Question

A scarlet compound $A$ is treated with cone. $HN{O}_3$ 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. $HN{O}_3$ in the presence of $Mn{\left(N{O}_3\right)}_2$ produces a pink coloured solution due to the formation of $D$. Identify $A,B,C$ and $D$. Write the reaction sequence.

Answer 1

$\underset{\text{ Scarlet }}{A\left(P{b}_3{O}_4\right)}\stackrel{C.HN{O}_3}{\to }\underset{\text{ Chocolate }}{B\left(Pb{O}_2\right)}\stackrel{\text{ Filtered }}{\to }\underset{\text{ Filtrate }}{Pb{\left(N{O}_3\right)}_2}$
Filtrate is neutralised with $NaOH$ and on reaction with $KI$, gives yellow ppt of $Pb{I}_2$.
$Pb{\left(N{O}_3\right)}_2+2KI⟶\underset{\text{ Yellow }}{Pb{l}_2\downarrow }+2KN{O}_3$
$Pb{O}_2$ on warming with conc. $HN{O}_3$ in presence of $Mn{\left(N{O}_3\right)}_2$ produced pink solution due to formation of $Pb{\left(Mn{O}_4\right)}_2$ (I).
$5Pb{O}_2+2Mn{\left(N{O}_3\right)}_2+4HN{O}_3⟶Pb{\left(Mn{O}_4\right)}_2$
$+4Pb{\left(N{O}_3\right)}_2+2H_{2}O$
$\Rightarrow A=P{b}_3{O}_4,B=Pb{O}_2,C=Pb{I}_2$ and $D=Pb{\left(Mn{O}_4\right)}_2.$