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3: Solving Problems Using a Scientific Approach 36

3: Solving Problems Using a Scientific Approach 36

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Chapter Review

54. Write the complete orbital diagram for each of the

following elements, using boxes to represent orbitals

and arrows to represent electrons.

a. magnesium, Z ϭ 12

b. argon, Z ϭ 18



c. lithium, Z ϭ 3

d. arsenic, Z ϭ 33



F 55. The “Chemistry in Focus” segment A Magnetic Mo-



ment discusses the ability to levitate a frog in a magnetic field because electrons, when sensing a strong

magnetic field, respond by opposing it. This is called

diamagnetism. Atoms that are diamagnetic have all

paired electrons. Which columns among the representative elements in the periodic table consist of diamagnetic atoms? Consider orbital diagrams when answering this question.

56. For each of the following, give an atom and its complete electron configuration that would be expected

to have the indicated number of valence electrons.

a. one

b. three



c. five

d. seven



11.10 Electron Configurations and the Periodic Table

QUESTIONS

57. Why do we believe that the valence electrons of calcium and potassium reside in the 4s orbital rather

than in the 3d orbital?

58. Would you expect the valence electrons of rubidium

and strontium to reside in the 5s or the 4d orbitals?

Why?

PROBLEMS



c. strontium, Z ϭ 38

d. chlorine, Z ϭ 17



60. To which element does each of the following abbreviated electron configurations refer?

a. [Ne]3s23p1

b. [Ar]4s1



c. [Ar]4s23d104p5

d. [Kr]5s24d105p2



61. Using the symbol of the previous noble gas to indicate the core electrons, write the electron configuration for each of the following elements.

a. scandium, Z ϭ 21

b. yttrium, Z ϭ 39



c. lanthanum, Z ϭ 57

d. actinium, Z ϭ 89



62. Using the symbol of the previous noble gas to indicate the core electrons, write the valence shell electron configuration for each of the following elements.

a.

b.

c.

d.



phosphorus, Z ϭ 15

chlorine, Z ϭ 17

magnesium, Z ϭ 12

zinc, Z ϭ 30



63. How many 3d electrons are found in each of the following elements?

a. nickel, Z ϭ 28

b. vanadium, Z ϭ 23



c. manganese, Z ϭ 25

d. iron, Z ϭ 26



64. How many 4d electrons are found in each of the following elements?

a. yttrium, Z ϭ 39

b. zirconium, Z ϭ 40



c. strontium, Z ϭ 38

d. cadmium, Z ϭ 48



65. For each of the following elements, indicate which

set of orbitals is filled last.

a. radium, Z ϭ 88

b. iodine, Z ϭ 53



c. gold, Z ϭ 79

d. lead, Z ϭ 82



66. For each of the following elements, indicate which

set of orbitals is being filled last.

a. plutonium, Z ϭ 94

b. nobelium, Z ϭ 102



c. praseodymium, Z ϭ 59

d. radon, Z ϭ 86



67. Write the valence shell electron configuration of each

of the following elements, basing your answer on the

element’s location on the periodic table.

a. rubidium, Z ϭ 37

b. barium, Z ϭ 56



c. titanium, Z ϭ 22

d. germanium, Z ϭ 32



68. The “Chemistry in Focus” segment The Chemistry of

Bohrium discusses element 107, bohrium (Bh). What

is the expected electron configuration of Bh?



11.11 Atomic Properties and the Periodic Table

QUESTIONS



59. Using the symbol of the previous noble gas to indicate the core electrons, write the electron configuration for each of the following elements.

a. arsenic, Z ϭ 33

b. titanium, Z ϭ 22



355



69. What are some of the physical properties that distinguish the metallic elements from the nonmetals? Are

these properties absolute, or do some nonmetallic elements exhibit some metallic properties (and vice

versa)?

70. What types of ions do the metals and the nonmetallic elements form? Do the metals lose or gain electrons in doing this? Do the nonmetallic elements

gain or lose electrons in doing this?

71. Give some similarities that exist among the elements

of Group 1.

72. Give some similarities that exist among the elements

of Group 7.

73. Which of the following elements most easily gives up

electrons during reactions: Li, K, or Cs? Explain your

choice.

74. Which elements in a given period (horizontal row) of

the periodic table lose electrons most easily? Why?

75. Where are the most nonmetallic elements located on

the periodic table? Why do these elements pull electrons from metallic elements so effectively during a

reaction?



All even-numbered Questions and Problems have answers in the back of this book and solutions in the Solutions Guide.



356 Chapter 11 Modern Atomic Theory

76. Why do the metallic elements of a given period (horizontal row) typically have much lower ionization

energies than do the nonmetallic elements of the

same period?

77. What are the metalloids? Where are the metalloids

found on the periodic table?

F 78. The “Chemistry in Focus” segment Fireworks dis-



cusses some of the chemicals that give rise to the colors of fireworks. How do these colors support the existence of quantized energy levels in atoms?

PROBLEMS

79. In each of the following groups, which element is

least reactive?

a. Group 1

b. Group 7



c. Group 2

d. Group 6



80. In each of the following sets of elements, which element would be expected to have the highest ionization energy?

a. Cs, K, Li

b. Ba, Sr, Ca



c. I, Br, Cl

d. Mg, Si, S



81. Arrange the following sets of elements in order of increasing atomic size.

a. Sn, Xe, Rb, Sr

b. Rn, He, Xe, Kr



c. Pb, Ba, Cs, At



82. In each of the following sets of elements, indicate

which element has the smallest atomic size.

a. Na, K, Rb

b. Na, Si, S



c. N, P, As

d. N, O, F



Additional Problems

83. Consider the bright line spectrum of hydrogen

shown in Figure 11.11. Which line in the spectrum

represents photons with the highest energy? With

the lowest energy?

84. The speed at which electromagnetic radiation moves

through a vacuum is called the

.

85. The portion of the electromagnetic spectrum between wavelengths of approximately 400 and 700

nanometers is called the

region.

86. A beam of light can be thought of as consisting of a

stream of light particles called

.

87. The lowest possible energy state of an atom is called

the

state.

88. The energy levels of hydrogen (and other atoms) are

, which means that only certain values of energy are allowed.

89. According to Bohr, the electron in the hydrogen

atom moved around the nucleus in circular paths

called

.

90. In the modern theory of the atom, a(n)

represents a region of space in which there is a high

probability of finding an electron.



91. Electrons found in the outermost principal energy

level of an atom are referred to as

electrons.

92. An element with partially filled d orbitals is called

a(n)

.

93. The

of electromagnetic radiation represents

the number of waves passing a given point in space

each second.

94. Only two electrons can occupy a given orbital in an

atom, and to be in the same orbital, they must have

opposite

.

95. One bit of evidence that the present theory of atomic

structure is “correct” lies in the magnetic properties

of matter. Atoms with unpaired electrons are attracted

by magnetic fields and thus are said to exhibit paramagnetism. The degree to which this effect is observed

is directly related to the number of unpaired electrons

present in the atom. On the basis of the electron orbital diagrams for the following elements, indicate

which atoms would be expected to be paramagnetic,

and tell how many unpaired electrons each atom

contains.

a. phosphorus, Z ϭ 15

b. iodine, Z ϭ 53

c. germanium, Z ϭ 32

96. Without referring to your textbook or a periodic

table, write the full electron configuration, the orbital box diagram, and the noble gas shorthand configuration for the elements with the following

atomic numbers.

a. Z ϭ 19

b. Z ϭ 22

c. Z ϭ 14



d. Z ϭ 26

e. Z ϭ 30



97. Without referring to your textbook or a periodic

table, write the full electron configuration, the orbital box diagram, and the noble gas shorthand configuration for the elements with the following

atomic numbers.

a. Z ϭ 21

b. Z ϭ 15

c. Z ϭ 36



d. Z ϭ 38

e. Z ϭ 30



98. Write the general valence configuration (for example,

ns1 for Group 1) for the group in which each of the

following elements is found.

a.

b.

c.

d.

e.



barium, Z ϭ 56

bromine, Z ϭ 35

tellurium, Z ϭ 52

potassium, Z ϭ 19

sulfur, Z ϭ 16



99. How many valence electrons does each of the following atoms have?

a.

b.

c.

d.



titanium, Z ϭ 22

iodine, Z ϭ 53

radium, Z ϭ 88

manganese, Z ϭ 25



All even-numbered Questions and Problems have answers in the back of this book and solutions in the Solutions Guide.



Chapter Review

100. In the text (Section 11.6) it was mentioned that current theories of atomic structure suggest that all

matter and all energy demonstrate both particlelike and wave-like properties under the appropriate

conditions, although the wave-like nature of matter

becomes apparent only in very small and very fastmoving particles. The relationship between wavelength (␭) observed for a particle and the mass and

velocity of that particle is called the de Broglie relationship. It is

␭ ϭ h/mv

in which h is Planck’s constant (6.63 ϫ 10Ϫ34 J ؒ s),* m

represents the mass of the particle in kilograms, and

v represents the velocity of the particle in meters per

second. Calculate the “de Broglie wavelength” for

each of the following, and use your numerical answers to explain why macroscopic (large) objects are

not ordinarily discussed in terms of their “wave-like”

properties.

a. an electron moving at 0.90 times the speed of light

b. a 150-g ball moving at a speed of 10. m/s

c. a 75-kg person walking at a speed of 2 km/h

101. Light waves move through space at a speed of

meters per second.

102. How do we know that the energy levels of the hydrogen atom are not continuous, as physicists originally

assumed?

103. How does the attractive force that the nucleus exerts

on an electron change with the principal energy level

of the electron?

104. Into how many sublevels is the third principal energy

level of hydrogen divided? What are the names of the

orbitals that constitute these sublevels? What are the

general shapes of these orbitals?

105. A student writes the electron configuration of carbon

(Z ϭ 6) as 1s32s3. Explain to him what is wrong with

this configuration.

106. Write three orbital designations that would be incorrect and explain why each is incorrect. For example,

1p would be an incorrect orbital designation because

there is no p subshell in the first orbit.

107. Why do we believe that the three electrons in the 2p

sublevel of nitrogen occupy different orbitals?

108. Write the full electron configuration (1s22s2, etc.) for

each of the following elements.

a. bromine, Z ϭ 35

b. xenon, Z ϭ 54



c. barium, Z ϭ 56

d. selenium, Z ϭ 34



357



109. Write the complete orbital diagram for each of the

following elements, using boxes to represent orbitals

and arrows to represent electrons.

a. scandium, Z ϭ 21

b. sulfur, Z ϭ 16



c. potassium, Z ϭ 19

d. nitrogen, Z ϭ 7



110. How many valence electrons does each of the following atoms have?

a. nitrogen, Z ϭ 7

b. chlorine, Z ϭ 17



c. sodium, Z ϭ 11

d. aluminum, Z ϭ 13



111. What name is given to the series of ten elements in

which the electrons are filling the 3d sublevel?

112. Using the symbol of the previous noble gas to indicate the core electrons, write the valence shell electron configuration for each of the following elements.

a. zirconium, Z ϭ 40

b. iodine, Z ϭ 53



c. germanium, Z ϭ 32

d. cesium, Z ϭ 55



113. Using the symbol of the previous noble gas to indicate core electrons, write the valence shell electron

configuration for each of the following elements.

a. titanium, Z ϭ 22

b. selenium, Z ϭ 34



c. antimony, Z ϭ 51

d. strontium, Z ϭ 38



114. Identify the element corresponding to each of the

following electron configurations.

a.

b.

c.

d.



1s22s22p63s23p64s23d104p4

[Ar]4s23d104p4

1s22s22p63s23p64s23d104p65s1

1s22s22p63s23p64s23d3



115. Write the shorthand valence shell electron configuration of each of the following elements, basing your

answer on the element’s location on the periodic

table.

a. nickel, Z ϭ 28

b. niobium, Z ϭ 41



c. hafnium, Z ϭ 72

d. astatine, Z ϭ 85



116. Metals have relatively (low/high) ionization energies,

whereas nonmetals have relatively (high/low) ionization energies.

117. In each of the following sets of elements, indicate

which element shows the most active chemical behavior.

a. B, Al, In



b. Na, Al, S



c. B, C, F



118. In each of the following sets of elements, indicate

which element has the smallest atomic size.

a. Ba, Ca, Ra



b. P, Si, Al



c. Rb, Cs, K



*Note that s is the abbreviation for “seconds.”



All even-numbered Questions and Problems have answers in the back of this book and solutions in the Solutions Guide.



12

Types of Chemical Bonds

Electronegativity

Bond Polarity and Dipole

Moments

12.4 Stable Electron

Configurations and Charges

on Ions

12.5 Ionic Bonding and Structures

of Ionic Compounds

12.6 Lewis Structures

12.7 Lewis Structures of

Molecules with Multiple

Bonds

12.8 Molecular Structure

12.9 Molecular Structure: The

VSEPR Model

12.10 Molecular Structure:

Molecules with Double

Bonds



Chemical Bonding



12.1

12.2

12.3



The ionic structure of boron. (Artem

Oganov/Stony Brook University, New York)



12.1 Types of Chemical Bonds

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359



T



he world around us is composed almost entirely of compounds and

mixtures of compounds. Rocks, coal, soil, petroleum, trees, and human beings are all complex mixtures of chemical compounds in which different

kinds of atoms are bound together. Most of the pure elements found in

the earth’s crust also contain many atoms bound together. In a gold

nugget each gold atom is bound to many other gold atoms, and in a diamond many carbon atoms are bonded very strongly to each other. Substances composed of unbound atoms do exist in nature, but they are very

rare. (Examples include the argon atoms in the atmosphere and the helium

atoms found in natural gas reserves.)

The manner in which atoms are bound together has a profound effect on the chemical and physical properties of substances. For example,

both graphite and diamond are composed solely of carbon atoms. However, graphite is a soft, slippery material used as a lubricant in locks, and

diamond is one of the hardest materials known, valuable both as a gemstone and in industrial cutting tools. Why do these materials, both composed solely of carbon atoms, have such different properties? The answer

lies in the different ways in which the carbon atoms are bound to each

other in these substances.

Molecular bonding and structure play the central role in determining

the course of chemical reactions, many of which are vital to our survival.

Most reactions in biological systems are very sensitive to the structures of

the participating molecules; in fact, very subtle differences in shape sometimes serve to channel the chemical reaction one way rather than another.

Molecules that act as drugs must have exactly the right structure to perform their functions correctly. Structure also plays a central role in our

senses of smell and taste. Substances have a particular odor because they

fit into the specially shaped receptors in our nasal passages. Taste is also

dependent on molecular shape, as we discuss in the “Chemistry in Focus”

on page 383.

To understand the behavior of natural materials, we must understand

the nature of chemical bonding and the factors that control the structures

of compounds. In this chapter, we will present various classes of compounds that illustrate the different types of bonds. We will then develop

models to describe the structure and bonding that characterize the materials found in nature.



12.1 Types of Chemical Bonds

OBJECTIVES:



A water molecule.



To learn about ionic and covalent bonds and explain how they are formed.

• To learn about the polar covalent bond.

What is a chemical bond? Although there are several possible ways to answer

this question, we will define a bond as a force that holds groups of two or

more atoms together and makes them function as a unit. For example, in

water the fundamental unit is the HOOOH molecule, which we describe as



360 Chapter 12 Chemical Bonding

being held together by the two OOH bonds. We can obtain information

about the strength of a bond by measuring the energy required to break the

bond, the bond energy.

Atoms can interact with one another in several ways to form aggregates. We will consider specific examples to illustrate the various types of

chemical bonds.

In Chapter 7 we saw that when solid sodium chloride is dissolved in water, the resulting solution conducts electricity, a fact that convinces chemists

that sodium chloride is composed of Naϩ and ClϪ ions. Thus, when sodium

and chlorine react to form sodium chloride, electrons are transferred from

the sodium atoms to the chlorine atoms to form Naϩ and ClϪ ions, which

then aggregate to form solid sodium chloride. The resulting solid sodium

chloride is a very sturdy material; it has a melting point of approximately

800 °C. The strong bonding forces present in sodium chloride result from the

attractions among the closely packed, oppositely charged ions. This is an example of ionic bonding. Ionic substances are formed when an atom that

loses electrons relatively easily reacts with an atom that has a high affinity

for electrons. In other words, an ionic compound results when a metal reacts with a nonmetal.

Metal



M



Nonmetal



ϩ



Ionic compound



Mϩ XϪ



X







We have seen that a bonding force develops when two very different

types of atoms react to form oppositely charged ions. But how does a bonding force develop between two identical atoms? Let’s explore this situation

by considering what happens when two hydrogen atoms are brought close

together, as shown in Figure 12.1. When hydrogen atoms are close together,

the two electrons are simultaneously attracted to both nuclei. Note in Figure 12.1b how the electron probability increases between the two nuclei indicating that the electrons are shared by the two nuclei.

The type of bonding we encounter in the hydrogen molecule and in

many other molecules where electrons are shared by nuclei is called covalent

bonding. Note that in the H2 molecule the electrons reside primarily in the

space between the two nuclei, where they are attracted simultaneously by

both protons. Although we will not go into detail about it here, the increased

attractive forces in this area lead to the formation of the H2 molecule from



When two hydrogen atoms come

close together, the two electrons

are attracted simultaneously by both

nuclei. This produces the bond.

Note the relatively large electron

probability between the nuclei

indicating sharing of the electrons.



Two separate

hydrogen atoms.



+



Figure 12.1

The formation of a bond

between two hydrogen atoms.



H atom

a



+

Hydrogen atoms

sufficiently far apart

to have no interaction



+



H atom



+



H2 molecule

b



12.2 Electronegativity



What the probability

map would like if the

two electrons in the

H—F bond were shared

equally.



H



The actual situation, where the shared

pair spends more time close to the fluorine

atom than to the hydrogen atom. This gives

fluorine a slight excess of negative charge

and the hydrogen a slight deficit of negative

charge (a slight positive charge).



F



Figure 12.2

Probability representations of the

electron sharing in HF.



Ionic and covalent bonds are

the extreme bond types.



a



361



H



F



δ+



δ−



b



the two separated hydrogen atoms. When we say that a bond is formed between the hydrogen atoms, we mean that the H2 molecule is more stable

than two separated hydrogen atoms by a certain quantity of energy (the

bond energy).

So far we have considered two extreme types of bonding. In ionic bonding, the participating atoms are so different that one or more electrons are

transferred to form oppositely charged ions. The bonding results from the attractions among these ions. In covalent bonding, two identical atoms share

electrons equally. The bonding results from the mutual attraction of the two

nuclei for the shared electrons. Between these extremes are intermediate

cases in which the atoms are not so different that electrons are completely

transferred but are different enough so that unequal sharing of electrons results, forming what is called a polar covalent bond. The hydrogen fluoride (HF) molecule contains this type of bond, which produces the following

charge distribution,

H¬F

␦ϩ



␦Ϫ



where ␦ (delta) is used to indicate a partial or fractional charge.

The most logical explanation for the development of bond polarity (the

partial positive and negative charges on the atoms in such molecules as HF)

is that the electrons in the bonds are not shared equally. For example, we can

account for the polarity of the HF molecule by assuming that the fluorine

atom has a stronger attraction than the hydrogen atom for the shared electrons (Figure 12.2). Because bond polarity has important chemical implications, we find it useful to assign a number that indicates an atom’s ability to

attract shared electrons. In the next section we show how this is done.



12.2 Electronegativity

OBJECTIVE:



To understand the nature of bonds and their relationship to electronegativity.

We saw in the previous section that when a metal and a nonmetal react, one

or more electrons are transferred from the metal to the nonmetal to give

ionic bonding. On the other hand, two identical atoms react to form a covalent bond in which electrons are shared equally. When different nonmetals

react, a bond forms in which electrons are shared unequally, giving a polar



362 Chapter 12 Chemical Bonding

Increasing electronegativity



H



Decreasing electronegativity



2.1



B



Be



Li



2.0



1.5



1.0



Na



Mg



0.9



1.2



S



P



Al



Si



1.5



1.8



2.1



4.0



3.5



3.0



2.5



F



O



N



C



2.5



Se



Cl

3.0



Br



Sc



Ti



V



Cr



Mn



Fe



Co



Ni



Cu



Zn



Ge



As



1.5



1.6



1.6



1.5



1.8



1.9



1.9



1.9



Ga



1.6



1.6



1.8



2.0



Nb



Mo



Tc



Ru



Rh



Pd



Ag



Sb



Te



2.2



1.9



Sn



1.9



2.2



In



1.8



2.2



Cd



1.6



1.7



1.7



1.8



1.9



2.1



W



Re



Os



Ir



Pt



Au



2.2



Tl



Pb



Bi



At



2.2



Hg



Po



2.2



2.4



1.7



1.9



1.9



1.8



1.9



1.9



2.0



2.2



K



Ca



0.8



1.0



1.3



Rb



Sr



Y



Zr



0.8



1.0



1.2



1.4



Cs



Ba



La–Lu



Hf



Ta



0.7



0.9



1.0–1.2



1.3



1.5



Fr



Ra



Ac



Th



Pa



U



Np–No



0.9



1.1



1.3



1.4



1.4



0.7



1.4–1.3



2.4



2.8



I

2.5



< 1.5



3.0–4.0



Figure 12.3

Electronegativity values for

selected elements. Note that

electronegativity generally

increases across a period and

decreases down a group. Note

also that metals have relatively

low electronegativity values and

that nonmetals have relatively

high values.



Table 12.1 The Relationship Between Electronegativity and Bond Type

Electronegativity Difference

Between the Bonding Atoms



Bond

Type



Covalent

Character



Ionic

Character



Zero



Covalent



Increases



2.0–2.9



covalent bond. The unequal sharing of electrons between two atoms is described by a property called electronegativity: the relative ability of an atom

in a molecule to attract shared electrons to itself.

Chemists determine electronegativity values for the elements (Figure 12.3) by measuring the polarities of the bonds between various atoms.

Note that electronegativity generally increases going from left to right across

a period and decreases going down a group for the representative elements.

The range of electronegativity values is from 4.0 for fluorine to 0.7 for cesium

and francium. Remember, the higher the atom’s electronegativity value, the

closer the shared electrons tend to be to that atom when it forms a bond.

The polarity of a bond depends on the difference between the electronegativity values of the atoms forming the bond. If the atoms have very

similar electronegativities, the electrons are shared almost equally and the

bond shows little polarity. If the atoms have very different electronegativity

values, a very polar bond is formed. In extreme cases one or more electrons

are actually transferred, forming ions and an ionic bond. For example, when

an element from Group 1 (electronegativity values of about 0.8) reacts with

an element from Group 7 (electronegativity values of about 3), ions are

formed and an ionic substance results.

The relationship between electronegativity and bond type is shown in

Table 12.1. The various types of bonds are summarized in Figure 12.4.



Decreases



1.5–1.9



T



T



Intermediate



Polar covalent



T



T



Large



Ionic



12.2 Electronegativity



δ+

a



+



δ−







c



b



A covalent bond formed

between identical atoms.



363



A polar covalent bond,

with both ionic and

covalent components.



An ionic bond, with

no electron sharing.



Figure 12.4

The three possible types of bonds.



EXAMPLE 12.1



Using Electronegativity to Determine Bond Polarity

Using the electronegativity values given in Figure 12.3, arrange the following bonds in order of increasing polarity: HOH, OOH, ClOH, SOH, and

FOH.

SOLUTION

The polarity of the bond increases as the difference in electronegativity increases. From the electronegativity values in Figure 12.3, the following variation in bond polarity is expected (the electronegativity value appears below

each element).

Electronegativity

Values



Difference in

Electronegativity Values



Bond

Type



HOH



(2.1)(2.1)



2.1 Ϫ 2.1 ϭ 0



Covalent



SOH



(2.5)(2.1)



2.5 Ϫ 2.1 ϭ 0.4



Polar covalent



ClOH



(3.0)(2.1)



3.0 Ϫ 2.1 ϭ 0.9



Polar covalent



OOH



(3.5)(2.1)



3.5 Ϫ 2.1 ϭ 1.4



Polar covalent



FOH



(4.0)(2.1)



4.0 Ϫ 2.1 ϭ 1.9



Polar covalent



Polarity

Increasing



Bond



Therefore, in order of increasing polarity, we have

HOH

Least polar



SOH



ClOH



OOH



FOH

Most polar



Self-Check EXERCISE 12.1 For each of the following pairs of bonds, choose the bond that will be more

polar.

a. HOP, HOC

b. OOF, OOI

c. NOO, SOO

d. NOH, SiOH

See Problems 12.17 through 12.20. ■



364 Chapter 12 Chemical Bonding



12.3 Bond Polarity and Dipole Moments

OBJECTIVE:



To understand bond polarity and how it is related to molecular polarity.

We saw in Section 12.1 that hydrogen fluoride has a positive end and a negative end. A molecule such as HF that has a center of positive charge and a

center of negative charge is said to have a dipole moment. The dipolar

character of a molecule is often represented by an arrow. This arrow points

toward the negative charge center, and its tail indicates the positive center of

charge:



δ+

H

O 2δ–

H

δ+



H



a



The charge distribution in the

water molecule. The oxygen has

a charge of 2δ– because it pulls

δ– of charge from each hydrogen atom (δ– ؉ δ– ‫ ؍‬2δ–).

Center of

positive

charge



Center of

negative

charge



b



The water molecule behaves as

if it had a positive end and a

negative end, as indicated by

the arrow.



Figure 12.5



δ+



δ−



δ+



δ−



δ−



δ+

δ−



δ+



δ−



δ+







δ−



δ+



δ+



δ−



a



Polar water molecules are

strongly attracted to positive

ions by their negative ends.



Figure 12.6



δ−



Any diatomic (two-atom) molecule that has a polar bond has a dipole

moment. Some polyatomic (more than two atoms) molecules also have dipole moments. For example, because the oxygen atom in the water molecule

has a greater electronegativity than the hydrogen atoms, the electrons are

not shared equally. This results in a charge distribution (Figure 12.5) that

causes the molecule to behave as though it had two centers of charge—one

positive and one negative. So the water molecule has a dipole moment.

The fact that the water molecule is polar (has a dipole moment) has a

profound impact on its properties. In fact, it is not overly dramatic to state

that the polarity of the water molecule is crucial to life as we know it on

earth. Because water molecules are polar, they can surround and attract both

positive and negative ions (Figure 12.6). These attractions allow ionic materials to dissolve in water. Also, the polarity of water molecules causes them

to attract each other strongly (Figure 12.7). This means that much energy is

required to change water from a liquid to a gas (the molecules must be separated from each other to undergo this change of state). Therefore, it is the

polarity of the water molecule that causes water to remain a liquid at the

temperatures on the earth’s surface. If it were nonpolar, water would be a gas

and the oceans would be empty.



δ+



+



F



δ+



δ−



b



They are also strongly attracted to negative ions by their

positive ends.



Figure 12.7

Polar water molecules are

strongly attracted to each other.



12.4 Stable Electron Configurations and Charges on Ions



365



12.4 Stable Electron Configurations

and Charges on Ions

OBJECTIVES:



To learn about stable electron configurations. • To learn to predict the

formulas of ionic compounds.

We have seen many times that when a metal and a nonmetal react to form

an ionic compound, the metal atom loses one or more electrons to the nonmetal. In Chapter 5, where binary ionic compounds were introduced, we saw

that in these reactions, Group 1 metals always form 1ϩ cations, Group 2

metals always form 2ϩ cations, and aluminum in Group 3 always forms a 3ϩ

cation. For the nonmetals, the Group 7 elements always form 1Ϫ anions,

and the Group 6 elements always form 2Ϫ anions. This is further illustrated

in Table 12.2.

Notice something very interesting about the ions in Table 12.2: they all

have the electron configuration of neon, a noble gas. That is, sodium loses its

one valence electron (the 3s) to form Naϩ, which has an [Ne] electron configuration. Likewise, Mg loses its two valence electrons to form Mg2ϩ, which also

has an [Ne] electron configuration. On the other hand, the nonmetal atoms

gain just the number of electrons needed for them to achieve the noble gas

electron configuration. The O atom gains two electrons and the F atom gains

one electron to give O2Ϫ and FϪ, respectively, both of which have the [Ne] electron configuration. We can summarize these observations as follows:



Electron Configurations of Ions

1. Representative (main-group) metals form ions by losing enough electrons

to achieve the configuration of the previous noble gas (that is, the noble

gas that occurs before the metal in question on the periodic table). For

example, note from the periodic table inside the front cover of the text that

neon is the noble gas previous to sodium and magnesium. Similarly, helium

is the noble gas previous to lithium and beryllium.

2. Nonmetals form ions by gaining enough electrons to achieve the

configuration of the next noble gas (that is, the noble gas that follows the

element in question on the periodic table). For example, note that neon is

the noble gas that follows oxygen and fluorine, and argon is the noble gas

that follows sulfur and chlorine.



Table 12.2 The Formation of Ions by Metals and Nonmetals

Electron Configuration

Group



Ion Formation



Atom



Ion

eϪ lost

S

2eϪ lost

S

3eϪ lost

S



1



Na S Naϩ ϩ eϪ



[Ne]3s1



2



Mg S Mg2ϩ ϩ 2eϪ



[Ne]3s2



3



Al S Al3ϩ ϩ 3e



6



Ϫ



O ϩ 2e S O



[He]2s 2p ϩ 2e S [He]2s 2p ϭ [Ne]



7



F ϩ eϪ S FϪ



[He]2s22p5 ϩ eϪ S [He]2s22p6 ϭ [Ne]



Ϫ







[Ne]3s23p1

2



4



Ϫ



[Ne]

[Ne]

[Ne]

2



6



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