For a hypothetical N₂ molecule suppress s–p mixing completely, so σ₂ₚ lies below the degenerate π₂ₚ pair. Fill all orbitals using Pauli and Hund rules, including the 1s core. Assume only the ordinary

Question

For a hypothetical N₂ molecule suppress s–p mixing completely, so σ₂ₚ lies below the degenerate π₂ₚ pair. Fill all orbitals using Pauli and Hund rules, including the 1s core. Assume only the ordinary one-electron electric-dipole inversion rule is being tested, so g↔u is allowed and g↔g or u↔u is forbidden. Consider I: HOMO→LUMO; II: σ₂ₛ→LUMO; III: σ*₂ₛ→LUMO; IV: HOMO→σ*₂ₚ. Which list contains exactly the inversion-allowed promotions?

TestHub · Accepted Answer

With the specified ordering N₂ has filled σ₂ₚ and π₂ₚ, so its HOMO is πᵤ and LUMO is π* g . Hence I is u→g and allowed. The σ₂ₛ orbital is g, so II is g→g and forbidden; σ*₂ₛ is u, making III u→g allowed. Finally σ*₂ₚ is u, so IV is u→u and forbidden. Therefore the allowed list is I and III, option C.

Chemistry - CHEMICAL BONDING Question with Solution | TestHub

Options:

Doubts & Discussion