1. Unit 4
Atomic Structure,
PES,
Periodicity

2. Unit 4 - Atomic Structure
Electromagnetic Radiation - energy equations
Bohr Model of the Atom
Quantum Model of the Atom
Electron Configurations
Spectroscopy
Mass spec and Average atomic mass
PES and Ionization energy
Periodic Trends and Justifications

3. Electromagnetic Spectrum
All frequencies of the Electromagnetic Spectrum are considered “LIGHT”.

4. Properties of Waves Wavelength (λ) Frequency (ν)
Wavelength and frequency are inversely proportional.

5. Visible Light spectrum
White light passed through a prism produces the continuous spectrum. When light that is emitted from excited electrons of a particular substance only some of the wavelengths are produced. That light passed through a prism creates an emission line spectrum.

6. Hydrogen Line Emission Spectrum
When Hydrogen gas absorbs energy the H atoms are excited. That excess energy is released and produces the Hydrogen Line Emission Spectrum. Here we are examining only those wavelengths produced that are in the visible region of the electromagnetic spectrum. The four spectral lines indicate that only those four energies (in the visible spectrum) are absorbed by the electron in the hydrogen atom. Energy is quantized.

7. H electrons absorb only these visible light energies
When electrons in their ground state absorb energy the electron jumps to a higher energy level. When the electron falls back down to a lower energy level (ground or other lower level) that excess energy is emitted as “light”.

8. Bohr model of the Atom
Emitted wavelengths of light are translated into electron energy levels.

9. Problems with Bohr model
Bohr’s model does not work for atoms other than Hydrogen.
hydrogen has only one electron
Bohr model does not work for multi-electron atoms
Neglects interactions between electrons
Does not explain why electrons orbiting the nucleus do not fall in to the nucleus when they emit light (lose energy)
electrons do not move around the nucleus in circular orbits

10. Quantum Mechanical Model of the atom
Mathematically he treated the electron like a standing wave.
Schrodinger’s equation contains a wave function.
Solutions to the wave function equation are called orbitals.
An orbital is NOT a Bohr orbit!
An orbital is a region of probability around the nucleus where an e- of a given energy is at least 90% likely to be found.

11. Quantum Mechanical Model of the atom
Heisenberg’s Uncertainty Principle - we cannot know both the position and momentum of a particle at a given time.
Therefore, we can not know the exact motion of an electron as it moves around the nucleus. So, electrons are not moving in well defined orbits, as in the Bohr model.

12. Quantum Mechanical Model
The propeller blade has the same probability of being anywhere in the blurry region, but you cannot tell its location at any instant. The electron cloud of an atom can be compared to a spinning airplane propeller.
The quantum model determines the allowed energies an electron can have and how likely it is to find the electron in various locations around the nucleus.

13. Orbitals
Energy levels are numbered (for ground state electrons, so far)
Orbitals are sublevels of those energy levels. The s orbitals are lowest energy, followed by p, then d, and finally f in increasing energy.

14. Write the full configuration, noble-gas notation, and orbital notation for the following:
Calcium
Calcium ion
Potassium
Potassium ion
Platinum
Zinc

15. Isoelectronics
Atoms and ions with the same electron configuration, but different numbers of protons in the nucleus.

16. Types of Spectroscopy Mass Spectrometry (or spectroscopy) PES
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
uses high energy X-rays to knock electrons out of atoms
data supports electron configurations
Students have a table summarizing spectroscopy in the student notes.

17. More types of Spectroscopy
UV/Vis spectroscopy (Spectophotometer or colorimeter)
excites valence electrons
used to identify an element or molecule and its concentration in solution
Beer’s Law IR
vibrates bonds in molecules
types of bonds/atoms/functional groups within a molecule
Microwaves
rotates molecules and atoms within polar molecules
determine molecular structure

18. Mass Spectrometry Data
92.58 %
7.42 %
Two isotopes are represented because there are 2 peaks. The element is Lithium. Lithium’s average atomic mass is amu, a weighted average between the Li-6 and Li-7.
Once you have exposed the percentages wait for students to calculate the average atomic mass in their notes. After they have finished go to the next slide.
How many isotopes are represented by this spectrum?
Which element does this spectrum represent?
Calculate the average atomic mass.

20. PES (Photoelectron spectroscopy) Data
Pay close attention to the Energy scale along the x axis. Sometimes Energy increases left to right, and sometimes the other way around!
The Energy is Ionization Energy. The higher the Energy number the closer the electrons are to the nucleus. The peak height indicates how many electrons are in that orbital. This is Lithium. It has 2 electrons in the 1s orbital and one electron in the 2s orbital. Notice peak #1 is half the height of peak #2. Peak #1 is lower energy than Peak #2; therefore, it is farther from the nucleus. The peak #1 2s electron requires less energy to remove than do the 1s electrons in Peak #2. Li
Which element does this spectrum represent?

21. PES Data
1s2
2p6
2s2
3p6
4s1
3s2
Let students try to guess which element is represented. Then click through the slide to reveal the energy levels and sublevels. K
Which element does this spectrum represent?

22. Notice how the Ionization energy here (in megaJoules per mole of electrons) decreases from left to right. Have students label these peaks in their notes. Answer is on next slide.

24. Run an absorbance spectrum of the colored solution.
Choose the wavelength with the maximum absorbance value.
Read absorbance using that wavelength for all known concentration solutions. This will produce the most linear Abs vs. conc. standard curve.
UV/Vis Spectroscopy

25. A = abc UV/Vis Spectroscopy
4. Measure Abs of unknown and from that determine concentration using the linear standard curve.
A is absorbance, a is molar absorptivity constant, b is path length, c is concentration. Absorbance is proportional to concentration because a and b are usually constant values.
A = abc

26. Periodic Trends

27. Coulomb’ Law
F = k q1q2 r2
F is the force of attraction between 2 charged particles
q1 and q2 are the charges of the 2 charged particles
r is the distance between the particles. Distance has a greater effect on force than does charge because the term is squared.
k is a mathematical constant

28. Explaining Periodic Trends
Effective nuclear charge (Zeff)- the more positive the nucleus, the more attractive force there is emanating (coming out) from the nucleus, drawing electrons in or holding them in place. Relate this to ENERGY whenever possible. Stronger attractive forces require more energy to overcome them. *This refers to the q terms in Coulomb’s Law.
Zeff = Z - S
Z = nuclear charge (atomic number)
S = number of shielding electrons (inner core e- = total e- - valence e-)
F = k q1q2 r2

29. How to calculate Zeff The Zeff for any valence e- = # valence e-
Ex: Calculate Zeff for a phosphorus atom.
Zeff = protons - 10 core e- = +5
This means that each valence e- feels attracted by 5 protons.
The Zeff for e- in the first energy level (n=1) is equal to the atomic number.
Ex. Calculate the Zeff for an e- in the first shell of a P atom.
Zeff = 15 protons - 0 core e- = +15
This means that the 1s electrons feel attracted by all 15 protons.

30. Let’s Calculate Zeff Calculate Zeff for valence electrons in Chlorine.
Calculate Zeff for a 1s electron in Magnesium.
Zeff = 17 protons - 10 core electrons = +7
Zeff, the effective nuclear force, indicates how strongly different electrons are attracted to the nucleus. The core electrons shield the outer electrons from the nucleus, effectively cancelling some of the positive charge felt by the outer electrons.
Zeff = 12 protons - 0 core electrons = +12

31. Explaining Periodic Trends
II. Distance - attractive forces dissipate (become weaker) with
increased distance. Distant electrons are held loosely and thus easily removed. Relate this to ENERGY whenever possible.
*This refers to the r term in the denominator of Coulomb’s Law
F = k q1q2 r2

32. Explaining Periodic Trends
III. Shielding
Electrons in the “core” (all non-valence electrons) effectively shield the nucleus’ attractive force on the valence electrons.
Use this ONLY when going up and down a group, NOT across a period!
Relate this to ENERGY whenever possible.
Shielding reduces attractive force between valence e- and the nucleus, making valence e- easier to remove. Reduces IE.
This can also be explained by Zeff.

33. Explaining Periodic Trends
IV. Electron - electron Repulsion
Minimize electron/electron repulsions—this puts the atom at a lower energy state and makes it more stable.
Relate this to ENERGY whenever possible.
Like charges repel.
More electrons in the same energy level or sublevel repel each other because of the increased charge (q1 and q2). The force of repulsion is increased. This can expand the e- cloud.

34. Practice justifying these trends
What It Is
Across a Period
Down a Group
Atomic radius
Justification
First ionization energy
Have students define the periodic trend. Then above “justification” write whether the trend is to increase or decrease. In the justification box they need to use the preferably the first two ways to explain the periodic trend. Coulomb’s Law and its terms are preferred over shielding and e-e- repulsions.

35. Periodic Trend Justifications
What It Is
Across a Period
Down a Group
Atomic radius
Essentially the distance from the nucleus to valence electrons
Decreases
Increases
Justification
Increased effective nuclear charge.
More p+ and e-, Coulombs law (numerator)
Added electrons are all in same energy level so distance term is same
Energy levels added
In Coulombs law the distance is squared (denominator) making forces of nuclear attraction less between nucleus and electron cloud
First ionization energy
Energy required to remove an electron from an atom in the gas phase
Valence e- more attracted to nucleus requiring more energy to remove.
Valence e- become closer to nucleus also requiring more energy to remove.
Energy levels are added, so in Coulombs law the distance term is squared (denominator.
Reduces force of attraction between nucleus and electron cloud.
Valence e- require less energy to remove.
Go over answers with students and have them fill in the rest of the table in their notes. Discuss answers next day.

36. Multiple Choice practice questions

37. 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. ( )/2
C. (65 x 30.9) + (63 x 69.1)
D. (0.309 x 65) + (0.691 x 63)

38. 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.

39. 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?
Nitrogen atoms have a half-filled p subshell.
There are more electron-electron repulsions in oxygen atoms than in nitrogen atoms.
Electrons in the p subshell of oxygen atoms provide more shielding than electrons in the p subshell of nitrogen atoms.
Nitrogen atoms have a smaller nuclear charge than oxygen atoms.

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

41. 3. ) An element Z has the electron configuration [Xe]6s2
3.) An element Z has the electron configuration [Xe]6s2. The oxide of this element will have which formula below?
Z2O
Z2O3 ZO
ZO2

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

43. 4.) Of the following, the species that has the correctly predicted largest radius is
A. Ar
B. Br -
C. K+
D. Sr2+

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

45. involving the first ionization energies of O and N is best
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.

46. 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.
Paired electrons in O exhibit electron-electron repulsions. Those repulsive forces make the first valence electron on O easier to remove (requires less energy) than the first valence electron removed from N.

47. 6.) Which of the following elements would have the largest second ionization energy?
Hydrogen
Fluorine
Sulfur
Potassium

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

49. Ionization Energies for element X (kJ mol-1)
First
Second
Third
Fourth
Fifth
580
1815
2740
11600
14800
7.) The ionization energies for element X are listed above. On the basis of the data, element X is most likely to be Na Mg Al Si

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

51. Free Response practice questions

52. 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.
Write the equation for the ionization of atomic fluorine that requires 1,681.0 kJ mol-1.
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.)
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.
Give students about 8 minutes to answer these questions on paper. Then check answers from the next slide.

53. Short Free Response Rubric - (3 points possible)

54. 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?

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

56. Free Response #1 Rubric

57. Free Response #1 Rubric

58. Free Response #2 2.) Answer the following problems about gases.
(a) The average atomic mass of naturally occurring neon is 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 g sample of naturally occurring neon.
Give students no more than 5 minutes to perform these calculations. Then check answers with next slide.

59. Free Response #2 Rubric - (9 points possible)

60. Free Response #2 (cont.)
(b) A major line in the emission spectrum of neon corresponds to a frequency of 4.34 × 1014 s−1.
Calculate the wavelength, in nanometers, of light that corresponds to this line.
Give students no more than 5 minutes to perform these calculations. Then check answers with next slide.

61. Free Response #2 Rubric

62. Free Response #2
(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.
O3 (g) → O2 (g) + O (g)
A molecule of O3(g) absorbs a photon with a frequency of 1.00 × 1015 s−1.
(i) How much energy, in joules, does the O3 (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 × 1015 s−1 have enough energy to break this bond? Support your answer with a calculation.
Give students no more than 5 minutes to perform these calculations. Then check answers with next slide.

63. Free Response #2 Rubric