Question

If the freezing point of a 0.01 molal aqueous solution of a cobalt (III) chloride-ammonia complex (which behaves as a strong electrolyte) is $-0.0558{}^\text{o}C$, the number of chloride(s) in the coordination sphere of the complex is
[${{K}_{f}}$ of water =$1.86\text{ }K\text{ }kg\text{ }mo{{l}^{-1}}$ ]

Answer 1

Attempt this question by using the concepts of colligative properties and since we are given information regarding freezing point, therefore we will use depression in freezing point. Freezing point depression is a colligative property of solutions that is proportional to the molality of the solute added in the solvent.
Formula used:
The formula required for this question is:-
$\Delta {{T}_{f}}~=\text{ }i\times {{K}_{f}}\times m$
where,
$\Delta {{T}_{f}}~=\text{ }$Depression in freezing point
i = Van’t Hoff factor,
${{K}_{f}}$= cryoscopic constant, and
m = molality of the solution.

Complete answer:
As we know that depression in freezing point is a colligative property of solutions that is proportional to the molality of the solute added in the solvent.
For the calculation of the number of chlorine ions in the complex, we need to find out the number of dissociated ions. This can be done with the help of Van’t Hoff Factor.
-Calculation of Van’t Hoff Factor:-
$\Delta {{T}_{f}}~=\text{ }i\times {{K}_{f}}\times m$
where,
$\Delta {{T}_{f}}~=\text{ }$Depression in freezing point
i = Van’t Hoff factor,
${{K}_{f}}$= cryoscopic constant, and
m = molality of the solution.
-The values given are as follows:-
${{K}_{f}}$= $1.86\text{ }K\text{ }kg\text{ }mo{{l}^{-1}}$
m =0.01
${{T}_{f}}(solution)=$ $-0.0558{}^\text{o}C$
${{T}_{f}}(\text{pure solvent)=}{{\text{0}}^{\circ }}C$
$\begin{aligned} & \Delta {{T}_{f}}~=\text{ }{{T}_{f}}(\text{pure solvent)-}{{T}_{f}}(solution\text{)} \\\ & \Rightarrow \Delta {{T}_{f}}={{0}^{\circ }}C-(-{{0.0558}^{\circ }}C) \\\ & \Rightarrow \Delta {{T}_{f}}=0.0558K \\\ \end{aligned}$
(Unit is written in Kelvin because difference between 2 temperature terms will remain same in any unit)
-On substituting all the values in$\Delta {{T}_{f}}~=\text{ }i\times {{K}_{f}}\times m$, we get:-
$\begin{aligned} & \Delta {{T}_{f}}~=\text{ }i\times {{K}_{f}}\times m \\\ & \text{Rearranging the formula:-} \\\ & \Rightarrow i=\dfrac{\Delta {{T}_{f}}}{{{K}_{f}}\times m} \\\ & \Rightarrow i=\dfrac{0.0558K}{1.86\text{ }K\text{ }kg\text{ }mo{{l}^{-1}}\times 0.01mol\text{ }k{{g}^{-1}}} \\\ & \Rightarrow i=3 \\\ \end{aligned}$
-Relation of Van’t Hoff factor with number of ions dissociated of a complex:-
$i=\alpha n+(1-\alpha )$
Since it is given that, the complex is a strong electrolyte therefore it will dissociate completely in the solution and hence degree of dissociation ($\alpha$) = 1.
Therefore, i = n
So number of ions dissociated (n) = 3
-Possibilities of cobalt (III) chloride-ammonia complex are as follows:-
(A) $[Co{{(N{{H}_{3}})}_{3}}C{{l}_{3}}]$: If it is dissolved in solution, no ion will be formed as it is a neutral complex sphere. So it is not the desired complex.
(B) $[Co{{(N{{H}_{3}})}_{4}}C{{l}_{2}}]Cl$: If it is dissolved in solution, number of ions formed will be equal to 2 (one ion of $C{{l}^{-}}$ and one complex sphere ion as ${{[Co{{(N{{H}_{3}})}_{4}}C{{l}_{2}}]}^{+}}$. So it is not the desired complex as well.
(C) $[Co{{(N{{H}_{3}})}_{5}}Cl]C{{l}_{2}}$: If it is dissolved in solution, number of ions formed will be equal to 3 (two ions of $C{{l}^{-}}$ and one complex sphere ion as ${{[Co{{(N{{H}_{3}})}_{5}}C{{l}_{2}}]}^{2+}}$. This is the desired complex as well.
(D) $[Co{{(N{{H}_{3}})}_{6}}]C{{l}_{3}}$: If it is dissolved in solution, number of ions formed will be equal to 4 (three ions of $C{{l}^{-}}$ and one complex sphere ion as ${{[Co{{(N{{H}_{3}})}_{6}}]}^{3+}}$. This is not the desired complex.
-So from the above data we conclude that, the complex is $[Co{{(N{{H}_{3}})}_{5}}Cl]C{{l}_{2}}$ and the number of $C{{l}^{-}}$ ions present inside the complex sphere= 1

Note:
-While solving such questions, always read the statements clearly as even small information can help in the solution such as strong electrolyte.
-Also we make all the possibilities in such a way that the oxidation state of Cobalt always remains +3.