Download Chapter 7 The Quantum Mechanical Model of the Atom

Chemistry: A Molecular Approach, 1st Ed.
Nivaldo Tro
Chapter 7
The Quantum-Mechanical
Model of the Atom
•Quantum Mechanics
•The Nature of Light
•Atomic Spectroscopy and the Bohr
Model
•The Wave Nature of Matter
•The de Broglie Wavelength
•Uncertainty Principle
•Quantum Mechanics and the Atom
•The Shapes of Atomic Orbitals
Anyone who is not shocked by quantum mechanics has not understood it
Neils Bohr
Tro 1/e Chap 7
Slide 2
The Behavior of the Very Small
• electrons are incredibly small
– a single speck of dust has more electrons than the number of people
who have ever lived on earth
• electron behavior determines the physical and chemical properties of atoms
• to understand this behavior we must understand how electrons exist within
atoms!
• directly observing electrons in the atom is impossible,
e.g. if you attempt to measure its position using light, the light itself
disturbs the electron changing its position
Tro 1/e Chap 7
Slide 3
A Theory that Explains Electron Behavior
• the quantum-mechanical model explains the manner
electrons exist and behave in atoms
• helps us understand and predict the properties of atoms
that are directly related to the behavior of the electrons
– why some elements are metals while others are nonmetals
– why some elements gain 1 electron when forming an anion,
while others gain 2
– why some elements are very reactive while others are
practically inert
– and other Periodic patterns we see in the properties of the
elements
Tro 1/e Chap 7
Slide 4
• Prior to the development of QM the nature of light was viewed as being
very different from that of subatomic particles such as electrons
• As QM developed, light was found to have many of the characteristics in
common with electrons
• Chief among these is the wave-particle duality of light – certain properties
are best described by thinking of it as a wave, while other properties are
best described by thinking of it as a particle
Tro 1/e Chap 7
Slide 5
The Nature of Light
its Wave Nature
• light is a form of electromagnetic radiation
– composed of perpendicular oscillating waves, one for the
electric field and one for the magnetic field
• an electric field is a region where an electrically charged particle
experiences a force
• a magnetic field is a region where an magnetized particle
experiences a force
• all electromagnetic waves move through space at the
same, constant speed
– 3.00 x 108 m/s in a vacuum = the speed of light, c
Tro 1/e Chap 7
Slide 6
Speed of Energy Transmission
Tro 1/e Chap 7
Slide 7
The Electromagnetic Spectrum
Insert fig 5.15
Tro 1/e Chap 7
Slide 8
Electromagnetic Radiation
• Electromagnetic waves consist of oscillating,
perpendicular electric and magnetic fields.
• The wavelength of radiation is the distance
between peaks in a wave. ()
• The frequency is the number of peaks that pass a
point in a second. ( )
Tro 1/e Chap 7
Slide 9
Electromagnetic Radiation
Tro 1/e Chap 7
Slide 10
Characterizing Waves
•
•
the amplitude is the height of the wave
– the distance from node to crest
• or node to trough
– the amplitude is a measure of how intense the light is – the larger the amplitude, the brighter the light
the wavelength, () is a measure of the distance covered by the wave
– the distance from one crest to the next
• or the distance from one trough to the next, or the distance between alternate nodes
Tro 1/e Chap 7
Slide 11
Characterizing Waves
•
the frequency, () is the number of waves that pass a point in a given period of time
– the number of waves = number of cycles
– units are hertz, (Hz) or cycles/s = s-1
• 1 Hz = 1 s-1
•
the total energy is proportional to the amplitude and frequency of the waves
– the larger the wave amplitude, the more force it has
– the more frequently the waves strike, the more total force there is
Tro 1/e Chap 7
Slide 12
The Relationship Between
Wavelength and Frequency
• for waves traveling at the same speed, the shorter
the wavelength, the more frequently they pass
• this means that the wavelength and frequency of
electromagnetic waves are inversely proportional
– since the speed of light is constant, if we know
wavelength we can find the frequency, and visa versa
c 
 s  
 m 
-1
m
s
Tro 1/e Chap 7
Slide 13
Frequency and Wavelength
• Given two of the following calculate the
third: , , c (velocity of propagation).
  c
  c/
  c/
 is wavelength measured in length units (m, cm, nm,
etc.)
 is frequency measured in Hz (s-1).
• c is the velocity
• velocity of light in vacuum = 3.0 x108 ms-1
Tro 1/e Chap 7
Slide 14
Example 7.1- Calculate the wavelength of red
light with a frequency of 4.62 x 1014 s-1
Given:  = 4.62 x 1014 s-1
Find: , (nm)
Concept Plan:  s-1)
 (m)
c


1 nm
 (nm)
109 m
Relationships: ∙ = c, 1 nm = 10-9 m
Solve:
c 3.00 108 m  s -1
7
 

6
.
49

10
m
14
1

4.62 10 s
1 nm
7
6.49 10 m  9  6.49 102 nm
10 m
Check: the unit is correct, the wavelength is appropriate
for red light
Tro 1/e Chap 7
Slide 15
Practice – Calculate the wavelength of a radio
signal with a frequency of 100.7 MHz
Tro 1/e Chap 7
Slide 16
Practice – Calculate the wavelength of a radio
signal with a frequency of 100.7 MHz
Given:  = 100.7 MHz
Find: , (m)
Concept Plan:  MHz)
6 -1
10 s
1 MHz
 (s-1)

c
 (m)

Relationships: ∙ = c, 1 MHz = 106 s-1
Solve:
106 s -1
100.7 MHz 
 1.007 108 s -1
1 MHz
c 3.00 108 m  s -1
 
 2.98 m
8
1

1.007 10 s
Check: the unit is correct, the wavelength is appropriate
for radiowaves
Tro 1/e Chap 7
Slide 17
Color
• the color of light is determined by its wavelength
– or frequency
• white light is a mixture of all the colors of visible light
– a spectrum
– RedOrangeYellowGreenBlueViolet
• when an object absorbs some of the wavelengths of
white light while reflecting others, it appears colored
– the observed color is predominantly the colors reflected
Tro 1/e Chap 7
Slide 18
Amplitude & Wavelength
Slide 19
Electromagnetic Spectrum
Tro 1/e Chap 7
Slide 20
The Electromagnetic Spectrum
• visible light comprises only a small fraction of
all the wavelengths of light – called the
electromagnetic spectrum
• short wavelength (high frequency) light has high
energy
– radiowave light has the lowest energy
– gamma ray light has the highest energy
• high energy electromagnetic radiation can
potentially damage biological molecules
– ionizing radiation
Tro 1/e Chap 7
Slide 21
Thermal Imaging using Infrared Light
Tro 1/e Chap 7
Slide 22
Electromagnetic Spectrum
• Recognize common units for , .
  wavelength
 frequency
• meters (m) radio
Hertz Hz s-1
• micrometers mm
(cycles per
• (10-6 m) microwaves
second)
• nanometers nm
megahertz
• (10-9 m) light
MHz (106 Hz)
• Å angstrom (10-10 m)
Tro 1/e Chap 7
Slide 23
Tro 1/e Chap 7
Slide 24
Interference
• the interaction between waves is called interference
• when waves interact so that they add to make a larger wave it is called
constructive interference
– waves are in-phase
• when waves interact so they cancel each other it is called destructive
interference
– waves are out-of-phase
Tro 1/e Chap 7
Slide 25
Diffraction
• when traveling waves encounter an
obstacle or opening in a barrier that is
about the same size as the
wavelength, they bend around it
– this is called diffraction
– traveling particles do not diffract
Tro 1/e Chap 7
Slide 26
Diffraction
•
•
the diffraction of light through two slits separated by a distance comparable to the wavelength results in an
interference pattern of the diffracted waves
an interference pattern is a characteristic of all light waves
Tro 1/e Chap 7
Slide 27
The Particle Nature of Light
• Prior to the 1900s light was thought to be purely a wave phenomenon
– Principally due to the discovery of diffraction
• This classical view began to be questioned….
Tro 1/e Chap 7
Slide 28
The Photoelectric Effect
• it was observed that many metals emit electrons when a light shines on their
surface
– this is called the Photoelectric Effect
• classic wave theory attributed this effect to the light energy being transferred
to the electron
• according to this theory, if the wavelength of light is made shorter, or the
light waves intensity made brighter, more electrons should be ejected
– remember: the energy of a wave is directly proportional to its amplitude
and its frequency
– if a dim light was used there would be a lag time before electrons were
emitted
• to give the electrons time to absorb enough energy
Tro 1/e Chap 7
Slide 29
The Photoelectric Effect
Tro 1/e Chap 7
Slide 30
The Photoelectric Effect
The Problem
• in experiments with the photoelectric effect, it was observed
that there was a maximum wavelength for electrons to be
emitted
– called the threshold frequency
– regardless of the intensity
• it was also observed that high frequency light with a dim
source caused electron emission without any lag time
Tro 1/e Chap 7
Slide 31
Photoelectric Effect
Tro 1/e Chap 7
Slide 32
Einstein’s Explanation
• Einstein proposed that the light energy was delivered to the
atoms in packets, called quanta or photons
• the energy of a photon of light was directly proportional to its
frequency
– inversely proportional to it wavelength
– the proportionality constant is called Planck’s Constant, (h)
and has the value 6.626 x 10-34 J∙s
E  h 
hc

Tro 1/e Chap 7
Slide 33
Define quantum, photon.
• Quantum- A packet of energy equal to h. The
smallest quantity of energy that can be emitted or
absorbed.
• Photon- A quantum of electromagnetic radiation.
• Thus light can be described as a particle (photon)
or as a wave with wavelength and frequency. This
is called wave-particle duality.
Tro 1/e Chap 7
Slide 34
Given one of the following,
calculate the other: E, .
• Ephoton = h
• where:
Ephoton is the energy of each photon (packet) of light.
h is Planck’s constant = 6.626x10-34 Js
 is the frequency of light in Hz (s-1).
Tro 1/e Chap 7
Slide 35
Example 7.2- Calculate the number of photons in a laser
pulse with wavelength 337 nm and total energy 3.83 mJ
Given:  = 337 nm, Epulse = 3.83 mJ
Find: number of photons
Concept Plan: nm)
E
 (m)
9
hc photon
10 m
1 nm
E photon 
E pulse

number
photons
E photon
Relationships: E=hc/, 1 nm = 10-9 m, 1 mJ = 10-3 J, Epulse/Ephoton = # photons
Solve:
9
10 m
 3.37 10 7 m
1 nm
6.626 1034 J  s 3.00 108 m  s -1
3.37 10 2 nm 
E photon 
hc



3

3.37 10
10 J
3.83 mJ 
 3.83 10 3 J
1 mJ
7
m
  5.8985 10
19
J
number of photons 
3.83 10 3 J
5.8985 10 3 J
 6.49 1015 photons
Tro 1/e Chap 7
Slide 36
Practice – What is the frequency of radiation
required to supply 1.0 x 102 J of energy from
8.5 x 1027 photons?
Tro 1/e Chap 7
Slide 37
What is the frequency of radiation required to supply
1.0 x 102 J of energy from 8.5 x 1027 photons?
Given: Etotal = 1.0 x 102 J, number of photons = 8.5 x 1027
Find: 
Concept Plan: number
E
 (s-1)
photon
photons E total
E photon

number of photons
h
Relationships: E=h, Etotal = Ephoton∙# photons
Solve:
2
E photon 

E photon
h

1.0 10 J
8.5 1027
 1.176 1026 J
1.176 10   1.8 10
26
6.626 10
Tro 1/e Chap 7
34
J
7 -1
s
Js
Slide 38
Ejected Electrons
• 1 photon at the threshold frequency has just
enough energy for an electron to escape the atom
– binding energy, f
• for higher frequencies, the electron absorbs more
energy than is necessary to escape
• this excess energy becomes kinetic energy of the
ejected electron
Kinetic Energy = Ephoton – Ebinding
KE = h - f
Tro 1/e Chap 7
Slide 39
• Although quantization of light explained the photoelectric
effect, the wave explanation of light continued to have
explanatory power
• What slowly emerged is what we now call the
wave-particle duality of light
Tro 1/e Chap 7
Slide 40
Exciting
e
Many ways:
• Heat, light, e- bombardment, chemical reactions
Tro 1/e Chap 7
Slide 41
Atomic Spectra
• when atoms or molecules absorb energy, that energy is
often released as light energy
– fireworks, neon lights, etc.
• when that light is passed through a prism, a pattern is
seen that is unique to that type of atom or molecule –
the pattern is called an emission spectrum
– non-continuous
– can be used to identify the material
• flame tests
Tro 1/e Chap 7
Hg
He
H
Slide 42
Emission Spectra
Tro 1/e Chap 7
Slide 43
Examples of Spectra
Oxygen spectrum
Neon spectrum
Tro 1/e Chap 7
Slide 44
Atomic Spectra
• Rydberg analyzed the spectrum of hydrogen and found
that it could be described with an equation that
involved an inverse square of integers
• Predicted wavelengths of emission spectrum for
hydrogen
1 
 1
 1.097 107 m -1  2  2 

m n 
1
Tro 1/e Chap 7
Slide 45
A Bar Code for Atoms
Flame Tests
Na
K
Li
Tro 1/e Chap 7
Ba
Slide 46
Emission vs. Absorption Spectra
Spectra of Mercury
Tro 1/e Chap 7
Slide 47
Bohr’s Model
• Bohr attempted to develop a
model for the atom that
explained atomic spectra…
Tro 1/e Chap 7
Slide 48
Bohr’s Model
• Neils Bohr proposed that the electrons could only have
very specific amounts of energy
– fixed amounts = quantized
• the electrons traveled in orbits that were a fixed
distance from the nucleus
– stationary states
– therefore the energy of the electron was proportional the
distance the orbital was from the nucleus
• electrons emitted radiation when they “jumped” from
an orbit with higher energy down to an orbit with lower
energy
– the distance between the orbits determined the energy of the
photon of light produced
Tro 1/e Chap 7
Slide 49
Bohr Model of H Atoms
e- is never observed
between states, only in
one state or the next
Tro 1/e Chap 7
Slide 50
Ephoton = DE = Ehigh-Elow
DE = h = hc/
light = DE/h
 = hc/DE
• Since Ehigh and Elow are discreet numbers, DE must be a discreet
number
• Therefore,
 and  must be discreet numbers, giving rise to single
frequencies and wavelengths of light. Hence, the line
spectra.
Tro 1/e Chap 7
Slide 51
Question
For the yellow line in the sodium spectrum,  =
589.0 nm. Calculate the energy in joules of a
photon emitted by an excited sodium atom.
 = 5.090  1014 /s
DE = 3.373  10-19 J
Tro 1/e Chap 7
Slide 52
Tro 1/e Chap 7
Slide 53
• Bohr’s model could not be applied to atoms with
2 or more eNeeded a new approach…
Tro 1/e Chap 7
Slide 54
Wave Behavior of Electrons
• de Broglie proposed that particles could have wave-like
character
• because it is so small, the wave character of electrons is
significant
• electron beams shot at slits show an interference
pattern
– A single e- produces an interference pattern
– the electron interferes with its own wave (not other e-)
Tro 1/e Chap 7
Slide 55
Electron Diffraction
however, electrons actually
if electrons behave like
present an interference
particles, there should
pattern, demonstrating the
only be two bright spots
behave like waves
on the target
Tro 1/e Chap 7
Slide 56
• Explains the existence of stationary states / energy levels in the
Bohr model
http://www.colorado.edu/physics/2000/quantumzone/debroglie
.html
• Prevents e- from crashing into the nucleus as predicted by
classical physics
Tro 1/e Chap 7
Slide 57
Wave Behavior of Electrons
• de Broglie predicted that the wavelength of a particle
was inversely proportional to its momentum
 m 
h

kg m 2
s
s2

mass (kg)  velocity (m  s -1 )
Tro 1/e Chap 7
Slide 58
Example 7.3- Calculate the wavelength of an electron
traveling at 2.65 x 106 m/s
Given: v = 2.65 x 106 m/s, m = 9.11 x 10-31 kg (back leaf)
Find: , m
Concept Plan:
m, v
 (m)

h
mv
Relationships: =h/mv
Solve:


34 kg  m
s
 6.626 10
2
h
s


 
mv

-31
6 m
9.1110 kg  2.65 10 s 


 2.74 10 10 m
2

Tro 1/e Chap 7

Slide 59
Practice - Determine the wavelength of a neutron
traveling at 1.00 x 102 m/s
(Massneutron = 1.675 x 10-24 g)
Tro 1/e Chap 7
Slide 60
Practice - Determine the wavelength of a neutron
traveling at 1.00 x 102 m/s
Given: v = 1.00 x 102 m/s, m = 1.675 x 10-24 g
Find: , m
Concept Plan:
m(g) 1 kg m (kg), v
h  (m)

103 g
mv
Relationships: =h/mv, 1 kg = 103 g
Solve:
1.675 1024 g 
1 kg
103 g
 1.675 1027 kg


34 kg  m
s
 6.626 10
2
h
s




mv

- 27
2 m
1.675 10 kg 1.00 10 s 


 3.96 10 9 m
2

Tro 1/e Chap 7

Slide 61
Practice - Determine the wavelength of a
bowling ball traveling at 1 m/s
(Mass = 1 kg)
1 x 10-24 nm
1 septillionth of a nanometer. This is so ridiculously small compared to the
size of the bowling ball itself that you'd never notice any wavelike stuff
going on; that's why we can generally ignore the effects of quantum
mechanics when we're talking about everyday objects
It's only at the molecular or atomic level that the waves begin to be large
enough (compared to the size of an atom) to have a noticeable effect.
Tro 1/e Chap 7
Slide 62
The Uncertainty Principle
• How can a single entity behave as both a particle and a wave?
• How does a single e- aimed at a double-slit produce an interference pattern?
• A possible hypothesis is the e- splits into 2, travels through both slots and
interferes with itself
Tro 1/e Chap 7
Slide 63
Testing the hypothesis…
any experiment
designed to observe the
electron results in
detection of a single
electron particle and no
interference pattern
Experiment designed to ‘watch’ which slit the e- travels through using a
laser beam, flash is seen from one Tro
slit1/eorChap
the7other…never both at once
Slide 64
• No matter how hard we try we can never see the interference pattern and
simultaneously determine which hole the e- goes through
• We have encountered the absolutely small and have no way of determining
what it is without disturbing it
• Wave nature and particle nature of the e- are complementary properties
• The more you know about one, the less you know about the other
Tro 1/e Chap 7
Slide 65
Uncertainty Principle
h
Dx  Dv 
4
1
 
m
• Heisenberg stated that the product of the uncertainties in both the position and
speed of a particle was inversely proportional to its mass
– x = position, Dx = uncertainty in position (related to particle nature)
– v = velocity, Dv = uncertainty in velocity (related to wave nature)
– m = mass
• the means that the more accurately you know the position of a small particle,
like an electron, the less you know about its speed
– and visa-versa
• Electron is observed EITHER as a particle OR a wave, never both at once
Tro 1/e Chap 7
Slide 66
Determinacy vs. Indeterminacy
• according to classical physics, particles move in a path determined by the
particle’s velocity, position, and forces acting on it
– determinacy = definite, predictable future
• because we cannot know both the position and velocity of an electron, we
cannot predict the path it will follow
– indeterminacy = indefinite future, can only predict probability
• the best we can do is to describe the probability an electron will be found in a
particular region using statistical functions
Tro 1/e Chap 7
Slide 67
Trajectory vs. Probability
Slide 68
Electron Energy
Ηψ  Eψ
• Position and velocity are complimentary
• electron energy and position are complimentary
– because kinetic energy is related to velocity, KE = ½mv2
• for an electron with a given energy, the best we can do is describe a region
in the atom of high probability of finding it – called an orbital
– a probability distribution map of a region where the electron is likely to
be found
– A plot of distance vs. y2 represents an orbital
• many of the properties of atoms are related to the energies of the electrons
Tro 1/e Chap 7
Slide 69
Wave Function, y
• calculations show that the size, shape and orientation in space of an orbital
are determined be three integer terms in the wave function
– added to quantize the energy of the electron
• these integers are called quantum numbers
– principal quantum number, n
– angular momentum quantum number, l
– magnetic quantum number, ml
Tro 1/e Chap 7
Slide 70
Principal Quantum Number, n
• characterizes the energy of the electron in a particular orbital
– corresponds to Bohr’s energy level
• n can be any integer  1
• the larger the value of n, the greater/less negative the energy
• energies are defined as being negative
– an electron would have E = 0 when it just escapes the atom
• the larger the value of n, the larger the orbital
• as n gets larger, the amount of energy between orbitals gets smaller
E n  -2.1810
-18
 1 
J 2  for an electron in H
n 
Tro 1/e Chap 7
Slide 71
Principal Energy Levels in Hydrogen
Tro 1/e Chap 7
Slide 72
Electron Transitions
• in order to transition to a higher energy state, the electron must
gain the correct amount of energy corresponding to the
difference in energy between the final and initial states
• electrons in high energy states are unstable and tend to lose
energy and transition to lower energy states
– energy released as a photon of light
• each line in the emission spectrum corresponds to the difference
in energy between two energy states
Slide 73
Predicting the Spectrum of Hydrogen
• the wavelengths of lines in the emission spectrum of
hydrogen can be predicted by calculating the
difference in energy between any two states
• for an electron in energy state n, there are (n – 1)
energy states it can transition to, therefore (n – 1) lines
it can generate
• both the Bohr and Quantum Mechanical Models can
predict these lines very accurately

E photonreleased  DE hydrogenelectron   E final  E initial


 1  
 1  
18 

18
    2.18 10 J
 
h 
   2.18 10 J
2
2
n
 
 n final  


initial

 



hc
Tro 1/e Chap 7
Slide 74
Hydrogen Energy Transitions
Slide 75
Example 7.7- Calculate the wavelength of light emitted when
the hydrogen electron transitions from n = 6 to n = 5
Given: ni = 6, nf = 5
Find: , m
Concept Plan: n , n
DEatom
i
f
 1 
E  R H  2 
n 
Ephoton
DEatom = -Ephoton
hc

E

Relationships: E=hc/, En = -2.18 x 10-18 J (1/n2)
1 
Solve: DE
18  1
 20


2
.
18

10
J



2
.
6
6
44

10
J
 2
atom
2 
5 6 
Ephoton = -(-2.6644 x 10-20 J)
= 2.6644 x 10-20 J


3.00 10 
hc 6.626 10
 
 7.46 10
E
2.6644 10 
34
8 m
s
Js
-20
6
m
J
Check: the unit is correct, the wavelength is in the infrared, which
is appropriate because less energy than 4→2 (in the visible)
Practice – Calculate the wavelength of light emitted when
the hydrogen electron transitions from n = 2 to n = 1
Tro 1/e Chap 7
Slide 77
Calculate the wavelength of light emitted when the hydrogen
electron transitions from n = 2 to n = 1
Given: ni = 2, nf = 1
Find: , m
Concept Plan: n , n
DEatom
i
f
 1 
E  R H  2 
n 
Ephoton
DEatom = -Ephoton
hc

E

Relationships: E=hc/, En = -2.18 x 10-18 J (1/n2)
1 
Solve:
18  1
DE atom  2.18 10 J 2  2   1.64 10 18 J
1 2 
Ephoton = -(-1.64 x 10-18 J) =
1.64 x 10-18 J


3.00 10 
hc 6.626 10
 
 1.2110
E
1.64 10 
34
8 m
s
Js
-18
7
m
J
Check: the unit is correct, the wavelength is in the UV, which is
appropriate because more energy than 3→2 (in the visible)
Shapes of Orbitals
• Shapes are important because covalent bonds depend on sharing e• Chemical bonds consists on overlapping orbitals
Tro 1/e Chap 7
Slide 79
The Shapes of Atomic Orbitals
• the l quantum number primarily determines the
shape of the orbital
• l can have integer values from 0 to (n – 1)
• each value of l is called by a particular letter that
designates the shape of the orbital
–
–
–
–
s orbitals are spherical (l = 0)
p orbitals are like two balloons tied at the knots
d orbitals are mainly like 4 balloons tied at the knot
f orbitals are mainly like 8 balloons tied at the knot
Tro 1/e Chap 7
Slide 80
l = 0, the s orbital
• each principal energy state
has 1 s orbital
• lowest energy orbital in a
principal energy state
• spherical
• number of nodes = (n – 1)
Tro 1/e Chap 7
Slide 81
Probability & Radial Distribution
Functions
•
y2 is the probability density (= prob. / unit vol.)
– the probability of finding an electron at a particular point in space
– for s orbital maximum at the nucleus?
– decreases as you move away from the nucleus
• the Radial Distribution function represents the total probability at a certain
distance from the nucleus
– maximum at most probable radius
• nodes in the functions are where the probability drops to 0
Slide 82
Probability Density Function
Probability
density at a point
Tro 1/e Chap 7
Slide 83
Radial Distribution Function
Radial
distribution
represents total
probability at
radius r
Tro 1/e Chap 7
Slide 84
2s and 3s
2s
n = 2,
l=0
3s
n = 3,
l=0
Slide 85
l = 1, p orbitals
• each principal energy state above n = 1 has 3 p
orbitals
– ml = -1, 0, +1
• each of the 3 orbitals point along a different axis
– px, py, pz
• 2nd lowest energy orbitals in a principal energy state
• two-lobed
• node at the nucleus, total of n nodes
Tro 1/e Chap 7
Slide 86
p orbitals
Tro 1/e Chap 7
Slide 87
l = 2, d orbitals
• each principal energy state above n = 2 has 5 d
orbitals
– ml = -2, -1, 0, +1, +2
• 4 of the 5 orbitals are aligned in a different plane
– the fifth is aligned with the z axis, dz squared
– dxy, dyz, dxz, dx squared – y squared
• 3rd lowest energy orbitals in a principal energy state
• mainly 4-lobed
– one is two-lobed with a toroid
• planar nodes
– higher principal levels also have spherical nodes
Tro 1/e Chap 7
Slide 88
d orbitals
Tro 1/e Chap 7
Slide 89
l = 3, f orbitals
• each principal energy state above n = 3 has 7 d
orbitals
– ml = -3, -2, -1, 0, +1, +2, +3
• 4th lowest energy orbitals in a principal energy state
• mainly 8-lobed
– some 2-lobed with a toroid
• planar nodes
– higher principal levels also have spherical nodes
Tro 1/e Chap 7
Slide 90
f orbitals
Tro 1/e Chap 7
Slide 91
Why are Atoms Spherical?
Tro 1/e Chap 7
Slide 92
Energy Shells and Subshells
Tro 1/e Chap 7
Slide 93