Boiling CuCl2 with Cu in conc. HCl gives : — d and f Block Elements Chemistry Question

💡 Solution & Explanation

Step 1: Copper(II) chloride (CuCl2) can be reduced by metallic copper in a concentrated hydrochloric acid (HCl) medium. Step 2: The copper(II) ions are reduced to copper(I), while the metallic copper is oxidized to copper(I), forming soluble chlorocuprate(I) complexes such as H[CuCl2] in solution. Step 3: Upon diluting this solution with water, white cuprous chloride (CuCl) precipitates. Thus, boiling CuCl2 with Cu in conc. HCl yields CuCl as the main copper(I) chloride product, which corresponds to option (a).