Question

Which of the following pairs of complexes will show approximately the same electrical conductance for their 0.001 M aqueous solutions?

• A. $CoCl_{3} \cdot 4NH_{3}$ and $PtCl_{4} \cdot 4NH_{3}$
• B. $CoCl_{3} \cdot 3NH_{3}$ and $PtCl_{4} \cdot 5NH_{3}$
• C. $CoCl_{3} \cdot 6NH_{3}$ and $PtCl_{4} \cdot 5NH_{3}$
• D. $CoCl_{3} \cdot 6NH_{3}$ and $PtCl_{4} \cdot 3NH_{3}$

Hint:

Which of the following pairs of complexes will show approximately the same electrical conductance for their 0.001 M aqueous solutions?

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Electrical conductance depends on the number of ions produced in solution.
Step 2: Evaluation
$CoCl_{3} \cdot 6NH_{3}$ is $[Co(NH_{3})_{6}]Cl_{3}$, which produces 4 ions: $[Co(NH_{3})_{6}]^{3+} + 3Cl^{-}$.
Step 3: Analysis
$PtCl_{4} \cdot 5NH_{3}$ is $[PtCl(NH_{3})_{5}]Cl_{3}$, which also produces 4 ions: $[PtCl(NH_{3})_{5}]^{3+} + 3Cl^{-}$.
Step 4: Conclusion
Since they produce the same number of ionic species, their conductance will be approximately the same.
Hence, the Answer is: (C)