Silver chloride dissolves in excess of `NH_4OH`. The cation present in solution is.

Options

A. `[Ag(NH_3)]^(oplus)`
B. `[Ag(NH_3)_4]^(oplus)`
C. `[Ag(NH_3)_2]^(oplus)`
D. `[Ag(NH_3)_6]^(oplus)`

Explanation

To solve the problem of determining the cation present in the solution when silver chloride (AgCl) dissolves in excess ammonium hydroxide (NH4OH), we can follow these steps:

### Step 1: Understand the Reaction
When silver chloride is treated with excess ammonium hydroxide, a complex ion is formed. The reaction can be represented as:

\[ \text{AgCl} + 2 \text{NH}_4\text{OH} \rightarrow [\text{Ag}(\text{NH}_3)_2]^+ + \text{Cl}^- + \text{H}_2\text{O} \]

### Step 2: Identify the Complex Ion
In this reaction, silver ions (Ag⁺) form a complex with ammonia (NH₃) present in the ammonium hydroxide. The complex ion formed is:

\[ [\text{Ag}(\text{NH}_3)_2]^+ \]

### Step 3: Determine the Cation
The cation present in the solution is the complex ion itself, which can be written as:

\[ \text{Ag(NH}_3\text{)}_2^+ \]

### Step 4: Analyze the Given Options
Now, let’s analyze the options provided in the question:
1. Ag(NH₃)⁺
2. Ag(NH₃)₄⁺
3. Ag(NH₃)₂²⁺
4. Ag(NH₃)₆⁺

From our analysis, the correct cation formed is:

\[ \text{Ag(NH}_3\text{)}_2^+ \]

### Conclusion
Thus, the correct answer is option 3: Ag(NH₃)₂⁺.

---