1. Theories of Chemical Bonding Theories of bonding: explanations for chemical bond, Lewis dot structures and the following. Valance-bond (VB) theory Hybridization of atomic orbitals Multiple covalent bonds Molecular orbital (MO) theory Delocalized electrons Bonding in metals Practice mental reasoning and verbal explanation

2. Energy of Interaction Between Two H Atoms Potential energy distance – 346 kJ mol –1 H – H bond +346 kJ mol –1 antibonding Energies of attraction and repulsion as functions of distance between two H atoms are shown here. The minimum of the attraction force occur at H–H bond length of 74 pm, at which, the antibonding orbital is +346 kJ mole –1 above 0, energy when H atoms are far apart. How does energy affect the two-atom system?

3. The Valence-bond Method Valence bond method considers the covalent bond as a result of overlap of atomic orbitals. Electrons stay in regions between the two atoms. Some bond examples s-s s-p s-d p-p p-d d-d H-H H-C H-Pd C-C Se-F Fe-Fe (?) Li-H H-N in Pd P-P H-F hydride But overlapping of simple atomic orbitals does not explain all the features. Thus, we have to take another look, or do something about atomic orbitals – hybridization . How does valence-bond approach explain the formation of chemical bonds?

4. Hybridization of Atomic Orbitals The solutions of Schrodinger equation led to these atomic orbitals. 1 s , 2 s , 2 p , 3 s , 3 p , 3 d , 4 s , 4 p , 4 d , 4 f , etc. However, overlap of these orbitals does not give a satisfactory explanation. In order to explain bonding, these orbitals are combined to form new set of orbitals – this method is called hybridization . During the lecture, these hybridized orbitals will be explained: s p 2 sp hybrid orbitals from mixing of a s and a p orbital sp 2 3 sp 2 hybrid orbitals from mixing of a s and 2 p orbital s p 3 fill in you explanation please sp 3 d 5 sp 3 d hybrid orbitals from mixing of a s and 3 p and a d orbital sp 3 d 2 ____________ Provide a description for hybrid orbitals sp, sp 2 , sp 3 , sp 3 d, and sp 3 d 2

5. The sp Hybrid Orbitals The sp hybrid orbitals: formation of two sp hybrid orbitals + + + - = + - + – + - = - + hybridization of s and p orbitals = 2 sp hybrid orbitals  _  _ __ __  __ __ __ Two sp hybrid orbitlas => Two states of Be

6. Bonds with sp Hybrid Orbitals Formations of bonds in these molecules are discussed during the lecture. Be prepared to do the same by yourself. Cl–Be–Cl H–C  C–H H–C  N : O=C=O Double and triple bonds involve pi  bonding, and the the application of valence bond method to  bonds will be discussed. You are expected to be able to draw pictures to show the  bonding.

7. A  Bond Overlap of 2 2 p orbitals for the formation of  bond Sigma (  ) bond is symmetric about axis. Pi (  ) electron distribution above and below axis with a nodal plane , on which probability of finding electron is zero;  bond is not as strong as sigma - less overlap. Nodal plane Bonding of C 2 H 4 C 2s 2p 2p 2p sp 2 sp 2 sp 2 2p How are pi bonds formed?

8. Triple Bonds in H-C  C-H H-C-C-H: three  bonds due to overlapping of 1s H – sp C ; sp C – sp C ; and sp C – 1s H . Two  bonds in HC  CH and HC  N triple bonds are due to overlapping of p orbitals results. Draw and describe how atomic orbitals overlap to form all bonds in acetylene, H–C  C–H p y over lap p x over lap H H sp hybrid orbitals Two nodal planes of  bonds are perpendicular to each other. in  bond in  bond C 2s 2p 2p 2p sp sp 2p 2p

9. Two  Bonds in H–C  C–H A triple bond consists of a sigma and two pi bonds. Overlaps of two sets of p orbitals form of two  bonds.

10. Bonding of CO 2 For CO 2 , the C atom forms a  bond and a  bond with each of two O atoms. The two nodal planes of the two  bonds are also perpendicular. During the lecture, I draw diagrams and explain the two  two  bonds in CO 2 . You are expected to be able to do the same, in a test. p y over lap in  bond Overlap p–p in  bonds p x over lap in  bond O=C=O or H 2 C=C=CH 2 Discuss the bonding of allene H 2 C=C=CH 2 See extra problems B17 in the handout Resonance structures : O – C  O : : O  C – O : . .

11. Bonding in CO 2 – another view Compare with H 2 C=C=CH 2

12. The sp 2 Hybrid Orbitals Ground state and excited state electronic configuration of B  _  _  _ __   _ __ __ The hybridization of a s and two p orbitals led to 3 sp 2 hybrid orbitals for bonding. Compounds involving sp2 hybrid orbitals: BF 3 , CO 3 2– , H 2 CO, H 2 C=CH 2 , NO 3 – , etc Nov. 25

13. An example of using sp 2 hybrid orbitals __ orbitals for bonding? Dipole moment = ____?

14. Bonding of H 2 C=CH 2 molecules Utilizing the sp 2 hybrid orbitals, each C atom form two H–C  bonds for a total of 4  H–C bonds. The C–C  bond is common to both C atoms. A C–C  bond is formed due to overlap of p orbitals from each of the C atoms. Hybrid orbitals ( sp 2 ) for H–C and C–C  bond Overlap of p orbital for C–C  bond C 2s 2p 2p 2p sp 2 sp 2 sp 2 2p

15. The sp 3 Hybridized Orbitals Ground state and excited state electronic configuration of C  _  _  _  _   _  _ __ The hybridization of a s and three p orbitals led to 4 sp 3 hybrid orbitals for bonding. Compounds involving sp 3 hybrid orbitals: CF 4 , CH 4 , : NH 3 , H 2 O :: , SiO 4 4– , SO 4 2– , ClO 4 – , etc

16. C 2s 2p 2p 2p sp 3 sp 3 sp 3 sp 3

17. The sp 3 d Hybrid Orbitals Hybridization of one s , three p , and a d orbitals results in 5 sp 3 d hybrid orbitals . The arrangement of these orbitals is a trigonal pyramid. Some structures due to these type of orbitals are PClF 4 , TeCl 4 E, and BrF 3 E 2 . How many unshared electron pairs are present in TeCl 4 and BrF 3 ? What are their shapes?

18. The sp 3 d 2 Hybrid Orbitals Hybridization of one s , three p , and two d orbitals results in 6 sp 3 d 2 hybrid orbitals . The arrangement of these orbitals is an octahedron. Compounds using these type of orbitals are shown here. AX 6 , AX 5 E, AX 4 E 2 AX 3 E 3 and AX 2 E 4 IOF 5 , IF 5 E, XeF 4 E 2 No known compounds of AX 3 E 3 and AX 2 E 4 are known or recognized, because they are predicted to have a T shape and linear shape respectively when the lone pairs of electrons are ignored.

19. Molecules with more than one central atom Describe the structure of CH 3 NCO. Draw the skeleton and add all valence electrons H 3 C – N – C – O Which Lewis dot structure is the most important (stable)? N = C = O H–C H H 120 o 109 o 180 o What hybridized orbitals are used for bonding in N and C? Why are the bond angles as indicated? No of  and  bonds = __, __? Give formal charges to all atoms in all structures. Take a new look at slide 22 in Bonding Basics Which structure is more stable, and why? N  C–O H–C H H N–C  O H–C H H

20. Why Molecular Orbital (MO) Theory Lewis dot and valence bond theories do not always give satisfactory account for various properties of molecules. For example, the dot and VB theory does not explain the fact that O 2 is paramagnetic and has a double bond. Dot and VB structures : O O : • O O • are unsatisfactory. MO theory, different from VB in that MO theory considers the orbitals of the whole molecules. However the approach of linear-combination-of-atomic-orbitals ( LCAO ) is usually used. There are other reasons, but it’s human nature to theorize. The theory is beautiful, and worth learning or teaching .

21. The Molecular Orbital (MO) Theory The two atoms in the H 2 molecule may be represented by A and B. Their s orbitals 1s A and 1s B respectively, are used for two MOs:  * = 1s A – 1s B  = 1s A + 1s B The energy levels of these AO and MO are represented by the diagram here , with the math hidden . For a molecule, there are certain orbitals each of which accommodates two electrons of opposite spin. The MO theory combines atomic orbitals (AO) to form MOs, & this method is called LCAO 1s A 1s B  MO  AO AO

22. MO for H 2 –type molecules: H 2 + , H 2 , H 2 – , He 2 + Generalize the technique of LCAO

23. Electronic configuration,  2 , for H 2 molecules Generalize the technique of LCAO

24. Electronic Configuration of H 2 -type Molecules From the previous theory, we can fill the M Os with electrons for the H 2 -type molecule: Molecule e-configuration Bond order bondlength H 2 + 1  (1  1 ) ½ 106 pm H 2 , He 2 2+ 1  2 1 74, ~75 H 2 – , He 2 + 1  2 1  ½ ~106, 108  H 2 2– , He 2 1  2 1  2 0 not formed Describe the relationships of bondlength & bondorder and e-configurations; learn to reason

25. Sigma MOs Formed Using p AOs Sigma MOs (  2p  2p * ) can be formed using p AOs, similar to VB theory . The gain in bonding orbital  2p (lower energy) is at the expense of the anti-bonding orbital  2p * (higher energy) Generalize the technique of LCAO

26. Pi  MOs from p AOs Generalize the technique of LCAO

27. MO Diagrams for O 2 and F 2 A full diagram of the energy level of molecular orbitals of O 2 and F 2 is shown here. The relative (approximate) height of these energies will be explained verbally during the lecture, and you are suppose to be able to do the same. Write the electronic configurations for O 2 , O 2 – , F 2 , F 2 – & Ne 2 .

28. The O 2 + , O 2 , O 2 – , & F 2 + , F 2 , F 2 – Molecules  O=O  Paramagnetic , bond length indicates double bond, electronic configuration agrees F–F Electronic configuration agree with single bond. For  O=O   _  _ See p. 457 for two MO energy-level diagrams

29. MO Energy Level Diagram for Be 2 – N 2 Due to close energy levels of 2 s and 2 p , the MO energy level diagram for Be 2 to N 2 differs from those of O 2 to F 2 . Reasons and explanation are given during the lecture. Hope you can do the same. Give electronic configurations for Be 2 + , Be 2 – , B 2 + , B 2 , B 2 – , C 2 + , C 2 , C 2 – , N 2 + , N 2 , N 2 – . See p. 457 for two MO energy-level diagrams

30. A more realistic energy level diagram for Be 2 – N 2 involving sp mixing, not in text _ _ _ 2 p _ 2 s Atomic orbital __  * 2p __ __  * 2p __  2p __ __  2p __  * 2s __  2s Molecular orbitals _ _ _ 2 p _ 2 s Atomic orbital This diagram from my CaCt website accounts for the sp mixing of the AO for the bonding consideration. This sp mixing effect is more detailed than that required for freshman chemistry (not to be tested). The sp mixing of AO gives stronger  2s bond and a weaker  2s bond. Thus, the split from 2s is not even. Effects on other bonds are also shown, but qualitatively.

31. Benzene The benzene structure has fascinated scientists for centuries. It’s bonding is particularly interesting. The C atom utilizes sp2 hybrid AO in the sigma bonds, and the remaining p AO overlap forming a ring of p bonds. Sigma  bonds are represented by lines, and the p orbitals for the  bonds are shown by balloon-shape blobs. Note the + and – signs of the p orbitals. Thus, we represent it by + – + + + – – – + +

32. More About Benzene Chem120 students may ignore this slide. The  and  * of C 6 H 6 are shown here; the symmetry is also interesting.

33. Delocalized electrons in Benzene and Ozone When p bonds are adjacent to each other or separated by on single C-C bonds, the p bonding electrons are delocalized. The delocalized electron path for benzene and ozone are shown here. These pictures represent p electron of the the structures formula contribute most to their structures. O O O CO 3 2– & have delocalized electrons

34. Joy on structure of benzene – a story Kathleen Londsdale (1903-1971) came from a very poor family in Kildare who moved to Essex when she was five. She studied physics, math and chemistry at school and went to college in London when she was 16, where she did extremely well. She was offered a place in the research team of William Bragg, so starting her life's work on X-ray crystallography. In 1929 she showed, by her analysis of hexamethyl benzene, that the benzene ring was flat. Benzene has fascinated scientists, and its precise structure was a matter of controversy till then. In 1945 she was elected the first ever woman Fellow of the Royal Society. She was also created a Dame Commander of the Order of the Brittish Empire in 1956.

35. MO for Heteronuclear Diatomic Molecules For heteronuclear diatomic molecules, the atomic orbitals are at different energy levels. Thus, the MO shifts are different from those of homonuclear diatomic molecules. The interactions of AO for MO for HF, LiF are similar, and explained in lectures. Chem120 students may ignore this slide.

36. Review Explain the bonding and anti-bonding orbitals, with a picture if possible Describe a  and a  bond, picture may be used. Give the electronic configurations and bond orders for N 2 , N 2 + , N 2 2+ , N 2 – , N 2 2– , O 2 , Draw the Lewis dot structure for ozone. Describe the molecular shape and justify for it. Explain the delocalized electrons of ozone. Use the MO theory to explain the fact that O 2 is paramagnetic , and has a double bond. A diagram of the MO energy levels will help. Give a few compounds that have the same number of electrons as O 3 .