Question

Arrange the following compounds in the decreasing order of the molar conductivities of their aqueous solutions.

• A. \(B>C>A>D\)
• B. \(D>A>C>B\)
• C. \(B>A>C>D\)
• D. \(A>B>D>C\)

Hint:

For coordination compounds, only the ions present outside the coordination sphere ionise in aqueous solution. More ionisation means higher molar conductivity.

Answer 1

The Correct Option is B

Step 1: Understanding molar conductivity.
Molar conductivity of an aqueous solution depends mainly on the number of ions produced by the compound in water.
Greater the number of ions formed in aqueous solution, greater will be the molar conductivity.
\[ \text{More number of ions} \Rightarrow \text{higher molar conductivity} \]

Step 2: Ionisation of compound A.
Compound A is:
\[ [\mathrm{Co(NH_3)_5Cl}]Cl_2 \] In aqueous solution, the two chloride ions outside the coordination sphere ionise completely:
\[ [\mathrm{Co(NH_3)_5Cl}]Cl_2 \rightarrow [\mathrm{Co(NH_3)_5Cl}]^{2+} + 2Cl^- \] Total number of ions produced:
\[ 1 + 2 = 3 \]

Step 3: Ionisation of compound B.
Compound B is:
\[ [\mathrm{Co(NH_3)_3Cl_3}] \] Here all chloride ions are inside the coordination sphere.
So, it does not produce counter ions in aqueous solution.
\[ [\mathrm{Co(NH_3)_3Cl_3}] \rightarrow \text{No ionisation} \] Total number of ions produced:
\[ 1 \] Hence, compound B has the lowest molar conductivity.

Step 4: Ionisation of compound C.
Compound C is:
\[ [\mathrm{Co(NH_3)_4Cl_2}]Cl \] Only one chloride ion is outside the coordination sphere, so it ionises in water:
\[ [\mathrm{Co(NH_3)_4Cl_2}]Cl \rightarrow [\mathrm{Co(NH_3)_4Cl_2}]^{+} + Cl^- \] Total number of ions produced:
\[ 1 + 1 = 2 \]

Step 5: Ionisation of compound D.
Compound D is:
\[ [\mathrm{Co(NH_3)_6}]Cl_3 \] Here three chloride ions are outside the coordination sphere, so they ionise completely in aqueous solution:
\[ [\mathrm{Co(NH_3)_6}]Cl_3 \rightarrow [\mathrm{Co(NH_3)_6}]^{3+} + 3Cl^- \] Total number of ions produced:
\[ 1 + 3 = 4 \]

Step 6: Comparing number of ions.
Now compare all compounds according to the number of ions produced:
\[ D = 4 \text{ ions} \] \[ A = 3 \text{ ions} \] \[ C = 2 \text{ ions} \] \[ B = 1 \text{ ion} \] Therefore, the decreasing order of molar conductivity is:
\[ D>A>C>B \]

Step 7: Final Answer.
Since compound D gives maximum ions and compound B gives minimum ions, the correct decreasing order is:
\[ \boxed{D>A>C>B} \] Hence, the correct answer is option (B).