1. TOPIC 14.1 FURTHER ASPECTS OF COVALENT BONDING AND STRUCTURE

2. ESSENTIAL IDEA Larger structures and more in-depth explanations of bonding systems often require more sophisticated concepts and theories of bonding. NATURE OF SCIENCE (2.7) Principle of Occam’s razor – bonding theories have been modified over time. Newer theories need to remain as simple as possible while maximizing explanatory power, for example the idea of formal charge.

3. INTERNATIONAL-MINDEDNESS How has ozone depletion changed over time? What have we done as a global community to reduce ozone depletion? To what extent is ozone depletion an example of both a success and a failure for solving an international environmental concern?

4. THEORY OF KNOWLEDGE Covalent bonding can be described using valence bond or molecular orbital theory. To what extent is having alternative ways of describing the same phenomena a strength or a weakness?

5. UNDERSTANDING/KEY IDEA 14.1.A Covalent bonds result from the overlap of atomic orbitals. A sigma bond is formed by the direct head-on/end-to-end overlap of atomic orbitals, resulting in electron density being concentrated between the nuclei of the bonding atoms. A pi bond is formed from the sideways overlap of atomic orbitals, resulting in electron density above and below the plane of the nuclei of the bonding atoms.

6. APPLICATION/SKILLS Be able to predict whether sigma or pi bonds are formed from the linear combination of atomic orbitals.

7. Sigma σ and Pi π bonds

8. SIGMA BONDS All single covalent bonds are sigma bonds. Sigma bonds form by the overlap of orbitals along the bond axis. The following form sigma bonds: s and s (H2) s and p (HCl) p and p (Cl2) Hybrid orbitals and s Hybrid orbitals and hybrid orbitals

9. Pi BONDS All double covalent bonds contain one pi and one sigma bond. All triple covalent bonds contain two pi and one sigma bond. The following form pi bonds: p and p sideways Pi bonds are weaker than sigma bonds since their electron density is farther away from the positive nucleus. The double bonds break more readily and are more reactive than those with only sigma bonds.

10. UNDERSTANDING/KEY IDEA 14.1.B Formal charge (FC) can be used to decide which Lewis structure is preferred from several. The FC is the charge an atom would have if all atoms in the molecule had the same electronegativity. FC = (number of valence electrons) – ½(Number of bonding electrons) – (Number of non-bonding electrons). The Lewis structure with the atoms having FC values closest to zero is preferred.

11. APPLICATION/SKILLS Be able to apply formal charge (FC) to ascertain which Lewis structure is preferred from different Lewis structure.

12. FORMAL CHARGE Sometimes it is possible to come up with more than one valid Lewis structure. This is most often seen in molecules that can have an expanded octet. One such example is sulfur dioxide, SO 2. Two resonance structures or the expanded octet.

13. FORMAL CHARGE Which structure is the most stable? The concept of formal charge treats covalent bonds as if they were purely covalent with equal electron distribution and ignores electronegativity. You basically count how many electrons belong to each atom in the Lewis structure and compare this with the number of valence electrons in the non-bonded atom. The difference is the formal charge.

14. FC = #valence electrons - number electrons assigned Things to remember. The valence electrons are counted as if the atom is nonbonded. Assigned electrons = ½ bonded electrons + e - in lone pairs FC = V-(1/2B + L) The lower the formal charge, the more stable the structure.

15. Back to SO 2 The formal charge for the resonance structure: Sulfur has 6 valence electrons, 3 bonding pairs and 1 lone pair (2e - ). FC = 6 – (3 +2) = +1 The left oxygen has 6 valence e -, 2 bonding pairs and 2 lone pairs (4e - ). FC = 6 – (2+4) = 0 The right oxygen has 6 valence e -, 1 bonding pair and 3 lone pairs (6e - ). FC = 6 – (1 + 6) = -1

16. The expanded octet structure: Sulfur has 6 valence e -, 4 bonding pairs and 1 lone pair (2e - ) so FC = 6 – (4+2) = 0 Both oxygens – 6 valence e -, 2 bonding pairs and 2 lone pairs (4e - ) so FC = 6 – (2 + 4) = 0

17. 0 0 0 0 +1 The more stable structure would be the expanded octet with the formal charge closest to zero. Formal charge does ignore electronegativity so the most stable structure should have the lowest formal charge and the negative values for formal charge on the most electronegative atom.

18. Example with N 2 O Figure out the best Lewis structure using FC.

19. UNDERSTANDING/KEY IDEA 14.1.C Exceptions to the octet rule include some species having incomplete octets and expanded octets. (This understanding was explained in detail in ppt 4.3. Incomplete octets involve B and Be and expanded octets encompass any structure with the trigonal bipyramidal and octahedral shapes.)

20. APPLICATION/SKILLS Be able to deduce the Lewis structure of molecules and ions showing all valence electrons for up to six electron pairs on each atom. (Covered in detail in ppt 4.3)

21. APPLICATION/SKILLS Be able to deduce using the VSEPR theory the electron domain geometry and molecular geometry with five and six electron domains and associated bond angles and polarity. (Covered in detail in ppt 4.3)

22. UNDERSTANDING/KEY IDEA 14.1.D Delocalization involves electrons that are shared by/between all atoms in a molecule or ion as opposed to being localized between a pair of atoms.

23. DELOCALIZATION Sometimes electrons show a tendency to be shared between more than one bonding position and are said to be delocalized. Since they are free from the constraints of a single bonding position, delocalized electrons spread themselves out making the molecule more stable. Delocalization happens when you have a double bond that can have more than one possible position.

24. PROPERTIES Delocalized electrons give rise to intermediate bond strength and length. This new length is in between a single and double bond. Bond Order – is the number of shared electron pairs divided by the number of bonding positions. The higher the bond order, the greater the electron density. Greater stability due to the fact that delocalization spreads out the electrons and minimizes the repulsions between them. Delocalized electrons allow electrical conductivity.

25. UNDERSTANDING/KEY IDEA 14.1.E Resonance involves using two or more Lewis structures to represent a particular molecule or ion. A resonance structure is one of two or more alternative Lewis structures for a molecule or ion that cannot be described fully with one Lewis structure alone.

26. RESONANCE Delocalization cannot be depicted by one Lewis structure so the concept of resonance is introduced. The actual structure of the species is an average of the number of Lewis structures that can be drawn as resonance structures. Benzene is a good example of this.

27. APPLICATION/SKILLS Be able to explain the wavelength of light required to dissociate oxygen and ozone.

28. OZONE Ozone, O 3, has a bent shape with a bond angle of 117 ◦ It has 2 resonance structures. www.chemwiki.ucdavis.edu The double bond consists of one pi and one sigma bond. The electrons in the pi bond are held less tightly so they become delocalized giving rise to the resonance structure. The bond order is 1.5 which means the length is intermediate and the strength is between a double and single bond.

29. OZONE The molecule is polar which is explained by formal charge and the uneven distribution of electrons. The lower part of the stratosphere, known as the ozone layer, contains 90% of the atmospheric ozone. Ozone levels are maintained through a cycle of reactions involving the formation and breakdown of oxygen and ozone.

30. Ref: schooltutoring.com

31. OZONE CYCLE There are 2 key steps in the ozone cycle. The O. represents a free radical that has an unpaired electron so it is highly reactive. Oxygen dissociation O 2(g) O. (g) + O. (g) light wavelength <242 nm Ozone dissociation 1. O 3(g) fast O. (g) + 2O 2(g) light wavelength <330 nm 2. O 3(g) + O. (g) slow 2O 2(g) exothermic reaction H=neg

32. The O 2 bonds are stronger and harder to break than the O 3 bonds of ozone. The stronger O 2 bonds require UV light energy of shorter wavelengths because they need the higher energy radiation to break the bonds. The bond energy of O2 with a bond order of 2 is 498 kJ/mol. The bond energy of O3 with a bond order of 1.5 is 364 kJ/mol.

33. The fact that ozone absorbs radiation of wavelengths of 200 nm to 315 nm is very important. This corresponds to the higher range of ultraviolet light, known as UV-B and UV-C which causes damage to living tissue. The ozone layer protects us from this radiation. The absorption of UV radiation by ozone is also a major source of heat in the stratosphere and is the reason why the temperature in the stratosphere rises with height.

34. APPLICATION/SKILLS Be able to describe the mechanism of the catalysis of ozone depletion when catalyzed by CFCs and NOx.

35. Ozone’s ability to absorb UV radiation also means that it is unstable. It reacts easily with compounds found in the stratosphere that have been created by human activity. There are two types of compounds that produce highly reactive free radicals that catalyze the decomposition of ozone to oxygen. Nitrogen oxides, NO x Chlorofluorocarbons, CFC’s

36. Nitrogen oxides Nitrogen monoxide is produced in vehicle engines. It is a free radical as it has an odd number of electrons. Nitrogen dioxide forms from the oxidation of NO and is also a free radical. The reactions of nitrogen oxides with ozone are: NO. (g) + O 3(g) NO 2. (g) + O 2(g) NO 2. (g) + O. (g) NO. (g) + O 2(g) NO. has acted as a catalyst because it regenerated during the reaction and the net change is the breakdown of O 3. O 3(g) + O. (g) 2O 2(g)

37. Chlorofluorocarbons Chlorofluorocarbons are widely used in aerosols, refrigerants, solvents and plastics due to their low reactivity and low toxicity in the troposphere. However, when they reach the stratosphere, the higher energy UV radiation breaks them down releasing free chlorine atoms which are reactive free radicals. The reaction of the CFC freon is: CCl 2 F 2(g) CClF 2. (g) + Cl. (g) Cl. (g) + O 3(g) O 2(g) + ClO. (g) ClO. (g) + O. (g) O 2(g) + Cl. (g)

38. Here the chlorine radical acts as the catalyst and the net reaction is again: O 3(g) + O. (g) 2O 2(g) These reactions upset the balance of the ozone cycle and lead to the thinning of the ozone layer. The UV radiation reaching the Earth is most pronounced in the polar regions. This has been a global concern since the 1970’s.

39. Citations International Baccalaureate Organization. Chemistry Guide, First assessment 2016. Updated 2015. Brown, Catrin, and Mike Ford. Higher Level Chemistry. 2nd ed. N.p.: Pearson Baccalaureate, 2014. Print. Most of the information found in this power point comes directly from this textbook. The power point has been made to directly complement the Higher Level Chemistry textbook by Catrin and Brown and is used for direct instructional purposes only.