The solution of `AgBr` in presence of large excess of `NH_(3)` contains mainly the cation.
(a) `NH_(4)^(o+)`
`Ag^(o+)`
(c ) `[Ag(NH_(3))]^(o+)`
(d) `[Ag(NH_(3))_(2)]^(o+)` .

Explanation

To solve the question regarding the solution of AgBr in the presence of a large excess of NH₃, we will follow these steps:

### Step 1: Understand the Reaction
When AgBr is dissolved in a large excess of ammonia (NH₃), a complex ion is formed. Ammonia acts as a ligand and coordinates with the silver ion (Ag⁺).

### Step 2: Identify the Products
The reaction can be represented as:
\[ \text{AgBr} + 2 \text{NH}_3 \rightarrow [\text{Ag(NH}_3)_2]^+ + \text{Br}^- \]
Here, Ag⁺ from AgBr forms a complex with two NH₃ molecules, resulting in the complex ion \([\text{Ag(NH}_3)_2]^+\) and bromide ion (Br⁻) as a byproduct.

### Step 3: Determine the Cation
From the reaction, we can see that the cation formed is the complex ion \([\text{Ag(NH}_3)_2]^+\). The bromide ion (Br⁻) is the anion and does not contribute to the cationic part of the solution.

### Step 4: Analyze the Options
Now, we will compare our findings with the options provided:
- (a) \( \text{NH}_4^+ \) - This is not formed in this reaction.
- (b) \( \text{Ag}^+ \) - While Ag⁺ is present, it is not the main species in the presence of excess NH₃.
- (c) \([\text{Ag(NH}_3)]^+\) - This is not the correct complex formed since we have two NH₃ ligands.
- (d) \([\text{Ag(NH}_3)_2]^+\) - This is the correct complex ion formed.

### Final Answer
The solution of AgBr in the presence of a large excess of NH₃ contains mainly the cation \([\text{Ag(NH}_3)_2]^+\). Thus, the correct answer is:
**(d) \([\text{Ag(NH}_3)_2]^+\)**.

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