Which one of the following statements is correct when $\mathrm{SO}{}_2$ is passed through acidified $\mathrm{K}{}_2\mathrm{Cr}{}_2\mathrm{O}{}_7$ solution?

Solution & Explanation

\textbf{Answer:} (d) Green $\mathrm{Cr}{}_2(\mathrm{SO}{}_4){}_3$ is formed. \textbf{Solution:} In acidic medium, dichromate is a strong oxidising agent. $\mathrm{SO}{}_2$ (where S is in the $+4$ state) is oxidised to sulphate, while $\mathrm{Cr}(\mathrm{VI})$ of the orange $\mathrm{Cr}{}_2\mathrm{O}{}_7{}^{2-}$ is reduced to green $\mathrm{Cr}(\mathrm{III})$: $\mathrm{Cr}{}_2\mathrm{O}{}_7{}^{2-}+3\mathrm{SO}{}_2+2\mathrm{H}{}^{+}\to 2\mathrm{Cr}{}^{3+}+3\mathrm{SO}{}_4{}^{2-}+\mathrm{H}{}_2\mathrm{O}$ The $\mathrm{Cr}{}^{3+}$ produced appears as green $\mathrm{Cr}{}_2(\mathrm{SO}{}_4){}_3$ in solution. So the orange colour changes to green (it is not decolourised), and it is $\mathrm{SO}{}_2$ that is oxidised, not reduced. Hence option (d) is correct.