Ammonia forms the complex `[Cu(NH_(3))_(4)]^(2+)` with copper ion in alkaline solution but not in acidic solution. This is due to the reason that :

Options

A. in alkaline solution insoluble `Cu(OH)_(2)` is precipitated which is soluble in excess of any alkali
B. `Cu(OH)_(2)` is amphoteric
C. in acidic solution hydration protects `Cu^(2+)` ion
D. in acidic solution, proton coordinates with `NH_(3)` forming `NH_(4)^(+)` ion and `NH_(3)` is not available

Explanation

To solve the question regarding why ammonia forms the complex \([Cu(NH_3)_4]^{2+}\) with copper ion in alkaline solution but not in acidic solution, we can break it down into clear steps:

### Step-by-Step Solution:

1. **Understanding the Reaction**:
- In alkaline conditions, copper ions (\(Cu^{2+}\)) react with ammonia (\(NH_3\)) to form the complex \([Cu(NH_3)_4]^{2+}\).
- In acidic conditions, the presence of protons (\(H^+\)) affects the reaction.

2. **Reaction in Alkaline Medium**:
- In alkaline medium, ammonia acts as a ligand and donates its lone pair of electrons to the copper ion.
- The reaction can be represented as:
\[ Cu^{2+} + 4 NH_3 \rightarrow [Cu(NH_3)_4]^{2+} \]
- Here, \(NH_3\) coordinates with \(Cu^{2+}\) to form the complex.

3. **Reaction in Acidic Medium**:
- In acidic medium, the protons (\(H^+\)) react with ammonia.
- The reaction can be represented as:
\[ NH_3 + H^+ \rightarrow NH_4^+ \]
- This means that ammonia is converted into ammonium ion (\(NH_4^+\)), which cannot act as a ligand to form the complex.

4. **Conclusion**:
- The formation of the complex \([Cu(NH_3)_4]^{2+}\) is not possible in acidic solution because ammonia is protonated to form \(NH_4^+\), thus it is no longer available to coordinate with the copper ion.

5. **Final Answer**:
- The reason why ammonia forms the complex \([Cu(NH_3)_4]^{2+}\) in alkaline solution but not in acidic solution is that in acidic conditions, ammonia is converted into ammonium ion (\(NH_4^+\)), which cannot coordinate with copper ions.