On passing ${CrO}_{2} {Cl}_{2}$ in water & then adding ${({CH}_{3} COO)}_{2} Pb$ , the ppt formed is

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${PbCl}_{2}$

${PbCrO}_{4}$

${CrCl}_{3}$

${Pb}_{2} {CrO}_{4}$

answer is A.

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Detailed Solution

$\begin{array}{l} {CrO}_{2} {Cl}_{2} + 2 H_{2} O \to H_{2} {CrO}_{4} + 2 HCl \\ H_{2} {CrO}_{4} + {({CH}_{3} COO)}_{2} {Pb}_{} \to {\underset{yellow ppt}{PbCrO}}_{} ↓ + 2 {CH}_{3} COOH \end{array}$

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