SolnsChap14EVENZum7e

Mo theory can handle odd electron species without any

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Unformatted text preview: andle odd electron species without any modifications. From the MO electron configurations, the bond order is 2.5 for NO and 2 for NO− (see Exercise 14.37d for the electron configuration of NO−). Therefore, NO should have the stronger bond (and it does). In addition, hybrid orbital theory does not predict that NO− is 40. If we calculate a nonzero bond order for a molecule, then we predict that it can exist (is stable). a. H2+: (σ1s)1 H2: (σ1s) 2 − H2 : (σ1s) (σ1s*) B.O. = (1 − 0)/2 = 1/2, stable 2 B.O. = (2 − 0)/2 = 1, stable 1 B.O. = (2 − 1)/2 = 1/2, stable H22−: (σ1s)2(σ1s*)2 B.O. = (2 − 2)/2 = 0, not stable b. N22−: (σ2s)2(σ2s*)2(π2p)4(σ2p)2(π2p*)2 2− 2 2 2 4 O2 : (σ2s) (σ2s*) (σ2p) (π2p) (π2p*) 4 B.O. = (8 − 4)/2 = 2, stable B.O. = (8 − 6)/2 = 1, stable F22−: (σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)4(σ2p*)2 B.O. = (8 − 8)/2 = 0, not stable c. Be2: 2 B2: 2 (σ2s) (σ2s*) (π2p) Ne2: 42. (σ2s)2(σ2s*)2 2 2 B.O. = (2 − 2)/2 = 0, not stable 2 B.O. = (4 − 2)/2 = 1, stable 2 4 (σ2s) (σ2s*) (σ2p) (π2p) (π2p*)4(σ2p*)2B.O. = (8 − 8)/2 = 0, not stable H2: (σ1s)2 B2: (σ2s)2(σ2s*)2(π2p)2 C22−: (σ2s)2(σ2s*)2(π2p)4(σ2p)2 OF: (σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)3 The bond strength will weaken if the electron removed comes from a bonding orbital. Of the molecules listed, H2, B2, and C22− would be expected to have their bond strength weaken as an electron is removed. OF has the electron removed from an antibonding orbital, so its bond strength increases. CHAPTER 14 COVALENT BONDING: ORBITALS 584 44. The electron configurations are (assuming the same orbital order as that for N2): CO: (σ2s)2(σ2s*)2(π2p)4(σ2p)2 B.O. = (8 – 2 )/2 = 3, 0 unpaired e− CO+: (σ2s)2(σ2s*)2(π2p)4(σ2p)1 B.O. = (7 – 2 )/2 = 2.5, 1 unpaired e− CO2+: (σ2s)2(σ2s*)2(π2p)4 B.O. = (6 – 2 )/2 = 2, 0 unpaired e− Because bond order is directly proportional to bond energy and, in turn, inversely proportional to bond length, the correct bond length order...
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