Chapter 9 Bonding and Molecular Structure: Orbital Hybridization and Molecular Orbitals - 9-3 Molecular Orbital Theory - Review & Check for Section 9-3 - Page 365: 1

B

Nitrogen has 7 electrons in total, oxygen has 8. Configuration: $\sigma_{1s}\ \sigma_{1s}^*\ \sigma_{2s}\ \sigma_{2s}^*\ 3×\sigma_{2p}\ 3×\sigma_{2p}^*$ There are 10 electrons in bonding orbitals and 5 in antibonding so: $B.O.=(N.\ Bonding -N.\ Antibonding)/2$ B.o. = 5/2=2.5