Question

A scarlet compound $A$ is treated with cone. $HNO _{3}$ to give a chocolate brown precipitate $B$. The precipitate is filtered and the filtrate is neutralised with $NaOH$. Addition of $KI$ to the resulting solution gives a yellow ppt $C$. The brown ppt $B$ on warming with cone. $H N O_{3}$ in the presence of $M n\left(N O_{3}\right)_{2}$ produces a pink coloured solution due to the formation of $D$. Identify $A, B, C$ and $D$. Write the reaction sequence.

Answer 1

$\underset{\text { Scarlet }}{A\left( Pb _{3} O _{4}\right) }\xrightarrow{C . HNO _{3}} \underset{\text { Chocolate }}{B\left( PbO _{2}\right)} \xrightarrow{\text { Filtered }} \underset{\text { Filtrate }}{ Pb \left( NO _{3}\right)_{2}}$
Filtrate is neutralised with $NaOH$ and on reaction with $KI$, gives yellow ppt of $PbI _{2}$.
$Pb \left( NO _{3}\right)_{2}+2 KI \longrightarrow \underset{\text { Yellow }}{ Pbl _{2} \downarrow }+2 KNO _{3}$
$PbO _{2}$ on warming with conc. $HNO _{3}$ in presence of $Mn \left( NO _{3}\right)_{2}$ produced pink solution due to formation of $Pb \left( MnO _{4}\right)_{2}$ (I).
$5 PbO _{2}+2 Mn \left( NO _{3}\right)_{2}+4 HNO _{3} \longrightarrow Pb \left( MnO _{4}\right)_{2} $
$+4 Pb \left( NO _{3}\right)_{2}+2 H _{2} O $
$\Rightarrow A= Pb _{3} O _{4}, B= PbO _{2}, C= PbI _{2} $ and $ D= Pb \left( MnO _{4}\right)_{2} .$