1

 Atomic structure explains chemical properties and patterns of chemical reactivity.
 Chemical reactions involve electrons. Knowing where the electrons are, how many, and what their energy levels helps explain many chemical phenomena.
 Spectroscopy is used to explore atomic structure. Because of this, we start with a discussion of the nature of light.
Slide 2

Particles or Waves??


Light must be made of particles because it…
 travels in a vacuum

 reflects off of objects exerts force (on the tails of comets)
Light must consist of waves because it…
 reflects like waves

 refracts and diffracts exhibits interference
By the end of the 19 th century, scientists had concluded that light is composed of WAVES!
Slide 3







Visible light is just one form of electromagnetic radiation
Light propagates in space as a wave
In vacuum, speed of light is constant and given the symbol “c”, c = 3.00 x 10 8 m/s
Light waves have amplitude, frequency, and wavelength.

= wavelength - distance between consecutive crests,

= frequency, # of crests that pass a given point in one second (SI Unit is s -1 or Hz) c =
 and

= c/

Slide 4

James Maxwell proposed in the mid1800’s that light is composed of perpendicular electric and magnetic waves
Slide 5

are distances – S.I. unit is the meter
Which of the above waves has the higher frequency?
Slide 6

Slide 7

Slide 8

wavelength/frequency conversion
 Calculate the frequency of visible light having a wavelength of 485 nm?

Remember to use S.I. units in your calculations!

= c /

= (3.00 x 10 8 m/s)
(4.85 x 10 -7 m)
= 6.19 x 10 14 s -1
 What colour of visible light is this?
Slide 9

wavelength & frequency conversion
 CO
2 absorbs light with a wavelength of 0.018 mm.
Which frequency is this?
c
 
1

3
0.018 10 m

1.7 10 s

1
Remember that 1 Hertz (Hz)

1 s -1
 What is the wavelength of WFAE at 90.7 MHz?
c

90.7 10 s

1

1

3.31 m
Slide 10





In 1900, Max Planck turned the world of physics on its head by resurrecting the particle theory of light.
Planck was trying to explain the “ultraviolet catastrophe” while studying blackbody radiation.
He proposed that light is composed of particles
(quanta) each carrying a fixed amount of energy.
The amount of energy per quantum is directly proportional to the frequency of the light:
E h or E
 hc

, h
  
34 
Slide 11

 The wavelength of maximum visual acuity in humans is 550 nm. (green light)

What is the energy of a single photon having this wavelength?
 Ans. 3.61 x 10 -19 J per photon

What is the energy of a mole of photons at 550 nm?
 Ans. 218 kJ/mol
Slide 12

a) b) c) d)
X-Rays
Red light
Green light
Radio waves




Rank the above in order of … increasing wavelength decreasing energy increasing frequency
Slide 13

“Line Spectra” vs “Continuous Spectra”





Historic work (1800’s) involving light emitted by pure elements in gas phase, subjected to very high voltages.
The light was passed through a prism or diffraction grating to produce a spectrum.
Each element emitted only certain wavelengths of light – a LINE SPECTRUM instead of the more familiar
“continuous” spectrum seen in the rainbow.
Balmer and Rydberg described the lines mathematically, for hydrogen.
Hydrogen’s emission spectrum has several lines in the visible region – called the “Balmer series”
Slide 14

contains all wavelengths of light – a rainbow
Slide 15

Slide 16




Neils Bohr developed a mathematical model that could explain the observation of atomic line spectra.
He proposed that electrons orbit the nucleus in certain
“allowed” orbits - or energy levels . That is, the electron’s energy is QUANTIZED (not continuous).
Working on a model of single-electron atoms, Bohr derived an equation to calculate the energy of an electron in the n th orbit of such an atom:
E n
 
2.18 10

18

Z
2


 n
2

 where Z = atomic number and n = electron energy level
Slide 17





The electron in hydrogen occupies discrete energy levels.
The atom does not radiate energy when the electron is in an energy level.
When an electron falls to a lower energy level, a quantum of radiation is released with energy equal to the difference between energy levels.
An electron can jump to higher energy levels if the atom absorbs a quantum of radiation with sufficient energy.
Slide 18

Slide 19

 Show that an electron transition from n = 4 to n = 2 results in the emission of visible light. Calculate the wavelength of the light emitted.
 Step 1: Calculate E
4

E
4
= -1.36 x 10 -19 J
 E
2
= - 5.45 x 10 -19 J
& E
2
 Step 2: Find
D
E
4

2

D
E = E
2
– E
4
= 4.09 x 10 -19 J
 Step 3: Calculate



= 4.86 x 10 -7 m = 486 nm
Slide 20



Although it works great for single-electron atoms, Bohr’s model fails for atoms with 2 or more electrons!
It was a huge leap forward, but was fundamentally flawed.


Ultimately, the failure of
Bohr’s model lay in the fact that he treated the electron as a charged particle orbiting the nucleus like a planet around the sun.
Electrons are more complicated …
Slide 21





Einstein’s famous equation, E = mc 2 , suggests that energy and mass are related – one can convert matter directly into energy.
Louis de Broglie made the leap that if light can behave as wave/particles, then so can matter!
He showed that the wavelength of a baseball is negligible
(as expected), but that the wavelength of an electron was on the order of magnitude equal to that of electromagnetic radiation!
This is what Bohr had missed – an atomic model must make use of the wave-nature of electrons to be complete!
Slide 22

Slide 23

The more precisely the position is determined, the less precisely the momentum is known in this instant, and vice versa.


It is impossible to determine BOTH the position and momentum (velocity) of an electron - the act of observing the electron changes it.
The electron exists in nature as a waveparticle (whatever that is ) until an experimenter observes it and imposes one of the two natures upon it!
Slide 24

- Niels Bohr to Werner Heisenberg
25

The Heisenburg uncertainty principle is the catch 22 of physics. Most modern physics and chemistry is largely based on the study of electrons. In order to view an electron we must use light to observe it, but in the process of observing we change the electron’s momentary existence because the light necessary to view it takes the form of a photon particle. The photon particle travels at approximately the same velocity as the electron particle, so that when it impacts the electron it alters its position as when two asteroids collide into each other at similar velocities but different angles.
Therefore, according to Heisenburg, it is impossible to know decisively an electron’s position and momentum.
Slide 26

Heisenburg proposed that this inability to perceive reality on an atomic level is not the result of technological ignorance, but rather a fact of the laws of nature. In other words, Heisenburg believed that no matter how technologically advanced we become we would never be able to fully perceive and record atomic reality. This means that an electron’s orbital path around the nucleus of an atom can only be statistically approximated.
This is why the quantum atomic model has a cloud of electrons, not a few electrons following exact elliptical orbits as in Bohr's atomic model. Each electron in the quantum atomic model is a calculated probability, so the cloud describes not the exact orbit of millions of electrons, but rather the probability of the position of only a few electrons at any given time.
Slide 27





Schr ödinger applied de Broglie’s concept of matter waves to the electron
He explained the quantum energies of the electron orbits in the Bohr model of the atom as vibration frequencies of electron "matter waves" around the atom's nucleus.
He provided a mathematical equation (wave function) to describe electron waves.
Physicists had difficulty explaining the
MEANING of the wave function itself, but it turns out that its square gives the probability of finding the electron in a particular region of space in the atom.
Slide 28




These regions of probability are called orbitals, and this concept replaces that of the electron “orbit”.
Instead of electrons orbiting like planets around the sun,
Schrodinger’s picture shows an “electron cloud” surrounding the nucleus and doesn’t state anything about the path or position of electrons within that cloud.
Orbital “pictures” shown in texts represent regions of space where the probability of finding an electron is 90% or greater.
Slide 29





Schrodinger's wave equation used three constants
(“quantum numbers”) that were used to describe orbitals
– regions of space where an electron has a probability of being found.
His equation used 3 quantum numbers to describe the size & energy, shape, and orientations of the orbitals in atoms.
The three quantum numbers: n, l and m l
.
This set of three numbers provides us with an “orbital address” for an electron within an atom – we can describe the orbital in which an electron is found if we know these three quantum numbers.
Slide 30

 n can have values > 1, integers only.
 For element atoms on the periodic table in their ground states , 1
 n

6
 As n increases, energy of electron increases and size of orbital increases
 Larger n means greater probability of greater distance from the nucleus.
 Electrons with same value of n are said to be in the same “electron shell”
Slide 31

l


The values for l
(lower case letter L, script) are limited by the principal energy level – they range from 0 to “n–1”
Example: If n = 3, l can be 0, 1 or 2
 l gives information about shape of orbital l
= 0 s orbital (found on all principle energy levels) l
= 1 p orbitals (found on level 2 and higher) l
= 2 l
= 3 d orbitals (found on level 3 and higher) f orbitals (found on level 4 and higher)
Slide 32

Graphs show
Probability vs Distance from nucleus s Orbitals on three energy levels
NODE
Slide 33

Slide 34

Slide 35

l
 The values of m l are limited by the value of l for a given orbital. They include positive and negative integers from l through + l
, inclusive


For a p orbital, l
= 1 , so m l can be:

1, 0, 1 m l gives information about orbitals in space orientation of
 The # of allowed values gives the number of orbitals of this type on a given energy level. (3, in the case of p orbitals)
Slide 36

s
A spinning negative charge creates a magnetic field.
The direction of spin d determines the direction of the field.




A fourth quantum number, m s describes electron spin
(either + ½ or
–½
)
Each electron in atom has a unique set of these four quantum numbers.
Electrons in orbitals with same n and l values are said to be in the same subshell.
Electrons with all three numbers the same, n, l
, and m l , are in the same orbital.
Slide 37

 Electrons have negative charge and repel each other. How are the electrons in an atom distributed?
 Wolfgang Pauli proposed that no two electrons in a given atom can be described by the same four quantum numbers!
 The first three quantum numbers determine an orbital – therefore spins must be opposite!
 Practical result is that each orbital can hold a maximum of two electrons, with opposite spins.
Slide 38

 Which of the following sets of quantum numbers are NOT allowed in the hydrogen atoms? For those that are incorrect, state what is wrong.
 What is the maximum number of electrons in an atom that can have these quantum numbers?
n
2
1
8
1 l
1
1
7
0 m l m s
-1 + ½
0 ½
-6 + ½
2 ½
 n = 4



 n = 5, m l
= +1 n = 5, m s
= + ½ n = 2, l
= 1 n = 3, l
= 2
 n = 3, m l
= 4
Slide 39




Electrons orbitals are defined by their quantum numbers, n, l and m l
.
Each electron in an atom has a unique set of 4 quantum numbers.
No two electrons can have the same “address”, i.e., the same 4 quantum #’s
 Rules define how multiple electrons will be distributed among the possible energy levels
Slide 40





In the ground state, the electrons occupy the lowest available energy levels. An atom is in an excited state if one or more electrons are in higher energy orbitals.
In atoms with more than one electron the lower energy orbitals get filled by electrons first!
This is the Aufbau principle, which is named after the
German word which means “to build up”.
When one describes the locations of the electrons in an atom, start with the lowest energy electron and work up to the highest energy electron.
Slide 41




A set of orbitals is said to be “degenerate” if the orbitals possess the same energies. For example, all three “2p” orbitals on energy level 2 are degenerate. All five “3d” orbitals on energy level 3 are also degenerate.
When filling a set of degenerate orbitals, Hund said that electrons should be left unpaired as long as possible so as to minimize electron-electron repulsions within the orbitals.
When each degenerate orbital has one electron, electrons will then pair, spins opposed, until that subshell is filled.
Slide 42

 Electrons distribute to lower energy levels until they are filled, before occupying higher levels. (Aufbau Principle)
 Electrons will spread out as much as possible within a subshell. (Hund’s Rule)
 Electrons will pair up, two to an orbital, spins opposed, until that sub-shell is filled. (Pauli
Exclusion)
 Every electron will have unique set of 4 quantum numbers.
Slide 43






An electron configuration is a way of describing the locations of electrons within an atom.
Usually written for the ground state of an atom: when its electrons occupy orbitals giving the lowest possible overall energy for the atom
Each subshell is designated by “n” and the type of orbital. Such as: 1s 3p 4d
The number of electrons in an orbital is shown as a superscript: 1s 2 3p 3 4d 9
How does one determine the order of filling of the orbitals in an atom?? Which orbitals get filled first??
Slide 44

Slide 45

Simply follow the order of elements in the periodic table!
Slide 46

Start at the top and follow the arrows:
1s,
2s,
2p, 3s,
3p, 4s,
3d, 4p, 5s…
Slide 47






Helium has 2 electrons in the 1s orbital, He: 1s 2
Carbon has 6 electrons, C: 1s 2 2s 2 2p 2
Calcium has 20 electrons, Ca: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2
Calcium cation, Ca 2+ : 1s 2 2s 2 2p 6 3s 2 3p 6
Sulfide anion, S 2
–
: 1s 2 2s 2 2p 6 3s 2 3p 6
 Notice that the superscripts add up to the total number of electrons on the atom or ion! Each of these examples represents the “ground state” of the atom/ion because the electrons are in the lowest possible energy levels.
Slide 48

Slide 49

 The core electrons can be indicated by a short form: use the last noble gas to indicate filled shells.
 Find the last noble gas before the element under consideration. Start the electron configuration with the symbol for this noble gas element and just tack on the extra electrons:
 Ca: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2
 Ca: [Ar] 4s 2
Ar: 1s 2 2s 2 2p 6 3s 2 3p 6
Slide 50




As atoms get larger, their outermost electrons become closer in energy
Large atoms have
“exceptions” to the
Aufbau Principle in the locations of their valence electrons. We will memorize two: Cr and Cu
Transition metal cations are also strange in their electron configurations


Cr: [Ar] 4s 1 3d 5
Cu: [Ar] 4s 1 3d 10

In both cases, one of the 4s electrons is promoted to the
“3d” subshell.


Fe 2+ : [Ar] 3d 6
Cu 2+ : [Ar] 3d 9

Notice that the 4s electrons are lost before the 3d electrons!
Slide 51




An atom is said to be paramagnetic if it possesses unpaired electrons.
These electrons will result in the atom possessing a net magnetic field
It will then be attracted into an external magnetic field


An atom is diamagnetic if all its electrons are paired – it will NOT possess a net magnetic field because each electron’s field will be cancelled
These atoms are NOT attracted into an external magnetic field

What is “ferromagnetism”?
Slide 52

See Kotz & Treichel page 290
Slide 53


“Periodicity” refers to similarities in behavior and reactivity due to similar outer shell electron configurations.
 All the Alkali Metals have one unpaired valence electron; all the Noble Gases have completely filled subshells
 We will examine periodic trends in atomic radius, ionization energy, electronegativity, and electron affinity
Slide 54





Valence electrons are those electrons in the highest principal energy level within an atom - the s and p electrons in the outermost shell of an atom in its ground state.
These are the electrons involved in forming bonds with other atoms during chemical reactions.
Most common oxidation states (ion charge) for the element can be derived from valence electrons
Completely filled, half filled and empty sub-shells have special stability (not sure why!?)
Slide 55




The “valence configuration” is that part of an electron configuration that describes the valence electrons.
For example, sodium’s valence configuration is just “3s 1 ”.
The valence configuration of bromine is “4s 2 4d 5 ”.
We leave out the “3d 10 ” electrons because they are not in the outermost principal energy level.
Slide 56

 Properties of elements and their ions are determined by electron configurations
 Individual atoms cannot be accurately measured.
 Atomic sizes are determined from bond lengths of compounds with various atoms
Br-Br
2.28 A r = 1.14 A
Br-C
1.91 A r carbon
= 1.91-1.14
= 0.77 A
Slide 57

 Atomic radii increase within a group (column) as the principal quantum number of outermost shell increases
 Electrons of outermost shells have greater probability of being farther from the nucleus
 Atomic size decreases across a row (period) from Left to Right, because the effective nuclear charge increases.

There are more and more protons in the nuclei attracting more and more electrons!
Slide 58

Atomic Radii:
Atoms get larger as you move down a group
Atoms get smaller as you go across a period
Slide 59

Slide 60

 First Ionization energy: energy required to remove an electron from the ground state atom in the gaseous state to make a cation
 1st ionization of Mg:
Mg

Mg + I.E. = 738 J/mol
[Ne]3s 2

[Ne]3s 1
 2nd ionization of Mg
Mg +

Mg 2+ 1450 J/mol
[Ne]3s 1

[Ne]
Slide 61

 First ionization energies decrease down a group as radii increase: Na > K > Cs
 First ionization energies increase across a row as radii decrease: Li < F < Ne
 For similar reasons that explained trends in atomic radii.
 For a given element, IE
2 always > IE
1 because of the same # protons in nucleus holding onto fewer electrons.
Slide 62

Slide 63

Trends in First
Ionization Energies
Slide 64




Lowest IE
1 are in the lower left corner of periodic table (Fr and Cs)
Irregularities in IE
1 arise from special trends electron configurations.
Half filled sub-shells are especially stable: every orbital has only one electron.


Oxygen has slightly lower
IE
1 than nitrogen because it takes more energy to disrupt N’s half-filled orbital configuration
Boron has slightly lower
IE
1 than Beryllium because its first electron is removed from the psubshell further from the nucleus than the first electron removed in Be
Slide 65

 Energy change when a gas phase atom gains an electron to become an anion.
Cl
(g)
+ e-

Cl
–
(g)
EA = – 349 kJ/mol



The more negative an EA, the greater the affinity for electrons!
Electron affinities “tend” to become more negative across a period and less negative moving down a group
EA least negative for alkaline earth and noble gas elements: they have full or half full s- or p-orbitals and it would take a lot of energy to add another electron
Slide 66

Slide 67

Slide 68

Slide 69

 Cations : radius is always smaller than the neutral atom.


Loss of electrons (to make cation) leaves fewer electrons attracted to same # of protons
OR….. loss of outermost shell altogether as in group 1A [Ne]3s 1

[Ne] for sodium
 Anions : radius is always larger than neutral atom.
 Same # protons pulling on more electrons; also electrons repel each other.
Slide 70

Slide 71



Isoelectronic atoms or ions have same electron configurations – the same number of electrons in same orbitals
Na + , Ne and F are isoelectronic: 1s 2 2s 2 2p 6
 For isoelectronic atoms and ions, the species with the most protons will be smallest
 It has greater nuclear attraction for the same number of electrons, so the electron shells will be pulled closer to nucleus.
Slide 72

Ionic Bonds
Polar Covalent Bonds
Non-Polar Covalent Bonds
73

 ELECTRONEGATIVITY is the attraction an atom has for electrons in a chemical bond.
 A POLAR BOND is one between atoms with different electronegativities - one atom attracts electrons more strongly than the other. As a result, one atom acquires a partial positive charge and the other a partial negative charge.
Slide 74

More Definitions
 A DIPOLE is a separation of charge. A polar bond is an example of a dipole.
 A DIPOLE MOMENT is the magnitude of a dipole - it depends on the size of the charges involved.
 A POLAR MOLECULE is one that possesses an overall dipole moment.
Slide 75


IONIC bonds are characterized as…
 bonds between metals and non-metals
 bonds between cations and anions
 bonds involving a transfer of electrons (from a metal to a non-metal)
 The best classification of an ionic bond uses the concept of electronegativity .
Slide 76

Classifying Chemical Bonds
 An ionic bond forms between two atoms with a large difference in electronegativity - usually stated as being greater than about 1.7 Paulings.
 e.g. NaCl: difference is 2.0 Paulings
 e.g. CuO: difference is 1.7 Paulings
Slide 77

Classifying Chemical Bonds

COVALENT bonds are usually described as…
 bonds between two non-metal atoms
 bonds involving a sharing of electrons
 Using the concept of electronegativity, we can see that there are two types of covalent bonds.
Slide 78

Classifying Chemical Bonds
 POLAR COVALENT BONDS result from unequal sharing of electrons by atoms that creates a small dipole.
 This type of bond occurs when there is a small but significant difference in electronegativity between 0.4 and 1.7 Paulings.
 e.g. N-H: diff is 0.9 Paulings
 e.g. Cu-S: diff is 0.7 Paulings
Slide 79

Classifying Chemical Bonds
 NON-POLAR COVALENT BONDS result from an equal sharing of electrons.
 There is no dipole created since the electrons are shared equally.
 This bond forms between atoms with similar (or the same) electronegativities - differences between 0 and 0.4 Paulings.
Slide 80

 Classify the following chemical bonds. Show your work.

Li-F

Br-Br

Fe-O

Al-Cl

C-S

C-H
Slide 81

 A diatomic molecule (e.g. HF) will be polar if its bond is polar.
 Larger molecules are polar only when the following criteria are met:
 they possess at least one polar bond
 their geometries do not result in the dipoles cancelling each other out! (e.g. CO
2
)
Slide 82