The correct ionic radii order is

Options

A. `N^(3-)gtO^(2-)gtF^(-)gtNa^(+)`
B. `N^(3-)gtNa^(+)gtO^(2-)gtF^(-)`
C. `Na^(+)gtO^(2-)gtN^(3-)gtF^(-)`
D. `O^(2-)gtF^(-)gtNa^(+)gtN^(3-)`

Explanation

To determine the correct order of ionic radii for the given species \(N^{3-}\), \(O^{2-}\), \(F^{-}\), and \(Na^{+}\), we will follow these steps:

### Step 1: Identify the number of electrons and protons in each ion
- **For \(N^{3-}\)**:
- Atomic number of nitrogen (N) = 7 (7 protons)
- As \(N^{3-}\) has gained 3 electrons, it has \(7 + 3 = 10\) electrons.

- **For \(O^{2-}\)**:
- Atomic number of oxygen (O) = 8 (8 protons)
- As \(O^{2-}\) has gained 2 electrons, it has \(8 + 2 = 10\) electrons.

- **For \(F^{-}\)**:
- Atomic number of fluorine (F) = 9 (9 protons)
- As \(F^{-}\) has gained 1 electron, it has \(9 + 1 = 10\) electrons.

- **For \(Na^{+}\)**:
- Atomic number of sodium (Na) = 11 (11 protons)
- As \(Na^{+}\) has lost 1 electron, it has \(11 - 1 = 10\) electrons.

### Step 2: Identify that all species are isoelectronic
All four species \(N^{3-}\), \(O^{2-}\), \(F^{-}\), and \(Na^{+}\) have the same number of electrons (10 electrons), making them isoelectronic.

### Step 3: Determine the relationship between ionic size and atomic number
For isoelectronic species, the ionic size is inversely proportional to the number of protons (atomic number). This means that the more protons an ion has, the smaller its ionic radius will be.

### Step 4: Compare the ionic sizes based on the number of protons
- **\(N^{3-}\)**: 7 protons (largest radius)
- **\(O^{2-}\)**: 8 protons (smaller than \(N^{3-}\))
- **\(F^{-}\)**: 9 protons (smaller than \(O^{2-}\))
- **\(Na^{+}\)**: 11 protons (smallest radius)

### Step 5: Write the order of ionic radii from largest to smallest
Based on the above analysis, the correct order of ionic radii is:
\[ N^{3-} > O^{2-} > F^{-} > Na^{+} \]

### Conclusion
Thus, the correct ionic radii order is \(N^{3-} > O^{2-} > F^{-} > Na^{+}\).

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