1. Atomic theory

2. BIG IDEA 1
The chemical elements are fundamental building materials of matter, and all matter can be understood in terms of arrangements of atoms. These atoms retain their identity in chemical reactions.

3. HISTORY REVISITED (9:42)
In 1909 Millikan was able to determine the magnitude of the electron charge, 9.11 x kg.
In a Millikan experiment, the charge on 4 oil drops were found to be:
3.33 Coulombs
8.88 Coulombs
6.66 Coulombs
11.10 Coulombs
What is the charge on the electron according to this experiment?
Calculate smallest common factor.
3.33 Coulombs/3.33 = 1
So the charge on the electron is 3.33 = 3 electrons
The charge is then 1.11 Coulombs
(Coulombs are the unit for electric charge.)
Millikan’s Oil Drop Experiment

4. Review What is a group/family? What is a period?
Why is the periodic table set up like it is?
Name the groups.
Quick Check page 333.

5. Electromagnetic Radiation
Calculate the energy of a wave with a wavelength of 450 nm and 700 nm.
Which of these waves would be closer to the visible red part of the spectrum?
Planck’s constant is x J· s
Constructive and
destructive interference

6. de Broglie’s equation
This equation relates wavelength to atomic mass. Remember that the dual nature of light is that it behaves as both a wave and a particle.
m = hc/λ or
λ = h/mv (for a particle not moving at c)
What is the wavelength of an electron (mass = 9.11 x 10-31kg) traveling at 5.31 x 106 m/s?
Remember that 1 J = 1 kg· m2/s2

7. What does isoelectronic mean?
It means that when an element gains or loses enough electrons to achieve the identical electron configuration of the nearest noble gas, that makes it isoelectronic with the noble gas.
Why do elements of the same chemical family have similar chemical properties?
Where are the most metallic elements on the table?
Where are the most non-metallic elements on the table?
Name the semi-metals.

8. Atomic size is determined by the distance between the nuclei.
In nonmetals (diatomics) the atomic size is defined as the covalent radius, represented by half the distance between two identical nuclei in the molecule.
In metals, the radius is measured as half the distance between identical nuclei.
Therefore both are determined by measuring half the bond length between identical atoms.
Why does atomic radii decrease across a period and increase down a group?

9. Learning Objectives
1.5: The student is able to explain the distribution of electrons in an atom or ion based upon data.
1.6: The student is able to analyze data relating to electron energies for patterns and relationships.
Electron configurations provide a method for describing the distribution of electrons in an atom or ion.
There are three quantum numbers that specify something different about the orbitals and electrons, n, l and m.

10. “n” is the principal quantum number
“n” is the principal quantum number. It indicates the relative size of the atomic orbital. 1s2 the principal quantum number is 1. “l” is the angular momentum quantum number and it represents the number of orbital shapes or sublevels that exist in that energy level. When n=1, there is only one orbital type/shape and that’s the s orbital. n=2 has two types, n=3 has three types and so on. l determines the subshell occupied by an electron. l = 0 is the s orbital, l = 1 is the p orbital, l = 2 is the d orbital and l = 3 is the f orbital. For principal quantum level n = 5, determine the number of allowed subshells (different values of l), and give the designation of each: For n = 5, the allowed values of l run from 0 to 4 (n-1 = 5-1). Thus the subshells and their designations are l = 0 l = 1 l=2 l = 3 l = 4 5s 5p 5d 5f 5g

11. “m” is the magnetic quantum number and it gives information on the orientation in space of a given orbital. ml has integral values between l and –l including zero. The value of ml is related to the orientation of the orbital in space relative to the other orbitals in the atom. ms is the spin of the electron designated by + ½ and - ½. No two electrons in the same orbital can have the same spin.
The p orbitals can have orientations of px, py and pz
The Pauli exclusion principle states that no two electrons in the same atom can be described by the same set of four quantum numbers.
YOU can’t do that!

12. Which of the following sets of quantum numbers is possible for a 3p electron?
n = 3, l = 0, ml = -1 ms = +1/2
n = 2, l = 1, ml = ms = -1/2
n = 1, l = 2, ml = ms = -1/2
n = 3, l = 1, ml = ms = -1/2

13. What is the electron configuration for iron?
1s22s22p63s23p64s23d6
What about copper?
1s22s22p63s23p64s13d10
In the case of copper and chromium, it takes less energy to completely fill the d sublevel and so an electron will be removed from the s sublevel.

14. Paramagnetism and O2 (5:00)
Paramagnetism refers to the magnetic state of an atom with one or more unpaired electrons. The unpaired electrons are attracted by a magnetic field due to the electrons' magnetic dipole moments. Hund's Rule states that electrons must occupy every orbital singly before any orbital is doubly occupied. This may leave the atom with many unpaired electrons. Because unpaired electrons can spin in either direction, they display magnetic moments in any direction. This capability allows paramagnetic atoms to be attracted to magnetic fields. Diatomic oxygen, O2 is a good example of paramagnetism (described via molecular orbital theory). The following video shows liquid oxygen attracted into a magnetic field created by a strong magnet. Atoms with the most unpaired orbitals will be the most paramagnetic.

15. Ionization energies

16. 1.9 The student is able to predict and/or justify trends in atomic properties based on location on the periodic table and/or shell model. Do’s and Don’ts of Explaining Periodic Trends
Use Zeff to justify trends across a period. Where Zeff is the effective nuclear charge and Z represents the atomic number, which is, effectively the number of protons.
Z-the number of screening electrons. The screening electrons are found in energy levels below the valence shell.
The reduction in nuclear charge felt by a valence electron is often referred to as the shielding effect.
Use increased distance (greater value of n) to justify trends down a group.

17. Explain why atomic radii decrease across a period:
The effective nuclear charge, Zeff increases the attraction of the nucleus and therefore pulls the electron cloud closer to the nucleus resulting in a smaller atomic radius. The increased number of energy levels (n) increases the distance over which the nucleus must pull and therefore reduces the attraction for electrons. Full energy levels (not sublevels) provide some shielding between the nucleus and valence electrons.

18. Explain why ionization energy increases as atomic number increases in any given period.
The effective nuclear charge, Zeff, increases the attraction of the nucleus and therefore holds the electrons more tightly.
A drop in IE occurs between groups II and III because the p electrons do not penetrate the nuclear region as greatly as s electrons do and are not as tightly held.
A drop in IE occurs between groups V and VI because the increased repulsion created by the first pairing of electrons outweighs the increase in Zeff and less energy is required to remove an electron.
Oxygen has a pair of electrons in the p orbital and nitrogen does not have paired electrons.

19. Explain why electronegativity increases as atomic numbers increase in any given period.
The effective nuclear charge Zeff increases the attraction of the nucleus and therefore increases it strengthens the attraction for the electrons.
The increased number of energy levels (n) increases the distance over which the nucleus must pull and therefore reduces the attraction for electrons.
Full energy levels provide some shielding between the nucleus and valence electrons.

20. Ionic radius Positive ions are smaller than their respective neutral atoms
Positive metal ions result from the loss of electrons. In many cases this means the farthest electrons are now in a smaller principal energy level (n) than the original neutral atom.
As electrons are lost the ratio of protons to electrons increases and thus the electrons are held closer and more strength.

21. Negative ions are larger than their respective neutral atoms
Negative nonmetal ions result from the addition of valence electrons. The primary explanation is the change in the proton to electron ration. As electrons are added, the proton to electron ratio decreases and the electrons are not as closely held.
Increased electron/electron repulsions also play a role in expanding the electron cloud.

22. Metals are more reactive as you move down a group.
Because metals react by losing electrons, a loosely held electron will result in a more reactive metal. This is directly tied to ionization energy. With an increased number of energy levels (n) comes increased distance from the nuclear attraction and thus a more loosely held electron available for reacting.

23. Nonmetals are more reactive as you move up a group
Because nonmetals tend to gain electrons, a strong nuclear attraction will result in a more reactive non-metal. This means that an atom with the highest Zeff and the least number of energy levels should be the most reactive nonmetal because its nucleus exerts the strongest pull.

24. How to recognize the difference between two species:
Locate both elements on the periodic table and state the principal energy level (n) and the sublevel containing the valence electrons for each element.
Do they have the same or different n values?
If they have the same n, argue with Zeff.
If they have different n values, argue with n versus n distances.

25. Each electron in an atom has a different ionization energy, which can be qualitatively explained through Coulomb’s law.
Based on Coulomb’s law, the force between two charged particles is proportional to the magnitude of each of the two charges (Q1, Q2), inversely proportional to the square of the distance, r, between them.
F = kQ1 Q2 ÷ r2
The relationship between the force on two charged objects, (like protons and electrons) the charges and the distance between them is Coulomb’s law or the inverse square law.
The constant, k is dependent on the medium the objects are immersed in. For air, it is approximately 9.0 x 109 N • m2 / C2
How does this relate to ionization energies and atomic radii?

26. The first ionization energy is the minimum amount of energy required to remove the least tightly held electron. The relative magnitude of the IE can be estimated with Coulomb’s law. The farther an electron is from the nucleus, the lower its ionization energy. When comparing two species with the same arrangement of electrons, the higher the nuclear charge, the higher the ionization energy of an electron in a given subshell. Compare ionization energies of nitrogen vs. oxygen. Nitrogen has an EV of 10 and oxygen an EV of 15—why?
An equation for the ionization of something like Li would be:
Li kJ/mol  Li+ + 1e- OR Li  Li+ + 1e-
What can you deduce about an element with a low IE1? What about one with a high IE1?
What is the difference between ionization energy, activation energy, free energy and lattice energy?

27. Lattice energy
Lattice energy is the energy change that occurs in the conversion of an ionic solid to widely separated gaseous ions.

28. What is the difference between Spectrometry and Spectroscopy?
• Spectroscopy is the science of studying the interaction between matter and radiated energy while spectrometry is the method used to acquire a quantitative measurement of the spectrum.
• Spectroscopy does not generate any results. It is the theoretical approach of science. Spectrometry is the practical application where the results are generated.

29. Mass spectrometer (8:00)
The mass spectrometer is an instrument which can measure the masses and relative concentrations of atoms and molecules. It makes use of the basic magnetic force on a moving charged particle. The particles with the least mass and highest charge will be deflected the most.
1.   A small sample is ionized, usually to cations by loss of an electron.   The Ion Source 2.   The ions are sorted and separated according to their mass and charge.   The Mass Analyzer 3.   The separated ions are then measured, and the results displayed on a chart.   The Detector

30. Spectroscopy and Astronomy
Spectroscopy is the study of the interaction between matter and radiated energy. This can be interpreted as the science of studying the interactions of matter and radiation. To understand spectroscopy, one must first understand spectrum. The visible light is a form of electromagnetic waves. There are other forms of EM waves such as X-Rays, Microwaves, Radio waves, Infrared and Ultraviolet rays. The energy of these waves is dependent on the wavelength or the frequency of the wave. High frequency waves have high amounts of energies, and low frequency waves have low amounts of energies. The light waves are made up of small packets of waves or energy known as photons. For a monochromatic ray, the energy of a photon is fixed. The electromagnetic spectrum is the plot of the intensity versus the frequency of the photons. When a beam of waves having the whole range of wavelengths is passed through some liquid or gas, the bonds or electrons in these materials absorb certain photons from the beam. It is due to the quantum mechanical effect that only photons with certain energies get absorbed. This can be understood using the energy level diagrams of atoms and molecules. Spectroscopy is studying the incident spectrums, emitted spectrums and absorbed spectrums of materials.

31. Spectrophotometry (13:05)
A spectrophotometer is employed to measure the amount of light that a sample absorbs. The instrument operates by passing a beam of light through a sample and measuring the intensity of light reaching a detector.
The beam of light consists of a stream of photons. When a photon encounters an analyte molecule (the analyte is the molecule being studied), there is a chance the analyte will absorb the photon. This absorption reduces the number of photons in the beam of light, thereby reducing the intensity of the light beam.
In most applications, one wishes to relate the amount of light absorbed to the concentration of the absorbing molecule.

32. What is the concentration of a solution that has an absorbance of 0.14?
If you had 10 mL of this solution and diluted it to 100 mL, what is the molarity of the original 10 mL of solution if the absorbance value is .05?
The original solution is 10 times more concentrated than the diluted solution. So, at .05 A, concentration is approximately .05 mg/L so multiply by 10 to get .50 mg/L

33. NMR Nuclear Magnetic Resonance
This shows three different types of H atoms with regard to the NMR. The H on the –OH group appears as a singlet, the CH3 – and CH2- result in a triplet and quartet respectively. The graph is signal intensity vs. chemical shift.
Radio waves are the longest waves with the least amount of energy.
When radio waves interact with atoms, there is not enough energy to impact the position of the nucleus of an atom or the bonds of a molecule. However, it can cause a change in nuclear spin (analogous to electron spin). By spinning the nucleus, a magnetic moment is created which enables the observation of a NMR. This spectrum can reveal the structure of a molecule by showing a bonding pattern. This pattern can be compared to already identified patterns to figure out what the substance is.

34. PES
1.7: The student is able to describe the electronic structure of the atom, using PES data, ionization energy data, and/or Coulomb’s law to construct explanations of how the energies of electrons within shells in atoms vary.
1.8: The student is able to explain the distribution of electrons using Coulomb’s law to analyze measured energies.
Photoelectron spectroscopy provides direct evidence for the shell mode. It can be used to determine the energy needed to eject electrons from the material. These energies provide a method to determine the shell structure of an atom. It measures the energies of core electrons in pure samples of an element.
Practice problems on page 366 of Edvantage Science.

35. Infrared Spectroscopy
IR Spectroscopy is used to identify parts of organic molecules by measuring the energy of vibration of the atoms in the molecule.

36. Visible and UV Spectroscopy
UV-visible spectroscopy is a technique that readily allows one to determine the concentrations of substances and therefore enables scientists to study the rates of reactions, and determine rate equations for reactions, from which a mechanism can be proposed. As such UV spectroscopy is used extensively in teaching, research and analytical laboratories for the quantitative analysis of all molecules that absorb ultraviolet and visible electromagnetic radiation
Colour in organic compounds is associated with unsaturation.
A carbon-carbon double bond in ethene, for example, absorbs energy as electrons in π bonds are promoted to higher energy levels. Bonds formed between carbon atoms in ethene are the result of overlap between atomic orbitals, producing bonding and antibonding molecular orbitals.

37. Explain why ionization energy decreases as atomic number increases down a column or group
The increase number of energy levels (n) increases the distance over which the nucleus must pull and therefore reduces the attraction for electrons.
Full energy levels (not full sublevels) provide some shielding between the nucleus and valence electrons.

38. In multi-electron atoms and ions, the electrons can be thought of as being in “shells” or “subshells” as indicated by the relatively close ionization energies associated with some groups of electrons. Inner electrons are called core electrons, and outer electrons are called valence electrons.
Core electrons “shield” the valence electrons from the full electrostatic attraction of the nucleus. Zeff

39. PRACTICE
The radius of a Mg atom is more than the radius of an ion of Mg. Explain why this is so.
1 pt. for indicating a loss of electrons and 1 pt. for indicating the outermost electrons are in different sublevels.
LiF has an lattice energy (U) of kJ/mol and NaCl has a lattice energy of -756 kJ/mol. Why the difference? Explain in terms of Coulombic attractions.
P.S. using a dot diagram would NOT work with this as these are ionic not molecular. 1 pt. for understanding what lattice energy is and 1 pt. for explaining the size or charge.

40. Ionization energies
Explain the difference between Li and Be with regard to:
Their first ionization energies.
Their second ionization energies.
Be has more protons and is smaller than Li, thus the electrons are held more tightly by the nuclear charge. 1pt.
The outermost electrons for Be are further away from the nucleus then lithium’s so it takes relatively less energy to pull the electron away. 1 pt.
Predict whether the first ionization energy of atomic He is greater than, less than or equal to the first ionization of Li or Be. Justify your prediction.
1st EI = kJ/mol 2nd EI = kJ/mol
First EI
kJ/mol
Second EI Li
520.3
7298.5 Be
899.5
1752.2 He ?

41. Place the following atoms, P, Kr, Mg, Li in order of increasing first ionization energies.
Li<Mg<P<Kr
What elements have the following electron configurations? [Noble gas] ns2 nd5?
Mn, Tc, Re
Place the following atoms, Cl, F, Na, C, in order of decreasing electron affinities.
Cl>F>C>Na Note that this is not an expected trend but the lower energy released when an electron is added to fluorine is attributed to the small size of the 2-p orbitals which causes strong electron-electron repulsions.
Which of the following elements has the lowest reducing ability- Li, Cs, Na or K?
Li—it does not want to gain an electron and has the lowest number of energy levels.

42. Relative Abundance
Based on the information given in the table, calculate the average atomic mass.
Did you know that Carbon 14 is only detectable in organisms that have died recently? That’s why if you were to study the remains of a wooly mammoth you would not find any carbon-14.
Isotope
Relative Abundance
C-12
98.90%
C-13
1.10%
C-14
Less than 0.1%
(trace amount)

43. Calculate the percent abundance of each isotope
The average atomic mass of boron is amu. There are two common isotopes of naturally occurring boron as indicated in the table.
Calculate the percent abundance of each isotope
Calculate the number of B-11 atoms in a gram sample of naturally occurring boron.
The weighted averages of B-10 and B-11 equals The sum of the fractional abundances must be 1. So if x =abundance of B-10 then the abundance of B-11 must be 1-x
x ( ) + (1-x)( ) = amu
A major line in the emission spectrum of boron corresponds to a frequency of 550 THz. Calculate the wavelength and energy of the wave.
Isotope
Mass (amu)
B-10
B-11

44. Which peak would represent the first electron that would be removed from a magnesium atom?
How much energy is required to remove this first electron?
Once the first electron is removed from an atom of magnesium what happens to the amount of energy required to remove the second? Justify your answer.
Using Coulomb’s Law, explain why atoms of As would have lower first ionization energies than P.