The correct decreasing order of oxidation state of the underlined atom in each molecule is:
- A.₄O₁₀ > ₃ > H₂
- B.₂O₅ > ₂O₃ > H₂✓
- C.₂ > ₂O₃ > ₃
- D.₄O₆ > ₂O₇ > ₃
Correct Answer
(B) ₂O₅ > ₂O₃ > H₂
Solution & Explanation
Concept: assign the oxidation state of the underlined atom in each species (overall charge = sum of oxidation states; O = −2, H = +1 in these neutral molecules), then compare. The correct option is (2): N₂O₅ > Al₂O₃ > H₂S, with the underlined atoms N, Al and S. - N in N₂O₅: 2N + 5(−2) = 0 → 2N = +10 → N = +5. - Al in Al₂O₃: 2Al + 3(−2) = 0 → 2Al = +6 → Al = +3. - S in H₂S: 2(+1) + S = 0 → S = −2. Order: +5 > +3 > −2, a correct decreasing order. ✓ Why the others fail: - (1) underlined S in SO₃ = +6 would exceed P (+5 in P₄O₁₀), so the order written is wrong. - (3) underlined S in SO₂ = +4 is not the largest as claimed when ranked against the others. - (4) underlined Cl in Cl₂O₇ = +7 exceeds P, breaking the stated order. Hence the answer is B — N₂O₅ > Al₂O₃ > H₂S.
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