Chemistry - Complete P-block Question with Solution | TestHub

Sulfur dioxide gas turns acidified potassium dichromate (VI) solution from orange to green. Chromium is reduced from (VI) to (III) after reaction.

${\mathrm{SO}}_2+\mathrm{dil} {\mathrm{H}}_2{\mathrm{SO}}_4⟶{\mathrm{SO}}_3\left(\mathrm{g}\right)$

${\mathrm{SO}}_3+{\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7\underset{{\mathrm{H}}_2{\mathrm{SO}}_4}{\overset{\mathrm{dil}}{\to }}{\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3$

Hence, On treating a compound with warm dil. ${\mathrm{H}}_2{\mathrm{SO}}_4$, gas $\mathrm{X}$ is evolved which turns ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ paper acidified with dil. ${\mathrm{H}}_2{\mathrm{SO}}_4$ to a green compound $\mathrm{Y}$. $\mathrm{X}$ and $\mathrm{Y}$ respectively are ${\mathrm{SO}}_3$ and ${\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3$.