Both `Co^(3+) and Pt^(4+)` have a coordination number of six. Which of the following pair of complexes will show approximately the same electrical conductance for their `0.001` M aqueous solution ?

Options

A. `CoCl_(3).4NH_(3) and PtCl_(4). 4NH_(3)`
B. `CoCl_(3). 3NH_(3) and PtCl_(4). 5NH_(3)`
C. `CoCl_(3). 6NH_(3) and PtCl_(4).5NH_(3)`
D. `CoCl_(3). 6NH_(3) and PtCl_(4). 3NH_(3)`

Answer

To determine which pair of complexes will show approximately the same electrical conductance for their 0.001 M aqueous solution, we need to analyze the number of ions produced when each complex is dissolved in water. The electrical conductance of a solution is directly proportional to the number of ions present in the solution.

### Step-by-Step Solution:

1. **Identify the complexes**: We have two metal ions, Co^(3+) and Pt^(4+), and we need to analyze the following complexes:
- CoCl3·4NH3
- PtCl4·4NH3
- CoCl3·3NH3
- PtCl4·5NH3
- CoCl3·6NH3
- PtCl4·5NH3
- CoCl3·6NH3
- PtCl4·3NH3

2. **Analyze the first pair: CoCl3·4NH3 and PtCl4·4NH3**:
- For CoCl3·4NH3:
- The complex ion formed is [Co(NH3)4Cl2]^+ + 2Cl^−.
- Total ions = 1 (complex ion) + 2 (chloride ions) = 3 ions.
- For PtCl4·4NH3:
- The complex ion formed is [Pt(NH3)4Cl2]^+ + 2Cl^−.
- Total ions = 1 (complex ion) + 2 (chloride ions) = 3 ions.
- **Conclusion**: Both produce 3 ions.

3. **Analyze the second pair: CoCl3·3NH3 and PtCl4·5NH3**:
- For CoCl3·3NH3:
- The complex ion formed is [Co(NH3)3Cl3]^+.
- Total ions = 1 (complex ion) + 3 (chloride ions) = 4 ions.
- For PtCl4·5NH3:
- The complex ion formed is [Pt(NH3)5Cl]^+ + 3Cl^−.
- Total ions = 1 (complex ion) + 3 (chloride ions) = 4 ions.
- **Conclusion**: Both produce 4 ions.

4. **Analyze the third pair: CoCl3·6NH3 and PtCl4·5NH3**:
- For CoCl3·6NH3:
- The complex ion formed is [Co(NH3)6]^3+ + 3Cl^−.
- Total ions = 1 (complex ion) + 3 (chloride ions) = 4 ions.
- For PtCl4·5NH3:
- The complex ion formed is [Pt(NH3)5Cl]^+ + 3Cl^−.
- Total ions = 1 (complex ion) + 3 (chloride ions) = 4 ions.
- **Conclusion**: Both produce 4 ions.

5. **Analyze the fourth pair: CoCl3·6NH3 and PtCl4·3NH3**:
- For CoCl3·6NH3:
- The complex ion formed is [Co(NH3)6]^3+ + 3Cl^−.
- Total ions = 1 (complex ion) + 3 (chloride ions) = 4 ions.
- For PtCl4·3NH3:
- The complex ion formed is [Pt(NH3)3Cl3]^+ + 3Cl^−.
- Total ions = 1 (complex ion) + 3 (chloride ions) = 4 ions.
- **Conclusion**: Both produce 4 ions.

### Final Conclusion:
The pairs of complexes that will show approximately the same electrical conductance for their 0.001 M aqueous solution are:
- CoCl3·4NH3 and PtCl4·4NH3 (3 ions)
- CoCl3·3NH3 and PtCl4·5NH3 (4 ions)
- CoCl3·6NH3 and PtCl4·5NH3 (4 ions)