To determine the order of ionic sizes for the isoelectronic ions \( \text{Na}^+, \text{Mg}^{2+}, \text{Al}^{3+}, \text{Si}^{4+} \), we will follow these steps:
### Step 1: Identify the number of electrons in each ion
All four ions are isoelectronic, meaning they have the same number of electrons.
- **Na\(^+\)**: Sodium has 11 electrons, but loses one to become Na\(^+\), resulting in 10 electrons.
- **Mg\(^{2+}\)**: Magnesium has 12 electrons, but loses two to become Mg\(^{2+}\), resulting in 10 electrons.
- **Al\(^{3+}\)**: Aluminum has 13 electrons, but loses three to become Al\(^{3+}\), resulting in 10 electrons.
- **Si\(^{4+}\)**: Silicon has 14 electrons, but loses four to become Si\(^{4+}\), resulting in 10 electrons.
### Step 2: Determine the number of protons in each ion
Next, we need to consider the number of protons in each ion, as this affects the ionic size.
- **Na\(^+\)**: 11 protons
- **Mg\(^{2+}\)**: 12 protons
- **Al\(^{3+}\)**: 13 protons
- **Si\(^{4+}\)**: 14 protons
### Step 3: Understand the relationship between nuclear charge and ionic size
The ionic radius is inversely proportional to the nuclear charge (number of protons). This means that as the number of protons increases, the ionic radius decreases.
### Step 4: Compare the ionic sizes based on proton count
- **Si\(^{4+}\)** has 14 protons, so it will have the smallest ionic radius.
- **Al\(^{3+}\)** has 13 protons, so it will be larger than Si\(^{4+}\) but smaller than the others.
- **Mg\(^{2+}\)** has 12 protons, making it larger than Al\(^{3+}\).
- **Na\(^+\)** has 11 protons, so it will have the largest ionic radius.
### Step 5: Write the final order of ionic sizes
Based on the above analysis, the order of ionic sizes from smallest to largest is:
\[ \text{Si}^{4+} < \text{Al}^{3+} < \text{Mg}^{2+} < \text{Na}^+ \]
### Final Answer
The order of ionic sizes will be:
\[ \text{Si}^{4+} < \text{Al}^{3+} < \text{Mg}^{2+} < \text{Na}^+ \]
