The inorganic iodide (A) is PH4I
(i) On heating with a solution of KOH, it gives a gas (B) which is phosphine and the solution of a compound (C) which is potassium iodide.
PH4I+KOHΔ→PH3+H2O+KI
(ii) PH3 on ignition in air gives a compound (D) which is P2O5 and water.
2PH3+4O2⟶P2O5+3H2O
(iii) Copper sulphate is finally reduced to the metal on passing phosphine through its solution.
3CuSO4+2PH3⟶Cu3P2+3H2SO4
Cu3P2+5CuSO4+8H2O⟶8Cu+2H3PO4+5H2SO4
(iv) A precipitate of compound (E) (which is Cu2I2) is formed on reaction of KI with CuSO4 solution.
2CuSO4+4KI⟶Cu2I2↓+2K2SO4+I2