Lewis Structures –Bond Pairs and Lone Pairs In Lewis structures with five or six pairs of electrons around the central atom we need to distinguish between lone pairs and bonding pairs. Lone pairs on average are located closer to the central atom than bonding pairs. Why? Coulombs law would suggest that the magnitude of electron pair repulsions will vary and be strongest when two lone pairs are involved.

Order of Lone Pair Repulsions The magnitude of electron repulsions follows the order Lone pair : lone pair ˃ lone pair : bond pair ˃ bond pair : bond pair We attempt (as does nature!) to place the lone pairs so as to minimize repulsive interactions. We maximize the angles between lone pairs and other pairs where possible.

Class Example 3. Draw Lewis structures for SF 4, IF 5 and ClF 3. Determine the shape of each of these molecules and which, if any, of these molecules are polar.

Single and Multiple Bonds – Strengths and Lengths We have drawn Lewis structures with single, double and triple bonds. The Lewis structures do reflect physical reality (at least when properly drawn!). Multiple bonds between a given pair of atoms (e.g. C and O) are shorter than single bonds. Multiple bonds between a given pair of atoms require more energy to break than do single bonds.

Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 10 Slide 5 of 48

Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 10 Slide 6 of 48

Bond Energies and Enthalpies of Reaction – ΔH’s Bond energies can be combined to calculate very crude values for heats of reaction. Care must be taken to use bond energies for single bonds, double bonds or triple bonds as appropriate. Often one must draw Lewis structures to decide what types of bonds to consider. ΔH values calculated in this way are very approximate since bond energies for a particular atom pair vary between molecules.

Bond Energies Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 10 Slide 8 of 48 FIGURE Some bond energies compared

Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 10Slide 9 of 48 Calculating an Enthalpy of Reaction from Bond Energies. ΔH rxn =  ΔH(bond breakage) +  ΔH(bond formation) ≈  BE(reactants) -  BE(products) Does this equation look familiar?

Copyright  2011 Pearson Canada Inc

Class Example 4. Use bond energy values from Petrucci in order to calculate the enthalpy change for the reaction N 2 (g) + 3H 2 (g) → 2 NH 3 (g). Compare this crude ΔH value to that calculated from tabulated heat of formation data. Key ideas: Bond breaking is an endothermic process. Bond formation is an exothermic process.

Class Example 5. Draw a Lewis structure or OPCl 3 using (a) four single bonds and (b) one double bond between P and O. Assign formal charges to all atoms in both structures. Which structure is more reasonable? Determine the shape of the molecule. Is this molecule polar?

Chemical Bonds – Molecular Orbitals The electronic structures of atoms and molecules have many features in common. Individual atoms usually possess unpaired electrons. These atoms are often chemically unstable. Two such atoms (or more!) can combine to form molecules with (usually) no unpaired electrons. The process involves the formation of chemical bonds and is highly exothermic.

Chemical Bonds – Energetics (Homonuclear Diatomics) H(g) + H(g) → H:H(g)ΔH = -436 kJ N(g) + N(g)→ :N:::N:(g)ΔH = -946 kJ Note on signs: The above reactions are highly exothermic. Energy is released when the atoms combine to form molecules. The bond energy tells us how much energy is required to break a mole of bonds (usually!). The bond energy of the bond in N 2 is +946 kJ mol -1.

Atomic and Molecular Orbitals In isolated atoms most electrons are found in pairs in a number of different atomic orbitals. When atoms combine valence shell electrons are rearranged. Two electrons from different atoms can, for example, “pair up” to form a single covalent bond where the bonding molecular orbital is associated with more than one atom (and often more than two atoms!).

Molecular Orbitals – Wave Properties of Electrons We will need to consider the wave properties of electrons when discussing molecular orbitals. By analogy to constructive and destructive interference in conventional waves we will see atomic orbitals combining constructively to form bonding molecular orbitals and combining destructively to form anti-bonding molecular orbitals.

Bonding Theory Objectives A detailed bonding theory should account quantitatively for experimentally observed molecular shapes, bond distances, electrical polarity, bond strength and so on. It should also have predictive capability. A quantitative treatment is reserved for higher level courses. In a qualitative manner we begin by reminding ourselves that coulombic interactions (between particles with wave character) are key. Simplest example – the H 2 molecule.

What a Bonding Theory Should Do Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 11Slide 18 of 57 FIGURE 11-1

The H 2 Molecule – Coulombic Interactions In physics you may already have calculated the size of coulombic forces and associated potential energies. For static point charges this is a simple exercise. For a group of charged particles in rapid motion (and wavelike!) life is more complex. Nevertheless the charged particles are “capable” of finding the lowest energy configuration. Why? Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 11Slide 19 of 57

Energy of interaction of two hydrogen atoms plotted for internuclear separations from zero to infinity FIGURE 11-2 Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 11Slide 20 of 57

Valence Bond Theory Provides a simple picture of covalent bond formation from overlapping atomic orbitals (each containing one electron). One assumes that both core electrons and lone pair valence shell electrons keep the orbital locations that they had in the separated atoms. The valence bond method does account for some common valences. We’ll consider O, S and N atoms reacting with H atoms.

Reaction of O, S and N with H Atoms The complete electron configurations for O, N and H are: Oxygen1s 2 2s 2 2p x 2 2p y 1 2p z 1 Nitrogen 1s 2 2s 2 2p x 1 2p y 1 2p z 1 Hydrogen 1s 1 We can “pair up” all electrons if O combines with two H atoms and N combines with three H atoms to form H 2 O and NH 3 respectively.

Reaction of O, S and N with H Atoms The complete electron configurations for S could be written as Sulfur 1s 2 2s 2 2p 6 3p x 2 3p y 1 3p z 1 Again, we can “pair up” all electrons if S and two H atoms combine to form H 2 S. The valence bond picture suggests that all bond angles in H 2 O, NH 3 and H 2 S should be 90 o. This is close to the value seen in H 2 S (92 o ) but significantly underestimates bond angles in H 2 O (105 o ) and NH 3 (107 o ).

Bonding in H 2 S represented by atomic orbital overlap FIGURE 11-3 Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 11Slide 24 of 57

The Methane “Problem” The ground state configurations for C can be written as 1s 2 2s 2 2p x 1 2p y 1 Using the valence bond picture and the concept of paired electrons in molecular orbitals we might expect C to react with H atoms to form CH 2. The CH 2 molecule does form but is unstable (a transient species). However, carbon “happily” reacts with H to form the methane, CH 4.

Methane and Hybridization By experiment, as previously discussed, methane has a regular tetrahedral geometry – four equal bond distances and all bond angles of o. The regular geometry of methane and its ability to form four bonds can be explained using the concept of hybridization. How have we explained carbon’s tendency to form four bonds previously?

Methane and Hybridization – cont’d: In the hybridization picture we imagine methane being formed from C and H atoms in three steps. In the first step we take a ground state C atom and excite one electron (from the 2s orbital) to form the lowest lying ( or first) excited state. Carbon Ground State: 1s 2 2s 2 2p x 1 2p y 1 Carbon Excited State: 1s 2 2s 1 2p x 1 2p y 1 2p z 1

Methane and Hybridization – cont’d: In the second step we imagine “combining” the single occupied 2s orbital and the three occupied 3p orbitals in the excited to form four equivalent sp 3 hybrid orbitals (each containing a single unpaired electron). In step three the “hybridized C atom” reacts with four H atoms to form a CH 4 molecule. The process is represented on the next few slides.

Hybridization of Atomic Orbitals Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 11Slide 29 of 57

The sp 3 hybridization scheme FIGURE 11-6 Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 11Slide 30 of 57

Bonding and structure of CH 4 FIGURE 11-7 Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 11Slide 31 of 57

Organic Compounds and Structures: An Overview Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 26Slide 32 of 75 FIGURE 26-1 Representations of the methane molecule

Organic Chemistry The next two slides illustrate what starts to happen when two or more sp 3 hybridized carbons are linked. The chemistry of carbon is infinitely varied and organic compounds are part of all of living things, important energy sources, key pharmaceuticals and so on.

The ethane molecule C 2 H 6 FIGURE 26-2 Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 26Slide 34 of 75

The propane molecule, C 3 H 8 FIGURE 26-3 Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 26Slide 35 of 75