Problems and questions How is a molecule or polyatomic ion held together? Why are atoms distributed at strange angles? Why are molecules not flat?

Transcription

1 1 Cocaine

2 2 Problems and questions ow is a molecule or polyatomic ion held together? Why are atoms distributed at strange angles? Why are molecules not flat? Can we predict the structure? ow is structure related to chemical and physical properties?

3 3 There are 2 extreme forms of connecting or bonding atoms: Ionic Ionic complete complete transfer of 1 or more electrons from one atom to another Covalent some valence electrons shared between atoms Most bonds are somewhere in between.

4 4 Essentially complete electron transfer from an element of low IE (metal) to an element of high affinity for electrons (nonmetal) 2 Na(s) ) + Cl 2 (g) ---> 2 Na Cl - Therefore, ionic compds.. exist primarily between metals at left of periodic table (Grps( 1A and 2A and transition metals) and nonmetals at right (O and halogens).

5 5 The bond arises from the mutual attraction of 2 nuclei for the same electrons. Electron sharing results. (Screen 9.5) A + B A B Bond is a balance of attractive and repulsive forces.

6 6! Objectives are to understand: 1. valence e-e distribution in molecules and ions. 2. molecular structures 3. bond properties and their effect on molecular properties.

7 7 " " G. N. Lewis Electron distribution is depicted with Lewis electron dot structures Valence electrons are distributed as shared or BOND PAIRS and unshared or PAIRS. or LONE

8 8 # Valence electrons are distributed as shared or BOND PAIRS and unshared or LONE PAIRS. unshared or LONE PAIRS. Cl shared or bond pair lone pair (LP) This is called a LEWIS ELECTRON DOT structure.

9 9 A bond can result from a head-to to-head overlap of atomic orbitals on neighboring atoms. Cl + Cl Overlap of (1s) and Cl (2p) Note that each atom has a single, unpaired electron.

10 $ 10 core and Electrons are divided between core valence electrons B 1s 2 2s 2 2p 1 Core = [e], valence = 2s 2 2p 1 Br [Ar[ Ar] ] 3d 10 4s 2 4p 5 Core = [Ar[ Ar] ] 3d 10, valence = 4s 2 4p 5

11 %" 11 No. of valence electrons of a main group atom = Group number or Groups 1A-4A (14), no. of bond pairs = group number. or Groups 5A (15)-7A (17), BP s = 8 - Grp.. No.

12 %" 12 No. of valence electrons of an atom = Group number or Groups 1A-4A (14), no. of bond pairs = group number or Groups 5A (15)-7A (17), BP s = 8 - Grp.. No. Except for (and sometimes atoms of 3rd and higher periods), BP s + LP s = 4 This observation is called the &&%'

13 "("" 13 Ammonia, N 3 1. Decide on the central atom; never. Central atom is atom of lowest affinity for electrons. Therefore, N is central 2. Count valence electrons = 1 and N = 5 Total = (3 x 1) + 5 = 8 electrons / 4 pairs

14 Building a Dot Structure orm a single bond between the central atom and each surrounding atom N 4. Remaining electrons form LONE PAIRS to complete octet as needed. 3 BOND PAIRS and 1 LONE PAIR. Note that N has a share in 4 pairs (8 electrons), while shares 1 pair. N

15 (")( +, * 15 Step 1. Central atom = S Step 2. Count valence electrons S = 6 3 x O = 3 x 6 = 18 Negative charge = 2 TOTAL = 26 e-e or 13 pairs Step 3. orm bonds 10 pairs of electrons are now left. O O S O

16 (")( +, * 16 Remaining pairs become lone pairs, first on outside atoms and then on central atom. O O S O Each atom is surrounded by an octet of electrons.

17 -) Central atom = 2. Valence electrons = or pairs 3. orm bonds. O C O This leaves 6 pairs. 4. Place lone pairs on outer atoms. O C O

18 -) Place lone pairs on outer atoms. O C O 5. So that C has an octet, we shall form DOUBLE BONDS between C and O. O C O O C O The second bonding pair forms a pi () bond.

19 Double and even triple bonds are commonly observed for C, N, P, O, and S 2 CO 19 SO 3 C 2 4 O C O

20 (""-)( + 1. Central atom = S 2. Valence electrons = 18 or 9 pairs 20 O S O 3. orm double bond so that S has an octet but note that there are two ways of doing this. bring in left pair O S O OR bring in right pair

21 bring in left pair Sulfur Dioxide, SO 2 O S OR bring in right pair O 21 This leads to the following structures. O S O O These equivalent structures are called RESONANCE STRUCTURES.. The true electronic structure is a YBRID of the two. S O

22 '). + / + 22

23 23 '). + / + 1. Number of valence electrons = 24 e-e 2. Draw sigma bonds. O N C N

24 24 '). + / + 3. Place remaining electron pairs in the molecule. O N C N

25 25 '). + / + 4. Complete C atom octet with double bond. O N C N

26 $ %" 26 Usually occurs with B and elements of higher periods. B 3 S 4

27 &" 27 Central atom = Valence electrons = or electron pairs = Assemble dot structure B The B atom has a share in only 6 pairs of electrons (or 3 pairs). B atom in many molecules is electron deficient.

28 28 (""&")( 0 Central atom = Valence electrons = or pairs. orm sigma bonds and distribute electron pairs. S 5 pairs around the S atom. A common occurrence outside the 2nd period.

29 29 Atoms in molecules often bear a charge (+ or -). The predominant resonance structure of a molecule is the one with charges as close to 0 as possible. 1" /,.3#/

30 -) ( 1 / 2 ) ( 4 ) - 4 = 0 O C O 4 - ( 1 / 2 ) ( 8 ) - 0 = 0

31 31 "# + Yellow = negative & red = positive Relative size = relative charge

32 &7 )(, (1/2)(2) - 6 = (1/2)(6) - 2 = 0 S C N 4 - (1/2)(8) - 0 = 0

33 &7 )(, 33 S C N S C N S C N Which is the most important resonance form?

34 "# (, 34 All atoms negative, but most on the S S C N

35 35 &") * +1 B B What if we form a B B double bond to satisfy the B atom octet? -1

36 36 & 1" * Calc d partial charges in B 3 is negative and B is positive

37 '% 37 &%8

38 '% &%8 38 VSEPR Valence Shell Electron Pair Repulsion theory. Most important factor in determining geometry is relative repulsion between electron pairs. Molecule adopts the shape that minimizes the electron pair repulsions.

39 39 # "935+

40 No. of e- Pairs Around Central Atom Example Geometry 40 2 Be linear 180Þ 3 B planar trigonal 120Þ 4 C 109 Þ tetrahedral

41 No. of e- Pairs Around Central Atom Example Geometry 41 2 Be linear 180Þ 3 B planar trigonal 120Þ 4 C 109 Þ tetrahedral

42 No. of e- Pairs Around Central Atom Example Geometry 42 2 Be linear 180Þ 3 B planar trigonal 120Þ 4 C 109 Þ tetrahedral

43 No. of e- Pairs Around Central Atom Example Geometry 43 2 Be linear 180Þ 3 B planar trigonal 120Þ 4 C 109 Þ tetrahedral

44 44 # "935+

45 ("" 7 $(#% Ammonia, N 3 1. Draw electron dot structure 2. Count BP s and LP s = 4 3. The 4 electron pairs are at the corners of a tetrahedron. N 45 N lone pair of electrons in tetrahedral position

46 ("" 7 $(#% 46 Ammonia, N 3 There are 4 electron pairs at the corners of a tetrahedron. N N lone pair of electrons in tetrahedral position The ELECTRON PAIR GEOMETRY is tetrahedral.

47 ("" 7 $(#% 47 Ammonia, N 3 The electron pair geometry is tetrahedral. N lone pair of electrons in tetrahedral position The MOLECULAR GEOMETRY the positions of the atoms is PYRAMIDAL. positions of the atoms is PYRAMIDAL.

48 ("" 7 $(#% 48 Water, 2 O 1. Draw electron dot structure 2. Count BP s and LP s = 4 O 3. The 4 electron pairs are at the corners of a tetrahedron. O The electron pair geometry is is TETRAEDRAL.

49 ("" 7 $(#% 49 Water, 2 O O O The electron pair geometry is is TETRAEDRAL The molecular geometry is is BENT.

50 "# "935* 50

51 ("" 7 $(#% ormaldehyde, C 2 O 1. Draw electron dot structure O 2. Count BP s and LP s at C 3. There are 3 electron lumps around C at the corners of a planar triangle. C 51 O C The electron pair geometry is is PLANAR TRIGONAL with 120 o bond angles.

52 ("" 7 $(#% 52 ormaldehyde, C 2 O O C The molecular geometry is is also planar trigonal. The electron pair geometry is is PLANAR TRIGONAL

53 ("" 7 $(#% 53 Methanol, C 3 O Define -C- and C-O- C bond angles -C- = 109 o C-O- = 109 o In both cases the atom is surrounded by 4 electron pairs. C O

54 ("" 7 $(#% Acetonitrile,, C 3 CN Define unique bond angles -C- = 109 o C-C-N N = 180 o 109 C C One C is surrounded by 4 electron lumps and the other by 2 lumps 180 N 54

55 55 #7) 1 C C 2 O 3 C C C C C O C C N 4 5

56 #7 56

57 ( "": : & & 0 # 57 Often occurs with Group 3A elements and with those of 3rd period and higher.

58 58 2" Consider boron trifluoride,, B 3 The B atom is surrounded by only 3 electron pairs. Bond angles are 120 o B Geometry described as planar trigonal

59 59 2":;<# " 90Þ Trigonal bipyramid P 120Þ 5 electron pairs

60 60 (""&")( 0 Number of valence electrons = 34 Central atom = S Dot structure S Electron pair geometry --> trigonal bipyramid (because there are 5 pairs around the S) 90Þ S 120Þ

61 61 (""&")( 0 Lone pair is in the equator because it requires more room. 90Þ S 120Þ S

62 62 " # "9350

63 63 2":;<# " 90Þ S Octahedron 90Þ 6 electron pairs

64 64 " (- # "9350

65 #2 65 What is the effect of bonding and structure on molecular properties? Buckyball in IV-protease

66 = : 2 66 Double bond Single bond 7 Triple bond

67 ractional bond orders occur in molecules with resonance structures. Consider NO 2 - N N O O O O The N O N O bond order = Bond order = Total # of e - pairs used for a type of bond Total # of bonds of that type Bond order = 3 e - pairs in N O bonds 2 N O bonds

68 68 Bond order is proportional to two important bond properties: (a) bond strength (b) bond length 414 kj 123 pm 110 pm 745 kj

69 69 Bond length is the distance between the nuclei of two bonded atoms.

70 Bond length depends on size of bonded atoms. 70 Cl Bond distances measured using CAChe software. In Angstrom units where 1 A = pm. I

71 Bond length depends on bond order. 71 Bond distances measured using CAChe software. In In Angstrom units where 1 A = pm.

72 ( measured by the energy req d to break a bond. See Table

73 measured by the energy req d to break a bond. See Table 9.9. BOND STRENGT (kj/mol) 436 C C 346 C=C 602 C C 835 N N 945 ( 73 The GREATER the number of bonds (bond order) the IGER the bond strength and the SORTER the bond.

74 ( 74 Bond Order Length Strength O O O=O pm kj/mol O O O ?

75 75 ' Estimate the energy of the reaction + Cl Cl ----> 2 Cl Net energy = rxn = = energy required to break bonds - energy evolved when bonds are made = 436 kj/mol Cl Cl = 242 kj/mol Cl = 432 kj/mol

76 76 ' Estimate the energy of the reaction + Cl Cl ----> 2 Cl = kj/mol Cl Cl = kj/mol Cl Cl = kj/mol Sum of - + Cl-Cl bond energies = 436 kj kj = +678 kj 2 mol -Cl bond energies = 864 kj Net = = +678 kj kj = -186 kj

77 77 Estimate the energy of the reaction 2 O O ----> O=O + 2 O Is the reaction exo- or endothermic? Which is larger: ' A) energy req d to break bonds B) or energy evolved on making bonds?

78 78 ' 2 O O ----> O=O + 2 O Energy required to break bonds: break 4 mol of O bonds = 4 (463 kj) break 2 mol O O bonds = 2 (146 kj) TOTAL ENERGY to break bonds = 2144 kj TOTAL ENERGY evolved on making O=O bonds and 4 O- O bonds bonds = 2350 kj

79 79 ' 2 O O ----> O=O + 2 O Net energy = kj kj = kj The reaction is exothermic! More energy is evolved on making bonds than is expended in breaking bonds.

80 80 "#7 Boiling point = 100 C Boiling point = -161 C Why do water and methane differ so much in their boiling points? Why do ionic compounds dissolve in water?

81 > 7" + )) :? 81

82 82 #7 Cl is # % because it has a positive end and a negative end. +δ -δ Cl Cl has a greater share in bonding electrons than does. Cl has slight negative charge (-δ) and has slight positive charge (+ δ)

83 #7 83 This model, calc d using CAChe software, shows that is + (red) and Cl is - (yellow). Calc d charge is + or (On CD see PARTCRG folder in MODELS.)

84 +δ -δ Cl BOND pure bond real bond #7 Due to the bond polarity, the Cl bond energy is GREATER than expected for a pure covalent bond. ENERGY 339 kj/mol calc d 432 kj/mol measured 84 Difference = 92 kj. This difference is proportional to the difference in &% &$&8, χ.

85 85 7) )χ χ is a measure of the ability of an atom in a molecule to attract electrons to itself. Concept proposed by Linus Pauling

86 " 86 The only person to receive two unshared Nobel prizes (for Peace and Chemistry). Chemistry areas: bonding, electronegativity, protein structure

87 7 "939 87

88 has maximum χ. C N O S P Si Atom with lowest χ is the center atom in most molecules. Relative values of χ determine BOND POLARITY (and point of attack on a molecule). Cl 7) χ See igure

89 89 #7 Which bond is more polar (or DIPOLAR)? O O χ O is more polar than O -δ O +δ +δ -δ O and polarity is reversed.

90 "#7 90 Molecules such such as Cl and 2 O can be POLAR (or dipolar). They have a # &. The polar Cl molecule will turn to align with an electric field. igure 9.15

91 91 "#7 The magnitude of the dipole is given in Debye units. Named for Peter Debye ( ). Rec d 1936 Nobel prize for work on x-ray diffraction and dipole moments.

92 2 92 Why are some molecules polar but others are not?

93 "#7 93 Molecules will be polar if a) bonds are polar AND b) the molecule is NOT symmetric All above are NOT polar

94 # 2? 94 Compare CO 2 and 2 O. Which one is polar?

95 - 95 O C O CO 2 is NOT polar even though the CO bonds are polar. CO 2 is symmetrical. Positive C atom is reason CO O gives 2 CO

96 96 Microwave oven A" + #7

97 97 # 2? Consider AB 3 molecules: B 3, Cl 2 CO, and N 3.

98 "#7) * 98 B B atom is positive and atoms are negative. B bonds in B 3 are polar. But molecule is symmetrical and NOT polar

99 "#7) + 99 B B atom is positive but & atoms are negative. B and B B bonds in B 2 are polar. But molecule is NOT symmetrical and is polar.

100 ) 0 )#? 100 C Methane is symmetrical and is NOT polar.

101 * #? 101 C C bond is very polar. Molecule is not symmetrical and so is polar.

102 (" "7 102 C bonds are MUC more polar than C bonds. Because both C C bonds are on same side of molecule, molecule is POLAR.

103 (" "7 103 C bonds are MUC more polar than C bonds. Because both C C bonds are on opposing ends of molecule, molecule is NOT POLAR. ends of molecule, molecule is NOT POLAR