Consider the following conversions:
(i) `O_((g))+e^(-) to O_((g))^(-),DeltaH_(1)`
(ii) `F_((g))+e^(-) to F_((g))^(-),DeltaH_(2)`
(iii) `Cl_((g))+e^(-) to Cl_((g))^(-),DeltaH_(3)`
(iv) `O_((g))^(-) +e^(-) to O_((g))^(2-),DeltaH_(4)`
the according to given information the incorrect statement is :

Options

A. `DeltaH_(3)` is more negative than `DeltaH_(1) and DeltaH_(2)`
B. `DeltaH_(1)` is less negative than `DeltaH_(2)`
C. `DeltaH_(1),DeltaH_(2) and DeltaH_(3)` are negative whereas `DeltaH_(4)` is positive
D. `DeltaH_(1) and DeltaH_(3)` are negative whereas `DeltaH_(2) and DeltaH_(4)` positive

Explanation

To solve the question, we need to analyze the given conversions and their associated enthalpy changes (ΔH). Let's break down each step:

### Step 1: Understand the conversions
1. **Conversion (i)**: \( O(g) + e^- \rightarrow O(g)^- \) with enthalpy change \( \Delta H_1 \)
2. **Conversion (ii)**: \( F(g) + e^- \rightarrow F(g)^- \) with enthalpy change \( \Delta H_2 \)
3. **Conversion (iii)**: \( Cl(g) + e^- \rightarrow Cl(g)^- \) with enthalpy change \( \Delta H_3 \)
4. **Conversion (iv)**: \( O(g)^- + e^- \rightarrow O(g)^{2-} \) with enthalpy change \( \Delta H_4 \)

### Step 2: Analyze the trends in electron affinity
- **Electron Affinity Trend**: Generally, as we move from left to right across a period, the electron affinity becomes more negative (i.e., it increases). Conversely, as we move down a group, the electron affinity tends to become less negative (i.e., it decreases).
- **Reason**: This is due to the increase in effective nuclear charge (Z_eff) as we move across a period, making it easier to add electrons. However, as we go down a group, the atomic size increases, making it harder to attract additional electrons.

### Step 3: Compare the enthalpy changes
- **ΔH1 (O)**: Oxygen has a higher electron affinity than fluorine because it has fewer electrons and less inter-electronic repulsion.
- **ΔH2 (F)**: Fluorine has a high electron affinity but is slightly less than that of oxygen due to its smaller size and increased electron-electron repulsion.
- **ΔH3 (Cl)**: Chlorine has a lower electron affinity than both oxygen and fluorine because it is larger in size and has more electron-electron repulsion.
- **ΔH4**: The addition of a second electron to \( O(g)^- \) is always positive due to the increased repulsion from the already present electron.

### Step 4: Determine the incorrect statement
Now we need to evaluate the given statements:
1. **Statement 1**: \( \Delta H_3 \) is more negative than \( \Delta H_1 \) and \( \Delta H_2 \). (Incorrect, since \( \Delta H_3 \) is less negative)
2. **Statement 2**: \( \Delta H_1 \) is less negative than \( \Delta H_2 \). (Correct, since \( \Delta H_1 \) is more negative than \( \Delta H_2 \))
3. **Statement 3**: \( \Delta H_1, \Delta H_2, \Delta H_3 \) are negative while \( \Delta H_4 \) is positive. (Correct, since \( \Delta H_4 \) is indeed positive)
4. **Statement 4**: \( \Delta H_1, \Delta H_3 \) are negative while \( \Delta H_2 \) is positive. (Incorrect, since \( \Delta H_2 \) is negative)

### Conclusion
The incorrect statement is **Statement 1**: \( \Delta H_3 \) is more negative than \( \Delta H_1 \) and \( \Delta H_2 \).