Chapter 10: Chemical Bonding II. Bonding Theories

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1 Chapter 10: Chemical Bonding II Dr. Chris Kozak Memorial University of Newfoundland, Canada Bonding Theories Previously, we saw how the shapes of molecules can be predicted from the orientation of electron domains (bonding or lone-pair electrons) around a central atom. How exactly do these electron domains fit in with what we discussed in atomic orbital theory? What we will now learn is that these atomic orbitals can be used as a starting point to discuss the location of electrons in molecules, or molecular orbitals. These molecular orbitals are the foundation of current bonding theory. Of course, there are many levels of sophistication to these theories and the exact level of complexity we choose depends on what we need to know. Lewis theory has its shortcomings. It does not explain conduction or semiconductors. More sophisticated approaches are required. Slide 2 1

2 What a Bonding Theory Should Do Bring atoms together from a distance. e - are attracted to both nuclei. e - are repelled by each other. Nuclei are repelled by each other. Plot the total potential energy verses distance. -ve energies correspond to net attractive forces. +ve energies correspond to net repulsive forces. Slide 3 Potential Energy Diagram Slide 4 2

3 Introduction to the Valence-Bond Method A covalent bond forms when the orbitals of two atoms overlap and the overlap region, which is between the nuclei, is occupied by a pair of electrons. The two wave functions are in phase so the amplitude increases between the nuclei. A localized model of bonding. Slide 5 Central Themes to V-B Theory A set of overlapping orbitals has a maximum of two electrons that must have opposite spins. The greater the orbital overlap, the stronger (more stable) the bond. The valence atomic orbitals in a molecule are different from those in isolated atoms. There is a hybridization (mixture of properties) of atomic orbitals to form molecular orbitals. Slide 6 3

4 Orbital Overlap and Spin Pairing Hydrogen, H 2 Hydrogen fluoride, HF Fluorine, F 2 Slide 7 Bonding in H 2 S This proposes a 90 o H-S-H angle! But VSEPR predicts ~ o (actual is o ) How is this possible? Orbital Hybridization! Note + and regions of the orbitals (a result of the AO Ψ where + and amplitudes exist). This phase of the AOs is important in forming MOs, since constructive and destructive interference occur. Slide 8 4

5 Example Using the Valence-Bond Method to Describe a Molecular Structure. Describe the phosphine molecule, PH 3, by the valencebond method. Identify valence electrons: Slide 9 Example Sketch the orbitals: Overlap the orbitals: A good approximation, but it isn t truly accurate Describe the shape: Trigonal pyramidal Slide 10 5

6 Hybridization of Atomic Orbitals C C Not really an excited state, but instead consider the 2s and 2p orbitals now become degenerate Hybridization does this! Slide 11 Key Points Hybrid Orbitals The number of hybrid orbitals obtained equals the number of atomic orbitals mixed. The type of hybrid orbitals obtained varies with the types of atomic orbitals mixed. Types of Hybrid Orbitals (These give the 5 main shapes!) sp sp 2 sp 3 sp 3 d?? sp 3 d 2?? Trigonal bipyramidal Linear Trigonal planar Octahedral Tetrahedral See sp 3 d and sp 3 d 2 used in older textbooks but new theory shows this isn t really true Slide 12 6

7 sp Hybridization The hybridized orbitals possess a weighted average of the energies of the s and p orbitals combined Slide 13 Orbitals in Beryllium (Linear Molecules) There are two p orbitals left over The p z and p y orbitals are unused Slide 14 7

8 Orbitals in Beryllium sp Hybrid atomic orbitals hybrid orbitals Slide 15 sp Hybridization in BeCl 2 orbital box diagrams with orbital contours Slide 16 8

9 sp 2 Hybridization Note that sp 2 orbitals are slightly higher in energy than sp orbitals Two p orbitals are now combined with a s orbital, therefore, the weighted average is higher Slide 17 Orbitals in Boron (Trigonal Planar) There is one p orbital left over, in this case, p z is still unused Slide 18 9

10 sp 2 Hybridization in BF 3 The vacant p orbital is why BF 3 can act as a Lewis acid This is where electrons from a Lewis base may reside to form a coordinate covalent bond Slide 19 sp 3 Hybridization This allows for a tetrahedral arrangement around a central atom like C. Slide 20 10

11 sp 3 Hybridization The four sp 3 orbitals are now degenerate Four electron pairs of equal energy are now possible in molecules like CH 4 Slide 21 Orbitals in Methane (Tetrahedral) All valence orbitals are used Slide 22 11

12 Bonding in Methane H-C-H bond angles are o An ideal tetrahedron Changing a H atom to another atom or to a lone pair causes a change in the bond angles Slide 23 sp 3 Hybridization in Nitrogen One sp 3 orbital is doubly occupied a lone pair! Can only make three covalent bonds now Slide 24 12

13 sp 3 Hybridization in NH 3 Similar to CH 4, but the lone pair occupies a different volume of space than a H atom The shape is similar, but the angles are not exactly o Slide 25 Bonding in NH 3 (Trigonal Pyramidal) Slide 26 13

14 Draw the Lewis structure for ethenoxyethene, CH 2 CHOCHCH 2, depicted to the right. What is the hybridiza?on of the carbon and oxygen atoms indicated by the numbers 1, 2, and 3, respec?vely?. A. sp 3, sp 3, sp 3 B. sp 3, sp 2, sp 3 C. sp 3, sp 3, sp 2 D. sp 3, sp 2, sp 2 E. sp 2, sp 2, sp 2 F. sp 2, sp 2, sp 3 G. sp 2, sp 2, sp % 14.3% 14.3% 14.3% 14.3% 14.3% 14.3% A. B. C. D. E. F. G. sp 3 d and sp 3 d 2 Hybridization Slide 28 14

15 The sp 3 d hybrid orbitals in PCl 5 Hybrid orbital theory explains expanded octets by using a d orbital to form hybrids A simple but often debated theory. We will see the basics of a more comprehensive theory (Molecular Orbital Theory) later Slide 29 The sp 3 d 2 hybrid orbitals in SF 6 Can use two d orbitals, three p orbitals and one s orbital to give us an octahedral geometry, sp 3 d 2 hybrid Slide 30 15

16 Hybrid Orbitals and VSEPR Write a plausible Lewis structure. Use VSEPR to predict electron geometry. Select the appropriate hybridization. Slide 31 Postulating Hybrid Orbitals in a Molecule PROBLEM: Use partial orbital diagrams to describe mixing of the atomic orbitals of the central atom leads to hybrid orbitals in each of the following: (a) Methanol, CH 3 OH (b) Sulfur tetrafluoride, SF 4 PLAN: Use the Lewis structures to ascertain the arrangement of groups and shape of each molecule. Postulate the hybrid orbitals. Use partial orbital box diagrams to indicate the hybrid for the central atoms. SOLUTION: (a) CH 3 OH H H C H O H The groups around C are arranged as a tetrahedron. O also has a tetrahedral arrangement with 2 nonbonding e - pairs. Slide 32 16

17 Postulating Hybrid Orbitals in a Molecule 2p sp 3 2p sp 3 hybridized 2s single C atom hybridized C atom 2s single O atom O atom (b) SF 4 has a seesaw shape with 4 bonding and 1 nonbonding e - pairs. F F F S F 3d 3d 3p 3s S atom sp 3 d hybridized S atom Slide 33 Different Bonding Types σ-bond (sigma bond) Orbital overlap along the internuclear axis Present in single bonds π-bond (pi bond) Orbital overlap above and below internuclear axis Present in double and triple bonds (along with a σ-bond) Slide 34 17

18 Properties of σ vs π Bonds Symmetry is important in deciding the type of overlap A σ bond has the highest degree of overlap (stronger) A π bond has a weaker degree of overlap Due to symmetry (no node visible along bond axis), σ bonds can rotate! A π bond has a node along the bond axis and cannot rotate! A single bond has only a σ overlap A double bond has one σ and one π overlap A triple bond has one σ and two π overlaps Transition metals may have quadruple bonds! (σ, π and δ overlaps, but not part of Chem 1050) Slide 35 Multiple Covalent Bonds Ethylene has a double bond in its Lewis structure. VSEPR says trigonal planar at carbon. sp 2 hybrid and a singly occupied p-orbital Slide 36 18

19 Ethylene The π bond means the molecule is rigid and cannot rotate along the C=C axis without breaking the electron density Slide 37 Acetylene Acetylene, C 2 H 2, has a triple bond. VSEPR says linear at carbon. Also rigid along the C C axis (two π bonds!) Slide 38 19

20 Molecular Orbital Theory Atomic orbitals are isolated on atoms. Molecular orbitals span two or more atoms. LCAO Linear Combination of Atomic Orbitals. Ψ 1 = φ 1 + φ 2 Ψ 2 = φ 1 - φ 2 For larger molecules, these MOs become very complicated functions requiring lots of computing power to solve Slide 39 What is Molecular Orbital Theory? Scientists choose the model that best helps them answer a particular question If we want to know shape, use VSEPR along with Valence-Bond and Hybrid- Orbital Theories But, VB doesn t explain magnetic & spectral properties (colours of compounds!) and only considers localized electron density MO theory considers the TOTAL electron densities and bonding properties according to symmetry (constructive and destructive orbital overlaps) In MO theory, the molecule is described as a collection of nuclei with the electron orbitals delocalized over the entire molecule Slide 40 20

21 How does it work? Key ideas: formation of MOs from AOs, energy and shape of MOs and how they are populated by electrons Approximations are used: Mathematically combine (add or subtract) the atomic orbitals (specifically, the atomic wave functions) of nearby atoms to form MOs (molecular wave functions) Addition of two wavefunctions gives a bonding MO (increases probability of finding electrons between the nuclei giving a bond) Subtraction of two wavefunctions gives an antibonding MO (decreases probability of finding electrons between two nuclei to zero, a node) This means number of MOs will always be the same as the number of AOs Slide 41 How does it work? Bonding MOs are lower in energy than their antibonding MOs AOs must be of similar energy and orientation (the symmetry) to combine Bonding and antibonding MOs most commonly combine in two types of symmetry, σ (sigma) and π (pi) bonds Electrons fill MOs just as they do AOs, following Aufbau, Pauli and Hund rules MOs are best represented in a Molecular Orbital Diagram The MOs are labelled according to their bonding symmetry (σ or π) and the orbitals from which they are made (1s, 2p 3d etc.) Antibonding orbitals are denoted by a superscript star (*) Eg. 1s 1 + 1s 1 = (σ 1s ) 2 1s 1 1s 1 = (σ 1s *) 2 Slide 42 21

22 Combining Atomic Orbitals Sigma antibonding MO ( sigma star orbital) Sigma bonding MO Slide 43 Molecular Orbitals of Hydrogen Two AOs give two MOs Antibonding orbital is higher in energy than its bonding orbital Slide 44 22

23 Basic Ideas Concerning MOs Number of MOs = Number of AOs. Bonding and antibonding MOs formed from AOs. e - fill the lowest energy MO first (Aufbau Process). Pauli exclusion principle is followed (only two electrons per orbital, singly occupied orbitals have parallel spin). Hund s rule is followed (degenerate orbitals filled singly first) Slide 45 Bond Order Stable species have more electrons in bonding orbitals than antibonding (means a bond is present) Bond Order = No. e- in bonding MOs - No. e- in antibonding MOs 2 Slide 46 23

24 Diatomic Molecules of the First-Period BO = (e - bond - e- antibond )/2 BO = (1-0)/2 = ½ H 2 + BO = (2-0)/2 = 1 H 2 BO He 2 + = (2-1)/2 = ½ BO = (2-2)/2 = 0 He 2 Slide 47 Molecular Orbitals of the Second Period First period use only 1s orbitals. Second period have 2s and 2p orbitals available. p orbital overlap: End-on overlap is best sigma bond (σ). Side-on overlap is good pi bond (π). Slide 48 24

25 Molecular Orbitals of the Second Period Bonding symmetry (in phase) Antibonding symmetry (anti phase) Slide 49 Combining p orbitals Slide 50 25

26 Expected MO Diagram of C 2 Expected MO suggests C 2 is paramagnetic. It isn t! Slide 51 Modified MO Diagram of C 2 This is more accurate. Why do π 2p and σ 2p orbitals change place? s-p orbital mixing (consider it a type of hybridization) Slide 52 26

27 MO Diagrams of 2 nd Period Diatomics Do not memorize these, but know how to fill in the electrons Z 7 Z 8 The MO diagram for O 2 predicts a paramagnetic (2 upe) molecule, whereas VSEPR and VB theory do not Slide 53 Contours and energies of s and p MOs through combinations of 2p atomic orbitals. Slide 54 27

28 Relative MO energy levels for Period 2 homonuclear diatomic molecules. without 2s-2p mixing with 2s-2p mixing MO energy levels for O 2, F 2, and Ne 2 MO energy levels for B 2, C 2, and N 2 Slide 55 Orbital Mixing Explained s-p mixing means s & p orbitals of the same atom both contribute to the bonding. When does s-p mixing occur? O, F, and Xe all have > 1/2 filled 2p orbitals Having > 1/2 filled 2p orbitals raises the energies of these orbitals due to e - - e - repulsion If 2 electrons are forced to be in the same orbital, their energies go up. (This is the basis of Hund s Rule for orbital filling.) s-p mixing only occurs when the s and p atomic orbitals are close in energy ( 1/2 filled 2p orbitals, as in Li 2 to N 2 ). If they are close in E, then can hybridize! Slide 56 28

29 MO occupancy and molecular properties for B 2 through Ne 2 Slide 57 MO diagram for He 2 + and He 2. σ* 1s σ* 1s Energy 1s 1s Energy 1s 1s σ 1s σ 1s AO of He MO of He 2 + AO of He + AO of He MO of He 2 AO of He He 2 + bond order = He 2 bond order = Slide 58 29

30 SAMPLE PROBLEM Predicting Stability of Species Using MO Diagrams PROBLEM: Use MO diagrams to predict whether H 2 + and H 2 - exist. Determine their bond orders and electron configurations. PLAN: Use H 2 as a model and accommodate the number of electrons in bonding and antibonding orbitals. Find the bond order. SOLUTION: bond order bond order σ* σ* 1s 1s 1s 1s AO of H σ MO of H 2 + AO of H configuration is AO of H AO of H - σ configuration is MO of H - 2 Slide 59 Bonding in s-block homonuclear diatomic molecules. Energy σ* 2s σ* 2s 2s 2s 2s 2s σ 2s σ 2s Li 2 Be 2 Li 2 bond order = Be 2 bond order = Slide 60 30

31 SAMPLE PROBLEM Using MO Theory to Explain Bond Properties PROBLEM: As the following data show, removing an electron from N 2 forms an ion with a weaker, longer bond than in the parent molecules, whereas the ion formed from O 2 has a stronger, shorter bond: N 2 N 2 + O 2 O 2 + Bond energy (kj/mol) Bond length (pm) Explain these facts with diagrams that show the sequence and occupancy of MOs. PLAN: Find the number of valence electrons for each species, draw the MO diagrams, calculate bond orders, and then compare the results. SOLUTION: N 2 has 10 valence electrons, so N 2 + has 9. O 2 has 12 valence electrons, so O 2 + has 11. Slide 61 SAMPLE PROBLEM Using MO Theory to Explain Bond Properties N 2 N 2 + O 2 O 2 + σ* 2p σ* 2p π* 2p π* 2p σ 2p π 2p π 2p σ 2p σ* 2s σ* 2s bond orders: σ 2s σ 2s Slide 62 31

32 The MO diagram for HF σ* Energy 1s 2p x 2p y 2p σ AO of H 2s MO of HF 2s AO of F Slide 63 σ* 2p The MO diagram for CO 2p π* 2p Energy σ 2p π 2p 2p Lewis structure C O 2s σ* 2s AO of C MO of CO σ 2s 2s AO of O Slide 64 32

33 σ* 2p The MO diagram for NO π* 2p Energy 2p σ 2p 2p possible Lewis structures π 2p 0 0 N O σ* 2s s AO of N MO of NO σ 2s 2s AO of O N O Slide 65 33

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