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 corresponding enthalpy changes (ΔH). The reactions represent the process of electron affinity, which is the energy change when an electron is added to a neutral atom.

### Step-by-Step Solution:

1. **Understanding Electron Affinity**:
- Electron affinity is defined as the energy change when an electron is added to a neutral atom to form a negative ion. A more negative ΔH value indicates a greater tendency to gain an electron.

2. **Analyzing the Reactions**:
- (i) `O_(g) + e^(-) → O_(g)^(-), ΔH_1`: This reaction represents the addition of an electron to oxygen.
- (ii) `F_(g) + e^(-) → F_(g)^(-), ΔH_2`: This reaction represents the addition of an electron to fluorine.
- (iii) `Cl_(g) + e^(-) → Cl_(g)^(-), ΔH_3`: This reaction represents the addition of an electron to chlorine.
- (iv) `O_(g)^(-) + e^(-) → O_(g)^(2-), ΔH_4`: This reaction represents the addition of an electron to the negatively charged oxygen ion.

3. **Comparing Electron Affinities**:
- **Statement 1**: ΔH3 is more negative than ΔH1 and ΔH2.
- This is true because chlorine has a higher electron affinity than both oxygen and fluorine due to its larger atomic size, which allows it to accommodate an additional electron more easily despite being less electronegative.

- **Statement 2**: ΔH1 is less negative than ΔH2.
- This is also true since fluorine has a higher electron affinity than oxygen, meaning that the energy released (ΔH) when fluorine gains an electron is greater (more negative) than that of oxygen.

4. **Analyzing the Fourth Reaction**:
- **Statement 3**: ΔH1, ΔH2, and ΔH3 are negative whereas ΔH4 is positive.
- This statement is true. The addition of an electron to a negatively charged ion (O^(-)) is endothermic (requires energy), resulting in a positive ΔH4. This is because negatively charged species repel incoming electrons.

5. **Final Evaluation**:
- **Statement 4**: ΔH1 and ΔH3 are negative whereas ΔH2 and ΔH4 are positive.
- This statement is incorrect. While ΔH1 and ΔH3 are indeed negative, ΔH2 is also negative, as fluorine has a high electron affinity.

### Conclusion:
The correct statements are:
1. ΔH3 is more negative than ΔH1 and ΔH2.
2. ΔH1 is less negative than ΔH2.
3. ΔH1, ΔH2, and ΔH3 are negative whereas ΔH4 is positive.

Thus, the answer is statements 1, 2, and 3 are correct.