Question 5

**Analysis of each statement (using the standard MO diagram for homonuclear diatomics):** **(A)** Ne₂: 16 valence electrons → σ2s² σ*2s² σ2p² π2p⁴ π*2p⁴ σ*2p². Bond order = (8 − 8)/2 = 0. **Correct statement.** **(B)** F₂: 14 valence electrons → σ2s² σ*2s² σ2p² π2p⁴ π*2p⁴. The highest occupied MO is the π*2p (antibonding π) orbital, not σ-type. **Incorrect statement.** **(C)** O₂ has bond order 2 (σ2s² σ*2s² σ2p² π2p⁴ π*2p²). Removing an electron from the antibonding π* orbital gives O₂⁺ with bond order 2.5. Higher bond order → stronger bond, so bond energy of O₂⁺ is *greater* than that of O₂. **Incorrect statement.** **(D)** Li₂ (bond order 1, large Li atoms, ~267 pm) vs B₂ (bond order 1, smaller B atoms, ~159 pm). Li₂ indeed has the larger bond length. **Correct statement.** The INCORRECT statements are (B) and (C).

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