Find out total number of compound (s) in which at least half of `Cl^(-)` are ionizable `CrCl_(3).6NH_(3), CrCl_(3).5NH_(3),CrCl_(3).4NH_(3),CrCl_(3).3NH_(3),PtCl_(4).6NH_(3),PtCl_(4).5NH_(3), PtCl_(4).4NH_(3),PtCl_(4).3NH_(3),PtCl_(4).2NH_(3)`

Answer

To solve the problem of finding the total number of compounds in which at least half of Cl⁻ ions are ionizable from the given complexes, we will analyze each compound step by step.

### Step-by-Step Solution

1. **Identify the Compounds**: The compounds given are:
- CrCl₃.6NH₃
- CrCl₃.5NH₃
- CrCl₃.4NH₃
- CrCl₃.3NH₃
- PtCl₄.6NH₃
- PtCl₄.5NH₃
- PtCl₄.4NH₃
- PtCl₄.3NH₃
- PtCl₄.2NH₃

2. **Analyze Each Compound**:
- **CrCl₃.6NH₃**:
- Ionization: CrCl₃.6NH₃ → [Cr(NH₃)₆]³⁺ + 3Cl⁻
- All 3 Cl⁻ ions are ionizable. (Ionizable Cl⁻ = 3)

- **CrCl₃.5NH₃**:
- Ionization: CrCl₃.5NH₃ → [Cr(NH₃)₅Cl]²⁺ + 2Cl⁻
- 2 out of 3 Cl⁻ ions are ionizable. (Ionizable Cl⁻ = 2)

- **CrCl₃.4NH₃**:
- Ionization: CrCl₃.4NH₃ → [Cr(NH₃)₄Cl₂]⁺ + 2Cl⁻
- 2 out of 3 Cl⁻ ions are ionizable. (Ionizable Cl⁻ = 2)

- **CrCl₃.3NH₃**:
- Ionization: CrCl₃.3NH₃ → [Cr(NH₃)₃Cl₃] + 0Cl⁻
- No Cl⁻ ions are ionizable. (Ionizable Cl⁻ = 0)

- **PtCl₄.6NH₃**:
- Ionization: PtCl₄.6NH₃ → [Pt(NH₃)₆]⁴⁺ + 4Cl⁻
- All 4 Cl⁻ ions are ionizable. (Ionizable Cl⁻ = 4)

- **PtCl₄.5NH₃**:
- Ionization: PtCl₄.5NH₃ → [Pt(NH₃)₅Cl]³⁺ + 3Cl⁻
- 3 out of 4 Cl⁻ ions are ionizable. (Ionizable Cl⁻ = 3)

- **PtCl₄.4NH₃**:
- Ionization: PtCl₄.4NH₃ → [Pt(NH₃)₄Cl₂]²⁺ + 2Cl⁻
- 2 out of 4 Cl⁻ ions are ionizable. (Ionizable Cl⁻ = 2)

- **PtCl₄.3NH₃**:
- Ionization: PtCl₄.3NH₃ → [Pt(NH₃)₃Cl₃] + 0Cl⁻
- No Cl⁻ ions are ionizable. (Ionizable Cl⁻ = 0)

- **PtCl₄.2NH₃**:
- Ionization: PtCl₄.2NH₃ → [Pt(NH₃)₂Cl₄] + 0Cl⁻
- No Cl⁻ ions are ionizable. (Ionizable Cl⁻ = 0)

3. **Count the Compounds with At Least Half Ionizable Cl⁻**:
- **CrCl₃.6NH₃**: 3 ionizable Cl⁻ (Yes)
- **CrCl₃.5NH₃**: 2 ionizable Cl⁻ (Yes)
- **CrCl₃.4NH₃**: 2 ionizable Cl⁻ (Yes)
- **CrCl₃.3NH₃**: 0 ionizable Cl⁻ (No)
- **PtCl₄.6NH₃**: 4 ionizable Cl⁻ (Yes)
- **PtCl₄.5NH₃**: 3 ionizable Cl⁻ (Yes)
- **PtCl₄.4NH₃**: 2 ionizable Cl⁻ (Yes)
- **PtCl₄.3NH₃**: 0 ionizable Cl⁻ (No)
- **PtCl₄.2NH₃**: 0 ionizable Cl⁻ (No)

4. **Total Count**:
- The compounds with at least half of Cl⁻ ions ionizable are:
- CrCl₃.6NH₃
- CrCl₃.5NH₃
- CrCl₃.4NH₃
- PtCl₄.6NH₃
- PtCl₄.5NH₃
- PtCl₄.4NH₃
- Total = 6 compounds.

### Final Answer:
The total number of compounds in which at least half of Cl⁻ are ionizable is **6**.