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(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 copper sulphate solution. Identify (A) to (E) and give chemical equations for reactions at steps (i) to (iv). |
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Answer» SOLUTION :(a) Since the gas (B) reduces copper sulphate solution finally to copper metal, therefore, it must be a good reducing agent. Further since the gas (B) along with a solution of compound (C) are obtained when the inorganic iodide (A) is heated whith a solution of KOH, therefore, gas (B) must be phosphonium iodide `(PH_(4)I)`. (b) Since the gas (B) on ignition in air gives a compound (D) and water, therefore, compound (D) must be phosphorus pentoxide `(P_(2)O_(5))`. (c) Since compound (C), i.e., KI on treatment with `CuSO_(4)` solution gives a precipitate of compound (E), therefore, compound (E) must be cuprous iodide `(Cu_(2)I_(2))`. (d) The chemical equations for the REACTIONS at steps (i) to (iv) are as follows : (i) `underset((A))(PH_(4)I)+KOH overset(Delta)rarr underset((B))(PH_(3)) + underset((C))(KI)+H_(2)O` (ii) `underset((B))(2PH_(3))+4O_(2) rarr underset((D))(P_(2)O_(5)) + 3 H_(2)O` (iii) `3 CuSO_(4) + underset((B))(2PH_(3)) rarr underset("Copper phosphide")(Cu_(3)P_(2))+2H_(2)SO_(4)` `Cu_(3)P_(2)+3CuSO_(4) + 6H_(2)O rarr 6 Cu darr + I_(2) + 2 K_(2)SO_(4)` (iv) `2CuSO_(4) + 4KI rarr underset((E))(Cu_(2)I_(2)) darr + I_(2) + 2K_(2)SO_(4)` Thus, inorganic compound (A) is `PH_(4)I`, gas (B) is `PH_(3)`, compound (C) is KI, compound (D) is `P_(2)O_(5_` and compound (E) is `Cu_(2)I_(2)`. |
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