Question

given PbS,So32-,H2S, i2 moist, co2, so42-, number of entities ocidised by O3=p Na2SO3+dilHCl= on heating gives gas Q find P and Q

Answer 1

P = 3 and Q = SO₂

Answer 2

Solution:

We note that among the given substances, the ones having sulfur in a “reduced” state (i.e. liable to further oxidation) are:

• PbS: Here S is in the –II state.
• H₂S: Sulfur is –II.
• S₂O₃²⁻ (thiosulfate): Although it contains two S atoms (one in a higher and one in a lower oxidation state), the S(–I) centre is oxidizable to sulfate.

Ozone (O₃) being a strong oxidant will oxidize these three sulfur–species. Thus we take
  P = 3.

Next, consider the reaction of sodium sulfite with dilute HCl. The reaction is

$$\ce{Na2SO3 + 2HCl -> Na2SO4 + SO2 ^ + H2O}$$

This shows that acidification of Na₂SO₃ liberates sulfur dioxide gas. Hence,

  Q = SO₂.

---

Core Minimal Explanation:

1. Oxidation by O₃: PbS, H₂S, and S₂O₃²⁻ (via its S(–I) centre) are oxidized → Total = 3; hence P = 3.
2. Reaction of Na₂SO₃ + Dil. HCl: Produces SO₂ gas → Q = SO₂.