Copper(II) chloride forms brown crystals soluble in water. — Inorganic Chemistry — Solid State Chemistry Question
Simple experiments with copper(II) chloride
Copper(II) chloride forms brown crystals soluble in water.
The color of the solutions depends on concentration. Explain this fact.
Model Answer
A diluted solution of copper chloride is blue due to [Cu(H2O)6]2+ ions. Upon the concentration of the solution its color changes to green, as the substitution of coordinated water molecules by chloride-ions occurs:
[Cu(H2O)6]2+ + 4 Cl– →← [CuCl4]2– + 6 H2O.
Draw the structure of at least three copper containing species that can be present in the solution.
Model Answer
See: Bernd M. Rode and Saiful M. Islam. Structure of aqueous copper chloride solutions: results from Monte Carlo simulations at various concentrations // J. Chem. Soc. Faraday Trans., 1992, 88(3), 417-422.
To the aliquots of the copper chloride(II) solution in the test tubes (a) zinc powder and the solutions of (b) sodium iodide, (c) sodium nitrate, (d) sodium sulfide were added dropwise. What changes happen if they are? Provide equations. In what cases copper is completely or partly reduced?
Model Answer
(a) CuCl2 + Zn → Cu↓ + ZnCl2 (red precipitate – copper),
(b) 2 CuCl2 + 4 NaI → 2 CuI↓ + I2↓ + 4 NaCl (grey precipitate – mixture of copper iodide and iodine),
(c) No noticeable changes,
(d) CuCl2 + Na2S → CuS↓ + 2 NaCl (black precipitate).
Copper is completely reduced in (a), and partly reduced in (b) and (d). Surprisingly, copper sulfide CuS is a mixed sulfide-disulfide of copper(I) and copper(II).
Suggest a synthetic route to copper chloride(II) starting from copper(II) sulfate aqueous solution.
Model Answer
A possible synthetic route is the reduction of copper by zinc followed by chlorination.
CuSO4 + Zn → ZnSO4 + Cu
Cu + Cl2 → CuCl2.
Another way is metathesis reaction with barium chloride with crystallization of hydrated copper chloride. The dehydration can be achieved by heating with thionyl chloride:
CuSO4 + BaCl2 → BaSO4 + CuCl2 (from solution CuCl2 ⋅ 2 H2O forms).
CuCl2 ⋅ 2 H2O + 2 SOCl2 → CuCl2 + 2 SO2 + 4 HCl