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(C) $PbO_2$ is prepared by the action of nitric acid on red lead $(Pb_3O_4)$.$(a)$ $PbO + 2HCl \longrightarrow PbCl_2 + H_2O$ (This reaction does not

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the reaction of $Pb_3O_4$ with $HNO_3$

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(C) $PbO_2$ is prepared by the action of nitric acid on red lead $(Pb_3O_4)$.$(a)$ $PbO + 2HCl \longrightarrow PbCl_2 + H_2O$ (This reaction does not produce $PbO_2$)$(b)$ $2Pb(NO_3)_2 \xrightarrow{200^{\circ}C} 2PbO + 4NO_2 + O_2$ (This reaction produces $PbO$)$(c)$ $Pb_3O_4 + 4HNO_3 \longrightarrow 2Pb(NO_3)_2 + PbO_2 + 2H_2O$ (This reaction produces $PbO_2$)$(d)$ $Pb +$ air (contains $O_2, H_2O$ and $CO_2$) forms a protective layer of $PbCO_3$ on the surface.

$PbO_2$ is obtained from

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