Question

On treating a compound with warm dilute ${\mathrm{H}}_2{\mathrm{SO}}_4$ gas X is evolved which turns ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ paper acidified with dilute ${\mathrm{H}}_2{\mathrm{SO}}_4$ to a green compound Y. X and Y respectively are

• A. $\mathrm{X}={\mathrm{SO}}_2,\mathrm{Y}={\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3$
• B. $\mathrm{X}={\mathrm{SO}}_2,\mathrm{Y}={\mathrm{Cr}}_2{\mathrm{O}}_3$
• C. $\mathrm{X}={\mathrm{SO}}_3,\mathrm{Y}={\mathrm{Cr}}_2{\mathrm{O}}_3$
• D. $\mathrm{X}={\mathrm{SO}}_3,\mathrm{Y}={\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3$

Answer 1

The correct option is A

$\mathrm{X}={\mathrm{SO}}_2,\mathrm{Y}={\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3$

The explanation for the correct option:

Option (A): $\mathrm{X}={\mathrm{SO}}_2,\mathrm{Y}={\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3$

1. Sulfur dioxide is the gas that turns ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ paper acidified with dilute ${\mathrm{H}}_2{\mathrm{SO}}_4$ to a green compound.
2. It is formed when sulfur is reacted with sulfuric acid.$\underset{\mathrm{Sulfur}}{\mathrm{S}\left(\mathrm{s}\right)}+\underset{\mathrm{Sulfuric}\mathrm{acid}}{2{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{l}\right)}\to \underset{\mathrm{Sulfur}\mathrm{dioxide}}{3{\mathrm{SO}}_2\left(\mathrm{g}\right)}+2\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}$
3. Due to the production of chromium $\left(\mathrm{III}\right)$sulfate, ${\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3$the orange-colored dichromate solution will turn green.$\underset{\mathrm{Potassium}\mathrm{dichromate}}{{\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7\left(\mathrm{s}\right)}+\underset{\mathrm{Sulfuric}\mathrm{acid}}{{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{l}\right)}+\underset{\mathrm{Sulfur}\mathrm{dioxide}}{3{\mathrm{SO}}_2\left(\mathrm{g}\right)}\to \underset{\mathrm{Potassium}\mathrm{sulfate}}{{\mathrm{K}}_2{\mathrm{SO}}_4\left(\mathrm{s}\right)}+\underset{\mathrm{Chromium}\mathrm{sulfate}\left(\mathrm{green}\right)}{{\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3\left(\mathrm{s}\right)}+\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}$
4. A redox reaction has occurred, Sulfur dioxide serves as the reducing agent and potassium dichromate as the oxidizing agent.
5. This process is considered an experimental confirmation of the presence of sulfur dioxide gas.
6. Therefore, X is ${\mathrm{SO}}_2$ and Y is ${\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3$
7. Hence, it is the correct option.

Therefore, $\mathrm{X}={\mathrm{SO}}_2,\mathrm{Y}={\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3$