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On heating, lead $(II)$ nitrate gives a brown gas $(A)$ . The $(A)$ on cooling changes to a colourless solid/liquid $(B)$ .on heating with $NO$ changes to a blue solid $(C)$ . oxidation number of nitrogen in solid $(C)$ is

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answer is B.

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Detailed Solution

$Pb {({NO}_{3})}_{2} \overset{Δ}{⟶} PbO + 2 {NO}_{2}_{Brown gas} + \frac{1}{2} O_{2} (g) (A)$

$2 {NO}_{2} (g) \overset{Cooling}{⟶} \underset{(B)}{N_{2} O_{4}}$

$N_{2} O_{4} + 2 NO \overset{}{\overset{∆}{\to}} \underset{\begin{matrix} Blue solid \\ (C) \end{matrix}}{2 N_{2} O_{3}}$

$O . S .$ of nitrogen in $N_{2} O_{3}$ is +3

$\begin{array}{l} N_{2} O_{3} : 2 x + 3 (- 2) = 0 \\ x = + 3 . \end{array}$

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