chemistry the molecular nature of matter and change 3 rd edition chapter 7 lecture notes: quantum...
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CHEMISTRYThe Molecular Nature of Matter and Change 3rd Edition
Chapter 7 Lecture Notes:
Quantum Theory and Atomic Structure
Chem 150 - Ken Marr - Winter 2006
Welcome to Chem 150!! Below are a few due dates and other useful
information
1. Do the Prelab Preparation for tomorrow's lab activity, Atomic Spectrum of Hydrogen. Turn in the prelab questions at the start of lab and complete in your lab notebook the following sections of the report for this lab exercise: Title, Introduction, Materials/Methods and Data Tables.
2. The completed report for lab 1 is due on Monday January 9, 2005.
3. Due Friday January 6, 2006: ALE 1
Quantum Theory and Atomic Structure
7.1 The Nature of Light
7.2 Atomic Spectra
7.3 The Wave-Particle Duality of Matter and Energy
7.4 The Quantum-Mechanical Model of the Atom
Section 7.1 The Nature of Light (Electromagnetic
Radiation)
• Light consists of waves with electrical and magnetic components
• Waves have a specific Frequency and Wavelength
» Symbol and Units of Each?
c = = 3.00 X 108 m/s
C = 2.99792 X 108 m/s
Figure 7.1
Frequency and
Wavelength
c =
Figure 7.2
Amplitude (Intensity) of a Wave
Figure 7.3 Regions of the Electromagnetic Spectrum
Increasing Frequency, S-1
Increasing Wavelength
7-7
Sample Problem 7.1
SOLUTION:PLAN:
Interconverting Wavelength and Frequency
PROBLEM: A dental hygienist uses x-rays (= 1.00A) to take a series of dental radiographs while the patient listens to a radio station ( = 325cm) and looks out the window at the blue sky (= 473nm). What is the frequency (in s-1) of the electromagnetic radiation from each source? (Assume that the radiation travels at the speed of light, 3.00x108m/s.)
wavelength in units given
wavelength in m
frequency (s-1 or Hz)
1A = 10-10m1cm = 10-2m1nm = 10-9m
= c/
Use c = 1.00A
325cm
473nm
10-10m1A
10-2m1cm
10-9m1nm
= 1.00x10-10m
= 325x10-2m
= 473x10-9m
=3x108m/s
1.00x10-10m= 3x1018s-1
=
=
3x108m/s
325x10-2m= 9.23x107s-1
3x108m/s
473x10-9m= 6.34x1014s-1
Practice Problems:
Interconverting Frequency and Wavelength
1. Calculate the frequency in hertz of green light with a wavelength of 550 nm.
2. Calculate the broadcast wavelength in meters of an FM radio station that broadcasts at 104.3 MHz.
Answers:
1. 5.4 x 1014 hertz
2. 2.876 m
Wave-Particle Duality of Light: in some cases light behaves as waves, in other times as
photons (particles)
1. Evidence for Wave Behavior of light» Refraction of light» Diffraction of light
2. Evidence for Particle Behavior of light» Blackbody Radiation» Photoelectric Effect
Refraction of Light
Diffraction of Light
Fig. 7.4 Different Behaviors of Waves and Particles
Speed changes when pebble enters H2O
Evidence for the wave
nature of light Diffraction of
Light—
Diffraction of Light—
7-10
Figure 7.6
Blackbody Radiation
E = h Blackbody Radiation
Evidence for the Particle Behavior of Light
~ 1000 K emits a soft red glow
~ 1500 K brighter & more orange
~ 2000 K brighter & white in color
Ephoton = h
Blackbody RadiationEvidence for Particle Behavior of
Light
1. Only specific colors of light are emitted when blackbodies (heated solids) are heated
~ 1000 K emits a soft red glow
~ 1500 K brighter and more orange
~ 2000 K brighter and white in color
2. Max Planck’s (1900): Atoms can only absorb or give off specific packets or quanta of light energy.
• These packet of energy are called photons.
Particle Nature of LightMax Planck (1900)
• EMR is emitted as weightless packets of energy called photons
• Each photon has its own energy and frequency, Ephoton = h
h = Planck’s constant = 6.626 x 10-34 J.s
Photoelectric Effect:
Evidence for Particle Behavior of Light
• Light of a certain minimum frequency (color) is needed to dislodge electrons from a metal plate.
•Wave theory predicts a wave of a minimum amplitude.
Einstein’s Explanation of the Photoelectric Effect (1905)
1. Light intensity is due to the number of photons striking the metal per second, not the amplitude
2. A photon of some minimum energy must be absorbed by the metal
E photon
= h
Relationship between Energy of Light and
Wavelength
1. Derive an equation that relates E and from the following equations: c = and E = h
2. Use this equation to Answer the following questions.....
a. Microwave ovens emit light of = 3.00 mm. Calculate the energy of each photon emitted from a microwave oven.
Ans. 6.63 x 10-23 J/photonb. How many photons of light are needed for a microwave oven
to raise the temperature of a cup of water (236 g) from 20.0 oC to 100.0 oC?
Ans. 1.19 x 1027 photons
Section 7.2 Atomic Spectra
1. Continuous Spectrum • Sunlight or from object heated to a very high
temperature (e.g. light filament)
2. Atomic Spectrum • Also called line, bright line or emission spectrum• Due to an atom’s electron(s) excited by electricity or
heat falling from a higher to a lower energy level—more about this later!!
7-13
Figure 7.8
The line spectra of several elements
Continuous Spectrum
Line Spectra
Rydberg Equation Predicts the Hydrogen Spectrum
Rydberg Equation • Empirically derived to fit hydrogen’s atomic spectrum• Predicts ’s of invisible line spectra
e.g. Hydrogen’s Ultraviolet line spectrum (nL = 1)
R = 1.096776 x 107 m-1 n = 1, 2, 3, 4, …
nn
12
i
2
f
11R
L H
7-14
= RRydberg equation -1
1
n22
1
n12
R is the Rydberg constant = 1.096776 m-1
Figure 7.9 Three series of spectral lines of atomic hydrogen
for the visible series, n1 = 2 and n2 = 3, 4, 5, ...
Using the Rydberg Equation
Practice Exercise:
Calculate the wavelength in nm and determine the color of the line in the visible spectrum of hydrogen for which nL = 2 and nH = 3.
Ans. 656.4 nm Color????
1st The Good News….Niels Bohr Planetary model of the atom
explains Hydrogen's Spectrum (1913)
1. An atom’s energy is quantized because electrons can only move in fixed orbits (energy levels) around the nucleus
2. Orbits are quantizedi.e. Each orbit can only have a certain radius
3. An electron can only move to another energy level (orbit) when the energy absorbed or emitted equals the difference in energy between the two energy levels
• Line spectra result as electrons emit light as they fall from a higher to lower energy level
Bohr’s Explanation of the Three series of Spectral Lines of the Hydrogen Spectrum
7-15
Figure 7.10
Quantum staircase
Animation of Bohr’s Planetary Model
1. Animation (Flash)
2. Animation (QuickTime)
Bohr’s Equation Derived from the Ideas of Planck, Einstein &
Classical Physics
1. Eelectron = ELower - EHigher or Eelectron = Efinal - Einitial
2. Eelectron = -2.18 x 10-19 J (1/n2Lower - 1/n2
higher)
But…… E = hc/ , substitution yields…
3. 1/= 1.10 x107 m-1 (1/n2Lower - 1/n2
higher)• Bohr’s Constant is within 0.05 % of the Rydberg Constant
• Equation provides a theoretical explanation of Hydrogen’s Atomic Spectrum
1/= 1.10 x107 m-1 (1/n2Lower - 1/n2
higher)
Bohr’s Equation Accurately Predicts
the Ionization Energy of Hydrogen
Use Bohr’s equation to calculate the ionization energy for
a.) one hydrogen atom
b.) one mole of hydrogen atoms
1/ = 1.10 x107 m-1 (1/n2Lower - 1/n2
higher)
Energy + H (g) H+(g) + e-
Answers: a.) 2.18 x 10-18J/atom ; b.) 1.31 x 103 kJ/mole
Now the Bad News…
Bohr’s Model is Incorrect!!
1. Closer inspection of spectral lines shows shows that they are not all single lines
• Bohr’s model doesn’t account for the extra lines
2. Only works for atoms or ions with one electron• Bohr’s model doesn’t account for presence of electron-
electron repulsions and electron-nucleus attractions in atoms with more than one electron.
3. Electrons do not orbit around the nucleus!!!• A new model is needed• Would you believe that electrons behave as waves and as
particles????
Section 7.3 The Wave-Particle Duality of Matter
Electron Diffraction: Evidence that electrons behave as waves!
Davisson & Germer (1927)
Electrons are diffracted by solids just like X-rays!
Hence, electrons behave as waves!
X-Ray tube Aluminum
X-Ray diffraction pattern of Aluminum
Source of electrons Aluminum Electron diffraction
pattern of Aluminum
7-23
Figure 7.14
Comparing the diffraction patterns of x-rays and electrons
Wave- Particle Duality of Matter and Energy
1. Matter behaves as if it moves like a wave!!
2. Only small, fast objects (e.g. e-, p+ , n0) have a measurable me = 9.11x10-31 kg; mp = mn = 1.67x10-27 kg
3. Louis DeBroglie (1924) combined
E = mc2 and E = hc / to yield
matter
= h/mu m = mass; u = velocity
4. DeBroglie too small to measure for heavy, slow objects
7-21
Table 7.1 The de Broglie Wavelengths of Several Objects
Substance Mass (g) Speed (m/s) (m)
slow electron
fast electron
alpha particle
one-gram mass
baseball
Earth
9x10-28
9x10-28
6.6x10-24
1.0
142
6.0x1027
1.0
5.9x106
1.5x107
0.01
25.0
3.0x104
7x10-4
1x10-10
7x10-15
7x10-29
2x10-34
4x10-63
h /mu
Locating an Electron....an uncertain affair!!
1. Orbital• Region in space where an electron wave is most likely to be
found
2. Exact location of an electron can’t be determined
3. Can only determine the probability of finding an electron....why?
• Electrons behave as waves!!
• In order to “see” the position of an electron we must probe it with radiation which changes its position and/or velocity
H e i s e n b e r g U n c e r t a i n t y P r i n c i p l e
1 . B o t h t h e v e l o c i t y a n d p o s i t i o n o f a n e l e c t r o n c a n n o t b e d e t e r m i n e d s i m u l t a n e o u s l y
x = u n c e r t a i n t y i n p o s i t i o n ; u = u n c e r t a i n t y i n v e l o c i t y ; m = m a s s o f o b j e c t
2 . C a n o n l y d e t e r m i n e t h e p r o b a b i l i t y o f f i n d i n g a n e l e c t r o n
» o r b i t a l s a r e r e g i o n s i n s p a c e w h e r e a n e l e c t r o n w i l l m o s t l i k e l y b e f o u n d
3 . S e e s a m p l e p r o b l e m 7 . 4
4
h umX
7-27
Sample Problem 7.4
SOLUTION:
PLAN:
Applying the Uncertainty Principle
PROBLEM: An electron moving near an atomic nucleus has a speed 6x106 ± 1% m/s. What is the uncertainty in its position (x)?
The uncertainty (x) is given as ±1%(0.01) of 6x106m/s. Once we calculate this, plug it into the uncertainty equation.
u = (0.01)(6x106m/s) = 6x4m/s
x * m u ? h
4
x ?4 (9.11x10-31kg)(6x104m/s)
6.626x10-34kg*m2/s= 10-9m
Section 7.4
Quantum Mechanical Model of the Atom: Electron Waves in Atoms
1. Electrons are standing waves• Peaks and troughs only move up and down• Similar to how guitar strings move
2. Orbitals• Are areas in space where electron waves are most
likely to be found• Orbitals are made of electron waves
Quantum Mechanics and Atomic Orbitals
• Erin Schrodinger (1926) developed a mathematical equation called a wave function to describe the energy of electrons
• The square of the wave function gives the probability of finding an electron at any point in space, thus producing a map of an orbital
7-28
The Schrödinger Equation
H = E
d2dy2
d2dx2
d2dz2
+ +82m
h2(E-V(x,y,z)(x,y,z) = 0+
how changes in space
mass of electron
total quantized energy of the atomic system
potential energy at x,y,zwave function
Atomic Orbital
An area in space where an electron wave is most likely to be found outside of the nucleus
7-30
Quantum Numbers and Atomic Orbitals
An atomic orbital is specified by three quantum numbers.
n the principal quantum number - a positive integer
l the angular momentum quantum number - an integer from 0 to n-1
ml the magnetic moment quantum number - an integer from -l to +l
Orbitals are Identified by 3 Quantum Numbers
1. Principle Quantum Number, n (n = 1,2,3…)• Determines the orbital’s size and energy (I.e. which energy
level the electron occupies)
• Relates to the average distance of the e- to the nucleus
2. Secondary Quantum Number, l• Determines the orbital’s shape or sublevel : s, p, d or f
• l = 0 to n-1
• Orbitals with the same values for n and l are called sublevels
Orbitals are Identified by 3 Quantum Numbers
3. Magnetic Quantum Number, ml
• Determines the orbital’s orientation in space
• ml = -l, …, 0 , …+l
• ml represents the orbital within the sublevel.
S - sublevel has 1 orbital
p - sublevel has 3 orbitals
d - sublevel has 5 orbitals
F - sublevel has 7 orbitals
7-31
Table 7.2 The Hierarchy of Quantum Numbers for Atomic Orbitals
Name, Symbol(Property) Allowed Values Quantum Numbers
Principal, n(size, energy)
Angular momentum, l
(shape)
Magnetic, ml
(orientation)
Positive integer(1, 2, 3, ...)
0 to n-1
-l,…,0,…,+l
1
0
0
2
0 1
0
3
0 1 2
0
0-1 +1 -1 0 +1
0 +1 +2-1-2
n = Principal quantum Number (size and energy of orbital)
l = Angular momentum Q.N. (shape of orbital)
ml = magnetic Q.N. (orientation of orbital)
Relationship between Angular momentum Q.N. , l, and sublevel names: s, p, d and f
Value of l Sublevel
0 s
1 p
2 d
3 f f
4 g
5 h
Sublevels only used by electrons in the excited state
Summary of Relationships Between n, l and ml
ENERGY LEVEL n
Sublevels l
(0 to n-1)
Orbitals ml
(-l to +l)
1
2
3
4
7-32
Sample Problem 7.5
SOLUTION:
PLAN:
Determining Quantum Numbers for an Energy Level
PROBLEM: What values of the angular momentum (l) and magnetic (ml) quantum numbers are allowed for a principal quantum number (n) of 3? How many orbitals are allowed for n = 3?
Follow the rules for allowable quantum numbers found in the text.
l values can be integers from 0 to n-1; ml can be integers from -l through 0 to + l.
For n = 3, l = 0, 1, 2
For l = 0 ml = 0
For l = 1 ml = -1, 0, or +1
For l = 2 ml = -2, -1, 0, +1, or +2
There are 9 ml values and therefore 9 orbitals with n = 3.
7-33
Sample Problem 7.6
SOLUTION:
PLAN:
Determining Sublevel Names and Orbital Quantum Numbers
PROBLEM: Give the name, magnetic quantum numbers, and number of orbitals for each sublevel with the following quantum numbers:
(a) n = 3, l = 2 (b) n = 2, l = 0 (c) n = 5, l = 1 (d) n = 4, l = 3
Combine the n value and l designation to name the sublevel. Knowing l, we can find ml and the number of orbitals.
n l sublevel name possible ml values # of orbitals
(a)
(b)
(c)
(d)
3
2
5
4
2
0
1
3
3d
2s
5p
4f
-2, -1, 0, 1, 2
0
-1, 0, 1
-3, -2, -1, 0, 1, 2, 3
3
1
3
7
Practice Makes Perfect?
1. What is the subshell (e.g. 1s, 2s, 2p, etc.) corresponding to the following values for n and l?
a. n = 2, l = 1
b. n = 4, l = 0
c. n = 3, l = 2
d. n = 5, l = 3
e. n = 3, l =3
Practice Makes Perfect?
2. Which of the following sets of quantum numbers are not possible?
a. n = 2, l = 1, m l = 0
b. n = 2, l = 2, m l = 1
c. n = 2, l = 1, m l = -2
d. n = 3, l = 2, m l = -2
e. n = 0, l = 0, m l = 0
The Relationship between the 4 Quantum Numbers, Energy Levels, Sublevels and
Orbitals
See figure 6.15, page 239 in Brady (Transp.)
Practice Makes Perfect?
1. What subshells are found in the 4th shell?
2. Which subshell is higher in energy?a. 3s or 3p
b. 4p or 4d
c. 3p or 4p
1s orbital 2s orbital 3s orbitalShapes of orbitals
As the value for n increases, the electron is more likely to be found further from the nucleus
Fig. 7.18
Shapes of the three orbitals in the 2p sublevel: 2px 2py 2pz
Note that the three orbitals are mutually perpendicular to each other (fig. D), thus contributing to an atoms overall spherical shape
An accurate representation of the 2pz orbital
Stylized shape of 2pz used in most texts
Fig. 7.19 c-g Shapes of the five orbitals in the 3d sublevel
Note that the relative positions of the five orbitals in the 3d sublevel contribute to the overall spherical shape of an atom (fig. H)
Fig. 7.20
One of the possible seven
orbitals of the 4f sublevel
Since only the s, p, and d sublevels are commonly involved with bonding, we will not be concerned with the shapes of the orbitals of the f-sublevel