On heating, lead (II) nitrate gives a brown gas (A). The gas (A) on cooling changes to a colourless solid/liquid (B). (B) on heating with NO changes to a blue solid (C). The oxidation number of nitrogen in solid (C) is :
Solution
Pb(NO<sub>3</sub>
)<sub>2</sub> $\to$ PbO + NO<sub>2</sub>[(A) Brown gas]
+ O<sub>2</sub>
<br><br>2NO<sub>2</sub> $\buildrel {cooling} \over
\longrightarrow$ N<sub>2</sub>O<sub>4</sub>[(C) colourless solid]
<br><br> N<sub>2</sub>O<sub>4</sub> + NO $\buildrel \Delta \over
\longrightarrow$ N<sub>2</sub>O<sub>3</sub> [(C) Blue Solid]
<br><br>Let oxidation state of N in N<sub>2</sub>O<sub>3</sub> is x.
<br><br>$\therefore$ 2x + 3 (–2) = 0
<br><br>$\Rightarrow$ x = +3
About this question
Subject: Chemistry · Chapter: p-Block Elements · Topic: Group 13: Boron Family
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