`Li^(+),Mg^(2+),K^(+),Al^(3+)` (Arrange in increasing order of radii)

Answer

To arrange the ions \( \text{Li}^+, \text{Mg}^{2+}, \text{K}^+, \text{Al}^{3+} \) in increasing order of their ionic radii, we need to consider the charge of the ions and their positions in the periodic table. Here’s a step-by-step solution:

### Step 1: Understand Ionic Charge
The ionic size is inversely proportional to the charge of the ion. The greater the positive charge, the smaller the ionic radius. The ions we have are:
- \( \text{Li}^+ \) (1+ charge)
- \( \text{Mg}^{2+} \) (2+ charge)
- \( \text{K}^+ \) (1+ charge)
- \( \text{Al}^{3+} \) (3+ charge)

### Step 2: Compare the Charges
From the charges, we can see that:
- \( \text{Al}^{3+} \) has the highest positive charge, thus it will have the smallest radius.
- \( \text{Mg}^{2+} \) has the next highest charge, so it will be larger than \( \text{Al}^{3+} \) but smaller than the others.
- \( \text{Li}^+ \) and \( \text{K}^+ \) both have a 1+ charge, but since \( \text{K} \) is below \( \text{Li} \) in the periodic table, \( \text{K}^+ \) will be larger than \( \text{Li}^+ \).

### Step 3: Arrange in Order
Based on the analysis above, we can arrange the ions in increasing order of their ionic radii:
1. \( \text{Al}^{3+} \) (smallest)
2. \( \text{Mg}^{2+} \)
3. \( \text{Li}^+ \)
4. \( \text{K}^+ \) (largest)

### Final Order
Thus, the increasing order of ionic radii is:
\[ \text{Al}^{3+} < \text{Mg}^{2+} < \text{Li}^+ < \text{K}^+ \]

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