Correct Answer - B
`SO_(2)` gas can readily oxidise acidified `KMnO_(4)` solution because `KMnO_(4)` is an oxidising agent and `SO_(2)` act as reducing agent.
`2MnO_(4)^(-) + 5SO_(2) +2H_(2)O to 2Mn^(2+) + 5SO_(4)^(2-) + 4H^(+)`
While other options such as `NO_(2)` (strong oxidising agent), `CO_(2)` (neither oxidising agent nor reducing agent) cannot decolourise acidified `KMnO_(4)` solution.
