Identify `(A)` to `( E)`.
(a) An inorganic iodide `(A)` on heating with a solution of `KOH` gives a gas `(B)` and the solution of a compound `( C)`.
(b) The gas `(B)` on ignition air gives a compound `(D)` and water.
( C) Copper sulphate is reduced to the metal on passing `(B)` through the solution.
(d) A precipitate of the compound `( E)` is formed on reaction of `( C)` with copper sulphate solution.

Answer

(A)`PH_(4)I`; (B)`PH_(3)`; ( C)`KI` ; (D)`P_(4)O_(10)` ; (E)`Cu_(2)I_(2)`

Explanation

`(A)` is `PH_(4)I`. The given changes are:
(i)`underset((A))(PH_(4)l)+KOH to underset((B))(PH_(3))+H_(2)O`
(ii)`underset((B))(4PH_(3))+8O_(2) to underset((D))(P_(4)O_(10))+6H_(2)O`
(iii)`4Cu^(2+) + PH_(3)+ 4H_(2)O to H_(3)PO_(4) + 4Cu darr + 8H^(+)`
(iv)`2CuSO_(4)+4KI to underset((E))(Cu_(2)l_(2))darr+2K_(2)SO_(4) + I_(2)`