Statement I: The size `H^(-)` is greater than that of `F^(-)`.
Statement II: The `e//p` ratio in `H^(-)` is 2 while that in `F^(-)` is `10//9 ` 

Options

A. If both Statement I and Statement II are true and Statement II is the correct explanation of Statement I.
B. If both Statement I and Statement II are true but Statement II is not the correct explanation for Statement I.
C. If Statement I is true but Statement II is false.
D. If Statement I is false but Statement II is true.

Answer

To solve the question, we need to analyze both statements regarding the sizes of the ions \( H^{-} \) and \( F^{-} \) and their electron-to-proton (e/p) ratios.

### Step 1: Compare the sizes of \( H^{-} \) and \( F^{-} \)

- **Understanding Ion Sizes**: The size of an ion is influenced by the number of electrons relative to the number of protons. More electrons compared to protons generally lead to a larger ionic size due to increased electron-electron repulsion.

- **For \( H^{-} \)**:
- It has 1 proton and 2 electrons.
- The electron-to-proton ratio (e/p) for \( H^{-} \) is \( \frac{2}{1} = 2 \).

- **For \( F^{-} \)**:
- It has 9 protons and 10 electrons.
- The electron-to-proton ratio (e/p) for \( F^{-} \) is \( \frac{10}{9} \approx 1.11 \).

### Step 2: Analyze the e/p ratios

- **Comparing e/p Ratios**:
- The e/p ratio for \( H^{-} \) is 2, while for \( F^{-} \) it is approximately 1.11.
- Since the e/p ratio for \( H^{-} \) is greater than that of \( F^{-} \), we can conclude that \( H^{-} \) will have a larger ionic size.

### Step 3: Conclusion on the statements

- **Statement I**: "The size \( H^{-} \) is greater than that of \( F^{-} \)" is **True**.
- **Statement II**: "The e/p ratio in \( H^{-} \) is 2 while that in \( F^{-} \) is \( \frac{10}{9} \)" is also **True** and provides a correct explanation for Statement I.

### Final Answer

Both statements are true and Statement II correctly explains Statement I.