Which of the following is the smallest in size?

Options

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

Answer

To determine which of the given species is the smallest in size, we need to analyze the options provided: \(N^{3-}\), \(F^{-}\), \(O^{2-}\), and \(Na^{+}\).

### Step-by-Step Solution:

1. **Identify the Species**:
We have the following four species:
- \(N^{3-}\) (Nitride ion)
- \(F^{-}\) (Fluoride ion)
- \(O^{2-}\) (Oxide ion)
- \(Na^{+}\) (Sodium ion)

2. **Determine the Number of Electrons**:
- \(N^{3-}\): Nitrogen has 7 protons, so \(N^{3-}\) has \(7 + 3 = 10\) electrons.
- \(F^{-}\): Fluorine has 9 protons, so \(F^{-}\) has \(9 + 1 = 10\) electrons.
- \(O^{2-}\): Oxygen has 8 protons, so \(O^{2-}\) has \(8 + 2 = 10\) electrons.
- \(Na^{+}\): Sodium has 11 protons, so \(Na^{+}\) has \(11 - 1 = 10\) electrons.

3. **Identify Isoelectronic Species**:
All four species are isoelectronic, meaning they all have the same number of electrons (10 electrons).

4. **Consider Effective Nuclear Charge**:
The size of isoelectronic species can be compared based on their effective nuclear charge (\(Z_{eff}\)). The effective nuclear charge is the net positive charge experienced by an electron in a multi-electron atom.

5. **Calculate the Nuclear Charge**:
- \(N^{3-}\): Nuclear charge = +7
- \(F^{-}\): Nuclear charge = +9
- \(O^{2-}\): Nuclear charge = +8
- \(Na^{+}\): Nuclear charge = +11

6. **Relate Nuclear Charge to Size**:
The greater the nuclear charge, the stronger the attraction between the nucleus and the electrons, which results in a smaller ionic size. Therefore, the order of size will be inversely related to the nuclear charge.

7. **Determine the Smallest Ion**:
Since \(Na^{+}\) has the highest nuclear charge (+11), it will have the smallest size among the given ions.

### Conclusion:
The smallest ion among the options provided is \(Na^{+}\).