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Question

(i) An inorganic iodide (A) on heating with a solution of KOH gives a gas (B) and the solution of a compound (C).
(ii) The gas (B) on ignition in air gives a compound (D) and water.
(iii) Copper sulphate is finally reduced to the metal on passing (B) through its solution.
(iv) A precipitate of compound (E) is formed on reaction of (C) with CuSO4 solution.Identify (A) to (E) and explain reactions (i) to (iv).

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Solution

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+KOHΔPH3+H2O+KI
(ii) PH3 on ignition in air gives a compound (D) which is P2O5 and water.
2PH3+4O2P2O5+3H2O
(iii) Copper sulphate is finally reduced to the metal on passing phosphine through its solution.
3CuSO4+2PH3Cu3P2+3H2SO4
Cu3P2+5CuSO4+8H2O8Cu+2H3PO4+5H2SO4
(iv) A precipitate of compound (E) (which is Cu2I2) is formed on reaction of KI with CuSO4 solution.
2CuSO4+4KICu2I2+2K2SO4+I2

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