Statement-1: `Na^(+)` and `AI^(3+)` are isoelectronic but the magnitude of ionic radius or `AI^(3+)` is less than that of `Na^(+)`.
Statement-2: The magnitude of effective nuclear charge of the outer most shell electrons in `AI^(3+)` is greater than that of `Na^(+)`.

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 problem, we need to analyze the two statements provided regarding the ionic radii of `Na^(+)` and `Al^(3+)`, as well as the effective nuclear charge experienced by the outermost electrons in these ions.

### Step-by-Step Solution:

1. **Identify the Ionic Species**:
- Sodium ion (`Na^(+)`) has lost one electron from its neutral atom (which has 11 electrons), resulting in 10 electrons.
- Aluminum ion (`Al^(3+)`) has lost three electrons from its neutral atom (which has 13 electrons), also resulting in 10 electrons.

2. **Determine Isoelectronic Nature**:
- Since both `Na^(+)` and `Al^(3+)` have the same number of electrons (10), they are classified as isoelectronic species.

3. **Analyze Nuclear Charge**:
- The nuclear charge of sodium is +11 (from 11 protons), and for aluminum, it is +13 (from 13 protons).
- When these ions are formed, the effective nuclear charge (Z_eff) acting on the remaining electrons differs due to the number of protons in the nucleus.

4. **Calculate Effective Nuclear Charge**:
- For `Na^(+)`: The effective nuclear charge on the 10 electrons is +11.
- For `Al^(3+)`: The effective nuclear charge on the 10 electrons is +13.
- Since `Al^(3+)` has a higher nuclear charge (more protons), the effective nuclear charge experienced by the outermost electrons in `Al^(3+)` is greater than that in `Na^(+)`.

5. **Compare Ionic Radii**:
- The ionic radius is influenced by the effective nuclear charge. A higher effective nuclear charge pulls the electrons closer to the nucleus, resulting in a smaller ionic radius.
- Therefore, `Al^(3+)`, with a greater effective nuclear charge, will have a smaller ionic radius compared to `Na^(+)`.

6. **Conclusion**:
- Statement-1: `Na^(+)` and `Al^(3+)` are isoelectronic, and the ionic radius of `Al^(3+)` is indeed less than that of `Na^(+)` (True).
- Statement-2: The effective nuclear charge of the outermost shell electrons in `Al^(3+)` is greater than that of `Na^(+)` (True).

Both statements are correct, and Statement-2 correctly explains Statement-1.

### Final Answer:
Both Statement-1 and Statement-2 are true, and Statement-2 is the correct explanation for Statement-1.

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