Week 8 Key Practice Problems

Week 8 Key Practice Problems - CHEMISTRY 201X Practice...

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CHEMISTRY 201X October 8-12, 2007 Practice Problems (62) The molecular orbital energy diagram for O 2 and F 2 (the one with no 2 s -2 p hybridization) is also appropriate to describe the molecular orbital energies for heteroatomic molecules like CO and NO. (a) Use this diagram, and draw the molecular orbitals for the CO molecule. CO has 10 valence electrons, which fills 5 MOs. In this molecule, O is more electronegative that C, so the bonding MOs have more O character; the antibonding MOs have more C character. σ 2 s σ * 2 s σ * 2 p σ 2 p π 2 p π * 2 p CO (b) Use the diagram to predict the bond orders and numbers of unpaired electrons for BN, CN , NO + , NO, NO , BF, and BO. BN: 3+5 = 8 electrons; ( σ 2s ) 2 ( σ * 2s ) 2 ( σ 2p ) 2 ( π 2p ) 2 ; 2 unpaired electrons; Bond order = (6 2) / 2 = 2 (double bond) CN : 4+5+1 = 10 electrons; ( σ 2s ) 2 ( σ * 2s ) 2 ( σ 2p ) 2 ( π 2p ) 4 ; 0 unpaired electrons; Bond order = (8 2) / 2 = 3 (triple bond) NO + : 5+6 1 = 10 electrons; ( σ 2s ) 2 ( σ * 2s ) 2 ( σ 2p ) 2 ( π 2p ) 4 ; 0 unpaired electrons; Bond order = (8 2) / 2 = 3 (triple bond) NO: 5+6 = 11 electrons; ( σ 2s ) 2 ( σ * 2s ) 2 ( σ 2p ) 2 ( π 2p ) 4 ( π∗ 2p ) 1 ; 1 unpaired electron; Bond order = (8 3) / 2 = 2.5 (between double and triple bond) NO : 5+6+1 = 12 electrons; ( σ 2s ) 2 ( σ * 2s ) 2 ( σ 2p ) 2 ( π 2p ) 4 ( π∗ 2p ) 2 ; 2 unpaired e’s; Bond order = (8 4) / 2 = 2 (double bond) BF: 3+7 = 10 electrons; ( σ 2s ) 2 ( σ * 2s ) 2 ( σ 2p ) 2 ( π 2p ) 4 ; 0 unpaired electrons;
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CHEMISTRY 201X October 8-12, 2007 Practice Problems Bond order = (8 2) / 2 = 3 (triple bond) BO: 3+6 = 9 electrons; ( σ 2s ) 2 ( σ * 2s ) 2 ( σ
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This homework help was uploaded on 03/27/2008 for the course CHEM 201 taught by Professor Miller during the Fall '07 term at Iowa State.

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Week 8 Key Practice Problems - CHEMISTRY 201X Practice...

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