Arrange the following as stated: Increasing order of ionic size
`N^(3-),Na^(oplus),F^(ɵ),O^(2-),Mg^(2+)`

Explanation

To arrange the ions \(N^{3-}, Na^{+}, F^{-}, O^{2-}, Mg^{2+}\) in increasing order of ionic size, we need to consider the following factors:

1. **Ionic Charge**: The ionic size is influenced by the charge of the ion. Anions (negatively charged ions) tend to be larger than their neutral atoms because the addition of electrons increases electron-electron repulsion. Conversely, cations (positively charged ions) are smaller than their neutral atoms because the loss of electrons leads to a greater effective nuclear charge acting on the remaining electrons, pulling them closer to the nucleus.

2. **Isoelectronic Series**: Ions that have the same number of electrons are said to be isoelectronic. In this case, \(N^{3-}, O^{2-}, F^{-}, Na^{+},\) and \(Mg^{2+}\) are all isoelectronic with 10 electrons. However, the size will vary based on the nuclear charge (number of protons) in the nucleus. The greater the number of protons, the smaller the ionic size due to increased nuclear attraction.

### Step-by-Step Solution:

1. **Identify the number of electrons and protons**:
- \(N^{3-}\): 10 electrons (7 protons)
- \(O^{2-}\): 10 electrons (8 protons)
- \(F^{-}\): 10 electrons (9 protons)
- \(Na^{+}\): 10 electrons (11 protons)
- \(Mg^{2+}\): 10 electrons (12 protons)

2. **Analyze ionic sizes based on charge and nuclear protons**:
- \(N^{3-}\) has the least protons (7) and the most electron-electron repulsion, making it the largest.
- \(O^{2-}\) has 8 protons, so it is smaller than \(N^{3-}\).
- \(F^{-}\) has 9 protons, making it smaller than \(O^{2-}\).
- \(Na^{+}\) has 11 protons, making it smaller than \(F^{-}\).
- \(Mg^{2+}\) has the most protons (12), so it is the smallest.

3. **Arrange the ions in increasing order of size**:
- \(Mg^{2+} < Na^{+} < F^{-} < O^{2-} < N^{3-}\)

### Final Answer:
The increasing order of ionic size is:
\[ Mg^{2+} < Na^{+} < F^{-} < O^{2-} < N^{3-} \]