MCQ
A red coloured mixed oxide $(X)$ on treatment with cone. $HNO_3$ gives a compound $(Y).$ $(Y)$ with $HCl ,$ produces a chloride compound $(Z)$ which can also be produced by treating $(X)$ with cone. $HCl.$ Compounds $(X) , (Y),$ and $(Z)$ will be
  • A
    $Mn_3O_4, MnO_2, MnCl_2$
  • $Pb_3O_4, PbO_2, PbCl_2$
  • C
    $Fe_3O_4, Fe_2O_3, FeCl_2$
  • D
    $Fe_3O_4, Fe_2O_3, FeCl_3$

Answer

Correct option: B.
$Pb_3O_4, PbO_2, PbCl_2$
b
$\mathrm{Pb}_{3} \mathrm{O}_{4}+4 \mathrm{HN} \mathrm{O}_{3} \rightarrow 2 \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}+\mathrm{P} \mathrm{bO}_{2}+2 \mathrm{H}_{2} \mathrm{O}$

$\mathrm{PbO}_{2}+\mathrm{HCl} \rightarrow \mathrm{PbCl}_{2}+\mathrm{H}_{2} \mathrm{O}+\mathrm{Cl}_{2}$

$\mathrm{P} \mathrm{b}_{3} \mathrm{O}_{4}+8 \mathrm{HCl} \rightarrow 4 \mathrm{H}_{2} \mathrm{O}+3 \mathrm{P} \mathrm{bCl}_{2}+\mathrm{Cl}_{2}$

$\mathrm{X}, \mathrm{Y}$ and $\mathrm{Z}$ are $\mathrm{Pb}_{3} \mathrm{O}_{4}, \mathrm{P} \mathrm{bO}_{2}, \mathrm{P} \mathrm{bCl}_{2}$ respectively.

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