An octahedral cobalt(III) compound has the empirical formula $\mathrm{CoCl}{}_3\cdot 5\mathrm{NH}{}_3$. On adding excess $\mathrm{AgNO}{}_3$ solution to one mole of it, two moles of $\mathrm{AgCl}$ precipitate immediately. Its formula is

Only ionisable (free) chloride precipitates, so two chlorides are outside the coordination sphere and one is bound. Octahedral means coordination number 6, so five $\mathrm{NH}{}_3$ plus one $\mathrm{Cl}{}^{-}$ gives $[\mathrm{Co}(\mathrm{NH}{}_3){}_5\mathrm{Cl}]\mathrm{Cl}{}_2$. The others have coordination number 7 or 5, or would give one or three $\mathrm{AgCl}$.