Ncert Exercises & Solutions

When 1 mole of ${CrCl}_{3} 6 H_{2} O$ is treated with excess of ${AgNO}_{3},$ 3 moles of $AgCl$ are obtained. The formula of the complex is:

1. [CrCl₃(H₂O)₃].3H₂O
2. [CrCl₂(H₂O)₄]Cl.2H₂O
3. [CrCl₃(H₂O)₅]Cl₂.H₂O
4. [Cr(H₂O)₆]Cl₃

Hint: Complex dissociate and generate 4 ions in the solution

1 mole of AgNO₃, precipitates one free chloride ion (

{C l}^{-}

).

Here, 3 moles of AgCI are precipitated by an excess of AgNO₃. Hence, there must be three free Cl $^{-}$ ions

So, the formula of the complex can be $[C r {(H_{2} O)}_{6}] C l_{3}$ and the correct choice is option 4th