AP Chemistry
Chapter 6 Electronic Structure of Atoms
Chapter 6. Electronic Structure of Atoms
6.1 The Wave Nature of Light
•
Electromagnetic radiation is characterized by its wave nature.
•
All waves have a characteristic wavelength, lambda), and amplitude, A.
•
The frequency,  are Hertz (1 Hz = 1 s–1).
•
The speed of a wave is given by its frequency multiplied by its wavelength.
•
For light, speed, c =  moves through a vacuum with a speed of approximately 3.00 x 108 m/s.
•
The electromagnetic spectrum is a display of the various types of electromagnetic radiation arranged in
order of increasing wavelength. E.g. visible radiation has wavelengths between 400 nm (violet) and 750
nm (red).
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AP Chemistry
Chapter 6 Electronic Structure of Atoms
Sample Exercise 6.1 (p. 214)
Two electromagnetic waves are represented below.
a)
Which wave has the higher frequency?
b)
If one wave represents visible light and the other represents infrared
radiation, which wave is which?
Practice Exercise 6.1
Figure 06.05-01UN
If one of the waves represents blue light and the other red light, which would be which?
Sample Exercise 6.2 (p. 214)
The yellow light given off by a sodium vapor lamp used for public lighting has a wavelength of 589 nm. What
is the frequency of this radiation?
(5.09 x 1014 s-1)
Practice Exercise 6.2
a) A laser used in eye surgery to fuse detached retinas produces radiation with a wavelength of 640.0 nm.
Calculate the frequency of this radiation.
(4.688 x 1014 s-1)
b) An FM radio station broadcasts electromagnetic radiation at a frequency of 103.4 MHz.
Calculate the wavelength of this radiation.
(2.901 m)
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AP Chemistry
Chapter 6 Electronic Structure of Atoms
6.2 Quantized Energy and Photons – Read about black body radiation and the photoelectric effect
in Section 6.2 in either your textbook or the earlier edition (blue), then complete the following problems.
Sample Exercise 6.3 (p. 217)
Calculate the energy of one photon of yellow light whose wavelength is 589 nm. (3.37 x 10-19 J)
If one photon of radiant energy supplies 3.37 x 10-19 J, then how many photons will one mole of these photons
supply?
(2.03 x 105 J/mol)
Note: this amount is the magnitude at which chemical reactions take place. Radiation  break bonds,
i.e. photochemical reactions.
Practice Exercise 6.3
a)
A laser emits light with a frequency of 4.69 x 1014s-1. What is the energy of one photon of the
radiation from this laser?
(3.11 x 10-19 J)
b)
If the laser emits a burst or pulse of energy containing
5.0 x 1017 photons of this radiation, what is the total energy of
that pulse? (0.16 J)
c)
If the laser emits 1.3 x 10-2 J of energy during a pulse, how many photons are emitted during the pulse?
(4.2 x 1016 photons)
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AP Chemistry
Chapter 6 Electronic Structure of Atoms
6.3 Line Spectra and the Bohr Model
Line Spectra
•
spectrum that contains radiation of only specific wavelengths; this is called a line spectrum.
•
Rutherford assumed the electrons orbited the nucleus analogous to planets around the sun.
• However, a charged particle moving in a circular path should lose energy.
• This means that the atom should be unstable according to Rutherford’s theory.
•
Bohr noted the line spectra of certain elements and assumed the electrons were confined to specific energy
states. These were called orbits.
•
Bohr model is based on three postulates:
1. Electrons in an atom can only occupy certain orbits (corresponding to certain energies).
2. Electrons in permitted orbits have specific, “allowed” energies;
these energies will not be radiated from the atom.
3. 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 = h.
Bohr’s Model
The Energy States of the Hydrogen Atom
•
•
•
The first orbit in the Bohr model has n = 1 and is closest to the nucleus.
The furthest orbit in the Bohr model has n close to infinity and corresponds to E = 0.
Electrons in the Bohr model can only move between orbits by absorbing and emitting energy in quanta
(∆E =Ef – Ei = h.
• The ground state = the lowest energy state (Ei).
• An electron in a higher energy state is said to be in an excited state (Ef).
 1 
 1 
E  (hcRH ) 2  (2.18 1018 J) 2 
n 
n 
where n = principal quantum number (i.e. n = 1,2,3,…) and RH is the Rydberg contstant.
The product hcRH = 2.18 x 10-18 J.
•

The amount of energy absorbed or emitted by moving between energy states is given by
 1
1 
 E  E f  E i  h  2.18 1018 J

n 2 n 2 

 f
i 
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
AP Chemistry
Chapter 6 Electronic Structure of Atoms
Sample Exercise 6.4 (p. 222)
Using the figure above, predict which of the following electronic transitions produces the longest wavelength
spectral line: n = 2 to n = 1, n = 3 to n = 2, or n = 4 to n = 3.
Practice Exercise 6.4
Indicate whether each of the following electronic transitions emits energy or requires the absorption of energy:
a)
n = 3 to n = 1;
b)
n = 2 to n = 4.
Limitations of the Bohr Model
•
The Bohr Model has several limitations:
• It cannot explain the spectra of atoms other than hydrogen.
• Electrons do not move about the nucleus in circular orbits.
•
However the model introduces two important ideas:
• Electrons exist only in certain energy levels described by quantum numbers.
• Energy gain or loss is involved in moving an electron from one energy level to another.
Note: Niels Bohr won the Nobel Prize in 1922 for his atomic model.
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AP Chemistry
Chapter 6 Electronic Structure of Atoms
6.4 The Wave Behavior of Matter
•
•
Louis de Broglie presented the idea that if light can have material properties, matter should exhibit wave
properties. Using Einstein’s and Planck’s equations, deBroglie derived :
h
m
The momentum, mv, is a particle property, whereas  is a wave property.
• Matter waves is the term used to describe wave characteristics of material particles.
• Therefore, in one equation de Broglie summarized the concepts of waves and particles as they apply to
low-mass, high-speed objects.
• As a consequence of deBroglie’s discovery, we now have techniques such as X-ray diffraction and
electron microscopy to study small objects.
Sample Exercise 6.5 (p. 223) Note: You will need the following conversion factor: 1 J = kg m2
s2
What is the wavelength of an electron with a velocity of 5.97 x 106 m/s?
(0.122 nm or 1.22 Å)
Practice Exercise 6.5
Calculate the velocity of a neutron whose de Broglie wavelength is 500. pm. (mass of a neutron = 1 amu =
1.66053873 x 10-24 g)
(799 m/s)
The Uncertainty Principle
•
Heisenberg’s uncertainty principle: We cannot determine the exact position, direction of motion, and
speed of subatomic particles simultaneously.
•
For electrons: We cannot determine their momentum and position simultaneously.
6.5 Quantum Mechanics and Atomic Orbitals
•
•
•
•
Schrödinger proposed an equation containing both wave and particle terms:
Solving the equation leads to wave functions, .
The wave function gives the shape of the electron’s orbital.
• The square of the wave function, , gives the probability of finding the electron.
• That is,  gives the electron density for the atom.
•  is called the probability density.
Electron density is another way of expressing probability.
• A region of high electron density is one where there is a high probability of finding an electron.
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AP Chemistry
Chapter 6 Electronic Structure of Atoms
Orbitals and Quantum Numbers
•
•
•
If we solve the Schrödinger equation we get wave functions and energies for the wave functions.
We call  orbitals.
Schrödinger’s equation requires three quantum numbers:
•
Principal quantum number, n. This is the same as Bohr’s n.
• As n becomes larger, the atom becomes larger and the electron is further from the nucleus.
•
Angular momentum quantum number, l. This quantum number depends on the value of n.
• The values of l begin at 0 and increase to n – 1.
• We usually use letters for l (s, p, d, and f for l = 0, 1, 2, and 3). Usually we refer to the s, p, d, and f
orbitals. l defines the shape of the orbital.
• This quantum number defines the shape of the orbital.
•
Magnetic quantum number, ml.
• This quantum number depends on l.
• The magnetic quantum number has integer values between –l and +l.
• There are (2l+1) possible values of ml.
• For example, for l = 1, there are (21+1) = 3 values of ml : 0, +1, and -1.
• Consequently, for l = 1, there are 3 orbitals: px, py and pz.
• Magnetic quantum numbers give the three-dimensional orientation of each orbital.
•
A collection of orbitals with the same value of n is called an electron shell.
• There are n2 orbitals in a shell described by a the n value.
• For example, for n = 3, there are 32 = 9 orbitals.
• A set of orbitals with the same n and l is called a subshell.
• Each subshell is designated by a number and a letter.
• For example, 3p orbitals have n = 3 and l = 1.
• There are n types of subshells in a shell described by a the n value.
• For example, for n = 3, there are 3 subshells: 3s, 3p and 3d.
•
Orbitals can be ranked in terms of energy to yield an Aufbau diagram.
• Note that this Aufbau diagram is for a single electron system.
As n increases note that the spacing between energy levels becomes smaller.
•
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AP Chemistry
Chapter 6 Electronic Structure of Atoms
Sample Exercise 6.6 (p. 228)
a) Without referring to Table 6.2, predict the number of subshells in the fourth shell, that is, for n = 4.
b) Give the label for each of these subshells.
c) How many orbitals are in each subshell?
Practice Exercise 6.6
a) What is the designation for the subshell with n = 5 and l = 1?
b) How many orbitals are in this subshell?
c) Indicate the values of ml for each of these orbitals.
6.6 Representation of Orbitals; 6.7 Many-Electron Atoms; 6.8 Electron
Configurations; 6.9 Electronic Configurations and the Periodic Table:
Read these sections in your textbook, then complete the following problems:
Sample Exercise 6.7 (p. 237)
Draw the orbital diagram representation for the electron configuration of oxygen, atomic number 8. How many
unpaired electrons does an oxygen atom possess?
Practice Exercise 6.7
a) Write the electron configuration of phosphorus, element 15.
b) How many unpaired electrons does a phosphorus atom possess?
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AP Chemistry
Chapter 6 Electronic Structure of Atoms
Note: Condensed Electron Configurations = Noble Gas configurations
Sample Exercise 6.8 (p. 241)
What is the characteristic valence shell electron configuration of the group 7A elements, the halogens?
Practice Exercise 6.8
What family of elements is characterized by having an ns2np2 outer-electron configuration?
Sample Exercise 6.9 (p. 241)
a) Write the complete electron configuration for bismuth, element number 83.
b) Write the condensed electron configuration for this element, showing the appropriate noble-gas core.
c) How many unpaired electrons does each atom of bismuth possess?
Practice Exercise 6.9
Use the periodic table to write the condensed electron configurations for the following atoms:
a) Co (atomic number 27)
b) Te (atomic number 52)
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AP Chemistry
Chapter 6 Electronic Structure of Atoms
Integrative Exercise 6 (p. 243)
Boron, atomic number 5, occurs naturally as two isotopes, 10B and 11B, with natural abundances of 19.9% and
80.1% respectively.
a) In what ways do the two isotopes differ from each other?
Does the electronic configurations of 10B differ from that of
11
B?
b) Draw the orbital diagram for an atom of 11B.
Which electrons are the valence electrons (the ones involved in chemical reactions)?
c)
Indicate three major ways in which the 1s electrons in boron different from the 2s electrons.
d) Elemental boron reacts with fluorine to form BF3, a gas.
Write a balanced chemical equation for the reaction of solid boron with fluorine gas.
f) When BCl3, also a gas at room temperature, comes into contact with water, it reacts to form hydrochloric
acid and boric acid, H3BO3, a very weak acid in water. Write a balanced net ionic equation for this reaction.
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