Consider the following reaction:
(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)`
then according to given information the correct statement is/are:

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 problem, we need to analyze the given reactions and their associated enthalpies of electron gain (ΔH). Here’s a step-by-step breakdown:

### Step 1: Understand the Reactions
We have four reactions involving the addition of electrons to different atoms:
1. \( O_{(g)} + e^{-} \rightarrow O_{(g)}^{-} \) with enthalpy change \( \Delta H_1 \)
2. \( F_{(g)} + e^{-} \rightarrow F_{(g)}^{-} \) with enthalpy change \( \Delta H_2 \)
3. \( Cl_{(g)} + e^{-} \rightarrow Cl_{(g)}^{-} \) with enthalpy change \( \Delta H_3 \)
4. \( O_{(g)}^{-} + e^{-} \rightarrow O_{(g)}^{2-} \) with enthalpy change \( \Delta H_4 \)

### Step 2: Analyze Electron Gain Enthalpy
- **Electron Gain Enthalpy**: When an electron is added to an atom, energy is released, making the enthalpy change negative.
- As we move down a group in the periodic table, the electron gain enthalpy becomes less negative (i.e., it increases).
- As we move from left to right across a period, the electron gain enthalpy becomes more negative.

### Step 3: Compare the Elements
- **Oxygen (O)**: Has a high electron affinity but is less stable when gaining a second electron (due to electron-electron repulsion).
- **Fluorine (F)**: Has the highest electron affinity among the halogens.
- **Chlorine (Cl)**: Has a lower electron affinity than fluorine but is still negative.

From this, we can conclude:
- \( \Delta H_1 \) (O) is negative but less negative than \( \Delta H_2 \) (F).
- \( \Delta H_3 \) (Cl) is more negative than \( \Delta H_1 \) and \( \Delta H_2 \).
- \( \Delta H_4 \) (adding a second electron to O) is positive because adding a second electron to a negatively charged ion requires energy.

### Step 4: Summarize the Findings
1. \( \Delta H_1, \Delta H_2, \Delta H_3 \) are negative.
2. \( \Delta H_4 \) is positive.
3. \( \Delta H_3 \) is more negative than both \( \Delta H_1 \) and \( \Delta H_2 \).
4. \( \Delta H_1 \) is less negative than \( \Delta H_2 \).

### Step 5: Evaluate the Statements
Based on the analysis:
- **Statement 1**: \( \Delta H_3 \) is more negative than \( \Delta H_1 \) and \( \Delta H_2 \) - **True**
- **Statement 2**: \( \Delta H_1 \) is less negative than \( \Delta H_2 \) - **True**
- **Statement 3**: \( \Delta H_1, \Delta H_2, \Delta H_3 \) are negative whereas \( \Delta H_4 \) is positive - **True**
- **Statement 4**: \( \Delta H_1 \) and \( \Delta H_3 \) are negative whereas \( \Delta H_2 \) and \( \Delta H_4 \) are positive - **False** (since \( \Delta H_2 \) is also negative).

### Conclusion
The correct statements are:
1. \( \Delta H_3 \) is more negative than \( \Delta H_1 \) and \( \Delta H_2 \).
2. \( \Delta H_1 \) is less negative than \( \Delta H_2 \).
3. \( \Delta H_1, \Delta H_2, \Delta H_3 \) are negative whereas \( \Delta H_4 \) is positive.