The dissolution of `Al(OH)_(3)` by a solution of `NaOH` results in the formation of

Options

A. `[Al(H_(2)O)_(4)(OH)_(2)]^(+)`
B. `[Al(H_(2)O)_(3)(OH)_(3)]`
C. `[Al(H_(2)O)_(2)(OH)_(4)]^(-)`
D. `[Al(H_(2)O)_(6)(OH)_(3)]`

Explanation

To solve the question regarding the dissolution of `Al(OH)₃` by a solution of `NaOH`, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactants in this reaction are aluminum hydroxide, `Al(OH)₃`, and sodium hydroxide, `NaOH`.

2. **Understand the Reaction**:
When `Al(OH)₃` is treated with `NaOH`, it undergoes a reaction where aluminum hydroxide dissolves in the alkaline solution.

3. **Formation of Complex Ion**:
The dissolution leads to the formation of a complex ion. The hydroxide ions from `NaOH` interact with aluminum hydroxide to form the tetrahydroxoaluminate ion, `Al(OH)₄⁻`.

4. **Write the Chemical Equation**:
The balanced chemical equation for the reaction can be written as:
\[ Al(OH)_3 + OH^- \rightarrow Al(OH)_4^- \]
This indicates that one hydroxide ion from `NaOH` reacts with `Al(OH)₃` to form the complex ion.

5. **Consider the Aqueous Medium**:
Since the reaction occurs in an aqueous medium, water molecules can also be involved in the coordination sphere of the aluminum ion, leading to the formation of an octahedral complex.

6. **Final Product**:
The final product of the reaction is the tetrahydroxoaluminate ion, `Al(OH)₄⁻`, along with sodium ions, `Na⁺`, which remain in solution.

7. **IUPAC Naming**:
The IUPAC name for the complex can be derived as follows:
- The complex ion `Al(OH)₄⁻` is called tetrahydroxoaluminate.
- In the presence of sodium ions, the full name becomes sodium tetrahydroxoaluminate.

### Final Answer:
The dissolution of `Al(OH)₃` by a solution of `NaOH` results in the formation of the complex ion `Al(OH)₄⁻`, which can be referred to as sodium tetrahydroxoaluminate.

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