Question

$A, B$ and $C$ are three complexes of chromium (III) with the empirical formula $H _{12} O _{6} Cl _{3} Cr$. All the three complexes have water and chloride ion as ligands.
Complex $A$ does not react with concentrated $H _{2} SO _{4}$, whereas complexes $B$ and $C$ lose $6.75 \%$ and $13.5 \%$ of their original mass, respectively, on treatment with concentrated $H _{2} SO _{4}$. Identify $A, B$ and $C$.

Answer 1

A has no water molecules of crystallisation.
Hence, $A$ is $\left[ Cr \left( H _{2} O _{6}\right] Cl _{3}\right.$.
Both $B$ and $C$ loses weight with concentrated $H _{2} SO _{4}$, therefore, both $B$ and $C$ have some water molecules of crystallisation.
Moreover, weight loss with $C$ is just double of the same with $B$ indicates that number of water molecules of crystallisation of $C$ is double of the same for $B$. Therefore, $B$ has one and $C$ has two water molecules of crystallisation.
$B=\left[ Cr \left( H _{2} O \right)_{5} Cl \right] Cl _{2} \cdot H _{2} O , C =\left[ Cr \left( H _{2} O \right)_{4} Cl _{2}\right] Cl \cdot 2 H _{2} O$