141
Summer Chemistry 2008
Silberberg Chapter 7
Dr. Ben Tovrog
7-1
Silberberg Chapter 7 Problem Assignment
These problems are for your practice. They will not be collected or graded. Easy and Medium: 7.2, 7.7, 7.9, 7.11, 7.13, 7.16, 7.20, 7.23, 7.25, 7.31, 7.34, 7.37, 7.41, 7.47, 7.48, 7.49, 7.51, 7.55, 7.57, 7.59, 7.66, 7.74. Hard 7.69, 7.71, 7.81, 7.97
7-2
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
7-3
The Wave Nature of Light
Figure 7.1
Frequency and Wavelength
c =
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Figure 7.2
Amplitude (intensity) of a wave.
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Figure 7.3
Regions of the electromagnetic spectrum.
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Sample Problem 7.1 PROBLEM:
Interconverting Wavelength and Frequency
o
A dental hygienist uses x-rays (= 1.00A) to take a series of dental radiographs while the patient listens to a radio station ( = 325 cm) and looks out the window at the blue sky (= 473 nm). What is the frequency (in s-1) of the electromagnetic radiation from each source? (Assume that the radiation travels at the speed of light, 3.00x108 m/s.)
SOLUTION: o -10 -10 1.00A 10 o m = 1.00x10 m 1A 3x108 m/s = 3x1018 s-1 = 1.00x10-10 m -2 325 cm 10 m = 325x10-2 m 1 cm 3x108 m/s = = 9.23x107 s-1 325x10-2 m 10-9 m 473nm = 473x10-9 m 1 nm 3x108 m/s = 6.34x1014 s-1 = 473x10-9 m
PLAN:
Use c =
wavelength in units given
1 A = 10-10 m 1 cm = 10-2 m 1 nm = 10-9 m
wavelength in m
o
= c/
frequency (s-1 or Hz)
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Wavelength, Frequency, Energy
1. An FM station broadcasts classical music at 93.5 MHz (megahertz). Find the wavelength (in m, nm, and A) for this frequency.
2. Rank the following photons in terms of decreasing energy a) IR (v = 6.5 x 10 13 s -1), b) microwave (v = 6.5 x 10 11 s -1), c) UV (v = 8.0 x 10 15 s -1 )
3. Covalent bonds in a molecule absorb radiation in the IR region and vibrate at characteristic frequencies. a) The C-O bond in an organic compound absorbs radiation of wavelength 9.6 m. What is the corresponding frequency? b) The H-Cl bond has a frequency of vibration of 8.652 x 10 13 Hz. What is the corresponding wavelength.
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Figure 7.4
Different behaviors of waves and particles.
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The diffraction pattern caused by light passing through two adjacent slits. Figure 7.5
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Plank And Einstein Quantum Theory and Photons
While radiation has wave properties and appears to continuously vary over the visible spectrum, Plank determined that energy is emitted in only discrete levels, i.e. E = n*h* E is the energy of the radiation is its frequency n is a positive integer (1,2,3, etc) called a quantum number h is Plank's constant = 6.626 x 10-34 J*s.
Einstein's photoelectric effects extends quantum theory and proposes that light has particle characteristics in addition to the well-known wave properties. These light "particles" are called photons
These breakthroughs seem interesting but not powerful. In fact, they form the basis for emission spectra, fireworks, atomic structure, molecular structure, reactivity of molecules and atoms, modern diagnostic methods like PET/MRI, explain the structure of the periodic table, paved the way for semiconductors, computers, cell phones, TVs, i-Pods
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Figure 7.7
Demonstration of the photoelectric effect.
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Calculating the Energy of Radiation from Its Wavelength PROBLEM: A cook uses a microwave oven to heat a meal. The wavelength of the radiation is 1.20cm. What is the energy of one photon of this microwave radiation? PLAN: After converting cm to m, we can use the energy equation, E = h combined with = c/ to find the energy.
SOLUTION: E= E = hc/ 6.626X10-34J*s x 3x108m/s = 1.66x10-23J
Sample Problem 7.2
1.20cm
10-2m
cm
Followup Problem 7.2: Calculate the energies of one photon of the following radiation wavelengths: ultraviolet (1 x 10 -8 m), visible (5 x 10-7 m), and infared (1x 10-4 m).
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Figure 7.8
The line spectra of several elements.
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Rydberg equation
1
=
R
1
n12
-
1
n22
R is the Rydberg constant = 1.096776x107 m-1
Figure 7.9
Three series of spectral lines of atomic hydrogen.
for the visible series, n1 = 2 and n2 = 3, 4, 5, ...
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Figure 7.10
Quantum staircase.
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Figure 7.11 The Bohr explanation of the three series of spectral lines.
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Figure B7.1
Flame tests.
strontium 38Sr
copper 29Cu
Figure B7.2
Emission and absorption spectra of sodium atoms.
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7-19
Working with the Rydberg Equation
1. Use the Rydberg equation to calculate the wavelength (in A) of the photon absorbed when a hydrogen atom undergoes a transition from n=1 to n=3.
2. An electron in the n =5 level of an H atom emits a photon of wavelength 1281 nm. To what energy level does the electron move?
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Figure 7.15
CLASSICAL THEORY Matter particulate, massive Energy continuous, wavelike
Summary of the major observations and theories leading from classical theory to quantum theory.
Since matter is discontinuous and particulate perhaps energy is discontinuous and particulate.
Observation
blackbody radiation photoelectric effect atomic line spectra
Theory
Planck: Energy is quantized; only certain values allowed Einstein: Light has particulate behavior (photons) Bohr: Energy of atoms is quantized; photon emitted when electron changes orbit.
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Figure 7.15 continued Since energy is wavelike perhaps matter is wavelike Observation Theory
Davisson/Germer: electron diffraction by metal All crystal
deBroglie: matter travels in waves; energy of atom is quantized due to wave motion of electrons
Since matter has mass perhaps energy has mass Observation Compton: photon wavelength increases (momentum decreases) after colliding with electron Theory Einstein/deBroglie: Mass and energy are equivalent; particles have wavelength and photons have momentum. QUANTUM THEORY
Energy same as Matter particulate, massive, wavelike
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h /mu
Table 7.1
The de Broglie Wavelengths of Several Objects
(m) 7x10-4
1x10-10 7x10-15 7x10-29
Substance slow electron
fast electron alpha particle one-gram mass
Mass (g) 9x10-28
9x10-28 6.6x10-24 1.0
Speed (m/s) 1.0
5.9x106 1.5x107 0.01
baseball
Earth
142
6.0x1027
25.0
3.0x104
2x10-34
4x10-63
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Sample Problem 7.3 Calculating the de Broglie Wavelength of an Electron
PROBLEM: Find the deBroglie wavelength of an electron with a speed of 1.00x106m/s (electron mass = 9.11x10-31kg; h = 6.626x10-34 kg*m2/s).
PLAN:
Knowing the mass and the speed of the electron allows to use the equation = h/mu to find the wavelength.
6.626x10-34kg*m2/s = 7.27x10-10m
SOLUTION:
=
9.11x10-31kg x 1.00x106m/s Follow-up Problem 7.3:
What is the speed of an electron that has a deBroglie wavelength of 100. nm.? Problem 7.42: How fast must a 142 g baseball travel in order to have a deBroglie wavelength that is equal to that of an x-ray photon with = 100. pm
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The Heisenberg Uncertainty Principle
x * m u
h 4
It is impossible to know the exact position and momentum of a particle simultaneously. Practically speaking this has the largest implications for electrons. Rearranging we have: x > h
4m u
The very small value of the electron mass causes x to be large see sample problem 7.4
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Sample Problem 7.4
Applying the Uncertainty Principle
PROBLEM: An electron moving near an atomic nucleus has a speed 6x10 6 1% m/s. What is the uncertainty in its position (x)?
PLAN: The uncertainty (x) is given as 1%(0.01) of 6x106m/s. Once we calculate this, plug it into the uncertainty equation.
SOLUTION:
u = (0.01)(6x106m/s) = 6x104m/s h x * m u 4 x
6.626x10-34kg*m2/s 4 (9.11x10-31kg)(6x104m/s) = 9.52x10-9m
The diameter of the atom is only 10 -10 m, so the uncertainly in the electrons position is 10 times that of the atom diameter. Good luck finding it. This is why in the next few slides we will talk about diffuse atomic orbital shapes and probabilities of electron locations.
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The Schrdinger Equation
H = E
wave function mass of electron potential energy at x,y,z
d2 d2 d2 82m + + + (E-V(x,y,z)(x,y,z) = 0 2 2 2 2 dx dy dz h
how changes in space total quantized energy of the atomic system
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Figure 7.16
Electron probability in the ground-state H atom.
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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
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Table 7.2 The Hierarchy of Quantum Numbers for Atomic Orbitals
Name, Symbol (Property) Allowed Values
Principal, n Positive integer (size, energy) (1, 2, 3, ...)
Quantum Numbers
1
2
3
Angular momentum, l 0 to n-1 (shape)
0
0
1
0
1
2
0 Magnetic, ml -l,...,0,...,+l (orientation)
0
0
-1 0 +1
-1 0 +1
-2
-1
0
+1 +2
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Sample Problem 7.5
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?
PLAN: 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.
SOLUTION: 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.
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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
Sample Problem 7.6
PLAN: Combine the n value and l designation to name the sublevel. Knowing l, we can find ml and the number of orbitals.
SOLUTION:
n
(a) (b) (c) 3 2 5
l
2 0 1
sublevel name possible ml values # of orbitals
3d 2s 5p -2, -1, 0, 1, 2 0 -1, 0, 1
5
1 3
(d)
4
3
4f
-3, -2, -1, 0, 1, 2, 3
7
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Figure 7.17
1s
2s
3s
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Figure 7.18
The 2p orbitals.
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Figure 7.19
The 3d orbitals.
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Figure 7.19 continued
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Figure 7.20
One of the seven possible 4f orbitals.
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More Orbitals
1. How many orbitals in an atom can have each of the following designations? a. 5f n=5, L=3, Ml= -3 -2 -1 0 1 2 3 b. 4p n=4, L=1, Ml= -1 0 1 c. 5d n=5, L=3, Ml= -3 -2 -1 0 1 2 d. N = 2. L=1, Ml=3... total=4
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Give all possible ml values for orbitals that have each of the following: L = 3 Ml= -3 -2 -1 0 1 2 n=2 n = 6, l = 1.
More and More Orbitals
1. Are the following quantum number combinations allowed? If not, show two ways to correct them.
n = 1, l = 0, ml = 0. n = 2, l = 2, ml = +1 n = 7, l = 1, ml = +2 n = 3, l = 1, ml = -2. OK L can't = N. EX: n=2, L=0, Ml=0 Ml: can only be -1, 0, +1
1. a. b. c.
For each of the following sublevels, give the n and l values and the number of orbitals. 6g 4s 3d
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