1. Chapter 9 Covalent Bonding: Orbitals

2. Schroedinger An atomic orbital is the energy state of an electron bound to an atomic nucleus Energy state changes when it is bonded to another atom Hybridization is the combining of wave functions to give new wave functions http://www.mikeblaber.org/oldwine/chm1045/ notes/Geometry/Hybrid/Geom05.htm

3. 9.1 Hybridization and the Localized Electron Model Localized electron model - a molecule is a collection of atoms bound together by sharing electrons

4. A modification of the LE model is needed to account for the type and arrangement of bonds actually seen in molecules A molecule is more than the sum of its parts. It tries to achieve minimum energy

5. Hybridization – mixing of “native” atomic orbitals to form special orbitals for bonding

6. Localized model Write the Lewis structure Determine the best arrangement using VSEPR Determine the hybrid atomic orbitals necessary (apply to peripheral and central atoms)

7. Electron pairs (lone pairs) not involved in bonding may occupy hybrid orbitals. Atoms adjust to minimize energy and electron pair repulsions. The whole molecule is considered rather than the individual atom. Effective pairs – lone, single or multiple bonds

8. sp 3 hybridization Energy between 2s and 2p The new orbitals are four identically shaped orbitals (Fig. 9.3 p. 405) They are oriented towards the corners of a tetrahedron Account for known structure

9. One 2s orbital and three 2p orbitals combine to form four new sp 3 orbitals (s 1 p 3 ) Assume bonding only involves valence electrons The total number of electrons and their arrangement in the molecule is important Atomic orbitals are arranged to accommodate the best electron arrangement for the molecule as a whole

10. sp 3 hybridization gives 4 identical orbitals

12. *Whenever a set of equivalency tetrahedral atomic orbitals is required by an atom, the atom adopts a set of sp 3 orbitals. Ex. C has 2s 2 2p 2 valance electrons

13. sp 2 hybridization For three bonds or effective pairs Trigonal planar shape Combine one s and two p orbitals The plane is determined by the p orbitals (x,y,z)

14. sp 2 hybridization gives three identical orbitals whenever an atom is surrounded by three effective pairs, a set of sp 2 hybrid orbitals is required.

15. One p orbital remains unchanged

16. Sigma(s) bond the electron pair is shared in an area centered by the line running between the atoms formed from orbitals whose lobes point towards each other

17.  (pi) bonds sharing an electron pair in the space above and below the sigma bond Unhybridized p orbitals form pi bonds use the 2p orbital perpendicular to the sp 2 hybrid

18. So… A double bond consists of a sigma bond and a pi bond

19. Ethylene (C 2 H 4 )

20. sp hybridization One s and one p orbital 180 0 apart Two effective pairs around an atom Linear Two p orbitals remain unchanged; they are used for pi bonds

22. dsp 3 hybridization When a central atom exceeds the octet rule One d, one s, and three p orbitals When five electron pairs need to be represented Trigonal bipyramidal arrangement

23. d 2 sp 3 Six electrons pairs around the central atom Octahedral arrangement

25. Summary Hybrid orbitals demo Problems p. 431 # 24, 29 hybrid orbitals P. 432 # 33, 35, 37 sigma, pi bonds

26. 9.2 The Molecular Orbital Model LE model problems: Incorrectly assumes electrons are localized Needs the concept of resonance Unpaired electrons are a problem Gives no information about bond energies

27. Molecular Orbital Model (We don’t know exactly where electrons are, so we can’t accurately predict interactions) Compare predictions with experimental observations. MO’s are like atomic orbitals Hold two electrons with opposite spins Square of the MO wave function indicates electron probability Always the same number of atomic orbitals as the sum of the atoms that formed them

29. Ex. 2H  H 2 Each H has a 1s atomic orbital Quantum mechanical equations give two molecular orbitals MO 2 = 1 SA – 1 SB MO 1 = 1 SA + 1 SB

30. The greatest probability for MO 1 is between the nuclei. MO 2 is on the outside of each nuclei. Distribution is called sigma (like LE) molecular orbitals. (Atomic orbitals no longer exist.) MO 1 is lower in energy than 1s atomic, MO 2 is higher Electrons in MO 1 favor bonding, MO 2 are anti-bonding.

31. Bonding are lower energy since they are attracted by both nuclei. MO designations include shape, original (parent) atomic orbitals and bonding/anti- bonding properties.

32. For 2p: The overlapping pair produces one bonding and one anti-bonding sigma molecular orbital The parallel pairs produce two bonding and two antibonding p MO’s.

33. Ex. For H MO 1 =  1s MO 2 =  1s* (* designates anti-bonding) Molecular electron configurations – written like electron configurations H 2 (  1s ) 2

34. Bond Order difference between the number of bonding and the number of anti-bonding electrons divided by two. (Two is used because bonds represent pairs of electrons.) Is an indication of bond strength Larger bond order means greater bond strength A value of zero is unstable

35. Problems p. 433 # 39,41

36. 9.3 Bonding in Homonuclear Diatomic Molecules Two identical atoms Consider elements in period 2 Only the valence electrons contribute to molecular orbitals. (Core levels are smaller and the electrons are considered to be localized.) You can determine if a molecule is stable or if it exists based in bond order. Metals also form metallic bonds of many atoms.

38. Magnetic properties Most substances do not exhibit magnetism unless they are in a magnetic field. A. Paramagnetism – a substance is attracted into the inducing magnetic field

39. B. Diamagnetism – a substance is repelled from the inducing field.

40. A substance is weighed both with and without a magnetic field. If the weight increases, it is paramagnetic.

41. Unpaired electrons – paramagnetic properties (oxygen) Paired electrons – diamagnetic properties If both conditions exist, there is a net paramagnetic effect since this is stronger.

42. Boron needs to use s-p mixing to explain its magnetic properties. Energy of  2p and  2p are reversed.

43. Summary As bond order (predicted) increases, bond energy increases and bond length decreases. Bond order is not associated with particular bond energy. N 2 has a triple bond, bond order 3. In explosives, much energy is released from N containing compounds as N 2 is formed.

44. O 2 is paramagnetic as predicted from MO theory. LE theory predicts diamagnetic.

45. bond order = (bonding – anti bonding) / 2 You need to get electron configurations using MO’s and then use bond order to determine the strongest bond. Samples Problems p. 433 #47, 48

46. 9.4 Bonding in Heteronuclear Diatomic molecules Heteronuclear – different atoms If the atoms are close on the periodic table, use the MO for homonuclear molecules. When the atoms are very different, a new energy diagram is necessary for each molecule.

48. The sigma bond would show greater probability close to the atom with an orbital of lower energy. This would produce a polar bond.

49. 9.5 Combing the localized Electron and Molecular Orbital Models LE : Electrons are localized Resonance is necessary to explain some bond angles More than one valid Lewis structure can be drawn for some molecules

50. MO : allows for delocalization Combine LE and MO for molecules that require resonance structures.

51. Ex. A double bond that could be in more than one location becomes one  and one  bond to give the most physically accurate description of the molecule. LE describes the  MO describes the 

52. Ex. Benzene C 6 H 6 6 equivalent C-C bonds All the atoms are in the same plane sp 2 orbitals – sigma bond p orbitals perpendicular to the plane for  orbitals Gives delocalized  bonding

53. Additional exercises p. 447 # 51, 55, 57 Challenge # 61