1. Saturday Study Session 2 2 nd Class Atomic Structure and Bonding

2. Opening Activity TrendWhat It IsAcross a PeriodDown a Group Atomic radius Justification First ionization energy Justification

3. Content Crash Course Atomic Structure Instrumentation –Mass Spec –PES Electron Configurations Periodic Trends Intramolecular Bonds

4. Atomic Structure Small, dense nucleus –Protons (+) and neutrons (no charge) –Protons + neutrons = mass number Remaining atomic volume –Electrons (-) –Number of electrons in a neutral atom = atomic number Isotopes have the same atomic number, but different mass numbers. Useful equations: E = hv c = λν

5. 1.) Naturally occurring copper is composed of two isotopes, copper-63 and copper-65. If copper is 69.1% copper-63 and 30.9% copper-65, the average atomic mass could be calculated as A. [(0.309 x 65) + (0.691 x 63)]/2 B. (63 + 65)/2 C. (65 x 30.9) + (63 x 69.1) D. (0.309 x 65) + (0.691 x 63)

6. Question 1 Answer D Clue: Average atomic mass is calculated as the sum of the products of the relative abundance and atomic mass for each isotope of a given element.

7. Instrumentation Mass Spec –Separates atoms and molecules according to mass Substance is heated in a vacuum and ionized Acceleration through a magnetic field causes ions to be separated according to mass PES –Photoelectron spectroscopy Photons knock electrons out of atoms

8. Mass Spec Data How many isotopes are represented by this spectrum? Which element does this spectrum represent?

9. PES Data Which element does this spectrum represent?

10. PES Data Which element does this spectrum represent?

11. 2.) The photoelectron spectra show the energy required to remove a 1s electron from a nitrogen atom and from an oxygen atom. Which of the following statements best accounts for the peak in the upper spectrum being to the right of the peak in the lower spectrum? A.Nitrogen atoms have a half-filled p subshell. B.There are more electron-electron repulsions in oxygen atoms than in nitrogen atoms. C.Electrons in the p subshell of oxygen atoms provide more shielding than electrons in the p subshell of nitrogen atoms. D.Nitrogen atoms have a smaller nuclear charge than oxygen atoms.

12. Question 2 Answer D Clue: Less effective nuclear charge means less energy required to remove an electron.

13. Electron Configurations Atoms –# electrons = atomic # –Fill according to Aufbau Principle, Hund’s Rule, and Pauli Exclusion Principle –Use periodic table to “cheat” Orbital notation Noble gas notation Ions –Cations lose electrons –Anions gain electrons –Isoelectronic = same # of electrons

14. 3.) An element Z has the electron configuration [Xe]6s 2. The oxide of this element will have which formula below? A.Z 2 O B.Z 2 O 3 C.ZO D.ZO 2

15. Question 3 Answer C Clue: The electron configuration indicates 2 valence electrons, which would be given up to form a +2 ion.

16. Periodic Trends TrendWhat It IsAcross a PeriodDown a Group Atomic radius Essentially the distance from the nucleus to valence electrons DecreasesIncreases Justification Increased effective nuclear charge Stronger force of attraction between nucleus and electron cloud Justification Increased principal energy level Less strong force of attraction between nucleus and electron cloud First ionization energy Energy required to remove an electron from an atom in the gas phase IncreasesDecreases Justification Increased effective nuclear charge Stronger force of attraction between nucleus and electrons More energy required to remove an electron Justification Increased principal energy level Less strong force of attraction between nucleus and electrons Less energy required to remove an electron

17. 4.) Of the following, the species that has the correctly predicted largest radius is A. Ar B. Br - C. K + D. Sr 2+

18. Question 4 Answer B Clue: Anions exhibit greater radii because of additional electrons and decreased effective nuclear charge.

19. 5.) The observation involving the first ionization energies of O and N is best explained by which of the following? A. Differences in electronegativity between O and N cause the first ionization energy of O to be less than N. B. Repulsion between the two paired electrons in the O orbital cause the first ionization energy of O to be less than N. C. Greater effective nuclear charge in N than in O causes the first ionization energy of O to be less than N. D. Greater atomic radius in N than in O causes the first ionization energy of O to be less than N.

20. Question 5 Answer B Clue: Comparing electron configurations of O and N should allow you to see that O has paired electrons while N does not.

21. 6.) Which of the following elements would have the largest second ionization energy? A.Hydrogen B.Fluorine C.Sulfur D.Potassium

22. Question 6 Answer D Clue: While both H and K have one valence electron, only K has more than one electron period.

23. 7.) The ionization energies for element X are listed above. On the basis of the data, element X is most likely to be A.Na B.Mg C.Al D.Si Ionization Energies for element X (kJ mol -1 ) FirstSecondThirdFourthFifth 580181527401160014800

24. Question 7 Answer C Clue: The huge jump between third and fourth ionization energies means we’re dealing with three valence electrons.

25. Intramolecular Bonds Bonds BETWEEN atoms Not to be confused with intermolecular forces (IMFs) Result in chemical change Ionic –Large differences in electronegativity –Usually metal and non-metal Covalent –Small differences in electronegativity –Usually non-metals Polar (H-F); dipole moments DO NOT cancel Non-polar (F-F); dipole moments DO cancel out

26. 8.) Which of the following particulate diagrams best shows the formation of water vapor from hydrogen gas and oxygen gas in a rigid container at 125º C.

27. Question 8 Answer C Clue: 6H 2 (g) + 3O 2 (g)  6H 2 O (g) Or 2H 2 (g) + O 2 (g)  2H 2 O (g)

28. 9.) In which of the following processes are covalent bonds broken? A.I 2 (s)  I 2 (g) B.C (diamond)  C (g) C.Fe (s)  Fe (l) D.CO 2 (s)  CO 2 (g)

29. Question 9 Answer B Clue: Phase changes involve the overcoming of IMFs, not INTRAMOLECULAR bonds. Diamonds involve network covalent bonds.

30. Short Free Response 1.) Using principles of atomic structure and the information in the table above, answer the following questions about atomic fluorine, oxygen, and xenon. (a)Write the equation for the ionization of atomic fluorine that requires 1,681.0 kJ mol -1. (b)Account for the fact that the first ionization energy of atomic fluorine is greater than that of atomic oxygen. (You must discuss both atoms in your response.) (c)Predict whether the first ionization energy of atomic xenon is greater than, less than, or equal to the first ionization energy of atomic fluorine. Justify your prediction.

31. Short Free Response Rubric (3 points possible)

32. Free Response #1 1.) The table above shows the first three ionization energies for atoms of four elements from the third period of the periodic table. The elements are numbered randomly. Use the information in the table to answer the following questions. (a) Which element is most metallic in character? Explain your reasoning. (b) Identify element 3. Explain your reasoning. (c) Write the complete electron configuration for an atom of element 3. (d) What is the expected oxidation state for the most common ion of element 2? (e) What is the chemical symbol for element 2? (f) A neutral atom of which of the four elements has the smallest radius?

33. Free Response #1 Rubric (8 points possible)

34. Free Response #1 Rubric

36. Free Response #2 2.) Answer the following problems about gases. (a) The average atomic mass of naturally occurring neon is 20.18 amu. There are two common isotopes of naturally occurring neon as indicated in the table above. (i) Using the information above, calculate the percent abundance of each isotope. (ii) Calculate the number of Ne-22 atoms in a 12.55 g sample of naturally occurring neon. (b) A major line in the emission spectrum of neon corresponds to a frequency of 4.34 × 10 14 s −1. Calculate the wavelength, in nanometers, of light that corresponds to this line. (c) In the upper atmosphere, ozone molecules decompose as they absorb ultraviolet (UV) radiation, as shown by the equation below. Ozone serves to block harmful ultraviolet radiation that comes from the Sun. O 3 (g)  O 2 (g) + O (g) (d) A molecule of O 3 (g) absorbs a photon with a frequency of 1.00 × 10 15 s −1. (i) How much energy, in joules, does the O 3 (g) molecule absorb per photon? (ii) The minimum energy needed to break an oxygen-oxygen bond in ozone is 387 kJ mol -1. Does a photon with a frequency of 1.00 × 10 15 s −1 have enough energy to break this bond? Support your answer with a calculation.

37. Free Response #2 Rubric (9 points possible)

38. Free Response #2 Rubric