Complex A has a composition of H₁₂O₆Cl₃Cr. If the complex on treatment with conc. H₂SO₄ loses 13.5% of its original…

🧠 Translate the Mass Loss to a Number of Water Molecules

Total mass of $\mathrm{CrCl_3 \cdot 6H_2O}$: $52 + 3(35) + 6(18) = 265$ g/mol.

Mass lost on conc. $\mathrm{H_2SO_4}$ treatment: $13.5\% \times 265 = 35.78$ g/mol $\approx 36$ g/mol.

$36 / 18 = 2$ → 2 water molecules are lost.

🗺️ Why Conc. $\mathrm{H_2SO_4}$ Tells Us About Coordination Sphere

Concentrated $\mathrm{H_2SO_4}$ is a powerful dehydrating agent — but it can only abstract lattice water (water of crystallization, outside the coordination sphere). Water tightly bound to Cr(III) inside the inner sphere is not removed.

Lost = 2 → 2 lattice waters, leaving 4 inside the coordination sphere.

That fixes the inner sphere: $[\mathrm{Cr(H_2O)_4Cl_2}]^+$ — Cr(III) is octahedral with CN = 6, so the remaining 2 ligand slots are filled by 2 chlorides.

Full formula: $[\mathrm{Cr(H_2O)_4Cl_2}]\mathrm{Cl}\cdot 2\mathrm{H_2O}$.

⚡ The Three Hydrate Isomers of $\mathrm{CrCl_3 \cdot 6H_2O}$

| Isomer | AgCl with AgNO₃ | $\mathrm{H_2SO_4}$ loses | Colour | |---|---|---|---| | $[\mathrm{Cr(H_2O)_6}]\mathrm{Cl_3}$ | 3 mol | 0 | violet | | $[\mathrm{Cr(H_2O)_5Cl}]\mathrm{Cl_2}\cdot\mathrm{H_2O}$ | 2 mol | 1 | blue-green | | $[\mathrm{Cr(H_2O)_4Cl_2}]\mathrm{Cl}\cdot 2\mathrm{H_2O}$ | 1 mol | 2 ✓ | green |

⚠️ Don't Count Inner-Sphere Water as "Lost"

$\mathrm{H_2SO_4}$ does not strip coordinated water from Cr(III) — that bond is too strong. Only the loose lattice water is removed, which is why the mass-loss method is diagnostic for hydrate isomerism.

$\boxed{\text{Answer: (4) } [\mathrm{Cr(H_2O)_4Cl_2}]\mathrm{Cl}\cdot 2\mathrm{H_2O}}$