On heating, lead (II) nitrate gives a brown gas (A). The gas (A) on cooling changes to a colourless solid/liquid (B).…

Step 1: Identify the brown gas (A) Heating heavy metal nitrates, notably Lead(II) nitrate ($\ce{Pb(NO3)2}$), initiates thermal decomposition producing lead oxide, oxygen, and nitrogen dioxide gas. $\ce{2Pb(NO3)2(s) ->[\Delta] 2PbO(s) + 4NO2(g) + O2(g)}$ Nitrogen dioxide ($\ce{NO2}$) is distinctly known as an aggressively brown gas. Thus, (A) is $\ce{NO2}$.

Step 2: Identify the colourless substance (B) When the brown $\ce{NO2}$ gas is actively cooled, its odd-electron structure forces it to swiftly dimerize into stable Dinitrogen tetroxide ($\ce{N2O4}$), structurally shifting into a fully colourless liquid or solid depending precisely on temperature. $\ce{2NO2 (Brown) <=> N2O4 (Colourless)}$ Thus, (B) is $\ce{N2O4}$.

Step 3: Identify the blue solid (C) When liquid $\ce{N2O4}$ is fundamentally heated or reacted with Nitric oxide ($\ce{NO}$ gas) at severely low temperatures, it chemically fuses to produce Dinitrogen trioxide ($\ce{N2O3}$), which beautifully condenses into a deep blue liquid or solid. $\ce{N2O4 + 2NO <=> 2N2O3}$ Thus, (C) is $\ce{N2O3}$.

Step 4: Determine the Oxidation Number In Dinitrogen trioxide ($\ce{N2O3}$), establishing the oxidation state of Nitrogen (let it be $x$): $2(x) + 3(-2) = 0$ $2x - 6 = 0 \implies x = +3$

Conclusion: The oxidation state intrinsically held by Nitrogen in the blue solid $\ce{N2O3}$ is exactly $+3$.

Key Points to Remember:

• Thermal decomposition of $\ce{Pb(NO3)2}$ provides extremely pure $\ce{NO2}$ (brown gas).
• $\ce{NO2}$ dimerizes actively on cooling precisely to $\ce{N2O4}$ (colourless).
• Equimolar combination of $\ce{NO}$ and $\ce{NO2}$ (or interacting $\ce{NO}$ heavily with $\ce{N2O4}$) remarkably produces $\ce{N2O3}$ (blue solid/liquid), harboring Nitrogen precisely in the $+3$ oxidation state.