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Chemistry, The Central Science , 11th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten. Chapter 7 (Zumdahl) Electronic Structure of Atoms. John D. Bookstaver St. Charles Community College Cottleville, MO. Waves.

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chapter 7 zumdahl electronic structure of atoms

Chemistry, The Central Science, 11th edition

Theodore L. Brown; H. Eugene LeMay, Jr.;

and Bruce E. Bursten

Chapter 7 (Zumdahl)Electronic Structureof Atoms

John D. Bookstaver

St. Charles Community College

Cottleville, MO

© 2009, Prentice-Hall, Inc.

waves
Waves
  • To understand the electronic structure of atoms, one must understand the nature of electromagnetic (EM) radiation.
  • The distance between corresponding points on adjacent waves is the wavelength ().

© 2009, Prentice-Hall, Inc.

waves3
Waves
  • The number of waves passing a given point per unit of time is the frequency().
  • For waves traveling at the same velocity, the longer the wavelength, the smaller the frequency.
  • Frequency is measured in the following units:

s-1, = waves/s = Hz (hertz)

© 2009, Prentice-Hall, Inc.

electromagnetic radiation
Electromagnetic Radiation
  • All electromagnetic radiation travels at the same velocity: the speed of light (c),

3.00  108 m/s.

  • Therefore,

c = 

© 2009, Prentice-Hall, Inc.

sample exercise 7 1 p 292
Sample Exercise 7.1 (p 292)
  • (Copy from text into notes.)
  • Follow as I review the book solution to this problem.

© 2009, Prentice-Hall, Inc.

slide6
Practice Problems: Text, p 337,#39-40
  • #39: The laser in an audio compact disc player uses light with a wavelength of 7.80 x 102 nm. Calculate the frequency of this light.
  • Solution: label your given(s) & your unknown.
  •  = given = unknown
  • Identify the connection between given(s) and unknown. (c= )
  • Recognize that you know the value of c for all EM radiation ( c =3.00 x 108 m/s)
  • Substitute in all given/known values & solve for the unknown. (NOTE: Convert nm to m first!)

© 2009, Prentice-Hall, Inc.

the nature of energy
The Nature of Energy
  • The wave nature of light does not explain how an object can glow when its temperature increases.
  • Max Planck explained it by assuming that energy comes in packets called quanta.

© 2009, Prentice-Hall, Inc.

the nature of energy8
The Nature of Energy
  • Einstein used this assumption to explain the photoelectric effect.
  • He concluded that energy is proportional to frequency:

E = h

where h is Planck’s constant, 6.626  10−34 J-s.

  • He also concluded that wave energy has mass!

© 2009, Prentice-Hall, Inc.

the nature of energy9
The Nature of Energy
  • Therefore, if one knows the wavelength of light, one can calculate the energy in one photon, or packet, of that light (also known as a “quantum” of energy):

c = 

E = h

© 2009, Prentice-Hall, Inc.

sample exercise 7 2 p 293 294
Sample Exercise 7.2p 293-294
  • Copy this problem and solution.
  • Follow along as we walk thru the solution together.
  • The blue color in fireworks is often achieved by heating copper(I) chloride (CuCl) to about 1200 C. Then the compound emits blue light having a wavelength of 450 nm. What is the increment of energy (the quantum) that is emitted by CuCl?

© 2009, Prentice-Hall, Inc.

slide11

Solution: label your given(s) & your unknown.

  •  = given E= unknown
  • Identify the connection between given(s) and unknown.
    • There is not a direct connection. No equation we have includes both E and .
    • However, we know that E = h.
    • We also know the value of h (=6.626 x 10−34 J-s).
    • If we know the value of , we can calculate E. Can we find ? (is there a connection between  and?
    • YES! c= .
  • Substitute given values into this second equation and solve for . (Don’t forget to convert from nm to m first!)
  • When  is obtained, substitute it and the value of h into the first equation and solve for E.

© 2009, Prentice-Hall, Inc.

practice problems p 337 41 44
Practice Problemsp 337, #41-44

© 2009, Prentice-Hall, Inc.

the nature of energy13
The Nature of Energy

Another mystery in the early 20th century involved the emission spectra observed from energy emitted by atoms and molecules.

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the nature of energy14
The Nature of Energy
  • For atoms and molecules one does not observe a continuous spectrum, as one gets from a white light source.
  • Only a line spectrum of discrete wavelengths is observed.

© 2009, Prentice-Hall, Inc.

the nature of energy15
The Nature of Energy
  • Niels Bohr adopted Planck’s assumption and explained these phenomena in this way:
    • Electrons in an atom can only occupy certain orbits (corresponding to certain energies).

© 2009, Prentice-Hall, Inc.

the nature of energy16
The Nature of Energy
  • Niels Bohr adopted Planck’s assumption and explained these phenomena in this way:
    • Electrons in permitted orbits have specific, “allowed” energies; these energies will not be radiated from the atom.

© 2009, Prentice-Hall, Inc.

the nature of energy17
The Nature of Energy
  • Niels Bohr adopted Planck’s assumption and explained these phenomena in this way:
    • Energy is only absorbed or emitted in such a way as to move an electron from one “allowed” energy state to another; the energy is defined by

E = nh

Where n=main energy level

© 2009, Prentice-Hall, Inc.

solving for e energy if are unknown
Solving for e- energy if  & are unknown
  • The energy of an electron in a hydrogen atom is
  • E= (-2.178 x 10-18 J) Z2n2

[

]

Where Z= nuclear charge (atomic #) and

n = main energy level and

RH = the Rydberg constant, 2.18  10−18 J

© 2009, Prentice-Hall, Inc.

the nature of energy19

12

nf2

( )

-

E = −RH

12

ni2

The Nature of Energy

The energy absorbed or emitted from the process of electron promotion or demotion can be calculated by the equation:

ni and nf are the initial and final energy levels of the electron.

And Z = 1 (atomic # of hydrogen!)

© 2009, Prentice-Hall, Inc.

sample exercise
Sample Exercise
  • P 338 #51a.
  • Calculate the energy emitted when the following transition occurs in a hydrogen atom: n=3  n= 2

© 2009, Prentice-Hall, Inc.

solution
Solution
  • Our givens are
    • nfinal (level 2)
    • ninitial (level 3)
  • Our unknown is the energy released (E)
  • Identify the connection between given(s) & unknown:E= -RH(1/nf2-1/ni2)
  • Substitute given values & solve for unknown!

© 2009, Prentice-Hall, Inc.

calculating of light from energy released
Calculating  of light from energy released
  • Continue w/#51 from p 338
  • Calculate the wavelength of light emitted when the following transition occurs in the hydrogen atom. What type of EM radiation is emitted?
  • Givens: E (from prior slide)
  • Unknown: 
  • Connection: E = h to solve for 
  • then use c= to solve for 

Substitute & Solve:

© 2009, Prentice-Hall, Inc.

alternative approach
ALTERNATIVE APPROACH
  • Derive formula that includes all the variables you are using: , , & E
  • Use both eqns E = h & c= 
  • Set both eqns equal to 
    • E =  & c= 

h 

  • Set both eqns equal to one another
    • E =c

h 

  • Isolate , since this is your unknown.
  •  = hc

E

4. Substitute values (c, h, E) to solve for .

© 2009, Prentice-Hall, Inc.

the wave nature of matter

h

mv

 =

The Wave Nature of Matter
  • Louis de Broglie posited that if light can have material properties, matter should exhibit wave properties.
  • He demonstrated that the relationship between mass and wavelength was

This formula is on the AP exam formula sheet

Please note: the base units for Joules are kg●m2/s2 . This will be important for calculations

© 2009, Prentice-Hall, Inc.

wave particle duality
Wave-Particle Duality
  • Site with Cool Visuals & Explanation of Wave Interference, Including electron waves

© 2009, Prentice-Hall, Inc.

sample exercise from brown lemay p 223
Sample Exercise (from Brown/Lemay) p 223
  • What is the wavelength of an electron moving with a speed of 5.97 x 106 m/s? The mass of the electron is 9.11 x 10-31 kg.

SOLUTION:

Givens: melectron = 9.11 x 10-31 kg

velectron = 5.97 x 106 m/s

Unknown = 

CONNECTION between givens & unknown:

 = h

mv

Substitute given & constant values, solve for .

© 2009, Prentice-Hall, Inc.

sample exercise cont
Sample Exercise, cont.

Substitute given & constant values, solve for .

 = h

mv

  • = 6.626 x 10-34 kg●m2/s

(9.11 x 10-31 kg)(5.97 x 106 m/s)

 = 1.22 x 10-10

Where did the exponent go?

J= kg●m2/s2

© 2009, Prentice-Hall, Inc.

practice problems
Practice Problems

6.41) Determine the wavelength of the following objects

  • An 85-kg person skiing at 50 km/hr
  • A 10.0 g bullet fired at 250 m/s.
  • A lithium atom moing at 2.5 x 105 m
  • An ozone molecule in the upper atmosphere moving at 550 m/s.

© 2009, Prentice-Hall, Inc.

practice problems30
Practice Problems:
  • 6.42) Among the elementary subatomic particles of physics is the muon, which decays within a few nanoseconds after formation. The muon has a mass 206.8 times that of an electron. Calculate the deBroglie wavelength associated with a muon traveling at 8.85 x 105 cm/s.

© 2009, Prentice-Hall, Inc.

the uncertainty principle

h

4

(x) (mv) 

The Uncertainty Principle
  • Heisenberg showed that the more precisely the momentum of a particle is known, the less precisely is its position known:
  • In many cases, our uncertainty of the whereabouts of an electron is greater than the size of the atom itself!

© 2009, Prentice-Hall, Inc.

sample exercise p 338 59
Sample ExerciseP 338, # 59
  • Using Heisenberg uncertainty principle, calculate ∆x for each of the following:

a) electron with a v = 0.100 m/s

© 2009, Prentice-Hall, Inc.

practice problems33
Practice Problems
  • Complete p 338 #59 & 60

© 2009, Prentice-Hall, Inc.

quantum mechanics
Quantum Mechanics
  • Erwin Schrödinger developed a mathematical treatment into which both the wave and particle nature of matter could be incorporated.
  • It is known as quantum mechanics.

© 2009, Prentice-Hall, Inc.

quantum mechanics35
Quantum Mechanics
  • The wave equation is designated with a lower case Greek psi ().
  • The square of the wave equation, 2, gives a probability density map of where an electron has a certain statistical likelihood of being at any given instant in time.

© 2009, Prentice-Hall, Inc.

quantum numbers
Quantum Numbers
  • Solving the wave equation gives a set of wave functions, or orbitals, and their corresponding energies.
  • Each orbital describes a spatial distribution of electron density.
  • An orbital is described by a set of three quantum numbers.

© 2009, Prentice-Hall, Inc.

principal quantum number n
Principal Quantum Number (n)
  • The principal quantum number, n, describes the energy level on which the orbital resides.
  • The values of n are integers ≥ 1.

© 2009, Prentice-Hall, Inc.

angular momentum quantum number l
Angular Momentum Quantum Number (l)
  • This quantum number defines the shape of the orbital.
  • Allowed values of l are integers ranging from 0 to n −1.
  • We use letter designations to communicate the different values of l and, therefore, the shapes and types of orbitals.

© 2009, Prentice-Hall, Inc.

angular momentum quantum number l39
Angular Momentum Quantum Number (l)

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magnetic quantum number m l
Magnetic Quantum Number (ml)
  • The magnetic quantum number describes the three-dimensional orientation of the orbital.
  • Allowed values of ml are integers ranging from -l to l:

−l ≤ ml≤ l.

  • Therefore, on any given energy level, there can be up to 1 s orbital, 3 p orbitals, 5 d orbitals, 7 f orbitals, etc.

© 2009, Prentice-Hall, Inc.

magnetic quantum number m l41
Magnetic Quantum Number (ml)
  • Orbitals with the same value of n form a shell.
  • Different orbital types within a shell are subshells.

© 2009, Prentice-Hall, Inc.

s orbitals
s Orbitals
  • The value of l for s orbitals is 0.
  • They are spherical in shape.
  • The radius of the sphere increases with the value of n.

© 2009, Prentice-Hall, Inc.

s orbitals43
s Orbitals

Observing a graph of probabilities of finding an electron versus distance from the nucleus, we see that s orbitals possess n−1 nodes, or regions where there is 0 probability of finding an electron.

© 2009, Prentice-Hall, Inc.

p orbitals
p Orbitals
  • The value of l for p orbitals is 1.
  • They have two lobes with a node between them.

© 2009, Prentice-Hall, Inc.

d orbitals
d Orbitals
  • The value of l for a d orbital is 2.
  • Four of the five d orbitals have 4 lobes; the other resembles a p orbital with a doughnut around the center.

© 2009, Prentice-Hall, Inc.

energies of orbitals
Energies of Orbitals
  • For a one-electron hydrogen atom, orbitals on the same energy level have the same energy.
  • That is, they are degenerate.

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energies of orbitals47
Energies of Orbitals
  • As the number of electrons increases, though, so does the repulsion between them.
  • Therefore, in many-electron atoms, orbitals on the same energy level are no longer degenerate.

© 2009, Prentice-Hall, Inc.

spin quantum number m s
Spin Quantum Number, ms
  • In the 1920s, it was discovered that two electrons in the same orbital do not have exactly the same energy.
  • The “spin” of an electron describes its magnetic field, which affects its energy.

© 2009, Prentice-Hall, Inc.

spin quantum number m s49
Spin Quantum Number, ms
  • This led to a fourth quantum number, the spin quantum number, ms.
  • The spin quantum number has only 2 allowed values: +1/2 and −1/2.

© 2009, Prentice-Hall, Inc.

pauli exclusion principle
Pauli Exclusion Principle
  • No two electrons in the same atom can have exactly the same energy.
  • Therefore, no two electrons in the same atom can have identical sets of quantum numbers.

© 2009, Prentice-Hall, Inc.

sample exercise 7 6 p 308
Sample Exercise: 7.6 p 308

© 2009, Prentice-Hall, Inc.

electron configurations
Electron Configurations
  • This shows the distribution of all electrons in an atom.
  • Each component consists of
    • A number denoting the energy level,

© 2009, Prentice-Hall, Inc.

electron configurations53
Electron Configurations
  • This shows the distribution of all electrons in an atom
  • Each component consists of
    • A number denoting the energy level,
    • A letter denoting the type of orbital,

© 2009, Prentice-Hall, Inc.

electron configurations54
Electron Configurations
  • This shows the distribution of all electrons in an atom.
  • Each component consists of
    • A number denoting the energy level,
    • A letter denoting the type of orbital,
    • A superscript denoting the number of electrons in those orbitals.

© 2009, Prentice-Hall, Inc.

orbital diagrams
Orbital Diagrams
  • Each box in the diagram represents one orbital.
  • Half-arrows represent the electrons.
  • The direction of the arrow represents the relative spin of the electron.

© 2009, Prentice-Hall, Inc.

hund s rule
Hund’s Rule

“For degenerate orbitals, the lowest energy is attained when the number of electrons with the same spin is maximized.”

Another translation: “Don’t pair electrons in degenerate orbitals until there is 1 electron in each orbital.”

© 2009, Prentice-Hall, Inc.

periodic table
Periodic Table
  • We fill orbitals in increasing order of energy.
  • Different blocks on the periodic table (shaded in different colors in this chart) correspond to different types of orbitals.

© 2009, Prentice-Hall, Inc.

some anomalies
Some Anomalies

Some irregularities occur when there are enough electrons to half-fill s and d orbitals on a given row.

© 2009, Prentice-Hall, Inc.

some anomalies59
Some Anomalies

For instance, the electron configuration for copper is

[Ar] 4s1 3d5

rather than the expected

[Ar] 4s2 3d4.

© 2009, Prentice-Hall, Inc.

some anomalies60
Some Anomalies
  • This occurs because the 4s and 3d orbitals are very close in energy.
  • These anomalies occur in f-block atoms, as well.

© 2009, Prentice-Hall, Inc.