Of the complex `[NiNH_(3))_(5)Br]CI`, the ionization isomer will give colour with `AgNO_(3)`

Options

A. White
B. Red
C. Yellow
D. Blue

Answer

To solve the question regarding the complex ion `[Ni(NH₃)₅Br]Cl`, we need to analyze the ionization isomers and their behavior when treated with `AgNO₃`. Here’s a step-by-step solution:

### Step 1: Identify the complex and its components
The complex given is `[Ni(NH₃)₅Br]Cl`. This indicates that the nickel ion (Ni²⁺) is coordinated with five ammonia (NH₃) ligands and one bromide (Br⁻) ligand, while the chloride ion (Cl⁻) is outside the coordination sphere.

### Step 2: Determine the ionization isomers
Ionization isomers occur when the anions and cations can be interchanged. In this case, we can have two forms:
1. `[Ni(NH₃)₅Br]Cl` - where Cl⁻ is the counter ion.
2. `[Ni(NH₃)₅Cl]Br` - where Br⁻ is the counter ion.

### Step 3: Dissociation in solution
When the complex `[Ni(NH₃)₅Br]Cl` is dissolved in water, it dissociates into:
- `[Ni(NH₃)₅Br]⁺` and `Cl⁻`

For the other isomer `[Ni(NH₃)₅Cl]Br`, it dissociates into:
- `[Ni(NH₃)₅Cl]⁺` and `Br⁻`

### Step 4: Reaction with AgNO₃
When `AgNO₃` is added to the solution, it reacts with the anion present:
- In the first isomer `[Ni(NH₃)₅Br]Cl`, the anion is `Cl⁻`, which does not form a precipitate with `AgNO₃`.
- In the second isomer `[Ni(NH₃)₅Cl]Br`, the anion is `Br⁻`, which reacts with `Ag⁺` from `AgNO₃` to form `AgBr`, a yellow precipitate.

### Conclusion
Thus, the ionization isomer that will give color (yellow precipitate) with `AgNO₃` is `[Ni(NH₃)₅Cl]Br` because it contains the `Br⁻` ion.

### Final Answer
The ionization isomer that will give color with `AgNO₃` is `[Ni(NH₃)₅Cl]Br`.

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