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9 Molecular Geometry Solutions to Exercises difference between heteroatomics increases the energy difference between the 2s AO on one atom and the 2p AO on the other, rendering the "no interaction" MO diagram in Figure 9.43 appropriate. Solve. CN: B.O. = (7 - 2) / 2 = 2.5 CN+: B.O. = (6 2) / 2 = 2.0 CN-: B.O. = (8 2) / 2 = 3.0 σ₂ₛ (a) CN⁻ has the highest bond order and therefore the strongest C-N bond. (b) CN and CN+. CN has an odd number of valence electrons, so it must have an unpaired electron. The electron configuration for CN is shown in the diagram. Removing one electron from the π2p MOs to form CN+ produces an ion with two unpaired electrons. Adding one electron to the π2p MOs of CN to form CN⁻ produces an ion with all electrons paired. 9.84 (a) The bond order of NO is [1/2 (8 3)] = 2.5. The electron that is lost is in an antibonding molecular orbital, so the bond order in NO+ is 3.0. The increase in bond order is the driving force for the formation of NO+. (b) To form NO-, an electron is added to an antibonding orbital, and the new bond order is [1/2 (8 4)] = 2. The order of increasing bond order and bond strength is: