Q.$\text{CrCl}_3.6\text{H}_2\text{O}$ is known to exist as three different coloured compounds: $[\text{Cr}(\text{H}_2\text{O})_6]\text{Cl}_3$ (violet), $[\text{Cr}(\text{H}_2\text{O})_5\text{Cl}]\text{Cl}_2.\text{H}_2\text{O}$ (grey-green), and $[\text{Cr}(\text{H}_2\text{O})_4\text{Cl}_2]\text{Cl}.2\text{H}_2\text{O}$ (dark green). Name this type of isomerism, and state how many moles of $\text{AgCl}$ each would precipitate with excess $\text{AgNO}_3$.

[!TLDR] Hydrate (solvate) isomerism — free water/chloride trade places. [!ANSWER] Violet form: 3 mol AgCl. Grey-gr …

These three compounds all share the identical overall formula $\text{CrCl}_3.6\text{H}_2\text{O}$ but differ in how many water molecules are coordinated directly to chromium (inside the coordination sphere) versus present as free water of crystallization, with a corresponding number of chloride ions moving between being coordinated and being free counter ions — this is hydrate isomerism (a specific case of ionization/solvate isomerism where water is the group exchanging places). In $[\text{Cr}(\text{H}_2\text{O})_6]\text{Cl}_3$ (violet), all six water molecules are coordinated and all three chlorides are free, ionizable counter ions, so treatment with excess $\text{AgNO}_3$ precipitates all three as $\text{AgCl}$: 3 mol AgCl. In $[\text{Cr}(\text{H}_2\text{O})_5\text{Cl}]\text{Cl}_2.\text{H}_2\text{O}$ (grey-green), one water has been displaced from the coordination sphere by one chloride (which is now coordinated and non-ionizable), leaving only two chlorides free and one water molecule as water of crystallization: 2 mol AgCl. In $[\text{Cr}(\text{H}_2\text{O})_4\tex …

Identify how many chloride ions are written outside the coordination-sphere brackets (free, ionizable) in each formula — that numb …

Assuming all 3 chlorides always precipitate in every isomer, since the overall formula 'CrCl3.6H2O' looks identical for all three — the precipitation result depends entirely on which specific str …