Question

Suppose 0.010 mol of each of the following compounds is

dissolved (separately) in 1.0 L water: ${\mathbf{BaCl}}_2\mathbf{,}{\mathbf{K}}_4\mathbf{\left[}\mathbf{Fe}{\mathbf{\left(}\mathbf{CN}\mathbf{\right)}}_6\mathbf{\right]}\mathbf{,}\mathbf{\left[}\mathbf{Cr}{\mathbf{\left(}{\mathbf{NH}}_3\mathbf{\right)}}_4{\mathbf{Cl}}_2\mathbf{\right]}\mathbf{Cl}\mathbf{,}\mathbf{\left[}\mathbf{Fe}{\mathbf{\left(}{\mathbf{NH}}_3\mathbf{\right)}}_3{\mathbf{Cl}}_3\mathbf{\right]}$. Rank the resulting four solutions in order of conductivity, from lowest to highest.

Short answer

$\left[\text{Fe}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}{\text{Cl}}_{\text{3}}\right]<\left[\text{Cr}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}{\text{Cl}}_{\text{2}}\right]{\text{Cl\hspace{0.17em}<BaCl}}_2{\text{<K}}_{\text{4}}\left[\text{Fe}\left({\text{CN}}_{\text{6}}\right)\right]$

Step-by-step solution

Conductivity

Conductivity, also known as “specific conductance (conductance at the specified condition),” is the reciprocal of resistivity $\left(\text{ρ}\right)$and is represented as follows:

$\mathbf{Conductivity}\mathbf{=}\frac{1}{\rho }$

Order of conductivity

The coordination compound dissolved in water that forms more ions has higher conductivity than the one which forms fewer ions.

BaCl₂ forms three ions in the solution.

${\text{K}}_{\text{4}}\left[\text{Fe}\left({\text{CN}}_{\text{6}}\right)\right]$forms four ions in the solution.

$\left[\text{Cr}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}{\text{Cl}}_{\text{2}}\right]\text{Cl}$ forms two ions in the solution.

$\left[\text{Fe}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}{\text{Cl}}_{\text{3}}\right]$, not form any ions in the solution.

Therefore the conductivity order from the lowest to highest of the given compound is as follows:

$\left[\text{Fe}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}{\text{Cl}}_{\text{3}}\right]<\left[\text{Cr}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}{\text{Cl}}_{\text{2}}\right]{\text{Cl<BaCl}}_2{\text{<K}}_{\text{4}}\left[\text{Fe}\left({\text{CN}}_{\text{6}}\right)\right]$