Ch 11 Valence Bond And Molecular Orbital Theories PDF

Ch. 11 – Quantum Mechanical Theories of Covalent Bonding Chemistry The Molecular Nature of Matter and Change Ninth Edition Martin S. Silberberg and Patricia G. Amateis ©McGraw-Hill Education. All rights reserved. Authorized only for instructor use in the classroom. No reproduction or further distrib...

Ch. 11 – Quantum Mechanical Theories of Covalent Bonding Chemistry The Molecular Nature of Matter and Change Ninth Edition Martin S. Silberberg and Patricia G. Amateis ©McGraw-Hill Education. All rights reserved. Authorized only for instructor use in the classroom. No reproduction or further distribution permitted without the prior written consent of McGraw-Hill Education. 11.1 Valence Bond (VB) Theory Any quantum mechanical treatment of the covalent bond must be able to explain: – what the covalent bond means in terms of molecular orbitals; – must be able to explain the bond angles observed experimentally According to the Valence Bond theory: – A covalent bond forms when the valence atomic orbitals of two reacting atoms overlap and a pair of electrons occupy the overlap region. The overlapping atomic orbitals represent the molecular orbital, now extending over 2 atoms. – The space formed by the overlapping orbitals can accommodate a maximum of two electrons and these electrons must have opposite (paired) spins. The Pauli principle continues to hold for molecular orbitals. – The greater the orbital overlap, the stronger the bond. – Extent of orbital overlap depends on orbital shape and direction of overlap. ©McGraw-Hill Education. Orbital Orientation and Overlap ©McGraw-Hill Education. Valence Bond Theory The Schrödinger equation cannot be solved exactly for multibody quantum systems such as molecules: 𝐻𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒 Ψ𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒 = 𝐸𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒 Ψ𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒 While the Hamiltonian operator Hmolecule can be set to include all the interactions, both attractive and repulsive, between electrons and multiple nuclei, the equation cannot be solved because it contains too many variables. In the Valence Bond theory, approximations are made to solve the Schrödinger equation. First approximation is that a chemical bond is localized between 2 atoms, therefore the number of nuclei is 2 and the number of electrons involved in the bond is 2 with opposite spins. A second approximation in VB theory is that the molecular orbital function is simply constructed as an algebraic linear combination of products of atomic orbitals for each participating electron: Ψ𝐻2 = Ψ𝐻(1𝑠)−1 𝑒1− Ψ𝐻(1𝑠)−2 𝑒2− + Ψ𝐻(1𝑠)−1 (𝑒2− )Ψ𝐻(1𝑠)−2 (𝑒1− ) https://chem.libretexts.org/LibreTexts/University_of_California_Davis/UCD_Chem_107B%3A_Physi cal_Chemistry_for_Life_Scientists/Chapters/5%3A_The_Chemical_Bond/5.2%3A_Valence_Bond_Th eory © 2017 Pearson Education, Inc. Orbital Diagram for the Formation of H2S VB predicts bond angle = 90° Actual bond angle = 92° VSEPR predicts 109.5 bond angle and tetrahedral electron geometry. © 2017 Pearson Education, Inc. Valence Bond Theory: Hybridization One of the limitations in VB theory is that the number of partially filled or empty atomic orbitals did not predict the number of bonds or orientation of bonds for some elements, especially C, Si, Ge. C = 2s22px12py12pz0 would predict two bonds that are 90° apart, rather than four bonds that are 109.5° apart. To adjust for these inconsistencies, Linus Pauling postulated that the valence atomic orbitals could hybridize right before bonding takes place in the transition state. Linus Pauling (1901-1994) Nobel Prize in Chemistry (1954) for hybridization and covalent bonding. © 2017 Pearson Education, Inc. Unhybridized C Orbitals Predict the Wrong Bonding and Geometry CH2 sextet C never displays a valence of 2 in Chemistry! © 2017 Pearson Education, Inc. VB Theory and Orbital Hybridization The orbitals that form during covalent bonding are different from the atomic orbitals in the isolated atoms. If no change occurred, we could not account for the molecular shapes and valences that are observed experimentally. Atomic orbitals “mix” or hybridize when bonding occurs to form hybrid orbitals. The spatial orientations of these hybrid orbitals correspond with observed molecular shapes. This is how VB theory bridges to VSEPR theory. Mixing of atomic orbitals into hybrid orbitals is a mathematical technique, well beyond the scope of this course. ©McGraw-Hill Education. Features of Hybrid Orbitals The number of hybrid orbitals formed equals the number of atomic orbitals mixed. The type of hybrid orbitals formed varies with the types of atomic orbitals mixed. The shape of the hybrid orbitals is the same for all resulting hybrid orbitals. The shape and orientation of a hybrid orbital maximizes overlap with the other atom in the bond and gives the correct bond angles. The energy of the electrons in hybrid orbitals is the same (degenerate) and intermediate between the energies of the electrons in the mixing atomic orbitals. We postulate the type of hybrid orbitals in a molecule after we observe its shape experimentally. ©McGraw-Hill Education. sp Hybrid Orbitals and the Bonding in BeCl2 E ©McGraw-Hill Education. sp Hybridization: linear geometry in VSEPR Figure 11.2 ©McGraw-Hill Education. The sp2 Hybrid Orbitals in BF3 The label sp2 shows only the number and type of atomic orbitals used in mixing, not the number of electrons. E B ground state: [He] 2s2 2p1 ©McGraw-Hill Education. sp2 Hybridization: trigonal planar geometry The resulting three sp2 hybrid orbitals from mixing are in the same plane and 120 apart. It explains the trigonal planar shape of the molecule and the bond angle. The empty, unhybridized p atomic orbital is perpendicular to the molecular plane. ©McGraw-Hill Education. The sp3 Hybrid Orbitals in CH4 Tetrahedral geometry E C ground state: [He] 2s2 2p2 ©McGraw-Hill Education. The sp3 Hybrid Orbitals in NH3 Trigonal pyramidal geometry E N ground state: [He] 2s2 2p3 ©McGraw-Hill Education. The sp3 Hybrid Orbitals in H2O Bent (V-shaped) geometry E O ground state: [He] 2s2 2p4 ©McGraw-Hill Education. From Molecular Formula to Hybrid Orbitals Figure 11.8 ©McGraw-Hill Education. Sample Problem 11.1 – Problem and Plan Postulating Hybrid Orbitals in a Molecule PROBLEM: Use partial orbital diagrams to describe how mixing of the atomic orbitals of the central atom(s) leads to hybrid orbitals in each of the following: (a) Methanol, CH3OH (b) Sulfur tetrafluoride, SF4 PLAN: We use the molecular formula to draw the Lewis structure and determine the electron-group arrangement around each central atom. We then postulate the type of hybrid orbitals required and write a partial orbital diagram. ADDITIONAL PRACTICE: SiCl4, NO2, PCl3 ©McGraw-Hill Education. Sample Problem 11.1 – CH3OH SOLUTION: (a) CH3OH: The Lewis structure of CH3OH shows that the C and O atoms each have four electron groups, with a tetrahedral electron-group arrangement around both atoms: The C and O atoms have each mixed one 2s and three 2p to become sp3 hybridized. The C atom has four halffilled sp3orbitals to form the four bonds: The O atom has two half-filled sp3 orbitals that are used to form two bonds and two sp3 orbitals filled with lone pairs: ©McGraw-Hill Education. Limitations of the Hybridization Model Hybridization is not always consistent with observed molecular shapes. – This is particularly true for the bonding of larger atoms (period 3 or higher). – The bond angle of 92° is close to the 90° angle of the atomic (unhybridized) p orbitals. Bond length, atomic size, and electrostatic repulsions influence shape. Larger atoms form longer bonds to H, which decreases repulsions; thus, overlap of unhybridized orbitals adequately explains these shapes. d-Orbitals do not hybridize effectively with s and p orbitals, which are much lower in energy and more stable. ©McGraw-Hill Education. 11.2 Types of Molecular Orbitals A sigma (σ) bond is formed by end-to-end overlap of orbitals. – All single bonds are σ bonds. – In any multiple bond, one of them is always a σ bond. A pi (p) bond is formed by sideways overlap of orbitals. – A p bond is weaker than a σ bond because sideways overlap is less effective than end-to-end overlap. – Pi (p) bonds always accompany one σ bond in a multiple bond. A double bond consists of one σ bond and one p bond. A triple bond consists of one σ bond and two p bonds. ©McGraw-Hill Education. Sigma (σ) Bond A sigma (σ) bond is formed by end-to-end overlap of orbitals. Sigma bonds are the most common bonds. All types of atomic orbitals, whether hybridized or not, are able to form sigma bonds. Hybrid orbitals always overlap the larger lobe in a sigma bond. Here is an incomplete list of sigma bonds: ©McGraw-Hill Education. Sigma (σ) Bond A sigma bond has its highest electron density along the bond axis. Note that the overlap is best when the resulting molecular orbital has cylindrical symmetry. Banana type of σ molecular orbitals are very rare in Chemistry. ©McGraw-Hill Education. Pi (π) Bond A pi (π) bond is formed by side-to-side overlap of 2 lobes of atomic orbitals that have 2 or more lobes: p-p, p-d, d-d, etc. An s orbital cannot form a π bond. Pi (π) bonds occur in double or triple bonds and accompany 1 sigma bond. The resulting π molecular orbital has the highest electron density split in two lobes, on above and one below the nodal plane. In the nodal plane, the electron density is zero, similar to the atomic orbitals that form the π bond. ©McGraw-Hill Education. Carbon atoms in sp3 hybridization: C2H6 In ethane (C2H6), both carbon atoms adopt sp3 hybridization. All bond angles are close to 109.5, all shapes are tetrahedral around the carbon atoms. All bonds are σ bonds. ©McGraw-Hill Education. Carbon atoms in sp2 hybridization: C2H4 Carbon adopts sp2 hybridization in C2H4. How is it possible? Note that the pi (π) bond plane is perpendicular to the molecular plane. ©McGraw-Hill Education. Carbon atoms in sp hybridization: C2H2 How does Carbon adopt sp hybridization? ©McGraw-Hill Education. Electron Density in Ethane, Ethylene, and Acetylene A double bond is less than twice as strong as a single bond, because a p bond is weaker than a σ bond. ©McGraw-Hill Education. Sample Problem 11.2 Describing the Types of Bonds in Molecules PROBLEM: Describe the types of bonds and orbitals in acetone, (CH3)2CO. PLAN: We use the shape around each central atom to postulate the hybrid orbitals involved and use unhybridized orbitals to form the C═O bond. SOLUTION: The shapes are tetrahedral around each C of the two CH3(methyl) groups and trigonal planar around the middle C. Thus, the middle C has three sp2 orbitals and one unhybridized p orbital. Each of the two methyl C atoms has four sp3orbitals. Three of these form σ bonds with the 1s orbitals of H atoms; the fourth forms a σ bond with an sp2 orbital of the middle C. Thus, two of the three sp2 orbitals of the middle C form σ bonds to the other two C atoms. ©McGraw-Hill Education. Sample Problem 11.2 –Solution, Cont’d The O atom is also sp2 hybridized and has an unhybridized p orbital that can form a π bond. Two of the O atom’s sp2 orbitals hold lone pairs, and the third forms a σ bond with the third sp2 orbital of the middle C atom. The unhybridized, half-filled 2p orbitals of C and O form a π bond. The σ and π bonds constitute the C═O bond: ©McGraw-Hill Education. Bond Rotation Because the orbitals that form the σ bond point along the internuclear axis, rotation around that bond does not require breaking the overlap between the orbitals. The σ molecular orbital continues to maintain the shape and symmetry in the rotation. The orbitals that form the π bond interact above and below the internuclear axis, so rotation around the axis requires the breaking of the overlap between the orbitals. Rotation does not happen around a multiple bond. © 2017 Pearson Education, Inc. Lack of Bond Rotation in a p bond creates cis-trans isomers Isomers = compounds with the same molecular formula but different structures and properties © 2017 Pearson Education, Inc. Cis-Trans Isomers The cis isomer is polar while the trans isomer is not. The boiling point of the cis isomer is 13ºC higher than that of the trans isomer, therefore the 2 isomers can be separated by physical methods. ©McGraw-Hill Education. Carbon atoms in various hybridizations 1) C in sp3: when all bonds formed by a C atom are σ bonds. (Different C atoms in a polyatomic molecule may be in different hybridization states.) 2) C in sp2: when a C atom is involved in only 1 double bond. 3) C in sp: when a C atom is involved in 1 triple bond, or in 2 adjacent double bonds (CO2). ©McGraw-Hill Education. Molecular Orbital (MO) Theory The Nobel Prize in Chemistry 1966 was awarded to Robert S. Mulliken "for his fundamental work concerning chemical bonds and the electronic structure of molecules by the molecular orbital method.“ Robert S. Mulliken (1896-1986) Affiliation at the time of the award: University of Chicago, Chicago, IL, USA Coined the term “orbital” Friedrich Hund (1896-1997) Early contributions to the Molecular Orbital theory Hund’s rule of maximum spin multiplicity in populating empty orbitals Quantum tunneling effect © 2017 Pearson Education, Inc. Molecular Orbital (MO) Theory The Schrödinger equation cannot be solved exactly for multibody quantum systems such as molecules. 𝐻𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒 Ψ𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒 𝑗 = 𝐸𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒 𝑗 Ψ𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒 𝑗 In Molecular Orbital theory, as different from the Valence Bond theory, the wave functions of the molecule are computed from all the atomic orbitals (not just the valence orbitals), of all the atoms making up the molecule, not just 2 atoms at a time making up a localized bond. This approach then gives molecular orbitals that could possibly be delocalized over the entire molecule. https://chem.libretexts.org/LibreTexts/University_of_California_Davis/UCD_Che m_107B%3A_Physical_Chemistry_for_Life_Scientists/Chapters/5%3A_The_Che mical_Bond/5.5%3A_Molecular_Orbital_Theory © 2017 Pearson Education, Inc. Linear Combination of Atomic Orbitals (LCAO) – J.E. Lennard Jones, 1929 The wave functions of the molecule are guessed (set up) as weighted sums of the atomic orbitals of the atoms; this is called the linear combination of atomic orbitals (LCAO) method: Ψ𝑗 = ෍ 𝑐𝑖𝑗 𝜓𝑖 𝑖 𝑤ℎ𝑒𝑟𝑒 𝜓𝑖 𝑎𝑟𝑒 𝑎𝑡𝑜𝑚𝑖𝑐 𝑜𝑟𝑏𝑖𝑡𝑎𝑙𝑠 and 𝑐𝑖𝑗 𝑎𝑟𝑒 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑠 𝑡ℎ𝑎𝑡 𝑛𝑒𝑒𝑑 𝑡𝑜 𝑏𝑒 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒𝑑. Because the orbitals are wave functions, the waves can combine either constructively or destructively. The number of resulting molecular orbitals equals the number of combining atomic orbitals. We start with good guesses from our experience as to what the orbital should look like. Then we test and tweak the coefficients until the energy of the molecular orbital is minimized. In this treatment, the electrons belong to the whole molecule, so the orbitals belong to the whole molecule (delocalization of the MO). © 2017 Pearson Education, Inc. Molecular Orbitals: Bonding and Antibonding When the wave functions combine constructively, the resulting molecular orbital has less energy than the original atomic orbitals; it is called a bonding molecular orbital. Notation: σ, π – Most of the electron density is localized between the nuclei When the wave functions combine destructively, the resulting molecular orbital has more energy than the original atomic orbitals; it is called an antibonding molecular orbital. Notation: σ*, π* – Most of the electron density is localized outside the nuclei – At least one node forms between nuclei © 2017 Pearson Education, Inc. An Analogy Between Light Waves and Atomic Wave Functions Figure 11.15 ©McGraw-Hill Education. Molecular Orbital Diagram for H2 A MO diagram, just like an atomic orbital diagram, shows the relative energy, number of electrons and spin orientation in each MO. The MO diagram also shows the AOs from which each MO is formed. ©McGraw-Hill Education. Molecular Orbitals: Energy diagram and stability Electrons in bonding MOs have lower energy than the atomic orbitals. Electrons in antibonding MOs are destabilizing. – Higher in energy than atomic orbitals – Electron density located outside the internuclear axis – Electrons in antibonding orbitals cancel stability gained by electrons in bonding orbitals. https://chem.libretexts.org/LibreTexts/University_of_California_Davis/ UCD_Chem_107B%3A_Physical_Chemistry_for_Life_Scientists/Chap ters/5%3A_The_Chemical_Bond/5.6%3A_Diatomic_Molecules © 2017 Pearson Education, Inc. Bond Order, Magnetic Properties Double, triple bonds have no meaning in MO theory. We need another concept. Bond order = difference between number of electrons in bonding and antibonding orbitals, divided by 2: – – – – It is the bond multiplicity equivalent in the MO theory. May be a fraction. Higher bond order means stronger and shorter bonds. If bond order = 0, then bond is unstable compared to individual atoms, and no bond will form. 2−0 𝐻2 𝐵𝑜𝑛𝑑 𝑂𝑟𝑑𝑒𝑟 = =1 2 A substance will be paramagnetic if its MO diagram has unpaired electrons. If all electrons paired, it is diamagnetic. © 2017 Pearson Education, Inc. He2: Bond Order = 0 does not exist as molecule © 2017 Pearson Education, Inc. He2+: Bond Order = 0.5 © 2017 Pearson Education, Inc. 𝑯− 𝟐 anion: Bond Order = 0.5 Practice exercise 10.9: Use MOT energy diagram to predict the bond order in H2+. Is the H2+ bond stronger or weaker than the H2 bond? © 2017 Pearson Education, Inc. Sample Problem 11.3 – Problem, Plan and Solution Predicting Stability of Species Using MO Diagrams PROBLEM: Use an MO diagram to find the bond order and predict whether H ​ 2+ exists. If it exists, write its electron configuration. PLAN: Since the 1s AOs form the MOs, the MO diagram is similar to the one for H2. We find the number of electrons and distribute them one at a time to the MOs in order of increasing energy. We obtain the bond order with Equation 11.1 and write the electron configuration as described in the text. SOLUTION: H2 has two e−, so H2+ has only one, which enters the bonding MO. The bond order is ½(1-0)= ½ so we predict that H2+ exists. The electron configuration is (1s)1. ©McGraw-Hill Education. Sample Problem 11.3 – Solution, Cont’d ©McGraw-Hill Education. Li2(g): Bond Order = 1 © 2017 Pearson Education, Inc. Be2(g): Bond Order = 0 © 2017 Pearson Education, Inc.  Molecular Orbitals of p Atomic Orbitals - + - © 2017 Pearson Education, Inc. - + + - p Molecular Orbitals of p Atomic Orbitals © 2017 Pearson Education, Inc. Relative Energies of Molecular Orbitals in period-2 homo-diatomic molecules Energy ordering of MO cannot be guessed. It is determined computationally. No single order works for all species. Differences are caused by the degree of 2s-2p mixing in various molecular orbitals. © 2017 Pearson Education, Inc. MO Occupancy and Molecular Properties for B2 through Ne2 The aufbau principle works here, too: molecular orbitals are populated with electrons in the order of increasing energy. Hund’s rule of maximum spin multiplicity applies here, too. Pauli’s exclusion rule: not more than 2 electrons per molecular orbital, with opposite spins. ©McGraw-Hill Education. O2 Molecular oxygen O2 is paramagnetic. Paramagnetic molecule has unpaired electrons. Neither Lewis theory nor valence bond theory predict this result. https://www.youtube.com/watch?time_continue=5&v=Lt4P6ctf06Q © 2017 Pearson Education, Inc. O2 as Described by Lewis, VBT and MOT Only MOT explains the paramagnetism of O2 (Lennard-Jones, 1929). J.E. Lennard-Jones (1894-1954) © 2017 Pearson Education, Inc. Sample Problem 11.4 – Problem and Plan Using MO Theory to Explain Bond Properties PROBLEM: Explain the following data with diagrams showing the occupancy of MOs: N2 N2 + O2 O2+ Bond energy (kJ/mol) 945 841 498 623 Bond length (pm) 110 112 121 112 PLAN: The data show that removing an electron from each parent molecule has opposite effects: N2+ has a weaker longer bond than N2, but O2+ has a stronger, shorter bond than O2. We determine the valence electrons in each species, draw the sequence of MO energy levels (showing orbital mixing in N2 but not in O2), and fill them with electrons. We then calculate bond orders, which relate directly to bond energy and inversely to bond length. ©McGraw-Hill Education. Sample Problem 11.4 - Solution SOLUTION: Determining the valence electrons: N has 5 valence e−, so N2 has 10 and N2+ has 9. O has 6 valence e−, so O2 has 12 and O2+ has 11. Drawing and filling the MO diagrams: Calculating bond orders: 1 1 1 1 8 − 2 = 3, 7 − 2 = 2.5, 8 − 4 = 2, 8 − 3 = 2.5 2 2 2 2 ©McGraw-Hill Education. Heteronuclear Diatomic Molecules and Ions When the combining atomic orbitals are identical and of equal energy, the contribution of each atomic orbital to the molecular orbital is equal. When the combining atomic orbitals are different types and energies, the atomic orbital closest in energy to the molecular orbital contributes more to the molecular orbital. The more electronegative an atom is, the lower in energy its orbitals are. Lower-energy atomic orbitals contribute more to the bonding MOs. Higher-energy atomic orbitals contribute more to the antibonding MOs. Nonbonding MOs remain localized on the atom donating its atomic orbitals. © 2017 Pearson Education, Inc. The MO Diagram for NO The Lewis structure, strictly obeying the octet rule, is: : 𝑁ሶ = 𝑂:ሷ Oxygen is more electronegative than Nitrogen. The energies of the atomic 2s and 2p sublevels are not equal. ©McGraw-Hill Education. The lone, unpaired electron is not localized on the N atom, but delocalized over the entire molecule in an antibonding molecular orbital. Reactive radicals have the unpaired electron in a bonding molecular orbital. The MO Diagram for HF Note the large energy difference between the H(1s) and F(2p) atomic orbitals. The Fluorine electronegativity is so high compared to H that the occupied F(2p) orbitals essentially do not interact with the H(1s), resulting in nonbonding molecular orbitals close to the initial energy of the atomic orbitals. ©McGraw-Hill Education.