Which one of the following complexes will consume more equivalent of aqueous solution of `Ag(NO_(3))` ?

Options

A. `Na_(3)[CrCl_(6)]`
B. `[Cr(H_(2)O)_(5)Cl]Cl_(2)`
C. `[Cr(H_(2)O)_(6)]Cl_(3)`
D. `Na_(2)[CrCl_(5)(H_(2)O)]`

Answer

To determine which complex will consume more equivalents of aqueous Ag(NO₃)₃, we need to analyze the number of chloride ions (Cl⁻) released by each complex. The more Cl⁻ ions present, the more AgCl precipitate will form when reacted with AgNO₃.

### Step-by-Step Solution:

1. **Identify the Complexes**: We need to analyze the given complexes to see how many Cl⁻ ions each one releases into the solution.

2. **Complex 1: Na₃[CrCl₆]**
- This complex dissociates into 3 Na⁺ ions and 1 CrCl₆³⁻ ion.
- The CrCl₆³⁻ ion contains 6 Cl⁻ ions.
- **Total Cl⁻ ions = 6**

3. **Complex 2: [Cr(H₂O)₅Cl]Cl₂**
- This complex dissociates into 1 [Cr(H₂O)₅Cl]²⁺ ion and 2 Cl⁻ ions.
- **Total Cl⁻ ions = 2**

4. **Complex 3: [Cr(H₂O)₆]Cl₃**
- This complex dissociates into 1 [Cr(H₂O)₆]³⁺ ion and 3 Cl⁻ ions.
- **Total Cl⁻ ions = 3**

5. **Complex 4: Na₂[CrCl₅(H₂O)]**
- This complex dissociates into 2 Na⁺ ions and 1 [CrCl₅(H₂O)]²⁻ ion.
- The [CrCl₅(H₂O)]²⁻ ion contains 5 Cl⁻ ions.
- **Total Cl⁻ ions = 5**

6. **Comparison of Cl⁻ Ions**:
- Complex 1: 6 Cl⁻
- Complex 2: 2 Cl⁻
- Complex 3: 3 Cl⁻
- Complex 4: 5 Cl⁻

7. **Conclusion**: The complex that releases the most Cl⁻ ions is Complex 1: Na₃[CrCl₆], which releases 6 Cl⁻ ions. Therefore, it will consume the most equivalents of Ag(NO₃)₃, leading to the formation of AgCl precipitate.

### Final Answer:
The complex that will consume more equivalents of aqueous Ag(NO₃)₃ is **Na₃[CrCl₆]**.