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