chem_TE_ch05.fm Page 133 Saturday, April 15, 2006 11:29 AM
5.2
1
Connecting to Your World
Does this scene look natural to
you? Surprisingly, it is. Arrangements like this are rare in nature because
they are unstable. Unstable arrangements, whether the grains of sand in a
sandcastle or the rock formation shown here, tend to become
more stable by losing energy. If this rock were to tumble
over, it would end up at a lower height. It would have
less energy than before, but its position would be
more stable. In this section, you will learn that
energy and stability play an important role in
determining how electrons are configured in an
atom.
Electron Configurations
Try to balance a pencil on its point. Each time you try, the pencil falls over.
At the end of its fall, its energy has decreased. In most natural phenomena,
change proceeds toward the lowest possible energy. In an atom, electrons
and the nucleus interact to make the most stable arrangement possible.
The ways in which electrons are arranged in various orbitals around the
nuclei of atoms are called electron configurations.
Three rules—the aufbau principle, the Pauli exclusion principle,
and Hund’s rule—tell you how to find the electron configurations of atoms.
The three rules are as follows.
Guide for Reading
Key Concepts
• What are the three rules
for writing the electron
configurations of elements?
• Why do actual electron
configurations for some
elements differ from those
assigned using the aufbau
principle?
Vocabulary
electron configurations
aufbau principle
Pauli exclusion principle
Hund’s rule
Reading Strategy
Building Vocabulary As you
read the section, write a definition
of each vocabulary term in your
own words.
6p
5p
5s
5d
5.2.1 Describe how to write the electron configuration for an atom.
5.2.2 Explain why the actual electron configurations for some
elements differ from those predicted by the aufbau principle.
Guide for Reading
L2
Build Vocabulary
Word Forms Explain that a configuration is an arrangement of parts. An
electron configuration is the arrangement of electrons in orbitals around
the nucleus of an atom
L2
2
INSTRUCT
4f
4d
4p
3d
4s
Objectives
SQ3R Survey, Question, Read, Recite,
Review Encourage students to apply
SQ3R as they read this section. Model
how to employ it for the first two pages.
the orbitals of lowest energy first. Look at the aufbau diagram in Figure 5.7.
Each box represents an atomic orbital.
6s
FOCUS
Reading Strategy
Aufbau Principle According to the aufbau principle, electrons occupy
Increasing Energy
5.2
Electron Arrangement in Atoms
3p
3s
2p
2s
1s
Figure 5.7 This aufbau
diagram shows the energy
levels of the various atomic
orbitals. Orbitals of greater
energy are higher on the
diagram. Using Tables
Which is of higher energy, a
4d or a 5s orbital?
Section 5.2 Electron Arrangement in Atoms 133
Because of gravity, rocks, such as the
one shown, are less stable than rocks on
level ground. Explain that electrons have
a comparable ground level. Ask, What
role do energy and stability play in
the way that electrons are configured
in an atom? (In the most stable atoms,
electrons occupy the lowest possible
energy orbitals.)
Electron Configurations
Use Visuals
Section Resources
Print
• Guided Reading and Study Workbook,
Section 5.2
• Core Teaching Resources, Section 5.2
Review
Technology
• Interactive Textbook with ChemASAP,
Simulation 2, Problem-Solving 5.9,
Assessment 5.2
• Go Online, Section 5.2
L1
Figure 5.7 Ask, Do all the orbitals at
energy level n = 4 have the same
energy? (No, 4p orbitals have higher
energy than 4s, and 4d have higher
energy than 4p.) Which has higher
energy, 4s or 3d ? (3d)
• Small-Scale Chemistry Laboratory Manual, Lab 7
• Transparencies, T58–T60
Answers to...
Figure 5.7 4d
Electrons in Atoms
133
chem_TE_ch05.fm Page 134 Thursday, April 14, 2005 7:20 AM
Section 5.2 (continued)
The orbitals for any sublevel of a principal energy level are always of
equal energy. Further, within a principal energy level the s sublevel is
always the lowest-energy sublevel. Yet the range of energy levels within a
principal energy level can overlap the energy levels of another principal
level. Notice again in Figure 5.7 that the filling of atomic orbitals does not
follow a simple pattern beyond the second energy level. For example, the
4s orbital is lower in energy than a 3d orbital.
Pauli Exclusion Principle According to the Pauli exclusion principle, an
atomic orbital may describe at most two electrons. For example, either one
or two electrons can occupy an s orbital or a p orbital. To occupy the same
orbital, two electrons must have opposite spins; that is, the electron spins
must be paired. Spin is a quantum mechanical property of electrons and
may be thought of as clockwise or counterclockwise. A vertical arrow indicates an electron and its direction of spin (↑ or ↓). An orbital containing
paired electrons is written as
.
Download a worksheet on Electron
Configuration for students to complete, and find additional teacher
support from NSTA SciLinks.
L2
Discuss
Hund’s Rule When you use the aufbau diagram to decide how electrons
occupy orbitals of equal energy, one electron enters each orbital until all
the orbitals contain one electron with the same spin direction. Hund’s rule
states that electrons occupy orbitals of the same energy in a way that makes
the number of electrons with the same spin direction as large as possible.
For example, three electrons would occupy three orbitals of equal energy as
follows:
. Second electrons then occupy each orbital so that their
spins are paired with the first electron in the orbital. Thus each orbital can
eventually have two electrons with paired spins.
Develop the electron configurations
for several of the simpler elements.
Introduce each rule governing the process as needed. Begin with hydrogen.
Use the aufbau diagram to explain that
electrons enter orbitals of lowest
energy first. Show how the orbital
notation (1s1) describes the energy
level, the orbital, and the number of
electrons. Repeat the process for
helium. Then continue with lithium,
beryllium, and boron. Apply the Pauli
exclusion principle to explain why
additional orbitals must be used. When
you reach carbon, explain and apply
Hund’s rule. Complete the configurations for the second period elements.
Table 5.3
Electron Configurations for Some Selected Elements
Orbital filling
Element
1s
2s
2px
2py
2pz
3s
Electron
configuration
H
1s1
He
1s 2
Li
1s 22s1
C
1s 22s 22p 2
L2
N
1s 22s 22p 3
Purpose Students gain practice in
determining the correct order for filling orbitals.
Procedure Have students work with a
partner to develop the electron configurations for the third period elements.
When the exercise is complete, ask students to compare the configurations of
second and third period elements in
preparation for future discussions of
the periodic table.
O
CLASS
Activity
Writing Electron
Configurations
Simulation 2 Fill atomic
orbitals to build the ground
state of several atoms.
F
1s 22s 22p 5
Ne
1s 22s 22p 6
Na
1s 22s 22p 63s 1
with ChemASAP
134 Chapter 5
Differentiated Instruction
L3
Gifted and Talented
Students may be interested in learning that
an electron in an orbital is described
precisely by four quantum numbers, the
first being the principle quantum number, n.
Direct them to a college text and ask them
to name and explain the three other
134 Chapter 5
1s 22s 22p 4
quantum numbers. (The azimuthal quantum
number, l, defines the shape of the orbital; the
magnetic quantum number, m, describes the
orientation of the orbital in space; and the
spin quantum number, ms, denotes the
direction of electron spin, + or –.)
chem_TE_ch05.fm Page 135 Thursday, April 14, 2005 7:31 AM
Look at the orbital filling diagrams of the atoms listed in Table 5.3. An
oxygen atom contains eight electrons. The orbital of lowest energy, 1s,
has one electron, then a second electron of opposite spin. The next
orbital to fill is 2s. It also has one electron, then a second electron of
opposite spin. One electron then occupies each of the three 2p orbitals of
equal energy. The remaining electron now pairs with an electron occupying one of the 2p orbitals. The other two 2p orbitals remain only half
filled, with one electron each.
A convenient shorthand method for showing the electron configuration of an atom involves writing the energy level and the symbol for every
sublevel occupied by an electron. You indicate the number of electrons
occupying that sublevel with a superscript. For hydrogen, with one electron in a 1s orbital, the electron configuration is written 1s1. For helium,
with two electrons in a 1s orbital, the configuration is 1s 2. For oxygen,
with two electrons in a 1s orbital, two electrons in a 2s orbital, and four
electrons in 2p orbitals, it is 1s 2 2s 2 2p 4. Note that the sum of the superscripts equals the number of electrons in the atom.
When the configurations are written, the sublevels within the same
principal energy level are generally written together. This is not always
the same order as given on the aufbau diagram. The 3d sublevel, for
example, is written before the 4s sublevel, even though the aufbau diagram shows the 4s sublevel to have lower energy.
For: Links on Electron
Configuration
Visit: www.SciLinks.org
Web Code: cdn-1052
Answers
Writing Electron Configurations
Phosphorus, an element used in matches, has an atomic number
of 15. Write the electron configuration of a phosphorus atom.
Solve Apply concepts to this situation.
Using Figure 5.7 on page 133, place electrons
in the orbital with the lowest energy (1s) first,
then continue placing electrons in each
orbital with the next higher energy.
The electron configuration of phosphorus is
1s 2 2s 2 2p 6 3s 2 3p 3.
The superscripts add up to the number of
electrons. When the configurations are written, the sublevels within the same principal
energy level are written together. This is not
always the same order as given on the aufbau
diagram.
Practice Problems
8. Write the electron configuration for each atom.
a. carbon
b. argon
c. nickel
9. Write the electron configuration for each atom. How
many unpaired electrons does each atom have?
a. boron
b. silicon
Students may ask how two negative
electrons can occupy the same orbital,
or why an orbital never contains more
than two electrons. The answer lies in
the quantum property known as spin.
A spinning electron acts as a tiny magnet, with a north pole at one end and a
south pole at the other. If electrons
with opposite spins enter one orbital,
the attraction between their opposite
magnetic poles counteracts some of
the electrical repulsive force. If a third
electron tries to enter the orbital, its
spin will always be the same as one of
the existing electrons, and it will be
repelled.
CONCEPTUAL PROBLEM 5.1
CONCEPTUAL PROBLEM 5.1
Analyze Identify the relevant concepts.
Phosphorus has 15 electrons. There is a maximum of two electrons per orbital. Electrons
do not pair up within an energy sublevel
(orbitals of equal energy) until each orbital
already has one electron.
L2
Discuss
Problem Solving 5.9
Solve Problem 9 with the help
of an interactive guided tutorial.
with ChemASAP
8. a. 1s22s22p2
b. 1s22s22p63s23p6
c. 1s22s22p63s23p63d84s2
9. a. 1s22s22p1; one unpaired
electron
b. 1s22s22p63s23p2; two unpaired
electrons
Practice Problems Plus
L2
What are the electron configurations
for atoms of the following elements?
How many unpaired electrons does
each atom have?
a. neon (1s22s22p6; no unpaired
electrons) b. sulfur (1s22s22p63s23p4;
two unpaired electrons)
Students may find it helpful to use a
diagonal diagram to write electron configurations until they become familiar
with the order in which sublevels fill.
Section 5.2 Electron Arrangement in Atoms 135
Differentiated Instruction
L1
Encourage students to draw the aufbau diagram (Figure 5.7) on a 5 x 7 card and keep it
handy when writing electron configurations.
Also encourage them to make flash cards,
Special Needs
each with the name and atomic number of
an element on one side and its electron configuration on the other side. Pairs of students
can use these to quiz each other.
Electrons in Atoms
135
chem_TE_ch05.fm Page 136 Saturday, April 15, 2006 11:32 AM
Section 5.2 (continued)
Exceptional Electron Configurations
Copper, shown in Figure 5.8, has an electron configuration that is an exception to the aufbau principle. You can obtain correct electron configurations
for the elements up to vanadium (atomic number 23) by following the
aufbau diagram for orbital filling. If you were to continue in that fashion,
however, you would assign chromium and copper the following incorrect
configurations:
Exceptional Electron
Configurations
L2
Discuss
Among the transition elements there
are some exceptions to the filling rules.
Exceptions can be explained by the
atom’ s tendency to keep its energy as
low as possible. These exceptions help
explain the unexpected chemical
behavior of transition elements.
3
ASSESS
Evaluate Understanding
Cr 1s 2 2s 2 2p 6 3s 2 3p 6 3d 4 4s 2
Cu 1s 2 2s 2 2p 6 3s 2 3p 6 3d 9 4s 2
The correct electron configurations are as follows:
Cr 1s 2 2s 2 2p 6 3s 2 3p 6 3d 5 4s 1
Cu 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 1
Figure 5.8 Copper is a good
conductor of electricity and is
commonly used in electrical
wiring.
L2
Make a set of small cards, each with the
symbol and atomic number of an element. Choose elements from throughout the periodic table. Have students
choose cards and write the electron
configurations for the elements.
L1
Reteach
Have students work in small groups,
using the set of element cards to practice writing more electron configurations. Have them refer to Figure 5.7, if
necessary.
[icon] Writing Activity
Students are likely to write that the
two magnets would push each
other apart. In the same way, electrons with the same spin would
push apart and be unable to occupy
the same orbital.
These arrangements give chromium a half-filled d sublevel and copper a
filled d sublevel. Filled energy sublevels are more stable than partially filled
sublevels.
Some actual electron configurations differ from those
assigned using the aufbau principle because half-filled sublevels are not
as stable as filled sublevels, but they are more stable than other configurations. This tendency overcomes the small difference between the energies of the 3d and 4s sublevels in copper and chromium.
Exceptions to the aufbau principle are due to subtle electron-electron
interactions in orbitals with very similar energies. At higher principal quantum numbers, energy differences between some sublevels (such as 5f and
6d, for example) are even smaller than in the chromium and copper examples. As a result, there are other exceptions to the aufbau principle.
Although it is worth knowing that exceptions to the aufbau principle occur,
it is more important to understand the general rules for determining electron configurations in the many cases where the aufbau rule applies.
Checkpoint How are energy differences and exceptions to the aufbau
principle related?
5.2 Section Assessment
10.
Key Concept What are the three rules for writing the electron configuration of elements?
11.
Key Concept Explain why the actual electron
configurations for some elements differ from those
assigned using the aufbau principle.
12. Use Figure 5.7 to arrange the following sublevels in
Modeling the Pauli Exclusion Principle Write a
brief description of how trying to place two bar
magnets pointing in the same direction alongside
each other is like trying to place two electrons into
the same orbital.
order of decreasing energy: 2p, 4s, 3s, 3d, and 3p.
13. Why does one electron in a potassium atom go
into the fourth energy level instead of squeezing
into the third energy level along with the eight
already there?
Assessment 5.2 Test yourself on
the concepts in Section 5.2.
with ChemASAP
136 Chapter 5
If your class subscribes to the Interactive Textbook, use it to review key
concepts in Section 5.2.
with ChemASAP
Answers to...
Checkpoint Exceptions to the
aufbau principle are due to subtle
electronic interactions in orbitals with
similar energies.
136 Chapter 5
Section 5.2 Assessment
10. aufbau principle, Pauli exclusion principle, Hund’s rule
11. Half-filled sublevels and filled sublevels are
more stable than other configurations.
12. 3d, 4s, 3p, 3s, 2p
13. The 3s and 3p orbitals are already filled,
so the last electron must go to the next
higher energy orbital, which is 4s.
chem_TE_ch05.fm Page 137 Saturday, April 15, 2006 11:36 AM
Small-Scale
LAB
Small-Scale
LAB
Atomic Emission Spectra
L2
Objective After completing this
activity, students will be able to:
• build a simple spectroscope and use
it to determine wavelength, frequency, and energy of emission lines
Atomic Emission Spectra
Purpose
To build a spectroscope and use it to measure the wavelengths, frequencies, and energies of atomic emission
lines
Materials
• cereal box
• tape
• pencil
• black construction paper
• diffraction grating
• ruler
• scissors
• white notebook paper
Procedure
Tape together two 2.0 cm 10 cm strips of black construction paper so that they are parallel and form a narrow
slit about 2 mm wide. Remove the top of a cereal box and
tape the construction paper slit as shown. Cover the rest
of the opening with white notebook paper. Cut a square
hole (approximately 2 cm per side) and tape a diffraction
grating over the hole as shown. Point the spectroscope
toward a fluorescent light. Tape up any light leaks. Your lab
partner should mark the exact positions of all the colored
emission lines you see on the notebook paper. Measure
the distances between the violet line and the other lines
you have marked.
Analyze
1. List the number of distinct lines that you see as well as
their colors.
2. Each line you see has a property called its wavelength.
The prominent violet line has a wavelength of 436 nm
and the prominent green line is 546 nm. How many
mm apart are these lines on the paper? By how many
nm do their wavelengths differ? How many nanometers of wavelength are represented by each millimeter
you measured?
Diffraction grating
Black
construction
paper
Cereal
box
3. Using the nm/mm value you calculated in Step 2 and
the mm distance you measured for each line from the
violet reference line, calculate the wavelengths of all
the other lines you see.
4. Each wavelength corresponds to another property of
light called its frequency. Use the wavelength value of
each line to calculate its frequency given that ν c/ λ
where c 2.998 1017 nm/s (2.998 10 8 m/s).
5. The energy (E) of a quantum of light an atom emits is
related to its frequency (ν) by E h ν. Use the
frequency value for each line and h 6.63 1034 J·s
to calculate its corresponding energy.
You’re The Chemist
1. Design It! Design and carry out an experiment to
measure the longest and shortest wavelengths you can
see in daylight. Use your spectroscope to observe light
from daylight reflected off a white piece of paper.
Caution: Do not look directly at the sun! Describe the
differences in daylight and fluorescent light.
2. Design It! Design and carry out an experiment to
determine the effect of colored filters on the spectrum
of fluorescent light or daylight. For each filter tell which
colors are transmitted and which are absorbed.
3. Analyze It! Use your spectroscope to observe various
atomic emission discharge tubes provided by your
teacher. Note and record the lines you see and measure
their wavelengths.
Notebook
paper
Slit
Spectrum
appears here
Small-Scale Lab 137
You’re The Chemist
1. Sunlight has more red and more violet and
lacks the distinct mercury emission lines of
fluorescent light. Typical measurements are:
Violet line to edge of red = 59 mm; 436 nm +
(59 mm × 3.67 nm/mm) = 653 nm
Violet line to edge of violet = – 4 mm. 436
nm – (4 × 3.67) nm/mm = 421 nm
2. Translucent colored file tabs from an office supply store work well as filters. A yellow filter
transmits green, yellow, orange and red and
absorbs violet and blue. A green filter transmits
blue, green and yellow and absorbs violet,
orange and red. A blue filter transmits violet,
blue and green and absorbs yellow, orange and
red. A red filter transmits yellow, orange and
red and absorbs violet, blue and green.
3. Answers will vary. Provide students with gas
discharge tubes containing hydrogen and
noble gases.
Prep Time 10 minutes
Class Time 20 minutes
Teaching Tips Have a completed spectroscope for students to examine.
Have students bring in their own cereal
boxes. Use holographic diffraction gratings of about 700 lines per millimeter.
The diffraction grating needs to be positioned so that the lines on the grating
are parallel to the slit. Very narrow slits
do not let in enough light to see a
bright spectrum.
Expected Outcome Students calculate
the wavelength, frequency, and energy
of emissions from a fluorescent light.
Analyze
1. Most fluorescent lights display 5 lines:
violet, blue, green, yellow and red. All
fluorescent lights display violet, green
and yellow. Older lights will display
only two distinct violet and blue lines.
A diffuse yellow line is also evident.
2. A typical box will yield a measurement of 30 mm between the violet
and green lines; 546 – 436 = 110 nm;
110 nm/30 mm = 3.67 nm/mm
3.Blue: 436 nm + (16 mm × 3.67 nm/
mm) = 495 nm; Green: 436 nm + (30
mm × 3.67 nm/mm) = 546 nm; Yellow: 436 nm + (42 × 3.67 nm/mm) =
590 nm; Red: 436 nm + (50 × 3.67
nm/mm) = 620 nm
4. Violet: ν = 3.00 × 1017 nm/s /436 nm =
6.88 × 1014 s–1; Blue: ν = 3.00 × 1017
nm/s /495 nm = 6.06 × 1014 s–1; Green:
ν = 3.00 × 1017 nm/s/546 nm = 5.49 ×
1014 s–1; Yellow: ν = 3.00 × 1017 nm/s/
590 nm = 5.08 × 1014 s–1; Red: ν = 3.00
× 1017 nm/s/620 nm = 4.84 × 1014 s–1
5. Violet: 6.63 × 10–34 J⋅s × 6.88 × 1014
s–1 = 4.56 × 10–19 J; Blue: 6.63 × 10–34
J⋅s × 6.06 × 1014 s–1 = 4.02 × 10–19 J;
Green: 6.63 × 10–34 J⋅s × 5.49 × 1014
s–1 = 3.64 × 10–19 J; Yellow: 6.63 ×
10–34 J⋅s × 5.08 × 1014 s–1 = 3.37 ×
10–19 J; Red: 6.63 × 10–34 J⋅s × 4.84 ×
1014 s–1 = 3.21 × 10–19 J
Electrons in Atoms
137