NH4NO3⟶ heat X+2H2O2PbNO32⟶ heat Y+PbO+O2Y⇌heatcoolZX, Y, and Z respectively are

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${NH}_{4} {NO}_{3} \overset{heat}{⟶} X + 2 H_{2} O$
$2 Pb {({NO}_{3})}_{2} \overset{heat}{⟶} Y + PbO + O_{2}$
$Y ⇌_{heat}^{cool} Z$
X, Y, and Z respectively are

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$N_{2}, {NO}_{2}, NO$

$N_{2} O, {NO}_{2}, N_{2} O_{4}$

${NO}_{2}, N_{2} O, N_{2} O_{4}$

$NO, {NO}_{2}, N_{2} O$

answer is A.

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Detailed Solution

${NH}_{4} {NO}_{3}$ on heating liberates $N_{2} O, P b {({NO}_{3})}_{2}$ on heating gives ${NO}_{2}, N_{2} O$ dimerises to give $N_{2} O_{4}$

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