Question

A six coordinate complex of formula $CrC{{l}_{3}}.6{{H}_{2}}O$ has green color. A $0.1M$ solution of the complex when treated with excess of $AgN{{O}_{3}}$ gave $28.7g$ of white precipitate. The formula of complex would be ___________.
A.) $\left[ Cr{{\left( {{H}_{2}}O \right)}_{6}} \right]C{{l}_{3}}$
B.) $\left[ Cr{{\left( {{H}_{2}}O \right)}_{5}}Cl \right]C{{l}_{2}}.{{H}_{2}}O$
C.) $\left[ Cr{{\left( {{H}_{2}}O \right)}_{4}}Cl \right]C{{l}_{2}}.2{{H}_{2}}O$
D.) $\left[ Cr{{\left( {{H}_{2}}O \right)}_{3}}C{{l}_{3}} \right].3{{H}_{2}}O$

Answer 1

To predict and give the appropriate option, we need to refer Werner’s theory and observation of coordination compounds and specifically their primary and secondary valences where in a series of binary compounds like $Cobalt\left( III \right ) Chloride$ with ammonia, it was found that some of the chloride ions precipitated as $AgCl$ on adding excess of silver nitrate $\left( AgN{{O}_{3}} \right)$ solution in cold though some still remained in the solution.

Complete Solution :
The information already given to us:
A six coordinate complex of empirical formula $CrC{{l}_{3}}.6{{H}_{2}}O$ has green color; and concentration of solution of the complex when treated with excess of $AgN{{O}_{3}}$ which gave $28.7g$ of white precipitate is $0.1M$.
- Keeping Werner’s observations in mind, we know that the white precipitate formed is Silver chloride $\left( AgCl \right)$
Total atomic mass of Silver chloride $\left( AgCl \right ) = 107.8 + 35.5 = 143.2u$
Thus, Total number of moles of Silver chloride $\left( AgCl \right)$precipitated$= \dfrac{Given\text{ }mass\text{ }}{molar\text{ }mass}=\dfrac{28.7}{143.2}=0.2$

Since, the concentration of the solution of the complex which gives out $0.2$ moles of Silver chloride $\left( AgCl \right)$ is $0.1M$, this would mean that there are $2\left( C{{l}^{-}} \right)$ ions that are ionisable or are out of coordination sphere of the complex.
- Therefore, the correct formula of the complex would be $\left[ Cr{{\left( {{H}_{2}}O \right)}_{5}}Cl \right]C{{l}_{2}}.{{H}_{2}}O$ and thus, the correct option would be (B) $\left[ Cr{{\left( {{H}_{2}}O \right)}_{5}}Cl \right]C{{l}_{2}}.{{H}_{2}}O$.
So, the correct answer is “Option B”.

Note: One is advised to remember that considering the above mentioned observation and necessity, Werner proposed the term secondary valence for the number of groups that are directly bonded to the central metal ion or central metal atom.