Chemistry

Designating Isotopes

The isotopes of hydrogen are unusual in that they have distinct names. www.scilinks.org
Isotopes are usually identified by specifying their mass number. There Topic: Isotopes
are two methods for specifying isotopes. In the first method, the mass Code: HC60820
number is written with a hyphen after the name of the element. Tritium,
for example, is written as hydrogen-3. We will refer to this method as
hyphen notation. The uranium isotope used as fuel for nuclear power
plants has a mass number of 235 and is therefore known as uranium-
235. The second method shows the composition of a nucleus using the
isotope’s nuclear symbol. For example, uranium-235 is written as 29352U.
The superscript indicates the mass number (protons + neutrons) and
the subscript indicates the atomic number (number of protons). The
number of neutrons is found by subtracting the atomic number from the
mass number.

mass number − atomic number = number of neutrons
235 (protons + neutrons) − 92 protons = 143 neutrons

Thus, a uranium-235 nucleus is made up of 92 protons and 143 neutrons.
Table 3 gives the names, symbols, and compositions of the isotopes of

hydrogen and helium. Nuclide is a general term for a specific isotope of
an element. We could say that Table 3 lists the compositions of five dif-
ferent nuclides, three hydrogen nuclides and two helium nuclides.

TABLE 3 Isotopes of Hydrogen and Helium

Isotope Nuclear Number of Number of Number of
Hydrogen-1 (protium) symbol protons electrons neutrons
Hydrogen-2 (deuterium) 1 1 0
Hydrogen-3 (tritium) 11H 1 1 1
Helium-3 21H 1 1 2
Helium-4 31H 2 2 1
32He 2 2 2
42He

SAMPLE PROBLEM A
How many protons, electrons, and neutrons are there in an atom of chlorine-37?

SOLUTION

1 ANALYZE Given: name and mass number of chlorine-37
Unknown: numbers of protons, electrons, and neutrons

2 PLAN atomic number = number of protons = number of electrons
mass number = number of neutrons + number of protons

A T O M S : T H E B U I L D I N G B L O C K S O F M A T T E R 79

3 COMPUTE The mass number of chlorine-37 is 37. Consulting the periodic table reveals that chlorine’s
atomic number is 17. The number of neutrons can be found by subtracting the atomic
number from the mass number.

mass number of chlorine-37 − atomic number of chlorine =
number of neutrons in chlorine-37

mass number − atomic number = 37 (protons plus neutrons) − 17 protons
= 20 neutrons

An atom of chlorine-37 is made up of 17 electrons, 17 protons, and 20 neutrons.

4 EVALUATE The number of protons in a neutral atom equals the number of electrons. And the sum
of the protons and neutrons equals the given mass number.

PRACTICE Answers in Appendix E

1. How many protons, electrons, and neutrons make up an atom of Go to go.hrw.com for
bromine-80? more practice problems
that ask you to work
2. Write the nuclear symbol for carbon-13. with numbers of
subatomic particles.
3. Write the hyphen notation for the isotope with 15 electrons and
15 neutrons. Keyword: HC6ATMX

Relative Atomic Masses

Masses of atoms expressed in grams are very small. As we shall see,
an atom of oxygen-16, for example, has a mass of 2.656 × 10–23 g. For
most chemical calculations it is more convenient to use relative atomic
masses. As you read in Chapter 2, scientists use standards of measure-
ment that are constant and are the same everywhere. In order to set up
a relative scale of atomic mass, one atom has been arbitrarily chosen as
the standard and assigned a mass value. The masses of all other atoms
are expressed in relation to this defined standard.

The standard used by scientists to compare units of atomic mass is the
carbon-12 atom. It has been arbitrarily assigned a mass of exactly 12 atom-
ic mass units, or 12 amu. One atomic mass unit, or 1 amu, is exactly 1/12
the mass of a carbon-12 atom. The atomic mass of any other atom is
determined by comparing it with the mass of the carbon-12 atom.
The hydrogen-1 atom has an atomic mass of about 1/12 that of the
carbon-12 atom, or about 1 amu. The precise value of the atomic mass
of a hydrogen-1 atom is 1.007 825 amu. An oxygen-16 atom has about
16/12 (or 4/3) the mass of a carbon-12 atom. Careful measurements
show the atomic mass of oxygen-16 to be 15.994 915 amu. The mass
of a magnesium-24 atom is found to be slightly less than twice that of a
carbon-12 atom. Its atomic mass is 23.985 042 amu.

80 C H A P T E R 3

Some additional examples of the atomic masses of the naturally Discovery of Element 43
occurring isotopes of several elements are given in Table 4 on the next
page. Isotopes of an element may occur naturally, or they may be made The discovery of element 43, tech-
in the laboratory (artificial isotopes). Although isotopes have different netium, is credited to Carlo Perrier and
masses, they do not differ significantly in their chemical behavior. Emilio Segrè, who artificially produced
it in 1937. However, in 1925, a German
The masses of subatomic particles can also be expressed on the atom- chemist named Ida Tacke reported the
ic mass scale (see Table 1). The mass of the electron is 0.000 5486 amu, discovery of element 43, which she
that of the proton is 1.007 276 amu, and that of the neutron is called masurium, in niobium ores. At
1.008 665 amu. Note that the proton and neutron masses are close to but the time, her discovery was not accept-
not equal to 1 amu. You have learned that the mass number is the total ed because it was thought technetium
number of protons and neutrons that make up the nucleus of an atom. could not occur naturally. Recent stud-
You can now see that the mass number and relative atomic mass of a ies confirm that Tacke and coworkers
given nuclide are quite close to each other. They are not identical probably did discover element 43.
because the proton and neutron masses deviate slightly from 1 amu and
the atomic masses include electrons. Also, as you will read in Chapter 21,
a small amount of mass is changed to energy in the creation of a nucleus
from its protons and neutrons.

Average Atomic Masses of Elements

Most elements occur naturally as mixtures of isotopes, as indicated in 81
Table 4. The percentage of each isotope in the naturally occurring ele-
ment on Earth is nearly always the same, no matter where the element
is found. The percentage at which each of an element’s isotopes occurs
in nature is taken into account when calculating the element’s average
atomic mass. Average atomic mass is the weighted average of the atom-
ic masses of the naturally occurring isotopes of an element.

The following is a simple example of how to calculate a weighted
average. Suppose you have a box containing two sizes of marbles. If
25% of the marbles have masses of 2.00 g each and 75% have masses of
3.00 g each, how is the weighted average calculated? You could count
the number of each type of marble, calculate the total mass of the mix-
ture, and divide by the total number of marbles. If you had 100 marbles,
the calculations would be as follows.

25 marbles × 2.00 g = 50 g
75 marbles × 3.00 g = 225 g

Adding these masses gives the total mass of the marbles.

50 g + 225 g = 275 g

Dividing the total mass by 100 gives an average marble mass of 2.75 g.
A simpler method is to multiply the mass of each marble by the dec-

imal fraction representing its percentage in the mixture. Then add the
products.

25% = 0.25 75% = 0.75
(2.00 g × 0.25) + (3.00 g × 0.75) = 2.75 g

ATOMS: THE BUILDING BLOCKS OF MATTER

TABLE 4 Atomic Masses and Abundances of Several Naturally Occurring Isotopes

Isotope Mass Percentage natural Atomic mass Average
number abundance (amu) atomic mass
of element (amu)

Hydrogen-1 1 99.9885 1.007 825 1.007 94
Hydrogen-2 2 0.0115 2.014 102

Carbon-12 12 98.93 12 (by definition) 12.0107
Carbon-13 13 1.07 13.003 355

Oxygen-16 16 99.757 15.994 915 15.9994
Oxygen-17 17 0.038 16.999 132
Oxygen-18 18 0.205 17.999 160

Copper-63 63 69.15 62.929 601 63.546
Copper-65 65 30.85 64.927 794

Cesium-133 133 100 132.905 447 132.905

Uranium-234 234 0.0054 234.040 945 238.029
Uranium-235 235 0.7204 235.043 922
Uranium-238 238 99.2742 238.050 784

Calculating Average Atomic Mass

The average atomic mass of an element depends on both the mass and
the relative abundance of each of the element’s isotopes. For example,
naturally occurring copper consists of 69.15% copper-63, which has an
atomic mass of 62.929 601 amu, and 30.85% copper-65, which has an
atomic mass of 64.927 794 amu. The average atomic mass of copper can
be calculated by multiplying the atomic mass of each isotope by its rel-
ative abundance (expressed in decimal form) and adding the results.

0.6915 × 62.929 601 amu + 0.3085 × 64.927 794 amu = 63.55 amu

The calculated average atomic mass of naturally occurring copper is
63.55 amu.

The average atomic mass is included for the elements listed in
Table 4. As illustrated in the table, most atomic masses are known to
four or more significant figures. In this book, an element’s atomic mass is
usually rounded to two decimal places before it is used in a calculation.

Relating Mass to Numbers of Atoms

The relative atomic mass scale makes it possible to know how many
atoms of an element are present in a sample of the element with a mea-
surable mass. Three very important concepts—the mole, Avogadro’s
number, and molar mass—provide the basis for relating masses in
grams to numbers of atoms.

82 C H A P T E R 3

The Mole (a)

The mole is the SI unit for amount of substance. A mole (abbreviated (b)
mol) is the amount of a substance that contains as many particles as
there are atoms in exactly 12 g of carbon-12. The mole is a counting (c)
unit, just like a dozen is. We don’t usually order 12 or 24 ears of corn; FIGURE 10 Shown is approxi-
we order one dozen or two dozen. Similarly, a chemist may want 1 mol mately one molar mass of each
of carbon, or 2 mol of iron, or 2.567 mol of calcium. In the sections that of three elements: (a) carbon
follow, you will see how the mole relates to masses of atoms and (graphite), (b) iron (nails), and
compounds. (c) copper (wire).

Avogadro’s Number

The number of particles in a mole has been experimentally deter-
mined in a number of ways. The best modern value is 6.022 141 79 ×
1023. This means that exactly 12 g of carbon-12 contains 6.022 141 79 ×
1023 carbon-12 atoms. The number of particles in a mole is known as
Avogadro’s number, named for the nineteenth-century Italian scien-
tist Amedeo Avogadro, whose ideas were crucial in explaining the
relationship between mass and numbers of atoms. Avogadro’s num-
ber—6.022 141 79 × 1023—is the number of particles in exactly one
mole of a pure substance. For most purposes, Avogadro’s number is
rounded to 6.022 × 1023.

To get a sense of how large Avogadro’s number is, consider the fol-
lowing: If every person living on Earth (6 billion people) worked to
count the atoms in one mole of an element, and if each person counted
continuously at a rate of one atom per second, it would take about
3 million years for all the atoms to be counted.

Molar Mass

An alternative definition of mole is the amount of a substance that con-
tains Avogadro’s number of particles. Can you figure out the approxi-
mate mass of one mole of helium atoms? You know that a mole of
carbon-12 atoms has a mass of exactly 12 g and that a carbon-12 atom
has an atomic mass of 12 amu. The atomic mass of a helium atom is
4.00 amu, which is about one-third the mass of a carbon-12 atom. It fol-
lows that a mole of helium atoms will have about one-third the mass of
a mole of carbon-12 atoms. Thus, one mole of helium has a mass of
about 4.00 g.

The mass of one mole of a pure substance is called the molar mass of
that substance. Molar mass is usually written in units of g/mol. The molar
mass of an element is numerically equal to the atomic mass of the ele-
ment in atomic mass units (which can be found in the periodic table). For
example, the molar mass of lithium, Li, is 6.94 g/mol, while the molar
mass of mercury, Hg, is 200.59 g/mol (rounding each value to two deci-
mal places).

The molar mass of an element contains one mole of atoms. For exam-
ple, 4.00 g of helium, 6.94 g of lithium, and 200.59 g of mercury all con-
tain a mole of atoms. Figure 10 shows molar masses of three common
elements.

A T O M S : T H E B U I L D I N G B L O C K S O F M A T T E R 83

Mass of element molar mass Amount ‫ ؍‬1 mol ؋ Number of atoms
in grams ‫ ؍‬of element ؋ of element 6.022 ؋ 1023 atoms of element

1 mol in moles ؋ 6.022 ؋ 1023 atoms ‫؍‬
1 mol
1 mol
؋ molar mass ‫؍‬

of element

FIGURE 11 The diagram shows Gram/Mole Conversions
the relationship between mass in
grams, amount in moles, and number Chemists use molar mass as a conversion factor in chemical calcula-
of atoms of an element in a sample. tions. For example, the molar mass of helium is 4.00 g He/mol He. To
find how many grams of helium there are in two moles of helium, mul-
tiply by the molar mass.

2.00 mol He × 4.00 g He = 8.00 g He
1 mol He

Figure 11 shows how to use molar mass, moles, and Avogadro’s number
to relate mass in grams, amount in moles, and number of atoms of an
element.

SAMPLE PROBLEM B For more help, go to the Math Tutor at the end of this chapter.

What is the mass in grams of 3.50 mol of the element copper, Cu?

SOLUTION

1 ANALYZE Given: 3.50 mol Cu
Unknown: mass of Cu in grams

2 PLAN amount of Cu in moles ⎯→ mass of Cu in grams

According to Figure 11, the mass of an element in grams can be calculated by multiplying
the amount of the element in moles by the element’s molar mass.

3 COMPUTE moles Cu × grams Cu = grams Cu
moles Cu

The molar mass of copper from the periodic table is rounded to 63.55 g/mol.

3.50 mol Cu × 63.55 g Cu = 222 g Cu

1 mol Cu

4 EVALUATE Because the amount of copper in moles was given to three significant figures, the answer
was rounded to three significant figures. The size of the answer is reasonable because it is
somewhat more than 3.5 times 60.

84 C H A P T E R 3

PRACTICE Answers in Appendix E

1. What is the mass in grams of 2.25 mol of the element iron, Fe?

2. What is the mass in grams of 0.375 mol of the element potassium, K? Go to go.hrw.com for
3. What is the mass in grams of 0.0135 mol of the element sodium, Na? more practice problems
4. What is the mass in grams of 16.3 mol of the element nickel, Ni? that ask you to convert
from amount in moles
to mass.

Keyword: HC6ATMX

SAMPLE PROBLEM C For more help, go to the Math Tutor at the end of this chapter.

A chemist produced 11.9 g of aluminum, Al. How many moles of aluminum were produced?

SOLUTION

1 ANALYZE Given: 11.9 g Al
Unknown: amount of Al in moles

2 PLAN mass of Al in grams ⎯→ amount of Al in moles

As shown in Figure 11, amount in moles can be obtained by dividing mass in grams by
molar mass, which is mathematically the same as multiplying mass in grams by the reciprocal
of molar mass.

grams Al × moles Al = moles Al

grams Al

3 COMPUTE The molar mass of aluminum from the periodic table is rounded to 26.98 g/mol.

11.9 g Al × 1 mol Al = 0.441 mol Al
26.98 g Al

4 EVALUATE The answer is correctly given to three significant figures. The answer is reasonable because
11.9 g is somewhat less than half of 26.98 g.

PRACTICE Answers in Appendix E

1. How many moles of calcium, Ca, are in 5.00 g of calcium? Go to go.hrw.com for
2. How many moles of gold, Au, are in 3.60 × 10−5 g of gold? more practice problems
3. How many moles of zinc, Zn, are in 0.535 g of zinc? that ask you to convert
from mass to amount in
moles.

Keyword: HC6ATMX

A T O M S : T H E B U I L D I N G B L O C K S O F M A T T E R 85

Conversions with Avogadro’s Number

Figure 11 shows that Avogadro’s number can be used to find the num-
ber of atoms of an element from the amount in moles or to find the
amount of an element in moles from the number of atoms. While these
types of problems are less common in chemistry than converting
between amount in moles and mass in grams, they are useful in demon-
strating the meaning of Avogadro’s number. Note that in these calcula-
tions, Avogadro’s number is expressed in units of atoms per mole.

SAMPLE PROBLEM D For more help, go to the Math Tutor at the end of this chapter.

How many moles of silver, Ag, are in 3.01 ؋ 1023 atoms of silver?

SOLUTION

1 ANALYZE Given: 3.01 × 1023 atoms of Ag
Unknown: amount of Ag in moles

2 PLAN number of atoms of Ag ⎯→ amount of Ag in moles

From Figure 11, we know that number of atoms is converted to amount in moles by
dividing by Avogadro’s number. This is equivalent to multiplying numbers of atoms
by the reciprocal of Avogadro’s number.

Ag atoms × moles Ag = moles Ag

Avogadro’s number of Ag atoms

3 COMPUTE 3.01 × 1023 Ag atoms × 1 mol Ag = 0.500 mol Ag
4 EVALUATE
6.022 × 1023 Ag atoms

The answer is correct—units cancel correctly and the number of atoms is one-half of
Avogadro’s number.

PRACTICE Answers in Appendix E

1. How many moles of lead, Pb, are in 1.50 × 1012 atoms of lead? Go to go.hrw.com for
2. How many moles of tin, Sn, are in 2500 atoms of tin? more practice problems
3. How many atoms of aluminum, Al, are in 2.75 mol of aluminum? that ask you to convert
between atoms and
moles.

Keyword: HC6ATMX

SAMPLE PROBLEM E For more help, go to the Math Tutor at the end of this chapter.

What is the mass in grams of 1.20 ؋ 108 atoms of copper, Cu?

SOLUTION

1 ANALYZE Given: 1.20 × 108 atoms of Cu
Unknown: mass of Cu in grams

86 C H A P T E R 3

2 PLAN number of atoms of Cu ⎯→ amount of Cu in moles ⎯→ mass of Cu in grams

As indicated in Figure 11, the given number of atoms must first be converted to amount in
moles by dividing by Avogadro’s number. Amount in moles is then multiplied by molar
mass to yield mass in grams.

Cu atoms × moles Cu × grams Cu = grams Cu

Avogadro’s number of Cu atoms moles Cu

3 COMPUTE The molar mass of copper from the periodic table is rounded to 63.55 g/mol.

1.20 × 108 Cu atoms × 1 mol Cu × 63.55 g Cu = 1.27 × 10−14 g Cu

6.022 × 1023 Cu atoms 1 mol Cu

4 EVALUATE Units cancel correctly to give the answer in grams. The size of the answer is reasonable—
108 has been divided by about 1024 and multiplied by about 102.

PRACTICE Answers in Appendix E

1. What is the mass in grams of 7.5 × 1015 atoms of nickel, Ni? Go to go.hrw.com for
2. How many atoms of sulfur, S, are in 4.00 g of sulfur? more practice problems
that ask you to convert
3. What mass of gold, Au, contains the same number of atoms as among atoms, grams,
9.0 g of aluminum, Al? and moles.

Keyword: HC6ATMX

SECTION REVIEW 4. To two decimal places, what is the relative atomic
mass and the molar mass of the element potas-
1. Define each of the following: sium, K?

a. atomic number e. mole 5. Determine the mass in grams of the following:

b. mass number f. Avogadro’s number a. 2.00 mol N
b. 3.01 × 1023 atoms Cl
c. relative atomic mass g. molar mass 6. Determine the amount in moles of the following:

d. average atomic mass h. isotope a. 12.15 g Mg
b. 1.50 × 1023 atoms F
2. Determine the number of protons, electrons, and
neutrons in each of the following isotopes: Critical Thinking

a. sodium-23 c. 6249Cu 7. ANALYZING DATA Beaker A contains 2.06 mol of
copper, and Beaker B contains 222 grams of silver.
b. calcium-40 d. 10487Ag Which beaker contains the larger mass? Which
beaker has the larger number of atoms?
3. Write the nuclear symbol and hyphen notation for

each of the following isotopes:

a. mass number of 28 and atomic number of 14

b. 26 protons and 30 neutrons

A T O M S : T H E B U I L D I N G B L O C K S O F M A T T E R 87

CHAPTER HIGHLIGHTS

The Atom: From Philosophical Idea to Scientific Theory

Vocabulary • The idea of atoms has been around since the time of the
law of conservation of mass
law of definite proportions ancient Greeks. In the nineteenth century, John Dalton pro-
law of multiple proportions posed a scientific theory of atoms that can still be used to
explain properties of most chemicals today.

• Matter and its mass cannot be created or destroyed in chemi-

cal reactions.

• The mass ratios of the elements that make up a given com-

pound are always the same, regardless of how much of the
compound there is or how it was formed.

• If two or more different compounds are composed of the same

two elements, then the ratio of the masses of the second el-
ement combined with a certain mass of the first element can
be expressed as a ratio of small whole numbers.

The Structure of the Atom • Cathode-ray tubes supplied evidence of the existence of elec-

Vocabulary trons, which are negatively charged subatomic particles that
atom
nuclear forces have relatively little mass.

• Rutherford found evidence for the existence of the atomic

nucleus by bombarding gold foil with a beam of positively

charged particles.

• Atomic nuclei are composed of protons, which have an electric

charge of +1, and (in all but one case) neutrons, which have no

electric charge.

• Atomic nuclei have radii of about 0.001 pm (pm = picometers;

1 pm × 10−12 m), and atoms have radii of about 40–270 pm.

Counting Atoms • The atomic number of an element is equal to the number of

Vocabulary protons of an atom of that element.
atomic number
isotope • The mass number is equal to the total number of protons and
mass number
nuclide neutrons that make up the nucleus of an atom of that element.
atomic mass unit
average atomic mass • The relative atomic mass unit (amu) is based on the carbon-12
mole
Avogadro’s number atom and is a convenient unit for measuring the mass of atoms.
molar mass It equals 1.660 540 × 10−24 g.

• The average atomic mass of an element is found by calculating

the weighted average of the atomic masses of the naturally

occurring isotopes of the element.

• Avogadro’s number is equal to approximately 6.022 × 1023. A

sample that contains a number of particles equal to Avogadro’s

number contains a mole of those particles.

88 C H A P T E R 3

CHAPTER REVIEW

For more practice, go to the Problem Bank in Appendix D. 9. a. What is the atomic number of an element?

The Atom: From Philosophical b. What is the mass number of an isotope?
Idea to Scientific Theory
c. In the nuclear symbol for deuterium, 12H,
SECTION 1 REVIEW identify the atomic number and the mass

1. Explain each of the following in terms of number.
Dalton’s atomic theory:
a. the law of conservation of mass 10. What is a nuclide?
b. the law of definite proportions
c. the law of multiple proportions 11. Use the periodic table and the information that

2. According to the law of conservation of mass, if follows to write the hyphen notation for each
element A has an atomic mass of 2 mass units
and element B has an atomic mass of 3 mass isotope described.
units, what mass would be expected for com-
pound AB? for compound A2B3? a. atomic number = 2, mass number = 4

The Structure of the Atom b. atomic number = 8, mass number = 16

SECTION 2 REVIEW c. atomic number = 19, mass number = 39

3. a. What is an atom? 12. a. What nuclide is used as the standard in the
b. What two regions make up all atoms?
relative scale for atomic masses?
4. Describe at least four properties of electrons
that were determined based on the experiments b. What is its assigned atomic mass?
of Thomson and Millikan.
13. What is the atomic mass of an atom if its mass is
5. Summarize Rutherford’s model of the atom, and
explain how he developed this model based on approximately equal to the following?
the results of his famous gold-foil experiment. ⎯1⎯
a. 3 that of carbon-12
6. What number uniquely identifies an element?
b. 4.5 times as much as carbon-12

14. a. What is the definition of a mole?

b. What is the abbreviation for mole?

c. How many particles are in one mole?

d. What name is given to the number of parti-

cles in a mole?

15. a. What is the molar mass of an element?

b. To two decimal places, write the molar mass-

es of carbon, neon, iron, and uranium.

16. Suppose you have a sample of an element.

a. How is the mass in grams of the element

Counting Atoms converted to amount in moles?

b. How is the mass in grams of the element

converted to number of atoms?

SECTION 3 REVIEW PRACTICE PROBLEMS

7. a. What are isotopes? 17. What is the mass in grams of each of the follow-
b. How are the isotopes of a particular element
alike? ing? (Hint: See Sample Problems B and E.)
c. How are they different?
a. 1.00 mol Li d. 1.00 molar mass Fe
8. Copy and complete the following table concern-
ing the three isotopes of silicon, Si. b. 1.00 mol Al e. 6.022 × 1023 atoms C
(Hint: See Sample Problem A.)
c. 1.00 molar mass Ca f. 6.022 × 1023 atoms Ag

18. How many moles of atoms are there in each of

Number Number of Number of the following? (Hint: See Sample Problems C
of protons electrons neutrons
Isotope and D.)
c. 3.25 × 105 g Pb
Si-28 a. 6.022 × 1023 atoms Ne d. 4.50 × 10−12 g O
Si-29 b. 3.011 × 1023 atoms Mg
Si-30

A T O M S : T H E B U I L D I N G B L O C K S O F M A T T E R 89

CHAPTER REVIEW

19. Three isotopes of argon occur in nature—1368Ar, 25. Copy and complete the following table concern-
1388Ar, and 1408Ar. Calculate the average atomic ing the properties of subatomic particles.
mass of argon to two decimal places, given the
Mass Actual Relative
following relative atomic masses and abun- Particle Symbol number mass charge

dances of each of the isotopes: argon-36 Electron
Proton
(35.97 amu; 0.337%), argon-38 (37.96 amu; Neutron

0.063%), and argon-40 (39.96 amu; 99.600%).

20. Naturally occurring boron is 80.20% boron-11

(atomic mass = 11.01 amu) and 19.80% of some

other isotopic form of boron. What must the 26. a. How is an atomic mass unit (amu) related to

atomic mass of this second isotope be in order the mass of one carbon-12 atom?

to account for the 10.81 amu average atomic b. What is the relative atomic mass of an atom?

mass of boron? (Write the answer to two deci- 27. a. What is the nucleus of an atom?

mal places.) b. Who is credited with the discovery of the

21. How many atoms are there in each of the atomic nucleus?

following? c. Identify the two kinds of particles that make

a. 1.50 mol Na c. 7.02 g Si up the nucleus.

b. 6.755 mol Pb 28. How many moles of atoms are there in each of

22. What is the mass in grams of each of the the following?

following? a. 40.1 g Ca e. 2.65 g Fe

a. 3.011 × 1023 atoms F e. 25 atoms W b. 11.5 g Na f. 0.007 50 g Ag
b. 1.50 × 1023 atoms Mg f. 1 atom Au c. 5.87 g Ni g. 2.25 × 1025 atoms Zn
c. 4.50 × 1012 atoms Cl
d. 8.42 × 1018 atoms Br d. 150 g S h. 50 atoms Ba

29. State the law of multiple proportions, and give

23. Determine the number of atoms in each of the an example of two compounds that illustrate

following: the law.

a. 5.40 g B d. 0.025 50 g Pt 30. What is the approximate atomic mass of an
b. 0.250 mol S e. 1.00 × 10−10 g Au
atom if its mass is

c. 0.0384 mol K a. 12 times that of carbon-12?
1⎯⎯
b. 2 that of carbon-12?

31. What is an electron?

MIXED REVIEW CRITICAL THINKING

24. Determine the mass in grams of each of the 32. Organizing Ideas Using two chemical com-
following: pounds as an example, describe the difference
a. 3.00 mol Al between the law of definite proportions and the
b. 2.56 × 1024 atoms Li law of multiple proportions.
c. 1.38 mol N
d. 4.86 × 1024 atoms Au 33. Constructing Models As described in
e. 6.50 mol Cu Section 2, the structure of the atom was deter-
f. 2.57 × 108 mol S mined from observations made in painstaking
g. 1.05 × 1018 atoms Hg experimental research. Suppose a series of
experiments revealed that when an electric cur-
rent is passed through gas at low pressure, the
surface of the cathode-ray tube opposite the

90 C H A P T E R 3

CHAPTER REVIEW

anode glows. In addition, a paddle wheel placed ALTERNATIVE ASSESSMENT
in the tube rolls from the anode toward the
cathode when the current is on. 40. Observe a cathode-ray tube in operation, and
a. In which direction do particles pass through write a description of your observations.

the gas? 41. Performance Assessment Using colored clay,
b. What charge do the particles possess? build a model of the nucleus of each of carbon’s
three naturally occurring isotopes: carbon-12,
34. Analyzing Data Osmium is the element with carbon-13, and carbon-14. Specify the number
the greatest density, 22.58 g/cm3. How does the of electrons that would surround each nucleus.
density of osmium compare to the density of a
typical nucleus of 2 × 108 metric tons/cm3?
(1 metric ton = 1000 kg)

USING THE HANDBOOK Graphing Calculator Calculating Numbers
of Protons, Electrons, and Neutrons
35. Group 14 of the Elements Handbook describes Go to go.hrw.com for a graphing calculator
the reactions that produce CO and CO2. exercise that asks you to calculate numbers
Review this section to answer the following: of protons, electrons, and neutrons.
a. When a fuel burns, what determines whether
CO or CO2 will be produced? Keyword: HC6ATMX
b. What happens in the body if hemoglobin
picks up CO?
c. Why is CO poisoning most likely to occur in
homes that are well sealed during cold winter
months?

RESEARCH & WRITING

36. Prepare a report on the series of experiments
conducted by Sir James Chadwick that led to
the discovery of the neutron.

37. Write a report on the contributions of Amedeo
Avogadro that led to the determination of the
value of Avogadro’s number.

38. Trace the development of the electron micro-
scope, and cite some of its many uses.

39. The study of atomic structure and the nucleus
produced a new field of medicine called nuclear
medicine. Describe the use of radioactive
tracers to detect and treat diseases.

A T O M S : T H E B U I L D I N G B L O C K S O F M A T T E R 91

Math Tutor CONVERSION FACTORS

Most calculations in chemistry require that all measurements of the same quantity
(mass, length, volume, temperature, and so on) be expressed in the same unit. To
change the units of a quantity, you can multiply the quantity by a conversion factor.
With SI units, such conversions are easy because units of the same quantity are related
by multiples of 10, 100, 1000, or 1 million. Suppose you want to convert a given
amount in milliliters to liters. You can use the relationship 1 L = 1000 mL. From this
relationship, you can derive the following conversion factors.

1⎯000 mL and ⎯1 L
1 L 1000 mL

The correct strategy is to multiply the given amount (in mL) by the conversion factor
that allows milliliter units to cancel out and liter units to remain. Using the second
conversion factor will give you the units you want.

These conversion factors are based on an exact definition (1000 mL = 1 L exactly),
so significant figures do not apply to these factors. The number of significant figures in
a converted measurement depends on the certainty of the measurement you start with.

SAMPLE 1 SAMPLE 2
A sample of aluminum has a mass of 0.087 g.
What is the sample’s mass in milligrams? A sample of a mineral has 4.08 ؋ 10؊5 mol of

Based on SI prefixes, you know that 1 g = vanadium per kilogram of mass. How many micro-
1000 mg. Therefore, the possible conversion fac-
tors are moles of vanadium per kilogram does the mineral

1⎯000 mg and ⎯1 g contain? ⎯1⎯
1 g 1000 mg 1 000 000
The prefix micro- specifies or 1 × 10−6 of
The first conversion factor cancels grams, leav-
ing milligrams. the base unit.

0.087 g × 1⎯000 mg = 87 mg So, 1 mmol = 1 × 10−6 mol. The possible conver-
1g
sion factors are
Notice that the values 0.087 g and 87 mg each
have two significant figures. 1⎯×11m0m−⎯6oml ol and 1⎯× 10−⎯6 mol
1 mmol
PRACTICE PROBLEMS
1. Express each of the following measurements The first conversion factor will allow moles to
in the units indicated. cancel and micromoles to remain.
a. 2250 mg in grams
b. 59.3 kL in liters 4.08 × 10−5 mol × 1⎯×11m0m−⎯6oml ol = 40.8 mmol

Notice that the values 4.08 × 10−5 mol and
40.8 mmol each have three significant figures.

2. Use scientific notation to express each of the
following measurements in the units indicated.

a. 0.000 072 g in micrograms
b. 3.98 × 106 m in kilometers

92 C H A P T E R 3

Standardized Test Prep

Answer the following items on a separate piece of paper. 7.How many neutrons are present in an atom of
tin that has an atomic number of 50 and a mass
MULTIPLE CHOICE number of 119?
A. 50
1.A chemical compound always has the same el- B. 69
ements in the same proportions by mass regard- C. 119
less of the source of the compound. This is a D. 169
statement of
A. the law of multiple proportions. 8.What is the mass of 1.50 mol of sodium, Na?
B. the law of isotopes. A. 0.652 g
C. the law of definite proportions. B. 0.478 g
D. the law of conservation of mass. C. 11.0 g
D. 34.5 g
2.An important result of Rutherford’s experi-
ments with gold foil was to establish that 9.How many moles of carbon are in a 28.0 g
A. atoms have mass. sample?
B. electrons have a negative charge. A. 336 mol
C. neutrons are uncharged particles. B. 72.0 mol
D. the atom is mostly empty space. C. 2.33 mol
D. 0.500 mol
3.Which subatomic particle has a charge of +1?
A. electron SHORT ANSWER
B. neutron
C. proton 10.Which atom has more neutrons, potassium-40 or
D. meson argon-40?

4.Which particle has the least mass? 11.What is the mass of 1.20 × 1023 atoms of
A. electron phosphorus?
B. neutron
C. proton EXTENDED RESPONSE
D. All have the same mass.
12.Cathode rays emitted by a piece of silver and a
5.Cathode rays are composed of piece of copper illustrate identical properties.
A. alpha particles. What is the significance of this observation?
B. electrons.
C. protons. 13.A student believed that she had discovered a
D. neutrons. new element and named it mythium. Analysis
found it contained two isotopes. The composi-
6. The atomic number of an element is the same tion of the isotopes was 19.9% of atomic mass
as the number of 10.013 and 80.1% of atomic mass 11.009. What
A. protons. is the average atomic mass, and do you think
B. neutrons. mythium was a new element?
C. protons + electrons.
D. protons + neutrons.

Choose the best possible answer
for each question, even if you think there is another
possible answer that is not given.

A T O M S : T H E B U I L D I N G B L O C K S O F M A T T E R 93

CHAPTER LAB ? INQUIRY MICRO-
LAB
LAB

Conservation of Mass

OBJECTIVES BACKGROUND
• Observe the signs of a chemical reaction. The law of conservation of mass states that matter
• Compare masses of reactants and products. is neither created nor destroyed during a chemical
• Design experiments. reaction. Therefore, the mass of a system should
• Relate observations to the law of remain constant during any chemical process. In this
experiment, you will determine whether mass is con-
conservation of mass. served by examining a simple chemical reaction and
MATERIALS comparing the mass of the system before the reac-
• 2 L plastic soda bottle tion with its mass after the reaction.
• 5% acetic acid solution (vinegar)
• balance SAFETY
• clear plastic cups, 2
• graduated cylinder For review of safety, please see Safety in the
• hook-insert cap for bottle Chemistry Laboratory in the front of your book.
• microplunger
• sodium hydrogen carbonate (baking soda) PREPARATION
1. Make two data tables in your lab notebook, one
FIGURE A Slowly add the vinegar to prevent the for Part I and another for Part II. In each table,
reaction from getting out of control. create three columns labeled “Initial mass (g),”
“Final mass (g),” and “Change in mass (g).”
94 C H A P T E R 3 Each table should also have space for observa-
tions of the reaction.

PROCEDURE—PART I
1. Obtain a microplunger, and tap it down into
a sample of baking soda until the bulb end is
packed with a plug of the powder (4–5 mL of
baking soda should be enough to pack the bulb).

2. Hold the microplunger over a plastic cup, and
squeeze the sides of the microplunger to loosen
the plug of baking soda so that it falls into the cup.

3. Use a graduated cylinder to measure 100 mL
of vinegar, and pour it into a second plastic cup.

4. Place the two cups side by side on the balance
pan, and measure the total mass of the system

(before reaction) to the nearest 0.01 g. Record ANALYSIS AND INTERPRETATION—
the mass in your data table. PART I

5. Add the vinegar to the baking soda a little at 1. Drawing Conclusions: What evidence was there
a time to prevent the reaction from getting out that a chemical reaction occurred?
of control, as shown in Figure A. Allow the
vinegar to slowly run down the inside of the 2. Organizing Data: How did the final mass of the
cup. Observe and record your observations system compare with the initial mass of the
about the reaction. system?

6. When the reaction is complete, place both cups 3. Resolving Discrepancies: Does your answer
on the balance, and determine the total final mass to the previous question show that the law
of the system to the nearest 0.01 g. Calculate any of conservation of mass was violated? (Hint:
change in mass. Record both the final mass and Another way to express the law of conservation
any change in mass in your data table. of mass is to say that the mass of all of the
products equals the mass of all of the reactants.)
7. Examine the plastic bottle and the hook-insert What do you think might cause the mass
cap. Try to develop a modified procedure that difference?
will test the law of conservation of mass more
accurately than the procedure in Part I. ANALYSIS AND INTERPRETATION—
PART II
8. In your notebook, write the answers to items 1
through 3 in Analysis and Interpretation—Part I. 1. Drawing Conclusions: Was there any new
evidence in Part II indicating that a chemical
PROCEDURE—PART II reaction occurred?
9. Your teacher should approve the procedure you
designed in Procedure—Part I, step 7. Implement 2. Organizing Ideas: Identify the state of matter
your procedure with the same chemicals and for each reactant in Part II. Identify the state
quantities you used in Part I, but use the bottle of matter for each product.
and hook-insert cap in place of the two cups.
Record your data in your data table. CONCLUSIONS
1. Relating Ideas: What is the difference between
10. If you were successful in step 9 and your results the system in Part I and the system in Part II?
reflect the conservation of mass, proceed to What change led to the improved results in
complete the experiment. If not, find a lab group Part II?
that was successful, and discuss with them what
they did and why they did it. Your group should 2. Evaluating Methods: Why did the procedure for
then test the other group’s procedure to deter- Part II work better than the procedure for Part I?
mine whether their results are reproducible.
EXTENSIONS
CLEANUP AND DISPOSAL 1. Applying Models: When a log burns, the result-
11. Clean your lab station. Clean all equip- ing ash obviously has less mass than the
unburned log did. Explain whether this loss of
ment, and return it to its proper place. mass violates the law of conservation of mass.
Dispose of chemicals and solutions in
the containers designated by your teacher. 2. Designing Experiments: Design a procedure that
Do not pour any chemicals down the drain would test the law of conservation of mass for
or throw anything in the trash unless your the burning log described in Extension item 1.
teacher directs you to do so. Wash your hands
thoroughly after all work is finished and before
you leave the lab.

A T O M S : T H E B U I L D I N G B L O C K S O F M A T T E R 95

CHAPTER 4

Arrangement of
Electrons in Atoms

The emission of light is fundamentally
related to the behavior of electrons.

Neon Walkway

The Development of a SECTION 1
New Atomic Model
OBJECTIVES
T he Rutherford model of the atom was an improvement over previ-
Explain the mathematical
ous models, but it was incomplete. It did not explain how the atom’s relationship among the speed,
negatively charged electrons are distributed in the space surrounding its wavelength, and frequency of
positively charged nucleus. After all, it was well known that oppositely electromagnetic radiation.
charged particles attract each other. So what prevented the negative
electrons from being drawn into the positive nucleus? Discuss the dual wave-particle
nature of light.
In the early twentieth century, a new atomic model evolved as a
result of investigations into the absorption and emission of light by mat- Discuss the significance of
ter. The studies revealed a relationship between light and an atom’s the photoelectric effect and
electrons.This new understanding led directly to a revolutionary view of the line-emission spectrum of
the nature of energy, matter, and atomic structure. hydrogen to the development
of the atomic model.
Properties of Light
Describe the Bohr model
of the hydrogen atom.

Before 1900, scientists thought light behaved solely as a wave. This
belief changed when it was later discovered that light also has particle-
like characteristics. Still, many of light’s properties can be described in
terms of waves. A quick review of these wavelike properties will help
you understand the basic theory of light as it existed at the beginning of
the twentieth century.

The Wave Description of Light www.scilinks.org
Topic: Electromagnetic
Visible light is a kind of electromagnetic radiation, which is a form of
energy that exhibits wavelike behavior as it travels through space. Other Spectrum
kinds of electromagnetic radiation include X rays, ultraviolet and Code: HC60482
infrared light, microwaves, and radio waves. Together, all the forms of
electromagnetic radiation form the electromagnetic spectrum. The elec-
tromagnetic spectrum is represented in Figure 1 on the next page.
All forms of electromagnetic radiation move at a constant speed of
3.00 × 108 meters per second (m/s) through a vacuum and at slightly
slower speeds through matter. Because air is mostly empty space, the
value of 3.00 × 108 m/s is also light’s approximate speed through air.

The significant feature of wave motion is its repetitive nature, which
can be characterized by the measurable properties of wavelength and
frequency. Wavelength (λ) is the distance between corresponding points
on adjacent waves. The unit for wavelength is a distance unit. Depending
on the type of electromagnetic radiation, it may be expressed in meters,

ARRANGEMENT OF ELECTRONS IN ATOMS 97

Visible spectrum FIGURE 1 Electromagnetic
radiation travels in the form of
Violet Blue Green Yellow Orange Red waves covering a wide range of
600 nm 700 nm wavelengths and frequencies.
400 nm 500 nm This range is known as the elec-
tromagnetic spectrum. Only a
small portion of the spectrum,
from 400 nm to 700 nm, is visi-
ble to the human eye.

␥ rays X rays Ultraviolet Infrared Microwave Radio waves

Radar TV Short Long
wave wave

FM

10–2 nm 10–1 nm 100 nm 101 nm 102 nm 103 nm 10–3 cm 10–2 cm 10–1 cm 100 cm 101 cm 1 m 101 m 102 m 103 m 104 m

Wavelength, ␭

1019 Hz 1018 Hz 1017 Hz 1016 Hz 1015 Hz 1014 Hz 1013 Hz 1012 Hz 1011 Hz 1010 Hz 109 Hz 100 MHz 10 MHz 1 MHz 100 KHz

Frequency, ν

Electromagnetic spectrum

centimeters, or nanometers, as shown in Figure 1. Frequency (ν) is
defined as the number of waves that pass a given point in a specific time,
usually one second. Frequency is expressed in waves/second. One
wave/second is called a hertz (Hz), named for Heinrich Hertz, who was
a pioneer in the study of electromagnetic radiation. Figure 2 illustrates
the properties of wavelength and frequency for a familiar kind of wave,
a wave on the surface of water. The wave in Figure 2a has a longer wave-
length and a lower frequency than the wave in Figure 2b.

To the beach

λ

FIGURE 2 The distance between (a) λ
any two corresponding points on (b)
one of these water waves, such as
from crest to crest, is the wave’s
wavelength, λ. We can measure the
wave’s frequency, ν, by observing
how often the water level rises and
falls at a given point, such as at the
post.

98 C H A P T E R 4

Frequency and wavelength are mathematically related to each other. Light
For electromagnetic radiation, this relationship is written as follows.

c = λν

In the equation, c is the speed of light (in m/s), λ is the wavelength of
the electromagnetic wave (in m), and ν is the frequency of the electro-
magnetic wave (in s−1). Because c is the same for all electromagnetic
radiation, the product λν is a constant. Consequently, we know that λ is
inversely proportional to ν. In other words, as the wavelength of light
decreases, its frequency increases, and vice versa.

The Photoelectric Effect Stream of electrons
Anode
In the early 1900s, scientists conducted two experiments involving inter- Cathode
actions of light and matter that could not be explained by the wave theory
of light. One experiment involved a phenomenon known as the photo- (metal plate)
electric effect. The photoelectric effect refers to the emission of electrons Voltage source
from a metal when light shines on the metal, as illustrated in Figure 3.
FIGURE 3 The photoelectric
The mystery of the photoelectric effect involved the frequency of the effect: electromagnetic radiation
light striking the metal. For a given metal, no electrons were emitted if strikes the surface of the metal,
the light’s frequency was below a certain minimum—regardless of the ejecting electrons from the metal
light’s intensity. Light was known to be a form of energy, capable of and causing an electric current.
knocking loose an electron from a metal. But the wave theory of light
predicted that light of any frequency could supply enough energy to
eject an electron. Scientists couldn’t explain why the light had to be of
a minimum frequency in order for the photoelectric effect to occur.

The Particle Description of Light

The explanation of the photoelectric effect dates back to 1900, when
German physicist Max Planck was studying the emission of light by hot
objects. He proposed that a hot object does not emit electromagnetic
energy continuously, as would be expected if the energy emitted were in
the form of waves. Instead, Planck suggested that the object emits ener-
gy in small, specific packets called quanta. A quantum of energy is the
minimum quantity of energy that can be lost or gained by an atom.
Planck proposed the following relationship between a quantum of ener-
gy and the frequency of radiation.

E = hν

In the equation, E is the energy, in joules, of a quantum of radiation, ν www.scilinks.org
is the frequency, in s−1, of the radiation emitted, and h is a fundamental Topic: Photoelectric Effect
physical constant now known as Planck’s constant; h = 6.626 × 10−34 J• s. Code: HC61138

In 1905, Albert Einstein expanded on Planck’s theory by introducing

the radical idea that electromagnetic radiation has a dual wave-particle

nature. While light exhibits many wavelike properties, it can also be

A R R A N G E M E N T O F E L E C T R O N S I N A T O M S 99

thought of as a stream of particles. Each particle of light carries a quan-
tum of energy. Einstein called these particles photons. A photon is a
particle of electromagnetic radiation having zero mass and carrying a
quantum of energy. The energy of a particular photon depends on the
frequency of the radiation.

Ephoton = hν

Einstein explained the photoelectric effect by proposing that elec-
tromagnetic radiation is absorbed by matter only in whole numbers of
photons. In order for an electron to be ejected from a metal surface, the
electron must be struck by a single photon possessing at least the mini-
mum energy required to knock the electron loose. According to the
equation Ephoton = hν, this minimum energy corresponds to a minimum
frequency. If a photon’s frequency is below the minimum, then the elec-
tron remains bound to the metal surface. Electrons in different metals
are bound more or less tightly, so different metals require different
minimum frequencies to exhibit the photoelectric effect.

FIGURE 4 Excited neon atoms The Hydrogen-Atom
emit light when electrons in higher Emission-Line Spectrum
energy levels fall back to the ground
state or to a lower-energy excited When current is passed through a gas at low pressure, the potential ener-
state. gy of the gas atoms increases. The lowest energy state of an atom is its
ground state. A state in which an atom has a higher potential energy than
100 C H A P T E R 4 it has in its ground state is an excited state. There are many possible excit-
ed states, each with a unique energy, but only one ground state energy
for atoms of a given element.When an excited atom returns to its ground
state or a lower energy excited state, it gives off the energy it gained in
the form of electromagnetic radiation. The production of colored light in
neon signs, as shown in Figure 4, is a familiar example of this process.

When investigators passed electric current through a vacuum tube
containing hydrogen gas at low pressure, they observed the emission of a
characteristic pinkish glow. When a narrow beam of the emitted light was
shined through a prism, it was separated into four specific colors of the vis-
ible spectrum. The four bands of light were part of what is known
as hydrogen’s emission-line spectrum. The production of hydrogen’s
emission-line spectrum is illustrated in Figure 5. Additional series of lines
were discovered in the ultraviolet and infrared regions of hydrogen’s
emission-line spectrum. The wavelengths of some of the spectral series
are shown in Figure 6. They are known as the Lyman, Balmer, and
Paschen series, after their discoverers.

Classical theory predicted that the hydrogen atoms would be excited
by whatever amount of energy was added to them. Scientists had thus
expected to observe the emission of a continuous range of frequencies of
electromagnetic radiation, that is, a continuous spectrum. Why had the
hydrogen atoms given off only specific frequencies of light? Attempts to
explain this observation led to an entirely new atomic theory called
quantum theory.

397 nm FIGURE 5 Excited hydrogen
410 nm atoms emit a pinkish glow, as is
434 nm shown in this diagram. When the
486 nm visible portion of the emitted light
656 nm is passed through a prism, it is sepa-
rated into specific wavelengths that
Slits Prism are part of hydrogen’s emission-line
spectrum. The line at 397 nm is in
the ultraviolet and is not visible to
the human eye.

Current is passed through
a glass tube containing
hydrogen at low pressure.
The line at 397 nm
is in the ultraviolet and is
not visible to the human eye.

Lyman series Balmer series Paschen series
(ultraviolet) (visible) (infrared)

ea dc b a cb a
dc b 1800 2000
0 200 400 600 800 1000 1200 1400 1600

Wavelength (nm)

FIGURE 6 A series of specific wavelengths of emitted light makes up hydro-
gen’s emission-line spectrum. The letters below the lines label hydrogen’s vari-
ous energy-level transitions. Niels Bohr’s model of the hydrogen atom
provided an explanation for these transitions.

Whenever an excited hydrogen atom falls to its ground state or to a E2
lower-energy excited state, it emits a photon of radiation. The energy of
this photon (Ephoton = hν) is equal to the difference in energy between Ephoton = E2 – E1 = hν
the atom’s initial state and its final state, as illustrated in Figure 7. The
fact that hydrogen atoms emit only specific frequencies of light indicated E1
that the energy differences between the atoms’ energy states were fixed. FIGURE 7 When an excited atom
This suggested that the electron of a hydrogen atom exists only in very with energy E2 falls back to energy
specific energy states. E1, it releases a photon that has
energy E2 − E1 = Ephoton = hν.
In the late nineteenth century, a mathematical formula that related
the various wavelengths of hydrogen’s emission-line spectrum was dis-
covered. The challenge facing scientists was to provide a model of the
hydrogen atom that accounted for this relationship.

A R R A N G E M E N T O F E L E C T R O N S I N A T O M S 101

Fireflies Bohr Model of the Hydrogen Atom

What kinds of reactions produce light? The puzzle of the hydrogen-atom spectrum was solved in 1913 by the
In this chapter, you are learning how Danish physicist Niels Bohr. He proposed a hydrogen-atom model that
excited atoms can produce light. In linked the atom’s electron to photon emission. According to the model,
parts of the United States, summer is the electron can circle the nucleus only in allowed paths, or orbits.
accompanied by the appearance of fire- When the electron is in one of these orbits, the atom has a definite, fixed
flies, or lightning bugs. What makes energy. The electron—and therefore the hydrogen atom—is in its low-
them glow? A bioluminescent chemical est energy state when it is in the orbit closest to the nucleus. This orbit
reaction that involves luciferin, is separated from the nucleus by a large empty space where the electron
luciferase (an enzyme), adenosine cannot exist. The energy of the electron is higher when the electron is in
triphosphate (ATP), and oxygen takes orbits that are successively farther from the nucleus.
place in the firefly and produces the
characteristic yellow-green glow. The electron orbits, or atomic energy levels, in Bohr’s model can be
Unlike most reactions that produce compared to the rungs of a ladder. When you are standing on a ladder,
light, bioluminescent reactions do not your feet are on one rung or another. The amount of potential energy
generate energy in the form of heat. that you possess corresponds to standing on the first rung, the second
rung, and so forth. Your energy cannot correspond to standing between
two rungs because you cannot stand in midair. In the same way, an elec-
tron can be in one orbit or another, but not in between.

How does Bohr’s model of the hydrogen atom explain the observed
spectral lines? While in a given orbit, the electron is neither gaining nor
losing energy. It can, however, move to a higher-energy orbit by gaining
an amount of energy equal to the difference in energy between the
higher-energy orbit and the initial lower-energy orbit. When a hydrogen
atom is in an excited state, its electron is in one of the higher-energy
orbits. When the electron falls to a lower energy level, a photon is emit-
ted, and the process is called emission. The photon’s energy is equal to
the energy difference between the initial higher energy level and the
final lower energy level. Energy must be added to an atom in order to
move an electron from a lower energy level to a higher energy level.
This process is called absorption. Absorption and emission of radiation
in Bohr’s model of the hydrogen atom are illustrated in Figure 8. The
energy of each absorbed or emitted photon corresponds to a particular
frequency of emitted radiation, Ephoton = hν.

Based on the different wavelengths of the hydrogen emission-line
spectrum, Bohr calculated the allowed energy levels for the hydrogen

e– Ephoton = E3 – E1

Ephoton = E2 – E1 E3 E3

FIGURE 8 (a) Absorption and (b) e– E2 E2
emission of a photon by a hydrogen E1 E1
atom according to Bohr’s model. The
frequencies of light that can be Nucleus Nucleus
absorbed and emitted are restricted
because the electron can only be in (a) Absorption (b) Emission
orbits corresponding to the energies
E1, E2, E3, and so forth.

102 C H A P T E R 4

ab c E FIGURE 9 This energy-state dia-
Paschen series E6 gram for a hydrogen atom shows
E5 some of the energy transitions for
abcd E4 the Lyman, Balmer, and Paschen
Balmer series spectral series. Bohr’s model of the
E3 atom accounted mathematically for
the energy of each of the transitions
E2 shown.

Energy

ab c de E1
Lyman series

atom. He then related the possible energy-level changes to the lines in
the hydrogen emission-line spectrum. The five lines in the Lyman series,
for example, were shown to be the result of electrons dropping from
energy levels E6, E5, E4, E3, and E2 to the ground-state energy level E1.

Bohr’s calculated values agreed with the experimentally observed
values for the lines in each series. The origins of three of the series of
lines in hydrogen’s emission-line spectrum are shown in Figure 9.

Bohr’s model of the hydrogen atom explained observed spectral lines
so well that many scientists concluded that the model could be applied to
all atoms. It was soon recognized, however, that Bohr’s approach did not
explain the spectra of atoms with more than one electron. Nor did Bohr’s
theory explain the chemical behavior of atoms.

SECTION REVIEW 5. Describe the Bohr model of the hydrogen atom.

1. What was the major shortcoming of Rutherford’s Critical Thinking
model of the atom?
6. INTERPRETING GRAPHICS Use the diagram in
2. Write and label the equation that relates the speed, Figure 9 to answer the following:
wavelength, and frequency of electromagnetic
radiation. a. Characterize each of the following as absorp-
tion or emission: an electron moves from E2 to
3. Define the following: E1; an electron moves from E1 to E3; and an
electron moves from E6 to E3.
a. electromagnetic radiation b. wavelength
b. Which energy-level change above emits or
c. frequency d. quantum e. photon absorbs the highest energy? the lowest energy?

4. What is meant by the dual wave-particle nature
of light?

A R R A N G E M E N T O F E L E C T R O N S I N A T O M S 103

SECTION 2 The Quantum Model
of the Atom
OBJECTIVES
T o the scientists of the early twentieth century, Bohr’s model of the
Discuss Louis de Broglie’s role
in the development of the hydrogen atom contradicted common sense. Why did hydrogen’s
quantum model of the atom. electron exist around the nucleus only in certain allowed orbits with
definite energies? Why couldn’t the electron exist in a limitless number
Compare and contrast the of orbits with slightly different energies? To explain why atomic energy
Bohr model and the quantum states are quantized, scientists had to change the way they viewed the
model of the atom. nature of the electron.

Explain how the Heisenberg Electrons as Waves
uncertainty principle and the
Schrödinger wave equation The investigations into the photoelectric effect and hydrogen’s
led to the idea of atomic emission-line spectrum revealed that light could behave as both a wave
orbitals. and a particle. Could electrons have a dual wave-particle nature as well?
In 1924, the French scientist Louis de Broglie asked himself this very
List the four quantum question. And the answer that he proposed led to a revolution in our
numbers and describe basic understanding of matter.
their significance.
De Broglie pointed out that in many ways the behavior of electrons
Relate the number of in Bohr’s quantized orbits was similar to the known behavior of waves.
sublevels corresponding For example, scientists at the time knew that any wave confined to a
to each of an atom’s main space can have only certain frequencies. De Broglie suggested that elec-
energy levels, the number of trons be considered waves confined to the space around an atomic
orbitals per sublevel, and the nucleus. It followed that the electron waves could exist only at specific
number of orbitals per main frequencies. And according to the relationship E = hν, these frequencies
energy level. corresponded to specific energies—the quantized energies of Bohr’s
orbits.
104 C H A P T E R 4
Other aspects of de Broglie’s hypothesis that electrons have wave-
like properties were soon confirmed by experiments. Investigators
demonstrated that electrons, like light waves, can be bent, or diffracted.
Diffraction refers to the bending of a wave as it passes by the edge of an
object or through a small opening. Diffraction experiments and other
investigations also showed that electron beams, like waves, can interfere
with each other. Interference occurs when waves overlap (see the Quick
Lab in this section). This overlapping results in a reduction of energy in
some areas and an increase of energy in others. The effects of diffrac-
tion and interference can be seen in Figure 10.

FIGURE 10 Diffraction patterns
produced by (a) a beam of electrons
passed through a substance and
(b) a beam of visible light passed
through a tiny aperture. Each pat-
tern shows the results of bent waves
that have interfered with each other.
The bright areas correspond to areas
of increased energy, while the dark
areas correspond to areas of
decreased energy.

(a) (b)

The Heisenberg Uncertainty
Principle

The idea of electrons having a dual wave-particle nature troubled scien-
tists. If electrons are both particles and waves, then where are they in the
atom? To answer this question, it is important to consider a proposal first
made in 1927 by the German theoretical physicist Werner Heisenberg.

Heisenberg’s idea involved the detection of electrons. Electrons are
detected by their interaction with photons. Because photons have about
the same energy as electrons, any attempt to locate a specific electron
with a photon knocks the electron off its course. As a result, there is
always a basic uncertainty in trying to locate an electron (or any other
particle). The Heisenberg uncertainty principle states that it is impossi-
ble to determine simultaneously both the position and velocity of an elec-
tron or any other particle. Although it was difficult for scientists to
accept this fact at the time, it has proven to be one of the fundamental
principles of our present understanding of light and matter.

The Schrödinger Wave Equation

In 1926, the Austrian physicist Erwin Schrödinger used the hypothesis
that electrons have a dual wave-particle nature to develop an equation
that treated electrons in atoms as waves. Unlike Bohr’s theory, which
assumed quantization as a fact, quantization of electron energies was
a natural outcome of Schrödinger’s equation. Only waves of specific
energies, and therefore frequencies, provided solutions to the equa-
tion. Together with the Heisenberg uncertainty principle, the
Schrödinger wave equation laid the foundation for modern quantum
theory. Quantum theory describes mathematically the wave properties
of electrons and other very small particles.

A R R A N G E M E N T O F E L E C T R O N S I N A T O M S 105

Wear safety goggles and an apron.

The Wave Nature of Light: Interference Materials

• scissors

Question about 50 cm from the projec- • manila folders
tion screen, as shown in the • thumbtack
Does light show the wave diagram. Adjust the distance to • masking tape
property of interference form a sharp image on the pro- • aluminum foil
when a beam of light is pro- jection screen. • white poster board or
jected through a pinhole
onto a screen? Discussion cardboard
• flashlight
Procedure 1. Did you observe interference

Record all your observations. patterns on the screen?

1. To make the pinhole screen, 2. As a result of your observa-

cut a 20 cm × 20 cm square tions, what do you conclude
from a manila folder. In the about the nature of light?
center of the square, cut a

2 cm square hole. Cut a

7 cm × 7 cm square of alu-

minum foil. Using a thumb-

tack, make a pinhole in the

center of the foil square. Tape

the aluminum foil over the

2 cm square hole, making sure

the pinhole is centered as

shown in the diagram. 1cm

2. Use white poster board to

make a projection screen

35 cm × 35 cm. Image

3. In a dark room, center the light 50 cm

beam from a flashlight on the
pinhole. Hold the flashlight
about 1 cm from the pinhole.
The pinhole screen should be

Solutions to the Schrödinger wave equation are known as wave func-
tions. Based on the Heisenberg uncertainty principle, the early devel-
opers of quantum theory determined that wave functions give only the
probability of finding an electron at a given place around the nucleus.
Thus, electrons do not travel around the nucleus in neat orbits, as Bohr
had postulated. Instead, they exist in certain regions called orbitals. An
orbital is a three-dimensional region around the nucleus that indicates
the probable location of an electron.

Figure 11 illustrates two ways of picturing one type of atomic orbital.
As you will see later in this section, atomic orbitals have different
shapes and sizes.

106 C H A P T E R 4

z y z y FIGURE 11 Two ways of showing
(a) x (b) x a simple atomic orbital are present-
ed. In (a) the probability of finding
the electron is proportional to the
density of the cloud. Shown in (b) is
a surface within which the electron
can be found a certain percentage
of the time, conventionally 90%.

Atomic Orbitals and
Quantum Numbers

In the Bohr atomic model, electrons of increasing energy occupy orbits Energy n=6
farther and farther from the nucleus. According to the Schrödinger n=5
equation, electrons in atomic orbitals also have quantized energies. An n=4
electron’s energy level is not the only characteristic of an orbital that is n=3
indicated by solving the Schrödinger equation.
n=2
In order to completely describe orbitals, scientists use quantum num-
bers. Quantum numbers specify the properties of atomic orbitals and the n=1
properties of electrons in orbitals. The first three quantum numbers
result from solutions to the Schrödinger equation. They indicate the FIGURE 12 The main energy lev-
main energy level, the shape, and the orientation of an orbital. The els of an atom are represented by
fourth, the spin quantum number, describes a fundamental state of the the principal quantum number, n.
electron that occupies the orbital. As you read the following descrip-
tions of the quantum numbers, refer to the appropriate columns in
Table 2.

Principal Quantum Number

The principal quantum number, symbolized by n, indicates the main
energy level occupied by the electron. Values of n are positive integers
only—1, 2, 3, and so on. As n increases, the electron’s energy and its
average distance from the nucleus increase (see Figure 12). For exam-
ple, an electron for which n = 1 occupies the first, or lowest, main ener-
gy level and is located closest to the nucleus. As you will see, more than
one electron can have the same n value. These electrons are sometimes
said to be in the same electron shell. The total number of orbitals that
exist in a given shell, or main energy level, is equal to n2.

Angular Momentum Quantum Number

Except at the first main energy level, orbitals of different shapes—
known as sublevels—exist for a given value of n. The angular momentum
quantum number, symbolized by l, indicates the shape of the orbital. For
a specific main energy level, the number of orbital shapes possible is
equal to n. The values of l allowed are zero and all positive integers less

A R R A N G E M E N T O F E L E C T R O N S I N A T O M S 107

TABLE 1 Orbital Letter Designations According to
Values of l

l Letter
0s
1p
2d
3f

FIGURE 13 The orbitals s, p, and than or equal to n − 1. For example, orbitals for which n = 2 can have
d have different shapes. Each one of two shapes corresponding to l = 0 and l = 1. Depending on its
of the orbitals shown occupies a value of l, an orbital is assigned a letter, as shown in Table 1.
different region of space around
the nucleus. As shown in Figure 13, s orbitals are spherical, p orbitals have dumb-
bell shapes, and d orbitals are more complex. (The f orbital shapes are
z even more complex.) In the first energy level, n = 1, there is only one
y sublevel possible—an s orbital. As mentioned, the second energy level,
n = 2, has two sublevels—the s and p orbitals. The third energy level,
n = 3, has three sublevels—the s, p, and d orbitals. The fourth energy
level, n = 4, has four sublevels—the s, p, d, and f orbitals. In an nth main
energy level, there are n sublevels.

Each atomic orbital is designated by the principal quantum number
followed by the letter of the sublevel. For example, the 1s sublevel is the
s orbital in the first main energy level, while the 2p sublevel is the set of
three p orbitals in the second main energy level. On the other hand, a 4d
orbital is part of the d sublevel in the fourth main energy level. How
would you designate the p sublevel in the third main energy level? How
many other sublevels are in the third main energy level with this one?

Magnetic Quantum Number

Atomic orbitals can have the same shape but different orientations
around the nucleus. The magnetic quantum number, symbolized by m,
indicates the orientation of an orbital around the nucleus. Values of m are
whole numbers, including zero, from −l to +l. Because an s orbital is
spherical and is centered around the nucleus, it has only one possible
orientation. This orientation corresponds to a magnetic quantum num-

z z
y y

x x x

s orbital p orbital d orbital

108 C H A P T E R 4

z z z
y y y

x xx

px orbital py orbital pz orbital

ber of m = 0. There is therefore only one s orbital in each s sublevel. As FIGURE 14 The subscripts x,
shown in Figure 14, the lobes of a p orbital extend along the x, y, or z y, and z indicate the three different
orientations of p orbitals. The inter-
axis of a three-dimensional coordinate system. There are therefore section of the x, y, and z axes indi-
cates the location of the center of
three p orbitals in each p sublevel, which are designated as px , py , and the nucleus.
pz orbitals. The three p orbitals occupy different regions of space and
those regions are related to values of m = −1, m = 0, and m = +1.

There are five different d orbitals in each d sublevel (see

Figure 15). The five different orientations, including one with a differ-
ent shape, correspond to values of m = −2, m = −1, m = 0, m = +1, and
m = +2. There are seven different f orbitals in each f sublevel.

z z z
y y y

x xx

dx2–y2 orbital dxy orbital dyz orbital
z z
y
y
x
x FIGURE 15 The five different ori-
dxz orbital dz2 orbital entations of the d orbitals. Four have
the same shape but different orien-
tations. The fifth has a different
shape and a different orientation
than the others. Each orbital occu-
pies a different region of space.

A R R A N G E M E N T O F E L E C T R O N S I N A T O M S 109

TABLE 2 Quantum Number Relationships in Atomic Structure

Principal Sublevels in Number of Number of Number of Number of
quantum main energy orbitals electrons electrons
number: level per orbitals per per main
main energy (n sublevels) sublevel sublevel energy level
level (n) per main (2n2)
2
energy level 2
(n2) 2 8
6
1 s1 1 18
2
2 s1 4 6 32
p3 10

s1 2
6
3 p3 9 10
14
d5

s1

4 p3 16
d5

f7

CROSS-DISCIPLINARY As you can see in Table 2, the total number of orbitals in a main
energy level increases with the value of n. In fact, the number of orbitals
Go to go.hrw.com for for a full- at each main energy level equals the square of the principal quantum
length article on spintronics. number, n2. What is the total number of orbitals in the third energy
level? Specify each of the sublevels using the orbital designations
Keyword: HC6ARRX you’ve learned so far.

Spin Quantum Number

An electron in an orbital behaves in some ways like Earth spinning on
an axis. The electron exists in one of two possible spin states, which cre-
ates a magnetic field. To account for the magnetic properties of the elec-
tron, theoreticians of the early twentieth century created the spin
quantum number. The spin quantum number has only two possible
values—(+ /1 2, − /1 2)—which indicate the two fundamental spin states of an
electron in an orbital. A single orbital can hold a maximum of two elec-
trons, but the two electrons must have opposite spin states.

SECTION REVIEW 3. Describe briefly what specific information is given
by each of the four quantum numbers.
1. Define the following:
a. main energy levels Critical Thinking
b. quantum numbers
4. INFERRING RELATIONSHIPS What are the possible
2. a. List the four quantum numbers. values of the magnetic quantum number m for f
b. What general information about atomic orbitals orbitals? What is the maximum number of elec-
is provided by the quantum numbers? trons that can exist in 4f orbitals?

110 C H A P T E R 4

Electron SECTION 3
Configurations
OBJECTIVES
T he quantum model of the atom improves on the Bohr model because
List the total number of elec-
it describes the arrangements of electrons in atoms other than hydrogen. trons needed to fully occupy
The arrangement of electrons in an atom is known as the atom’s electron each main energy level.
configuration. Because atoms of different elements have different num-
bers of electrons, a unique electron configuration exists for the atoms of State the Aufbau principle,
each element. Like all systems in nature, electrons in atoms tend to the Pauli exclusion principle,
assume arrangements that have the lowest possible energies. The lowest- and Hund’s rule.
energy arrangement of the electrons for each element is called the ele-
ment’s ground-state electron configuration. A few simple rules, combined Describe the electron
with the quantum number relationships discussed in Section 2, allow us to configurations for the
determine these ground-state electron configurations. atoms of any element using
orbital notation, electron-
configuration notation, and,
when appropriate, noble-gas
notation.

Rules Governing Electron
Configurations

To build up electron configurations for the ground state 6d 7s 6p 6d 5f
of any particular atom, first the energy levels of the 7s5f 6s 5p 5d 4f
orbitals are determined. Then electrons are added to the 5s 4p 4d
orbitals, one by one, according to three basic rules. Energy 6p 4s 3p 3d
(Remember that real atoms are not built up by adding 65sd4f 3s
protons and electrons one at a time.) 2p
5p 2s
The first rule shows the order in which electrons occu- 4d5s
py orbitals. According to the Aufbau principle, an elec- 4p
tron occupies the lowest-energy orbital that can receive it. 4s 3d
Figure 16 shows the atomic orbitals in order of increas- 3p
ing energy. The orbital with the lowest energy is the
1s orbital. In a ground-state hydrogen atom, the electron 3s
is in this orbital. The 2s orbital is the next highest in
energy, then the 2p orbitals. Beginning with the third 2p
main energy level, n = 3, the energies of the sublevels in 2s
different main energy levels begin to overlap.
1s
Note in the figure, for example, that the 4s sublevel is 1s
lower in energy than the 3d sublevel. Therefore, the
4s orbital is filled before any electrons enter the FIGURE 16 The order of increasing energy for
3d orbitals. (Less energy is required for two electrons to atomic sublevels is shown on the vertical axis. Each
pair up in the 4s orbital than for those two electrons to individual box represents an orbital.

A R R A N G E M E N T O F E L E C T R O N S I N A T O M S 111

1s orbital occupy a 3d orbital.) Once the 3d orbitals are fully occupied, which
sublevel will be occupied next?
FIGURE 17 According to the
Pauli exclusion principle, an orbital The second rule reflects the importance of the spin quantum num-
can hold two electrons of opposite ber. According to the Pauli exclusion principle, no two electrons in the
spin states. In this electron configu- same atom can have the same set of four quantum numbers. The princi-
ration of a helium atom, each arrow pal, angular momentum, and magnetic quantum numbers specify the
represents one of the atom’s two energy, shape, and orientation of an orbital. The two values of the spin
electrons. The direction of the arrow quantum number reflect the fact that for two electrons to occupy the
indicates the electron’s spin state. same orbital, they must have opposite spin states (see Figure 17).

The third rule requires placing as many unpaired electrons as possible
in separate orbitals in the same sublevel. In this way, electron-electron
repulsion is minimized so that the electron arrangements have the low-
est energy possible. According to Hund’s rule, orbitals of equal energy are
each occupied by one electron before any orbital is occupied by a second
electron, and all electrons in singly occupied orbitals must have the same
spin state. Applying this rule shows, for example, that one electron will
enter each of the three p orbitals in a main energy level before a second
electron enters any of them. This is illustrated in Figure 18. What is the
maximum number of unpaired electrons in a d sublevel?

FIGURE 18 The figure shows how (a) (b) (c)
(a) two, (b) three, and (c) four elec-
trons fill the p sublevel of a given
main energy level according
to Hund’s rule.

Representing Electron
Configurations

Three methods, or notations, are used to indicate electron configura-
tions. Two of these notations will be discussed in the next two sections
for the first-period elements, hydrogen and helium. The third notation
is used mostly with elements of the third period and higher. It will be
discussed in the section on third-period elements.

In a ground-state hydrogen atom, the single electron is in the lowest-
energy orbital, the 1s orbital. The electron can be in either one of its
two spin states. Helium has two electrons, which are paired in the
1s orbital.

Orbital Notation

In orbital notation, an unoccupied orbital is represented by a line, ,
with the orbital’s name written underneath the line.An orbital containing
one electron is represented as ↑ . An orbital containing two electrons
is represented as ↑↓ , showing the electrons paired and with opposite
spin states. The lines are labeled with the principal quantum number and

112 C H A P T E R 4

sublevel letter. For example, the orbital notations for hydrogen and
helium are written as follows.

H ↑ He ↑↓
1s 1s

Electron-Configuration Notation

Electron-configuration notation eliminates the lines and arrows of

orbital notation. Instead, the number of electrons in a sublevel is shown

by adding a superscript to the sublevel designation. The hydrogen con-
figuration is represented by 1s1. The superscript indicates that one elec-

tron is present in hydrogen’s 1s orbital. The helium configuration is
represented by 1s2. Here the superscript indicates that there are two

electrons in helium’s 1s orbital.

SAMPLE PROBLEM A For more help, go to the Math Tutor at the end of Chapter 5.

The electron configuration of boron is 1s22s22p1. How many electrons are present in an atom of boron?
What is the atomic number for boron? Write the orbital notation for boron.

SOLUTION The number of electrons in a boron atom is equal to the sum of the superscripts in its
electron-configuration notation: 2 + 2 + 1 = 5 electrons. The number of protons equals the
number of electrons in a neutral atom. So we know that boron has 5 protons and thus has an
atomic number of 5. To write the orbital notation, first draw the lines representing orbitals.

1s 2s egfgh

2p

Next, add arrows showing the electron locations. The first two electrons occupy n = 1 energy
level and fill the 1s orbital.

↑↓

1s 2s egfgh

2p

The next three electrons occupy the n = 2 main energy level. Two of these occupy the lower-
energy 2s orbital. The third occupies a higher-energy p orbital.

↑↓ ↑↓ ↑

1s 2s egfgh

2p

PRACTICE Answers in Appendix E

1. The electron configuration of nitrogen is 1s22s22p3. How many Go to go.hrw.com for
electrons are present in a nitrogen atom? What is the atomic num- more practice problems
ber of nitrogen? Write the orbital notation for nitrogen. that deal with electron
configurations.
2. The electron configuration of fluorine is 1s22s22p5. What is the
atomic number of fluorine? How many of its p orbitals are filled? Keyword: HC6ARRX

A R R A N G E M E N T O F E L E C T R O N S I N A T O M S 113

The Noble Decade

By the late nineteenth century, the science of chemistry Rayleigh was at a loss to explain his discovery. Finally, in
had begun to be organized. In 1860, the First International 1892, he published a letter in Nature magazine to appeal
Congress of Chemistry established the field’s first stan- to his colleagues for an explanation. A month later, he
dards. And Dmitri Mendeleev’s periodic table of elements received a reply from a Scottish chemist named William
gave chemists across the globe a systematic understand- Ramsay. Ramsay related that he too had been stumped by
ing of matter’s building blocks. But many important the density difference between chemical and atmospheric
findings—including the discovery of a family of rare, nitrogen. Rayleigh decided to report his findings to the
unreactive gases that were unlike any substances known Royal Society.
at the time—were yet to come.
A Chemist’s Approach
Cross-Disciplinary Correspondence With Rayleigh’s permission, Ramsay attempted to remove
In 1888, the British physicist Lord Rayleigh encountered a all known components from a sample of air and to ana-
small but significant discrepancy in the results of one of lyze what, if anything, remained. Having removed water
his experiments. In an effort to redetermine the atomic vapor, carbon dioxide, and oxygen from the air, Ramsay
mass of nitrogen, he measured the densities of several repeatedly passed the sample over hot magnesium. The
samples of nitrogen gas. Each sample had been prepared nitrogen reacted with the magnesium to form solid mag-
by a different method. All samples that had been isolated nesium nitride. As a result, all of the then-known compo-
from chemical reactions exhibited similar densities. But nents of air were removed. What remained was a
they were about one-tenth of a percent less dense than minuscule portion of a mysterious gas.
the nitrogen isolated from air, which at the time was
believed to be a mixture of nitrogen, oxygen, water vapor, Ramsay tried to cause the gas to react with chemically
and carbon dioxide. active substances, such as hydrogen, sodium, and caustic
soda, but the gas remained unaltered. He decided to name
this new atmospheric component argon (Greek for “inert”
or “idle”).

This excerpt from Lord Rayleigh’s letter was originally pub- Periodic Problems
lished in Nature magazine in 1892. Rayleigh and Ramsay were sure that they had discovered a
new element. But this created a problem. Their calculations
indicated that argon had an atomic mass of about 40.
However, as it appeared in 1894, the periodic table had no
space for such an element. The elements with atomic masses
closest to that of argon were chlorine and potassium.
Unfortunately, the chemical properties of the families of each
of these elements were completely dissimilar to those of the
strange gas.

Ramsay contemplated argon’s lack of reactivity. He knew
that Mendeleev had created the periodic table on the basis
of valence, or the number of atomic partners an element
bonds with in forming a compound. Because Ramsay could
not cause argon to form any compounds, he assigned it a
valence of zero. And because the valence of the elements in

114 C H A P T E R 4

Groups III IV V VI VII VIII I II III IV V VI VII 0 I II
bbaaaaa aa
Periods bbbbb b

1 H He

2 Li Be

3 B C N O F Ne Na Mg

4 Al Si P S Cl Ar K Ca

5 Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr Rb Sr

6 Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe Cs Ba

7 La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn Fr Ra

8 Ac

In 1893, Scottish chemist William Transition elements Main-group elements
Ramsay isolated a previously unknown
component of the atmosphere. This version of the periodic table shows how it looked after the discovery of the
noble gases. The placement of the Group 1 and 2 elements at the far right of the
table shows clearly how the noble gases fit in between the chlorine family and the
potassium family of elements. The 0 above the noble-gas family indicates the zero
valency of the gases.

the families of both chlorine and potassium was one, Questions
perhaps argon fit in between them.
1. What evidence led Ramsay to report that the mysteri-
Ramsay’s insight that argon merited a new spot ous gas was inert?
between the halogen family and the alkali metal family on
the periodic table was correct. And as Ramsay would soon 2. What property of argon caused Ramsay to propose a
confirm, his newly discovered gas was indeed one of a new column in the periodic table?
previously unknown family of elements.

New Neighbors www.scilinks.org
In 1895, Ramsay isolated a light, inert gas from a mineral Topic: William Ramsay
called cleveite. Physical analysis revealed that the gas was Code: HC61666
the same as one that had been identified in the sun in
1868—helium. Helium was the second zero-valent ele-
ment found on Earth, and its discovery made chemists
aware that the periodic table had been missing a whole
column of elements.

Over the next three years, Ramsay and his assistant,
Morris Travers, identified three more inert gases present in
the atmosphere: neon (Greek for “new”), krypton (“hid-
den”), and xenon (“stranger”). Finally in 1900, German
chemist Friedrich Ernst Dorn discovered radon, the last of
the new family of elements known today as the noble
gases. For his discovery, Ramsay received the Nobel Prize
in 1904.

A R R A N G E M E N T O F E L E C T R O N S I N A T O M S 115

1s Elements of the Second Period
2s 2p
3s 3p 3d In the first-period elements, hydrogen and helium, electrons occupy the
4s 4p 4d 4f orbital of the first main energy level. The ground-state configurations in
5s 5p 5d 5f Table 3 illustrate how the Aufbau principle, the Pauli exclusion principle,
6s 6p 6d and Hund’s rule are applied to atoms of elements in the second period.
7s 7p Figure 19 provides a pattern to help you remember the order in which
orbitals are filled according to the Aufbau principle.
FIGURE 19 Follow the diagonal
arrows from the top to get the order According to the Aufbau principle, after the 1s orbital is filled, the
in which atomic orbitals are filled next electron occupies the s sublevel in the second main energy level.
according to the Aufbau principle. Thus, lithium, Li, has a configuration of 1s22s1. The electron occupying
the 2s level of a lithium atom is in the atom’s highest, or outermost, occu-
pied level. The highest-occupied energy level is the electron-containing
main energy level with the highest principal quantum number. The two
electrons in the 1s sublevel of lithium are no longer in the outermost
main energy level. They have become inner-shell electrons, which are
electrons that are not in the highest-occupied energy level.

The fourth electron in an atom of beryllium, Be, must complete the
pair in the 2s sublevel because this sublevel is of lower energy than the
2p sublevel. With the 2s sublevel filled, the 2p sublevel, which has three
vacant orbitals of equal energy, can be occupied. One of the three p
orbitals is occupied by a single electron in an atom of boron, B. Two of
the three p orbitals are occupied by unpaired electrons in an atom of
carbon, C. And all three p orbitals are occupied by unpaired electrons
in an atom of nitrogen, N. Hund’s rule applies here, as is shown in the
orbital notations in Table 3.

According to the Aufbau principle, the next electron must pair with
another electron in one of the 2p orbitals rather than enter the third main
energy level. The Pauli exclusion principle allows the electron to pair with

TABLE 3 Electron Configurations of Atoms of Second-Period
Elements Showing Two Notations

Name Symbol 1s Orbital notation Electron-
configuration
2p notation
2s 1s22s1
1s22s2
Lithium Li ↑↓ ↑ 1s22s22p1
1s22s22p2
Beryllium Be ↑↓ ↑↓ ↑ 1s22s22p3
Boron B ↑↓ ↑↓ ↑ ↑ 1s22s22p4
↑ 1s22s22p5
Carbon C ↑↓ ↑↓ ↑ ↑ ↑↓ 1s22s22p6

Nitrogen N ↑↓ ↑↓ ↑ ↑

Oxygen O ↑↓ ↑↓ ↑↓ ↑

Fluorine F ↑↓ ↑↓ ↑↓ ↑↓

Neon Ne ↑↓ ↑↓ ↑↓ ↑↓

116 C H A P T E R 4

one of the electrons occupying the 2p orbitals as long as the spins of the
paired electrons are opposite. Thus, atoms of oxygen, O, have the config-
uration 1s22s22p4. Oxygen’s orbital notation is shown in Table 3.

Two 2p orbitals are filled in fluorine, F, and all three are filled in neon,
Ne. Atoms such as those of neon, which have the s and p sublevels of
their highest occupied level filled with eight electrons, are said to have
an octet of electrons. Examine the periodic table inside the back cover
of the text. Notice that neon is the last element in the second period.

Elements of the Third Period

After the outer octet is filled in neon, the next electron enters the s sub-
level in the n = 3 main energy level. Thus, atoms of sodium, Na, have
the configuration 1s22s22p63s1. Compare the configuration of a sodium
atom with that of an atom of neon in Table 3. Notice that the first
10 electrons in a sodium atom have the same configuration as a neon
atom, 1s22s22p6. In fact, the first 10 electrons in an atom of each of the
third-period elements have the same configuration as neon. This simi-
larity allows us to use a shorthand notation for the electron configura-
tions of the third-period elements.

Noble-Gas Notation

Neon is a member of the Group 18 elements. The Group 18 elements
(helium, neon, argon, krypton, xenon, and radon) are called the noble
gases. To simplify sodium’s notation, the symbol for neon, enclosed in
square brackets, is used to represent the complete neon configuration:
[Ne] = 1s22s22p6. This allows us to write sodium’s electron configuration
as [Ne]3s1, which is called sodium’s noble-gas notation. Table 4 shows
the electron configuration of each of the third-period elements using
noble-gas notation.

TABLE 4 Electron Configurations of Atoms of Third-Period Elements

Name Symbol Atomic Number of electrons in sublevels Noble-gas
number 1s 2s 2p 3s 3p notation
*[Ne]3s1
Sodium Na 11 226 1 [Ne]3s2
[Ne]3s23p1
Magnesium Mg 12 226 2 [Ne]3s23p2
[Ne]3s23p3
Aluminum Al 13 226 21 [Ne]3s23p4
[Ne]3s23p5
Silicon Si 14 226 22 [Ne]3s23p6

Phosphorus P 15 2 2 6 2 3

Sulfur S 16 2 2 6 2 4

Chlorine Cl 17 226 25

Argon Ar 18 226 26

*[Ne] = 1s22s22p6

A R R A N G E M E N T O F E L E C T R O N S I N A T O M S 117

The last element in the third period is argon, Ar, which is a noble gas.
As in neon, the highest-occupied energy level of argon has an octet of
electrons, [Ne]3s23p6. In fact, each noble gas other than He has an elec-
tron octet in its highest energy level. A noble-gas configuration refers to
an outer main energy level occupied, in most cases, by eight electrons.

Elements of the Fourth Period

The electron configurations of atoms in the fourth-period elements
are shown in Table 5. The period begins by filling the 4s orbital, the
empty orbital of lowest energy. Thus, the first element in the fourth
period is potassium, K, which has the electron configuration [Ar]4s1.
The next element is calcium, Ca, which has the electron configuration
[Ar]4s2.

With the 4s sublevel filled, the 4p and 3d sublevels are the next avail-
able vacant orbitals. Figure 16 shows that the 3d sublevel is lower in
energy than the 4p sublevel. Therefore, the five 3d orbitals are next to

TABLE 5 Electron Configuration of Atoms of Elements in the Fourth Period

Name Symbol Atomic Number of electrons Noble-gas
number in sublevels above 2p notation
3s 3p 3d 4s 4p *[Ar]4s1
[Ar]4s2
Potassium K 19 26 1 [Ar]3d 14s2
[Ar]3d 24s2
Calcium Ca 20 26 2 [Ar]3d 34s2
[Ar]3d 54s1
Scandium Sc 21 26 12 [Ar]3d 54s2
[Ar]3d 64s2
Titanium Ti 22 26 22 [Ar]3d 74s2
[Ar]3d 84s2
Vanadium V 23 26 32 [Ar]3d 104s1
[Ar]3d 104s2
Chromium Cr 24 26 51 [Ar]3d 104s24p1
[Ar]3d 104s24p2
Manganese Mn 25 26 52 [Ar]3d 104s24p3
[Ar]3d 104s24p4
Iron Fe 26 26 62 [Ar]3d 104s24p5
[Ar]3d 104s24p6
Cobalt Co 27 26 72

Nickel Ni 28 26 82

Copper Cu 29 2 6 10 1

Zinc Zn 30 2 6 10 2

Gallium Ga 31 2 6 10 2 1

Germanium Ge 32 2 6 10 2 2

Arsenic As 33 2 6 10 2 3

Selenium Se 34 2 6 10 2 4

Bromine Br 35 2 6 10 2 5

Krypton Kr 36 2 6 10 2 6

*[Ar] = 1s22s22p63s23p6

118 C H A P T E R 4

be filled. A total of 10 electrons can occupy the 3d orbitals. These are
filled successively in the 10 elements from scandium (atomic number
21) to zinc (atomic number 30).

Scandium, Sc, has the electron configuration [Ar]3d14s2. Titanium, Ti,
has the configuration [Ar]3d24s2. And vanadium, V, has the configura-
tion [Ar]3d34s2. Up to this point, three electrons with the same spin have
been added to three separate d orbitals, as required by Hund’s rule.

Surprisingly, chromium, Cr, has the electron configuration [Ar]3d54s1.
Not only did the added electron go into the fourth 3d orbital, but an
electron also moved from the 4s orbital into the fifth 3d orbital, leaving
the 4s orbital with a single electron. Chromium’s electron configuration
is contrary to what is expected according to the Aufbau principle.
However, in reality the [Ar]3d54s1 configuration is of lower energy than
a [Ar]3d44s2 configuration. For chromium, having six orbitals, all with
unpaired electrons, is a more stable arrangement than having four
unpaired electrons in the 3d orbitals and forcing two electrons to pair
up in the 4s orbital. On the other hand, for tungsten, W, which is in the
same group as chromium, having four electrons in the 5d orbitals and
two electrons paired in the 6s orbital is the most stable arrrangement.
Unfortunately, there is no simple explanation for such deviations from
the expected order given in Figure 19.

Manganese, Mn, has the electron configuration [Ar]3d54s2. The
added electron goes to the 4s orbital, completely filling this orbital
while leaving the 3d orbitals still half-filled. Beginning with the next ele-
ment, electrons continue to pair in the d orbitals. Thus, iron, Fe, has the
configuration [Ar]3d64s2; cobalt, Co, has the configuration [Ar]3d74s2;
and nickel, Ni, has the configuration [Ar]3d84s2. Next is copper, Cu, in
which an electron moves from the 4s orbital to pair with the electron in
the fifth 3d orbital. The result is an electron configuration of
[Ar]3d104s1—the lowest-energy configuration for Cu. As with Cr, there
is no simple explanation for this deviation from the expected order.

In atoms of zinc, Zn, the 4s sublevel is filled to give the electron con-
figuration [Ar]3d104s2. In atoms of the next six elements, electrons add
one by one to the three 4p orbitals. According to Hund’s rule, one elec-
tron is added to each of the three 4p orbitals before electrons are paired
in any 4p orbital.

Elements of the Fifth Period

In the 18 elements of the fifth period, sublevels fill in a similar manner
as in elements of the fourth period. However, they start at the 5s orbital
instead of the 4s. Successive electrons are added first to the 5s orbital,
then to the 4d orbitals, and finally to the 5p orbitals. This can be seen in
Table 6. There are occasional deviations from the predicted configura-
tions here also. The deviations differ from those for fourth-period ele-
ments, but in each case the preferred configuration has the lowest
possible energy.

A R R A N G E M E N T O F E L E C T R O N S I N A T O M S 119

TABLE 6 Electron Configurations of Atoms of Elements in the Fifth Period

Name Symbol Atomic Number of electrons Noble-gas
number in sublevels above 3d notation
4s 4p 4d 5s 5p *[Kr]5s1
[Kr]5s2
Rubidium Rb 37 26 1 [Kr]4d 15s2
[Kr]4d 25s2
Strontium Sr 38 26 2 [Kr]4d 45s1
[Kr]4d 55s1
Yttrium Y 39 26 12 [Kr]4d 65s1
[Kr]4d 75s1
Zirconium Zr 40 26 22 [Kr]4d 85s1
[Kr]4d 10
Niobium Nb 41 26 41 [Kr]4d 105s1
[Kr]4d 105s2
Molybdenum Mo 42 26 51 [Kr]4d 105s25p1
[Kr]4d 105s25p2
Technetium Tc 43 26 61 [Kr]4d 105s25p3
[Kr]4d 105s25p4
Ruthenium Ru 44 26 71 [Kr]4d 105s25p5
[Kr]4d 105s25p6
Rhodium Rh 45 26 81

Palladium Pd 46 2 6 10

Silver Ag 47 2 6 10 1

Cadmium Cd 48 2 6 10 2

Indium In 49 2 6 10 2 1

Tin Sn 50 2 6 10 2 2

Antimony Sb 51 2 6 10 2 3

Tellurium Te 52 2 6 10 2 4

Iodine I 53 2 6 10 2 5

Xenon Xe 54 2 6 10 2 6

*[Kr] = 1s22s22p63s23p63d 104s24p6

SAMPLE PROBLEM B For more help, go to the Math Tutor at the end of Chapter 5.

a. Write both the complete electron-configuration notation and the noble-gas notation for iron, Fe.

b. How many electron-containing orbitals are in an atom of iron? How many of these orbitals are com-
pletely filled? How many unpaired electrons are there in an atom of iron? In which sublevel are the
unpaired electrons located?

SOLUTION a. The complete electron-configuration notation of iron is 1s22s22p63s23p63d64s2. The
periodic table inside the back cover of the text reveals that 1s22s22p63s23p6 is the

electron configuration of the noble gas argon, Ar. Therefore, as shown in Table 5,
iron’s noble-gas notation is [Ar]3d64s2.

b. An iron atom has 15 orbitals that contain electrons. They consist of one 1s orbital, one
2s orbital, three 2p orbitals, one 3s orbital, three 3p orbitals, five 3d orbitals, and one
4s orbital. Eleven of these orbitals are filled, and there are four unpaired electrons.
They are located in the 3d sublevel. The notation 3d6 represents

3d ↑↓ ↑ ↑ ↑ ↑

120 C H A P T E R 4

PRACTICE Answers in Appendix E

1. a. Write both the complete electron-configuration notation and
the noble-gas notation for iodine, I. How many inner-shell elec-
trons does an iodine atom contain?

b. How many electron-containing orbitals are in an atom of
iodine? How many of these orbitals are filled? How many
unpaired electrons are there in an atom of iodine?

2. a. Write the noble-gas notation for tin, Sn. How many unpaired
electrons are there in an atom of tin?

b. How many electron-containing d orbitals are there in an atom
of tin? Name the element in the fourth period whose atoms
have the same number of electrons in their highest energy lev-
els that tin’s atoms do.

3. a. Write the complete electron configuration for the element
with atomic number 25. You may use the diagram shown in
Figure 19.

b. Identify the element described in item 3a.

4. a. How many orbitals are completely filled in an atom of the ele- Go to go.hrw.com for
ment with atomic number 18? Write the complete electron con- more practice problems
figuration for this element. that deal with electron
configurations.
b. Identify the element described in item 4a.
Keyword: HC6ARRX

Elements of the Sixth
and Seventh Periods

The sixth period consists of 32 elements. It is much longer than the
periods that precede it in the periodic table. To build up electron con-
figurations for elements of this period, electrons are added first to the
6s orbital in cesium, Cs, and barium, Ba. Then, in lanthanum, La, an
electron is added to the 5d orbital.

With the next element, cerium, Ce, the 4f orbitals begin to fill,
giving cerium atoms a configuration of [Xe]4f 15d16s2. In the next 13
elements, the 4f orbitals are filled. Next the 5d orbitals are filled and
the period is completed by filling the 6p orbitals. Because the 4f and
the 5d orbitals are very close in energy, numerous deviations from the
simple rules occur as these orbitals are filled. The electron configura-
tions of the sixth-period elements can be found in the periodic table
inside the back cover of the text.

The seventh period is incomplete and consists largely of synthetic
elements, which will be discussed in Chapter 21.

A R R A N G E M E N T O F E L E C T R O N S I N A T O M S 121

SAMPLE PROBLEM C For more help, go to the Math Tutor at the end of Chapter 5.

a. Write both the complete electron-configuration notation and the noble-gas notation
for a rubidium atom.

b. Identify the elements in the second, third, and fourth periods that have the same number of highest-
energy-level electrons as rubidium.

SOLUTION a. 1s22s22p63s23p63d104s24p65s1, [Kr]5s1

b. Rubidium has one electron in its highest energy level (the fifth). The elements with the
same outermost configuration are, in the second period, lithium, Li; in the third period,
sodium, Na; and in the fourth period, potassium, K.

PRACTICE Answers in Appendix E

1. a. Write both the complete electron-configuration notation and
the noble-gas notation for a barium atom.

b. Identify the elements in the second, third, fourth, and fifth peri-
ods that have the same number of highest-energy-level elec-
trons as barium.

2. a. Write the noble-gas notation for a gold atom. Go to go.hrw.com for
more practice problems
b. Identify the elements in the sixth period that have one that deal with electron
unpaired electron in their 6s sublevel. configurations.

Keyword: HC6ARRX

SECTION REVIEW 5. Identify the elements having the following
electron configurations:
1. a. What is an atom’s electron configuration? a. 1s 22s 22p 63s 23p 3
b. What three principles guide the electron b. [Ar]4s 1
configuration of an atom?
c. contains four electrons in its third and outer
2. What three methods are used to represent main energy level
the arrangement of electrons in atoms?
d. contains one set of paired and three unpaired
3. What is an octet of electrons? Which electrons in its fourth and outer main energy level
elements contain an octet of electrons?
Critical Thinking
4. Write the complete electron-configuration
notation, the noble-gas notation, and the 6. RELATING IDEAS Write the electron configuration
orbital notation for the following elements: for the third-period elements Al, Si, P, S, and Cl. Is
there a relationship between the group number of
a. carbon b. neon c. sulfur each element and the number of electrons in the
outermost energy level?

122 C H A P T E R 4

CHAPTER HIGHLIGHTS

The Development of a New Atomic Model

Vocabulary • In the early twentieth century, light was determined to have
electromagnetic radiation
electromagnetic spectrum a dual wave-particle nature.
wavelength
frequency • Quantum theory was developed to explain observations such
photoelectric effect
quantum as the photoelectric effect and the line-emission spectrum of
photon
ground state hydrogen.
excited state
line-emission spectrum • Quantum theory states that electrons can exist only at specific
continuous spectrum
atomic energy levels.

• When an electron moves from one main energy level to a main

energy level of lower energy, a photon is emitted. The photon’s

energy equals the energy difference between the two levels.

• An electron in an atom can move from one main energy level

to a higher main energy level only by absorbing an amount of

energy exactly equal to the difference between the two levels.

The Quantum Model of the Atom

Vocabulary • In the early twentieth century, electrons were determined
Heisenberg uncertainty principle
quantum theory to have a dual wave-particle nature.
orbital
quantum number • The Heisenberg uncertainty principle states that it is impossi-
principal quantum number
angular momentum quantum number ble to determine simultaneously the position and velocity of
magnetic quantum number an electron or any other particle.
spin quantum number
• Quantization of electron energies is a natural outcome of the

Schrödinger wave equation, which describes the properties of
an atom’s electrons.

• An orbital, a three-dimensional region around the nucleus,

shows the region in space where an electron is most likely to
be found.

• The four quantum numbers that describe the properties of

electrons in atomic orbitals are the principal quantum number,
the angular momentum quantum number, the magnetic quan-
tum number, and the spin quantum number.

Electron Configurations • The ground-state electron configuration of an atom can be

Vocabulary written by using the Aufbau principle, Hund’s rule, and the
electron configuration Pauli exclusion principle.
Aufbau principle
Pauli exclusion principle • Electron configurations can be depicted by using different
Hund’s rule
noble gas types of notation. In this book, three types of notation are
noble-gas configuration used: orbital notation, electron-configuration notation, and
noble-gas notation.

• Electron configurations of some atoms, such as chromium,

deviate from the predictions of the Aufbau principle, but the
ground-state configuration that results is the configuration
with the minimum possible energy.

A R R A N G E M E N T O F E L E C T R O N S I N A T O M S 123

CHAPTER REVIEW

The Development of a New 12. Using the two equations E = hv and c = λv,
Atomic Model derive an equation expressing E in terms of h, c,
and λ.
SECTION 1 REVIEW
13. How long would it take a radio wave whose fre-
1. a. List five examples of electromagnetic quency is 7.25 × 105 Hz to travel from Mars to
radiation. Earth if the distance between the two planets is
approximately 8.00 × 107 km?
b. What is the speed of all forms of electro-
magnetic radiation in a vacuum? 14. Cobalt-60 is an artificial radioisotope that is
produced in a nuclear reactor and is used as a
2. Prepare a two-column table. List the properties gamma-ray source in the treatment of certain
of light that can best be explained by the wave types of cancer. If the wavelength of the gamma
theory in one column. List those best explained radiation from a cobalt-60 source is 1.00 ×
by the particle theory in the second column. 10−3 nm, calculate the energy of a photon of this
You may want to consult a physics textbook for radiation.
reference.
The Quantum Model
3. What are the frequency and wavelength ranges of the Atom
of visible light?
SECTION 2 REVIEW
4. List the colors of light in the visible spectrum in
order of increasing frequency. 15. Describe two major shortcomings of Bohr’s
model of the atom.
5. In the early twentieth century, what two experi-
ments involving light and matter could not be 16. a. What is the principal quantum number?
explained by the wave theory of light? b. How is it symbolized?
c. What are shells?
6. a. How are the wavelength and frequency of d. How does n relate to the number of elec-
electromagnetic radiation related? trons allowed per main energy level?

b. How are the energy and frequency of elec- 17. a. What information is given by the angular
tromagnetic radiation related? momentum quantum number?

c. How are the energy and wavelength of elec- b. What are sublevels, or subshells?
tromagnetic radiation related? 18. For each of the following values of n, indicate

7. Which theory of light—the wave or particle the numbers and types of sublevels possible for
theory—best explains the following phenomena? that main energy level. (Hint: See Table 2.)
a. the interference of light a. n = 1
b. the photoelectric effect b. n = 2
c. the emission of electromagnetic radiation c. n = 3
by an excited atom d. n = 4
e. n = 7 (number only)
8. Distinguish between the ground state and an 19. a. What information is given by the magnetic
excited state of an atom.
quantum number?
9. According to Bohr’s model of the hydrogen b. How many orbital orientations are possible
atom, how is hydrogen’s emission spectrum
produced? in each of the s, p, d, and f sublevels?
c. Explain and illustrate the notation for distin-
PRACTICE PROBLEMS
guishing between the different p orbitals in a
10. Determine the frequency of light whose wave- sublevel.
length is 4.257 × 10−7 cm.

11. Determine the energy in joules of a photon
whose frequency is 3.55 × 1017 Hz.

124 C H A P T E R 4

CHAPTER REVIEW

20. a. What is the relationship between n and the 31. Write the orbital notation for the following el-
total number of orbitals in a main energy ements. (Hint: See Sample Problem A.)
level? a. P
b. B
b. How many total orbitals are contained in the c. Na
third main energy level? in the fifth? d. O

21. a. What information is given by the spin quan- 32. Write the electron-configuration notation for
tum number? the element whose atoms contain the following
number of electrons:
b. What are the possible values for this quan- a. 3
tum number? b. 6
c. 8
22. How many electrons could be contained in the d. 13
following main energy levels with n equal to the
number provided? 33. Given that the electron configuration for oxy-
a. 1 gen is 1s22s22p4, answer the following questions:
b. 3 a. How many electrons are in each oxygen atom?
c. 4 b. What is the atomic number of this element?
d. 6 c. Write the orbital notation for oxygen’s elec-
e. 7 tron configuration.
d. How many unpaired electrons does oxygen
PRACTICE PROBLEMS have?
e. What is the highest occupied energy level?
23. Sketch the shape of an s orbital and a p orbital. f. How many inner-shell electrons does the
24. How does a 2s orbital differ from a 1s orbital? atom contain?
25. How do a 2px and a 2py orbital differ? g. In which orbital(s) are these inner-shell elec-
trons located?
Electron Configurations
34. a. What are the noble gases?
SECTION 3 REVIEW b. What is a noble-gas configuration?
c. How does noble-gas notation simplify writing
26. a. In your own words, state the Aufbau principle. an atom’s electron configuration?
b. Explain the meaning of this principle in
terms of an atom with many electrons. 35. Write the noble-gas notation for the electron
configuration of each of the elements below.
27. a. In your own words, state Hund’s rule. (Hint: See Sample Problem B.)
b. What is the basis for this rule? a. Cl
b. Ca
28. a. In your own words, state the Pauli exclusion c. Se
principle.
36. a. What information is given by the noble-gas
b. What is the significance of the spin quantum notation [Ne]3s2?
number?
b. What element does this represent?
29. a. What is meant by the highest occupied energy 37. Write both the complete electron-configuration
level in an atom?
notation and the noble-gas notation for each
b. What are inner-shell electrons? of the elements below. (Hint: See Sample
30. Determine the highest occupied energy level in Problem C.)
a. Na
the following elements: b. Sr
a. He c. P
b. Be
c. Al
d. Ca
e. Sn

A R R A N G E M E N T O F E L E C T R O N S I N A T O M S 125

CHAPTER REVIEW

38. Identify each of the following atoms on the c. Write the orbital notation for this element.

basis of its electron configuration: d. How many unpaired electrons does an atom
a. 1s22s22p1
b. 1s22s22p5 of phosphorus have?
c. [Ne]3s2
d. [Ne]3s23p2 e. What is its highest occupied energy level?
e. [Ne]3s23p5
f. [Ar]4s1 f. How many inner-shell electrons does the
g. [Ar]3d64s2
atom contain?

g. In which orbital(s) are these inner-shell elec-

trons located?

47. What is the frequency of a radio wave whose
energy is 1.55 × 10−24 J per photon?

PRACTICE PROBLEMS 48. Write the noble-gas notation for the electron

39. List the order in which orbitals generally fill, configurations of each of the following elements:

from the 1s to the 7p orbital. a. Hf d. At

40. Write the noble-gas notation for the electron b. Sc e. Ac

configurations of each of the following elements: c. Fe f. Zn

a. As e. Sn 49. Describe the major similarities and differences

b. Pb f. Xe between Schrödinger’s model of the atom and

c. Lr g. La the model proposed by Bohr.

d. Hg 50. When sodium is heated, a yellow spectral
line whose energy is 3.37 × 10−19 J per photon is
41. How do the electron configurations of chromi-
produced.
um and copper contradict the Aufbau principle?
a. What is the frequency of this light?

b. What is the wavelength of this light?

MIXED REVIEW 51. a. What is an orbital?

b. Describe an orbital in terms of an electron

42. a. Which has a longer wavelength: green light cloud.

or yellow light?

b. Which has a higher frequency: an X ray or a CRITICAL THINKING

microwave? 52. Inferring Relationships In the emission spec-
trum of hydrogen shown in Figure 5, each col-
c. Which travels at a greater speed: ultraviolet ored line is produced by the emission of photons
with specific energies. Substances also produce
light or infrared light? absorption spectra when electromagnetic radia-
tion passes through them. Certain wavelengths
43. Write both the complete electron-configuration are absorbed. Using the diagram below, predict
what the wavelengths of the absorption lines will
and noble-gas notation for each of the following: be when white light (all of the colors of the visi-
ble spectrum) is passed through hydrogen gas.
a. Ar b. Br c. Al

44. Given the speed of light as 3.00 × 108 m/s, calcu-

late the wavelength of the electromagnetic ra-

diation whose frequency is 7.500 × 1012 Hz.

45. a. What is the electromagnetic spectrum?

b. What units can be used to express wavelength?

c. What unit is used to express frequencies of

electromagnetic waves?

46. Given that the electron configuration for phos-

phorus is 1s22s22p63s23p3, answer the following

questions:

a. How many electrons are in each atom?

b. What is the atomic number of this element? 300 nm Hydrogen absorption spectrum 700 nm

126 C H A P T E R 4

CHAPTER REVIEW

53. Applying Models In discussions of the photo- RESEARCH & WRITING
electric effect, the minimum energy needed to
remove an electron from the metal is called the 57. Neon signs do not always contain neon gas. The
threshold energy and is a characteristic of the various colored lights produced by the signs are
metal. For example, chromium, Cr, will emit due to the emission of a variety of low-pressure
electrons when the wavelength of the radiation gases in different tubes. Research other kinds of
is 284 nm or less. Calculate the threshold energy gases used in neon signs, and list the colors that
for chromium. (Hint: You will need to use the they emit.
two equations that describe the relationships
between wavelength, frequency, speed of light, 58. Prepare a report about the photoelectric effect,
and Planck’s constant.) and cite some of its practical uses. Explain the
basic operation of each device or technique
54. Analyzing Information Four electrons in an mentioned.
atom have the four sets of quantum numbers
given below. Which electrons are in the same ALTERNATIVE ASSESSMENT
orbital? Explain your answer.
a. 1, 0, 0, −__ 59. Performance A spectroscope is a device used
b. 1, 0, 0, +__ to produce and analyze spectra. Construct a
c. 2, 1, 1, +__ simple spectroscope, and determine the absorp-
d. 2, 1, 0, +__ tion spectra of several elemental gases. (Your
teacher will provide you with the gas discharge
55. Relating Ideas Which of the sets of quantum tubes containing samples of different gases.)
numbers below are possible? Which are impos-
sible? Explain your choices. Graphing Calculator Calculating
a. 2, 2, 1, +__ Quantum Number Relationships
b. 2, 0, 0, −__ Go to go.hrw.com for a graphing calculator
c. 2, 0, 1, −__ exercise that asks you to calculate quantum
number relationships.
USING THE HANDBOOK
Keyword: HC6ARRX
56. Sections 1 and 2 of the Elements Handbook
contain information on an analytical test and a
technological application for Group 1 and 2 el-
ements. The test and application are based on
the emission of light from atoms. Review these
sections to answer the following:
a. What analytical technique utilizes the emis-
sion of light from excited atoms?
b. What elements in Groups 1 and 2 can be
identified by this technique?
c. What types of compounds are used to pro-
vide color in fireworks?
d. What wavelengths within the visible spec-
trum would most likely contain emission
lines for barium?

A R R A N G E M E N T O F E L E C T R O N S I N A T O M S 127

Math Tutor WEIGHTED AVERAGES AND ATOMIC MASS

You have learned that the mass of a proton is about 1 amu and that a neutron is only
slightly heavier. Because atomic nuclei consist of whole numbers of protons and neu-
trons, you might expect that the atomic mass of an element would be very near a
whole number. However, if you look at the periodic table, you will see that the atomic
masses of many elements lie somewhere between whole numbers. In fact, the atomic
masses listed on the table are average atomic masses. The atomic masses are averages
because most elements occur in nature as a specific mixture of isotopes. For example,
75.76% of chlorine atoms have a mass of 34.969 amu, and 24.24% have a mass of
36.966 amu. If the isotopes were in a 1:1 ratio, you could simply add the masses of
the two isotopes together and divide by 2. However, to account for the differing abun-
dance of the isotopes, you must calculate a weighted average. For chlorine, the weight-
ed average is 35.45 amu. The following two examples demonstrate how weighted
averages are calculated.

SAMPLE 1 SAMPLE 2

Naturally occurring silver consists of 51.839% Naturally occurring magnesium consists of
Ag-107 (atomic mass 106.905 093) and 48.161% 78.99% Mg-24 (atomic mass 23.985 042), 10.00%
Ag-109 (atomic mass 108.904 756). What is the Mg-25 (atomic mass 24.985 837), and 11.01%
average atomic mass of silver? Mg-26 (atomic mass 25.982 593). What is the
average atomic mass of magnesium?
To find average atomic mass, convert each per-
centage to a decimal equivalent and multiply by Again, convert each percentage to a decimal
the atomic mass of the isotope. and multiply by the atomic mass of the isotope to
get the mass contributed by each isotope.
0.518 39 × 106.905 093 amu = 55.419 amu
0.481 61 × 108.904 756 amu = 52.450 amu 0.7899 × 23.985 042 amu = 18.95 amu
0.1000 × 24.985 837 amu = 2.499 amu
107.869 amu 0.1101 × 25.982 593 amu = 2.861 amu

Adding the masses contributed by each isotope 24.31 amu
gives an average atomic mass of 107.869 amu.
Note that this value for the average atomic mass Adding the masses contributed by each isotope
of silver is very near the one given in the periodic gives an average atomic mass of 24.31 amu.
table.
2. The element silicon occurs as a mixture of
PRACTICE PROBLEMS three isotopes: 92.22% Si-28, 4.69% Si-29, and
3.09% Si-30. The atomic masses of these three
1. Rubidium occurs naturally as a mixture of isotopes are as follows: Si-28 = 27.976 926
two isotopes, 72.17% Rb-85 (atomic mass amu, Si-29 = 28.976 495 amu, and Si-30 =
84.911 792 amu) and 27.83% Rb-87 (atomic 29.973 770 amu.
mass 86.909 186 amu). What is the average
atomic mass of rubidium? Find the average atomic mass of silicon.

128 C H A P T E R 4