1. Chemistry XXI Unit 2 How do we determine structure? M4. Inferring Charge Distribution Analyzing the distribution of electrons in molecules. M3. Predicting Geometry Predicting the three dimensional geometry of molecules. M2. Looking for Patterns Deducing atom connectivity based on atomic structure. M1. Analyzing Light-Matter Interactions Using spectroscopy to derive structural information. The central goal of this unit is to help you develop ways of thinking that can be used to predict the atomic and molecular structure of substances.

2. Chemistry XXI Unit 2 How do we determine structure? Module 2: Looking for Patterns Central goal: To use experimental spectroscopic data, together with information about atomic composition and electron configurations, to deduce atom connectivity in a given molecule.

3. Chemistry XXI The Challenge Analysis What is this? How can we determine what is bonded to what in the molecules of a given substance? How can we predict how many bonds may exist between the different atoms? Molecular structure is the fundamental clue to determine the identity and properties of substances. The structure of a molecule is determined by both atomic composition and atom connectivity. Aspirin C9H8O4C9H8O4

4. Chemistry XXI The idea that molecules are comprised of atoms connected to each other in some way and arranged in particular geometries in space has been very useful to explain and predict the properties of chemical substances. A Useful Model Octane DNA Water What holds the atoms together?

5. Chemistry XXI A Bonding Model Atoms are assumed to be “bonded” or connected to each other in molecular compounds. How do we model it? 1.Pairs of atoms in molecules are attracted to each other through forces that we call “covalent bonds.” The formation of these bonds is induced by electrostatic attraction between electrons and protons in the different atoms.

6. Chemistry XXI Other Interactions Protons and electrons have an additional property called “spin.” Spin generates an additional force between electrons (magnetic) Magnetic force < Electric force When electrons are confined to the same region of the space, like in between two nuclei, “spin pairing” reduces the energy of the system.

7. Chemistry XXI Covalent Bonding  Increased “electron density” between atoms;  Electron delocalization.  Electron spin pairing Bond formation is favored by: 2 e - per bond Electrons do not behave as “classical” particles. We cannot know their position and velocity simultaneously: Heisenberg Uncertainty Principle Electron Densities Measure of probabilities

8. Chemistry XXI Dynamic Nature Chemical bonds are not static entities. The atomic nuclei and electrons involved are constantly moving. The atoms in a bond vibrate around their equilibrium positions with a frequency that depends on the strength of the force between the atoms and their masses.

9. Chemistry XXI Analyzing Vibrations Molecules experience a wide variety of vibrational motions, characteristic of their component atoms and the bonds between them. Symmetrical Stretching Antisymmetrical Stretching ScissoringRocking Vibrational states are “quantized” and transitions between them can be induced by providing the right  E. EE Energy h   IR

10. Chemistry XXI The absorption of IR radiation at specific frequencies can then be used to detect the presence of specific bonds. IR Spectrophotometer Infrared Spectroscopy Formaldehyde Wavenumber (1  )

11. Chemistry XXI Wavenumber = 1/ (cm -1 ) Increasing Energy Absorption Bands O-H C-H C-C C-H

12. Chemistry XXI IR + MS The combination of IR Spectroscopy and Mass Spectrometry is a powerful tool of analysis. O-HO-H C-HC-H C-O M CH3+CH3+ CH3O+CH3O+ The elemental analysis of a poison that causes blindness reveals the formula CH 4 O. What can we say about the molecular structure of this compound? Let’s think Methanol 15 31 32

13. Chemistry XXI IR/MS breath analysis can be used to explore fat metabolism for people on a diet. The metabolism of fat produces a volatile compound with the formula C 3 H 6 O. Let’s Think C-H C=O C-C What is the molecular structure of this compound? 1/ 15 43 58 Acetone

14. Chemistry XXI Cl P Cl Cl F F H F Bonding Capacity Inferences about atom connectivity are facilitated by analyzing the bonding patterns of different types of atoms. Analyze the structure of these molecules and identify bonding patterns for different atoms. How is bonding capacity related to the position of the atom in the periodic table? Let’s think

15. Chemistry XXI Valence Many nonmetallic elements seem to have fixed bonding capacities (valence) or number of bonds they can form: C  4 bonds N,P  3 bonds O,S  2 bonds H, F,Cl  1 bond Molecular compounds result from the combination of nonmetallic elements. Valence is a periodic property (elements from the same group behave similarly) 4 3 2 1 0 Valence Nonmetallic elements

16. Chemistry XXI Let’s apply what we have learned to solve a “prototypical” identification problem. Let’s Think A substance found in most salad dressings may be responsible for the development of “stomach ulcers.” The substance is normally found dissolved in water and is most likely a liquid in its pure form. Propose a method to isolate (separate) the substance from salad dressing. Once isolated, what would you do?

17. Chemistry XXI Elemental analysis reveals that the empirical formula of the compound is CH 2 O. Let’s Think O-H C-H C=O C-C C-O What is the molecular structure of this compound? 1/ 15 45 43 + + + M Acetic Acid 60

18. Chemistry XXI In order to better infer and predict the molecular structure of a compound, we need to understand why different atoms form different numbers of bonds. Covalent bonding is the result of interactions between electrons and protons in the bonded atoms. Thus, exploring the internal structure of single atoms may provide important clues. Looking into Atoms Data on: Atomic Radius and Ionization Energy are very revealing. e-e-

19. Chemistry XXI Atomic Radius The size of atoms is not well defined given that we cannot know the position of electrons with precision. Atomic radius is derived indirectly by measuring distance between bonded atoms in molecules. d r(O) = d/2 Diffraction Pattern X-Ray Crystallography Structural Information

20. Chemistry XXI Atomic Radius Periodic Behavior R increases Li Na K Rb Ne Ar Kr He 1 pm = 1 x 10 -12 m = 1 x 10 -3 nm

21. Chemistry XXI Ionization Energy Insight into atomic structure can also be gained by analyzing the energy required to remove a first electron from a gaseous atom: M (g)  M + (g) + e - e-e- LiNaK Rb Ne Ar Kr He Periodic Behavior IE increases

22. Chemistry XXI Let’s Think Li Na KRb Ne Ar Kr He Li Na K Rb Ne Ar Kr He What do the periodic trends for atomic radius and first ionization energy suggest about the electron structure of different atoms? What atomic model would allow us to explain these trends?

23. Chemistry XXI Shell Model The trends in atomic radius and first ionization energy may be explained if we assume that electrons are arranged in shells. Shell  # of e - n = 1  2 e - n = 2  8 e - n = 3  8 e - n = 4  18 e - He Li Ne Na H H E 0

24. Chemistry XXI Looking Deeper Additional clues about electron structure can be derived using photoelectron spectroscopy. KE h Ionization Energy = h - KE The n th Ionization Energy is the energy needed to remove the n th electron from an atom. This technique can be used to determine the number of electrons in each shell. X-ray

25. Chemistry XXI Photoelectron Spectroscopy MJ/mol What are these results telling us about electron configurations in the different atoms? Let’s think Ionization Energy in

26. Chemistry XXI Modified Shell Model The trends in the results from photoelectron spectroscopy suggest that electrons arrange in subshells within a shell. ShellSubshell# of e - n = 1n = 11 s2 e-2 e- n = 2n = 22 s2 e-2 e- 2 p6 e-6 e- n = 3n = 33 s2 e-2 e- 3 p6 e-6 e- 3 d10 e - Electrons in an atom occupy different energy levels. In the ground state of an atom, the lower energy levels are occupied first (1s  2s  2p  3s).

27. Chemistry XXI Electron Configurations H  1s 1 He  1s 2 Li  1s 2 2s 1 = [He] 2s 1 Be  1s 2 2s 2 = [He] 2s 2 B  1s 2 2s 2 2p 1 = [He] 2s 2 2p 1 How are electrons distributed in N, Ne, Na, P, Ar, and Sc? What periodic trends do you observe? Let’s think

28. Chemistry XXI Recognizing Trends s 1 s 2 d 1 d 2 d 3 d 4 d 5 d 6 d 7 d 8 d 9 d 10 p 1 p 2 p 3 p 4 p 5 p 6 Ge  ? [Ar]4s 2 3d 10 4p 2

29. Chemistry XXI Let’s Think Write the electron configuration of Al and sketch is expected photoelectron spectrum. Al  [Ne]3s 2 3p 1

30. Chemistry XXI Explaining Valence The bonding capacity or valence of different atoms can be explained based on their electron configurations. Consider carbon:C  [He] 2s 2 2p 2 We can divide its electrons into two main groups: CORE electrons: In lower energy levels; spend more time close to the nucleus; not very accessible. Core e - [He] VALENCE electrons: In high energy levels; farther from the nucleus; exposed to interaction with other atoms. Valence e - 2s 2 2p 2

31. Chemistry XXI [ ] [ ] s 2 p 2 [ ] s 2 p 1 [ ] s 2 p 3 [ ] s 2 p 4 [ ] s 2 p 5 3 4 5 6 7 # valence e - Valence Electrons 4 3 2 1Valency The number of covalent bonds that each atom forms is determined by the number of valence electrons that the atom must share to have a full shell. Valency can be explained if we assume that:

32. Chemistry XXI A Powerful Rule OCTET RULE: When atoms combine, the most stable structures are those in which each atom has eight valence electrons (full valence shell). Lewis electron-dot symbol Consider the fluorine atom: + If two atoms of fluorine form a single covalent bond, each of them achieve a full shell configuration: 8 e-8 e- 8 e-8 e- F

33. Chemistry XXI O 8 O2O2 In the atmosphere, nitrogen (N 2 ) and oxygen (O 2 ) exist as diatomic molecules. Why is so? How many bonds exist between the atoms in these molecules? N2N2 ? When two atoms of nonmetallic elements combine, their valence electrons are reorganized. The number of covalent bonds that are formed are determined by the most stable electron configurations (full valence shell). Let’s Think O N

34. Chemistry XXI Assess what you know Let′s apply!

35. Chemistry XXI Retention Time Breath analysis of mouth air using GC/MS reveals the presence of three compounds assumed to be responsible for halitosis (bad breath). Let′s apply! A comparison search in a MS database allows to identify the substances as: 1.H 2 S 2.CH 4 S 3.C 2 H 6 S Propose a molecular structure for the first two compounds. Justify your decisions based on electron configurations of the atoms involved.

36. Chemistry XXI Let′s apply! H  1s 1 C  [He] 2s 2 p 2 S  [Ne] 3s 2 3p 4 H C S H 2 S  SHSHH  H Full shell 2 e - Full shell 8 e - CH 4 S  CH H H HS  Dihydrogen Sulfide  CH H H HS  Methyl Mercaptam

37. Chemistry XXI Let′s apply! Based on your previous analysis and the additional information provided, propose a molecular structure for the third compound: C 2 H 6 S. C-S C-H 15 47 C-H 62 Dimethyl Sulfide (DMS)

38. Chemistry XXI Describe one important application of what you have learned in this module.

39. Chemistry XXI Summary Looking for Patterns The formation of covalent bonds between atoms is the result of electrostatic interactions between electrons and protons in the bonding atoms. Molecules experience characteristic vibrational motions. The absorption of EM radiation at specific frequencies can then be used to detect the presence of specific bonds.

40. Chemistry XXI Models of atomic structure can be used to make predictions about electron configurations. In turn, electron configurations allow us to make predictions about bonding behavior. Summary Looking for Patterns When two atoms combine, their valence electrons are reorganized. The most stable structures tend to be those in which each atom has eight valence electrons (full valence shell; octet rule). CH H H HS

41. Chemistry XXI For next class, Investigate how we can use the octet rule to derive the molecular structure of simple compounds. How can we use the molecular structure to predict the molecular geometry of a molecule?