Midterm WEDNESDAY Ch 8-10.5 Final Exam will be the last day of class. Sorry.

1. Midterm WEDNESDAYCh 8-10.5Final Exam will be the last day of class. Sorry.

2. Lewis Structures of Simple Molecules H H C H H .. .. F .. .. .. F C F .. .. .. .. F H H .. CH4 .. H C C O H Methane H H Ethyl Alcohol (Ethanol) .. .. .. .. O K+ .. Cl .. .. .. .. .. .. O O .. CF4 .. KClO3 Potassium Chlorate Carbon Tetrafluoride

3. Lewis Structures of Simple Molecules Resonance Structures -III Nitrate .. .. .. O N .. .. .. .. .. O O .. .. .. O .. .. O N N .. .. .. .. .. .. .. O O .. .. .. .. O O

4. Resonance: Delocalized Electron-Pair Bonding - I Ozone : O3 .. .. .. .. .. O O .. .. .. .. .. .. O O O O .. II I Resonance Hybrid Structure .. O .. .. .. .. O O One pair of electron’s resonances between the two locations!!

5. Resonance: Delocalized Electron-Pair Bonding - II H H C C H H C C C C H H C C H H C C H H H C C C H H C H H C H C C H C Benzene Resonance Structure H

7. Postulating the Hybrid Orbitals in a Molecule Problem: Describe how mixing of atomic orbitals on the central atoms leads to the hybrid orbitals in the following: a) Methyl amine, CH3NH2b) Xenon tetrafluoride, XeF4 Plan: From the Lewis structure and molecular shape, we know the number and arrangement of electron groups around the central atoms, from which we postulate the type of hybrid orbitals involved. Then we write the partial orbital diagram for each central atom before and after the orbitals are hybridized.

8. Postulating the Hybrid Orbitals in a Molecule Problem: Describe how mixing of atomic orbitals on the central atoms leads to the hybrid orbitals in the following: a) Methyl amine, CH3NH2b) Xenon tetrafluoride, XeF4 Plan: From the Lewis structure and molecular shape, we know the number and arrangement of electron groups around the central atoms, from which we postulate the type of hybrid orbitals involved. Then we write the partial orbital diagram for each central atom before and after the orbitals are hybridized. Solution: a) For CH3NH2: The shape is tetrahedral around the C and N atoms. Therefore, each central atom is sp3hybridized. The carbon atom has four half-filled sp3 orbitals: 2s 2p sp3 Isolated Carbon Atom Hybridized Carbon Atom

9. The N atom has three half-filled sp3orbitals and one filled with a lone pair. 2s sp3 2p .. H C N H H H H

10. b) The Xenon atom has filled 5 s and 5 p orbitals with the 5 d orbitals empty. Isolated Xe atom 5 d 5 s 5 p Hybridized Xe atom: sp3d2 5 d

11. .. .. b) continued:For XeF4. for Xenon, normally it has a full octet of electrons,which would mean an octahedral geometry, so to make the compound, two pairs must be broken up, and bonds made to the four fluorine atoms. If the two lone pairs are on the equatorial positions, they will be at 900 to each other, whereas if the two polar positions are chosen, the two electron groups will be 1800 from each other. Thereby minimizing the repulsion between the two electron groups. F F F F 1800 Xe Xe F F F F Square planar

12. Fig. 9.14

13. Figure 9.15: Electronegatives of the elements.

14. Fig. 9.16

15. Fig. 9.17

16. Determining Bond Polarity from Electronegativity Values Problem: (a)Indicate the polarity of the following bonds with a polarity arrow: O - H, O - Cl, C - N, P - N, N - S, C - Br, As - S (b) rank those bonds in order of increasing polarity. Plan: (a) We use Fig. 9.16 to find the EN values, and point the arrow toward the negative end. (b) Use the EN values. Solution: a) the EN of O = 3.5 and of H = 2.1: O - H the EN of O = 3.5 and of Cl = 3.0: O - Cl the EN of C = 2.5 and of P = 2.1: C - P the EN of P = 2.1 and of N = 3.0: P - N the EN of N = 3.0 and of S = 2.1: N - S the EN of C = 2.5 and of Br = 2.8: C - Br the EN of As = 2.0 and of O = 3.5: As - O b) C - Br < C - P < O - Cl < P - N < N - S < O - H < As - O 0.3 < 0.4 < 0.5 < 0.9 < 0.9 < 1.4 < 1.5

17. Fig. 9.18

18. Percent Ionic Character as a Function ofElectronegativity Difference (En) Fig. 9.19

19. .. .. Cl .. .. B .. .. .. .. Cl Cl Lewis Structures for Octet Rule Exceptions .. .. .. .. .. .. F .. .. .. .. F Cl .. .. F Each chlorine atom has 8 electrons associated. Boron has only 6! Each fluorine atom has 8 electrons associated. Chlorine has 10 electrons! . .. .. .. .. .. .. N .. .. .. .. Cl Be Cl .. O O Each chlorine atom has 8 electrons associated. The beryllium has only 4 electrons. NO2 is an odd electron atom. The nitrogen has 7 electrons.

20. Resonance Structures - Expanded Valence Shells .. .. O O .. .. .. .. H O S O H H O S O H .. .. O O .. .. .. .. .. .. .. .. .. .. .. .. .. .. F F F F .. .. .. .. .. .. .. F S F P .. .. F .. F .. .. .. .. .. .. .. .. F F .. F p = 10e- S = 12e- Sulfur hexafluoride Phosphorous pentafluoride .. .. .. .. .. Resonance Structures .. .. .. .. .. Sulfuric acid S = 12e-

21. Lewis Structures of Simple Molecules . . Sulfate . . -2 O . . . . Resonance Structures-V . . . . O S O . . . . -2 . . . . O o * o o o o O Plus 4 others for a total of 6 x o o o o o x O S O o * x o o o . . x o o -2 o o . . . . O x x . . . . o o O o o O S O o o . . . . . . . . O . . x = Sulfur electrons o = Oxygen electrons

22. Chemical Compounds and Bonds Chemical Bonds - The electrostatic forces that hold the atoms of elements together in the compound. Covalent Compounds - Electrons are shared between atoms of different elements to form Covalent Cpds. Ionic Compounds - Electrons are transferred from one atom to another to form Ionic Cpds. “Cations” - Metal atoms lose electrons to form “ + ” ions. “Anions” - Nonmetal atoms gain electrons to form “ - ” ions. Mono-atomic ions form binary ionic compounds

23. Figure 2.16: Molecular and structural formulas and molecular models for some compounds.

24. Ethanol

25. An ionic compound is composed of cations and anions. Ions are arranged in a repeating three-dimensional pattern, forming a crystal. The formula of an ionic compound gives the smallest possible integer number of ions in the substance (without writing charges) so that the combination is electrically neutral. The formula gives the formula unit of the compounds. A formula unit is not a molecule!

26. Fig. 2.18

27. Figure 2.19: A model of a portion of NaCl.

28. The Periodic Table of the Elements Most Probable Oxidation State +1 0 +2 +3 +_4 - 3 - 2 - 1 H He Li Be B C N O F Ne +3 +4 +5 +1 + 2 Na Mg Al Si P S Cl Ar K Ca Sc Ti V Cu Zn Ga Ge As Se Br Cr Mn Fe Co Ni Kr Rb Sr Zr Nb Ag I Mo Tc Ru Rh Pd Xe Y Cd In Sn Sb Te Ba Au Hg W Re Os Ir Pt Rn Cs La Hf Ta Tl Pb Bi Po At Fr Ra Ac Rf Du Sg Bo Ha Me +3 Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr +3

29. Fig. 2.20

31. Fig. 2.19

34. What is formula of the ionic compound of Mg2+ and N3-? • The common multiple of the charges is 6, so we need three Mg2+ and two N3-. The resulting formula is • Mg3N2

35. What is the formula of the ionic compound of Ca2+ and PO43-? • The common multiple of the charges is 6, so we need three Ca2+ and two PO43-. The resulting formula is • Ca3(PO4)2

36. Chemical nomenclature is the systematic naming of chemical compounds. Compounds that are not organic are called inorganic compounds. Carbon monoxide, carbon dioxide, carbonates, and cyanides are also classified as inorganic compounds.

37. Naming Inorganic Compounds Name the cation. Name the anion.

38. 2. Some main-group metals with high atomic number have more than one cation. One cation will have the charge of the group number minus 2; the second cation will have a charge equal to the group number Pb in Group IVA(14) has two ions: Pb2+ and Pb4+ Tl in Group IIIA(13) has two ions: Tl+ and Tl3+

39. 3. Most transition metals form more than one cation, of which one is +2. Zn and Cd form only the +2 ion. Ag forms only the +1 ion. 4. Nonmetal main-group elements form one monatomic anion with a charge equal to the group number minus 8. F in Group VIIA(17) forms the F- ion. S in Group VIA(16) forms the S2- ion. N in Group VA(15) forms the N3- ion.

40. Naming Monatomic Ions Monatomic cations are named after the element if the element forms only one cation.

41. If more than one cation forms: In the Stock system, the charge is written using a Roman numeral and is enclosed in parentheses. Cu2+ is copper(II). Cu+ is copper(I). • Fe3+ is iron(III) • Fe2+ is iron(II) • Hg2+ is mercury(II). • The second ion mercury forms is diatomic: Hg22+ is mercury(I).

42. Cr3+ is chromium(III). Cr2+ is chromium(II). Mn2+ is manganese(II). Co2+ is cobalt(II). Zinc forms only Zn2+, so it is called zinc ion. Cadmium forms only Cd2+, so it is called cadmium ion. Silver forms only Ag+, so it is called silver ion.

43. Polyatomic Ion An ion consisting of two or more atoms chemically bonded together and carrying an electrical charge. Table 2.5 lists common polyatomic ions.

44. Cations mercury(I) or mercurous Hg22+ ammonium NH4+ Anions peroxide O2- hydroxide OH- cyanide CN-

45. What are the names of the following ionic compounds? BaO Cr2(SO4)3 BaO is barium oxide. Cr2(SO4)3 is chromium(III) sulfate or chromic sulfate.

46. What are the chemical formulas for the following ionic compounds? potassium carbonate manganese(II) sulfate The ions K+ and CO32- form K2CO3 The ions Mn2+ and SO42- form MnSO4

47. Binary Molecular Compounds A compound composed of only two elements. Binary compound of a metal and a nonmetal are generally named using ionic rules.

48. Naming Binary MolecularCompounds We usually name the elements in the order given in the formula. Name the first element using the element name. Name the second element using the element root + -ide suffix.

49. Add a prefix to each name to indicate the number of atoms of that element. The prefix mono- is used only when needed to distinguish two compounds of the same two elements. The final vowel of the prefix is often dropped when followed by an element name that begins with a vowel. Oxygen is the most common example. N2O4 dinitrogen tetroxide (“a” is dropped) NO nitrogen monoxide (only one “o”) (also called nitric oxide)

50. Don’t use these when naming ionic compounds--they’re ONLY for covalent compounds!!