The correct order of the stoichiometries of $\mathrm{AgCl}$ formed when $\mathrm{AgNO}{}_3$ in excess is treated with the complexes $\mathrm{CoCl}{}_3\cdot 6\mathrm{NH}{}_3$, $\mathrm{CoCl}{}_3\cdot 5\mathrm{NH}{}_3$, $\mathrm{CoCl}{}_3\cdot 4\mathrm{NH}{}_3$ respectively is:

Solution & Explanation

\textbf{Answer:} (C) 3 $\mathrm{AgCl}$, 2 $\mathrm{AgCl}$, 1 $\mathrm{AgCl}$. \textbf{Solution:} By Werner's theory, only the chloride ions present in the ionisation sphere (outside the coordination sphere, held by primary valence) are free in solution and precipitate instantly with $\mathrm{Ag}{}^{+}$ as $\mathrm{AgCl}$. Chloride bound to the metal by secondary valence (inside the square brackets) does not precipitate. Writing each addition compound in true coordination form: $\mathrm{CoCl}{}_3\cdot 6\mathrm{NH}{}_3$ = $[\mathrm{Co}(\mathrm{NH}{}_3){}_6]\mathrm{Cl}{}_3$ → 3 ionisable $\mathrm{Cl}{}^{-}$ → 3 $\mathrm{AgCl}$ $\mathrm{CoCl}{}_3\cdot 5\mathrm{NH}{}_3$ = $[\mathrm{Co}(\mathrm{NH}{}_3){}_5\mathrm{Cl}]\mathrm{Cl}{}_2$ → 2 ionisable $\mathrm{Cl}{}^{-}$ → 2 $\mathrm{AgCl}$ $\mathrm{CoCl}{}_3\cdot 4\mathrm{NH}{}_3$ = $[\mathrm{Co}(\mathrm{NH}{}_3){}_4\mathrm{Cl}{}_2]\mathrm{Cl}$ → 1 ionisable $\mathrm{Cl}{}^{-}$ → 1 $\mathrm{AgCl}$ Thus the stoichiometries are 3, 2, 1 $\mathrm{AgCl}$ respectively, option (C).