0243Pb7an-09

0243Pb7an-09 - Answers to Problem 7 Chemistry 0243 2009(a...

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Answers to Problem 7, Chemistry 0243 - 2009 (a) there will be six MOs (b) Be H—H 2 p x 2 y 2 z 2 s ! B ! A (c)The bonding MO of H 2 interacts with 2 to give MOs 1 and 3 . The antibonding MO of H 2 interacts with 2 x to give MOs 2 and 4 . Neither the 2 y nor the 2 z orbital interacts with either the bonding or antibonding MO of H 2 (net zero, orthogonal interactions), and so are unchanged in the calculation. ! B ± 2 ! A ± 2 x 1 3 4 2 (d) Order the MOs by counting the nodes. MO 1 and 2 are net bonding, 3 and 4 are antibonding, and the residual 2 orbitals are nonbonding. Neutral BeH 2 will have four electrons, two from the pair of hydrogens and two from Be. The cation will have one fewer electron and the anion one more electron. Here are the electronic occupancies for BeH 2 , + BeH 2 , and BeH 2 . BeH 2 BeH 2 BeH 2 2 y 2 z 1 3 4 2 Energy 2 y 2 z 1 3 4 2 2 y 2 z 1 3 4 2 + .
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0243Pb7an-09 - Answers to Problem 7 Chemistry 0243 2009(a...

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