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

Strategy: Trace the reaction sequence from Pb(NO₃)₂ through intermediates to find compound C and its nitrogen oxidation state.

Step 1: Brown Gas A $\ce{2Pb(NO3)2 ->[\ \Delta\ ] 2PbO + 4NO2 + O2}$ A = $\ce{NO2}$ (brown gas, N in +4 oxidation state)

Step 2: Colourless Solid/Liquid B On cooling, $\ce{NO2}$ dimerises: $\ce{2NO2 \rightleftharpoons N2O4}$ B = $\ce{N2O4}$ (dinitrogen tetroxide, colourless solid/liquid). N in $\ce{N2O4}$: $2x + 4(-2) = 0 \Rightarrow x = +4$.

Step 3: Blue Solid C from B + NO $\ce{N2O4 + NO -> N2O3 + NO2}$ Wait — the correct reaction is: $\ce{N2O4 + NO -> 3NO2}$ However, $\ce{NO2}$ reacting with $\ce{NO}$ can give: $\ce{NO2 + NO -> N2O3}$ C = $\ce{N2O3}$ (dinitrogen trioxide), a blue solid.

Step 4: Oxidation Number in C In $\ce{N2O3}$: $2x + 3(-2) = 0 \Rightarrow x = +3$

Oxidation number of N in solid C = +3

$\boxed{\text{Answer: (C) +3}}$