KCET · Chemistry · d and f Block Elements
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 \(\mathrm{Na}_{2} \mathrm{SO}_{3}, \mathrm{SO}_{2}, \mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}\)
- 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 \(\mathrm{Na}_{2} \mathrm{SO}_{4}, \mathrm{SO}_{2}, \mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}\)
Answer & Solution
Correct Answer
(B) \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}, \mathrm{SO}_{2}, \mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}\)
Step-by-step Solution
Detailed explanation
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

The reactions are as follows

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