Assertion: `F^(-)` ion has highest hydrated radius among the other halide ions.
Reason: Ionic radius of `F^(-)` is smallest in the periodic table.

Options

A. If both assertion and reason are true annd the reason is the correct explanation of assertion
B. If both assertion and reason are true but reason is not the correct explanation of assertion
C. If assertion is true but the reason is false
D. if assertion is false but the reason is true

Answer

To solve the question, we need to analyze both the assertion and the reason provided.

### Step 1: Understand the Assertion
The assertion states that the `F^(-)` ion has the highest hydrated radius among other halide ions. Hydrated radius refers to the effective size of an ion when it is surrounded by water molecules.

### Step 2: Understand the Reason
The reason given is that the ionic radius of `F^(-)` is the smallest in the periodic table. The ionic radius is the measure of an ion's size. In the case of halide ions, we have `F^(-)`, `Cl^(-)`, `Br^(-)`, and `I^(-)`.

### Step 3: Analyze Ionic Radii of Halides
- The ionic radii of the halide ions increase as we move down the group in the periodic table:
- `F^(-)` < `Cl^(-)` < `Br^(-)` < `I^(-)`
- This means that `F^(-)` has the smallest ionic radius among the halides.

### Step 4: Relationship Between Ionic Radius and Hydration
Hydration energy is inversely proportional to the size of the ion. This means that smaller ions have higher hydration energy because they can attract water molecules more effectively due to their higher charge density.

### Step 5: Conclusion on Hydrated Radius
Since `F^(-)` has the smallest ionic radius, it will have the highest hydration energy. However, the hydrated radius is the effective size of the ion when hydrated, which can be counterintuitive. Smaller ions like `F^(-)` tend to have a smaller hydrated radius than larger ions, despite having higher hydration energy.

### Final Conclusion
- The assertion is **false** because `F^(-)` does not have the highest hydrated radius; it actually has a smaller hydrated radius compared to larger halide ions like `I^(-)`.
- The reason is **true** because `F^(-)` does indeed have the smallest ionic radius.

Thus, the correct answer is that the assertion is false, while the reason is true.