The products for the reaction of $KMnO_4$ with potassium iodide in acidic solutions are

$Mn^{2+}, H_2O, I_2$

The correct answer is Option (1) → $Mn^{2+}, H_2O, I_2$

In an acidic solution, KMnO₄ is a strong oxidising agent.

• $\mathrm{MnO_4^-}$​ is reduced to $\mathrm{Mn^{2+}}$
• $\mathrm{I^-}$ is oxidised to $\mathrm{I_2}$​
• Water is formed

The balanced reaction is:

$\mathrm{MnO_4^-} + 10\mathrm{I^-} + 16\mathrm{H^+} \rightarrow 2\mathrm{Mn^{2+}} + 5\mathrm{I_2} + 8\mathrm{H_2O}$

Therefore, the products are Mn²⁺, H₂O, and I₂