Question

A, B and C are three complexes of chromium (III) with the empirical formula $H_{12}{O}_{6}C{l}_{3}Cr$. All the three complexes have water and chloride ions as ligands. Complex A does not react with concentrated $H_{2}S{O}_4$, whereas complexes B and C lose 6.75% and 13.5% of their original weight respectively, on treatment with concentrated $H_{2}S{O}_4$.

Thus, loss in weight on treatment with concentrated $H_{2}S{O}_4$ is due to :

• A. $C{l}^{-}$ to lost as HCl
• B. $C{r}^{3+}$ is lost as $C{r}_2(S{O}_4{)}_3$
• C. $H_{2}O$ is lost due to absorption.
• D. $C{l}^{-}$ and lost as $C{l}_2$

Answer 1

Answer: (C)

$H_{2}O$ is lost due to absorption.

Sulphuric acid is a strong dehydrating agent.

It removes water present in the ionization sphere of the complexes.

The empirical formula, $H_{12}{O}_{6}C{l}_{3}Cr$ has weight of 266.5 g/mol.

In the compound A, all the water molecules are present in the coordination sphere.

Hence, on treatment with sulphuric acid, no water is lost.

On the other hand, compounds B and C contain one and two water molecules respectively.

Hence, on treatment with sulphuric acid, they lose 6.75% and 13.5% of their original weight, respectively.

Therefore, option C is correct.