Why do the following reactions proceed differently? `Pb_(3)O_(4)+8HClrarr3PbCl_(2)+Cl_(2)+4H_(2)O` and `Pb_(3)O_(3)+4HNO_(3)rarr 2Pb(NO_(3))_(2)+PbO_(2)+2H_(2)O`

Explanation

`Pb_(3)O_(4)` is actually a stoichiometric mixture of 2 mol of PbO and 1 mol of `PbO_(2)`. In `PbO_(2)`, lead is present in +4 oxidation state, whereas the stable oxidation state of lead in PbO is +2. `PbO_(2)` thus can act as an oxidant (oxidising agent) and, therefore, can oxidise `Cl^(-)` ion of HCl into chlorine. We may also keep in mind that PbO is a basic oxide. Therefore, the reaction
`Pb_(3)O_(4) + 8 HCl rarr 3 PbCl_(2) + Cl_(2) + 4H_(2)O` can be splitted into two reactions namely :
`2PbO + 4HCl rarr 2PbCl_(2) + 2H_(2)O` (acid-base reaction)
`overset(+4)(Pb)O_(2) + 4 overset(-1)(HCl) rarr overset(+2)(PbCl_(2)) + overset(0)(Cl_(2)) + 2H_(2)O` (redox reaction)
Since `HNO_(3)` itself is an oxidising agent therefore, it is unlikely that the reaction may occur between `PbO_(2) and HNO_(3)`. However. the acid-base reaction occurs between PbO and `IINO_(3)` as :
`2PbO + 4HNO_(3) rarr 2Pb (NO_(3))_(2) + 2H_(2)O`
It is the passive nature of `PbO_(2)` against `HNO_(3)` that makes the reaction different from the one that follows with HCl.