The element oxidized in the reaction described in pbo2 + 4hcl → 2h2o + pbcl2 + cl2 is a. pb. b. o. c. h. d. cl.

For an element to be oxidized, its oxidation number would have to increase.

An element or compound's oxidation number is mainly dependent on its charge, except in special cases.

In PbO₂, Oxygen's oxidation number is always -2, except in hydrogen peroxide where it is -1. Since PbO₂ is neutral, its total oxidation number must be 0. Since there are two oxygens with an oxidation number of -2, we know that Pb's oxidation number is +4, since +4 -2×2 = 0.
Using the same method we can find Pb's oxidation number in PbCl₂. Unless paired with another halogen, which it isn't, Cl in a compound has an oxidation number of -1, so Pb = +2 since +2 -1×2 = 0.
Pb's oxidation number has gone from +4 to +2, so it has been reduced, so A. Pb is not the answer.

Oxygen in a compound's oxidation number is always -2 unless in H₂O₂ and it isn't on either side of the equation, so its oxidation number doesn't change and it remains in a compound, so B. O is not the answer.

Hydrogen in a compound's oxidation number is always +1, except in hydrides when it is -1, but it isn't, so it doesn't change , so C. H is the answer.

Therefore D. Cl is correct, since its oxidation number increases from -1 in HCl to 0 in Cl2 (since elements not in a compound have an oxidation number of 0)

Hope this helps!

The answer is b metal denting