To solve this question, let's analyze the behavior of anhydrous aluminum chloride \((\text{AlCl}_3)\) when it is dissolved in water.
### Explanation
1. **Dimerization of Aluminum Chloride:**
- In the solid state, aluminum chloride exists as a dimer, \(\text{Al}_2\text{Cl}_6\). In this form, two \(\text{AlCl}_3\) units are linked together.
2. **Hydrolysis in Water:**
- When anhydrous aluminum chloride is added to water, it hydrolyzes. The \(\text{Al}^{3+}\) ions interact with water molecules, forming a complex. This process can be represented as:
\[
\text{Al}_2\text{Cl}_6 + 6\text{H}_2\text{O} \rightarrow 2[\text{Al(H}_2\text{O)}_6]^{3+} + 6\text{Cl}^-
\]
3. **Products of Hydrolysis:**
- The products of this hydrolysis are:
- Hexaaquaaluminum ion \([\text{Al(H}_2\text{O)}_6]^{3+}\)
- Chloride ions \(\text{Cl}^-\)
4. **Considering the Given Options:**
- **Option A:** \([Al(OH)_6]^{3-} + 3HCl\) – This is incorrect because aluminum does not form such a complex in aqueous solution at this concentration.
- **Option B:** \([\text{Al(H}_2\text{O)}_6]^{3+} + 3\text{HCl}\) – This is partially correct; it forms the hexaaquaaluminum ion but does not produce HCl directly in significant amounts.
- **Option C:** \(\text{Al}^{3+} + 3\text{Cl}^-\) – This represents the dissociation of the aluminum ion and chloride ions, which does occur.
- **Option D:** \(\text{Al}_2\text{O}_3 + 6\text{HCl}\) – This option is not relevant as it doesn't describe the process of dissolution in water.
### Conclusion
The correct answer is **C) \(\text{Al}^{3+} + 3\text{Cl}^-\)**.
This option correctly represents the ions present when anhydrous aluminum chloride is dissolved in water, forming aluminum ions and chloride ions, with the formation of the hexaaquaaluminum complex being implicit in the dissolution process.
