1. Use standard salt-analysis reactions of lead salts

   In qualitative analysis, lead commonly shows the following sequence:

   • Passing $\mathrm{H_2S}$ through a lead(II) salt solution gives black precipitate of lead sulfide: $\mathrm{Pb^{2+} + H_2S \rightarrow PbS\downarrow + 2H^+}$ So, $A = \mathrm{PbS}$.

2. Oxidation of lead sulfide

   Lead sulfide on treatment with an oxidizing acid such as concentrated $\mathrm{HNO_3}$ gets converted to lead sulfate: $\mathrm{PbS \xrightarrow[oxidation]{} PbSO_4}$ Thus, $B = \mathrm{PbSO_4}$.

3. Formation of lead chromate

   Lead sulfate can be converted into a soluble lead salt and then on adding chromate ions gives yellow precipitate of lead chromate: $\mathrm{Pb^{2+} + CrO_4^{2-} \rightarrow PbCrO_4\downarrow}$ Hence, $C = \mathrm{PbCrO_4}$.

4. Match with options

   Therefore, $A = \mathrm{PbS},\quad B = \mathrm{PbSO_4},\quad C = \mathrm{PbCrO_4}$

   This matches Option C.

5. Check other options briefly

   • A: starts with $\mathrm{PbCl_2}$, not consistent with black sulfide formation sequence.
   • B: final product is lead acetate, whereas chromate test gives $\mathrm{PbCrO_4}$.
   • D: contains $\mathrm{Pb(SO_4)_2}$, which is not the usual lead(II) sulfate species here.

So the correct choice is C.