To solve this problem, we need to understand the reaction between $SO_2$ and acidified $K_2Cr_2O_7. \\$When $SO_2$ is passed through acidified $K_2Cr_2O_7$ solution, a redox reaction occurs.$\\$The balanced chemical equation for the reaction is:$\\ K_2Cr_2O_7 + 3SO_2 + H_2SO_4 \rightarrow Cr_2(SO_4)_3 + K_2SO_4 + H_2O \\$In this reaction:$\\$• $SO_2$ acts as a reducing agent and is oxidized to $SO_4^{2-}. \\$• $Cr_2O_7^{2-}$ is reduced to $Cr^{3+}. \\$The reduction of $Cr_2O_7^{2-}$ to $Cr^{3+}$ results in the formation of green $Cr_2(SO_4)_3. \\$Therefore, the correct statement is:$\\$Green $Cr_2(SO_4)_3$ is formed.$\\$This corresponds to Option 4.