Coordination Compounds 2 Question 68
####70. $A, B$ 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 $C$.
$(1999,2 M)$
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Solution:
- A has no water molecules of crystallisation.
Hence, $A$ is $\left[Cr\left(H _2 O\right) _6\right] Cl _3$.
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.
$B=\left[Cr\left(H _2 O\right) _5 Cl\right] Cl _2 \cdot H _2 O, C=\left[Cr\left(H _2 O\right) _4 Cl _2\right] Cl \cdot 2 H _2 O$
