`O_((g))^(-)+e^(-)toO_((g))^(-2) /_\H_(1)`
`O_((g))+e^(-)toO_((g))^(-) /_\H_(2)`
`O_(2(g))toO_(2(g))^(+)+e^(-) /_\H_(3)`
`O_((g))toO_(2(g))^(+)+e^(-) /_\H_(4)`
`H_(2(g))toH_(2(g))^(+)+e^(-) /_\H_(5)`
`O_((g))toH_((g))^(+)+e^(-) /_\H_(6)`
`O_((g))+2e^(-)toO^(-2) /_\H_(7)`
Which of the following option (s) is/are correct?

Options

A. `|/_\H_(1)|gt|/_\H_(2)|`
B. `|/_\H_(4)|gt|/_\H_(3)|`
C. `|/_\H_(5)|gt|/_\H_(6)|`
D. `/_\H_(7)` is +ve

Explanation

To solve the question regarding the enthalpy changes (ΔH) for the given reactions, we will analyze each reaction step by step.

### Step 1: Analyze ΔH1 and ΔH2
1. **Reaction for ΔH1**: \( O^{-} + e^{-} \rightarrow O^{2-} \)
- This reaction involves the addition of an electron to \( O^{-} \) to form \( O^{2-} \). Since \( O^{2-} \) is more stable than \( O^{-} \), energy will be released. Therefore, ΔH1 is negative.
- Let's assume ΔH1 = -140 kJ/mol.

2. **Reaction for ΔH2**: \( O + e^{-} \rightarrow O^{-} \)
- This reaction involves the addition of an electron to a neutral oxygen atom to form \( O^{-} \). This process requires energy to overcome the repulsion between the electron and the nucleus. Therefore, ΔH2 is positive.
- Let's assume ΔH2 = +700 kJ/mol.

**Conclusion**: The magnitude of ΔH1 is less than the magnitude of ΔH2, making the first option correct.

### Step 2: Analyze ΔH3 and ΔH4
1. **Reaction for ΔH3**: \( O + O \rightarrow O_2 + e^{-} \)
- This reaction involves the formation of \( O_2 \) from two oxygen atoms. Since \( O_2 \) is more stable than two separate oxygen atoms, ΔH3 is negative.

2. **Reaction for ΔH4**: \( O \rightarrow O^{+} + e^{-} \)
- This reaction involves the ionization of an oxygen atom to form \( O^{+} \). This process requires energy, so ΔH4 is positive.

**Conclusion**: Since ΔH3 is negative and ΔH4 is positive, we can conclude that the magnitude of ΔH4 is greater than ΔH3, making the second option correct.

### Step 3: Analyze ΔH6
- The reaction for ΔH6 is not directly comparable to the previous reactions. Therefore, we cannot draw any conclusions about its value compared to the others.

### Step 4: Analyze ΔH7
1. **Reaction for ΔH7**: \( O + 2e^{-} \rightarrow O^{2-} \)
- This reaction is equivalent to the combination of the two previous reactions (ΔH1 and ΔH2). The enthalpy change for this reaction can be calculated as:
\[ ΔH7 = ΔH2 + ΔH1 = +700 kJ/mol - 140 kJ/mol = +560 kJ/mol \]
- Therefore, ΔH7 is positive.

**Conclusion**: This confirms that ΔH7 is consistent with the values calculated in previous steps.

### Final Answer
Based on the analysis:
- The first option is correct (ΔH1 < ΔH2).
- The second option is correct (ΔH4 > ΔH3).
- No conclusion can be drawn for ΔH6.
- ΔH7 is consistent with the previous calculations.