The ions `O^(-2),F^(-),Mg^(2+) and Al^(3+)` are isoelectronic. Their ionic radii show

Options

A. a decrease from `O^(2-)` to `F^(-)` and then increase from `Na^(+)` to `Al^(3+)`
B. a significant increase from `O^(2-)` to `Al^(3+)`
C. a significant decrease from `O^(2-)` to `Al^(3+)`
D. an increase from `O^(2-)` to `F^(-)` and then decrease from `Na^(+)` to `Al^(3+)`

Answer

To solve the problem regarding the ionic radii of the isoelectronic ions \( O^{2-}, F^{-}, Mg^{2+}, \) and \( Al^{3+} \), we will follow these steps:

### Step 1: Identify the number of electrons in each ion
All the given ions are isoelectronic, meaning they have the same number of electrons.
- \( O^{2-} \) has 8 protons and gains 2 electrons, resulting in a total of 10 electrons.
- \( F^{-} \) has 9 protons and gains 1 electron, resulting in a total of 10 electrons.
- \( Mg^{2+} \) has 12 protons and loses 2 electrons, resulting in a total of 10 electrons.
- \( Al^{3+} \) has 13 protons and loses 3 electrons, resulting in a total of 10 electrons.

### Step 2: Determine the nuclear charge (Z) for each ion
The nuclear charge is represented by the number of protons in the nucleus:
- For \( O^{2-} \): \( Z = 8 \)
- For \( F^{-} \): \( Z = 9 \)
- For \( Mg^{2+} \): \( Z = 12 \)
- For \( Al^{3+} \): \( Z = 13 \)

### Step 3: Calculate the Z/e ratio
The Z/e ratio indicates the effective nuclear charge experienced by the electrons. It is calculated as follows:
- For \( O^{2-} \): \( \frac{Z}{e} = \frac{8}{10} = 0.8 \)
- For \( F^{-} \): \( \frac{Z}{e} = \frac{9}{10} = 0.9 \)
- For \( Mg^{2+} \): \( \frac{Z}{e} = \frac{12}{10} = 1.2 \)
- For \( Al^{3+} \): \( \frac{Z}{e} = \frac{13}{10} = 1.3 \)

### Step 4: Analyze the relationship between ionic radius and Z/e ratio
The ionic radius is inversely proportional to the Z/e ratio. This means that as the Z/e ratio increases, the ionic radius decreases.

### Step 5: Order the ionic radii from largest to smallest
Based on the Z/e ratios calculated:
- \( O^{2-} \) has the smallest nuclear charge and thus the largest ionic radius.
- \( F^{-} \) has a slightly larger nuclear charge than \( O^{2-} \) and thus a smaller ionic radius.
- \( Mg^{2+} \) has a larger nuclear charge than \( F^{-} \) and thus an even smaller ionic radius.
- \( Al^{3+} \) has the largest nuclear charge and thus the smallest ionic radius.

The order of ionic radii from largest to smallest is:
1. \( O^{2-} \) (largest)
2. \( F^{-} \)
3. \( Mg^{2+} \)
4. \( Al^{3+} \) (smallest)

### Conclusion
The final order of ionic radii for the isoelectronic ions \( O^{2-}, F^{-}, Mg^{2+}, \) and \( Al^{3+} \) is:
\[ O^{2-} > F^{-} > Mg^{2+} > Al^{3+} \]