`B^(3+)` cannot exist in aqueous solution because of its :

Options

A. Strong reducing ability
B. Strong oxidizing ability
C. Small size and large charge
D. Large size and small charge

Explanation

To determine why \( B^{3+} \) cannot exist in aqueous solution, we will analyze the properties of boron and its ionization.

### Step-by-Step Solution:

1. **Understanding the Ion**:
- Boron (B) can lose three electrons to form the \( B^{3+} \) ion. This ion has a +3 oxidation state.

2. **Size of the Ion**:
- Boron is a small atom. When it loses three electrons, the resulting \( B^{3+} \) ion is also very small in size.

3. **Charge Density**:
- Charge density is defined as the charge of the ion divided by its radius. For \( B^{3+} \), the charge is +3, and due to its small size, the radius is also small.
- Therefore, the charge density of \( B^{3+} \) is very high.

4. **Implications of High Charge Density**:
- A high charge density leads to strong interactions with water molecules. Instead of remaining as a free ion in solution, \( B^{3+} \) tends to attract water molecules strongly.

5. **Hydrolysis**:
- Due to the strong interaction with water, \( B^{3+} \) undergoes hydrolysis. This means it reacts with water to form other species rather than existing as \( B^{3+} \) in solution.
- For example, \( BCl_3 \) can react with water to form boric acid (\( H_3BO_3 \)) and hydrochloric acid (\( HCl \)).

6. **Conclusion**:
- The combination of the small size of the \( B^{3+} \) ion and its high charge density leads to its inability to exist as a free ion in aqueous solution. Instead, it hydrolyzes and forms covalent compounds.

### Final Answer:
\( B^{3+} \) cannot exist in aqueous solution due to its small size and high charge density, which leads to hydrolysis and the formation of covalent compounds.

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