The radius of which is closest to that of the `Li^(+)` ions?

Options

A. `Na^(+)`
B. `Be^(2+)`
C. `Mg^(2+)`
D. `AI^(3+)`

Answer

To determine which ion has a radius closest to that of the `Li^(+)` ion, we will analyze the options provided based on their electronic configurations and the effects of ionic charges on atomic radii.

### Step-by-Step Solution:

1. **Understanding Lithium Ion (`Li^(+)`)**:
- Lithium (Li) has an atomic number of 3 and its electronic configuration is `1s² 2s¹`.
- When it loses one electron to form `Li^(+)`, its configuration becomes `1s²`, which means it has a total of 2 electrons.

2. **Analyzing the Options**:
- **Option A: Sodium Ion (`Na^(+)`)**:
- Sodium (Na) has an atomic number of 11 with the electronic configuration `1s² 2s² 2p⁶ 3s¹`.
- When it loses one electron to form `Na^(+)`, it has the configuration `1s² 2s² 2p⁶`, which is the same as neon (Ne).
- Since sodium has one more electron shell than lithium, its radius will be larger than that of `Li^(+)`. Thus, this option is incorrect.

- **Option B: Beryllium Ion (`Be^(2+)`)**:
- Beryllium (Be) has an atomic number of 4 and its configuration is `1s² 2s²`.
- When it loses two electrons to form `Be^(2+)`, it also has the configuration `1s²`.
- However, beryllium is smaller than lithium in its neutral form. The additional positive charge in `Be^(2+)` increases the nuclear attraction, making it smaller than `Li^(+)`. Thus, this option is also incorrect.

- **Option C: Magnesium Ion (`Mg^(2+)`)**:
- Magnesium (Mg) has an atomic number of 12 with the electronic configuration `1s² 2s² 2p⁶ 3s²`.
- When it loses two electrons to form `Mg^(2+)`, it has the configuration `1s² 2s² 2p⁶`, which is the same as neon (Ne).
- The neutral magnesium atom is larger than lithium, but the `Mg^(2+)` ion has a greater positive charge, which increases the nuclear attraction and reduces its size. Therefore, `Mg^(2+)` can be closer in size to `Li^(+)`. This option is a strong candidate.

- **Option D: Aluminium Ion (`Al^(3+)`)**:
- Aluminium (Al) has an atomic number of 13 with the electronic configuration `1s² 2s² 2p⁶ 3s² 3p¹`.
- When it loses three electrons to form `Al^(3+)`, it has the configuration `1s² 2s² 2p⁶`, which is also the same as neon (Ne).
- However, aluminium is larger than lithium in its neutral form, and losing three electrons makes it smaller, but it is still larger than `Li^(+)`. Thus, this option is incorrect.

3. **Conclusion**:
- After analyzing all options, the ion with the radius closest to `Li^(+)` is **Option C: `Mg^(2+)`**.

### Final Answer:
The radius closest to that of `Li^(+)` ions is `Mg^(2+)`.