Question

Hydrogen iodide cannot be prepared by the action of conc. sulfuric acid on potassium iodide because sulfuric acid is a
A. Oxidising agent
B. Reducing agent
C. A and B
D. None of these

Answer 1

Hint: As we move from fluorine to iodine among halogens the size of the halogen atom increases and reducing property of the halides also increases. A strong reducing agent does not give a stable product with a strong oxidising agent.

Complete Step by Step Answer:
Among the halogen hydrides hydrogen iodide is a strong reducing agent thus it can reduce other reagent and oxidize itself.
Again sulfuric acid is a strong oxidizing agent.
On reaction with potassium iodide with sulfuric acid hydrogen iodide is produced which reduces sulfuric acid to sulfur dioxide and oxidize itself to iodine gas. Thus ultimately no hydrogen iodide is produced.
Thus the correct option is A.

Additional Information: As we move from hydrogen fluoride to hydrogen iodide in halogen hydrides, the size of the halogen atom that is fluorine to iodine also increases. Thus the bond of the halogen atom with hydrogen atom becomes weaker and the halogen hydride can easily release the hydrogen atom and behaves as a strong acid. Thus among halogen hydrides hydrogen fluoride is the weakest and hydrogen iodide is the strongest halogen acid.

Note: Hydrogen iodide can be prepared by treating potassium iodide with concentrated phosphoric acid. As phosphoric acid is not a strong oxidizing agent so, no iodine is produced here.