JEE Mains · Chemistry · STD 12 - 4. d and f- block elements
\(\underset{\raise0.3em\hbox{\)\smash{\scriptscriptstyle-}\(}}{A} \xrightarrow{{4\,KO{H_1}{O_2}}}\mathop {2\underset{\raise0.3em\hbox{\)\smash{\scriptscriptstyle-}\(}}{B} }\limits_{(Green)} + 2{H_2}O\) \(3\underset{\raise0.3em\hbox{\)\smash{\scriptscriptstyle-}\(}}{B} \xrightarrow{{4\,HCl}}\mathop {2C}\limits_{(Purple)} + Mn{O_2} + 2{H_2}O\) \(3\underset{\raise0.3em\hbox{\)\smash{\scriptscriptstyle-}\(}}{C} \xrightarrow{{{H_2}O,KI}}2\underset{\raise0.3em\hbox{\)\smash{\scriptscriptstyle-}\(}}{A} + 2KOH + \underset{\raise0.3em\hbox{\)\smash{\scriptscriptstyle-}\(}}{D} \) In the above sequence of reactions, \(\underset{\raise0.3em\hbox{\)\smash{\scriptscriptstyle-}\(}}{A}\) and \(\underset{\raise0.3em\hbox{\)\smash{\scriptscriptstyle-}\(}}{D} \) , respectively are
- A \(KI\) and \(KMnO_4\)
- B \(MnO_2\) and \(KIO_3\)
- C \(KIO_3\) and \(MnO_2\)
- D \(KI\) and \(K_2MnO_4\)
Answer & Solution
Correct Answer
(B) \(MnO_2\) and \(KIO_3\)
Step-by-step Solution
Detailed explanation
\(Mn{O_2}\,\xrightarrow{{4KOH,\,{O_2}}}\mathop {2{K_2}Mn{O_4}}\limits_{(Green)} \, + \,2{H_2}O\) \(3{K_2}Mn{O_4}\,\xrightarrow{{4\,HCl}}\mathop {2KMn{O_4}}\limits_{(purple)} \, + \,Mn{O_2}\, + \,2{H_2}O\) \(2KMn{O_4}\,\xrightarrow{{{H_2}O,\,KI}}2Mn{O_2}\, + \,2KOH\, + \,KI{O_3}\)
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