Question

A scarlet compound (A) is treated with conc.$HN{O}_3$ 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 precipitate (C). The precipitate (B) on warming with conc.$HN{O}_3$ in the presence of $Mn(N{O}_3{)}_2$ produces a pink-coloured solution due to the formation of (D). Compounds (A), (B), (C) and (D) are identified as $P{b}_3{O}_4$, $Pb{O}_2$, $Pb{I}_2$ and $HMn{O}_4$ (pink).
If the given statement is true enter 1 else enter 0.

Answer 1

Hint: Take compound (A) to be $P{b}_3{O}_4$ and start running the reactions step by step to see what the products are. Also conc. $HN{O}_3$ acts as nitrating reagent and the permanganate ion ($Mn{O}_4^{-}$) ion mostly gives pink colour.

Complete step by step answer:
-For this question we will start by writing down the reactions involved in the sequence. We will begin from compound (A).
-The scarlet compound (A) is $P{b}_3{O}_4$. Now we will see what happens when it reacts with conc.$HN{O}_3$. The reaction is:
 $P{b}_3{O}_4\stackrel{Conc.HN{O}_3}{\to }Pb{O}_2\downarrow +2Pb(N{O}_3{)}_2+2H_{2}O$

Compound (A) $P{b}_3{O}_4$ reacts with conc.$HN{O}_3$, it leads to the formation of this chocolate brown compound (B) which is $Pb{O}_2$ and the filtrate is $Pb(N{O}_3{)}_2$. The precipitate obtained after this step is filtered and the filtrate is neutralised by the use of NaOH. To this KI is added which gives a yellow precipitate (C). The involved reaction is:
$Pb(N{O}_3{)}_2+KI\to Pb{I}_2\downarrow +2KN{O}_3$
  Hence the yellow precipitate (C) obtained is $Pb{I}_2$.

- The precipitate (B) $Pb{O}_2$ on warming with conc.$HN{O}_3$ in the presence of $Mn(N{O}_3{)}_2$ produces a pink-coloured solution due to the formation of (D). The reaction for this step is:
$5Pb{O}_2+6HN{O}_3+2Mn(N{O}_3{)}_2\to 2HMn{O}_4+2H_{2}O+5Pb(N{O}_3{)}_2$
The end result is a pink coloured solution due to formation of $HMn{O}_4$ which is compound (D).

-Hence, finally we can tell that:
   Compound (A) is: $P{b}_3{O}_4$
   Compound (B) is: $Pb{O}_2$
   Compound (C) is: $Pb{I}_2$
   Compound (D): $HMn{O}_4$
This tells us that the given statement is true and so we will enter 1.

Note: $P{b}_3{O}_4$ is a lead (ll, lV) oxide and it is named as Lead tetraoxide. It is a compound with mixed valency since it consists of both Pb(ll) and Pb(lV) in the ratio of (2 : 1). Because lead has some toxic effects, its use is limited. If consumed by mistake it can lead to poisoning. It is used in limited amounts as a paint primer for iron objects.