John C. Kotz State University of New York, College at Oneonta John C. Kotz Paul M. Treichel John Townsend Chapter 8 Bonding and Molecular Structure: Fundamental Concepts

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3 © 2009 Brooks/Cole - Cengage CHEMICAL BONDING Cocaine PLAY MOVIE

4 © 2009 Brooks/Cole - Cengage Chemical Bonding Problems and questions — How 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? How is structure related to chemical and physical properties?

5 © 2009 Brooks/Cole - Cengage Structure & Bonding NN triple bond. Molecule is unreactive Phosphorus is a tetrahedron of P atoms. Very reactive! Red phosphorus, a polymer. Used in matches. PLAY MOVIE

6 © 2009 Brooks/Cole - Cengage Forms of Chemical Bonds There are 2 extreme forms of connecting or bonding atoms:There are 2 extreme forms of connecting or bonding atoms: Ionic —complete transfer of 1 or more electrons from one atom to anotherIonic —complete transfer of 1 or more electrons from one atom to another Covalent —some valence electrons shared between atomsCovalent —some valence electrons shared between atoms Most bonds are somewhere in between.Most bonds are somewhere in between.

7 © 2009 Brooks/Cole - Cengage Ionic Compounds Metal of low IE Nonmetal of high EA 2 Na(s) + Cl 2 (g) f 2 Na Cl -

8 © 2009 Brooks/Cole - Cengage Covalent Bonding The bond arises from the mutual attraction of 2 nuclei for the same electrons. Electron sharing results. Bond is a balance of attractive and repulsive forces. PLAY MOVIE

9 © 2009 Brooks/Cole - Cengage Bond Formation A bond can result from a “head-to-head” overlap of atomic orbitals on neighboring atoms. Cl HH + Overlap of H (1s) and Cl (2p) Note that each atom has a single, unpaired electron.

10 © 2009 Brooks/Cole - Cengage Chemical Bonding: Objectives Objectives are to understand: 1. valence e- distribution in molecules and ions. 2. molecular structures 3. bond properties and their effect on molecular properties.

11 © 2009 Brooks/Cole - Cengage Electron Distribution in Molecules Electron distribution is depicted with Lewis electron dot structuresElectron distribution is depicted with Lewis electron dot structures Valence electrons are distributed as shared or BOND PAIRS and unshared or LONE PAIRS. Valence electrons are distributed as shared or BOND PAIRS and unshared or LONE PAIRS. G. N. Lewis

12 © 2009 Brooks/Cole - Cengage Bond and Lone Pairs Valence electrons are distributed as shared or BOND PAIRS and unshared or LONE PAIRS.Valence electrons are distributed as shared or BOND PAIRS and unshared or LONE PAIRS. HCl lone pair (LP) shared or bond pair This is called a LEWIS ELECTRON DOT structure.

13 © 2009 Brooks/Cole - Cengage Valence Electrons Electrons are divided between core and valence electrons B 1s 2 2s 2 2p 1 Core = [He], valence = 2s 2 2p 1 Br [Ar] 3d 10 4s 2 4p 5 Core = [Ar] 3d 10, valence = 4s 2 4p 5

14 © 2009 Brooks/Cole - Cengage Rules of the Game No. of valence electrons of a main group atom = Group number For Groups 1A-4A, no. of bond pairs = group number. For Groups 5A -7A, BP’s = 8 - Grp. No. For Groups 5A -7A, BP’s = 8 - Grp. No.

15 © 2009 Brooks/Cole - Cengage Rules of the Game No. of valence electrons of an atom = Group number For Groups 1A-4A, no. of bond pairs = group number For Groups 5A -7A, BP’s = 8 - Grp. No.For Groups 5A -7A, BP’s = 8 - Grp. No. Except for H (and sometimes atoms of 3rd and higher periods), BP’s + LP’s = 4 This observation is called the OCTET RULE

16 © 2009 Brooks/Cole - Cengage Building a Dot Structure Ammonia, NH 3 1. Decide on the central atom; never H. Central atom is atom of lowest affinity for electrons. Central atom is atom of lowest affinity for electrons. Therefore, N is central Therefore, N is central 2. Count valence electrons H = 1 and N = 5 H = 1 and N = 5 Total = (3 x 1) + 5 Total = (3 x 1) + 5 = 8 electrons / 4 pairs = 8 electrons / 4 pairs

17 © 2009 Brooks/Cole - Cengage 3.Form a single bond between the central atom and each surrounding atom H H H N Building a Dot Structure H H H 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 H shares 1 pair.

18 © 2009 Brooks/Cole - Cengage 10 pairs of electrons are now left. 10 pairs of electrons are now left. Sulfite ion, SO 3 2- Step 1. Central atom = S Step 2. Count valence electrons S = 6 3 x O = 3 x 6 = 18 3 x O = 3 x 6 = 18 Negative charge = 2 Negative charge = 2 TOTAL = 26 e- or 13 pairs Step 3. Form bonds

19 © 2009 Brooks/Cole - Cengage Sulfite ion, SO 3 2- Remaining pairs become lone pairs, first on outside atoms and then on central atom. OO O S Each atom is surrounded by an octet of electrons.

20 © 2009 Brooks/Cole - Cengage Carbon Dioxide, CO 2 1. Central atom = _______ 2. Valence electrons = __ or __ pairs 3. Form bonds. 4. Place lone pairs on outer atoms. This leaves 6 pairs.

21 © 2009 Brooks/Cole - Cengage Carbon Dioxide, CO 2 4. Place lone pairs on outer atoms. The second bonding pair forms a pi (π) bond. 5. So that C has an octet, we shall form DOUBLE BONDS between C and O.

22 © 2009 Brooks/Cole - Cengage Double and even triple bonds are commonly observed for C, N, P, O, and S H 2 CO SO 3 C2F4C2F4C2F4C2F4

23 © 2009 Brooks/Cole - Cengage Sulfur Dioxide, SO 2 1. Central atom = S 2. Valence electrons = 18 or 9 pairs bring in left pair OR bring in right pair 3. Form double bond so that S has an octet — but note that there are two ways of doing this. OS O

24 © 2009 Brooks/Cole - Cengage Sulfur Dioxide, SO 2 This leads to the following structures. These equivalent structures are called RESONANCE STRUCTURES. The true electronic structure is a HYBRID of the two. RESONANCE STRUCTURES. The true electronic structure is a HYBRID of the two.

25 © 2009 Brooks/Cole - Cengage Urea, (NH 2 ) 2 CO

26 © 2009 Brooks/Cole - Cengage Urea, (NH 2 ) 2 CO 1. Number of valence electrons = 24 e- 2. Draw sigma bonds.

27 © 2009 Brooks/Cole - Cengage 3. Place remaining electron pairs in the molecule. Urea, (NH 2 ) 2 CO

28 © 2009 Brooks/Cole - Cengage 4. Complete C atom octet with double bond. Urea, (NH 2 ) 2 CO

29 © 2009 Brooks/Cole - Cengage Atom Formal Charges 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. Formal charge = Group number – 1/2 (no. of bonding electrons) - (no. of LP electrons)Formal charge = Group number – 1/2 (no. of bonding electrons) - (no. of LP electrons)

30 © 2009 Brooks/Cole - Cengage OOC +4 - (1/2)(8) - 0 = (1/2)(4) - 4 = 0 Carbon Dioxide, CO 2

31 © 2009 Brooks/Cole - Cengage Calculated Partial Charges in CO 2 Yellow = negative & red = positive Relative size = relative charge Yellow = negative & red = positive Relative size = relative charge

32 © 2009 Brooks/Cole - Cengage Thiocyanate Ion, SCN (1/2)(2) - 6 = (1/2)(6) - 2 = (1/2)(8) - 0 = 0 SNC

33 © 2009 Brooks/Cole - Cengage Thiocyanate Ion, SCN - SNC SNC SNC Which is the most important resonance form?

34 © 2009 Brooks/Cole - Cengage Calculated Partial Charges in SCN - All atoms negative, but most on the S SNC

35 © 2009 Brooks/Cole - Cengage Violations of the Octet Rule Usually occurs with B and elements of higher periods. BF 3 SF 4

36 © 2009 Brooks/Cole - Cengage Boron Trifluoride Central atom = _____________Central atom = _____________ Valence electrons = __________ or electron pairs = __________Valence electrons = __________ or electron pairs = __________ Assemble dot structureAssemble dot structure The B atom has a share in only 6 pairs of electrons (or 3 pairs). B atom in many molecules is electron deficient.

37 © 2009 Brooks/Cole - Cengage Boron Trifluoride, BF 3 What if we form a B—F double bond to satisfy the B atom octet? F F F B ++ --

38 © 2009 Brooks/Cole - Cengage Is There a B=F Double Bond in BF 3 F is negative and B is positive Calc’d partial charges in BF 3

39 © 2009 Brooks/Cole - Cengage Sulfur Tetrafluoride, SF 4 Central atom =Central atom = Valence electrons = ___ or ___ pairs.Valence electrons = ___ or ___ pairs. Form sigma bonds and distribute electron pairs.Form sigma bonds and distribute electron pairs. 5 pairs around the S atom. A common occurrence outside the 2nd period.

MOLECULAR GEOMETRY

41 © 2009 Brooks/Cole - Cengage VSEPR VSEPR V alence S hell E lectron P air R epulsion theory.V alence S hell E lectron P air R epulsion theory. Most important factor in determining geometry is relative repulsion between electron pairs.Most important factor in determining geometry is relative repulsion between electron pairs. Molecule adopts the shape that minimizes the electron pair repulsions. MOLECULAR GEOMETRY PLAY MOVIE

42 © 2009 Brooks/Cole - Cengage Electron Pair Geometries See Active Figure 8.5

43 © 2009 Brooks/Cole - Cengage

44 © 2009 Brooks/Cole - Cengage PLAY MOVIE

45 © 2009 Brooks/Cole - Cengage PLAY MOVIE

46 © 2009 Brooks/Cole - Cengage PLAY MOVIE

47 © 2009 Brooks/Cole - Cengage Electron Pair Geometries See Active Figure 8.5

48 © 2009 Brooks/Cole - Cengage Structure Determination by VSEPR Ammonia, NH 3 1. Draw electron dot structure 2. Count BP’s and LP’s = 4 H H H N 3. The 4 electron pairs are at the corners of a tetrahedron.

49 © 2009 Brooks/Cole - Cengage Structure Determination by VSEPR Ammonia, NH 3 There are 4 electron pairs at the corners of a tetrahedron. The ELECTRON PAIR GEOMETRY is tetrahedral.

50 © 2009 Brooks/Cole - Cengage Ammonia, NH 3 The electron pair geometry is tetrahedral. Structure Determination by VSEPR The MOLECULAR GEOMETRY — the positions of the atoms — is PYRAMIDAL. PLAY MOVIE

51 © 2009 Brooks/Cole - Cengage Structure Determination by VSEPR Water, H 2 O 1. Draw electron dot structure The electron pair geometry is TETRAHEDRAL. 2. Count BP’s and LP’s = 4 3. The 4 electron pairs are at the corners of a tetrahedron.

52 © 2009 Brooks/Cole - Cengage Structure Determination by VSEPR Water, H 2 O The electron pair geometry is TETRAHEDRAL The molecular geometry is BENT.

53 © 2009 Brooks/Cole - Cengage Geometries for Four Electron Pairs See Figure 8.6

54 © 2009 Brooks/Cole - Cengage Structure Determination by VSEPR Formaldehyde, CH 2 O 1. Draw electron dot structure The electron pair geometry is PLANAR TRIGONAL with 120 o bond angles. CHH O C HH 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.

55 © 2009 Brooks/Cole - Cengage Structure Determination by VSEPR Formaldehyde, CH 2 O The electron pair geometry is PLANAR TRIGONAL The molecular geometry is also planar trigonal.

56 © 2009 Brooks/Cole - Cengage H-C-H = 109 o C-O-H = 109 o In both cases the atom is surrounded by 4 electron pairs. Structure Determination by VSEPR 109˚ Methanol, CH 3 OH Define H-C-H and C-O-H bond angles PLAY MOVIE

57 © 2009 Brooks/Cole - Cengage Structure Determination by VSEPR Acetonitrile, CH 3 CN H H H—C—CN 180˚ 109˚ H-C-H = 109 o C-C-N = 180 o One C is surrounded by 4 electron “lumps” and the other by 2 “lumps” Define unique bond angles

58 © 2009 Brooks/Cole - Cengage Phenylalanine, an amino acid

59 © 2009 Brooks/Cole - Cengage Phenylalanine

60 © 2009 Brooks/Cole - Cengage Structures with Central Atoms with More Than or Less Than 4 Electron Pairs Often occurs with Group 3A elements and with those of 3rd period and higher.

61 © 2009 Brooks/Cole - Cengage Boron Compounds Consider boron trifluoride, BF 3 Geometry described as planar trigonal The B atom is surrounded by only 3 electron pairs. Bond angles are 120 o

62 © 2009 Brooks/Cole - Cengage 5 electron pairs Compounds with 5 or 6 Pairs Around the Central Atom PLAY MOVIE

63 © 2009 Brooks/Cole - Cengage Molecular Geometries for Five Electron Pairs See Figure 8.8 All based on trigonal bipyramid

64 © 2009 Brooks/Cole - Cengage Number of valence electrons = 34Number of valence electrons = 34 Central atom = S Central atom = S Dot structureDot structure Sulfur Tetrafluoride, SF 4 Electron pair geometry is trigonal bipyramid (because there are 5 pairs around the S) Electron pair geometry is trigonal bipyramid (because there are 5 pairs around the S)

65 © 2009 Brooks/Cole - Cengage Lone pair is in the equator because it requires more room. Sulfur Tetrafluoride, SF 4

66 © 2009 Brooks/Cole - Cengage Molecular Geometries for Six Electron Pairs See Figure 8.8 All are based on the 8- sided octahedron

67 © 2009 Brooks/Cole - Cengage 6 electron pairs Compounds with 5 or 6 Pairs Around the Central Atom PLAY MOVIE

68 © 2009 Brooks/Cole - Cengage Bond Properties What is the effect of bonding and structure on molecular properties? Free rotation around C–C single bond No rotation around C=C double bond

69 © 2009 Brooks/Cole - Cengage Bond Order # of bonds between a pair of atoms # of bonds between a pair of atoms Bond Order # of bonds between a pair of atoms # of bonds between a pair of atoms Double bond Single bond Triple bond AcrylonitrileAcrylonitrile

70 © 2009 Brooks/Cole - Cengage Bond Order Fractional bond orders occur in molecules with resonance structures. Consider NO 2 - The N—O bond order = 1.5

71 © 2009 Brooks/Cole - Cengage Bond Order Bond order is proportional to two important bond properties: (a) bond strength (b)bond length 745 kJ 414 kJ 110 pm 123 pm

72 © 2009 Brooks/Cole - Cengage Bond Length Bond length is the distance between the nuclei of two bonded atoms. PLAY MOVIE

73 © 2009 Brooks/Cole - Cengage Bond Length Bond length depends on size of bonded atoms. H—F H—Cl H—I Bond distances measured in Angstrom units where 1 A = pm.

74 © 2009 Brooks/Cole - Cengage Bond length depends on bond order. Bond distances measured in Angstrom units where 1 A = pm. Bond Length

75 © 2009 Brooks/Cole - Cengage Bond Strength —measured by the energy req’d to break a bond. See Table 8.9 PLAY MOVIE

76 © 2009 Brooks/Cole - Cengage —measured by the energy req’d to break a bond. See Table 8.9. BOND Bond dissociation enthalpy (kJ/mol) H—H436 C—C346 C=C602 C  C 835 N  N945 The GREATER the number of bonds (bond order) the HIGHER the bond strength and the SHORTER the bond. Bond Strength

77 © 2009 Brooks/Cole - Cengage

78 © 2009 Brooks/Cole - Cengage BondOrderLengthStrength HO—OH O=O pm 210 kJ/mol ?? Bond Strength

79 © 2009 Brooks/Cole - Cengage Using Bond Dissociation Enthalpies Estimate the energy of the reaction H—H(g) + Cl—Cl(g) f 2 H—Cl(g) Net energy = ∆ r H = = energy required to break bonds - energy evolved when bonds are made H—H = 436 kJ/mol Cl—Cl = 242 kJ/mol H—Cl = 432 kJ/mol H—H = 436 kJ/mol Cl—Cl = 242 kJ/mol H—Cl = 432 kJ/mol PLAY MOVIE

80 © 2009 Brooks/Cole - Cengage Estimate the energy of the reaction H—H + Cl—Cl ----> 2 H—Cl H—H = 436 kJ/mol Cl—Cl = 242 kJ/mol H—Cl = 432 kJ/mol H—H = 436 kJ/mol Cl—Cl = 242 kJ/mol H—Cl = 432 kJ/mol Sum of H-H + Cl-Cl bond energies = 436 kJ kJ = +678 kJ Using Bond Dissociation Enthalpies 2 mol H-Cl bond energies = 864 kJ Net = ∆ r H = +678 kJ kJ = -186 kJ

81 © 2009 Brooks/Cole - Cengage Estimate the energy of the reaction 2 H—O—O—H f O=O + 2 H—O—H Is the reaction exo- or endothermic? Which is larger: A)energy req’d to break bonds B)or energy evolved on making bonds? Using Bond Dissociation Enthalpies

82 © 2009 Brooks/Cole - Cengage 2 H—O—O—H f O=O + 2 H—O—H Energy required to break bonds: break 4 mol of O—H bonds = 4 (463 kJ) break 2 mol O—O bonds = 2 (146 kJ) Using Bond Dissociation Enthalpies TOTAL ENERGY evolved on making O=O bonds and 4 O-H bonds bonds = 2350 kJ TOTAL ENERGY to break bonds = 2144 kJ

83 © 2009 Brooks/Cole - Cengage 2 H—O—O—H f O=O + 2 H—O—H More energy is evolved on making bonds than is expended in breaking bonds. The reaction is exothermic! Using Bond Dissociation Enthalpies Net energy = kJ kJ = kJ

84 © 2009 Brooks/Cole - Cengage Molecular Polarity Why do ionic compounds dissolve in water? Water Boiling point = 100 ˚C Methane Boiling point = -161 ˚C Why do water and methane differ so much in their boiling points?

85 © 2009 Brooks/Cole - Cengage Bond Polarity HCl is POLAR because it has a positive end and a negative end. Cl has a greater share in bonding electrons than does H. Cl has slight negative charge (-  ) and H has slight positive charge (+  )

86 © 2009 Brooks/Cole - Cengage Bond Polarity Three molecules with polar, covalent bonds.Three molecules with polar, covalent bonds. Each bond has one atom with a slight negative charge (-  ) and and another with a slight positive charge (+  )Each bond has one atom with a slight negative charge (-  ) and and another with a slight positive charge (+  )

87 © 2009 Brooks/Cole - Cengage This model, calc’d using CAChe software for molecular calculations, shows that H is + (red) and Cl is - (yellow). Calc’d charge is + or Bond Polarity

88 © 2009 Brooks/Cole - Cengage Due to the bond polarity, the H—Cl bond energy is GREATER than expected for a “pure” covalent bond. BONDENERGY “pure” bond339 kJ/mol calc’d real bond432 kJ/mol measured BONDENERGY “pure” bond339 kJ/mol calc’d real bond432 kJ/mol measured Difference = 92 kJ. This difference is proportional to the difference in ELECTRONEGATIVITY, . Bond Polarity

89 © 2009 Brooks/Cole - Cengage Electronegativity,   is a measure of the ability of an atom in a molecule to attract electrons to itself. Concept proposed by Linus Pauling

90 © 2009 Brooks/Cole - Cengage Linus Pauling, The only person to receive two unshared Nobel prizes (for Peace and Chemistry). Chemistry areas: bonding, electronegativity, protein structure PLAY MOVIE

91 © 2009 Brooks/Cole - Cengage Electronegativity See Figure 8.11

92 © 2009 Brooks/Cole - Cengage F has maximum .F has maximum . Atom with lowest  is the center atom in most molecules.Atom with lowest  is the center atom in most molecules. Relative values of  determine BOND POLARITY (and point of attack on a molecule).Relative values of  determine BOND POLARITY (and point of attack on a molecule). Electronegativity,  See Figure 8.11

93 © 2009 Brooks/Cole - Cengage Bond Polarity Which bond is more polar (or DIPOLAR)? O—HO—F O—HO—F   OH is more polar than OF OH is more polar than OF and polarity is “reversed.”

94 © 2009 Brooks/Cole - Cengage Molecular Polarity Molecules—such as HI and H 2 O— can be POLAR (or dipolar). They have a DIPOLE MOMENT. The polar HCl molecule will turn to align with an electric field.

95 © 2009 Brooks/Cole - Cengage Molecular Polarity 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.

96 © 2009 Brooks/Cole - Cengage Dipole Moments Why are some molecules polar but others are not?

97 © 2009 Brooks/Cole - Cengage Molecular Polarity Molecules will be polar if a)bonds are polar b)AND the molecule is NOT “symmetric” All above are NOT polar

98 © 2009 Brooks/Cole - Cengage Polar or Nonpolar? Compare CO 2 and H 2 O. Which one is polar?

99 © 2009 Brooks/Cole - Cengage Carbon Dioxide CO 2 is NOT polar even though the CO bonds are polar.CO 2 is NOT polar even though the CO bonds are polar. CO 2 is symmetrical.CO 2 is symmetrical Positive C atom is reason CO 2 and H 2 O react to give H 2 CO 3

100 © 2009 Brooks/Cole - Cengage Polar or Nonpolar? Consider AB 3 molecules: BF 3, Cl 2 CO, and NH 3.

101 © 2009 Brooks/Cole - Cengage Molecular Polarity, BF 3 B atom is positive and F atoms are negative. B—F bonds in BF 3 are polar. But molecule is symmetrical and NOT polar

102 © 2009 Brooks/Cole - Cengage Molecular Polarity, HBF 2 B atom is positive but H & F atoms are negative. B—F and B—H bonds in HBF 2 are polar. But molecule is NOT symmetrical and is polar.

103 © 2009 Brooks/Cole - Cengage Is Methane, CH 4, Polar? Methane is symmetrical and is NOT polar.

104 © 2009 Brooks/Cole - Cengage Is CH 3 F Polar? C—F bond is very polar. Molecule is not symmetrical and so is polar.

105 © 2009 Brooks/Cole - Cengage CH 4 … CCl 4 Polar or Not? Only CH 4 and CCl 4 are NOT polar. These are the only two molecules that are “symmetrical.”

106 © 2009 Brooks/Cole - Cengage Substituted Ethylene C—F bonds are MUCH more polar than C—H bonds.C—F bonds are MUCH more polar than C—H bonds. Because both C—F bonds are on same side of molecule, molecule is POLAR.Because both C—F bonds are on same side of molecule, molecule is POLAR.

107 © 2009 Brooks/Cole - Cengage Substituted Ethylene C—F bonds are MUCH more polar than C—H bonds.C—F bonds are MUCH more polar than C—H bonds. Because both C—F bonds are on opposing ends of molecule, molecule is NOT POLAR.Because both C—F bonds are on opposing ends of molecule, molecule is NOT POLAR.

108 © 2009 Brooks/Cole - Cengage Visualizing Charges and Polarity Electrostatic Potential Surfaces Electrostatic potential surfaces (EPS) can be used to visualize a) a)Where the charges lie in molecules b) b)The polarity of molecules See page 382 The HF molecule

109 © 2009 Brooks/Cole - Cengage Visualizing Charges and Polarity The boundary surface around the molecule is made up of all the points in space where the electron density is a given value (here e - /A 3 ). The colors indicate the potential experienced by a H + ion on the surface. More attraction (a negative site) is red, and repulsion (a positive site) is blue. FH

110 © 2009 Brooks/Cole - Cengage Visualizing Charges and Polarity As expected, the surface near O in H 2 O and the N is CH 3 -NH 2 is red because that is the more electronegative atom. H 2 O is polar - with O more negative and H more positive. CH 3 -NH 2 is also polar.

111 © 2009 Brooks/Cole - Cengage Visualizing Charges and Polarity The EP surfaces show BF 3 is not polar (but the B is slightly positively charged), whereas Cl 2 CO is polar with the O more negative than Cl. BF 3 Cl 2 CO

112 © 2009 Brooks/Cole - Cengage Visualizing Charges and Polarity The EP surfaces show cis-C 2 H 2 Cl 2 (left) is polar whereas trans-C 2 H 2 Cl 2 (right) is not polar.