Q1(b). Bond order and ionisation

Answer in brief

Both changes lower bond order by 1/2: O₂: 2→1.5; N₂: 3→2.5.

Key points

• Use BO=1/2(N_b-Nₐ). The filled 1s core orbitals cancel and are omitted from the sketch.
• O₂ has eight bonding and four antibonding valence electrons. Adding an electron to π*₂ₚ gives (8-5)/2=1.5.
• N₂ has eight bonding and two antibonding valence electrons. Removal from bonding σ₂ₚ gives (7-2)/2=2.5.
• The σ₂ₚ level lies below π₂ₚ in oxygen but above it in nitrogen. Both ionisations therefore weaken and lengthen the bond.