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Problem set 3 Crib 2011

Problem set 3 Crib 2011 - Chapter 2 Exercises 2.3b 2.15(C2...

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Chapter 2 : Exercises 2.3b, 2.15(C2+), 2.19b, 2.21a-b 2.3b Square Planar 2.15 C 2 - 1 g 2 1 u 2 1 u 4 2 g 1 HOMO is 2 g Bonding sigma orbital C 2 + 1 g 2 1 u 2 1 u 3 HOMO is 1 u Bonding Pi orbitals 2.19b N 2 + e - adds an electron into an antibonding orbital. This decreases the bond order from 3 to 2.5 and therefore will increase the bond distance.
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2.21a-b I. Rationalize the bond length order O 2 + < O 2 < O 2 - with an MO scheme, and predict the magnetic properties of each species (paramagnetic or diamagnetic). Use Figure 2.17 to see the MO scheme of O 2 . The cation O 2 + has 11 electrons and a bond order of 2.5. O 2 has 12 electrons and a bond order of 2. The anion O 2 - has 13 electrons and a bond order of 1.5. The lengths of the bonds have the order of O 2 + < O 2 < O 2 - because the bonds are getting weaker as electrons are added and as a result longer. All the complexes are paramagnetic as they have at least 1 unpaired electron.
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