Question

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

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

Answer 1

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

Moles of $C{l}^{-}$ precipiated as per the reaction:

When 1 mol of $CrC{l}_3.6H_{2}O$ is treated with the excess of $AgN{O}_3.3\text{mol of}AgCl$ are obtained.

$CrC{l}_3.6H_{2}O+AgN{O}_3\to AgCl$
$1mol$ excess $3mol$

Formula of the coordination complex

It means that when $CrC{l}_3.6H_{2}O$ complex dissociate ions complex is $\left[Cr\right(H_{2}O{)}_6]C{l}_3$

Reaction will be

$\left[Cr\right(H_{2}O{)}_6]C{l}_3\to [Cr(H_{2}O{)}_6{]}^{3+}+3C{l}^{-}$
$3C{l}^{-}+3AgN{O}_3\to 3AgCl$$+3N{O}_3^{-}$
Hence, the correct answer is option $\left(d\right)$.