Read the passage carefully and answer the Questions.

$KMnO_4$ is prepared by the fusion of $MnO_2$ with an alkali metal hydroxide and an oxidizing agent like $KNO_3$ to give a dark- green manganate ion which disproportionate to give permanganate as follows.

$2MnO_2+ 4KOH + O_2→2KMnO_4+2H_2O$

$3KMnO_4+ 4H^+→2KMnO_4 + MnO_2 + 2H_2O$

On heating $KMnO_4$ decomposes at 513 K to give $K_2MnO_4$. Permanganate ion is tetrahedral and diamagnetic. Acidified $KMnO_4$ acts a strong oxidizing agent which oxidizes oxalic acid, ferrous ions, nitrite ion and iodides.

$HCl$ is not used to make the medium acidic in oxidation reaction of $KMnO_4$ in acidic medium, because....

$KMnO_4$ oxidizes $HCl$ into $Cl_2$ which is also an oxidizing agent

The correct answer is Option (2) → $KMnO_4$ oxidizes $HCl$ into $Cl_2$ which is also an oxidizing agent **

• When KMnO₄ is used in acidic medium for redox reactions, H₂SO₄ is preferred over HCl.
• If HCl is used, KMnO₄ oxidizes Cl⁻ ions from HCl to produce Cl₂ gas:

$2KMnO_4 + 16HCl → 2MnCl_2 + 5Cl_2 + 8H_2O + 2KCl$

• The produced Cl₂ is itself a strong oxidizing agent, which interferes with the intended reaction, giving erroneous results.

Hence, HCl is avoided in acidic KMnO₄ reactions.