Give the number of total possible ionisation isomers in `[Pt(NH_(3))_(4)CI_(2)]Br_(2)` .

Answer

To determine the total number of possible ionization isomers for the complex \([Pt(NH_3)_4Cl_2]Br_2\), we will follow these steps:

### Step 1: Understand Ionization Isomerism
Ionization isomerism occurs when two or more compounds have the same molecular formula but produce different ions in solution. This typically involves the exchange of ligands with counter ions.

**Hint:** Recall that ionization isomers differ in the ions they release when dissolved in solution.

### Step 2: Identify the Components of the Complex
The complex \([Pt(NH_3)_4Cl_2]Br_2\) consists of:
- 4 ammonia (NH3) ligands
- 2 chloride (Cl) ligands
- 2 bromide (Br) counter ions

**Hint:** Count the ligands and counter ions in the complex to understand how they can be rearranged.

### Step 3: Determine Possible Ionization Isomers
In this complex, the bromide ions can be exchanged with the chloride ions. The possible ionization isomers can be formed by varying the arrangement of the ligands and the counter ions.

1. **Isomer 1:** \([Pt(NH_3)_4Cl_2]^{2+} + 2Br^{-}\)
- This is the original complex where the ions released are 2 bromide ions and the cation contains 2 chloride ions.

2. **Isomer 2:** \([Pt(NH_3)_4Br_2]^{2+} + 2Cl^{-}\)
- Here, the complex releases 2 chloride ions instead of bromide ions.

3. **Isomer 3:** \([Pt(NH_3)_4ClBr]^{+} + Cl^{-} + Br^{-}\)
- In this case, one chloride and one bromide ion are released along with the cation.

**Hint:** Think about how the ligands can be swapped with the counter ions to create different ionization isomers.

### Step 4: Count the Total Ionization Isomers
From the analysis, we have identified three distinct ionization isomers:
1. \([Pt(NH_3)_4Cl_2]^{2+} + 2Br^{-}\)
2. \([Pt(NH_3)_4Br_2]^{2+} + 2Cl^{-}\)
3. \([Pt(NH_3)_4ClBr]^{+} + Cl^{-} + Br^{-}\)

Thus, the total number of possible ionization isomers for the complex \([Pt(NH_3)_4Cl_2]Br_2\) is **3**.

**Final Answer:** 3 ionization isomers.