COMEDK · Chemistry · 9. Redox Reactions
When sulphur dioxide is passed in an acidified \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) solution, the oxidation state of sulphur is changed from
- A \({0}+4\) to 0
- B \(+4\) to \(+2\)
- C \(+4\) to \(+6\)
- D \(+6\) to \(+4\)
Answer & Solution
Correct Answer
(C) \(+4\) to \(+6\)
Step-by-step Solution
Detailed explanation
\(3 \mathrm{SO}_{2}+\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{K}_{2} \mathrm{SO}_{4}+\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}+\mathrm{H}_{2} \mathrm{O}\) O.N. of \(\mathrm{S}\) in \(\mathrm{SO}_{2}\) is…
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