
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.










Millikan’s Oil Drop Experiment


In 1909 Millikan was able to
determine the magnitude of the
electron charge, 9.11 x 10-31 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.)





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.
Constructive and
destructive interference



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 6.626 x 10-34 J· s





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






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.





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?


a.
b.
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.
“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 = 51). Thus the subshells and their designations are
l=0
l=1
l=2
l=3
l=4
5s
5p
5d
5f
5g




“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!





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 = 0 ms = -1/2
n = 1, l = 2, ml = 1 ms = -1/2
n = 3, l = 1, ml = 0 ms = -1/2
a.
b.
c.
d.
e.
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.

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.




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.
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.




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.



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.


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.


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.

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.

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.




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.





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?





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 + 520.3 kJ/mol  Li+ + 1e- OR Li  Li+ + 1eWhat 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?

Lattice energy is the energy change that occurs
in the conversion of an ionic solid to widely
separated gaseous ions.


• 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.


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

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 XRays, 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.



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.



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



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.
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.




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.

IR Spectroscopy is used to identify parts of
organic molecules by measuring the energy of
vibration of the atoms in the molecule.



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.


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.
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




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 -1007 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.







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 = 2372.5 kJ/mol 2nd EI = 5250.7
kJ/mol
First EI
kJ/mol
Second
EI
kJ/mol
Li
520.3
7298.5
Be
899.5
1752.2
He
?








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.


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)






The average atomic mass of boron
is 10.81 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 32.56 gram sample of naturally
occurring boron.
The weighted averages of B-10 and
B-11 equals 10.81. 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 (10.012938) + (1-x)(11.09305) =
10.81 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
10.012938
B-11
11.009305




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.