0243Pb8an-09

0243Pb8an-09 - y B + 2p y B 2p y 2p x C + 2p x C 2p x three...

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Answers to Problem 8, Chemistry 0243 - 2009 (a) Just bend the MOs of the linear compound to produce A --> D , the MO s of square H 4 . They can be ordered in energy by counting nodes. A B C D (b) A B C Atomic Orbitals of Carbon 2s 2p y 2p x 2p z interacts with does not interact - all "interactions" are net zero interacts with interacts with
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(c) There will be eight MO s for planar methane. A C B does not interact - all "interactions" are net zero ± 2s A + 2s A – 2s ± 2p
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Unformatted text preview: y B + 2p y B 2p y 2p x C + 2p x C 2p x three bonding three antibonding one nonbonding antibonding D does not interact either (d,e) just count nodes. A + 2s has none, B + 2p x and C + 2p y have one each and will be at the same energy. The nonbonding orbital must lie above B +2p x and C + 2p y . There are eight electrons (four from the four hydrogens and four from carbon) and they will go in to the orbitals as shown. A + 2s B + 2p y C + 2p x E nonbonding...
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0243Pb8an-09 - y B + 2p y B 2p y 2p x C + 2p x C 2p x three...

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