Chem 14A

1. Chem 14A Monday, July 9, 2007

2. Lecture Outline Multiple Bonds, Bond Length, Bond Strength Resonance and Lewis Dot Structures Formal Charge Exceptions to the Octet Rule Expanded Valence Shells Dipoles and Electronegativity Polarization Power % Ionic character

3. Multiple Bonds  On Friday we began to discuss multiple bonds Multiple Bonds: Double or Triple Covalent Bonds Quadruple and higher-order bonds also exist, but are outside the scope of this course

4. Multiple Bonds and Bond Length Bond length= distance between the nuclei of two atoms forming a covalent bond Single bond such as C-H Double bond such as O=O Triple bond such as N=N

5. Multiple Bonds and Bond Length For any given set of bonded atoms: single bond >double bond >triple bond in length For example:

6. Multiple Bonds and Bond Strength Bond energy is a a direct measure of bond strength In general, shorter bonds are stronger than longer bonds, so they exhibit a larger bond energy It is more difficult to break multiple bonds, since they are shorter and stronger, and also have a larger bond energy

7. Multiple Bonds and Resonance Now that we know a little more about multiple bonds, let’s discuss resonance and how it alters Lewis dot structures Resonance occurs when there is more than one form of a molecule that exists In order for us to accurately describe the molecule, we use resonance structures

8. Resonance For example, the nitrate anion exists in three different forms. It is not sufficient to draw just one Lewis structure for the molecule, so we draw all three possible structures to fully describe the molecule. These are called resonance structures.

9. Resonance The actual structure for molecules exhibiting resonance should be thought of as a mixture of all the resonance structures combined- a resonance hybrid If only one structure was correct we would expect to have two N-O single bond lengths and one N=O double bond length 120pm N-O single bond and 140pm N=O double bond The actual bond length is 124pm for each bond

11. Resonance How do we know if a molecule has resonance structures? The arrangement of atoms does not change in a resonance structure, only the location of the electrons If were are able to change the location of the molecule’s electrons (delocalize) without changing the arrangement of atoms, the molecule has resonance structures Most commonly seen in molecules with double bonds

12. Resonance For example, benzene, a common organic solvent exhibits resonance

13. Resonance Why is resonance important? Resonance makes an important contribution to molecular stability Having a resonance hybrid results in a lower energy value and greater stability for the molecule

14. Quick Exercise I. Carbonate anion CO32- helps buffer the pH of blood. Draw the Lewis Dot Structure and resonance structure(s) for carbonate anion.

15. Answer Quick Exercise I.

16. Formal Charge The charge on each atom in a Lewis structure, assuming the bonds are 100% covalent, meaning that the electrons are shared equally, 50:50 Formal Charge = V-(L+½S) V= valence electrons in the atom L= lone pair electrons S= bonding electrons

17. Formal Charge How do we use formal charge? When drawing Lewis structures, there can be multiple ways to draw a molecule that does not exhibit resonance Formal charges tell us which structure is “correct” in terms of the lowest energy structure, the most stabile form of the molecule

18. Formal Charge How do we calculate formal charge, and how does the answer give us the “best” structure? For example, let’s look at carbon dioxide, CO2 Formal charge will help show us why the structure of carbon dioxide is O=C=O and not C=O-O

19. Formal Charge Carbon Dioxide, CO2, Lewis structure First, the total valence electrons: C= 4e- O= 6e- x 2= 12e- Total valence electrons= 16

20. Formal Charge Now let’s add the bonding electrons to the Lewis structure, followed by the lone pairs, assuming connectivity of O=C=O

21. Formal Charge Now, we’ll find the formal charge on each atom in the Lewis structure Each O= 6 valence-[4 lone +1/2(4 bond)]=0 C= 4 valence-[0 lone+1/2(8 bond)]=0

22. Formal Charge Now, let’s look at the alternate structure C=O-O Total valence electrons= 16 Lewis structure:

23. Formal Charge To find the formal charge for each atom in C=O-O C= 4 valence - [2 lone + 3 bond]= -1 Otriple= 6 valence – [0 lone + 4 bond]= +2 Osingle = 6 valence- [6 lone + 1 bond]= -1 The structure with the lowest average formal charge is the most stabile, therefore this structure is not as good as O=C=O with FC=0

24. Formal Charge

25. Quick Exercise II. Draw the Lewis structure and determine the formal charge on each atom for the following molecules: OCl- BF4-

26. Answers Quick Exercise II. OCl- Total valence electrons= 14e- Formal Charge Cl= 7-[6+1]=0 O= 6-[6+1]= -1

27. Answers Quick Exercise II. BF4- Total valence electrons=3e+7(4)+1=32 e- Formal Charge B= 3-[0+4]= -1 F= 7-[6+1]=0

28. Exceptions to the octet rule H only holds two valence electrons (duet) Boron (B), Beryllium (Be), and elements below these ie Aluminum (Al) usually don’t complete an octet Second-row elements C,N,O…never exceed an octet Third-row elements S, P, Cl…frequently exceed an octet by accessing d-orbitals

29. Exceptions to the octet rule For electron-deficient compounds containing Be, B, Al…. Draw the Lewis structure as you normally would, keeping in mind that these elements frequently do not complete an octet After adding in the lone pairs, you will “run out” of electrons, don’t use double bonds to satisfy the octet with these elements

30. Exceptions to the octet rule For example, let’s look at BF3 and BeCl2 BF3 total valence electrons= 3+7(3)= 24 e- After adding the bonding electrons and completing the octet for each F, all of the electrons have been used, and B has only six electrons

31. Exceptions to the octet rule Now let’s look at BeCl2 BeCl2 total valence electrons= 2+7(2)= 16e-

32. Exceptions to the octet rule When drawing a Lewis structure for third row elements, such as S, P, Cl… remember that these elements frequently expand their valence Expanded valences allow for more than eight electrons in the valence, by accessing d-orbitals It is helpful to remember that double bonds usually do not form in fluorine-containing compounds

33. Exceptions to the octet rule How do we handle Lewis structures for elements that can expand their valence? Draw the Lewis structure as you normally would, keeping in mind that these elements frequently exceed an octet After adding in the lone pairs, you may have left over electrons, which are added to the element with an expandable valence, usually the central atom

34. Exceptions to the octet rule Let’s take a look at XeF2 XeF2 total valence electrons= 8+7(2)= 22e-

35. Exceptions to the octet rule Sometimes, just like before, more than one Lewis structure is valid Let’s look at perchlorate, ClO4- a waste product from rocket fuel that is currently a terrible environmental contaminant

36. Exceptions to the octet rule ClO4- total valence electrons= 7+6(4)+1=32 e- Begin by drawing Cl as the central atom, and completing an octet for each O The formal charge on each atom is given

37. Exceptions to the octet rule Now let’s look at an alternate structure We know that Cl can form double bonds, so this is another way to draw the structure Formal charges are lower overall, so this is “best”

38. Exceptions to the octet rule Let’s look at another expanded valence example, SF6 Total valence electrons= 6+7(6)= 48 e-

39. Quick Exercise III. Draw the Lewis structure(s) for each of the following compounds PCl5 SO42- I3- BH3

40. Answers Quick Exercise III. PCl5 Total valence electrons= 5+7(5)= 40

41. Answers Quick Exercise III. SO42- Total valence electrons= 6+6(4)+2=32 e- More resonance structures exist…

42. Answers Quick Exercise III. I3- Total valence electrons= 7(3)+1=22

43. Answers Quick Exercise III. BH3 Total valence electrons= 3+1(3)= 6

44. Exceptions to the octet rule Some atoms or molecules have an odd number of valence electrons, and cannot follow the octet rule Radicals= chemical species with one unpaired electron Biradical= chemical species with two unpaired electrons

45. Exceptions to the octet rule For example, oxygen is a biradical O 1s22s22px22py12pz1 Hydrogen peroxide forms the radical OOH during a chemical reaction, Let’s look at the Lewis Structure for OOH

46. Exceptions to the octet rule OOH Total valence electrons= 6(2)+1= 13e- Note that this is an odd number

47. Dipoles and Electronegativity Molecules that have uneven distributions of electrons, for example: HCl, where H has 1e- and Cl has 7e- have a net dipole moment Polar molecules= exhibit net dipole moment A dipole is a vector that represent the magnitude of charge difference between two atoms In chemistry, the dipole is written as an arrow pointing from the region with less electron density towards the region with more electron density

48. Dipoles and Electronegativity Water has more electron density around the oxygen atom O= 6e- H=1e- Dipole arrows are drawn starting with a “+” at the region with fewer electrons

49. Dipoles and Electronegativity Dipoles are a vector quantity, so they can cancel each other out This leads to a non-polar molecule

50. Dipoles and Electronegativity

51. Dipoles and Electronegativity A net dipole corresponds to polarity in a molecule, indicating regional differences in charge density symbolized by d+ and d-, at the + and – regions of the dipole arrow vector

52. Dipoles The HF molecule exhibits a dipole as indicated by the d+ and d- values When placed in a field, the d+ H will move to the negative side, and the d- F will move to the positive side

53. Dipoles and Electronegativity The ratio of dipole moments is used to calculate a molecule’s % ionic character %ionic character= measured dipole X-Y

54. Dipoles and Electronegativity Molecular differences in charge density (dipole moments) are due to differences in electronegativity Electronegativity= the ability of an atom to attract shared electrons to itself

55. Electronegativities of the Elements

56. Dipoles and Electronegativity The magnitude of the difference in electronegativity between two atoms determines whether a bond will be ionic, polar-covalent, or covalent The key idea is this: the greater the electronegativity difference between two atoms, the more ionic the bond.

57. Dipoles and Electronegativity

58. Dipoles and Electronegativity

59. Dipoles and Electronegativity The D value is 0 for covalent bonds, the electrons are evenly shared The D value is >0 for polar covalent bonds, the electrons are not equally shared

60. Dipoles and Electronegativity How do we use electronegativity data to determine bond covalency? Let’s compare a series of bonds using this table of atomic EN values

61. Dipoles and Electronegativity Example: Which of the following compounds is expected to demonstrate intermediate bonding behavior (i.e., polar covalent), Cl2, OH, or NaCl? OH is a polar covalent bond relative to Cl2 and NaCl

62. Polarizability Another factor influencing covalent vs. ionic character is polarizability Polarizability refers to the tendency for an atom’s electron cloud to shift when an attractive or repulsive force is applied For example, When Na+ and Cl- are located in the same vicinity, the electron cloud of Cl- shifts toward Na+ due to electronic attraction

63. Polarizability Polarization power, the ability to polarize, is strongly dependent on size or atomic radius of an ion Na+ has a greater polarization power than Cs+ I- has a greater polarization power than F- The more polarizing power an ion has, the more covalent character will be found in its bonding For example, NaI will have more covalent character than CsF

64. Quick Exercise IV. Arrange the following ions in order of increasing polarizing power K+, Mg2+, Al3+, Cs+ N3-, P3-, I-, At -

65. Answers Quick Exercise IV. Cations: Cs+ < K+ < Mg2+ < Al3+ Smaller cations have more polarization power Anions: N3- < P3- < I- < At – Larger anions have more polarization power