1 Covalent Bonding: Orbitals Chapter 09

2 The four bonds around C are of equal length and Energy

3 Can you explain this based on your knowledge of electron energy levels? 6 C 1s 2 2s 2 2p 2 Bonding from s is Different from bonding From p. In addition the Angles should be within 90 0 !

4 How to generate four equal orbitals?

5 Hint: A key to wave mechanics is superposition Which is creating new waves from interference of old ones

6 Let’s do some mixing

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10 Cross section of an sp 3 orbital.

11 An energy-level diagram showing the formation of four sp 3 orbitals.

12 Ground state of C1s 2 2s 2 2p 2 Promote electron at n=22s 1 2p 3 Hybridize at n=2sp 3 sp 3 sp 3 sp 3 s pxpx pypy pzpz spxpx pypy pzpz Hybridization sp 3 sp 3 4sp 3 orbitals of equal length, energy and in tetrahedral shape

13 Hybridization The mixing of atomic orbitals to form special orbitals for bonding. The atoms are responding as needed to give the minimum energy for the molecule.

14 Valence Bond Theory and NH 3 N – 1s 2 2s 2 2p 3 3 H – 1s 1 If use the 3 2p orbitals predict 90 0 Actual H-N-H bond angle is How do you explain this? 2s 2p x 2p y 2p z

15 2s 2px 2py 2pz Original Mix 1s and 3p And generate four Equivalent sp 3 Hybridized orbitals sp3 sp3 3 bonding orbitals Which can accommodate The 1s1 electron from hydrogen 1 sp 3 lone pair Consider the n=2 for N

16 The nitrogen atom in ammonia is sp 3 hybridized.

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18 An orbital energy-level diagram for sp 2 hybridization. Note that one p orbital remains unchanged.

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20 When an s and two p orbitals are mixed to form a set of three sp 2 orbitals, one p orbital remains unchanged and is perpendicular to the plane of the hybrid orbitals.

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22 Figure 9.13: (a) The orbitals used to form the bonds in ethylene. (b) The Lewis structure for ethylene.

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24 Sigma bond (  ) – electron density between the 2 atoms Pi bond (  ) – electron density above and below plane of nuclei of the bonding atoms

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26 The s bonds in ethylene. Note that for each bond the shared electron pair occupies the region directly between the atoms.

27 A sigma (  ) bond centers along the internuclear axis. A pi (  ) bond occupies the space above and below the internuclear axis.

28 (a)The orbitals used to form the bonds in ethylene. (b) The Lewis structure for ethylene.

29 The orbital energy-level diagram for the formation of sp hybrid orbitals on carbon.

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31 When one s orbital and one p orbital are hybridized, a set of two sp orbitals oriented at 180 degrees results.

32 The orbitals of an sp hybridized carbon atom.

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34 The orbital arrangement for an sp 2 hybridized oxygen atom to form CO 2. 8 O 1s 2 2s 2 2p 4

35 The hybrid orbitals in the CO 2 molecule.

36 (a) The orbitals used to form the bonds in carbon dioxide. Note that the carbon-oxygen double bonds each consist of one s bond and one p bond. (b) The Lewis structure for carbon dioxide. 2sp orbitals from C to form two double bonds

37 (a) An sp hybridized nitrogen atom. (b) The s bond in the N 2 molecule. (c) The two p bonds in N 2 are formed when electron pairs are shared between two sets of parallel p orbitals. (d) The total bonding picture for N 2.

38 Sigma (  ) and Pi Bonds (  ) Single bond 1 sigma bond Double bond 1 sigma bond and 1 pi bond Triple bond 1 sigma bond and 2 pi bonds How many  and  bonds are in the acetic acid (vinegar) molecule CH 3 COOH? C H H CH O OH  bonds = = 7  bonds =

39 How to generate more than 4 bonds? Remember the PCl 5, SF 6, etc… The s and 3p can generate 4 orbitals Include d orbitals to generate more!

40 s p p p d d d d d Hybridize 1s and 3p and 1d Result in 5 dsp 3 orbitals dsp 3 dsp 3 dsp 3 dsp 3 dsp

41 A set of dsp 3 hybrid orbitals on a phosphorus atom. Note that the set of five dsp 3 orbitals has a trigonal bipyramidal arrangement. (Each dsp 3 orbital also has a small lobe that is not shown in this diagram.)

42 (a) The PCl 5 molecule. (b) The orbitals used to form the bonds in PCl 5. The phosphorus uses a set of five dsp 3 orbitals to share electron pairs with sp 3 orbitals on the five chlorine atoms. The other sp 3 orbitals on each chlorine atom hold lone pairs.

43 s p p p d d d d d Hybridize 1s and 3p and 2d Result in 6 d 2 sp 3 orbitals d 2 sp 3 ds 2 p 3 d 2 sp 3 d 2 sp 3 d 2 sp 3 d 2 sp 3 How to form 6 orbitals?

44 The relationship of the number of effective pairs, their spatial arrangement, and the hybrid orbital set required.

45 The Localized Electron Model 4 Draw the Lewis structure(s) 4 Determine the arrangement of electron pairs (VSEPR model). 4 Specify the necessary hybrid orbitals.

46 Molecular Orbitals (MO) Analagous to atomic orbitals for atoms, MOs are the quantum mechanical solutions to the organization of valence electrons in molecules. Remember bonds are waves and wave may be arranged in constructive or destructive interference.

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48 Types of MOs bonding: lower in energy than the atomic orbitals from which it is composed. antibonding: higher in energy (unstable) than the atomic orbitals from which it is composed.

49 The combination of hydrogen 1s atomic orbitals to form molecular orbitals.

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51 Bonding and antibonding molecular orbitals (MOs).

52 (a) The molecular orbital energy-level diagram for the H 2 molecule. (b) The shapes of the molecular orbitals are obtained by squaring the wave functions for MO 1 and MO 2.

53 A molecular orbital energy- level diagram for the H 2 molecule.

54 The molecular orbital energy-level diagram for the H 2 - ion.

55 The molecular orbital energy-level diagram for the He 2 molecule.

56 The molecular orbital energy-level diagram for the Li 2 molecule.

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58 The expected molecular orbital energy-level diagram resulting from the combination of the 2p orbitals on two boron atoms.

59 The expected molecular orbital energy-level diagram for the B 2 molecule. However, B 2 is found to be Paramagnetic!

60 The correct molecular orbital energy-level diagram for the B 2 molecule. When p-s mixing is allowed, the energies of the  2p and π 2p orbitals are reversed. The two electrons from the B 2p orbitals now occupy separate, degenerate π 2p molecular orbitals and thus have parallel spins. Therefore, this diagram explains the observed paramagnetism of B 2.

61 (a) The three mutually perpendicular 2p orbitals on two adjacent boron atoms. Two pairs of parallel p orbitals can overlap as shown in (b) and (c), and the third pair can overlap head-on as shown in (d).

62 (a) The two p orbitals on the boron atom that overlap head-on produce two s molecular orbitals, one bonding and one antibonding. (b) Two p orbitals that lie parallel overlap to produce two p molecular orbitals, one bonding and one antibonding.

63 The molecular orbital energy-level diagrams, bond orders, bond energies, and bond lengths for the diatomic molecules B 2 through F 2. Note that for O 2 and F 2 the  2p orbital is lower in energy than the π 2p orbitals.

64 Paramagnetic Diamagnetic

65 Note that O 2 is paramagnetic When liquid oxygen is poured into the space between the poles of a strong magnet, it remains there until it boils away. This attraction of liquid oxygen for the magnetic field demonstrates the paramagnetism of the O 2 molecule.

66 Paramagnetism 4 unpaired electrons 4 attracted to induced magnetic field 4 much stronger than diamagnetism

67 Bond Order (BO) Difference between the number of bonding electrons and number of antibonding electrons divided by two.

68 bond order = 1 2 Number of electrons in bonding MOs Number of electrons in antibonding MOs ( - ) bond order ½10½

69 Molecular Orbital (MO) Configurations 1.The number of molecular orbitals (MOs) formed is always equal to the number of atomic orbitals combined. 2.The more stable the bonding MO, the less stable the corresponding antibonding MO. 3.The filling of MOs proceeds from low to high energies. 4.Each MO can accommodate up to two electrons. 5.Use Hund’s rule when adding electrons to MOs of the same energy. 6.The number of electrons in the MOs is equal to the sum of all the electrons on the bonding atoms.

70 Outcomes of MO Model 1.As bond order increases, bond energy increases and bond length decreases. 2.Bond order is not absolutely associated with a particular bond energy. 3.N 2 has a triple bond, and a correspondingly high bond energy. 4.O 2 is paramagnetic. This is predicted by the MO model, not by the LE model, which predicts diamagnetism.

71 The molecular orbital energy-level diagram for the NO molecule. We assume that orbital order is the same as that for N 2. The bond order is 2.5. Heteronuclear Molecules and similar?

72 Figure 9.42: The molecular orbital energy-level diagram for both the NO + and CN - ions.

73 A partial molecular orbital energy-level diagram for the HF molecule. Heteronuclear Molecules and similar?

74 The electron probability distribution in the bonding molecular orbital of the HF molecule. Note the greater electron density close to the fluorine atom.

75 Combining LE and MO Models  bonds can be described as being localized.  bonding must be treated as being delocalized.

76 Delocalized molecular orbitals are not confined between two adjacent bonding atoms, but actually extend over three or more atoms.

77 (a) The benzene molecule consists of a ring of six carbon atoms with one hydrogen atom bound to each carbon. (b) Two of the resonance structures for the benzene molecule.

78 The  bonding system in the benzene molecule.

79 (a) The p molecular orbital system in benzene is formed by combining the six p orbitals from the six sp 2 hybridized carbon atoms. (b) The electrons in the resulting p molecular orbitals are delocalized over the entire ring of carbon atoms, giving six equivalent bonds. A composite of these orbitals is represented here.

80 Electron density above and below the plane of the benzene molecule.

81 The resonance structures for O 3 and NO 3 -. Note that it is the double bond that occupies various positions in the resonance structures.

82 (a) The p orbitals used to form the π bonding system in the NO 3 - ion. (b) A representation of the delocalization of the electrons in the π molecular orbital system of the NO 3 - ion.

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