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:

We have lead(II) nitrate $$\text{Pb(NO}_3)_2$$. A well-known rule for the thermal decomposition of the nitrate of any divalent metal (except the very light alkali metals) is

$$2\,M(NO_3)_2 \;\longrightarrow\; 2\,MO \;+\; 4\,NO_2 \;+\; O_2$$

where $$M$$ represents the divalent metal. Substituting $$M = \text{Pb}$$ gives

$$2\,\text{Pb(NO}_3)_2 \;\longrightarrow\; 2\,\text{PbO} \;+\; 4\,NO_2 \;+\; O_2$$

The gas set free is $$NO_2$$, which is brown in colour, so

$$(A) = NO_2$$.

When this brown gas is cooled, the well-known equilibrium

$$2\,NO_2 \;\rightleftharpoons\; N_2O_4$$

shifts to the right (because lowering the temperature favours the dimer), producing colourless liquid/solid $$N_2O_4$$. Hence

$$(B) = N_2O_4$$.

Now $$(B)$$ is heated in the presence of nitric oxide $$NO$$. The reaction that takes place is

$$N_2O_4 + NO \;\longrightarrow\; N_2O_3$$

The product $$N_2O_3$$ is a blue solid at sufficiently low temperatures, so

$$(C) = N_2O_3$$.

We must now find the oxidation number of nitrogen in $$N_2O_3$$. Let the oxidation number of a single nitrogen atom be $$x$$. Using the rule “the sum of oxidation numbers in a neutral molecule is zero”, we write

$$2x + 3(-2) = 0$$

because oxygen has the fixed oxidation number $$-2$$. Simplifying step by step,

$$2x - 6 = 0$$

$$2x = +6$$

$$x = +3$$.

So, in the blue solid $$N_2O_3$$, each nitrogen atom is in the $$+3$$ oxidation state.

Hence, the correct answer is Option C.