Bonding: General Concepts A computer representation of K3C60, a superconducting substance formed by reacting potassium with buckminster fullerine (C60) Source: Photo Researchers, Inc. Copyright © Houghton Mifflin Company. All rights reserved. 13a–2 Two forms of carbon; graphite and diamond. Source: Grant Hellman Copyright © Houghton Mifflin Company. All rights reserved. 13a–3 Quartz grows in beautiful, regular crystals SiO2 Vs CO2 Copyright © Houghton Mifflin Company. All rights reserved. 13a–4 Figure 13.1: (a) The interaction of two hydrogen atoms (b) Energy profile as a function of the distance between the nuclei of the hydrogen atoms. Copyright © Houghton Mifflin Company. All rights reserved. 13a–5 Figure 13.1: (a) The interaction of two hydrogen atoms (b) Energy profile as a function of the distance between the nuclei of the hydrogen atoms. Copyright © Houghton Mifflin Company. All rights reserved. 13a–6 WHERE DO THE ELECTRONS GO? • Are they shared equally? • Are they more on one atom than the other? Copyright © Houghton Mifflin Company. All rights reserved. 13a–7 WHERE DO THE ELECTRONS GO? • Are they shared equally? • Are they more on one atom than the other? Copyright © Houghton Mifflin Company. All rights reserved. 13a–8 WHERE DO THE ELECTRONS GO? • Are they shared equally? • Are they more on one atom than the other? ANSWER: It depends who is pulling harder Copyright © Houghton Mifflin Company. All rights reserved. 13a–9 WHERE DO THE ELECTRONS GO? • Are they shared equally? • Are they more on one atom than the other? ANSWER: It depends who is pulling harder (“Electro negativity”) Copyright © Houghton Mifflin Company. All rights reserved. 13a–10 Copyright © Houghton Mifflin Company. All rights reserved. 13a–11 The HCL molecule has a dipole moment Copyright © Houghton Mifflin Company. All rights reserved. 13a–12 Pauling and his electronegativity Copyright © Houghton Mifflin Company. All rights reserved. 13a–13 Pauling and his electronegativity Copyright © Houghton Mifflin Company. All rights reserved. 13a–14 Figure 13.3: The Pauling electronegativity values as updated by A.L. Allred in 1961. Expect A-B bond energy to be average of A-A and B-B If it is not electro negativities must be different (Stronger bond – more ionic – larger EN difference) Copyright © Houghton Mifflin Company. All rights reserved. 13a–15 Figure 13.3: The Pauling electronegativity values as updated by A.L. Allred in 1961. (cont’d) Arbitrarily set F as 4 Copyright © Houghton Mifflin Company. All rights reserved. 13a–16 Copyright © Houghton Mifflin Company. All rights reserved. 13a–17 Copyright © Houghton Mifflin Company. All rights reserved. 13a–18 The HCL molecule has a dipole moment Copyright © Houghton Mifflin Company. All rights reserved. 13a–19 Copyright © Houghton Mifflin Company. All rights reserved. 13a–20 Copyright © Houghton Mifflin Company. All rights reserved. 13a–21 Copyright © Houghton Mifflin Company. All rights reserved. 13a–22 Copyright © Houghton Mifflin Company. All rights reserved. 13a–23 models Now lets consider more than two atoms in a molecule Copyright © Houghton Mifflin Company. All rights reserved. 13a–24 Linear molecules Copyright © Houghton Mifflin Company. All rights reserved. 13a–25 Planar molecules Copyright © Houghton Mifflin Company. All rights reserved. 13a–26 Tetrahedral molecules Copyright © Houghton Mifflin Company. All rights reserved. 13a–27 MODELS • These are only models Copyright © Houghton Mifflin Company. All rights reserved. 13a–28 MODELS • These are only models • But…. Models are very useful for describing properties. Copyright © Houghton Mifflin Company. All rights reserved. 13a–29 MODELS • These are only models • But…. Models are very useful for describing properties. Newton: particles Copyright © Houghton Mifflin Company. All rights reserved. 13a–30 MODELS • These are only models • But…. Models are very useful for describing properties. Newton: particles Huygens: waves Copyright © Houghton Mifflin Company. All rights reserved. 13a–31 models Copyright © Houghton Mifflin Company. All rights reserved. 13a–32 Ball-and-stick model of a protein segment illustrating the alpha helix. Source: Photo Researchers, Inc. Copyright © Houghton Mifflin Company. All rights reserved. 13a–33 The concept of individual bonds makes it much easier to deal with complex molecules such as DNA. Source: Photo Researchers, Inc. Copyright © Houghton Mifflin Company. All rights reserved. 13a–34 Copyright © Houghton Mifflin Company. All rights reserved. 13a–35 Polarity of Molecules Dipole Moments of Polyatomic Molecules Example: in CO2, each C-O dipole is canceled because the molecule is linear. In H2O, the H-O dipoles do not cancel because the molecule is bent. O Copyright © Houghton Mifflin Company. All rights reserved. C O 13a–36 Skeletal Structure • Hydrogen atoms are always terminal atoms. • Central atoms are generally those with the lowest electronegativity. • Carbon atoms are always central atoms. • Generally structures are compact and symmetrical. Copyright © Houghton Mifflin Company. All rights reserved. 13a–37 Skeletal Structure • Identify central and terminal atoms in the molecule C2H6O (ethyl alcohol or ethanol). NOTE: Terminal atoms are all bonded to only one other atoms. Central atoms are bonded to two or more other atoms Now where do the electrons go? Copyright © Houghton Mifflin Company. All rights reserved. H H H C C O H H H 13a–38 Writing Lewis Structures • All the valence e- of atoms must appear. • Usually, the e- are paired. • Usually, each atom requires an octet. – H only requires 2 e-. • Multiple bonds may be needed. – Readily formed by C, N, O, S, and P. Copyright © Houghton Mifflin Company. All rights reserved. 13a–39 Lewis Structures Draw Lewis structures for: CH4: H N H H H H C H H Copyright © Houghton Mifflin Company. All rights reserved. or or or or H O H NH3: H2O: H F HF: H F H O H H N H H H H C H H Exceptions to the Octet Rule • Molecules with an odd number of electrons. • Molecules in which an atom has less than an octet of electrons. • Molecules in which an atom has more than an octet of electrons. Copyright © Houghton Mifflin Company. All rights reserved. Resonance Forms • Lewis structures that differ only in the placement of electrons are resonance forms. For O3: O O O = O O =O • Experimentally, it is found that both bonds are 0.128 nm long. • The Lewis structure of O3 must show both resonance forms. Copyright © Houghton Mifflin Company. All rights reserved. SAMPLE PROBLEM 10.4 O Writing Resonance Structures PROBLEM: Write resonance structures for the nitrate ion, NO3-. SOLUTION: Nitrate has 1(5) + 3(6) + 1 = 24 valence e- O O O N N N O O O O O N does not have an octet; a pair of ewill move in to form a double bond. O O O O N N N O Copyright © Houghton Mifflin Company. All rights reserved. O O O O 13a–45 Draw the resonance hybrid structure of NO3- O O O N N O O O -1 -1 N O O O The three resonance structures O Hybrid structure. Averaging leads to partial N-O double bond character (solid + dotted lines) -1 N O Copyright © Houghton Mifflin Company. All rights reserved. O 13a–46 Resonance Forms C 2H 6O 20 ve’s H H H H H C O C H H C C O H H H H H Methyl ether Ethyl alcohol Copyright © Houghton Mifflin Company. All rights reserved. Benzene Fredrich August von Kekule (German chemist) said that he discovered the ring-shaped chemical structure of benzene because of a strange, reptilian dream he had in 1865: "I turned my chair to the fire and dozed. Again the atoms were gamboling before my eyes. ... My mental eye... could not distinguish larger structures, of manifold conformation; long rows, sometimes more closely fitted together; all twining and twisting in snakelike motion. But look! What was that? One of the snakes had seized hold of its own tail, and the form whirled mockingly before my eyes. As if by a flash of lighting I awoke... " ( From "Creativity, Beyond the Myth of Genius" by Robert Weisberg published by W. H. Freeman 1992 .) Although some scholars now believe that Kekule's dream was a hoax to avoid sharing credit for the discovery of the hexagonal shape of benzene, it still makes a wonderful story. Copyright © Houghton Mifflin Company. All rights reserved. 13a–48 Resonance Forms 30 ve’s C 6H 6 H H C C H C C H C C H H H Copyright © Houghton Mifflin Company. All rights reserved. H C C H C C H C C H H Nonequivalent Forms The molecule N2O has three resonance forms: N NO N =N=O N N O Which one contributes most to the overall electronic structure? Copyright © Houghton Mifflin Company. All rights reserved. Formal Charge: Selecting the Best Resonance Structure An atom “owns” all of its nonbonding electrons and half of its bonding electrons. Formal charge of atom = # valence e- - (# unshared electrons + 1/2 # shared electrons) B For OA # valence e- O # nonbonding # bonding e- # valence e- = 6 O =6 e- For OC =4 = 4 X 1/2 = 2 A For OB Formal charge = 0 O # nonbonding e- = 6 C # valence # bonding e- = 2 X 1/2 = 1 Formal charge = -1 e- =6 # nonbonding e- = 2 # bonding e- = 6 X 1/2 = 3 Formal charge = +1 Copyright © Houghton Mifflin Company. All rights reserved. 13a–51 Formal Charges The formal charge on an atom assumes that the bonding electrons are shared equally. Formal charge = # valence electrons - # nonbonding electrons - ½ # bonding electrons Copyright © Houghton Mifflin Company. All rights reserved. 0 N =N=O 0 +1 -1 N N O N N O -1 +1 -2 +1 +1 Formal charges in N2O: Formal Charges The most favored resonance form is the one that has: • The smallest formal charges. • No formal charges with the same sign on adjacent atoms. • Negative formal charges on the most electronegative atoms. Copyright © Houghton Mifflin Company. All rights reserved. Formal Charges The formal charge on an atom assumes that the bonding electrons are shared equally. Formal charge = # valence electrons - # nonbonding electrons - ½ # bonding electrons 0 N =N=O 0 +1 -1 N N O N N O -1 +1 -2 +1 +1 Formal charges in N2O: Most favorable Copyright © Houghton Mifflin Company. All rights reserved. Resonance (continued) EXAMPLE: NCO- has 3 possible resonance forms - N C O N C A N C O B O C formal charges -2 0 N C +1 O -1 0 N C 0 O 0 0 N C -1 O Forms B and C have negative formal charges on N and O; this makes them more important than form A. Form C has a negative charge on O which is the more electronegative element, therefore C contributes the most to the resonance hybrid. Copyright © Houghton Mifflin Company. All rights reserved. 13a–55 Electron-Deficient Atoms • Molecules containing Be and B can have less than an octet for these atoms. F = Be = F 0 0 0 F Be F • BeF2: +1 -2 +1 Most Favorable • The small Be atom can’t accomodate a large formal charge. Copyright © Houghton Mifflin Company. All rights reserved. Electron-Deficient Atoms +1 -1 BF3: 0 F =B F F B F F B F F 0 F 0 0 0 0 F Copyright © Houghton Mifflin Company. All rights reserved. F B= F F Most Favorable Electron-Rich Atoms 0 An atom that has a d subshell in the valence electron shell E can accomodate more than an octet of electrons. Nitrogen can’t accommodate more than 8 electrons. Copyright © Houghton Mifflin Company. All rights reserved. 4s 3s 2s 3p 2p N in NF3 1s 3d Electron-Rich Atoms An atom that 0 has a d subshell 4s 3d 3p in the valence 3s electron shell E 2p can accomodate 2s more than an octet of electrons. P can accommodate P in PF5 more than 8 electrons 1s Copyright © Houghton Mifflin Company. All rights reserved. Electron-Rich Atoms PF5 F F F P F F Copyright © Houghton Mifflin Company. All rights reserved. Exceptions to the Octet Rule • Expanded octets. P Cl F S P Cl Cl Cl F F F F •• Cl •• •• Cl F •• Cl •• •• Cl •• •• •• •• •• •• Copyright © Houghton Mifflin Company. All rights reserved. 13a–61 Molecular Shapes •Lewis structures give atomic connectivity: they tell us which atoms are physically connected to which. •The shape of a molecule is determined by its bond angles. Copyright © Houghton Mifflin Company. All rights reserved. 13a–62 Molecular Shapes In order to predict molecular shape, we assume the valence electrons repel each other. Therefore, the molecule adopts whichever 3D geometry minimizes this repulsion. We call this process Valence Shell Electron Pair Repulsion (VSEPR) theory. Copyright © Houghton Mifflin Company. All rights reserved. 13a–63 The VSEPR Model Predicting Molecular Geometries Copyright © Houghton Mifflin Company. All rights reserved. 13a–64 The VSEPR Model Predicting Molecular Geometries Copyright © Houghton Mifflin Company. All rights reserved. 13a–65 Molecular Shapes experimentally we find all Cl-C-Cl bond angles are 109.5. Therefore, the molecule cannot be planar. All Cl atoms are located at the vertices of a tetrahedron with the C at its center. Copyright © Houghton Mifflin Company. All rights reserved. 13a–66 The VSEPR Model Predicting Molecular Geometries Copyright © Houghton Mifflin Company. All rights reserved. 13a–67 The VSEPR Model Predicting Molecular Geometries To determine the electron pair geometry: •draw the Lewis structure •count the total number of electron pairs around the central atom •arrange the electron pairs in one of the above geometries to minimize e--e- repulsion •multiple bounds count as one bonding pair Copyright © Houghton Mifflin Company. All rights reserved. 13a–68 The VSEPR Model Predicting Molecular Geometries Copyright © Houghton Mifflin Company. All rights reserved. 13a–69 The VSEPR Model Predicting Molecular Geometries Copyright © Houghton Mifflin Company. All rights reserved. 13a–70 AB2 Linear AB3 Trigonal planar AB5 Trigonal bipyramidal Molecular Shapes AB2E Angular or Bent AB4 Tetrahedral AB4E Irregular tetrahedral (see saw) AB6 Octahedral Copyright © Houghton Mifflin Company. All rights reserved. AB3E Trigonal pyramidal AB3E2 T-shaped AB5E Square pyramidal AB2E2 Angular or Bent AB2E3 Linear AB4E2 Square planar 13a–71 The VSEPR Model The valence electrons in a molecule are the bonding pairs of electrons as well as the lone pairs. There are 11 shapes that are important to us: Number of atoms, formula Shapes (3 atoms, AB2) (4 atoms, AB3) (5 atoms, AB4) (6 atoms, AB5) (7 atoms, AB6) linear or bent trigonal planar, trigonal bipyramidal, or T-shaped tetrahedral, square planar, or see-saw trigonal bipyramidal or square pyramidal octahedral Copyright © Houghton Mifflin Company. All rights reserved. 13a–72 The VSEPR Model Copyright © Houghton Mifflin Company. All rights reserved. 13a–73 The VSEPR Model The Effect of Nonbonding Electrons and Multiple Bonds on Bond Angles By experiment, the H-X-H bond angle decreases on moving from C to N to O: H H C H H 109.5O H N H H 107O O H H 104.5O Since electrons in a bond are attracted by two nuclei, they do not repel as much as lone pairs. Therefore, the bond angle decreases as the number of lone pairs increase. Copyright © Houghton Mifflin Company. All rights reserved. 13a–74 The VSEPR Model The Effect of Nonbonding Electrons and Multiple Bonds on Bond Angles Similarly, electrons in multiple bonds repel more than electrons in single bonds. Cl 111.4o Cl Copyright © Houghton Mifflin Company. All rights reserved. C O 124.3o 13a–75 The VSEPR Model Molecules with Expanded Valence Shells Copyright © Houghton Mifflin Company. All rights reserved. 13a–76 The VSEPR Model Molecules with Expanded Valence Shells Copyright © Houghton Mifflin Company. All rights reserved. 13a–77 The VSEPR Model Molecules with More than One Central Atom In acetic acid, CH3COOH, there are three central atoms. We assign the geometry about each central atom separately. H O H C C O 3 4 H H Number of electron domains Electron-domain geometry Predicted bond angles Copyright © Houghton Mifflin Company. All rights reserved. 4 Tetrahedral Trigonal planar Tetrahedral 109.5o 120o 109.5o 13a–78 The VSEPR Model Molecules with More than One Central Atom In acetic acid, CH3COOH, there are three central atoms. We assign the geometry about each central atom separately. Copyright © Houghton Mifflin Company. All rights reserved. 13a–79 Figure 13.6: The carbon dioxide molecule • Nonpolar Molecules – Dipole moments are symmetrical and cancel out. Copyright © Houghton Mifflin Company. All rights reserved. 13a–82 Determining Molecular Polarity • Nonpolar Molecules – Dipole moments are symmetrical and cancel out. F BF3 B F Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem Copyright © Houghton Mifflin Company. All rights reserved. F 13a–83 Determining Molecular Polarity • Polar Molecules – Dipole moments are asymmetrical and don’t cancel . O H2O H Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem Copyright © Houghton Mifflin Company. All rights reserved. H net dipole moment 13a–84 Determining Molecular Polarity polar molecules have... – asymmetrical shape (lone pairs) or – asymmetrical atoms H CHCl3 Cl Cl Cl Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem Copyright © Houghton Mifflin Company. All rights reserved. net dipole moment 13a–85 Figure 13.5: (a) The structure and charge distribution of the ammonia molecule. (b) The dipole moment of the ammonia molecule oriented in an electric field. Copyright © Houghton Mifflin Company. All rights reserved. 13a–86 Copyright © Houghton Mifflin Company. All rights reserved. 13a–87 Sulfur has a partial positive charge Copyright © Houghton Mifflin Company. All rights reserved. 13a–88 Hydrogen atoms have a partial positive charge Copyright © Houghton Mifflin Company. All rights reserved. 13a–89 Hydrogen atoms and a small partial negative charge on the carbon Copyright © Houghton Mifflin Company. All rights reserved. 13a–90 Polar Bonds .. F N O Cl H H Polar H F Polar H B Polar Cl F F Cl Polar F Cl F H Cl F H F Nonpolar H C Xe F F Nonpolar C Cl Cl Nonpolar H H Polar A molecule has a zero dipole moment when bond dipoles cancel one another. Copyright © Houghton Mifflin Company. All rights reserved. 13a–91 Covalent Bond Energy • Covalent bond energy is measured by the energy required to break the bond. • The bond enthalpy, D(X-Y) is the average H for breaking one mole of X-Y bonds in the gas phase: C + O C O D(C-O) = H = 358 kJ • When one mole of X-Y bonds is formed, the enthalpy change is -D(X-Y). Copyright © Houghton Mifflin Company. All rights reserved. Bond Enthalpies and Bond Lengths As bond order increases, the bond enthalpy increases and the bond length decreases. D(C-C) = 348 kJ 0.154 nm D(C=C) = 614 kJ 0.134 nm D(CC) = 839 kJ 0.120 nm D(C-O) = 358 kJ 0.143 nm D(C=O) = 799 kJ 0.123 nm D(CO) = 1072 kJ 0.113 nm Copyright © Houghton Mifflin Company. All rights reserved. Copyright © Houghton Mifflin Company. All rights reserved. 13a–94 Bond Enthalpies and Hrxn Consider the reaction of H2 and O2 to form H2O: H H + H H + O =O H O H + H O H H for breaking bonds = 2 D(H–H) + D(O=O) H for forming bonds = 4-D(O–H) Hrxn = 2 D(H–H) + D(O=O) - 4 D(O–H) H = D(bonds broken) - D(bonds formed) Copyright © Houghton Mifflin Company. All rights reserved. Bond Enthalpies and Hrxn Estimate H for the combustion of CH4: H H C H + 2 O =O H O=C=O +2 H O H H = 4 D(C–H) + 2 D(O=O) - 2 D(C=O) - 4 D(O–H) = [ 4(413) + 2(495) - 2(799) - 4(463) ] kJ = -808 kJ Copyright © Houghton Mifflin Company. All rights reserved. Comparing fuels • Natural gas: CH4 + 2O2 → CO2 + 2 H2O ΔH=-808 kJ/mol • Coal: C + O2 → CO2 ΔH=-393.5 kJ/mol • Oil: C20H42 + 30½O2 → 20CO2 + 21 H2O ΔH=-13315 kJ/mol ΔH=-666 kJ/mol.CO2 Copyright © Houghton Mifflin Company. All rights reserved. 13a–97 Comparing fuels Production of 1 GigaJoule of energy releases: • Natural gas: 109 J x 0.044 kg/mol ÷ 808,000 J/mol = 54.5 kg CO2 • Coal: 112 kg CO2 • Oil: 66 kg CO2 Copyright © Houghton Mifflin Company. All rights reserved. 13a–98 N3 high energy density 3 Copyright © Houghton Mifflin Company. All rights reserved. 2 13a–99 ATP energy Copyright © Houghton Mifflin Company. All rights reserved. 13a–100 ATP energy Repulsion weakens these bonds Resonance! Copyright © Houghton Mifflin Company. All rights reserved. 13a–101