A white crystalline salt A reacts with dilute $\mathrm{HCl}$ to liberate a suffocating gas B and also forms a yellow precipitate. The gas B turns potassium dichromate acidified with dilute $\mathrm{H}_{2} \mathrm{SO}_{4}$ to a green coloured solution C. A, B and C are respectively

• A

  $\mathrm{Na}_{2} \mathrm{SO}_{3}, \mathrm{SO}_{2}, \mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}$
• Correct answer
  B

  $\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}, \mathrm{SO}_{2}, \mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}$
• C

  $\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{SO}_{2}, \mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}$
• D

  $\mathrm{Na}_{2} \mathrm{SO}_{4}, \mathrm{SO}_{2}, \mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}$

Solution

Gas B turns the colour of acidified $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ green thus it is $\mathrm{SO}_{2}$ and $\mathrm{SO}_{2}$ is obtained along with yellow precipitate when thio sulphate is treated with dilute acids. Thus, $\mathrm{A}$ is $\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}$, B is $\mathrm{SO}_{2}$ and $\mathrm{C}$ is $\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}$. The reactions are as follows

Hence, the correct answer is option: 'B'.