Question

Which one is the most likely structure of $CrC{l}_3.6H_{2}O$, if $1/3$ of total chlorine of the compound is precipitated by adding $AgN{O}_3$ to its aqueous solution?

• A. $CrC{l}_3\cdot 6H_{2}O$
• 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)

$CrC{l}_3\cdot 6H_{2}O$ has $3-Cl$ atoms but $\frac{1}{3}Cl$ atoms precipitated so one $1Cl$ 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.
$\left[\mathrm{Cr}{\left({\mathrm{H}}_2\mathrm{O}\right)}_4{\mathrm{Cl}}_2\right]\mathrm{Cl}(\mathrm{aq})\rightleftharpoons {\left[\mathrm{Cr}{\left({\mathrm{H}}_2\mathrm{O}\right)}_4{\mathrm{Cl}}_2\right]}^{+}(\mathrm{aq})+{\mathrm{Cl}}^{-}(\mathrm{aq})$
$A(g{)}^{+}+C(l{)}^{-}\to AgCl\downarrow$ (white)