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Identify (A) to ( E). (a) An inorganic...

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.

Text Solution

Verified by Experts

Gas `(B)` on on ignition gives water, therefore hydrogen is present in the gas.
An inorganic iodide with alkali `(KOH)` gives a gas `(B)`, an hydrogen compound, so `(A)` may be `NH_4 I` or `PH_4 I`. As `NH_3` does not reduce `CuSO_4`, therefore the compound `(A)` is `PH_4 I`.
(a) `underset ((A))(PH_4 I) + KOH rarr underset ((B)) (PH_3) + underset ((C))(K I) + H_2 O`
(b) `4 PH_3 + 8O_2 rarr P_4 O_10 + 6 H_2 O`
`P_4 O_10 +2H_2 O rarr underset((D) ("Metaphosphoric acid"))(4 HPO_3)`
( c) `3CuSO_4 + underset ((B)) (2 PH_3) rarr Cu_3 P_2 + 3H_2 SO_4`
`Cu_3P_2 rarr 6 Cu + P_4`
(d) `CuSO_4 + underset ((B))(KI) rarr underset ((E)) (Cu_2 I_2) + 2K_2 SO_4 + I_2`.
Hence `(A)` is `PH_4 I, (B) is PH_3, ( C) is KI, (D) is HPO_3 and ( E) is Cu_2 I_2`.
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