A worked solution for the ground state of reads:" has 16 electrons

PHYSICS · Multi-part · Fermi University Physics · Quantum Mechanics · Atomic, Molecular & Optical Physics · Molecular structure and spectra

Deriving the Spectrum, Ch The Born-Oppenheimer separation and the chemical bond

Problem

A worked solution for the ground state of O2\mathrm{O_2} reads:"O2\mathrm{O_2} has 16 electrons. Filling the molecular orbitals in the order σg2s σu∗2s σg2p πu2p πg∗2p σu∗2p\sigma_g2s\ \sigma_u^\ast2s\ \sigma_g2p\ \pi_u2p\ \pi_g^\ast2p\ \sigma_u^\ast2p (the 1s1s core orbitals are filled and cancel, and are ignored), the last two valence electrons available after (σg2s)2(σu∗2s)2(σg2p)2(πu2p)4(\sigma_g2s)^2(\sigma_u^\ast2s)^2(\sigma_g2p)^2(\pi_u2p)^4 both go into the πg,x∗2p\pi_{g,x}^\ast2p orbital, one with spin up and one with spin down, since the Pauli principle allows two electrons of opposite spin in any one orbital. Every orbital is therefore doubly occupied, all electron spins are paired, and O2\mathrm{O_2} should be diamagnetic." ((a)) Find the error in this reasoning. ((b)) Give the correct ground-state configuration of the outer electrons, the resulting term symbol, and say whether O2\mathrm{O_2} is paramagnetic or diamagnetic.

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