Consider the following reactions.
$$PbCl_{2}+K_{2}CrO_{4}\rightarrow A+2KCI$$
(Hot solution)
$$A+NaOH\rightleftharpoons B+Na_{2}CrO_{4}$$
$$PbSO_{4}+4CH_{3}COONH_{4}\rightarrow (NH_{4})_{2}SO_{4}+X$$
In the above reactions, A, Band X are respectively.

The salts of lead(II) often give characteristic precipitates that can dissolve in excess alkali or in the presence of complex-forming anions. We shall analyse each reaction one by one.

Case 1: $$PbCl_2 + K_2CrO_4 \;(\text{hot}) \rightarrow A + 2KCl$$

Potassium chromate supplies the $$CrO_4^{2-}$$ ion. Lead(II) chloride reacts with this anion to give the sparingly soluble, yellow precipitate lead(II) chromate:

$$Pb^{2+} + CrO_4^{2-} \rightarrow PbCrO_4 \downarrow$$

Therefore, $$A = PbCrO_4$$.

Case 2: $$A + NaOH \rightleftharpoons B + Na_2CrO_4$$

Substituting $$A = PbCrO_4$$ gives

$$PbCrO_4 + 4\,NaOH \rightleftharpoons Na_2[Pb(OH)_4] + Na_2CrO_4$$

Here the chromate ion is set free as soluble $$Na_2CrO_4$$, while lead(II) passes into solution as the plumbite complex:

$$B = Na_2[Pb(OH)_4]$$ (sodium plumbite).

Case 3: $$PbSO_4 + 4\,CH_3COONH_4 \rightarrow (NH_4)_2SO_4 + X$$

Ammonium acetate supplies the acetate ion $$CH_3COO^-$$ and the ammonium ion $$NH_4^+$$. Lead(II) sulfate is only sparingly soluble, but in the presence of excess acetate it forms the tetra-acetato plumbate(II) complex. The balanced equation is

$$PbSO_4 + 4\,NH_4CH_3COO \rightarrow (NH_4)_2SO_4 + (NH_4)_2[Pb(CH_3COO)_4]$$

Hence

$$X = (NH_4)_2[Pb(CH_3COO)_4]$$.

Collecting the results:

 $$A = PbCrO_4$$,  $$B = Na_2[Pb(OH)_4]$$,  $$X = (NH_4)_2[Pb(CH_3COO)_4]$$.

These species correspond to Option B.