Chemistry · Alkali Metals - Group 1

JEE Main 2026 — 21 January, Morning Shift — Question 48

Consider the following reactions.

PbCl2+K2CrO4→ A+2KCl\mathrm{PbCl}_{2}+\mathrm{K}_{2} \mathrm{CrO}_{4} \rightarrow \mathrm{~A}+2 \mathrm{KCl} (Hot solution)

A+NaOH⇌B+Na2CrO4\mathrm{A}+\mathrm{NaOH} \rightleftharpoons \mathrm{B}+\mathrm{Na}_{2} \mathrm{CrO}_{4}

PbSO4+4CH3COONH4→(NH4)2SO4+X\mathrm{PbSO}_{4}+4 \mathrm{CH}_{3} \mathrm{COONH}_{4} \rightarrow\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}+\mathrm{X}

In the above reactions, A,B\mathrm{A}, \mathrm{B} and X are respectively.

  1. Option A:

    Na2[ Pb(OH)2],PbCrO4\mathrm{Na}_{2}\left[\mathrm{~Pb}(\mathrm{OH})_{2}\right], \mathrm{PbCrO}_{4} and (NH4)2[ Pb(CH3COO)4]\left(\mathrm{NH}_{4}\right)_{2}\left[\mathrm{~Pb}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{4}\right]

  2. Option B:

    PbCrO4,Na2[ Pb(OH)4]\mathrm{PbCrO}_{4}, \mathrm{Na}_{2}\left[\mathrm{~Pb}(\mathrm{OH})_{4}\right] and [Pb(NH3)4]SO4\left[\mathrm{Pb}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{SO}_{4}

  3. Option C:

    Na2[ Pb(OH)2],PbCrO4\mathrm{Na}_{2}\left[\mathrm{~Pb}(\mathrm{OH})_{2}\right], \mathrm{PbCrO}_{4} and [Pb(NH3)4]SO4\left[\mathrm{Pb}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{SO}_{4}

  4. Option D:

    PbCrO4,Na2[ Pb(OH)4]\mathrm{PbCrO}_{4}, \mathrm{Na}_{2}\left[\mathrm{~Pb}(\mathrm{OH})_{4}\right] and (NH4)2[ Pb(CH3COO)4]\left(\mathrm{NH}_{4}\right)_{2}\left[\mathrm{~Pb}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{4}\right]

    Correct

Answer: D

Step-by-step solution

PbCl2 (Hot solution) +K2CrO4→PbCrO4 (A) +2KCl\underset{\text { (Hot solution) }}{\mathrm{PbCl}_{2}}+\mathrm{K}_{2} \mathrm{CrO}_{4} \rightarrow \underset{\text { (A) }}{\mathrm{PbCrO}_{4}}+2 \mathrm{KCl}

PbCrO4+4NaOH (excess) →Na2[ Pb(OH)4]+Na2CrO4\begin{aligned} \mathrm{PbCrO}_{4}+4 \mathrm{NaOH} & \text { (excess) } & \rightarrow \mathrm{Na}_{2}\left[\mathrm{~Pb}(\mathrm{OH})_{4}\right]+\mathrm{Na}_{2} \mathrm{CrO}_{4} \end{aligned} PbSO4+4CH3COONH4→(NH4)2[ Pb((CH3COO)4(X)]+(NH4)2SO4\begin{aligned} \mathrm{PbSO}_{4} & +4 \mathrm{CH}_{3} \mathrm{COONH}_{4} & \rightarrow\left(\mathrm{NH}_{4}\right)_{2}\left[\mathrm{~Pb}\left(\underset{(\mathrm{X})}{\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{4}}\right]+\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\right. \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Subject
Chemistry
Chapter
Alkali Metals - Group 1
Topic
Compounds of Sodium