Question

Which one is the most likely structure of $CrCl _{3} \cdot 6 H _{2} O$, if $1 / 3$ of total chlorine of the compound is precipitated by adding $AgNO _{3}$ to its aqueous solution?

• A. $CrCl _{3} \cdot 6 H _{2} O$
• B. $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_3 \mathrm{Cl}_3\right] \cdot 3 \mathrm{H}_2 \mathrm{O}$
• C. $\left[\mathrm{CrCl}_2\left(\mathrm{H}_2 \mathrm{O}\right)_4\right] \mathrm{Cl} \cdot 2 \mathrm{H}_2 \mathrm{O}$
• D. $\left[\mathrm{CrCl}\left(\mathrm{H}_2 \mathrm{O}\right)_5\right] \mathrm{Cl}_2 \cdot \mathrm{H}_2 \mathrm{O}$

Answer 1

Answer: (C)

$CrCl _{3} \cdot 6 H _{2} O$ has $3- Cl$ atoms but $\frac{1}{3} Cl$ atoms precipitated so one $1 Cl$ atom get precipitated.
According to Werner's theory, only those ions are precipitated which are attached to the metal atoms with ionic bonds and are present outside the co-ordination sphere.
$\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4 \mathrm{Cl}_2\right] \mathrm{Cl}(\mathrm{aq}) \rightleftharpoons\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4 \mathrm{Cl}_2\right]^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})$ $\mathrm{Ag}^{+}+\mathrm{Cl}^{-} \longrightarrow \mathrm{AgCl} \downarrow$ (white)