Question

When $1\,mol\,CrCl_{3}.6H_{2}O$ is treated with excess of $AgNO_{3}$ , $3\,mol$ of $AgCl$ are obtained. The formula of the complex is

• A. $\left[\left(CrCl\right)_{3} \left(H_{2} O\right)_{3}\right].3H_{2}O$
• B. $\left[\left(CrCl\right)_{2} \left(H_{2} O\right)_{4}\right]Cl.2H_{2}O$
• C. $\left[CrCl \left(H_{2} O\right)_{5}\right]\left(Cl\right)_{2}\cdot H_{2}O$
• D. $\left[Cr \left(H_{2} O\right)_{6}\right]\left(Cl\right)_{3}$

Answer 1

Answer: (D)

$1\,mol\,CrCl_{3}.6H_{2}O$ is treated with excess of $AgNO_{3}$ , $3\,mol$ of $AgCl$ are obtained. This indicates that there are three chloride ions present in ionisation sphere. So the formula of complex will be $\left[Cr \left(H_{2} O\right)_{6}\right]\left(Cl\right)_{3}$ .
$\left[Cr \left(H_{2} O\right)_{6}\right]\left(Cl\right)_{3}+3\left(AgNO\right)_{3} \rightarrow 3AgCl+\left[Cr \left(H_{2} O\right)_{6}\right]^{3 +}+3NO_{3}^{-}$