The ionic radii of `O^(2-),F^(-) , Na^(+) and Mg^(2+)` are in the order :

Options

A. `O^(2-) gt F^(-) gt Mg^(2+) gt Na^(+)`
B. `Mg^(2+) gt Na^(+) gt F^(-) gt O^(2-)`
C. `O^(2-) gt F^(-) gt Na^(+) gt Mg^(2+)`
D. `F^(-) gt O^(2-) gt Na^(+) gt Mg^(2+)`

Answer

To determine the order of ionic radii for the ions \( O^{2-}, F^{-}, Na^{+}, \) and \( Mg^{2+} \), we can follow these steps:

### Step 1: Identify the number of electrons
All four ions have the same number of electrons, which is 10. This is because:
- \( O^{2-} \) has 8 protons and gains 2 electrons.
- \( F^{-} \) has 9 protons and gains 1 electron.
- \( Na^{+} \) has 11 protons and loses 1 electron.
- \( Mg^{2+} \) has 12 protons and loses 2 electrons.

### Step 2: Understand the effect of charge on ionic radii
The ionic radius is influenced by the charge of the ion. Generally, the greater the positive charge, the smaller the ionic radius, and the greater the negative charge, the larger the ionic radius. This is due to the effective nuclear charge experienced by the electrons.

### Step 3: Compare the charges
- \( O^{2-} \) has a charge of -2.
- \( F^{-} \) has a charge of -1.
- \( Na^{+} \) has a charge of +1.
- \( Mg^{2+} \) has a charge of +2.

### Step 4: Determine the order of ionic radii
Since \( O^{2-} \) has the largest negative charge, it will have the largest ionic radius. Conversely, \( Mg^{2+} \), with the highest positive charge, will have the smallest ionic radius. Therefore, we can arrange the ions in order of decreasing ionic radii as follows:

\[ O^{2-} > F^{-} > Na^{+} > Mg^{2+} \]

### Conclusion
Thus, the order of ionic radii is:

\[ O^{2-} > F^{-} > Na^{+} > Mg^{2+} \]

This corresponds to option 3.

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