Download Quantum Theory of the Atom

A Study in Advanced Topics in Chemistry
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Welcome to 40S, or Grade 12 Chemistry.
In this version of the course, we will be studying and applying
advanced ideas in chemical behaviour.
The Units that we will be covering are:
 1. Atomic Structure – the study of the properties of atoms and
elements, bonding, and their patterns.
 2. Rates of Chemical Reaction – the study of how fast reactions
progress, what can control it, and their mathematical relationships.
 3. Thermochemistry – the study of the forms of energy that
determine properties of a chemical reaction.
 4. Chemical Equilibrium – the study of chemical reactions that can
occur in two directions. We will look at factors that aid, or inhibit
these reactions.
 5. Redox and Electrochemistry – the study of chemical reactions
where electrons play a role in the change in matter. This leads to the
development of chemically induced electrical current.
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We will also attempt to have as many laboratory experiments
as possible as a supplement to the theory that you will be
learning in class.
Our first step into 40S Chemistry will be studying Atomic
Structure.
Energy & the Electron
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In the study of the atom, there are some important concepts
that can help you develop your understanding of matter.
Ideas that we will investigate include:
 1. The Wave & Particle Nature of Light (energy)
 2. The Nature of Electrons in Atoms
 3. Bonding – Geometry and Energy
 4. Trends in the Elements on the Periodic Table
Fundamentals, Properties and Relationships in Light
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Light is a form of energy and the light we see is part of the
Electromagnetic (EM) Spectrum
Light has wave properties, including:
 Wavelength (λ) – distance between consecutive crests or troughs in
waves (measured in meters)
 Amplitude – height of the wave
 Frequency (f) – number of crests or troughs passing each second
(measured in s-1 or Hertz)
 Speed (c) – for light being 3.00 x 108 m/s
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The physical features of light are shown below:
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There is a relationship between these properties of a wave.
This is shown by the formula:
c = fλ
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We can use this relationship to solve for either one of the
wave characteristics.
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What is the wavelength of light with a frequency of 3.44 x 109
Hz?
c = fλ
Rearrange to get: λ =
𝑐
ν
3.00 ×108 𝑚/𝑠
λ=
3.44 × 109 𝑠 −1
= 8.72 x 10-2 m
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What is the frequency of light with a wavelength of 4.53 x 10-6
m?
c = fl
Rearrange to get: ν =
𝑐
λ
3.00 × 108 𝑚/𝑠
ν=
4.53 × 10−6 𝑚
= 6.62 x 1013 Hz
Light’s Alter Ego
Light has both wave and particle properties (a dual nature)
 Why? Well….
The wave model does not explain the observations of why
heated objects will only emit certain frequencies of light at a
given temperature.
 Max Planck (1856-1947) proposed that there needed to be a
minimum amount of energy that can be gained, or lost, by an
atom (this energy is called a quantum)
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Planck determined a relationship for energy and the
observations made:
Equantum = hν
Where h = 6.626 x 10-34 J·s
Theory states that matter can only absorb or emit energy in
whole number multiples of hν (1 hν, 2 hν, 3 hν, ...) i.e. No
partial multiples
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We can use the formula, E = hν to solve for:
 1. Energy of a particle, or the
 2. Frequency of the particle
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Remember that h is a constant!
Example:
 What is the amount of energy in a particle that has a frequency of
7.76 x 1014 Hz?
E = hf
E = (6.626 x 10-34 J·s)(7.76 x 1014 Hz)
= 5.14 x 10-19 J
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Example:
 What is the energy of a particle that has a wavelength of 566 nm?
This question is little bit of a different take on the problem, since it has
two issues to overcome:
1. It provides λ instead of ν, and
2. It states a wavelength in nm (nanometers)
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A nano- anything means that the measure is actually a very
small number.
As a metric prefix, it stands for 10-9.
So, if we have 600 nm, it basically means:
 600 x 10-9 m, or:
6.00 x 10-7 m
 Note: you can enter the value as 600 x 10-9 on your calculator and it
will work, but you are expected to show the correct Scientific
Notation value when writing it down.
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So now that we have a way to fix the number, what do we do
with it, since we actually need frequency?
Any ideas?
 We can change wavelength to frequency using c=λν.
c = λν
ν=
𝑐
λ
=
3.00
3.00 ×
× 10
1088 𝑚/𝑠
𝑚/𝑠
−7 𝑚
5.66
566 ×
×10
10−9
𝑚
= 5.30 x 1014 Hz
Remember that 566 x 10-9 = 5.66 x 10-7
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Now that we have the value for ν, we can solve for E:
𝐸 = ℎν
= (6.626 × 10−34 𝐽 ∙ 𝑠)(5.30 × 1014 𝐻𝑧)
= 5.31 × 10−19 𝐽
 So for radiation with a wavelength of 566 nm, the energy of a
particle is 5.31 x 10-19 J.
Understanding the Atom and the Electron
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A neat property of the elements is that each element has a
unique, what is called, “emission spectrum”.
In a nutshell, an emission spectrum is a pattern of light
radiation that is produced by an element after it has received
energy (for example, being heated).
This is an “Absorption Spectrum”
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With the understanding that light has behaviour of both a
particle and a wave, we can start to understand the emission
spectra of atoms.
One in particular, hydrogen (shown below)
The theory of Planck and Einstein states that there are only
certain allowable energy levels or states.
The lowest allowable state is called the ground state.
Recall that it was stated
that light’s properties
could not be explained
entirely by the Wave
Model.
 What was the evidence?
 A phenomenon was
known at the time where a
certain frequency of light
shined on a metal surface
will cause the
“photoelectric effect”

This effect results in the
release of electrons from
the metal’s surface.
 The apparent ability of
light to cause ejection, or
excitation of electrons
attracted the interest of
Albert Einstein.
 He proposed that light
existed as bundles of
energy called “photons”
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Elements have the ability to absorb
certain amounts of energy.
 When the atoms of an element
absorb enough energy, they become
“excited”. In this state it is actually
the electrons that become excited.
 When these electrons release this
energy to go back down to “ground”
state, they release it in the form of
radiation (light).
 Each element will display a particular
emission spectrum, so we can
actually identify elements by their
emission spectrum.

Neils Bohr developed a model for an atom. He also
developed a quantum model of hydrogen that helps
explain the visible spectrum of hydrogen.
 Although hydrogen has only one electron, it can have
many different excited states.
 There is a different energy level corresponding to
each possible orbit around the atom (the lowest
energy level for the orbit closest to the nucleus)
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Bohr defined each orbit around an atom as
having a Quantum state or number. The closest
one having a value of n=1.
Orbit
Quantum
Number
Orbit Radius Corresponding
(nm)
energy level
Relative
Energy
First
n=1
0.0529
1
E1
Second
n=2
0.212
2
E2 = 4E1
Third
n=3
0.476
3
E3 = 9E1
Fourth
n=4
0.846
4
E4 = 16E1
Fifth
n=5
1.32
5
E5 = 25E1
Sixth
n=6
1.90
6
E6 = 36E1
Seventh
n=7
2.59
7
E7 = 49E1
The emission spectrum that
we do see is only a part of
what is released by the atom
of hydrogen (the visible part
is called the Balmer Series).
 There are 2 other Series
corresponding to the
ultraviolet range (Lyman) and
infrared range (Paschen),
which we cannot see.
 we can see what levels of
electrons move from and to
using the diagram to the
right:
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A scientist in the mid-1920’s by the name of de Broglie
proposed an interesting idea.
Since waves can display particle-like behaviour, then particles
should show, wave-like behaviour.
His idea developed into the following relationship:
Where:
m = mass in kg
v = velocity in m/s
λ=
ℎ
𝑚𝑣
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Let’s see the de Broglie equation in an example.
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Example: What is the wavelength for a car of mass 910 kg and a velocity of 25
m/s?
λ=
=
ℎ
𝑚𝑣
6.626 × 10−34 𝐽∙𝑠
910 𝑘𝑔 25 𝑚/𝑠
= 2.9 x 10-38 m
What this answer tells us is the wavelength for objects such as this are
so small we would not be able to detect it, never mind perceive it.
1.
What is the frequency of green light with a wavelength of 540 nm?
2.
14 Hz
5.6
x
10
How much energy in joules is there in light with a frequency of 4.67 x 1017 Hz?
3.
What is the wavelength of a Panther with a mass of 85 kg and a velocity
3.09 x 10-16 J of 12
m/s?
6.5 x 10-37 m
Study of the Configuration of the Atom’s Electrons
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We are familiar with the electron and its properties and roles in
matter.
 1. Involved in formation of ions.
 2. Involved in bonding (covalent and ionic)
 3. Exists in a space that surrounds the nucleus of the atom in a simple path
called an orbital.
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We will spend the next few classes learning about theories to help
explain how bonding occurs between the different elements.
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Heisenberg’s Uncertainty Principle states:
 Electrons are constantly moving in the space that surrounds the
nucleus.
 We cannot know both the precise location of an electron around the
atom and its speed/velocity.
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Instead, we have a region we call an “orbital”.
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Orbital – a region that indicates the electron’s most probable
location around the nucleus.
 Each orbital can carry, at most, 2 electrons.
 There are 4 different types of orbitals:
a.
b.
c.
d.
s-orbital
p-orbital
d-orbital
f-orbital
Orbital Structure in an Element
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This is a special notation used to write the electron structure
for an element that utilizes the s, p, d, and f orbital names.
Each orbital type has a
unique shape to account for
the additional electrons.
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This is how it works: (it helps to have a Periodic
Table)
 We start with s orbitals, and these can go up to 2
electrons (1 orbital only)
 We can then add p orbitals and these can go up to 6
electrons (3 x p orbitals)
 Next are the d orbitals that can go up to 10 electrons (5
x d orbitals)
 Finally, the f orbitals can take up to 14 electrons (7 x d
orbitals)
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Knowing these, how does it help us write the
electron configuration for an element? How does
the pattern work?
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You can better observe
how to write an electron
configuration by seeing
where the different
orbitals are formed:
 s-orbital
 p-orbital
 d-orbital
 f-orbital
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Observe these examples:
 Starting with hydrogen, we know that it has only the 1 electron. So
its electron configuration is 1s1.
 For helium, which has 2 electrons, its configuration is 1s2.
 For lithium, which has 3 electrons, its configuration is 1s22s1 (why?)
 For boron, which has 5 electrons, its configuration is 1s22s22p1
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You can check your own work by adding up all of the
exponents that are written. They should equal the atomic
number of the element.
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You can easily see that writing the full electron
configuration will get really long when you look at
elements that are farther down the Periodic Table.
To help us with that issue, we have developed a
shorthand form.
In this form, we don’t write the full configuration from
the start, instead we start from the most recent Noble
Gas.
To account for the earlier ones, we just write the
symbol for that Noble Gas inside square brackets (e.g.
[Ne]).
So we can write the electron configurations for Na as:
 Full Configuration:
 Shorthand Configuration:
1𝑠 2 2𝑠 2 2𝑝6 3𝑠1
[𝑁𝑒]3𝑠1
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Write the full and shorthand electron configurations for the
following elements:
Full
 Carbon
 Chlorine
 Potassium
Shorthand
1𝑠 2 2𝑠 2 2𝑝2
1𝑠 2 2𝑠 2 2𝑝6 3𝑠 2 3𝑝5
1𝑠 2 2𝑠 2 2𝑝6 3𝑠 2 3𝑝6 4𝑠1
[𝐻𝑒]2𝑠 2 2𝑝2
[𝑁𝑒]3𝑠 2 3𝑝5
[𝐴𝑟]4𝑠1
Developing our Understanding about Bonding
Elements we know, have certain abilities to form chemical bonds.
The number of bonds is based on the number of “valence” electrons for
that element (the electrons in the outermost shell or orbital).
 We can illustrate this for each element by drawing what is called an
“electron dot diagram”.
 The principle behind this method is based on VSEPR Theory. This
stands for Valence Shell Electron Pair Repulsion, which for the
purposes of this diagram, keeps single electrons apart until we have to
pair them up.
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Here is how Dot Diagrams are drawn:
 1. Hydrogen
 2. Helium
 3. Lithium
H•
• He
•
• Li
 5. Boron
• Be
•
•
•
•
•
 4. Beryllium
We place electrons around the
symbol (in no particular order).
•B •
•
•
•
We add new electrons to a spot
where there is no electrons until
the element is surrounded.
We now add electrons to form
pairs.
We add electrons until
all electrons are paired.
In order to determine the number of valence electrons
that an element has, we need to use, or write, the
electron configuration for that element.
 The number of valence electrons is equal to the
number of electrons found in its highest orbital
(“principal quantum number”).
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 For example: Sn (tin) has 50 electrons
 its electron configuration is written as:
[Kr]5s24d105p2
So we write the dot diagram as:
So Sn has
4 valence
electrons
We havePrincipal
2 electrons
Highest
from eachNumber
orbital
Quantum
•
• Sn
•
•
 Example: Draw the electron dot diagrams for the following
elements:
▪ 1. Bismuth
▪ 2. Iodine
Introduction to Hybridization and Molecular Shapes
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Recall that ground state is the lowest possible energy state for
an element (and its electrons)
The arrangement of an element’s electrons is dictated by
three rules or principles:
 Aufbau Principle – each electron occupies the lowest energy orbital
possible.
 Pauli Exclusion Principle – two electrons may occupy the same
orbital as long as they have opposite “spins”.
 Hund’s Rule – states that electrons must fill empty orbitals before
pairing electrons of opposite spin.
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In the Orbital Diagram, we draw each orbital in an element
using a box:

A single box represents the s orbital
A triple box represents the p orbitals
A set of 5 boxes represent the d orbitals
A set of 7 boxes represent the f orbitals
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When we draw an orbital diagram, we fill in the boxes using
arrows to represent the electrons like we see below:
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Try drawing the diagram for nitrogen (atomic number 7):
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Covalent chemical bonding is based on the number of valence
electrons that are available to form that bond for the element.
We are used to elements having the ability to form bonds like
with carbon, where it can form up to 4 bonds (one for each
valence electron).
The standard rule for bond formation is to complete what is
called an “octet” (8 valence electrons).
Another requirement that has been made necessary is the
need for free electrons in order to form a chemical bond.
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How do orbitals affect bonding?
Consider the following compound: BeCl2
If you draw the box diagram for Be, you normally would get
this:
So you actually don’t have any free electrons to form bonds.
 So how is Be able to form bonds?
Its Influence on Molecular Shape
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How it works for many elements is what is
called “hybridization of orbitals”.
In this process, an element creates free
electrons by forming a hybrid orbital.
This occurs by combining orbitals of the
same quantum number.
For BeCl2, we see this:
We move one electron
from the pair to the
available space in
the next orbital type
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In the previous example, Be actually changes its bonding
orbital type to the combination of the orbitals combined: sp
The naming is based on the type of orbitals combined and
how many “boxes” are used in the formation of the hybrid.
 Other possible hybrid types: sp2, sp3, sp3d, sp3d2

What kind of orbitals would we need for AlCl3?
 Draw the Box Diagrams for both the non-hybrid Al and a hybrid Al.
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Each chemical molecule will have a particular
shape associated with it.
Hybridization of orbitals will cause the
formation of a variety of molecular shapes
that are very interesting:
Grab a text and open it to page 260.
 You will find a table showing a variety of chemical
molecules and their known shape.
 The type of hybridization does influence the
generated shape for that molecule.
Hybrid Type
Example(s)
sp
BeCl2
sp2
AlCl3
sp3
CH4, PH3, H2O
sp3d
NbBr5
sp3d2
SF6
For the central atom (the one where there is only one), the hybrid orbital is formed.
Studying the Properties of the Elements
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We know that the elements vary in their properties, however,
the elements will display similarities as well.
We know that elements within a Group will have the same
chemical properties (react the same).
Trends are also seen as we move through the elements both
down a Group (column) as well as across the Periodic Table.

We will look at the following properties:
 1. Atomic Radius
▪ This is the measure of the atom of an element from the center of its nucleus to
the edge of its electron cloud.
 2. Ionic Radius
▪ This is the measure of the ion-form of an element from its center to the edge
of the its electron cloud.
 3. Ionization Energy
▪ This is the measure of the amount of energy it takes to remove an electron
from the atom of that element.
 4. Electronegativity
▪ This is basically a relative measure of the strength of attraction for electrons in
that element.
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The Trends:
 1. Atomic Radius – we see is that the radius of an atom will:
▪ Increase in size as we move down a Group (atoms get bigger, more electron
levels/shells)
▪ Decrease in size as we move across the Period (due to a higher core charge,
pulls the electrons closer)
 2. Ionic Radius – the ionic radius will:
▪ Increase as we move down a Group (the number of
orbits increase just like in atoms).
▪ Decrease as we move across (same idea as before, core
charge increases), until will change the type of ion
(positive to negative), then the pattern resets.
▪ The key ideas here are that:
▪ 1. Negative ions are larger than their atom
Why?
▪ 2. Positive ions are smaller than their respective atom
 3. Ionization Energy pattern (cont’d)
▪ The highest ionization energies belong to the Noble Gases (since they will
NOT want to lose them).
▪ The lowest belong to the elements that typically form positive ions (easily lose
electrons).
 Ionization Energy will:
▪ Decrease as we move down a Group (electrons get further away from the
nucleus).
▪ Generally increase as we move across a Period (towards the Noble Gases).
 4. Electronegativity – the strength of an atoms’ attraction towards
electrons will generally:
▪ Increase as we move from left to right (Noble Gases have no affinity for
electrons).
▪ Decrease as we move from top to bottom (along a Group).

What is the role of electronegativity?
 These electronegativity values will dictate what kind of bond will
form between the atoms.

Remember that there are 2 main types of bonds:
 Covalent
 Ionic
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The way we use electronegativity is the difference between
the values:
 If the difference is 1.0 or less, it will be a covalent bond.
 If the difference is 2.0 or more, it will be an ionic bond.
 If the difference is between 1.0 and 2.0, it will be covalent with some
polar character
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Atomic Structure
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
1. The Electromagnetic (EM) Spectrum
2. Energy of a Particle (Planck)
3. de Broglie Equation
4. Ground State vs. Excited Stated
5. Spectra of the Elements
6. Quantum Number
7. Electron Configuration
8. Electron Dot Diagrams
 8. Electron Box Diagrams
 9. Hybridization of Orbitals
 10. Trends in the Periodic Table
▪ Atomic Radius
▪ Ionic Radius
▪ Ionization Energy
▪ Electronegativity