- A${K_2}Mn{O_4}$
- ✓$Mn{O_2}$
- C$Mn{(OH)_2}$
- D$M{n^{2 + }}$
$KMn{O_4}$ is first reduced to manganate and then to insoluble manganese dioxide. Colour changes first from purple to green and finally becomes colourless.
$2KMn{O_7} + 2KOH \to 2{K_2}Mn{O_4} + {H_2}O + O$
$2{K_2}Mn{O_4} + 2{H_2}O \to 2Mn{O_2} + 4KOH + 2O$
$\overline {2KMn{O_2} + {H_2}O\xrightarrow{{{\text{alkaline}}}}2Mn{O_2} + 2KOH + 3[O]} $
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$C{{H}_{3}}COOEt+{{H}_{2}}O\xrightarrow{{{H}^{+}}}C{{H}_{3}}COOH+EtOH$
Statement $I$: Aniline reacts with con. $\mathrm{H}_2 \mathrm{SO}_4$ followed by heating at $453-473 \mathrm{~K}$ gives $\mathrm{p}$ aminobenzene sulphonic acid, which gives blood red colour in the 'Lassaigne's test'.
Statement $II$: In Friedel - Craft's alkylation and acylation reactions, aniline forms salt with the $\mathrm{AlCl}_3$ catalyst. Due to this, nitrogen of aniline aquires a positive charge and acts as deactivating group.
In the light of the above statements, choose the correct answer from the options given below :