. This means that when one mole of O⁻ ions (which already has one extra electron) gains another electron, 844 kJ of energy is required (i.e., energy is absorbed).

2. **Writing the Reaction**:
- The reaction we are considering is:
\[ 2O + 2e^- \rightarrow 2O^- \]
- This reaction involves two oxygen atoms each gaining one electron to become two O⁻ ions.

3. **Calculating the Total Energy Released**:
- For the first electron affinity (for each oxygen atom):
- Energy released for 2 moles of O gaining 2 moles of electrons:
\[ \text{Energy released} = 2 \times (-142 \text{ kJ/mol}) = -284 \text{ kJ} \]
- The second electron affinity (for the O⁻ ions) is not involved in this reaction since we are only considering the first electron affinity for the formation of O⁻ from O and e⁻.

4. **Final Answer**:
- The total energy released when 2 O atoms gain 2 electrons to form 2 O⁻ ions is -284 kJ.

Thus, the energy released from the reaction \( 2O + 2e^- \rightarrow 2O^- \) is **-284 kJ**.