The order of increasing ionic radius of the following is:

Options

A. `K^(+) lt Li^(+) lt Mg^(2+) lt Al^(3+)`
B. `K^(+) lt Mg^(2+) lt Li^(+) lt Al^(3+)`
C. `Li^(+) lt K^(+) Mg^(2+) lt Al^(3+)`
D. `Al^(3+) lt Mg^(2+) lt Li^(+) lt K^(+)`

Answer

To determine the order of increasing ionic radius for the given ions, we need to analyze the ions based on their charges and their positions in the periodic table. The ions we are considering are: Al³⁺, Mg²⁺, K⁺, and Li⁺.

### Step-by-Step Solution:

1. **Identify the Ions and Their Charges**:
- Al³⁺ (Aluminum ion with a +3 charge)
- Mg²⁺ (Magnesium ion with a +2 charge)
- K⁺ (Potassium ion with a +1 charge)
- Li⁺ (Lithium ion with a +1 charge)

2. **Determine Isoelectronic Species**:
- Al³⁺ and Mg²⁺ are both isoelectronic with Ne (neon) since they both have 10 electrons (Al has 13 protons, loses 3 electrons; Mg has 12 protons, loses 2 electrons).
- K⁺ and Li⁺ are also isoelectronic with Ne, having 10 electrons as well.

3. **Compare Ionic Radii of Isoelectronic Species**:
- For isoelectronic species, the ionic radius decreases with an increase in positive charge due to the greater nuclear charge attracting the electrons more strongly.
- Therefore, between Al³⁺ and Mg²⁺: Al³⁺ has a greater positive charge than Mg²⁺, so Al³⁺ is smaller than Mg²⁺. Thus, Al³⁺ < Mg²⁺.

4. **Compare K⁺ and Li⁺**:
- Both K⁺ and Li⁺ have the same charge (+1) but are in different periods. As we move down a group in the periodic table, the ionic radius increases due to the addition of electron shells.
- Therefore, K⁺ is larger than Li⁺. Thus, Li⁺ < K⁺.

5. **Combine the Findings**:
- From the comparisons, we have:
- Al³⁺ < Mg²⁺ (Aluminum ion is smaller than Magnesium ion)
- Li⁺ < K⁺ (Lithium ion is smaller than Potassium ion)

6. **Final Order of Increasing Ionic Radius**:
- Combining all the comparisons, we find the order of increasing ionic radius:
- Al³⁺ < Mg²⁺ < Li⁺ < K⁺

### Conclusion:
The order of increasing ionic radius is: **Al³⁺ < Mg²⁺ < Li⁺ < K⁺**.