A scarlet compound (A) is treated with conc. $H N O_{3}$ 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. $H N O_{3}$ in the presence of $M n (N O_{3})_{2}$ produces a pink-coloured solution due to the formation of (D). Compounds (A), (B), (C) and (D) are identified as $P b_{3} O_{4}$ , $P b O_{2}$ , $P b I_{2}$ and $H M n O_{4}$ (pink).
If the given statement is true enter 1 else enter 0.

Question

A scarlet compound (A) is treated with conc. $H N O_{3}$ 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. $H N O_{3}$ in the presence of $M n (N O_{3})_{2}$ produces a pink-coloured solution due to the formation of (D). Compounds (A), (B), (C) and (D) are identified as $P b_{3} O_{4}$ , $P b O_{2}$ , $P b I_{2}$ and $H M n O_{4}$ (pink).
If the given statement is true enter 1 else enter 0.

Vedantu Content Team · Accepted Answer

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 ($MnO_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: $P{b_3}{O_4}\xrightarrow{{Conc.HN{O_3}}}Pb{O_2} \downarrow  + 2Pb{(N{O_3})_2} + 2{H_2}O$  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:$5Pb{O_2} + 6HN{O_3} + 2Mn{(N{O_3})_2} \to 2HMn{O_4} + 2{H_2}O + 5Pb{(N{O_3})_2}$ 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.

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