1. HSTMr.Watson Dr. S. M. Condren Atoms, Molecules & Ions Chapter 2 HST

2. Mr.Watson Quantum Corral http://www.almaden.ibm.com/vis/stm/corral.html

3. HSTMr.Watson Scanning Tunneling Microscope

4. HSTMr.Watson Scanning Tunneling Microscope

5. HSTMr.Watson Scanning Tunneling Microscope

6. HSTMr.Watson

7. HSTMr.Watson http://mrsec.wisc.edu/ Developed in collaboration with the Institute for Chemical Education and the Magnetic Microscopy Center University of Minnesota http://www.physics.umn.edu/groups/mmc/

8. HSTMr.Watson Pull Probe Strip Probe Sample Pull Probe Strip http://www.nsf.gov/mps/dmr/mrsec.htm

9. HSTMr.Watson (a) (b) NorthSouth (c) Which best represents the poles?

10. HSTMr.Watson Atoms & Molecules Atoms can exist alone or enter into chemical combination the smallest indivisible particle of an element Molecules a combination of atoms that has its own characteristic set of properties

11. HSTMr.Watson Law of Constant Composition A chemical compound always contains the same elements in the same proportions by mass.

12. HSTMr.Watson Law of Multiple Proportions the same elements can be combined to form different compounds by combining the elements in different proportions

13. HSTMr.Watson Dalton’s Atomic Theory Postulates proposed in 1803 know at least 2 for first exam

14. HSTMr.Watson Dalton’s Atomic Theory Postulate 1 An element is composed of tiny particles called atoms. All atoms of a given element show the same chemical properties.

15. HSTMr.Watson Dalton’s Atomic Theory Postulate 2 Atoms of different elements have different properties.

16. HSTMr.Watson Dalton’s Atomic Theory Postulate 3 Compounds are formed when atoms of two or more elements combine. In a given compound, the relative number of atoms of each kind are definite and constant.

17. HSTMr.Watson Dalton’s Atomic Theory Postulate 4 In an ordinary chemical reaction, no atom of any element disappears or is changed into an atom of another element. Chemical reactions involve changing the way in which the atoms are joined together.

18. HSTMr.Watson Radioactivity

19. HSTMr.Watson Radioactivity Alpha – helium-4 nucleus Beta – high energy electron Gamma – energy resulting from transitions from one nuclear energy level to another

20. HSTMr.Watson Alpha Radiation composed of 2 protons and 2 neutrons thus, helium-4 nucleus +2 charge mass of 4 amu creates element with atomic number 2 lower Ra 226  Rn 222 + He 4 (  )

21. HSTMr.Watson Beta Radiation composed of a high energy electron which was ejected from the nucleus “neutron” converted to “proton” very little mass -1 charge creates element with atomic number 1 higher U 239  Np 239 +  -1

22. HSTMr.Watson Gamma Radiation nucleus has energy levels energy released from nucleus as the nucleus changes from higher to lower energy levels no mass no charge Ni 60*  Ni 60 + 

23. HSTMr.Watson Cathode Ray Tube

24. HSTMr.Watson Thompson’s Charge/Mass Ratio

25. HSTMr.Watson Millikin’s Oil Drop

26. HSTMr.Watson Rutherford’s Gold Foil

27. HSTMr.Watson Rutherford’s Model of the Atom

28. HSTMr.Watson Rutherford’s Model of the Atom atom is composed mainly of vacant space all the positive charge and most of the mass is in a small area called the nucleus electrons are in the electron cloud surrounding the nucleus

29. HSTMr.Watson Structure of the Atom Composed of: protons neutrons electrons

30. HSTMr.Watson Structure of the Atom Composed of: protons neutrons electrons protons –found in nucleus –relative charge of +1 –relative mass of 1.0073 amu

31. HSTMr.Watson Structure of the Atom Composed of: protons neutrons electrons neutrons –found in nucleus –neutral charge –relative mass of 1.0087 amu

32. HSTMr.Watson Structure of the Atom Composed of: protons neutrons electrons –found in electron cloud –relative charge of -1 –relative mass of 0.00055 amu

33. HSTMr.Watson Size of Nucleus If the nucleus were 1” in diameter, the atom would be 1.5 miles in diameter.

34. HSTMr.Watson Ions charged single atom charged cluster of atoms

35. HSTMr.Watson Ions cations –positive ions anions –negative ions ionic compounds –combination of cations and anions –zero net charge

36. HSTMr.Watson Atomic number, Z the number of protons in the nucleus the number of electrons in a neutral atom the integer on the periodic table for each element

37. HSTMr.Watson Isotopes atoms of the same element which differ in the number of neutrons in the nucleus designated by mass number

38. HSTMr.Watson Mass Number, A integer representing the approximate mass of an atom equal to the sum of the number of protons and neutrons in the nucleus

39. HSTMr.Watson Masses of Atoms Carbon-12 Scale

40. HSTMr.Watson Isotopes of Hydrogen H-1, 1 H, protium 1 proton and no neutrons in nucleus only isotope of any element containing no neutrons in the nucleus most common isotope of hydrogen

41. HSTMr.Watson Isotopes of Hydrogen H-2 or D, 2 H, deuterium 1 proton and 1 neutron in nucleus

42. HSTMr.Watson Isotopes of Hydrogen H-3 or T, 3 H, tritium 1 proton and 2 neutrons in nucleus

43. HSTMr.Watson Isotopes of Oxygen O-16 8 protons, 8 neutrons, & 8 electrons O-17 8 protons, 9 neutrons, & 8 electrons O-18 8 protons, 10 neutrons, & 8 electrons

44. HSTMr.Watson The radioactive isotope 14 C has how many neutrons? 6, 8, other

45. HSTMr.Watson The identity of an element is determined by the number of which particle? protons, neutrons, electrons

46. HSTMr.Watson Mass Spectrometer

47. HSTMr.Watson Mass Spectra of Neon

48. HSTMr.Watson Measurement of Atomic Masses Mass Spectrometer a simulation is available at http://www.colby.edu/chemistry/ OChem/DEMOS/MassSpec.html

49. HSTMr.Watson Atomic Masses and Isotopic Abundances natural atomic masses = sum[(atomic mass of isotope) *(fractional isotopic abundance)]

50. HSTMr.Watson Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of 34.96885 and 36.96590 amu, respectively. The natural atomic mass of chlorine is 35.453 amu. What are the percent abundances of the two isotopes? let x = fraction Cl-35y = fraction Cl-37 x + y = 1y = 1 - x (AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.453 Thus: 34.96885*x + 36.96590*y = 35.453

51. HSTMr.Watson Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of 34.96885 and 36.96590 amu, respectively. The natural atomic mass of chlorine is 35.453 amu. What are the percent abundances of the two isotopes? let x = fraction Cl-35 y = fraction Cl-37 x + y = 1 y = 1 - x

52. HSTMr.Watson Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of 34.96885 and 36.96590 amu, respectively. The natural atomic mass of chlorine is 35.453 amu. What are the percent abundances of the two isotopes? let x = fraction Cl-35 y = fraction Cl-37 x + y = 1 y = 1 - x (AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.453

53. HSTMr.Watson Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of 34.96885 and 36.96590 amu, respectively. The natural atomic mass of chlorine is 35.453 amu. What are the percent abundances of the two isotopes? let x = fraction Cl-35 y = fraction Cl-37 x + y = 1 y = 1 - x (AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.453 34.96885*x + 36.96590*y = 35.453 34.96885*x + 36.96590*(1-x) = 35.453

54. HSTMr.Watson Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of 34.96885 and 36.96590 amu, respectively. The natural atomic mass of chlorine is 35.453 amu. What are the percent abundances of the two isotopes? let x = fraction Cl-35 y = fraction Cl-37 x + y = 1 y = 1 - x (AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.453 34.96885*x + 36.96590*y = 35.453 34.96885*x + 36.96590*(1-x) = 35.453 (34.96885 - 36.96590)x + 36.96590 = 35.453

55. HSTMr.Watson Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of 34.96885 and 36.96590 amu, respectively. The natural atomic mass of chlorine is 35.453 amu. What are the percent abundances of the two isotopes? let x = fraction Cl-35 y = fraction Cl-37 x + y = 1 y = 1 - x (AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.453 34.96885*x + 36.96590*y = 35.453 34.96885*x + 36.96590*(1-x) = 35.453 (34.96885 - 36.96590)x + 36.96590 = 35.453

56. HSTMr.Watson Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of 34.96885 and 36.96590 amu, respectively. The natural atomic mass of chlorine is 35.453 amu. What are the percent abundances of the two isotopes? let x = fraction Cl-35 y = fraction Cl-37 x + y = 1 y = 1 - x (AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.453 34.96885*x + 36.96590*y = 35.453 34.96885*x + 36.96590*(1-x) = 35.453 (34.96885 - 36.96590)x + 36.96590 = 35.453 (34.96885 - 36.96590)x = (35.453 - 36.96590)

57. HSTMr.Watson Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of 34.96885 and 36.96590 amu, respectively. The natural atomic mass of chlorine is 35.453 amu. What are the percent abundances of the two isotopes? let x = fraction Cl-35 y = fraction Cl-37 x + y = 1 y = 1 - x (AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.453 34.96885*x + 36.96590*y = 35.453 34.96885*x + 36.96590*(1-x) = 35.453 (34.96885 - 36.96590)x + 36.96590 = 35.453 (34.96885 - 36.96590)x = (35.453 - 36.96590) - 1.99705x = - 1.5129

58. HSTMr.Watson Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of 34.96885 and 36.96590 amu, respectively. The natural atomic mass of chlorine is 35.453 amu. What are the percent abundances of the two isotopes? let x = fraction Cl-35 y = fraction Cl-37 x + y = 1 y = 1 - x (AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.453 34.96885*x + 36.96590*y = 35.453 34.96885*x + 36.96590*(1-x) = 35.453 (34.96885 - 36.96590)x + 36.96590 = 35.453 (34.96885 - 36.96590)x = (35.453 - 36.96590) - 1.99705x = - 1.5129 1.99705x = 1.5129

59. HSTMr.Watson Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of 34.96885 and 36.96590 amu, respectively. The natural atomic mass of chlorine is 35.453 amu. What are the percent abundances of the two isotopes? let x = fraction Cl-35 y = fraction Cl-37 x + y = 1 y = 1 - x (AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.453 34.96885*x + 36.96590*y = 35.453 34.96885*x + 36.96590*(1-x) = 35.453 (34.96885 - 36.96590)x + 36.96590 = 35.453 (34.96885 - 36.96590)x = (35.453 - 36.96590) - 1.99705x = - 1.5129 1.99705x = 1.5129 x = 0.7553 75.53% Cl-35

60. HSTMr.Watson Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of 34.96885 and 36.96590 amu, respectively. The natural atomic mass of chlorine is 35.453 amu. What are the percent abundances of the two isotopes? let x = fraction Cl-35 y = fraction Cl-37 x + y = 1 y = 1 - x (AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.453 34.96885*x + 36.96590*y = 35.453 34.96885*x + 36.96590*(1-x) = 35.453 (34.96885 - 36.96590)x + 36.96590 = 35.453 (34.96885 - 36.96590)x = (35.453 - 36.96590) - 1.99705x = - 1.5129 1.99705x = 1.5129 x = 0.7553 75.53% Cl-35 y = 1 - x

61. HSTMr.Watson Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of 34.96885 and 36.96590 amu, respectively. The natural atomic mass of chlorine is 35.453 amu. What are the percent abundances of the two isotopes? let x = fraction Cl-35 y = fraction Cl-37 x + y = 1 y = 1 - x (AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.453 34.96885*x + 36.96590*y = 35.453 34.96885*x + 36.96590*(1-x) = 35.453 (34.96885 - 36.96590)x + 36.96590 = 35.453 (34.96885 - 36.96590)x = (35.453 - 36.96590) - 1.99705x = - 1.5129 1.99705x = 1.5129 x = 0.7553 75.53% Cl-35 y = 1 - x = 1.0000 - 0.7553

62. HSTMr.Watson Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of 34.96885 and 36.96590 amu, respectively. The natural atomic mass of chlorine is 35.453 amu. What are the percent abundances of the two isotopes? let x = fraction Cl-35 y = fraction Cl-37 x + y = 1 y = 1 - x (AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.453 34.96885*x + 36.96590*y = 35.453 34.96885*x + 36.96590*(1-x) = 35.453 (34.96885 - 36.96590)x + 36.96590 = 35.453 (34.96885 - 36.96590)x = (35.453 - 36.96590) - 1.99705x = - 1.5129 1.99705x = 1.5129 x = 0.7553 75.53% Cl-35 y = 1 - x = 1.0000 - 0.7553 = 0.2447

63. HSTMr.Watson Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of 34.96885 and 36.96590 amu, respectively. The natural atomic mass of chlorine is 35.453 amu. What are the percent abundances of the two isotopes? let x = fraction Cl-35 y = fraction Cl-37 x + y = 1 y = 1 - x (AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.453 34.96885*x + 36.96590*y = 35.453 34.96885*x + 36.96590*(1-x) = 35.453 (34.96885 - 36.96590)x + 36.96590 = 35.453 (34.96885 - 36.96590)x = (35.453 - 36.96590) - 1.99705x = - 1.5129 1.99705x = 1.5129 x = 0.7553 75.53% Cl-35 y = 1 - x = 1.0000 - 0.7553 = 0.2447 24.47% Cl-37

64. HSTMr.Watson Development of Periodic Table Newlands - English 1864 - Law of Octaves - every 8th element has similar properties

65. HSTMr.Watson Development of Periodic Table Dmitri Mendeleev - Russian 1869 - Periodic Law - allowed him to predict properties of unknown elements

66. HSTMr.Watson Mendeleev’s Periodic Table the elements are arranged according to increasing atomic weights

67. HSTMr.Watson Missing elements: 44, 68, 72, & 100 amu Mendeleev’s Periodic Table

68. HSTMr.Watson Properties of Ekasilicon

69. HSTMr.Watson Modern Periodic Table Moseley, Henry Gwyn Jeffreys 1887–1915, English physicist. Studied the relations among bright-line spectra of different elements. Derived the ATOMIC NUMBERS from the frequencies of vibration of X-rays emitted by each element. Moseley concluded that the atomic number is equal to the charge on the nucleus. This work explained discrepancies in Mendeleev’s Periodic Law.

70. HSTMr.Watson Modern Periodic Table the elements are arranged according to increasing atomic numbers

71. HSTMr.Watson Periodic Table of the Elements

72. HSTMr.Watson Organization of Periodic Table period - horizontal row group - vertical column

73. HSTMr.Watson Family Names Group IAalkali metals Group IIAalkaline earth metals Group VIIAhalogens Group VIIIAnoble gases transition metals inner transition metals lanthanum seriesrare earths actinium seriestrans-uranium series

74. HSTMr.Watson Types of Elements metals nonmetals metalloids - semimetals

75. HSTMr.Watson Elements, Compounds, and Formulas Elements can exist as single atoms or molecules Compounds combination of two or more elements molecular formulas for molecular compounds empirical formulas for ionic compounds

76. HSTMr.Watson Organic Compounds Organic Chemistry branch of chemistry in which carbon compounds and their reactions are studied. the chemistry of carbon-hydrogen compounds

77. HSTMr.Watson Inorganic Compounds Inorganic Chemistry field of chemistry in which are studied the chemical reactions and properties of all the chemical elements and their compounds, with the exception of the hydrocarbons (compounds composed of carbon and hydrogen) and their derivatives.

78. HSTMr.Watson Molecular and Structural Formulas

79. HSTMr.Watson Bulk Substances mainly ionic compounds –empirical formulas –structural formulas

80. HSTMr.Watson Models of Sodium Chloride NaCl “table salt”

81. HSTMr.Watson How many atoms are in the formula Al 2 (SO 4 ) 3 ? 3, 5, 17

82. HSTMr.Watson Naming Binary Molecular Compounds For compounds composed of two non- metallic elements, the more metallic element is listed first. To designate the multiplicity of an element, Greek prefixes are used: mono => 1; di => 2; tri => 3; tetra => 4; penta => 5; hexa => 6; hepta => 7; octa => 8

83. HSTMr.Watson Common Compounds H2OH2O water NH 3 ammonia N2ON2O nitrous oxide CO carbon monoxide CS 2 carbon disulfide SO 3 sulfur trioxide CCl 4 carbon tetrachloride PCl 5 phosphorus pentachloride SF 6 sulfur hexafluoride

84. HSTMr.Watson Alkanes - C n H 2n+2 methane - CH 4 ethane - C 2 H 6 propane - C 3 H 8 butanes - C 4 H 10 pentanes - C 5 H 12 hexanes - C 6 H 14 heptanes - C 7 H 16 octanes - C 8 H 18 nonanes - C 9 H 20 decanes - C 10 H 22

85. HSTMr.Watson Burning of Propane Gas

86. HSTMr.Watson Butanes

87. HSTMr.Watson Ionic Bonding Characteristics of compounds with ionic bonding: non-volatile, thus high melting points solids do not conduct electricity, but melts (liquid state) do many, but not all, are water soluble

88. HSTMr.Watson Ion Formation

89. HSTMr.Watson Valance Charge on Ions compounds have electrical neutrality metals form positive monatomic ions non-metals form negative monatomic ions

90. HSTMr.Watson Valence of Metal Ions Monatomic Ions Group IA=> +1 Group IIA=> +2 Maximum positive valence equals Group A #

91. HSTMr.Watson Valence of Non-Metal Ions Monatomic Ions Group VIA=> -2 Group VIIA=> -1 Maximum negative valence equals (8 - Group A #)

92. HSTMr.Watson Charges of Some Important Ions

93. HSTMr.Watson Polyatomic Ions more than one atom joined together have negative charge except for NH 4 + and its relatives negative charges range from -1 to -4

94. HSTMr.Watson Polyatomic Ions ammoniumNH 4 + perchlorateClO 4 1- cyanideCN 1- hydroxideOH 1- nitrateNO 3 1- sulfateSO 4 2- carbonateCO 3 2- phosphatePO 4 3-

95. HSTMr.Watson Names of Ionic Compounds 1. Name the metal first. If the metal has more than one oxidation state, the oxidation state is specified by Roman numerals in parentheses. 2. Then name the non-metal, changing the ending of the non-metal to -ide.

96. HSTMr.Watson Nomenclature NaCl sodium chloride Fe 2 O 3 iron(III) oxide N2O4N2O4 dinitrogen tetroxide KI potassium iodide Mg 3 N 2 magnesium nitride SO 3 sulfur trioxide

97. HSTMr.Watson Nomenclature NH 4 NO 3 ammonium nitrate KClO 4 potassium perchlorate CaCO 3 calcium carbonate NaOH sodium hydroxide

98. HSTMr.Watson Nomenclature Drill Available for PCs: –http://science.widener.edu/svb/pset/nomen_ b.htmlhttp://science.widener.edu/svb/pset/nomen_ b.html – in the Chemistry Resource Center –, Links http://en.wikipedia.org/wiki/IUPAC_nomen clature_of_organic_chemistry

99. HSTMr.Watson How many moles of ions are there per mole of Al 2 (SO 4 ) 3 ? 2, 3, 5

100. HSTMr.Watson Chemical Equation reactants products coefficients reactants -----> products

101. HSTMr.Watson Writing and Balancing Chemical Equations Write a word equation. Convert word equation into formula equation. Balance the formula equation by the use of prefixes (coefficients) to balance the number of each type of atom on the reactant and product sides of the equation.

102. HSTMr.Watson Example Hydrogen gas reacts with oxygen gas to produce water. Step 1. hydrogen + oxygen -----> water Step 2. H 2 + O 2 -----> H 2 O Step 3. 2 H 2 + O 2 -----> 2 H 2 O

103. HSTMr.Watson Example Iron(III) oxide reacts with carbon monoxide to produce the iron oxide (Fe 3 O 4 ) and carbon dioxide. iron(III) oxide + carbon monoxide -----> Fe 3 O 4 + carbon dioxide Fe 2 O 3 + CO -----> Fe 3 O 4 + CO 2 3 Fe 2 O 3 + CO -----> 2 Fe 3 O 4 + CO 2