The ionic radii of `N^(3-), O^(2-)` and `F^(-)` are respectively given by:

Options

A. `1.36, 1.71 and 1.40`
B. `1.71, 1.40 and 1.36`
C. `1.71, 1.36 and 1.40`
D. `1.36, 1.40 and 1.71`

Answer

To solve the problem of determining the ionic radii of the ions \(N^{3-}\), \(O^{2-}\), and \(F^{-}\), we will follow these steps:

### Step 1: Identify the Ionic Species
We have three ions:
- \(N^{3-}\)
- \(O^{2-}\)
- \(F^{-}\)

### Step 2: Determine the Number of Electrons
All three ions are isoelectronic, meaning they have the same number of electrons. Each of these ions has gained electrons to achieve a stable electron configuration:
- \(N^{3-}\) has 10 electrons (7 from nitrogen + 3 additional).
- \(O^{2-}\) has 10 electrons (8 from oxygen + 2 additional).
- \(F^{-}\) has 10 electrons (9 from fluorine + 1 additional).

### Step 3: Compare Atomic Numbers
Next, we need to consider the atomic numbers of the neutral atoms:
- Nitrogen (N) has an atomic number of 7.
- Oxygen (O) has an atomic number of 8.
- Fluorine (F) has an atomic number of 9.

### Step 4: Apply the Concept of Ionic Radii
The ionic radius is influenced by the effective nuclear charge experienced by the electrons. For isoelectronic species:
- The higher the atomic number (Z), the greater the positive charge in the nucleus, which pulls the electrons closer and results in a smaller ionic radius.

### Step 5: Determine the Order of Ionic Radii
Since \(N^{3-}\), \(O^{2-}\), and \(F^{-}\) are isoelectronic:
- \(F^{-}\) (with atomic number 9) will have the smallest ionic radius.
- \(O^{2-}\) (with atomic number 8) will be larger than \(F^{-}\) but smaller than \(N^{3-}\).
- \(N^{3-}\) (with atomic number 7) will have the largest ionic radius among the three.

Thus, the order of ionic radii from largest to smallest is:
1. \(N^{3-}\) (largest)
2. \(O^{2-}\)
3. \(F^{-}\) (smallest)

### Step 6: Assign Values to the Ionic Radii
Based on the common values for these ionic radii:
- \(r(N^{3-}) \approx 1.71 \, \text{Å}\)
- \(r(O^{2-}) \approx 1.40 \, \text{Å}\)
- \(r(F^{-}) \approx 1.36 \, \text{Å}\)

### Final Answer
The ionic radii of \(N^{3-}\), \(O^{2-}\), and \(F^{-}\) are respectively:
- \(1.71 \, \text{Å}\) (for \(N^{3-}\))
- \(1.40 \, \text{Å}\) (for \(O^{2-}\))
- \(1.36 \, \text{Å}\) (for \(F^{-}\))