Statement-1 : Fluorine is a stronger oxidising agent than chlorine because
Statement-2 : It has (i) low enthalpy of dissociation of F-F bond and (ii) high hydration energy.

Options

A. Statement-1 is True, Statement-2 is True,Statement-2 is a correct explanation for Statement-1.
B. Statement-1 is True, Statement-2 is True,Statement-2 is NOT a correct explalanation for Statement-2
C. Statement-1 is True, Statement-2 is False
D. Statement-1 is False, Statement-2 is True

Answer

To analyze the statements provided in the question, we will evaluate each statement step by step and determine their validity.

### Step 1: Understanding Statement-1
**Statement-1:** Fluorine is a stronger oxidizing agent than chlorine.

- **Explanation:** An oxidizing agent is a substance that can accept electrons and, in the process, gets reduced. The strength of an oxidizing agent is often related to its electronegativity and its ability to stabilize the negative charge that results from gaining electrons.

### Step 2: Understanding Statement-2
**Statement-2:** It has (i) low enthalpy of dissociation of F-F bond and (ii) high hydration energy.

- **Part (i):** Low enthalpy of dissociation of the F-F bond.
- **Explanation:** The F-F bond is relatively weak due to the small size of the fluorine atom and the repulsion between the lone pairs of electrons on the two fluorine atoms. This weak bond means that it requires less energy to break, allowing fluorine to more readily participate in redox reactions.

- **Part (ii):** High hydration energy.
- **Explanation:** Hydration energy is the energy released when ions are solvated by water molecules. Fluorine, being the most electronegative element, has a high hydration energy, which means that when it gains an electron to form F⁻, it is stabilized effectively in aqueous solution.

### Step 3: Correlating the Statements
- **Analysis:** Since fluorine has a low bond dissociation energy, it can easily dissociate to form F⁻ ions. Additionally, its high hydration energy means that the F⁻ ions are stabilized in solution, making fluorine a very effective oxidizing agent. Chlorine, while also an oxidizing agent, is not as effective as fluorine due to its higher bond dissociation energy and lower hydration energy compared to fluorine.

### Conclusion
- **Final Evaluation:** Both statements are true, and Statement-2 provides a correct explanation for Statement-1. Therefore, fluorine is indeed a stronger oxidizing agent than chlorine due to its low enthalpy of dissociation of the F-F bond and high hydration energy.

### Summary of the Solution
1. **Statement-1 is true:** Fluorine is a stronger oxidizing agent than chlorine.
2. **Statement-2 is true:** It has a low enthalpy of dissociation of the F-F bond and high hydration energy.
3. **Statement-2 explains Statement-1:** The properties of fluorine contribute to its strength as an oxidizing agent.