The reaction of acidified \(KMnO_4\) and \(NO_2\) gives:

\(Mn^{2+}\) and \(NO_3^-\)

The correct answer is option 3. \(Mn^{2+}\) and \(NO_3^-\).

In the reaction between acidified potassium permanganate (\(KMnO_4\)) and nitrogen dioxide (\(NO_2\)), a redox reaction occurs. Let's break down the reaction and explain the process:

The reaction is as follows:

\[ 5 \, \text{NO}_2 + 2 \, \text{MnO}_4^- + 6 \, \text{H}^+ \rightarrow 5 \, \text{NO}_3^- + 2 \, \text{Mn}^{2+} + 3 \, \text{H}_2\text{O}\]

\(KMnO_4\) (\(MnO_4^-\)) acts as an oxidizing agent. In the reaction, each manganese atom in \(KMnO_4\) changes from a +7 oxidation state to a +2 oxidation state. This reduction of manganese from a higher oxidation state to a lower one indicates that \(KMnO_4\) is being reduced. It undergoes reduction to form \(Mn^{2+}\).

Nitrogen dioxide (\(NO_2\)) acts as a reducing agent. In the reaction, each nitrogen atom in \(NO_2\) changes from an oxidation state of +4 to +5. This increase in the oxidation state of nitrogen indicates that \(NO_2\) is being oxidized. It undergoes oxidation to form \(NO_3^-\).

Overall, the \(KMnO_4\) is reduced to \(Mn^{2+}\) ions, and \(NO_2\) is oxidized to \(NO_3^-\) ions. The other products formed are water (\(H_2O\)) and hydrogen ions (\(H^+\)), which are also involved in the reaction.

Therefore, in the reaction of acidified \(KMnO_4\) and \(NO_2\), the products formed are \(Mn^{2+}\) and \(NO_3^-\).