Chapter 10TMHsiung ©2014Slide 1 of 80 Chapter Ten Chemical Bonding II : Molecular Shapes, Valence Bond Theory, and Molecular Orbital Theory

Chapter 10TMHsiung ©2014Slide 2 of 80 1.Artificial Sweeteners: Fooled by Molecular Shape 2.VSEPR Theory: The Five Basic Shapes 3.VSEPR Theory: The Effect of Lone Pairs 4.VSEPR Theory: Predicting Molecular Geometries 5.Molecular Shape and Polarity 6.Valence Bond Theory: Orbital Overlap as a Chemical Bond 7.Valence Bond Theory: Hybridization of Atomic Orbitals 8.Molecular Orbital Theory: Electron Delocalization Contents

Chapter 10TMHsiung ©2014Slide 3 of 80 1.Artificial Sweeteners: Fooled by Molecular Shape The sugar molecule (the key) enters the active site (the lock ─ sugar receptor protein) of taste cell, resulting in ion channels opened, nerve signal transmission, and reaching the brain a sweet taste. Artificial sweeteners such as aspartame and saccharin bind to the active more strongly than sugar. Similarities in the shape of sucrose and artificial sweeteners give those sweeteners the ability to stimulate a sweet taste sensation.

Chapter 10TMHsiung ©2014Slide 4 of 80 Valence Shell Electron Pair Repulsion (VSEPR) theory: A theory that allows prediction of the shapes of molecules or polyatomic ion based on the idea that electrons ˗ either as lone pairs or as bonding pairs ˗ repel one another. Electron geometry: The geometrical arrangement of electron groups in a molecule. Molecular geometry: The geometrical arrangement of atoms in a molecule. *****

Chapter 10TMHsiung ©2014Slide 5 of 80  VSEPR theory proceeding i)Write a best Lewis structure ii)Determine VSEPR notation: AX m E n : A:Central atoms X:Terminal atoms E:Lone pairs electrons H 2 O for example: AX2E2AX2E2 *****

Chapter 10TMHsiung ©2014Slide 6 of 80 iii)Determine the electron geometry An electron group can be: - either single bond or a multiple bond - a (resonance) hybrid bond - a lone pairs of electron - a unpaired single-electron Repulsion force in general: LP vs. LP > LP vs. BP > BP vs. BP * Lone Pairs (LP), Bonding Pairs (BP) Angle for repulsion forces: 90° > 120° > 180° For central (interior) atom belong to third-period or higher element with VSEPR notation such as AX 5, AX 4 E, AX 3 E 2, AX 6, AX 5 E, AX 4 E 2 require an expanded octet such as 3d orbital. Multiple bond occupy more space than single bond *****

Chapter 10TMHsiung ©2014Slide 7 of 80 iv)Determine the molecular geometry Structures for the central atom without lone-pair electrons (AX n type), electron geometry and molecular geometry are identical. Structures for the central atom with lone-pair electrons (AX n E m type) type), electron geometry and molecular geometry are different. *****

Chapter 10TMHsiung ©2014Slide 8 of 80 *****

Chapter 10TMHsiung ©2014Slide 9 of 80 *****

Chapter 10TMHsiung ©2014Slide 10 of 80 *Count only electron groups around the central atom. Each of the following is considered one electron group: a lone pair, a single bond, a double bond, a triple bond, or a single electron. *****

Chapter 10TMHsiung ©2014Slide 11 of 80 2.VSEPR Theory: The Five Basic Shapes (All electrons around the central atom are bonding group) Two Electron Groups (AX 2 ): Linear

Chapter 10TMHsiung ©2014Slide 12 of 80 Three Electron Groups (AX 3 ): Trigonal Planar *double bond contains more electron density than the single bond

Chapter 10TMHsiung ©2014Slide 13 of 80 Four Electron Groups (AX 4 ): Tetrahedral Five Electron Groups (AX 5 ): Trigonal Bipyramidal

Chapter 10TMHsiung ©2014Slide 14 of 80 Six Electron Groups (AX 6 ): Octahedral

Chapter 10TMHsiung ©2014Slide 15 of 80 Solution NO 3 − has 5 + 3(6) + 1 = 24 valence electrons. The Lewis structure has three resonance structures: Use any one of the resonance structures to determine the number of electron groups around the central atom. The nitrogen atom has three electron groups. The electron geometry is trigonal planar: The molecular geometry is also trigonal planar. Determine the molecular geometry of NO 3 −. Example 10.1VSEPR Theory and the Basic Shapes

Chapter 10TMHsiung ©2014Slide 16 of 80 3.VSEPR Theory: The Effect of Lone Pairs (Some electrons around the central atom are lone pairs) Four Electron Groups with Lone Pairs AX 3 E AX 2 E 2

Chapter 10TMHsiung ©2014Slide 17 of 80 *Effect of Lone Pairs on Molecular Geometry

Chapter 10TMHsiung ©2014Slide 18 of 80 Five Electron Groups with Lone Pairs AX 4 E AX 3 E 2 AX 2 E 3

Chapter 10TMHsiung ©2014Slide 19 of 80 Six Electron Groups with Lone Pairs AX 5 E AX 4 E 2

Chapter 10TMHsiung ©2014Slide 20 of 80 Solution Step 1PCl 3 has 26 valence electrons. Lewis structure for the molecule: Step 2The central atom (P) has four electron groups. Step 3Three of the four electron groups around P are bonding groups and one is a lone pair. Step 4The electron geometry is tetrahedral The molecular geometry is trigonal pyramidal Predict the geometry and bond angles of PCl 3. Example 10.2Predicting Molecular Geometries 4.VSEPR Theory: Predicting Molecular Geometries

Chapter 10TMHsiung ©2014Slide 21 of 80 Solution Step 1ICl 4 − has 36 valence electrons. Lewis structure for the molecule: Step 2The central atom (I) has six electron groups. Step 3Four of the six electron groups around I are bonding groups and two are lone pairs. Step 4The electron geometry is octahedral The molecular geometry is square planar. Predict the geometry and bond angles of ICl 4 −. Example 10.3Predicting Molecular Geometries

Chapter 10TMHsiung ©2014Slide 22 of 80 Representing Molecular Geometries on Paper Examples:

Chapter 10TMHsiung ©2014Slide 23 of 80 Predicting the Shapes of Larger Molecules Example:

Chapter 10TMHsiung ©2014Slide 24 of 80 Solution The Lewis structure of CH 3 OH. Three-dimensional sketch of the molecule: Predict the geometry about each interior atom in methanol (CH 3 OH) and make a sketch of the molecule. Example 10.4Predicting the Shape of Larger Molecules

Chapter 10TMHsiung ©2014Slide 25 of 80 VSEPR Notation Electron Geometry Molecular Geometry AX 2 Linear AX 3 Trigonal planar AX 4 Tetrahedral AX 5 Trigonal bipyramidal AX 6 Octahedral Memo for VSEPR Without lone-pair electrons *****

Chapter 10TMHsiung ©2014Slide 26 of 80 VSEPR Notation Electron Geometry Molecular Geometry AX 2 ETrigonal planarBent AX 3 ETetrahedralTrigonal pyramidal AX 2 E 2 TetrahedralBent AX 4 ETrigonal bipyramidalSeesaw AX 3 E 2 Trigonal bipyramidalT-shaped AX 2 E 3 Trigonal bipyramidalLinear AX 5 EOctahedralSquare pyramidal AX 4 E 2 OctahedralSquare planar With lone-pair electrons *****

Chapter 10TMHsiung ©2014Slide 27 of 80 5.Molecular Shape and Polarity Bond dipole versus Molecular dipole Bond dipole: A separation of positive and negative charge in an individual bond. Molecular dipole: For diatomic molecule: molecular dipole is identical to bond dipole. For a molecule consisted by three or more atoms, molecular dipole is estimated by the vector sum of individual bond dipole moment (net dipole moment).

Chapter 10TMHsiung ©2014Slide 28 of 80 Polar molecule versus Nonpolar molecule Polar molecule: A molecule in which the molecular dipole is nonzero. Nonpolar molecule: A molecule in which the molecular dipole is zero. Molecular polarity prediction Draw the Lewis structure for the molecule and determine its molecular geometry. Determine if the molecule contains polar bonds by electronegativity values. Determine if the polar bonds add together to form a net dipole moment.

Chapter 10TMHsiung ©2014Slide 29 of 80 Examples H 2 O Molecular geometry: bent (net) dipole moment:  = 1.84 D Polar molecule CO 2 Molecular geometry: linear (net) dipole moment:  = 0 D Nonpolar molecule *****

Chapter 10TMHsiung ©2014Slide 30 of 80 Solution Lewis structure: Determine if the molecule contains polar bonds. The electronegativities of nitrogen and hydrogen are 3.0 and 2.1, respectively. Determine if the polar bonds add together to form a net dipole moment. The three dipole moments sum to a net dipole moment. Ans: The molecule is polar. Determine if NH 3 is polar. Example 10.5Determining if a Molecule Is Polar

Chapter 10TMHsiung ©2014Slide 31 of 80 Polarity effects of the intermolecular forces Example 1: For H 2 O Example 2: Like dissolves like Polar molecules interact strongly with other polar molecules excluding the nonpolar molecules and separating into distinct regions.

Chapter 10TMHsiung ©2014Slide 32 of 80  Quantum-Mechanical Approximation Technique Perturbation theory (used in valence bond theory): A complex system (such as a molecule) is viewed as a simpler system (such as two atoms) that is slightly altered or perturbed by some additional force or interaction (such as the interaction between the two atoms). Variational method (used in molecular orbital theory): The energy of a trial function (educated function) within the Schrodinger equation is minimized.

Chapter 10TMHsiung ©2014Slide 33 of 80  Schrodinger equation revisited H  = E  H (Hamiltonian operator), a set of mathematical operations that represent the total energy (kinetic and potential) of the electron within the atom. E is the actual energy of the electron.  is the wave function, a mathematical function that describes the wavelike nature of the electron.  Perturbation theory: Approach by small changes to a known system in which Hamiltonian operator is modified.  Variational method: Approach by combining systems of comparable weighting in which wave function is modified.

Chapter 10TMHsiung ©2014Slide 34 of 80  Valence bond theory versus molecular orbital theory Valence bond theory (VB): An advanced model of chemical bonding in which electrons reside in quantum-mechanical orbitals localized on individual atoms that are a hybridized blend of standard atomic orbitals; chemical bonds result from an overlap of these orbitals. Molecular orbital theory (MO): An advanced model of chemical bonding in which electrons reside in molecular orbitals delocalized over the entire molecule. In the simplest version, the molecular orbitals are simply linear combinations of atomic orbitals.

Chapter 10TMHsiung ©2014Slide 35 of 80 6.Valence Bond Theory: Orbital Overlap as a Chemical Bond Valence bond theory describes that covalent bonds are formed when atomic orbitals on different atoms overlap. Simple Atomic Orbitals (AO’s) Overlap Bonding in H 2 for example A covalent bond is formed by the pairing of two electrons with opposing spins in the region of overlap of atomic orbitals between two atoms. This overlap region has a high electron charge density. The overall energy of the system is lowered.

Chapter 10TMHsiung ©2014Slide 36 of 80

Chapter 10TMHsiung ©2014Slide 37 of 80 Acceptable simple Atomic Orbitals (AO’s) Overlap Bonding in H 2 S for example Predicted H ˗ S ˗ H angle is 90 o, actual H ˗ S ˗ H angle is 92 o, therefore, the simple AO overlap is acceptable for H 2 S molecule.

Chapter 10TMHsiung ©2014Slide 38 of 80 Unacceptable simple Atomic Orbitals (AO’s) Overlap Example 1: CH 4 Example 2: NH 3 and H 2 O Ground-state electron configuration of C for example, it should form only 2 bonds Actually, the central atom of H 2 S, H 2 O, NH 3, and CH 4, are sp 3 hybridization C

Chapter 10TMHsiung ©2014Slide 39 of 80 7.Valence Bond Theory: Hybridization of Atomic Orbitals Hybridization: A mathematical procedure in which standard atomic orbitals are combined to form new, hybrid orbitals. Hybridizing is mixing different types of orbitals in the valence shell to make a new set of degenerate orbitals such as sp, sp 2, sp 3, sp 3 d, sp 3 d 2. Hybrid orbitals minimize the energy of the molecule by maximizing the orbital overlap in a bond. Those central atoms are available hybridized, however, those terminal atoms are supposed to be unhybridized. *****

Chapter 10TMHsiung ©2014Slide 40 of 80 General statements regarding hybridization Hybridization is employed for central atom only, thus, the hybrid orbital describes the electron geometry for central atom. Number of hybrid orbitals = Number of standard atomic orbitals combined = Number of σ bond + Number of lone pairs. Number of hybridization obitals of a central atom = 2 → sp; = 3 → sp 2 ; = 4 → sp 3 ; = 5 → sp 3 d; = 6 → sp 3 d 2. Hybrid orbitals may overlap with standard atomic orbitals or with other hybrid orbitals to form σ bond. Molecular geometry is described by the relative atomic position around central atom. *****

Chapter 10TMHsiung ©2014Slide 41 of 80 sp 3 hybridization (C for example) one s orbital with three p orbitals combine to form four sp 3 hybrid orbitals (degenerate).

Chapter 10TMHsiung ©2014Slide 42 of 80

Chapter 10TMHsiung ©2014Slide 43 of 80 Examples of sp 3 hybridization (for central atom) Central atom Mole- cule Standard orbitals Hybrid Orbital C sp 3 2p 2s sp 3 2s 2p sp 3 2p 2s N O Geometry lone σσσσ σσσ σσ *****

Chapter 10TMHsiung ©2014Slide 44 of 80 sp 2 hybridization (B for example) one s orbital with two p orbitals combine to form three sp 2 hybrid orbitals

Chapter 10TMHsiung ©2014Slide 45 of 80 Examples of sp 2 hybridization (for central atom) B 2p 2s 2p C sp 2 2p N 2s 2p sp 2 2p Central atom Mole- cule Standard orbitals Hybrid Orbital Unhybridized Orbital sp 2 σσσ σσσ σσ lone π π *****

Chapter 10TMHsiung ©2014Slide 46 of 80 sp hybridization (Be for example) one s orbital with one p orbitals combine to form two sp hybrid orbitals

Chapter 10TMHsiung ©2014Slide 47 of 80 Examples of sp hybridization (for central atom) Be sp 2p 2s 2p C sp 2p Central atom Mole- cule Standard orbitals Hybrid Orbital Unhybridized Orbital σσ σσ π π *****

Chapter 10TMHsiung ©2014Slide 48 of 80 About Multiple Covalent Bond σ (sigma) bond: The first covalent bond formed by end-to-end overlap of standard or hybridized orbitals between the bonded atoms: s + s, s + p, p + p (end-to-end), s + hybrid orbital p + hybrid orbital, hybrid orbital + hybrid orbital π (Pi) bond: The second (and third, if present) bond in a multiple bond, results from side-by-side overlap of unhybridized p orbitals: p + p (side-by-side) Summary: -Single bonds:one σ bond -Double bond:one σ bond and one π bond -Triple bond:one σ bond and two π bonds *****

Chapter 10TMHsiung ©2014Slide 49 of 80 Sigma Bonding and Pi Bonding

Chapter 10TMHsiung ©2014Slide 50 of 80 VB theory of bonding in ethylene (H 2 C=CH 2 ) example of sp 2 hybridization and a double bond A π-bond has two lobes (above and below plane), but is one bond, side-by-side overlap of 2p–2p Lewis structure

Chapter 10TMHsiung ©2014Slide 51 of 80 Continued All six atoms in C 2 H 4 lie in the same plane

Chapter 10TMHsiung ©2014Slide 52 of 80 VB theory of bonding in Acetylene (HC  CH) example of sp hybridization and a triple bond Two π-bonds from 2p–2p overlap forming a cylinder of π- electron density around the two carbon atoms Lewis structure

Chapter 10TMHsiung ©2014Slide 53 of 80 Continued

Chapter 10TMHsiung ©2014Slide 54 of 80 VB theory of bonding in Formaldehyde (H 2 C=O) example of sp 2 hybridization and a double bond Lewis structure

Chapter 10TMHsiung ©2014Slide 55 of 80 Continued Valence bond model σσ πσ

Chapter 10TMHsiung ©2014Slide 56 of 80 Bond Rotation *Rotation around  bond does not require breaking the bond, however,  bond interact above and below the internuclear axis, so rotation around the axis requires the breaking the bond. * In general, bond energy of  bonds are weaker than that of  bonds because the side-to-side orbital overlap tends to be less efficient than the end-to-end orbital overlap.

Chapter 10TMHsiung ©2014Slide 57 of 80 cis versus trans Isomers

Chapter 10TMHsiung ©2014Slide 58 of 80 Expanded Octet hybridization sp 3 d hybridization, AsF 5 for example

Chapter 10TMHsiung ©2014Slide 59 of 80 Continued

Chapter 10TMHsiung ©2014Slide 60 of 80 sp 3 d 2 hybridization, SF 6 for example

Chapter 10TMHsiung ©2014Slide 61 of 80 Continued

Chapter 10TMHsiung ©2014Slide 62 of 80 Number of σ + lone Hybridi- zation VSEPR notation Electron geometry Molecular geometry Example 2spAX 2 LinearlinearCl-Be-Cl 3 sp 2 AX 3 AX 2 E Trigonal planar Angular BCl 3 SO 2 4sp 3 AX 4 AX 3 E AX 2 E 2 Tetrahedral Trigonal pyramidal Angular CH4NH3H2OCH4NH3H2O 5 sp 3 dAX 5 AX 4 E AX 3 E 2 AX 2 E 3 Trigonal bipyramidal Seesaw T-shaped Linear PBr 5 SF 4 ClF 3 XeF 2 6 sp 3 d 2 AX 6 AX 5 E AX 4 E 2 Octahedral Square pyramidal Square planar SF 6 BrF 5 XeF 4 Example for hybridization/electron geometry types versus molecular geometry *****

Chapter 10TMHsiung ©2014Slide 63 of 80 Procedure for Hybridization and Bonding Scheme 1.Write the Lewis structure for the molecule. 2.Use VSEPR theory to predict the electron geometry about the central atom. 3.Select the correct hybridization for the central atom based on the electron geometry. 4.Sketch the molecule, beginning with the central atom and its orbitals. Show overlap with the appropriate orbitals on the terminal atoms. 5.Label all bonds using the σ or π notation followed by the type of overlapping orbitals. *****

Chapter 10TMHsiung ©2014Slide 64 of 80 Solution Step 1Lewis structure Step 2The leftmost carbon atom, tetrahedral electron geometry. The rightmost carbon atom, trigonal planar geometry. Step 3The leftmost carbon atom, sp 3 hybridized The rightmost carbon atom, sp 2 hybridized. Write a hybridization and bonding scheme for acetaldehyde, Example 10.7Hybridization and Bonding Scheme

Chapter 10TMHsiung ©2014Slide 65 of 80 Step 4The appropriate orbitals on the terminal atoms. Step 5Type of overlapping orbitals. Continued *****

Chapter 10TMHsiung ©2014Slide 66 of 80 8.Molecular Orbital Theory: Electron Delocalization Chemical Bond Molecular Orbital (MO): A model of chemical bonding in which electrons reside in molecular orbitals delocalized over the entire molecule. The molecular orbitals are linear combinations of atomic orbitals (LCAO). Because the orbitals are wave functions, the waves can combine either constructively or destructively.

Chapter 10TMHsiung ©2014Slide 67 of 80 MOs formed by combining two 1s AOs

Chapter 10TMHsiung ©2014Slide 68 of 80 MOs formed by combining two set 2p AOs σ 2p and σ 2p *: end-to-end overlap of AOs π 2p and π 2p *: side-by-side overlap of AOs

Chapter 10TMHsiung ©2014Slide 69 of 80 Summarizing LCAO–MO Theory: The total number of MOs formed from a particular set of AOs always equals the number of AOs in the set. When two AOs combine to form two MOs, one MO is lower in energy (the bonding MO) and the other is higher in energy (the antibonding MO). When assigning the electrons of a molecule to MOs, fill the lowest energy MOs first with a maximum of two spin-paired electrons per orbital. When assigning electrons to two MOs of the same energy, follow Hund’s rule—fill the orbitals singly first, with parallel spins, before pairing. *****

Chapter 10TMHsiung ©2014Slide 70 of 80 Applications of MOs Estimate the bond order: Bond Order (BO) = (Σ bonding e – － Σ antibonding e – )/2 Predict the existence of molecule Estimating bond length and bond energy Predicting magnetic properties *****

Chapter 10TMHsiung ©2014Slide 71 of 80 1st Period Homonuclear Diatomic MOs H 2 and He 2 for example: AOs of H (two 1s AOs) MOs of H 2 AOs of He (two 1s AOs) MOs of He 2 σ 1s σ 1s * σ 1s σ 1s * BO = (2−0)/2 = 1 H 2 molecule does exist Diamagnetic BO = (2−2)/2 = 0 He 2 molecule does not exist *****

Chapter 10TMHsiung ©2014Slide 72 of 80 2nd Period Homonuclear Diatomic MOs Effects of 2s–2p Mixing: Increasing energy difference, decreasing the degree of mixing.

Chapter 10TMHsiung ©2014Slide 73 of 80 Continued *****

Chapter 10TMHsiung ©2014Slide 74 of 80 Lewis structure For O 2 : Experiment showed O 2 is paramagnetic MO prove O 2 have unpaired electrons Predicting magnetic properties by MOs

Chapter 10TMHsiung ©2014Slide 75 of 80 2nd Period Heteronuclear Diatomic MOs NO for example Oxygen is more electronegative than nitrogen, so its atomic orbitals are lower in energy than nitrogen’s atomic orbitals. The lower energy atomic orbital makes a greater contribution to the bonding molecular orbital and the higher energy atomic orbital makes a greater contribution to the antibonding molecular orbital.

Chapter 10TMHsiung ©2014Slide 76 of 80 HF for example

Chapter 10TMHsiung ©2014Slide 77 of 80 Solution The N 2 − ion has 11 valence electrons (5 for each nitrogen atom plus 1 for the negative charge). Assign the electrons to the molecular orbitals beginning with the lowest energy orbitals and following Hund’s rule. The bond order is 2.5, bond order for N 2 molecule is 3, the bond is weaker. N 2 − ion has one unpaired electron and is therefore paramagnetic. Draw an MO energy diagram and determine the bond order for the N 2 − ion. Do you expect the bond to be stronger or weaker than in the N 2 molecule? Is N 2 − diamagnetic or paramagnetic? Example 10.10Molecular Orbital Theory

Chapter 10TMHsiung ©2014Slide 78 of 80 Solution Number of valence electrons = 4 (from C) + 5 (from N) + 1 (from negative charge) = 10 Write an energy level diagram Since the MO diagram has no unpaired electrons, the ion is diamagnetic. Use molecular orbital theory to determine the bond order of the CN − ion. Is the ion paramagnetic or diamagnetic? Example 10.11Molecular Orbital Theory for Heteronuclear Diatomic Molecules and Ions

Chapter 10TMHsiung ©2014Slide 79 of 80 Polyatomic Molecules Example: O 3 Lewis modelVB modelMO model Example: C 6 H 6 Lewis modelMO model

Chapter 10TMHsiung ©2014Slide 80 of 80 End of Chapter 10