Search any question & find its solution
Question:
Answered & Verified by Expert
Aqueous sulphite reacts with dilute sulphuric acid to form \(X(g)\). The liberated \(X(g)\) is passed into acidified \(\mathrm{KMnO}_4\) solution. What is the oxidation state of \(\mathrm{Mn}\) in the product formed?
Options:
Solution:
2367 Upvotes
Verified Answer
The correct answer is:
+2
Aqueous sulphite when reacts with sulphuric acid, it gives \(\mathrm{SO}_2(g)\) which when react (passed) with acidified \(\mathrm{KMnO}_4\) gives \(\mathrm{Mn}^{2+}\) ions as follows:
(i) \(\mathrm{SO}_3^{2-}+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \mathrm{SO}_4^{2-}+\mathrm{SO}_2+\mathrm{H}_2 \mathrm{O}\)
(ii) \(5 \mathrm{SO}_2+2 \mathrm{KMnO}_4+2 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{K}_2 \mathrm{SO}_4\)
\(+2 \mathrm{MnSO}_4+2 \mathrm{H}_2 \mathrm{SO}_4\)
\(\because\) Oxidation state of \(\mathrm{Mn}\) in \(\mathrm{MnSO}_4\) is +2. Hence, option (c) is the correct answer.
(i) \(\mathrm{SO}_3^{2-}+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \mathrm{SO}_4^{2-}+\mathrm{SO}_2+\mathrm{H}_2 \mathrm{O}\)
(ii) \(5 \mathrm{SO}_2+2 \mathrm{KMnO}_4+2 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{K}_2 \mathrm{SO}_4\)
\(+2 \mathrm{MnSO}_4+2 \mathrm{H}_2 \mathrm{SO}_4\)
\(\because\) Oxidation state of \(\mathrm{Mn}\) in \(\mathrm{MnSO}_4\) is +2. Hence, option (c) is the correct answer.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.