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CHEMICAL BONDING II

CHEMICAL BONDING II. MOLECULAR ORBITAL THEORY II. C 2 + MOLECULE. NO MOLECULE. BOND ORDER. Bond order = number of bonding e- minus number of antibonding e- 2 If the bond order is zero → no bond! divide by two because of pairs of electrons.

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CHEMICAL BONDING II

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  1. CHEMICAL BONDING II MOLECULAR ORBITAL THEORY II

  2. C2+ MOLECULE

  3. NO MOLECULE

  4. BOND ORDER • Bond order = number of bonding e- minus number of antibonding e- 2 • If the bond order is zero→ no bond! • divide by two because of pairs of electrons. • Bond order is an indication of strength. • Larger bond order means greater bond strength. • H2 molecule: Bond order = (2-0)/2 = 1 and He2: Bond order = (2-2)/2 = 0 This implies it is not stable.

  5. BOND ORDER • Calculate the Bond order for F2, C2+, and NO

  6. BOND ORDER • F: (10-8)/2 = 1 • C : (7-4)/2 = 1.5 • NO: (10-5)/2 = 2.5

  7. HOMONUCLEAR DIATOMIC MOLECULES • those composed of two identical atoms. • Li2 Bond order = (2-0)/2 = 1. • Li2 is a stable molecule. • Be2 Bond order = (2-2)/2 = 0. This is not more stable than two Be atoms, so no molecule forms.

  8. FILLING THE DIAGRAM • It gets slightly more complicated when we leave Be and move to 2p. • The filling order for p’s is pi, pi, sigma all bonding, followed by pi, pi, sigma all antibonding. • Hund's Rule and the Pauli Exclusion Principle still apply.

  9. General Energy Level Sequence for Filling Orbitals Using the MO Theory • σ1s2σ1s2* σ2s2 σ2s2* Π2pxy4 σ2p2 π2pxy4* σ2p2*

  10. MOLECULAR ORBITAL CONFIGURATIONS • For period 2 diatomic molecules up to and including N2: • σ1s2 σ1s2* σ2s2 σ2s2* π2pxy4 σ2p2 π2pxy4* σ2p2* • For period 2 diatomic molecules O2, F2, and Ne2 (hypothetical): π2pand σ2pchange order. • σ1s2 σ1s2* σ2s2 σ2s2* σ2p2π2pxy4π2pxy4* σ2p2*

  11. MOLECULAR ORBITAL CONFIGURATIONS • Write the molecular orbital configurations for F2, C2+, and NO.

  12. MOLECULAR ORBITAL CONFIGURATIONS • F2: σ1s2 σ1s2* σ2s2 σ2s2* σ2p2 π2pxy4 π2pxy4* • C2+ : σ1s2 σ1s2* σ2s2 σ2s2* σ2p2 π2pxy3 • NO: σ1s2 σ1s2* σ2s2 σ2s2* σ2p2π2pxy4π2pxy1*

  13. PRACTICE SIX • Predict the bonding of B2.

  14. PRACTICE SIX - ANSWER • B – 1s22s22p1 B B

  15. PARAMAGNETISM • One of the most useful parts of this model is its ability to accurately predict paramagnetism and diamagnetism as well as bond order. • B2 is paramagnetic. That means that the pi orbitals are of LOWER energy than the sigmas and Hund’s rule demands that the 2 electrons fill the 2 bonding pi orbitals singly first before paring.

  16. PRACTICE SEVEN • Write the appropriate energy diagram using the MO theory for the nitrogen molecule. Find the bond order for the molecule and indicate whether this substance is paramagnetic or diamagnetic.

  17. PRACTICE SEVEN • N2 electron configuration: • 1s22s22p3 • Bond order = (10-4)/2 = 6 • diamagnetic N N

  18. PRACTICE EIGHT • For the species O2, O2+, O2-, give the electron configuration and the bond order for each. Which has the strongest bond?

  19. PRACTICE EIGHT • O2 • σ1s2σ1s2* σ2s2 σ2s2* σ2p2 • π2pxy4 π2pxy2* • Bond order = (8-4)/2 = 2

  20. PRACTICE EIGHT

  21. PRACTICE EIGHT • O2- • σ1s2σ1s2* σ2s2 σ2s2* σ2p2 • π2pxy4 π2pxy3* • Bond order = (8-5)/2 = 1.5 • O2+ • σ1s2σ1s2* σ2s2 σ2s2* σ2p2 • π2pxy4 π2pxy1* • Bond order = (8-3)/2 = 2.5

  22. PRACTICE NINE • Use the molecular orbital model to predict the bond order and magnetism of each of the following molecules. • Ne2 P2

  23. PRACTICE NINE • Ne2 • Bond order = (8 - 8)/2 = 0 • Does not exist

  24. PRACTICE NINE • P2 • Bond order = (18 - 12)/2 = 3 • diamagnetic

  25. PRACTICE TEN • Use the MO Model to predict the magnetism and bond order of the NO+ and CN- ions.

  26. PRACTICE TEN • Each has the same number of valence electrons = 10 and total electrons = 14 • σ1s2 σ1s2* σ2s2 σ2s2* π2pxy4 σ2p2 • Bond order = (10-4)/2 = 3 • Both are diamagnetic.

  27. The Molecular Orbtial Energy-Level Diagrams, Bond Orders, Bond Energies, and Bond Lengths for the Diatomic Molecules B2 Through F2

  28. COMBINING LE AND MO MODELS • LE model assumes that electrons are localized. This is not the case for some molecules. This is apparent with molecules for which we can draw several valid Lewis structures. So resonance was constructed. • The best model is one with the simplicity of the LE model with the delocalization characteristics of the MO model. So we combine these two models to describe molecules that require resonance. In O3 and NO3-1, the double bond changes position in the resonance structure.

  29. COMBINING LE AND MO MODELS • Since a double bond requires one σ and one π bond, there is a σ bond between all bound atoms in each resonance structure. • It is the π bond that has different locations. • Conclusion: σ bonds in a molecule can be described as being localized with no apparent problems, but the π bonding must be treated as delocalized. So use LE to describe the σ bonds in structures with resonance, but use MO to describe π bonds in these same structures.

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