Which of the following gas readily de-colourises the acidified $\mathrm{KMnO}_4$ solution?

Acidified potassium permanganate ($\mathrm{KMnO}_4$) is a strong oxidizing agent. It reacts with reducing agents to get reduced, resulting in the loss of its characteristic purple color. Among the given gases, $\mathrm{SO}_2$ acts as a strong reducing agent. The reaction is as follows: $5\mathrm{SO}_2 + 2\mathrm{KMnO}_4 + 2\mathrm{H}_2\mathrm{O} \rightarrow \mathrm{K}_2\mathrm{SO}_4 + 2\mathrm{MnSO}_4 + 2\mathrm{H}_2\mathrm{SO}_4$ In this reaction, the oxidation state of sulfur increases from $+4$ to $+6$, and the oxidation state of manganese decreases from $+7$ to $+2$. The formation of colorless $\mathrm{Mn}^{2+}$ ions leads to the de-colourisation of the solution. $\mathrm{CO}_2$ and $\mathrm{P}_2\mathrm{O}_5$ are acidic oxides that do not act as reducing agents in this context. $\mathrm{NO}_2$ can act as a reducing agent in some conditions, but $\mathrm{SO}_2$ is the standard gas known for readily de-colourising acidified $\mathrm{KMnO}_4$ solution. Answer: $\mathrm{SO}_2$