CH 10: Molecular Geometry & chemical bonding theory

1. Vanessa N. Prasad-Permaul Valencia Community College CHM 1045 CH 10: Molecular Geometry & chemical bonding theory

2. Molecular Shapes: VSEPR • Molecular Geometry: the general shape of a molecule, as determined by the relative positions of the atomic nuclei. • Valence-Shell Electron-Pair Repulsion Theory: predicts the shapes of molecules and ions by assuming that the valence-shell electron pairs are arranged about each atom so that the electron pairs are kept as far away from one another as possible, thus minimizing electron-pair repulsions.

3. Molecular Shapes: VSEPR • Two Electron Groups: Electron groups point in opposite directions. • LINEAR ARRANGEMENT

4. Molecular Shapes: VSEPR • Three Electron Groups:Electron groups lie in the same plane and point to the corners of an equilateral triangle. • TRIGONAL PLANAR ARRANGEMENT • Trigonal planar geometry or bent (angular)

5. Molecular Shapes: VSEPR • Four Electron Groups: Electron groups point to the corners of a regular tetrahedron. • TETRAHEDRAL ARRANGEMENT

6. Molecular Shapes: VSEPR • The approximate shape of molecules is given by Valence-Shell Electron-Pair Repulsion (VSEPR). • Step #1: Count the total electron groups. • Step #2: Arrange electron groups to maximize separation. • Groups are collections of bond pairs between two atoms or a lone pair. • Groups do not compete equally for space: Lone Pair > Triple Bond > Double Bond > Single Bond

7. Molecular Shapes: VSEPR Table of the Molecular Geometry and Examples of Certain Compounds

8. Molecular Shapes: VSEPR • EXAMPLE 10.1 : Predict the geometry of the following • Molecules or ions, using the VSEPR method: • BeCl2 B. NO2- C. SiCl4 • Be = 2 electrons Cl = 7 electrons each = 14 • Total of 16 electrons • :Cl:Be:Cl: • 2 sets of electron pairs, 0 lone pairs  linear ¨ ¨ ¨ ¨

9. Molecular Shapes: VSEPR B. NO2-N = 5 electrons, O = 6 electrons each = 12 Total of 17 electrons plus 1 electrons = 18 e- :O:N::O: 3 sets of electron pairs, 1 lone pair  trigonal planar arrangement and a bent geometry ¨ ¨ ¨ ¨

10. Molecular Shapes: VSEPR C. SiCl4 Si = 4 electrons, Cl = 7 electrons each = 28 electrons Total of 32 electrons 4 electron pairs, 0 lone pairs tetrahedral

11. Molecular Shapes: VSEPR • EXERCISE 10.1 : Use the VSEPR method to predict the • geometry of the following ion and molecules: • ClO3- • OF2 • SiF4

12. Molecular Shapes: VSEPR Five Electron Groups: Electron groups point to the corners of a trigonal bipyramid.

13. Molecular Shapes: VSEPR Six Electron Groups: Electron groups point to the corners of a regular octahedron.

14. Molecular Shapes: VSEPR EXAMPLE 10.2 : What do you expect for the geometry of tellurium tetrachloride? Tellurium has 6 electron pairs in its valence shell Cl = 7 electrons each = 28 electrons Total of 34 electrons There are 4 bonding pairs and 1 lone pair of electrons  Trigonal bipyramidal arrangement and a seesaw geometry ¨

15. Molecular Shapes: VSEPR EXERCISE 10.2 : According to the VSEPR model, what molecular geometry would you predict for iodine trichloride?

16. Dipole Moment & Molecular Geometry Dipole Moment: a quantitative measure of the degree of charge separation in a molecule. + - H Cl • If a molecule has a dipole moment it is polar • A compound could contain polar bonds but the molecule could be non-polar because there is no dipole moment • Bond dipoles can cancel each other out O C O

17. Dipole moment • Based on electronegativity differences between atoms in a molecule • The most electronegative atom is partially negative • The less electronegative atom is partially positive • The dipole moment is the average of all dipoles in the molecule • Exception (C-H bonds do not have a dipole)

18. Molecular Shapes: VSEPR EXAMPLE 10.3 : Each of the following molecules has a non-zero dipole moment. Select the molecular geometry that is consistent with this information. Explain. SO2: linear or bent BENT because in linear geometry, the dipole moment would be zero. There are 2 electron pairs and 1 lone pair which is a trigonal planar arrangement giving a bent geometry.

19. Molecular Shapes: VSEPR PH3: trigonal planar or trigonal pyramidal In a trigonal planar geometry, the dipole moment is zero, therefore PH3 has a trigonal pyramidal geometry. Tetrahedral arrangement; 3 electron pairs and one lone pair.

20. Molecular Shapes: VSEPR EXERCISE 10.3 : Bromine trifluoride BrF3, has a nonzero dipole moment. Indicate the correct arrangement and molecular geometry is consistent with this information.

21. Molecular Shapes: VSEPR • EXERCISE 10.4 : Which of the following would be expected to have a dipole moment of zero on the basis of symmetry. Explain. • SOCl2 • SiF4 • OF2

22. Valence Bond Theory 1. Covalent bonds are formed by overlapping of atomic orbitals, each of which contains one electron of opposite spin. 2. Each of the bonded atoms maintains its own atomic orbitals, but the electron pair in the overlapping orbitals is shared by both atoms. 3. The greater the amount of orbital overlap, the stronger the bond.

23. Valence Bond Theory According to Valence Bond Theory: • An orbital on one atom comes to occupy a portion of the same region of space as an orbital on the other atom. The two orbitals are said to overlap. • The total number of electrons in both orbitals is no more than two.

24. Valence Bond Theory Combined hybrid orbitals Hybrid orbitals # of orbitals VSEPR geometry

25. Valence Bond Theory • sp hybrid:

26. Valence Bond Theory • sp2 hybrid:

27. Valence Bond Theory • sp2 hybrid (π bond):

28. Valence Bond Theory • sp3 hybrid:

29. Valence Bond Theory sp3d hybrid:

30. Valence Bond Theory • sp3d2 hybrid:

31. Valence Bond Theory • Steps to obtain the bonding description about an atom in a molecule: • Write the Lewis electron-dot formula • From the Lewis formula, use the VSEPR model to obtain arrangement • From the geometry, s what type of hybrid orbitals are required • Assign valence electrons to the hybrid oritals of this atom one at a time, pairing them only when necessary • Form the bonds by overlapping singly occupied orbitals of other atoms

32. Valence Bond Theory EXAMPLE 10.4: Describe the bonding in H2O according to valence bond theory. Using the VESPR theory, notice there are 3 sets of electron pairs, and two lone pairs giving a tetrahedral arrangement and a bent geometry

33. Valence Bond Theory EXERCISE 10.5: Using hybrid orbitals, describe the bonding in NH3 according to valence bond theory.

34. Valence Bond Theory EXAMPLE 10.5: Describe the bonding in XeF4 using hybrid orbitals. 4 single bonds, 2 lone pairs octahedral arrangement and square planar geometry.

35. Valence Bond Theory EXERCISE 10.6: Describe the bonding in PCl5 using hybrid orbitals.

36. Molecular Orbital Theory • Additive and subtractive combination of p orbitals leads to the formation of both sigma and pi orbitals.

37. Valence Bond Theory EXAMPLE 10.6: Describe the bonding on a given N atom in dinitrogen difluoride, N2F2, using valence bond theory.  :F N N F:  ¨ ¨ ¨ ¨ ¨ ¨

38. Valence Bond Theory EXERCISE 10.7: Describe the bonding on the carbon atom in carbon dioxide using valence bond theory.

39. Valence Bond Theory EXERCISE 10.8: Dinitrogen difluoride exists as cis and trans isomers. Write structural formulas for these isomers and explain using valence bond theory why they exist.

40. Molecular Orbital Theory • The molecular orbital (MO) model provides a better explanation of chemical and physical properties than the valence bond (VB) model. • Atomic Orbital: Probability of finding the electron within a given region of space in an atom. • Molecular Orbital: Probability of finding the electron within a given region of space in a molecule.

41. Molecular Orbital Theory • Additive combination of orbitals (s) is lower in energy than two isolated 1s orbitals and is called a bonding molecular orbital.

42. Molecular Orbital Theory • Subtractive combination of orbitals (s*) is higher in energy than two isolated 1s orbitals and is called an antibonding molecular orbital.

43. Molecular Orbital Theory • Bond Order is the number of electron pairs shared between atoms. • Bond Order is obtained by subtracting the number of antibonding electrons from the number of bonding electrons and dividing by 2. BO = Bonding electrons – antibonding electrons 2

44. B2 B has 5 electrons So B2 has 10 elec Core electrons don’t count toward BO *2p *2p 2p 2p *2s 2s *1s 1s

45. C2 Carbon has 6 electrons so C2 has 12 electrons *2p *2p 2p 2p *2s 2s *1s 1s

46. Molecular Orbital Theory • Molecular Orbital Diagram for H2:

47. Molecular Orbital Theory • Molecular Orbital Diagrams for H2– and He2:

49. Molecular Orbital Theory • Second-Row MO Energy Level Diagrams:

50. Valence Bond Theory EXAMPLE 10.7: Give the orbital diagram of the molecule. Is the molecular substance diamagnetic or paramagnetic? What is the electron configuration? What is the bond order?