PbO and `PbO_(2)` react with HCl according to following chemical equations
`2PbO+4HCl to2PbCl_(2)+2H_(2)O`
`PbO_(2)+4HCl toPbCl_(2)+Cl_(2)+2H_(2)O`
Why do these compounds differ n their reactivity?

Explanation

Writing the oxidation numeber of each element above its symbol in the following reactions
(a) `underset("Basic")(2overset(+2)(P)overset(-2)(bO))+underset("Acid")(4overset(+1)(H)overset(-1)(Cl))to2overset(+2)(Pb)overset(-1)(Cl_(2))+2overset(+1)(H_(2))overset(-2)(O)`
In this reation, oxidation number of each element remains same hence, it is not a redox reaction. In fact, it is an example of acid-base reaction.
(b) `overset(+4)(Pb)overset(-2)(O_(2))+4overset(+1)(H)overset(-1)(Cl)tooverset(+2)(Pb)overset(-1)(Cl_(2))+2overset(0)(Cl_(2))+2overset(+1)(H_(2))overset(-2)(O)`
In `PbO_(2)`, Pb is in +4 oxidation state. Due to inert part effect Pb in +2 oxidation state is more stable. So, Pb in +4 oxidation state `(PbO_(2))` acts as an oxidising agent
It oxidises `Cl^(-)` to `Cl_(2)` and itself gets reduced to `Pb^(2+)`.