Q10 / 180 · COMEDK 2025 (10 May Shift 1) · Chemistry

Arrange the following compounds in the decreasing order of the molar conductivities of their aqueous solutions. $\begin{array}{cccc}\mathrm{A} & \quad \mathrm{B} & \quad \mathrm{C} & \quad \mathrm{D} \\ {\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}] \mathrm{Cl}_2\right.} & \quad {\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_3 \mathrm{Cl}_3\right]} & \quad {\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right] \mathrm{Cl}} & \quad {\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_3}\end{array}$

  • A
    $\mathrm{B}>\mathrm{A}>\mathrm{C}>\mathrm{D}$
  • B
    $\mathrm{B}>\mathrm{C}>\mathrm{A}>\mathrm{D}$
  • Correct answer
    C
    $D>A>C>B$
  • D
    $\mathrm{A}>\mathrm{B}>\mathrm{D}>\mathrm{C}$

Solution

Molar conductivity of an electrolyte in aqueous solution depends on the number of ions produced upon dissociation. The greater the number of ions, the higher the molar conductivity.

Dissociation of the given complexes:
A: $[Co(NH_3)_5Cl]Cl_2 \rightarrow [Co(NH_3)_5Cl]^{2+} + 2Cl^{-}$. Total ions = 3.
B: $[Co(NH_3)_3Cl_3] \rightarrow [Co(NH_3)_3Cl_3]$. Total ions = 0 (non-electrolyte).
C: $[Co(NH_3)_4Cl_2]Cl \rightarrow [Co(NH_3)_4Cl_2]^{+} + Cl^{-}$. Total ions = 2.
D: $[Co(NH_3)_6]Cl_3 \rightarrow [Co(NH_3)_6]^{3+} + 3Cl^{-}$. Total ions = 4.

Comparing the number of ions produced:
D (4 ions) > A (3 ions) > C (2 ions) > B (0 ions).

Therefore, the decreasing order of molar conductivity is D > A > C > B.

Answer: $D>A>C>B$

Hence, the correct answer is option: 'C'.

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