Chapter 7 (Zumdahl) Electronic Structure of Atoms

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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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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

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 ().

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)

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

3.00  108 m/s.

• Therefore,

c = 

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

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.
•  = 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!)

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.

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!

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

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?

•  = 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.

Practice Problemsp 337, #41-44

The Nature of Energy

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

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.

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).

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.

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

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

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!)

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

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!

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:

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 .

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

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

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 .

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

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.

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.

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!

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

a) electron with a v = 0.100 m/s

Practice Problems
• Complete p 338 #59 & 60

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.

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.

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.

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.

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.

Angular Momentum Quantum Number (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.

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

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.

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.

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

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.

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

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.

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.

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.

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.

Sample Exercise: 7.6 p 308

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

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,

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.

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.

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.”

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.

Some Anomalies

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

Some Anomalies

For instance, the electron configuration for copper is

[Ar] 4s1 3d5

rather than the expected

[Ar] 4s2 3d4.