JEE Chemistry 2026 Question with Solution

Given: The reactions are:

$$PbCl_2 + K_2CrO_4 \rightarrow PbCrO_4 + 2KCl$$ $$PbCrO_4 + 4NaOH \rightarrow Na_2[Pb(OH)_4] + Na_2CrO_4$$ $$PbSO_4 + 4CH_3COONH_4 \rightarrow (NH_4)_2[Pb(CH_3COO)_4] + (NH_4)_2SO_4$$

Find: The identities of $A$, $B$ and $X$.

From the first reaction, lead chloride reacts with potassium chromate to form yellow precipitate lead chromate. Therefore, $A = PbCrO_4$.

From the second reaction, lead chromate reacts with excess sodium hydroxide to form soluble plumbite complex. Therefore, $B = Na_2[Pb(OH)_4]$.

From the third reaction, lead sulfate dissolves in ammonium acetate forming soluble lead acetate complex. Therefore, $X = (NH_4)_2[Pb(CH_3COO)_4]$.

Thus,

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

Therefore, the correct option is D.

Given: Three qualitative analysis reactions involving lead salts.

Find: The correct sequence of $A$, $B$ and $X$.

Step 1: Reaction 1 Lead chloride reacts with chromate ion to form lead chromate.

$$PbCl_2 + K_2CrO_4 \rightarrow PbCrO_4 \downarrow + 2KCl$$

So, $A = PbCrO_4$.

Step 2: Reaction 2 Lead chromate reacts with excess sodium hydroxide to form soluble hydroxo complex.

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

So, $B = Na_2[Pb(OH)_4]$.

Step 3: Reaction 3 Lead sulfate is soluble in ammonium acetate. It forms the soluble lead acetate complex.

$$PbSO_4 + 4CH_3COONH_4 \rightarrow (NH_4)_2[Pb(CH_3COO)_4] + (NH_4)_2SO_4$$

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

Hence the sequence is $PbCrO_4$, $Na_2[Pb(OH)_4]$ and $(NH_4)_2[Pb(CH_3COO)_4]$. Therefore, the correct option is D.