Consider the following reactions. \ PbCl_2+ K_2 CrO_4 -> A+2 KCl\ (Hot solution) \ A+ NaOH B+ Na_2 CrO_4\ PbSO_4+4 CH_3 COONH_4 -> ( NH_4)_2 SO_4+ X In the above reactions, A, B and X are respectively.

underset (Hot solution) PbCl_2+ K_2 CrO_4 -> underset (A) PbCrO_4+2 KCl PbCrO_4+4 NaOH (excess) \ -> Na_2[ Pb( OH)_4]+ Na_2 CrO_4 \ PbSO_4 +4 CH_3 COONH_4 \ -> ( NH_4)_2[ Pb(underset( X)( CH_3 COO)_4]+( NH_4)_2 SO_4.

Consider the following reactions PbCl 2 K 2 CrO 4 arrow... | null (Chemistry)

JEE Mains 2026 21st January Shift 1

Chemistry

Consider the following reactions.

\\ \mathrm{PbCl}_{2}+\mathrm{K}_{2} \mathrm{CrO}_{4} \rightarrow \mathrm{~A}+2 \mathrm{KCl}

\\ (Hot solution)

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

\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,

\mathrm{A}, \mathrm{B}

and X are respectively.

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

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

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

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

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Explanation

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

\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}\\

\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}

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