Question

When 1 mol $CrC{l}_3.6H_{2}O$ is treated with excess of $AgN{O}_3$, 3 mol of AgCl are obtained. The formula of the complex is:

• A. $\left[CrC{l}_3\right(H_{2}O{)}_3].3H_{2}O$
• B. $\left[CrC{l}_2\right(H_{2}O{)}_4]Cl.2H_{2}O$
• C. $\left[CrCl\right(H_{2}O{)}_5]C{l}_2.H_{2}O$
• D. $Cr\left[\right(H_{2}O{)}_6]C{l}_3$

Answer 1

Answer: (D)

$Cr\left[\right(H_{2}O{)}_6]C{l}_3$

When 1 mol $CrC{l}_3.6H_{2}O$ is treated with excess of $AgN{O}_3$, 3 mol of AgCl are obtained. It means that when $CrC{l}_3.6H_{2}O$ molecule dissociates in aqueous solution, all three chloride ions comes in solution. The formula of the complex is $\left[Cr\right(H_{2}O{)}_6]C{l}_3$.

$CrC{l}_3.6H_{2}O$ or $\left[Cr\right(H_{2}O{)}_6]C{l}_3\to [Cr(H_{2}O{)}_6{]}^{3+}+3C{l}^{-}$

$3C{l}^{-}+3AgN{O}_3\to 3AgCl\downarrow +3N{O}_3^{-}$