1. Principles of Chemistry, Chapt. 2: Atomic Structure and The Elements I.The Structure of Atoms protons, neutrons, and electrons II.Atomic Structure and Properties—the Elements atomic mass, atomic number, isotopes III.The Mole Concept: 6.02 x 10 23 IV.The Periodic Table 1 Homework: Chapt. 2 Problems 26, 29, 37, 43, 75

2. ~ 10 -10 meters = 1 angstrom (Å) _ + 10 -14 m Smeared out electron charge cloud + + + + + + + Protons and neutrons 2 Most of the mass is here Most of the Chemistry is here Atomic Theory in a single Slide

3. 3 STM Image: Oxygen atoms at the surface of Al 2 O 3 /Ni 3 Al(111) S. Addepalli, et al. Surf. Sci. 442 (1999) 3464 Electronic charge cloud surrounding the nucleus

4. 4 What’s inside the nucleus: ParticleMass (amu) Charge Proton (p + )1.007 amu +1 Neutron (n 0 )1.009 amu 0 What’s outside the nucleus: Electron (e-).00055 amu-1 Note: mass ratio of electron/proton (M p+ /M e- ) = 1836 For any atom: # of electrons = # of protons: Why?

5. Atomic Theory: Late 19 th Century Atomic theory—everything is made of atoms—generally accepted (thanks to Ludwig Boltzman, and others). Mendeleev/periodic table—accepted, but the basis for periodic behavior not understood What are atoms made of? How are they held together? 5

6. 6 Electrical behavior: “+” attracts “-” but like charges repel Atoms must contain smaller sub-units. Radioactive material Beam of , , and  Electrically charged plates β-particles (“–”) Gamma ray (γ) No charge, no deflection α-particle (“+” ) Heavier, deflected less than β – + Alpha particle  2 n 0 + 2p + Beta particel  electron (e-) Gamma  photon

7. Electric and magnetic fields deflect the beam. Gives mass/charge of e - = −5.60 x 10 -9 g/C Coulomb (C) = SI unit of charge Thomson (1897) discovered the e - : 7 “Cathode rays” Travel from cathode (-) to anode (+). Negative charge (e − ). Emitted by cathode metal atoms. fluorescent screen – high voltage + cathode ray

8. + _ + _ Essence of the Thompson Experiement (and old fashioned TV’s) Electric field exerts Force + plate repels +charged particles - Plate repels – charged particles F = Eq = ma d = displacement = ½ at 2 = Eq/m (t = L/V x ) Therefore, the greater the displacement, the lower the mass of the particle d x y Phosphor screen 8

9. Millikan (1911) studied electrically-charged oil drops. 9

10. Charge on each droplet was: n (−1.60 x 10 -19 C) with n = 1, 2, 3,… n (e - charge) Modern value = −1.60217653 x 10 -19 C. = −1 “atomic units”. These experiments give: 10 Modern value = 9.1093826 x 10 -28 g = (-1.60 x 10 -19 C)(-5.60 x 10 -9 g/C) = 8.96 x 10 -28 g m e = charge x mass charge

11. Atoms gain a positive charge when e - are lost. 11 Implies a positive fundamental particle. Hydrogen ions had the lowest mass. Hydrogen nuclei assumed to have “unit mass” protonsCalled protons. Modern science: m p = 1.67262129 x 10 -24 g m p ≈ 1800 x m e. Charge = -1 x (e - charge). = +1.602176462 x 10 -19 C = +1 atomic units

12.  How were p + and e - arranged? 12 Thompson: Ball of uniform positive charge, with small negative dots (e - ) stuck in it. The “plum-pudding” model. 1910 1910 Rutherford (former Thompson graduate student) fired α-particles at thin metal foils. Expected them to pass through with minor deflections.

13. Rutherford Scattering Experiment 13

14. Different Models of the Atom: different scattering results α particles “Plum pudding model” + and – charges evenly distrubted low, uniform density of matter No back scattering Rutherford’s explanation of results: Small regions of very high density + charge in the dense regions - Charges in region around it From wikipedia 14

15. Some Large Deflections were osbserved α particles Rutherford “It was about as credible as if you had fired a 15-inch shell at a piece of paper and it came back and hit you.” 15

16. ≈10,000 times smaller diameter than the entire atom. e - occupy the remaining space. α particles nucleus Most of the mass and all “+” charge is concentrated in a small core, the nucleus. 16

17. 17 Nucleus diameter~ 10 -4 Å = 10 -14 meters Mass ~ 10 -27 Kg Charge cloud Diameter ~ 1 Å Mass ~ 10 -30 kg

18. 18 Most Chemistry involves rearrangement of outermost electrons, not nuclei Example: H  1p +, 1 e- H + H  H 2 +

19. 19

20. 7 Å Epitaxial Al2O3(111) film on Ni 3 Al(111) (Kelber group): Grown in UHV Uniform No Charging S. Addepalli, et al. Surf. Sci. 442 (1999) 3464 STM Start with ordered films  growth studies Proceed to amorphous films on Si(100) Surface terminated by hexagonal array of O anions 20

21. Atomic mass > mass of all p + and e - in an atom. Rutherford proposed a neutral particle. 21 m n ≈ m p (0.1% larger). m n = 1.67492728 x 10 -24 g. Present in all atoms (except ‘normal’ H). neutrons Chadwick (1932) fired  -particles at Be atoms. Neutral particles, neutrons, were ejected:

22. ~ 10 -10 meters = 1 angstrom (Å) _ + 10 -14 m Smeared out electron charge cloud + + + + + + + Protons and neutrons 22

23. Nucleus Contains p + and n 0 Most of the atomic mass. Small (~10,000x smaller diameter than the atom). Positive (each p + has +1 charge). Small light particles surrounding the nucleus. Occupy most of the volume. Charge = -1. Atoms are neutral. Number of e − = Number of p + Electrons 23

24. 24 A neutron can decay into a proton and electron: n 0  p + + e - This can cause decay of a radioactive element, e.g., 14 6 C # of p + + n 0 Atomic No. (# e- = # p + Elemental symbol (carbon) Carbon with 6 protons and 8 neutrons is unstable (radioactive) Carbon with 6 protons and 6 neutrons is stable (non-radioactive 14 6 C 12 6 C radioctivestable

25. 25 An atom of 14 C can undergo decay to N as a neutron turns into a proton + an emitted electron 14 6 C 7 N + e- 1 p +  1 n 0 + an electron (emitted)

26.  Same element - same number of p + 26 Atomic number Atomic number (Z) = number of p + 1 amu = 1.66054 x 10 -24 g ParticleMass Mass Charge (g) (amu) (atomic units) e − 9.1093826 x 10 -28 0.000548579 −1 p + 1.67262129 x 10 -24 1.00728 +1 n 0 1.67492728 x 10 -24 1.00866 0 Atomic mass unit (amu) = (mass of C atom) that contains 6 p + and 6 n 0. 1 12

27.  Isotopes same element  Atoms of the same element with different A. equal numbers of p + different numbers of n 0 27 deuterium (D) tritium (T) Hydrogen isotopes:H1 p +, 0 n 0 1111 2121 H1 p +, 1 n 0 3131 H1 p +, 2 n 0

28. ISOTOPES: SAME Element, Different numbers of neutrons C 12 C 14 Carbon: atomic no. = 6  6 protons in the nucleus+ 6 electrons Atomic mass = 12 amu (12 gr/mole) Therefore, 6 protons + 6 neutrons Atomic mass = 14 amu Therefore, 6 protons+ 8 neutrons 28

29. Isotopes Display the same chemical reactivities (which depend mainly on the outer arrangement of the electrons) 12 C + O 2  CO 2 14 C +O 2  CO 2 Isotopes display different nuclear properties C 12 stable C 14 Radioactive: spontaneously emits electrons. Half-life ~ 5730 years 29

30. Isotopes and Moles (more on this later) and isotope abundance: 1 mole = 6.02 x 10 23 of anything! 1 mole of atoms = 6.02 x 10 23 atoms Molar Mass (in grams) = average atomic mass (in amu) 1 mole of H atoms = 1.008 gr. Why not 1.000 gr??  most atoms are 1 H, but some are 2 H (deuterium) 30

31. Average atomic mass of H = 1.008 amu 100 atoms have a mass of 100.8 amu # of 2 H atoms = n # of 1 H atoms = 100 –n (assume these are the only two isotopes that matter) Mass of 100 atoms = n x 2.000 +(100-n) x 1.000 = 100.8 amu 2n + 100-n = 100.8 n = 0.8 So, out of every 100 atoms, have 0.8 2 H atoms Out of every 1000 atoms, have 8 2 H atoms Natural abundance of “heavy hydrogen (deuterium) is then 0.8% 31

32.  Most elements occur as a mixture of isotopes. 32 Magnesium is a mixture of: 24 Mg 25 Mg 26 Mg number of p + 121212 number of n 0 121314 mass/ amu23.98524.98625.982

33. percent abundance  For most elements, the percent abundance of its isotopes are constant (everywhere on earth).  The periodic table lists an average atomic weight. 33 Example Boron occurs as a mixture of 2 isotopes, 10 B and 11 B. The abundance of 10 B is 19.91%. Calculate the atomic weight of boron.

34. 34 Atomic weight for B = 1.994 + 8.817 amu = 10.811 amu Atomic mass = Σ (fractional abundance)(isotope mass) (11.0093 amu) = 8.817 amu 11 B 80.09 100 (10.0129 amu) 10 B 19.91 100 = 1.994 amu % abundance of 11 B = 100% - 19.91% = 80.09% Boron occurs as a mixture of 2 isotopes, 10 B and 11 B. The abundance of 10 B is 19.91%. Calculate the atomic weight of boron.

35. B 10.811 Boron 5 Atomic number (Z) Symbol Name Atomic weight Periodic table: 35

36.  A counting unit – a familiar counting unit is a “dozen”: 36 1 dozen eggs= 12 eggs 1 dozen donuts= 12 donuts 1 dozen apples= 12 apples 1 mole (mol) = Number of atoms in 12 g of 12 C Latin for “heap” or “pile” 1 mol = 6.02214199 x 10 23 “units” Avogadro’s numberAvogadro’s number

37.  A green pea has a ¼-inch diameter. 48 peas/foot.  (48) 3 / ft 3 ≈ 1 x 10 5 peas/ft 3.  V of 1 mol ≈ (6.0 x 10 23 peas)/(1x 10 5 peas/ft 3 )  ≈ 6.0 x 10 18 ft 3 37 height = V / area, 1 mol would cover the U.S. to: U.S. surface area = 3.0 x 10 6 mi 2 = 8.4 x 10 13 ft 2 6.0 x 10 18 ft 3 8.4 x 10 13 ft 2 =7.1 x 10 4 ft = 14 miles !

38.  1 mole of an atom = atomic weight in grams. 38 1 Xe atom has mass = 131.29 amu 1 mol of Xe atoms has mass = 131.29 g and There are 6.022 x 10 23 atoms in 1 mol of He and 1 mol of Xe – but they have different masses. 1 He atom has mass = 4.0026 amu 1 mol of He has mass = 4.0026 g … 1 dozen eggs is much heavier than 1 dozen peas!

39. Example How many moles of copper are in a 320.0 g sample? Cu-atom mass = 63.546 g/mol (periodic table) Conversion factor: 1 mol Cu 63.546 g = 1 n Cu = 320.0 g x 1 mol Cu 63.546 g = 5.036 mol Cu n = number of moles 39

40.  Calculate the number of atoms in a 1.000 g sample of boron. 40 n B = (1.000 g) 1 mol B 10.81 g = 0.092507 mol B B atoms = (0.092507 mol B)(6.022  10 23 atoms/mol) = 5.571  10 22 B atoms

41. Dimensional Analysis and Problem Solving Special Homework Problem: Due Tues. Recitation Density = mass/volume Problem: Assume that a hydrogen atom has a spherical diameter of 1 angstrom Assume that the nucleus (1 proton) has a diameter of 10 -4 angstrom 1.Calculate the densities of the nucleus, and of the electron charge cloud in kg/m 3 2.Calculate the ratio of the two densities: R = d nucleus /d electron cloud Mass of proton = 1.67 x 10 -27 kg Mass of electron = 9 x 10 -31 kg 41

42.  Summarizes Atomic numbers. Atomic weights. Physical state (solid/liquid/gas). Type (metal/non-metal/metalloid).  Periodicity Elements with similar properties are arranged in vertical groups. 42

43. In the USA, “A” denotes a main group element… …”B” indicates a transition element. International system uses 1 … 18. The Periodic Table 43

44. Main group metal Transition metal Metalloid Nonmetal The Periodic Table 44

45. A period is a horizontal row Period number 45

46. A group is a vertical column Group 7A Halogens Group 8A Noble gases Group 2A Alkaline earth metals Group 1A Alkali metals (not H) 46

47.  Alkali metals (group 1A; 1)  Alkaline earth metals (group 2A; 2) 47 Grey … silvery white colored. Highly reactive. Never found as native metals. Form alkaline solutions.

48.  Transition Elements (groups 1B – 8B) Also called transition metals. Middle of table, periods 4 – 7. Includes the lanthanides & actinides. 48 Lanthanides and Actinides Listed separately at the bottom. Chemically very similar. Relatively rare on earth.  (old name: rare earth elements)

49.  Groups 3A to 6A Most abundant elements in the Earth’s crust and atmosphere. Most important elements for living organisms. 49 Halogens (group 7A; 17) Very reactive non metals. Form salts with metals. Colored elements. Noble gases (8A; 18) Very low reactivity. Colorless, odorless gases.

50.  Atoms are very small. 1 tsp of water contains 3x as many atoms as there are tsp of water in the Atlantic Ocean! 50 Impractical to use pounds and inches... Need a universal unit system The metric system. The SI system (Systeme International) - derived from the metric system.

51. A decimal system. Prefixes multiply or divide a unit by multiples of ten. 51 Prefix Factor Example megaM10 6 1 megaton = 1 x 10 6 tons kilok10 3 1 kilometer (km) = 1 x 10 3 meter (m) decid10 -1 1 deciliter (dL) = 1 x 10 -1 liter (L) centic10 -2 1 centimeter (cm) = 1 x 10 -2 m millim10 -3 1 milligram (mg) = 1 x 10 -3 gram (g) microμ 10 -6 1 micrometer (μm) = 1 x 10 -6 m nanon10 -9 1 nanogram (ng) = 1 x 10 -9 g picop10 -12 1 picometer (pm) = 1 x 10 -12 m femtof10 -15 1 femtogram (fg) = 1 x 10 -15 g

52. 1 pm = 1 x 10 -12 m ;1 cm= 1 x 10 -2 m 52 How many copper atoms lie across the diameter of a penny? A penny has a diameter of 1.90 cm, and a copper atom has a diameter of 256 pm. x 1 x 10 -2 m 1 cm = 7.42 x 10 7 Cu atoms 1 pm 1 x 10 -12 m x 1.90 cm = 1.90 x 10 10 pm Number of atoms across the diameter: 1.90 x 10 10 pm x 1 Cu atom 256 pm

53. Length Length1 kilometer= 0.62137 mile 1 inch= 2.54 cm (exactly) 1 angstrom (Å)= 1 x 10 -10 m Volume Volume1 liter (L) = 1000 cm 3 = 1000 mL = 1.056710 quarts 1 gallon= 4 quarts = 8 pints Mass Mass1 amu= 1.66054 x 10 -24 g 1 pound= 453.59237 g = 16 ounces 1 ton (metric)= 1000 kg 1 ton (US)= 2000 pounds 53

54. Example: What is the volume of a 2 gallon container in Liters? 1 gallon x 4 quarts/gallon = 4 quarts 4 quarts x 1Liter/1.057 quarts = 3.784 Liters (L) 54

55. 165 mg dL A patient’s blood cholesterol level measured 165 mg/dL. Express this value in g/L 1 mg = 1 x 10 -3 g ; 1 dL = 1 x 10 -1 L x 1 x10 -3 g 1 mg = 1.65 g/L x 1 dL 1 x10 -1 L 55

56.  All measurements involve some uncertainty. one  Reported numbers include one uncertain digit. 56 Consider a reported mass of 6.3492 g Last digit (“2”) is uncertain Close to 2, but may be 4, 1, 0 … significant figuresFive significant figures in this number.

57.  Read numbers from left to right. starting  Count all digits, starting with the 1 st non-zero digit.  Allexcept  All digits are significant except zeros used to position a decimal point (“placeholders”).  0.00024030  5 sig. figs.  (2.4030 x 10 -4 ) 57 placeholders significant

58. 58

59. Round 37.663147 to 3 significant figures. 1 st non-significant digit  Examine the 1 st non-significant digit. If it: 59 > 5, round up. < 5, round down. = 5, check the 2 nd non-significant digit.  round up if absent or odd; round down if even. last retained digit 1 st non- significant digit Rounds up to 37.7 2 nd non- significant digit