Which one of the following elements is unable to form `MF_6^(2-)` ion?

Options

A. Ga
B. Al
C. B
D. In

Explanation

To determine which element is unable to form the `MF_6^(2-)` ion, we need to analyze the elements in group 13 of the periodic table, also known as the boron family. The elements in this group include boron (B), aluminum (Al), gallium (Ga), and indium (In).

### Step-by-Step Solution:

1. **Identify the Elements**: The elements in question are boron (B), aluminum (Al), gallium (Ga), and indium (In).

2. **Understand the Ion Formation**: The ion `MF_6^(2-)` indicates that the metal (M) is surrounded by six fluoride ions (F). For this to occur, the metal must be able to achieve a +3 oxidation state, as each fluoride ion has a -1 charge.

3. **Analyze Boron (B)**:
- Boron has an atomic number of 5 and an electronic configuration of `1s^2 2s^2 2p^1`.
- It can only accommodate a maximum of 4 pairs of electrons due to its limited s and p orbitals.
- When boron forms a compound, it typically forms `BF_4^-` and does not have the capacity to form `BF_6^(2-)`.

4. **Analyze Aluminum (Al)**:
- Aluminum has an atomic number of 13 and can form `AlF_6^(3-)` by accepting more electrons due to its availability of d-orbitals in the third period.

5. **Analyze Gallium (Ga)**:
- Gallium has an atomic number of 31 and can also form `GaF_6^(3-)` due to similar reasons as aluminum.

6. **Analyze Indium (In)**:
- Indium has an atomic number of 49 and can form `InF_6^(3-)`, benefiting from the presence of d-orbitals.

7. **Conclusion**: Among the elements analyzed, boron is the only element that cannot form the `MF_6^(2-)` ion due to its inability to accommodate more than four pairs of electrons. Therefore, the correct answer is boron (B).

### Final Answer:
The element that is unable to form `MF_6^(2-)` ion is **Boron (B)**.