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 problem, we need to analyze the given conversions and their associated enthalpy changes (ΔH). The conversions involve the addition of electrons to gaseous atoms, which relates to their electron affinity. Let's break down the conversions step by step.

### Step 1: Understanding Electron Gain Enthalpy
- **Definition**: Electron gain enthalpy (ΔH) is the energy change when an electron is added to a neutral gaseous atom to form a negative ion. A more negative ΔH indicates a greater tendency to gain an electron.
- **General Trend**: As we move from left to right across a period, the electron affinity generally increases (more negative ΔH) due to increasing nuclear charge. However, there are exceptions due to atomic size and electron-electron repulsions.

### Step 2: Analyzing Each Conversion
1. **Conversion (i)**: \( O_{(g)} + e^- \rightarrow O_{(g)}^- \) with ΔH1
- Oxygen has a relatively high electron affinity, but due to its small size, adding an electron results in significant electron-electron repulsion.

2. **Conversion (ii)**: \( F_{(g)} + e^- \rightarrow F_{(g)}^- \) with ΔH2
- Fluorine has the highest electron affinity among the elements, resulting in a very negative ΔH2.

3. **Conversion (iii)**: \( Cl_{(g)} + e^- \rightarrow Cl_{(g)}^- \) with ΔH3
- Chlorine has a high electron affinity, but it is less than that of fluorine. Thus, ΔH3 is less negative than ΔH2.

4. **Conversion (iv)**: \( O_{(g)}^- + e^- \rightarrow O_{(g)}^{2-} \) with ΔH4
- Adding a second electron to the already negatively charged oxygen ion results in repulsion, making ΔH4 positive (energy must be supplied).

### Step 3: Comparing the Enthalpy Changes
- **Order of Enthalpy Changes**:
- ΔH2 (Fluorine) < ΔH3 (Chlorine) < ΔH1 (Oxygen) < ΔH4 (Oxygen with two electrons)
- This means ΔH1 is less negative than ΔH2, and ΔH4 is positive.

### Step 4: Identifying the Incorrect Statement
- The incorrect statement among the options provided is:
- "ΔH2 is positive" (This is incorrect because ΔH2 is negative, indicating that energy is released when an electron is added to fluorine).

### Conclusion
The incorrect statement is that ΔH2 is positive. In reality, ΔH2 is negative, indicating that energy is released when an electron is added to fluorine.

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