KCET · Chemistry · d and f Block Elements
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 \(+4\) 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
Acidified \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) solution oxidises \(\mathrm{SO}_{2}\) into \(\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}\).
\[
\begin{aligned}
3 \mathrm{SO}_{2}+\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+& \mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{K}_{2} \mathrm{SO}_{4} \\
&+\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}+\mathrm{H}_{2} \mathrm{O}
\end{aligned}
\]
Hence, oxidation state of sulphur changes from \(+4\) to \(+6\).
\[
\begin{aligned}
3 \mathrm{SO}_{2}+\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+& \mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{K}_{2} \mathrm{SO}_{4} \\
&+\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}+\mathrm{H}_{2} \mathrm{O}
\end{aligned}
\]
Hence, oxidation state of sulphur changes from \(+4\) to \(+6\).
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