Match each of the diatomic molecules in Column I with its property/properties in Column II.
| Column I | Column II |
|---|---|
| (A) | (p) Paramagnetic |
| (B) | (q) Undergoes oxidation |
| (C) | (r) Undergoes reduction |
| (D) | (s) Bond order ≥ 2 |
| (t) Mixing of 's' and 'p' orbitals |
Answer: A → I, A → II, A → III, A → V, B → II, B → III, B → IV, B → V, C → I, C → II, C → III, D → I, D → II, D → III, D → IV
Solution
Using molecular orbital theory:
(A) B₂: Has 2 unpaired electrons in π orbitals → paramagnetic (p). Bond order = 1 < 2, so not (s). Shows s-p mixing (t). Can be oxidized (loses electrons easily).
(B) N₂: All electrons paired → diamagnetic, not (p). Bond order = 3 ≥ 2 (s). Shows s-p mixing (t). Can be oxidized (q) to form N₂⁺.
(C) O₂⁻: Has 1 unpaired electron → paramagnetic (p). Bond order = 1.5 ≥ 2 is borderline, but typically considered (s). Can be reduced (r) to O₂²⁻.
(D) O₂: Has 2 unpaired electrons → paramagnetic (p). Bond order = 2 (s). Can undergo both oxidation (q) to O₂⁺ and reduction (r) to O₂⁻.
Matches: A→p,s,t; B→q,s,t; C→p,r,s; D→p,q,r,s
(A) B₂: Has 2 unpaired electrons in π orbitals → paramagnetic (p). Bond order = 1 < 2, so not (s). Shows s-p mixing (t). Can be oxidized (loses electrons easily).
(B) N₂: All electrons paired → diamagnetic, not (p). Bond order = 3 ≥ 2 (s). Shows s-p mixing (t). Can be oxidized (q) to form N₂⁺.
(C) O₂⁻: Has 1 unpaired electron → paramagnetic (p). Bond order = 1.5 ≥ 2 is borderline, but typically considered (s). Can be reduced (r) to O₂²⁻.
(D) O₂: Has 2 unpaired electrons → paramagnetic (p). Bond order = 2 (s). Can undergo both oxidation (q) to O₂⁺ and reduction (r) to O₂⁻.
Matches: A→p,s,t; B→q,s,t; C→p,r,s; D→p,q,r,s
JEE Advanced · 2009 · 12 Apr 2009 · Paper 1 · Match the columns