C H A P T E R

8Periodic Relationships Among the ElementsI N T R O D U C T I O N

MANY OF THE CHEMICAL PROPERTIES OF THE ELEMENTS CAN BE UN-

DERSTOOD IN TERMS OF THEIR ELECTRON CONFIGURATIONS. BECAUSE

ELECTRONS FILL ATOMIC ORBITALS IN A FAIRLY REGULAR FASHION, IT IS

NOT SURPRISING THAT ELEMENTS WITH SIMILAR ELECTRON CONFIGURA-

TIONS, SUCH AS SODIUM AND POTASSIUM, BEHAVE SIMILARLY IN MANY

RESPECTS AND THAT, IN GENERAL, THE PROPERTIES OF THE ELEMENTS

EXHIBIT OBSERVABLE TRENDS. CHEMISTS IN THE NINETEENTH CENTURY

RECOGNIZED PERIODIC TRENDS IN THE PHYSICAL AND CHEMICAL PROP-

ERTIES OF ELEMENTS LONG BEFORE QUANTUM THEORY CAME ONTO THE

SCENE. ALTHOUGH THESE CHEMISTS WERE NOT AWARE OF THE EXISTENCE

OF ELECTRONS AND PROTONS, THEIR EFFORTS TO SYSTEMATIZE THE CHEM-

ISTRY OF THE ELEMENTS WERE REMARKABLY SUCCESSFUL. THEIR MAIN

SOURCES OF INFORMATION WERE THE ATOMIC MASSES OF THE ELEMENTS

AND OTHER KNOWN PHYSICAL AND CHEMICAL PROPERTIES.

8.1 DEVELOPMENT OF THE PERIODIC TABLE

8.2 PERIODIC CLASSIFICATION OF THEELEMENTS

8.3 PERIODIC VARIATION IN PHYSICALPROPERTIES

8.4 IONIZATION ENERGY

8.5 ELECTRON AFFINITY

8.6 VARIATION IN CHEMICAL PROPERTIES OFTHE REPRESENTATIVE ELEMENTS

287

Appendix 1 explains the names andsymbols of the elements.

In the nineteenth century, when chemists had only a vague idea of atoms and mole-cules and did not know of the existence of electrons and protons, they devised the pe-riodic table using their knowledge of atomic masses. Accurate measurements of theatomic masses of many elements had already been made. Arranging elements accord-ing to their atomic masses in a periodic table seemed logical to those chemists, whofelt that chemical behavior should somehow be related to atomic mass.

In 1864 the English chemist John Newlands† noticed that when the elements werearranged in order of atomic mass, every eighth element had similar properties. Newlandsreferred to this peculiar relationship as the law of octaves. However, this “law” turnedout to be inadequate for elements beyond calcium, and Newlands’s work was not ac-cepted by the scientific community.

In 1869 the Russian chemist Dmitri Mendeleev‡ and the German chemist LotharMeyer§ independently proposed a much more extensive tabulation of the elements basedon the regular, periodic recurrence of properties. Table 8.1 shows an early version ofMendeleev’s periodic table. Mendeleev’s classification system was a great improve-ment over Newlands’s for two reasons. First, it grouped the elements together more ac-curately, according to their properties. Equally important, it made possible the predic-tion of the properties of several elements that had not yet been discovered. For example,Mendeleev proposed the existence of an unknown element that he called eka-aluminumand predicted a number of its properties. (Eka is a Sanskrit word meaning “first”; thuseka-aluminum would be the first element under aluminum in the same group.) Whengallium was discovered four years later, its properties matched the predicted proper-ties of eka-aluminum remarkably well:

EKA-ALUMINUM (Ea) GALLIUM (Ga)

Atomic mass 68 amu 69.9 amuMelting point Low 30.15°CDensity 5.9 g/cm3 5.94 g/cm3

Formula of oxide Ea2O3 Ga2O3

Mendeleev’s periodic table included 66 known elements. By 1900, some 30 more hadbeen added to the list, filling in some of the empty spaces. Figure 8.1 charts the dis-covery of the elements chronologically.

Although this periodic table was a celebrated success, the early versions had someglaring inconsistencies. For example, the atomic mass of argon (39.95 amu) is greaterthan that of potassium (39.10 amu). If elements were arranged solely according to in-creasing atomic mass, argon would appear in the position occupied by potassium inour modern periodic table (see the inside front cover). But no chemist would place ar-gon, an inert gas, in the same group as lithium and sodium, two very reactive metals.This and other discrepancies suggested that some fundamental property other than

288 PERIODIC RELATIONSHIPS AMONG THE ELEMENTS

8.1 DEVELOPMENT OF THE PERIODIC TABLE

†John Alexander Reina Newlands (1838–1898). English chemist. Newlands’s work was a step in the right direction in theclassification of the elements. Unfortunately, because of its shortcomings, he was subjected to much criticism, and evenridicule. At one meeting he was asked if he had ever examined the elements according to the order of their initial letters!Nevertheless, in 1887 Newlands was honored by the Royal Society of London for his contribution.‡Dmitri Ivanovich Mendeleev (1836–1907). Russian chemist. His work on the periodic classification of elements is regardedby many as the most significant achievement in chemistry in the nineteenth century.

§Julius Lothar Meyer (1830–1895). German chemist. In addition to his contribution to the periodic table, Meyer also dis-covered the chemical affinity of hemoglobin for oxygen.

FIGURE 8.1 A chronologicalchart of the discovery of the ele-ments. To date, 112 elementshave been identified.

atomic mass must be the basis of periodicity. This property turned out to be associatedwith atomic number, a concept unknown to Mendeleev and his contemporaries.

Using data from �-scattering experiments (see Section 2.2), Rutherford estimatedthe number of positive charges in the nucleus of a few elements, but the significanceof these numbers was overlooked for several more years. In 1913 a young English

8.1 DEVELOPMENT OF THE PERIODIC TABLE 289

TABLE 8.1 The Periodic Table as Drawn by Mendeleev*

GRUPPE I GRUPPE II GRUPPE III GRUPPE IV GRUPPE V GRUPPE VI GRUPPE VII GRUPPE VIII

— — — RH4 RH3 RH2 RH —

REIHEN R2O RO R2O3 RO2 R2O5 RO3 R2O7 RO4

1 H � 12 Li � 7 Be � 9,4 B � 11 C � 12 N � 14 O � 16 F � 19

3 Na � 23 Mg � 24 Al � 27,3 Si � 28 P � 31 S � 32 Cl � 35,54 K � 39 Ca � 40 – �44 Ti � 48 V � 51 Cr � 52 Mn � 55 Fe � 56, Co � 59,

Ni � 59, Cu � 63.5 (Cu � 63) Zn � 65 – �68 – �72 As � 75 Se � 78 Br � 806 Rb � 85 Sr � 87 ?Yt � 88 Zr � 90 Nb � 94 Mo � 96 –�100 Ru � 104, Rh � 104,

Pd � 106, Ag � 108.7 (Ag � 108) Cd � 112 In � 113 Sn � 118 Sb � 122 Te � 125 J � 1278 Cs � 133 Ba � 137 ?Di � 138 ?Ce � 140 – – – — — — —9 (–) – – – – –

10 – – ?Er � 178 ?La � 180 Ta � 182 W � 184 – Os � 195, Ir � 197,Pt � 198, Au � 199.

11 (Au � 199) Hg � 200 Ti � 204 Pb � 207 Bi � 208 – –12 – – – Th � 231 U � 240 –

*Spaces are left for the unknown elements with atomic masses 44, 68, 72, and 100.

120

100

80

60

40

20

0

Num

ber

of e

lem

ents

1650 1700 1750 1800 1850 1900 1950 2000

Year discovered

Ag As Au C Cu FeHg Pb S Sb Sn

Elements known prior to 1650:

Synthetic elements

Mendeleev’s firstperiodic table

physicist, Henry Moseley,† discovered a correlation between what he called atomicnumber and the frequency of X rays generated by bombarding an element with high-energy electrons. Moseley noticed that the frequencies of X rays emitted from the el-ements could be correlated by the equation

�v� � a(Z � b)

where v is the frequency of the emitted X rays and a and b are constants that are thesame for all the elements. Thus, from the square root of the measured frequency of theX rays emitted, we can determine the atomic number of the element.

With a few exceptions, Moseley found that atomic number increases in the sameorder as atomic mass. For example, calcium is the twentieth element in order of in-creasing atomic mass, and it has an atomic number of 20. The discrepancies that hadpuzzled earlier scientists now made sense. The atomic number of argon is 18 and thatof potassium is 19, so potassium should follow argon in the periodic table.

A modern periodic table usually shows the atomic number along with the elementsymbol. As you already know, the atomic number also indicates the number of elec-trons in the atoms of an element. Electron configurations of elements help to explainthe recurrence of physical and chemical properties. The importance and usefulness ofthe periodic table lie in the fact that we can use our understanding of the general prop-erties and trends within a group or a period to predict with considerable accuracy theproperties of any element, even though that element may be unfamiliar to us.

Figure 8.2 shows the periodic table together with the outermost ground-state electronconfigurations of the elements. (The electron configurations of the elements are alsogiven in Table 7.3.) Starting with hydrogen, we see that subshells are filled in the or-der shown in Figure 7.23. According to the type of subshell being filled, the elementscan be divided into categories— the representative elements, the noble gases, the tran-sition elements (or transition metals), the lanthanides, and the actinides. The repre-sentative elements (also called main group elements) are the elements in Groups 1Athrough 7A, all of which have incompletely filled s or p subshells of the highest prin-cipal quantum number. With the exception of helium, the noble gases (the Group 8Aelements) all have a completely filled p subshell. (The electron configurations are 1s2

for helium and ns2np6 for the other noble gases, where n is the principal quantum num-ber for the outermost shell.)

The transition metals are the elements in Groups 1B and 3B through 8B, whichhave incompletely filled d subshells, or readily produce cations with incompletely filledd subshells. (These metals are sometimes referred to as the d-block transition elements.)The nonsequential numbering of the transition metals in the periodic table (that is, 3B–8B, followed by 1B–2B) acknowledges a correspondence between the outer electronconfigurations of these elements and those of the representative elements. For exam-ple, scandium and gallium both have 3 outer electrons. However, because they are indifferent types of atomic orbitals, they are placed in different groups (3A and 3B). The

290 PERIODIC RELATIONSHIPS AMONG THE ELEMENTS

8.2 PERIODIC CLASSIFICATION OF THE ELEMENTS

†Henry Gwyn-Jeffreys Moseley (1887–1915). English physicist. Moseley discovered the relationship between X-ray spec-tra and atomic number. A lieutenant in the Royal Engineers, he was killed in action at the age of 28 during the British cam-paign in Gallipoli, Turkey.

metals iron (Fe), cobalt (Co), and nickel (Ni) do not fit this classification and are allplaced in Group 8B. The Group 2B elements, Zn, Cd, and Hg, are neither representa-tive elements nor transition metals. There is no special name for this group of metals.It should be noted that the designation of A and B groups is not universal. In Europethe practice is to use B for representative elements and A for transition metals, whichis just the opposite of the American convention. The International Union of Pure andApplied Chemistry (IUPAC) has recommended numbering the columns sequentiallywith Arabic numerals 1 through 18 (see Figure 8.2). The proposal has sparked muchcontroversy in the international chemistry community, and its merits and drawbackswill be deliberated for some time to come. In this text we will adhere to the Americandesignation.

The lanthanides and actinides are sometimes called f-block transition elements be-cause they have incompletely filled f subshells. Figure 8.3 distinguishes the groups ofelements discussed here.

A clear pattern emerges when we examine the electron configurations of the ele-ments in a particular group. The electron configurations for Groups 1A and 2A areshown in Table 8.2. All members of the Group 1A alkali metals have similar outer elec-tron configurations; each has a noble gas core and an ns1 outer electron. Similarly, theGroup 2A alkaline earth metals have a noble gas core and an outer electron configu-ration of ns2. The outer electrons of an atom, which are the ones involved in chemicalbonding, are often called valence electrons. The similarity of the outer electron con-figurations (that is, they have the same number and type of valence electrons) is whatmakes the elements in the same group resemble one another in chemical behavior. This

8.2 PERIODIC CLASSIFICATION OF THE ELEMENTS 291

98B

11A

22A

33B

44B

55B

8 1066B

77B

111B

122B

133A

144A

155A

166A

177A

188A

3Li2s1

11Na3s1

19K4s1

37Rb5s1

55Cs6s1

87Fr7s1

4Be2s2

12Mg3s2

20Ca4s2

38Sr5s2

56Ba6s2

88Ra7s2

21Sc

4s23d1

39Y

5s24d1

57La

6s25d1

89Ac

7s26d1

22Ti

4s23d2

40Zr

5s24d2

72Hf

6s25d2

23V

4s23d3

41Nb

5s14d4

73Ta

6s25d3

24Cr

4s13d5

42Mo

5s14d5

74W

6s25d4

25Mn

4s23d5

43Tc

5s24d5

75Re

6s25d5

26Fe

4s23d6

44Ru

5s14d7

76Os

6s25d6

27Co

4s23d7

45Rh

5s14d8

77Ir

6s25d7

104Rf

7s26d2

105Ha

7s26d3

106Sg

7s26d4

107Ns

7s26d5

108Hs

7s26d6

109Mt

7s26d7

110

7s26d8

111

7s26d9

112

7s26d10

28Ni

4s23d8

46Pd4d10

78Pt

6s15d9

29Cu

4s13d10

47Ag

5s14d10

79Au

6s15d10

30Zn

4s23d10

48Cd

5s24d10

80Hg

6s25d10

31Ga

4s24p1

49In

5s25p1

81Tl

6s26p1

32Ge

4s24p2

50Sn

5s25p2

82Pb

6s26p2

33As

4s24p3

51Sb

5s25p3

83Bi

6s26p3

34Se

4s24p4

52Te

5s25p4

84Po

6s26p4

35Br

4s24p5

53I

5s25p5

85At

6s26p5

13Al

3s23p1

14Si

3s23p2

15P

3s23p3

16S

3s23p4

17Cl

3s23p5

5B

2s22p1

6C

2s22p2

7N

2s22p3

8O

2s22p4

9F

2s22p5

10Ne

2s22p6

2He1s2

18Ar

3s23p6

36Kr

4s24p6

54Xe

5s25p6

86Rn

6s26p6

1

2

3

4

5

6

7

1H1s1

58Ce

6s24f15d1

90Th

7s26d2

59Pr

6s24f3

91Pa

7s25f26d1

60Nd

6s24f4

92U

7s25f36d1

61Pm

6s24f5

93Np

7s25f46d1

62Sm

6s24f6

94Pu

7s25f6

63Eu

6s24f7

95Am

7s25f7

64Gd

6s24f75d1

96Cm

7s25f76d1

65Tb

6s24f9

97Bk

7s25f9

66Dy

6s24f10

98Cf

7s25f10

67Ho

6s24f11

99Es

7s25f11

68Er

6s24f12

100Fm

7s25f12

69Tm

6s24f13

101Md

7s25f13

70Yb

6s24f14

102No

7s25f14

71Lu

6s24f145d1

103Lr

7s25f146d1

FIGURE 8.2 The ground-state electron configurations ofthe elements. For simplicity,only the configurations of theouter electrons are shown.

TABLE 8.2 Electron Configurations of Group 1A and Group 2A Elements

GROUP 1A GROUP 2A

Li [He]2s1 Be [He]2s2

Na [Ne]3s1 Mg [Ne]3s2

K [Ar]4s1 Ca [Ar]4s2

Rb [Kr]5s1 Sr [Kr]5s2

Cs [Xe]6s1 Ba [Xe]6s2

Fr [Rn]7s1 Ra [Rn]7s2

FIGURE 8.3 Classificationof the elements. Note thatthe Group 2B elements areoften classified as transitionmetals even though they donot exhibit the characteris-tics of the transition metals.

observation holds true for the other representative elements. Thus, for instance, thehalogens (the Group 7A elements) all have outer electron configurations of ns2np5, andthey have very similar properties. We must be careful, however, in predicting proper-ties for Groups 3A through 7A. For example, the elements in Group 4A all have thesame outer electron configuration, ns2np2, but there is some variation in chemical prop-erties among these elements: Carbon is a nonmetal, silicon and germanium are metal-loids, and tin and lead are metals.

As a group, the noble gases behave very similarly. With the exception of kryptonand xenon, the rest of these elements are totally inert chemically. The reason is thatthese elements all have completely filled outer ns and np subshells, a condition thatrepresents great stability. Although the outer electron configuration of the transitionmetals is not always the same within a group and there is no regular pattern in thechange of the electron configuration from one metal to the next in the same period, alltransition metals share many characteristics that set them apart from other elements.The reason is that these metals all have an incompletely filled d subshell. Likewise,the lanthanide (and the actinide) elements resemble one another because they have in-completely filled f subshells.

EXAMPLE 8.1

A neutral atom of a certain element has 15 electrons. Without consulting a periodictable, answer the following questions: (a) What is the ground-state electron config-uration of the element? (b) How should the element be classified? (c) Are the atomsof this element diamagnetic or paramagnetic?

292 PERIODIC RELATIONSHIPS AMONG THE ELEMENTS

Representativeelements

Noble gases

Transitionmetals

1H

3Li

11Na

19K

37Rb

55Cs

87Fr

20Ca

38Sr

56Ba

88Ra

21Sc

39Y

57La

89Ac

22Ti

40Zr

72Hf

104Rf

23V

41Nb

73Ta

105Ha

24Cr

42Mo

74W

106Sg

25Mn

43Tc

75Re

107Ns

26Fe

44Ru

76Os

108Hs

27Co

45Rh

77Ir

109Mt

28Ni

46Pd

78Pt

29Cu

47Ag

79Au

30Zn

48Cd

80Hg

31Ga

49In

81Tl

32Ge

50Sn

82Pb

33As

51Sb

83Bi

34Se

52Te

84Po

35Br

53I

85At

36Kr

54Xe

86Rn

13Al

14Si

15P

16S

17Cl

18Ar

5B

6C

7N

8O

9F

10Ne

2He

4Be

12Mg

58Ce

90Th

59Pr

91Pa

60Nd

92U

61Pm

93Np

62Sm

94Pu

63Eu

95Am

64Gd

96Cm

65Tb

97Bk

66Dy

98Cf

67Ho

99Es

68Er

100Fm

69Tm

101Md

70Yb

102No

71Lu

103Lr

11A

22A

33B

44B

55B

866B

77B

1098B

111B

122B

133A

144A

155A

166A

177A

188A

110 111 112

ZincCadiumMercury

Lanthanides

Actinides

Similar problem: 8.20.

Note that these two equations forthe combustion of sulfur have iden-tical stoichiometry. This correspon-

dence should not be surprising,since both equations describe the

same chemical system. In both casesa number of sulfur atoms react with

twice as many oxygen atoms.

Answer (a) Using the building-up principle and knowing the maximum capacityof s and p subshells, we can write the ground-state electron configuration of the el-ement as 1s22s22p63s23p3.(b) Since the 3p subshell is not completely filled, this is a representative element.Based on the information given, we cannot say whether it is a metal, a nonmetal,or a metalloid.(c) According to Hund’s rule, the three electrons in the 3p orbitals have parallelspins. Therefore, the atoms of this element are paramagnetic, with three unpairedspins. (Remember, we saw in Chapter 7 that any atom that contains an odd numberof electrons must be paramagnetic.)

PRACTICE EXERCISEA neutral atom of a certain element has 20 electrons. (a) Write the ground-state elec-tron configuration of the element, (b) classify the element, and (c) determine whetherthe atoms of this element are diamagnetic or paramagnetic.

REPRESENTING FREE ELEMENTS IN CHEMICAL EQUATIONS

Having classified the elements according to their ground-state electron configurations,we can now look at the way chemists represent metals, metalloids, and nonmetals asfree elements in chemical equations. Because metals do not exist in discrete molecu-lar units, we always use their empirical formulas in chemical equations. The empiricalformulas are the same as the symbols that represent the elements. For example, theempirical formula for iron is Fe, the same as the symbol for the element.

For nonmetals there is no single rule. Carbon, for example, exists as an extensivethree-dimensional network of atoms, and so we use its empirical formula (C) to rep-resent elemental carbon in chemical equations. But hydrogen, nitrogen, oxygen, andthe halogens exist as diatomic molecules, and so we use their molecular formulas (H2,N2, O2, F2, Cl2, Br2, I2) in equations. The stable form of phosphorus is molecular (P4),and so we use P4. For sulfur chemists often use the empirical formula (S) in chemicalequations, rather than S8 which is the stable form. Thus, instead of writing the equa-tion for the combustion of sulfur as

S8(s) � 8O2(g) 88n 8SO2(g)

we usually write

S(s) � O2(g) 88n SO2(g)

All the noble gases are monatomic species; thus we use their symbols: He, Ne, Ar, Kr,Xe, and Rn. The metalloids, like the metals, all have complex three-dimensional net-works, and we represent them, too, with their empirical formulas, that is, their sym-bols: B, Si, Ge, and so on.

ELECTRON CONFIGURATIONS OF CATIONS AND ANIONS

Because many ionic compounds are made up of monatomic anions and cations, it ishelpful to know how to write the electron configurations of these ionic species. Just asfor neutral atoms, we use the Pauli exclusion principle and Hund’s rule in writing theground-state electron configurations of cations and anions. We will group the ions intwo categories for discussion.

8.2 PERIODIC CLASSIFICATION OF THE ELEMENTS 293

Bear in mind that the order of elec-tron filling does not determine or

predict the order of electron removalfor transition metals.

Ions Derived from Representative Elements

Ions formed from neutral atoms of most representative elements have the noble-gasouter-electron configuration of ns2np6. In the formation of a cation from the neutralatom of a representative element, one or more electrons are removed from the highestoccupied n shell. Following are the electron configurations of some neutral atoms andtheir corresponding cations:

Na: [Ne]3s1 Na�: [Ne]Ca: [Ar]4s2 Ca2�: [Ar]Al: [Ne]3s23p1 Al3�: [Ne]

Note that each ion has a stable noble gas configuration.In the formation of an anion, one or more electrons are added to the highest par-

tially filled n shell. Consider the following examples:

H: 1s1 H�: 1s2 or [He]F: 1s22s22p5 F�: 1s22s22p6 or [Ne]O: 1s22s22p4 O2�: 1s22s22p6 or [Ne]N: 1s22s22p3 N3�: 1s22s22p6 or [Ne]

All of these anions also have stable noble gas configurations. Notice that F�, Na�, andNe (and Al3�, O2�, and N3�) have the same electron configuration. They are said tobe isoelectronic because they have the same number of electrons, and hence the sameground-state electron configuration. Thus H� and He are also isoelectronic.

Cations Derived from Transition Metals

In Section 7.10 we saw that in the first-row transition metals (Sc to Cu), the 4s orbitalis always filled before the 3d orbitals. Consider manganese, whose electron configu-ration is [Ar]4s23d5. When the Mn2� ion is formed, we might expect the two electronsto be removed from the 3d orbitals to yield [Ar]4s23d3. In fact, the electron config-uration of Mn2� is [Ar]3d5! The reason is that the electron-electron and electron-nucleus interactions in a neutral atom can be quite different from those in its ion. Thus,whereas the 4s orbital is always filled before the 3d orbital in Mn, electrons are re-moved from the 4s orbital in forming Mn2� because the 3d orbital is more stable thanthe 4s orbital in transition metal ions. Therefore, when a cation is formed from an atomof a transition metal, electrons are always removed first from the ns orbital and thenfrom the (n � 1)d orbitals.

Keep in mind that most transition metals can form more than one cation and thatfrequently the cations are not isoelectronic with the preceding noble gases.

As we have seen, the electron configurations of the elements show a periodic variationwith increasing atomic number. Consequently, there are also periodic variations in phys-ical and chemical behavior. In this section and the next two, we will examine somephysical properties of elements that are in the same group or period and additionalproperties that influence the chemical behavior of the elements. First, let’s look at theconcept of effective nuclear charge, which has a direct bearing on atomic size and onthe tendency for ionization.

294 PERIODIC RELATIONSHIPS AMONG THE ELEMENTS

8.3 PERIODIC VARIATION IN PHYSICAL PROPERTIES

FIGURE 8.4 (a) In metals suchas beryllium, the atomic radius isdefined as one-half the distancebetween the centers of two adja-cent atoms. (b) For elements thatexist as diatomic molecules, suchas iodine, the radius of the atomis defined as one-half the dis-tance between the centers of theatoms in the molecule.

EFFECTIVE NUCLEAR CHARGE

In Chapter 7, we discussed the shielding effect that electrons close to the nucleus haveon outer-shell electrons in many-electron atoms. The presence of shielding electronsreduces the electrostatic attraction between the positively charged protons in the nu-cleus and the outer electrons. Moreover, the repulsive forces between electrons in amany-electron atom further offset the attractive force exerted by the nucleus. The con-cept of effective nuclear charge allows us to account for the effects of shielding on pe-riodic properties.

Consider, for example, the helium atom, which has the ground-state electron con-figuration 1s2. Helium’s two protons give the nucleus a charge of �2, but the full at-tractive force of this charge on the two 1s electrons is partially offset by electron-elec-tron repulsion. Consequently we say that the 1s electrons shield each other from thenucleus. The effective nuclear charge (Zeff), which is the charge felt by an electron, isgiven by

Zeff � Z � �

where Z is the actual nuclear charge (that is, the atomic number of the element) and �(sigma) is called the shielding constant (also called the screening constant). The shield-ing constant is greater than zero but smaller than Z.

One way to illustrate electron shielding is to consider the amounts of energy re-quired to remove the two electrons from a helium atom. Measurements show that ittakes 2373 kJ of energy to remove the first electron from 1 mole of He atoms and5251 kJ of energy to remove the remaining electron from 1 mole of He� ions. The rea-son it takes so much more energy to remove the second electron is that with only oneelectron present, there is no shielding, and the electron feels the full effect of the �2nuclear charge.

For atoms with three or more electrons, the electrons in a given shell are shieldedby electrons in inner shells (that is, shells closer to the nucleus) but not by electronsin outer shells. Thus, in a neutral lithium atom, whose electron configuration is 1s22s1,the 2s electron is shielded by the two 1s electrons, but the 2s electron does not have ashielding effect on the 1s electrons. In addition, filled inner shells shield outer elec-trons more effectively than electrons in the same subshell shield each other.

ATOMIC RADIUS

A number of physical properties, including density, melting point, and boiling point,are related to the sizes of atoms, but atomic size is difficult to define. As we saw inChapter 7, the electron density in an atom extends far beyond the nucleus, but we nor-mally think of atomic size as the volume containing about 90 percent of the total elec-tron density around the nucleus. When we must be even more specific, we define thesize of an atom in terms of its atomic radius, which is one-half the distance betweenthe two nuclei in two adjacent metal atoms.

For atoms linked together to form an extensive three-dimensional network, atomicradius is simply one-half the distance between the nuclei in two neighboring atoms[Figure 8.4(a)]. For elements that exist as simple diatomic molecules, the atomic ra-dius is one-half the distance between the nuclei of the two atoms in a particular mol-ecule [Figure 8.4(b)].

Figure 8.5 shows the atomic radii of many elements according to their positionsin the periodic table, and Figure 8.6 plots the atomic radii of these elements against

8.3 PERIODIC VARIATION IN PHYSICAL PROPERTIES 295

(a)

(b)

FIGURE 8.5 Atomic radii (inpicometers) of representative ele-ments according to their positionsin the periodic table. Note thatthere is no general agreement onthe size of atomic radii. We fo-cus only on the trends in atomicradii, not on their precise values.

their atomic numbers. Periodic trends are clearly evident. In studying the trends, bearin mind that the atomic radius is determined to a large extent by the strength of the at-traction between the outer-shell electrons and the nucleus. The larger the effective nu-clear charge, the stronger the hold of the nucleus on these electrons, and the smallerthe atomic radius. Consider the second-period elements from Li to F, for example.Moving from left to right, we find that the number of electrons in the inner shell (1s2)remains constant while the nuclear charge increases. The electrons that are added tocounterbalance the increasing nuclear charge are ineffective in shielding one another.Consequently, the effective nuclear charge increases steadily while the principal quan-tum number remains constant (n � 2). For example, the outer 2s electron in lithium isshielded from the nucleus (which has 3 protons) by the two 1s electrons. As an ap-proximation, we assume that the shielding effect of the two 1s electrons is to canceltwo positive charges in the nucleus. Thus the 2s electron only feels the attraction ofone proton in the nucleus; the effective nuclear charge is �1. In beryllium (1s22s2),each of the 2s electrons is shielded by the inner two 1s electrons, which cancel two ofthe four positive charges in the nucleus. Because the 2s electrons do not shield eachother as effectively, the net result is that the effective nuclear charge of each 2s elec-tron is greater than �1. Thus as the effective nuclear charge increases, the atomic ra-dius decreases steadily from lithium to fluorine.

296 PERIODIC RELATIONSHIPS AMONG THE ELEMENTS

Incr

easi

ng a

tom

ic r

adiu

s

Increasing atomic radius

265

248

227

186

152

32 50

H He

8A7A6A5A4A3A2A1A

222

215

197

160

112

171

166

135

143

98

175

162

137

132

91

170

159

139

128

92

164

160

140

127

73

142

133

114

99

72

140

131

112

98

70

Li Be

Cs

Rb

K

Ba

Sr

Ca

Tl

In

Ga

Pb

Sn

Ge

Bi

Sb

As

Po

Te

Se

At

I

Br

Rn

Xe

Kr

Na Mg Al Si P S Cl Ar

B C N O F Ne

FIGURE 8.6 Plot of atomicradii (in picometers) of elementsagainst their atomic numbers.

Within a group of elements we find that atomic radius increases with increasingatomic number. For the alkali metals in Group 1A, the outermost electron resides inthe ns orbital. Since orbital size increases with the increasing principal quantum num-ber n, the size of the metal atoms increases from Li to Cs. We can apply the same rea-soning to the elements in other groups.

EXAMPLE 8.2

Referring to a periodic table, arrange the following atoms in order of increasing ra-dius: P, Si, N.

Answer Note that N and P are in the same group (Group 5A) and that N is aboveP. Therefore, the radius of N is smaller than that of P (atomic radius increases aswe go down a group). Both Si and P are in the third period, and Si is to the left ofP. Therefore, the radius of P is smaller than that of Si (atomic radius decreases aswe move from left to right across a period). Thus the order of increasing radius isN � P � Si.

PRACTICE EXERCISEArrange the following atoms in order of decreasing radius: C, Li, Be.

IONIC RADIUS

Ionic radius is the radius of a cation or an anion. It can be measured by X-ray dif-fraction (see Chapter 11). Ionic radius affects the physical and chemical properties of

8.3 PERIODIC VARIATION IN PHYSICAL PROPERTIES 297

Similar problems: 8.37, 8.38.

Li

Na

KRb

Cs

F

Cl

Br

I

Po

100

300

250

200

150

50

0

Ato

mic

rad

ius

(pm

)

90Atomic number

8070605040302010

FIGURE 8.7 Comparison ofatomic radii with ionic radii.(a) Alkali metals and alkali metalcations. (b) Halogens and halideions.

an ionic compound. For example, the three-dimensional structure of an ionic compounddepends on the relative sizes of its cations and anions.

When a neutral atom is converted to an ion, we expect a change in size. If theatom forms an anion, its size (or radius) increases, since the nuclear charge remainsthe same but the repulsion resulting from the additional electron(s) enlarges the do-main of the electron cloud. On the other hand, removing one or more electrons froman atom reduces electron-electron repulsion but the nuclear charge remains the same,so the electron cloud shrinks, and the cation is smaller than the atom. Figure 8.7 showsthe changes in size that result when alkali metals are converted to cations and halo-gens are converted to anions; Figure 8.8 shows the changes in size that occur when alithium atom reacts with a fluorine atom to form a LiF unit.

Figure 8.9 shows the radii of ions derived from the familiar elements, arrangedaccording to elements’ positions in the periodic table. We can see parallel trends be-tween atomic radii and ionic radii. For example, from top to bottom both the atomicradius and the ionic radius increase within a group. For ions derived from elements indifferent groups, a size comparison is meaningful only if the ions are isoelectronic. Ifwe examine isoelectronic ions, we find that cations are smaller than anions. For ex-ample, Na� is smaller than F�. Both ions have the same number of electrons, but Na

298 PERIODIC RELATIONSHIPS AMONG THE ELEMENTS

FIGURE 8.8 Changes in thesizes of Li and F when they reactto form LiF. +

Li F Li+ F –

100

300

250

200

150

50

0

Rad

ius

(pm

)

Atomic number605040302010

(a)

Li

Na

KRb

Cs

Cs+

Rb+

K+

Na+

Li+

100

300

250

200

150

50

0

Rad

ius

(pm

)

Atomic number605040302010

(b)

F–

Cl–

Br–

I–

I

BrCl

F

(Z � 11) has more protons than F (Z � 9). The larger effective nuclear charge of Na�

results in a smaller radius.Focusing on isoelectronic cations, we see that the radii of tripositive ions (ions

that bear three positive charges) are smaller than those of dipositive ions (ions that beartwo positive charges), which in turn are smaller than unipositive ions (ions that bearone positive charge). This trend is nicely illustrated by the sizes of three isoelectronicions in the third period: Al3�, Mg2�, and Na� (see Figure 8.9). The Al3� ion has thesame number of electrons as Mg2�, but it has one more proton. Thus the electron cloudin Al3� is pulled inward more than that in Mg2�. The smaller radius of Mg2� com-pared with that of Na� can be similarly explained. Turning to isoelectronic anions, wefind that the radius increases as we go from ions with uninegative charge (�) to thosewith dinegative charge (2�), and so on. Thus the oxide ion is larger than the fluorideion because oxygen has one fewer proton than fluorine; the electron cloud is spreadout more in O2�.

EXAMPLE 8.3

For each of the following pairs, indicate which one of the two species is larger:(a) N3� or F�; (b) Mg2� or Ca2�; (c) Fe2� or Fe3�.

Answer (a) N3� and F� are isoelectronic anions. Since N3� has only seven pro-tons and F� has nine, N3� is larger.(b) Both Mg and Ca belong to Group 2A (the alkaline metals). The Ca2� ion islarger than Mg2� because Ca’s valence electrons are in a larger shell (n � 4) thanare Mg’s (n � 3).

8.3 PERIODIC VARIATION IN PHYSICAL PROPERTIES 299

216

I–

195

221

198

Se2–

136140

O2–

171

169

148

133

95

Na+

60

Li+

31

Be2+

65

Mg2+

99

135

113

81 68

Ti4+

V5+

64

Cr3+

80

Mn2+

69

Ni2+

96

Cu+

59 72

Co2+

74

Zn2+

62

50

Al3+

Ga3+

6077

126

Ag+

137

Fe3+ Fe2+

97

Cd2+

81

In3+

71

Sn4+

62

Sb5+

110

Hg2+

95

Tl3+

84

Pb4+

K+

Sr2+Rb+

Ba2+Cs+

Ca2+

Sc3+

Au+

N3–

Te2–

181184

Br–

S2– Cl–

F –

FIGURE 8.9 The radii (in picometers) of ions of familiar elements arranged according to the el-ements’ positions in the periodic table.

FIGURE 8.10 The third-periodelements. The photograph of ar-gon, which is a colorless, odor-less gas, shows the color emittedby the gas from a dischargetube.

(c) Both ions have the same nuclear charge, but Fe2� has one more electron andhence greater electron-electron repulsion. The radius of Fe2� is larger.

PRACTICE EXERCISESelect the smaller ion in each of the following pairs: (a) K�, Li�; (b) Au�, Au3�;(c) P3�, N3�.

VARIATION OF PHYSICAL PROPERTIES ACROSS A PERIOD AND WITHIN A GROUP

From left to right across a period there is a transition from metals to metalloids to non-metals. Consider the third-period elements from sodium to argon (Figure 8.10). Sodium,the first element in the third period, is a very reactive metal, whereas chlorine, the sec-

300 PERIODIC RELATIONSHIPS AMONG THE ELEMENTS

Sodium (Na)

Aluminum (Al)

Phosphorous (P) Sulfur (S) Chlorine (Cl2) Argon (Ar)

Silicon (Si)

Magnesium (Mg)

Similar problems: 8.43, 8.45.

ond-to-last element of that period, is a very reactive nonmetal. In between, the ele-ments show a gradual transition from metallic properties to nonmetallic properties.Sodium, magnesium, and aluminum all have extensive three-dimensional atomic net-works, which are held together by forces characteristic of the metallic state. Silicon isa metalloid; it has a giant three-dimensional structure in which the Si atoms are heldtogether very strongly. Starting with phosphorus, the elements exist in simple, discretemolecular units (P4, S8, Cl2, and Ar) which have low melting points and boiling points.

Within a periodic group the physical properties vary more predictably, especiallyif the elements are in the same physical state. For example, the melting points of ar-gon and xenon are �189.2°C and �111.9°C, respectively. We can estimate the melt-ing point of the intermediate element krypton by taking the average of these two val-ues as follows:

melting point of Kr � � �150.6°C

This value is quite close to the actual melting point of �156.6°C.The Chemistry in Action essay on p. 302 illustrates one interesting application of

periodic group properties.

Not only is there a correlation between electron configuration and physical properties,but a close correlation also exists between electron configuration (a microscopic prop-erty) and chemical behavior (a macroscopic property). As we will see throughout thisbook, the chemical properties of any atom are determined by the configuration of theatom’s valence electrons. The stability of these outermost electrons is reflected directlyin the atom’s ionization energies. Ionization energy is the minimum energy (in kJ/mol)required to remove an electron from a gaseous atom in its ground state. In other words,ionization energy is the amount of energy in kilojoules needed to strip one mole ofelectrons from one mole of gaseous atoms. Gaseous atoms are specified in this defi-nition because an atom in the gas phase is virtually uninfluenced by its neighbors andso there are no intermolecular forces (that is, forces between molecules) to take intoaccount when measuring ionization energy.

The magnitude of ionization energy is a measure of how “tightly” the electron isheld in the atom. The higher the ionization energy, the more difficult it is to removethe electron. For a many-electron atom, the amount of energy required to remove thefirst electron from the atom in its ground state,

energy � X(g) 88n X�(g) � e� (8.1)

is called the first ionization energy (I1). In equation (8.1), X represents an atom of anyelement, e is an electron, and g denotes the gaseous state. The second ionization en-ergy (I2) and the third ionization energy (I3) are shown in the following equations:

energy � X�(g) 88n X2�(g) � e� second ionization

energy � X2�(g) 88n X3�(g) � e� third ionization

The pattern continues for the removal of subsequent electrons.When an electron is removed from a neutral atom, the repulsion among the re-

maining electrons decreases. Since the nuclear charge remains constant, more energy

[(�189.2°C) � (�111.9°C)]���

2

8.4 IONIZATION ENERGY 301

8.4 IONIZATION ENERGY

302 PERIODIC RELATIONSHIPS AMONG THE ELEMENTS

Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry Chem

istry in

Actio

n C

hem

istry in

Actio

n C

hem

istry in

Actio

n C

hem

istry in

Actio

n C

hem

istry in

Actio

n C

hem

istry in

Actio

n C

hem

istry in

Actio

n

Of the 112 known elements, 11 are gases under at-mospheric conditions. Six of these are the Group 8Aelements (the noble gases He, Ne, Ar, Kr, Xe, andRn), and the other five are hydrogen (H2), nitrogen(N2), oxygen (O2), fluorine (F2), and chlorine (Cl2).Curiously, only two elements are liquids at 25°C: mer-cury (Hg) and bromine (Br2).

We do not know the properties of all the knownelements because some of them have never been pre-pared in quantities large enough for investigation. Inthese cases we must rely on periodic trends to predicttheir properties. What are the chances, then, of dis-covering a third liquid element?

Let us look at francium (Fr), the last member ofGroup 1A, to see if it might be a liquid at 25°C. Allof francium’s isotopes are radioactive. The most sta-ble isotope is francium-223, which has a half-life of21 minutes. (Half-life is the time it takes for one-halfof the nuclei in any given amount of a radioactive sub-

stance to disintegrate.) This short half-life means thatonly very small traces of francium could possibly ex-ist on Earth. And although it is feasible to preparefrancium in the laboratory, no weighable quantity ofthe element has been prepared or isolated. Thus weknow very little about francium’s physical and chemi-cal properties. Yet we can use the group periodictrends to predict some of those properties.

Take francium’s melting point as an example. Theplot shows how the melting points of the alkali metalsvary with atomic number. From lithium to sodium, themelting point drops 81.4°; from sodium to potassium,34.6°; from potassium to rubidium, 24°; from rubid-ium to cesium, 11°. On the basis of this trend, we canpredict that the change from cesium to francium wouldbe about 5°. If so, the melting point of francium wouldbe 23°C, which would make it a liquid under atmos-pheric conditions.

The Third Liquid Element?

180

150

90

120

60

30

0 20 40 60 100Atomic number

80

Na

Li

K

RbCs

Fr

Mel

ting

poin

t (°C

)

False-colored image of francium-210 atoms created bybombarding gold with oxygen and trapped by laserbeams. The central spot is 1 mm in diameter and con-sists of about 10,000 atoms. The yellow light is the fluo-rescence of the Fr atoms induced by laser. The otherbright areas come from laser light scattered off the glasssurfaces.

A plot of the melting points of the alkali metals versustheir atomic numbers. By extrapolation, the melting pointof francium should be 23°C.

is needed to remove another electron from the positively charged ion. Thus, ionizationenergies always increase in the following order:

I1 � I2 � I3 � . . .

Table 8.3 lists the ionization energies of the first 20 elements. Ionization is always anendothermic process. By convention, energy absorbed by atoms (or ions) in the ion-ization process has a positive value. Thus ionization energies are all positive quanti-ties. Figure 8.11 shows the variation of the first ionization energy with atomic num-ber. The plot clearly exhibits the periodicity in the stability of the most loosely heldelectron. Note that, apart from small irregularities, the first ionization energies of ele-ments in a period increase with increasing atomic number. This trend is due to the in-crease in effective nuclear charge from left to right (as in the case of atomic radii vari-ation). A larger effective nuclear charge means a more tightly held outer electron, andhence a higher first ionization energy. A notable feature of Figure 8.11 is the peaks,which correspond to the noble gases. The high ionization energies of the noble gases,stemming from their stable ground-state electron configurations, account for the factthat most of them are chemically unreactive. In fact, helium (1s2) has the highest firstionization energy of all the elements.

At the bottom of the graph in Figure 8.11 are the Group 1A elements (the alkalimetals) which have the lowest first ionization energies. Each of these metals has onevalence electron (the outermost electron configuration is ns1) which is effectivelyshielded by the completely filled inner shells. Consequently, it is energetically easy toremove an electron from the atom of an alkali metal to form a unipositive ion (Li�,Na�, K�, . . .). Significantly, the electron configurations of these cations are isoelec-tronic with those noble gases just preceding them in the periodic table.

8.4 IONIZATION ENERGY 303

TABLE 8.3 The Ionization Energies (kJ/mol) of the First 20 Elements

Z ELEMENT FIRST SECOND THIRD FOURTH FIFTH SIXTH

1 H 13122 He 2373 52513 Li 520 7300 118154 Be 899 1757 14850 210055 B 801 2430 3660 25000 328206 C 1086 2350 4620 6220 38000 472617 N 1400 2860 4580 7500 9400 530008 O 1314 3390 5300 7470 11000 130009 F 1680 3370 6050 8400 11000 15200

10 Ne 2080 3950 6120 9370 12200 1500011 Na 495.9 4560 6900 9540 13400 1660012 Mg 738.1 1450 7730 10500 13600 1800013 Al 577.9 1820 2750 11600 14800 1840014 Si 786.3 1580 3230 4360 16000 2000015 P 1012 1904 2910 4960 6240 2100016 S 999.5 2250 3360 4660 6990 850017 Cl 1251 2297 3820 5160 6540 930018 Ar 1521 2666 3900 5770 7240 880019 K 418.7 3052 4410 5900 8000 960020 Ca 589.5 1145 4900 6500 8100 11000

FIGURE 8.11 Variation of thefirst ionization energy withatomic number. Note that the no-ble gases have high ionizationenergies, whereas the alkali met-als and alkaline earth metalshave low ionization energies.

The group 2A elements (the alkaline earth metals) have higher first ionization en-ergies than the alkali metals do. The alkaline earth metals have two valence electrons(the outermost electron configuration is ns2). Because these two s electrons do notshield each other well, the effective nuclear charge for an alkaline earth metal atom islarger than that for the preceding alkali metal. Most alkaline earth compounds containdipositive ions (Mg2�, Ca2�, Sr2�, Ba2�). The Be2� ion is isoelectronic with Li� andwith He, Mg2� is isoelectronic with Na� and with Ne, and so on.

As Figure 8.11 shows, metals have relatively low ionization energies compared tononmetals. The ionization energies of the metalloids generally fall between those ofmetals and nonmetals. The difference in ionization energies suggests why metals al-ways form cations and nonmetals form anions in ionic compounds. (The only impor-tant nonmetallic cation is the ammonium ion, NH4

�.) For a given group, ionization en-ergy decreases with increasing atomic number (that is, as we move down the group).Elements in the same group have similar outer electron configurations. However, asthe principal quantum number n increases, so does the average distance of a valenceelectron from the nucleus. A greater separation between the electron and the nucleusmeans a weaker attraction, so that it becomes increasingly easier to remove the firstelectron as we go from element to element down a group. Thus the metallic characterof the elements within a group increases from top to bottom. This trend is particularlynoticeable for elements in Groups 3A to 7A. For example, in Group 4A, carbon is anonmetal, silicon and germanium are metalloids, and tin and lead are metals.

Although the general trend in the periodic table is for first ionization energies toincrease from left to right, some irregularities do exist. The first exception occurs be-tween Group 2A and 3A elements in the same period (for example, between Be andB and between Mg and Al).The Group 3A elements have lower first ionization ener-gies than 2A elements because they all have a single electron in the outermost p sub-shell (ns2np1), which is well shielded by the inner electrons and the ns2 electrons.Therefore, less energy is needed to remove a single p electron than to remove a paireds electron from the same principal energy level. The second irregularity occurs between

304 PERIODIC RELATIONSHIPS AMONG THE ELEMENTS

500Li

2500

2000

1500

1000

0

Firs

t ion

izat

ion

ener

gy (

kJ/m

ol)

90Atomic number (Z )

10 20 30 40 50 60 70 80

He

Ne

Na

Ar

K

Kr

Rb

Xe

Cs

RnH

Similar problem: 8.53.

Groups 5A and 6A (for example, between N and O and between P and S). In the Group5A elements (ns2np3) the p electrons are in three separate orbitals according to Hund’srule. In Group 6A (ns2np4) the additional electron must be paired with one of the threep electrons. The proximity of two electrons in the same orbital results in greater elec-trostatic repulsion, which makes it easier to ionize an atom of the Group 6A element,even though the nuclear charge has increased by one unit. Thus the ionization energiesfor Group 6A elements are lower than those for Group 5A elements in the same pe-riod.

The following example compares the ionization energies of some elements.

EXAMPLE 8.4

(a) Which atom should have a smaller first ionization energy: oxygen or sulfur?(b) Which atom should have a higher second ionization energy: lithium or beryllium?

Answer (a) Oxygen and sulfur are members of Group 6A. They have the same va-lence electron configuration (ns2np4), but the 3p electron in sulfur is farther fromthe nucleus and experiences less nuclear attraction than the 2p electron in oxygen.Thus, following the general rule that the ionization energy of elements decreases aswe move down a periodic group, we predict that sulfur should have a smaller firstionization energy. Table 8.3 confirms our reasoning.(b) The electron configurations of Li and Be are 1s22s1 and 1s22s2, respectively. Forthe second ionization process we write

Li�(g) 88n Li2�(g)1s2 1s1

Be�(g) 88n Be2�(g)1s22s1 1s2

Since 1s electrons shield 2s electrons much more effectively than they shield eachother, we predict that it should be much easier to remove a 2s electron from Be�

than to remove a 1s electron from Li�. This is consistent with the data in Table 8.3.

PRACTICE EXERCISE(a) Which of the following atoms should have a larger first ionization energy: N orP? (b) Which of the following atoms should have a smaller second ionization en-ergy: Na or Mg?

Another property that greatly influences the chemical behavior of atoms is their abil-ity to accept one or more electrons. This property is called electron affinity, which isthe energy change that occurs when an electron is accepted by an atom in the gaseousstate to form an anion. The equation is

X(g) � e� 88n X�(g) (8.2)

The sign of the electron affinity is opposite to the one we use for ionization energy. Aswe saw in Section 8.5, a positive ionization energy means that energy must be suppliedto remove an electron. A positive electron affinity, on the other hand, signifies that en-

8.5 ELECTRON AFFINITY 305

8.5 ELECTRON AFFINITY

ergy is liberated when an electron is added to an atom. To clarify this apparent para-dox, let us consider the process in which a gaseous fluorine atom accepts an electron:

F(g) � e� 88n F�(g) H � �328 kJ/mol

The sign of the enthalpy change indicates that this is an exothermic process; however,the electron affinity of fluorine is assigned a value of �328 kJ/mol. Thus we can thinkof electron affinity as the energy that must be supplied to remove an electron from anegative ion. For the removal of an electron from a fluoride ion, we have

F�(g) 88n F(g) � e� H � �328 kJ/mol

Two features of electron affinity to remember are: (1) The electron affinity of an ele-ment is equal to the enthalpy change that accompanies the ionization process of its an-ion, and (2) a large positive electron affinity means that the negative ion is very stable(that is, the atom has a great tendency to accept an electron), just as a high ionizationenergy of an atom means that the atom is very stable.

Experimentally, electron affinity is determined by removing the additional electronfrom an anion. In contrast to ionization energies, however, electron affinities are diffi-cult to measure because the anions of many elements are unstable. Table 8.4 shows theelectron affinities of some representative elements and the noble gases, and Figure 8.12plots the electron affinities of the first 20 elements versus atomic number. The overalltrend is an increase in the tendency to accept electrons (electron affinity values becomemore positive) from left to right across a period. The electron affinities of metals aregenerally lower than those of nonmetals. The values vary little within a given group.The halogens (Group 7A) have the highest electron affinity values. This is not surpris-ing when we realize that by accepting an electron, each halogen atom assumes the sta-ble electron configuration of the noble gas immediately to its right. For example, theelectron configuration of F� is 1s22s22p6, or [Ne]; for Cl� it is [Ne]3s23p6 or [Ar]; andso on. Calculations show that the noble gases all have electron affinities of less thanzero. Thus the anions of these gases, if formed, would be inherently unstable.

The electron affinity of oxygen has a positive value (141 kJ/mol), which meansthat the process

306 PERIODIC RELATIONSHIPS AMONG THE ELEMENTS

TABLE 8.4 Electron Affinities (kJ/mol) of Some Representative Elements and the Noble Gases*

1A 2A 3A 4A 5A 6A 7A 8A

H He73 � 0Li Be B C N O F Ne60 0 27 122 0 141 328 � 0Na Mg Al Si P S Cl Ar53 0 44 134 72 200 349 � 0K Ca Ga Ge As Se Br Kr48 2.4 29 118 77 195 325 � 0Rb Sr In Sn Sb Te I Xe47 4.7 29 121 101 190 295 � 0Cs Ba Tl Pb Bi Po At Rn45 14 30 110 110 ? ? � 0

*The electron affinities of the noble gases, Be, and Mg have not been determined experimentally,but are believed to be close to zero or negative.

FIGURE 8.12 A plot of elec-tron affinity against atomic num-ber for the first 20 elements.

O(g) � e� 88n O�(g) H � �141 kJ

is favorable (exothermic). On the other hand, the electron affinity of the O� ion ishighly negative (�780 kJ/mol), which means the process

O�(g) � e� 88n O2�(g) H � 780 kJ

is endothermic even though the O2� ion is isoelectronic with the noble gas Ne. Thisprocess is unfavorable in the gas phase because the resulting increase in electron-elec-tron repulsion outweighs the stability gained by achieving a noble gas configuration.However, note that O2� is common in ionic compounds (for example, Li2O and MgO);in solids, the O2� ion is stabilized by the neighboring cations. We will study the sta-bility of ionic compounds in the next chapter.

The following example shows why the alkaline earth metals do not have a greattendency to accept electrons.

EXAMPLE 8.5

Why are the electron affinities of the alkaline earth metals, shown in Table 8.4, ei-ther negative or small positive values?

Answer The valence configuration of the alkaline earth metals is ns2. For theprocess

M(g) � e� 88n M�(g)

where M denotes a member of the Group 2A family, the extra electron must enterthe np subshell, which is effectively shielded by the two ns electrons (the np elec-

8.5 ELECTRON AFFINITY 307

Li

C

Na

F

Si

Cl

K

Ge

Br

Rb

Sn

I

Cs

40 5010 20 30 60Atomic number (Z )

0

400

300

200

100

Ele

ctro

n af

fini

ty (

kJ/m

ol)

Similar problem: 8.61.

trons are farther away from the nucleus than the ns electrons) and the inner elec-trons. Consequently, alkaline earth metals have little tendency to pick up an extraelectron.

PRACTICE EXERCISEIs it likely that Ar will form the anion Ar�?

Ionization energy and electron affinity help chemists understand the types of reactionsthat elements undergo and the nature of the elements’ compounds. On a conceptuallevel these two measures are related in a simple way: Ionization energy indexes the at-traction of an atom for its own electrons, whereas electron affinity expresses the at-traction of an atom for an additional electron from some other source. Together theygive us insight into the general attraction of an atom for electrons. With these conceptswe can survey the chemical behavior of the elements systematically, paying particularattention to the relationship between chemical properties and electron configuration.

We have seen that the metallic character of the elements decreases from left toright across a period and increases from top to bottom within a group. On the basis ofthese trends and the knowledge that metals usually have low ionization energies whilenonmetals usually have high electron affinities, we can frequently predict the outcomeof a reaction involving some of these elements.

GENERAL TRENDS IN CHEMICAL PROPERTIES

Before we study the elements in individual groups, let us look at some overall trends.We have said that elements in the same group resemble one another in chemical be-havior because they have similar outer electron configurations. This statement, althoughcorrect in the general sense, must be applied with caution. Chemists have long knownthat the first member of each group (the element in the second period from lithium tofluorine) differs from the rest of the members of the same group. Lithium, for exam-ple, exhibits many, but not all, of the properties characteristic of the alkali metals.Similarly, beryllium is a somewhat atypical member of Group 2A, and so on. The dif-ference can be attributed to the unusually small size of the first element in each group(see Figure 8.5).

Another trend in the chemical behavior of the representative elements is the di-agonal relationship. Diagonal relationships are similarities between pairs of elementsin different groups and periods of the periodic table. Specifically, the first three mem-bers of the second period (Li, Be, and B) exhibit many similarities to those elementslocated diagonally below them in the periodic table (Figure 8.13). The reason for thisphenomenon is the closeness of the charge densities of their cations. (Charge densityis the charge of an ion divided by its volume.) Elements with comparable charge den-sities react similarly with anions and therefore form the same type of compounds. Thusthe chemistry of lithium resembles that of magnesium in some ways; the same holdsfor beryllium and aluminum and for boron and silicon. Each of these pairs is said toexhibit a diagonal relationship. We will see a number of examples of this relationshiplater.

308 PERIODIC RELATIONSHIPS AMONG THE ELEMENTS

8.6 VARIATION IN CHEMICAL PROPERTIES OF THE REPRESENTATIVE ELEMENTS

Li

Na Al Si

B CBe

Mg

1A 2A 3A 4A

FIGURE 8.13 Diagonal rela-tionships in the periodic table.

Bear in mind that a comparison of the properties of elements in the same groupis most valid if we are dealing with elements of the same type with respect to theirmetallic character. This guideline applies to the elements in Groups 1A and 2A, whichare all metals, and to the elements in Groups 7A and 8A, which are all nonmetals. InGroups 3A through 6A, where the elements change either from nonmetals to metalsor from nonmetals to metalloids, it is natural to expect greater variation in chemicalproperties even though the members of the same group have similar outer electron con-figurations.

Now let us take a closer look at the chemical properties of the representative el-ements and the noble gases. (We will consider the chemistry of the transition metalsin Chapter 22.)

Hydrogen (1s1)

There is no totally suitable position for hydrogen in the periodic table. Traditionallyhydrogen is shown in Group 1A, but it really could be a class by itself. Like the alkalimetals, it has a single s valence electron and forms a unipositive ion (H�), which ishydrated in solution. On the other hand, hydrogen also forms the hydride ion (H�) inionic compounds such as NaH and CaH2. In this respect, hydrogen resembles the halo-gens, all of which form uninegative ions (F�, Cl�, Br�, and I�) in ionic compounds.Ionic hydrides react with water to produce hydrogen gas and the corresponding metalhydroxides:

2NaH(s) � 2H2O(l) 88n 2NaOH(aq) � H2(g)

CaH2(s) � 2H2O(l) 88n Ca(OH)2(s) � 2H2(g)

Of course, the most important compound of hydrogen is water, which forms when hy-drogen burns in air:

2H2(g) � O2(g) 88n 2H2O(l)

Group 1A Elements (ns1, n � 2)

Figure 8.14 shows the Group 1A elements, the alkali metals. All of these elements havelow ionization energies and therefore a great tendency to lose the single valence elec-tron. In fact, in the vast majority of their compounds they are unipositive ions. Thesemetals are so reactive that they are never found in the pure state in nature. They reactwith water to produce hydrogen gas and the corresponding metal hydroxide:

2M(s) � 2H2O(l) 88n 2MOH(aq) � H2(g)

where M denotes an alkali metal. When exposed to air, they gradually lose their shinyappearance as they combine with oxygen gas to form oxides. Lithium forms lithiumoxide (containing the O2� ion):

4Li(s) � O2(g) 88n 2Li2O(s)

The other alkali metals all form oxides and peroxides (containing the O22� ion). For

example,

2Na(s) � O2(g) 88n Na2O2(s)

Potassium, rubidium, and cesium also form superoxides (containing the O2� ion):

K(s) � O2(g) 88n KO2(s)

8.6 VARIATION IN CHEMICAL PROPERTIES OF THE REPRESENTATIVE ELEMENTS 309

K

Rb

Cs

Na

Li

FIGURE 8.14 The Group 1Aelements: the alkali metals.Francium (not shown) is radioac-tive.

The reason that different types of oxides are formed when alkali metals react with oxy-gen has to do with the stability of the oxides in the solid state. Since these oxides areall ionic compounds, their stability depends on how strongly the cations and anions at-tract one another. Lithium tends to form predominantly lithium oxide because this com-pound is more stable than lithium peroxide. The formation of other alkali metal oxidescan be explained similarly.

Group 2A Elements (ns2, n � 2)

Figure 8.15 shows the Group 2A elements. As a group, the alkaline earth metals aresomewhat less reactive than the alkali metals. Both the first and the second ionizationenergies decrease from beryllium to barium. Thus the tendency is to form M2� ions(where M denotes an alkaline earth metal atom), and hence the metallic character in-creases from top to bottom. Most beryllium compounds (BeH2 and beryllium halides,such as BeCl2) and some magnesium compounds (MgH2, for example) are molecularrather than ionic in nature.

The reactivities of alkaline earth metals with water vary quite markedly. Berylliumdoes not react with water; magnesium reacts slowly with steam; calcium, strontium,and barium are reactive enough to attack cold water:

Ba(s) � 2H2O(l) 88n Ba(OH)2(aq) � H2(g)

The reactivities of the alkaline earth metals toward oxygen also increase from Be toBa. Beryllium and magnesium form oxides (BeO and MgO) only at elevated temper-atures, whereas CaO, SrO, and BaO form at room temperature.

310 PERIODIC RELATIONSHIPS AMONG THE ELEMENTS

Lithium (Li) Sodium (Na)

Rubidium (Rb)Potassium (K) Cesium (Cs)

Ca

Sr

Ba

Mg

Be

FIGURE 8.15 The Group 2Aelements: the alkaline earth metals. Magnesium reacts with acids in aqueous solution, liberating hydrogen gas:

Mg(s) � 2H�(aq) 88n Mg2�(aq) � H2(g)

Calcium, strontium, and barium also react with aqueous acid solutions to produce hy-drogen gas. However, because these metals also attack water, two different reactionswill occur simultaneously.

The chemical properties of calcium and strontium provide an interesting exampleof periodic group similarity. Strontium-90, a radioactive isotope, is a major product ofan atomic bomb explosion. If an atomic bomb is exploded in the atmosphere, the stron-tium-90 formed will eventually settle on land and water, and it will reach our bodiesvia a relatively short food chain. For example, if cows eat contaminated grass and drinkcontaminated water, they will pass along strontium-90 in their milk. Because calciumand strontium are chemically similar, Sr2� ions can replace Ca2� ions in our bones.Constant exposure of the body to the high-energy radiation emitted by the strontium-90 isotopes can lead to anemia, leukemia, and other chronic illnesses.

Group 3A Elements (ns2np1, n � 2)

The first member of Group 3A, boron, is a metalloid; the rest are metals (Figure 8.16).Boron does not form binary ionic compounds and is unreactive toward oxygen gas andwater. The next element, aluminum, readily forms aluminum oxide when exposed toair:

4Al(s) � 3O2(g) 88n 2Al2O3(s)

Aluminum that has a protective coating of aluminum oxide is less reactive than ele-mental aluminum. Aluminum forms only tripositive ions. It reacts with hydrochloricacid as follows:

8.6 VARIATION IN CHEMICAL PROPERTIES OF THE REPRESENTATIVE ELEMENTS 311

Beryllium (Be)

Strontium (Sr)

Magnesium (Mg)

Barium (Ba)

Calcium (Ca)

Radium (Ra)

Ga

In

Tl

Al

B

FIGURE 8.16 The Group 3Aelements.

2Al(s) � 6H�(aq) 88n 2Al3�(aq) � 3H2(g)

The other Group 3A metallic elements form both unipositive and tripositive ions.Moving down the group, we find that the unipositive ion becomes more stable than thetripositive ion.

The metallic elements in Group 3A also form many molecular compounds. Forexample, aluminum reacts with hydrogen to form AlH3, which resembles BeH2 in itsproperties. (Here is an example of the diagonal relationship.) Thus, from left to rightacross the periodic table, we are seeing a gradual shift from metallic to nonmetalliccharacter in the representative elements.

Group 4A Elements (ns2np2, n � 2)

The first member of group 4A, carbon, is a nonmetal, and the next two members, sil-icon and germanium, are metalloids (Figure 8.17). These elements do not form ioniccompounds. The metallic elements of this group, tin and lead, do not react with water,but they do react with acids (hydrochloric acid, for example) to liberate hydrogen gas:

Sn(s) � 2H�(aq) 88n Sn2�(aq) � H2(g)

Pb(s) � 2H�(aq) 88n Pb2�(aq) � H2(g)

The Group 4A elements form compounds in both the �2 and �4 oxidation states.For carbon and silicon, the �4 oxidation state is the more stable one. For example, CO2

is more stable than CO, and SiO2 is a stable compound, but SiO does not exist undernormal conditions. As we move down the group, however, the trend in stability isreversed. In tin compounds the �4 oxidation state is only slightly more stable than the�2 oxidation state. In lead compounds the �2 oxidation state is unquestionably themore stable one. The outer electron configuration of lead is 6s26p2, and lead tends to

312 PERIODIC RELATIONSHIPS AMONG THE ELEMENTS

Ge

Sn

Pb

Si

C

Boron (B) Aluminum (Al)

Indium (In)Gallium (Ga)

FIGURE 8.17 The Group 4Aelements.

lose only the 6p electrons (to form Pb2�) rather than both the 6p and 6s electrons (toform Pb4�).

Group 5A Elements (ns2np3, n � 2)

In Group 5A, nitrogen and phosphorus are nonmetals, arsenic and antimony are met-alloids, and bismuth is a metal (Figure 8.18). Thus we expect a greater variation inproperties within the group.

Elemental nitrogen is a diatomic gas (N2). It forms a number of oxides (NO, N2O,NO2, N2O4, and N2O5), of which only N2O5 is a solid; the others are gases. Nitrogenhas a tendency to accept three electrons to form the nitride ion, N3� (thus achieving theelectron configuration 2s22p6, which is isoelectronic with neon). Most metallic nitrides(Li3N and Mg3N2, for example) are ionic compounds. Phosphorus exists as P4 mole-cules. It forms two solid oxides with the formulas P4O6 and P4O10. The importantoxoacids HNO3 and H3PO4 are formed when the following oxides react with water:

N2O5(s) � H2O(l) 88n 2HNO3(aq)

P4O10(s) � 6H2O(l) 88n 4H3PO4(aq)

Arsenic, antimony, and bismuth have extensive three-dimensional structures. Bismuthis a far less reactive metal than those in the preceding groups.

Group 6A Elements (ns2np4, n � 2)

The first three members of Group 6A (oxygen, sulfur, and selenium) are nonmetals,and the last two (tellurium and polonium) are metalloids (Figure 8.19). Oxygen is adiatomic gas; elemental sulfur and selenium have the molecular formulas S8 and Se8,respectively; tellurium and polonium have more extensive three-dimensional structures.

8.6 VARIATION IN CHEMICAL PROPERTIES OF THE REPRESENTATIVE ELEMENTS 313

Carbon (graphite) Carbon (diamond) Silicon (Si)

Germanium (Ge) Tin (Sn) Lead (Pb)

As

Sb

Bi

P

N

Se

Te

Po

S

O

FIGURE 8.18 The Group 5Aelements. Molecular nitrogen is acolorless, odorless gas.

314 PERIODIC RELATIONSHIPS AMONG THE ELEMENTS

FIGURE 8.19 The Group 6A elements sulfur, selenium, and tellurium. Molecular oxygen is acolorless, odorless gas. Polonium (not shown) is radioactive.

Nitrogen (N2)

Arsenic (As)

Sulfur (S8) Selenium (Se8) Tellurium (Te)

Antimony (Sb) Bismuth (Bi)

White and red phosphorous (P)

“Not

Availa

ble in

e-Te

xt Ve

rsion

”

FIGURE 8.20 The Group 7Aelements chlorine, bromine, andiodine. Fluorine is a greenish-yellow gas that attacks ordi-nary glassware. Astatine is radioactive.

(Polonium is a radioactive element that is difficult to study in the laboratory.) Oxygenhas a tendency to accept two electrons to form the oxide ion (O2�) in many ionic com-pounds. Sulfur, selenium, and tellurium also form dinegative anions (S2�, Se2�, andTe2�). The elements in this group (especially oxygen) form a large number of molec-ular compounds with nonmetals. The important compounds of sulfur are SO2, SO3,and H2S. Sulfuric acid is formed when sulfur trioxide reacts with water:

SO3(g) � H2O(l) 88n H2SO4(aq)

Group 7A Elements (ns2np5, n � 2)

All the halogens are nonmetals with the general formula X2, where X denotes a halo-gen element (Figure 8.20). Because of their great reactivity, the halogens are neverfound in the elemental form in nature. (The last member of Group 7A, astatine, is aradioactive element. Little is known about its properties.) Fluorine is so reactive thatit attacks water to generate oxygen:

2F2(g) � 2H2O(l) 88n 4HF(aq) � O2(g)

Actually the reaction between molecular fluorine and water is quite complex; the prod-ucts formed depend on reaction conditions. The reaction shown above is one of sev-eral possible changes.

The halogens have high ionization energies and high electron affinities. Anionsderived from the halogens (F�, Cl�, Br�, and I�) are called halides. They are iso-electronic with the noble gases immediately to their right in the periodic table. For ex-ample, F� is isoelectronic with Ne, Cl� with Ar, and so on. The vast majority of thealkali metal halides and alkaline earth metal halides are ionic compounds. The halo-gens also form many molecular compounds among themselves (such as ICl and BrF3)and with nonmetallic elements in other groups (such as NF3, PCl5, and SF6). The halo-gens react with hydrogen to form hydrogen halides:

H2(g) � X2(g) 88n 2HX(g)

When this reaction involves fluorine, it is explosive, but it becomes less and less vio-lent as we substitute chlorine, bromine, and iodine. The hydrogen halides dissolve inwater to form hydrohalic acids. Hydrofluoric acid (HF) is a weak acid (that is, it is a

8.6 VARIATION IN CHEMICAL PROPERTIES OF THE REPRESENTATIVE ELEMENTS 315

Br

I

At

Cl

F

FIGURE 8.21 All noble gasesare colorless and odorless. Thesepictures show the colors emittedby the gases from a dischargetube.

weak electrolyte), but the other hydrohalic acids (HCl, HBr, and HI) are all strong acids(strong electrolytes).

Group 8A Elements (ns2np6, n � 2)

All noble gases exist as monatomic species (Figure 8.21). Their atoms have completelyfilled outer ns and np subshells, which give them great stability. (Helium is 1s2.) TheGroup 8A ionization energies are among the highest of all elements, and these gaseshave no tendency to accept extra electrons. For years these elements were called inertgases, and rightly so. Until 1963 no one had been able to prepare a compound con-taining any of these elements. The British chemist Neil Bartlett† shattered chemists’long-held views of these elements when he exposed xenon to platinum hexafluoride,a strong oxidizing agent, and brought about the following reaction (Figure 8.22):

Xe(g) � PtF6(g) 88n XePtF6(s)

Since then, a number of xenon compounds (XeF4, XeO3, XeO4, XeOF4) and a fewkrypton compounds (KrF2, for example) have been prepared (Figure 8.23). Despite theimmense interest in the chemistry of the noble gases, however, their compounds do nothave any commercial applications, and they are not involved in natural biologicalprocesses. No compounds of helium, neon, and argon are known.

COMPARISON OF GROUP 1A AND GROUP 1B ELEMENTS

When we compare the Group 1A elements (alkali metals) and the Group 1B elements(copper, silver, and gold), we arrive at an interesting conclusion. Although the metalsin these two groups have similar outer electron configurations, with one electron in theoutermost s orbital, their chemical properties are quite different.

316 PERIODIC RELATIONSHIPS AMONG THE ELEMENTS

Helium (He) Neon (Ne) Argon (Ar) Krypton (Kr) Xenon (Xe)

†Neil Bartlett (1932– ). English chemist. Bartlett’s work is mainly in the preparation and study of compounds with un-usual oxidation states and in solid-state chemistry.

Ar

Kr

Xe

Ne

He

FIGURE 8.22 (a) Xenon gas(colorless) and PtF6 (red gas) sep-arated from each other. (b) Whenthe two gases are allowed tomix, a yellow-orange solid com-pound is formed.

The first ionization energies of Cu, Ag, and Au are 745 kJ/mol, 731 kJ/mol, and890 kJ/mol, respectively. Since these values are considerably larger than those of thealkali metals (see Table 8.3), the Group 1B elements are much less reactive. The higherionization energies of the Group 1B elements result from incomplete shielding of thenucleus by the inner d electrons (compared with the more effective shielding of thecompletely filled noble gas cores). Consequently the outer s electrons of these elementsare more strongly attracted by the nucleus. In fact, copper, silver, and gold are so un-reactive that they are usually found in the uncombined state in nature. The inertnessand rarity of these metals make them valuable in the manufacture of coins and in jew-elry. For this reason, these metals are also called “coinage metals.” The difference inchemical properties between the Group 2A elements (the alkaline earth metals) and theGroup 2B metals (zinc, cadmium, and mercury) can be explained in a similar way.

PROPERTIES OF OXIDES ACROSS A PERIOD

One way to compare the properties of the representative elements across a period is toexamine the properties of a series of similar compounds. Since oxygen combines withalmost all elements, we will compare the properties of oxides of the third-period ele-ments to see how metals differ from metalloids and nonmetals. Some elements in thethird period (P, S, and Cl) form several types of oxides, but for simplicity we will con-sider only those oxides in which the elements have the highest oxidation number. Table8.5 lists a few general characteristics of these oxides. We observed earlier that oxygen

8.6 VARIATION IN CHEMICAL PROPERTIES OF THE REPRESENTATIVE ELEMENTS 317

FIGURE 8.23 Crystals ofxenon tetrafluoride (XeF4).

(a) (b)

ABLE 8.5 Some Properties of Oxides of the Third-Period Elements

Na2O MgO Al2O3 SiO2 P4O10 SO3 Cl2O7

Type of compound Ionic MolecularStructure Extensive three-dimensional Discrete

molecular unitsMelting point (°C) 1275 2800 2045 1610 580 16.8 �91.5Boiling point (°C) ? 3600 2980 2230 ? 44.8 82Acid-base nature Basic Basic Amphoteric Acidic

m8888888m88

m8888888m88

m888888m888

m888888m8888

m8888888 m8888888

Note that this acid-base neutraliza-tion produces a salt but no water.

has a tendency to form the oxide ion. This tendency is greatly favored when oxygencombines with metals that have low ionization energies, namely, those in Groups 1Aand 2A, plus aluminum. Thus Na2O, MgO, and Al2O3 are ionic compounds, as indi-cated by their high melting points and boiling points. They have extensive three-dimensional structures in which each cation is surrounded by a specific number of an-ions, and vice versa. As the ionization energies of the elements increase from left toright, so does the molecular nature of the oxides that are formed. Silicon is a metal-loid; its oxide (SiO2) also has a huge three-dimensional network, although no ions arepresent. The oxides of phosphorus, sulfur, and chlorine are molecular compounds com-posed of small discrete units. The weak attractions among these molecules result inrelatively low melting points and boiling points.

Most oxides can be classified as acidic or basic depending on whether they pro-duce acids or bases when dissolved in water or react as acids or bases in certainprocesses. Some oxides are amphoteric, which means that they display both acidic andbasic properties. The first two oxides of the third period, Na2O and MgO, are basicoxides. For example, Na2O reacts with water to form the base sodium hydroxide:

Na2O(s) � H2O(l) 88n 2NaOH(aq)

Magnesium oxide is quite insoluble; it does not react with water to any appreciable ex-tent. However, it does react with acids in a manner that resembles an acid-base reaction:

MgO(s) � 2HCl(aq) 88n MgCl2(aq) � H2O(l)

Note that the products of this reaction are a salt (MgCl2) and water, the usual productsof an acid-base neutralization.

Aluminum oxide is even less soluble than magnesium oxide; it too does not re-act with water. However, it shows basic properties by reacting with acids:

Al2O3(s) � 6HCl(aq) 88n 2AlCl3(aq) � 3H2O(l)

It also exhibits acidic properties by reacting with bases:

Al2O3(s) � 2NaOH(aq) � 3H2O(l) 88n 2NaAl(OH)4(aq)

Thus Al2O3 is classified as an amphoteric oxide because it has properties of both acidsand bases. Other amphoteric oxides are ZnO, BeO, and Bi2O3.

Silicon dioxide is insoluble and does not react with water. It has acidic proper-ties, however, because it reacts with very concentrated bases:

SiO2(s) � 2NaOH(aq) 88n Na2SiO3(aq) � H2O(l)

For this reason, concentrated aqueous, strong bases such as NaOH(aq) should not bestored in Pyrex glassware, which is made of SiO2.

The remaining third-period oxides are acidic. They react with water to form phos-phoric acid (H3PO4), sulfuric acid (H2SO4), and perchloric acid (HClO4):

P4O10(s) � 6H2O(l) 88n 4H3PO4(aq)

SO3(g) � H2O(l) 88n H2SO4(aq)

Cl2O7(l) � H2O(l) 88n 2HClO4(aq)

This brief examination of oxides of the third-period elements shows that as themetallic character of the elements decreases from left to right across the period, theiroxides change from basic to amphoteric to acidic. Metallic oxides are usually basic,and most oxides of nonmetals are acidic. The intermediate properties of the oxides (as

318 PERIODIC RELATIONSHIPS AMONG THE ELEMENTS

shown by the amphoteric oxides) are exhibited by elements whose positions are inter-mediate within the period. Note also that since the metallic character of the elementsincreases from top to bottom within a group of representative elements, we would ex-pect oxides of elements with higher atomic numbers to be more basic than the lighterelements. This is indeed the case.

EXAMPLE 8.6

Classify the following oxides as acidic, basic, or amphoteric: (a) Rb2O, (b) BeO,(c) As2O5.

Answer (a) Since rubidium is an alkali metal, we would expect Rb2O to be a ba-sic oxide. This is indeed true, as shown by rubidium oxide’s reaction with water toform rubidium hydroxide:

Rb2O(s) � H2O(l) 88n 2RbOH(aq)

(b) Beryllium is an alkaline earth metal. However, because it is the first member ofGroup 2A, we expect that it may differ somewhat from the other members of thegroup. Furthermore, beryllium and aluminum exhibit a diagonal relationship, so thatBeO may resemble Al2O3 in properties. It turns out that BeO, like Al2O3, is an am-photeric oxide, as shown by its reactions with acids and bases:

BeO(s) � 2H�(aq) � 3H2O(l) 88n Be(H2O)2�4 (aq)

BeO(s) � 2OH�(aq) � H2O(l) 88n Be(OH)2�4 (aq)

(c) Since arsenic is a nonmetal, we expect As2O5 to be an acidic oxide. This pre-diction is correct, as shown by the formation of arsenic acid when As2O5 reacts withwater:

As2O5(s) � 3H2O(l) 88n 2H3AsO4(aq)

PRACTICE EXERCISEClassify the following oxides as acidic, basic, or amphoteric: (a) ZnO, (b) P4O10,(c) CaO.

8.6 VARIATION IN CHEMICAL PROPERTIES OF THE REPRESENTATIVE ELEMENTS 319

Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry Chem

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In the late 1800s John William Strutt, Third Baron ofRayleigh, who was a professor of physics at theCavendish Laboratory in Cambridge, England, accu-rately determined the atomic masses of a number ofelements, but he obtained a puzzling result with ni-trogen. One of his methods of preparing nitrogen wasby the thermal decomposition of ammonia:

2NH3(g) 88n N2(g) � 3H2(g)

Another method was to start with air and remove from

it oxygen, carbon dioxide, and water vapor.Invariably, the nitrogen from air was a little denser (byabout 0.5%) than the nitrogen from ammonia.

Lord Rayleigh’s work caught the attention of SirWilliam Ramsay, a professor of chemistry at theUniversity College, London. In 1898 Ramsay passednitrogen, which he had obtained from air byRayleigh’s procedure, over red-hot magnesium to con-vert it to magnesium nitride:

3Mg(s) � N2(g) 88n Mg3N2(s)

Discovery of the Noble Gases

Similar problem: 8.70.

1. Nineteenth-century chemists developed the periodic table by arranging elements in the in-creasing order of their atomic masses. Discrepancies in early versions of the periodic tablewere resolved by arranging the elements in order of their atomic numbers.

2. Electron configuration determines the properties of an element. The modern periodic tableclassifies the elements according to their atomic numbers, and thus also by their electronconfigurations. The configuration of the valence electrons directly affects the properties ofthe atoms of the representative elements.

3. Periodic variations in the physical properties of the elements reflect differences in atomicstructure. The metallic character of elements decreases across a period from metals through

320 PERIODIC RELATIONSHIPS AMONG THE ELEMENTS

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After all of the nitrogen had reacted with magnesium,Ramsey was left with an unknown gas that would notcombine with anything.

With the help of Sir William Crookes, the inven-tor of the discharge tube, Ramsay and Lord Rayleighfound that the emission spectrum of the gas did notmatch any of the known elements. The gas was a newelement! They determined its atomic mass to be39.95 amu and called it argon, which means “thelazy one” in Greek.

Once argon had been discovered, other noblegases were quickly identified. Also in 1898 Ramsayisolated helium from uranium ores (see Chemistry inAction essay on p. 255). From the atomic masses ofhelium and argon, their lack of chemical reactivity,and what was then known about the periodic table,Ramsay was convinced that there were other unreac-tive gases and that they were all members of one pe-riodic group. He and his student Morris Travers setout to find the unknown gases. They used a refriger-ation machine to first produce liquid air. Applying atechnique called fractional distillation, they then al-lowed the liquid air to warm up gradually and col-lected components that boiled off at different temper-atures. In this manner they analyzed and identifiedthree new elements—neon, krypton, and xenon—inonly three months. Three new elements in three monthsis a record that may never be broken!

The discovery of the noble gases helped to com-plete the periodic table. Their atomic masses sug-gested that these elements should be placed to theright of the halogens. The apparent discrepancy withthe position of argon was resolved by Moseley, as dis-cussed in the chapter.

Finally, the last member of the noble gases, radon,was discovered by the German chemist Frederick Dorn

in 1900. A radioactive element and the heaviest ele-mental gas known, radon’s discovery not only com-pleted the Group 8A elements, but also advanced ourunderstanding about the nature of radioactive decayand transmutation of elements.

Lord Rayleigh and Ramsay both won Nobel Prizesin 1904 for the discovery of argon. Lord Rayleigh re-ceived the prize in physics and Ramsay’s award wasin chemistry.

Sir William Ramsay (1852–1916).

SUMMARY OF FACTS AND CONCEPTS

the metalloids to nonmetals and increases from top to bottom within a particular group ofrepresentative elements.

4. Atomic radius varies periodically with the arrangement of the elements in the periodictable. It decreases from left to right and increases from top to bottom.

5. Ionization energy is a measure of the tendency of an atom to resist the loss of an electron.The higher the ionization energy, the stronger the attraction between the nucleus and an elec-tron. Electron affinity is a measure of the tendency of an atom to gain an electron. The morepositive the electron affinity, the greater the tendency for the atom to gain an electron. Metalsusually have low ionization energies, and nonmetals usually have high electron affinities.

6. Noble gases are very stable because their outer ns and np subshells are completely filled. Themetals among the representative elements (in Groups 1A, 2A, and 3A) tend to lose electronsuntil their cations become isoelectronic with the noble gases that precede them in the periodictable. The nonmetals in Groups 5A, 6A, and 7A tend to accept electrons until their anionsbecome isoelectronic with the noble gases that follow them in the periodic table.

QUESTIONS AND PROBLEMS 321

KEY WORDS

Amphoteric oxide, p. 318Atomic radius, p. 295Diagonal relationship, p. 308

Electron affinity, p. 305Ionic radius, p. 297Ionization energy, p. 301

Isoelectronic, p. 294Noble gases, p. 290Representative elements, p. 290

Valence electrons, p. 291

DEVELOPMENT OF THE PERIODIC TABLE

Review Questions

8.1 Briefly describe the significance of Mendeleev’s pe-riodic table.

8.2 What is Moseley’s contribution to the modern peri-odic table?

8.3 Describe the general layout of a modern periodictable.

8.4 What is the most important relationship among ele-ments in the same group in the periodic table?

PERIODIC CLASSIFICATION OF THE ELEMENTS

Review Questions

8.5 Which of the following elements are metals, non-metals, or metalloids? As, Xe, Fe, Li, B, Cl, Ba, P, I,Si.

8.6 Compare the physical and chemical properties of met-als and nonmetals.

8.7 Draw a rough sketch of a periodic table (no detailsare required). Indicate regions where metals, non-metals, and metalloids are located.

8.8 What is a representative element? Give names andsymbols of four representative elements.

8.9 Without referring to a periodic table, write the nameand give the symbol for an element in each of the fol-lowing groups: 1A, 2A, 3A, 4A, 5A, 6A, 7A, 8A,transition metals.

QUESTIONS AND PROBLEMS8.10 Indicate whether the following elements exist as

atomic species, molecular species, or extensive three-dimensional structures in their most stable states at25°C and 1 atm and write the molecular or empiricalformula for each one: phosphorus, iodine, magne-sium, neon, arsenic, sulfur, boron, selenium, and oxy-gen.

8.11 You are given a dark shiny solid and asked to deter-mine whether it is iodine or a metallic element.Suggest a nondestructive test that would allow you toarrive at the correct answer.

8.12 What are valence electrons? For representative ele-ments, the number of valence electrons of an elementis equal to its group number. Show that this is truefor the following elements: Al, Sr, K, Br, P, S, C.

8.13 Write the outer electron configurations for the (a) al-kali metals, (b) alkaline earth metals, (c) halogens,(d) noble gases.

8.14 Use the first-row transition metals (Sc to Cu) as anexample to illustrate the characteristics of the elec-tron configurations of transition metals.

8.15 How does the electron configuration of ions derivedfrom representative elements give them stability?

8.16 What do we mean when we say that two ions or anatom and an ion are isoelectronic?

8.17 What is wrong with the statement “The atoms of el-ement X are isoelectronic with the atoms of elementY”?

8.18 Give three examples of first-row transition metal (Scto Cu) ions whose electron configurations are repre-sented by the argon core.

Problems

8.19 In the periodic table, the element hydrogen is some-times grouped with the alkali metals (as in this book)and sometimes with the halogens. Explain why hy-drogen can resemble the Group 1A and the Group 7Aelements.

8.20 A neutral atom of a certain element has 17 electrons.Without consulting a periodic table, (a) write theground-state electron configuration of the element,(b) classify the element, (c) determine whether theatoms of this element are diamagnetic or paramag-netic.

8.21 Group the following electron configurations in pairsthat would represent similar chemical properties oftheir atoms:(a) 1s22s22p63s2

(b) 1s22s22p3

(c) 1s22s22p63s23p64s23d104p6

(d) 1s22s2

(e) 1s22s22p6

(f) 1s22s22p63s23p3

8.22 Group the following electron configurations in pairsthat would represent similar chemical properties oftheir atoms:(a) 1s22s22p5

(b) 1s22s1

(c) 1s22s22p6

(d) 1s22s22p63s23p5

(e) 1s22s22p63s23p64s1

(f) 1s22s22p63s23p64s23d104p6

8.23 Without referring to a periodic table, write the elec-tron configuration of elements with the followingatomic numbers: (a) 9, (b) 20, (c) 26, (d) 33. Classifythe elements.

8.24 Specify the group of the periodic table in which eachof the following elements is found: (a) [Ne]3s1,(b) [Ne]3s23p3, (c) [Ne]3s23p6, (d) [Ar]4s23d8.

8.25 A M2� ion derived from a metal in the first transi-tion metal series has four electrons in the 3d subshell.What element might M be?

8.26 A metal ion with a net �3 charge has five electronsin the 3d subshell. Identify the metal.

8.27 Write the ground-state electron configurations of thefollowing ions: (a) Li�, (b) H�, (c) N3�, (d) F�,(e) S2�, (f) Al3�, (g) Se2�, (h) Br�, (i) Rb�, (j) Sr2�,(k) Sn2�, (l) Te2�, (m) Ba2�, (n) Pb2�, (o) In3�,(p) Tl�, (q) Tl3�.

8.28 Write the ground-state electron configurations of thefollowing ions, which play important roles in bio-

chemical processes in our bodies: (a) Na�, (b) Mg2�,(c) Cl�, (d) K�, (e) Ca2�, (f) Fe2�, (g) Cu2�,(h) Zn2�.

8.29 Write the ground-state electron configurations of thefollowing transition metal ions: (a) Sc3�, (b) Ti4�,(c) V5�, (d) Cr3�, (e) Mn2�, (f) Fe2�, (g) Fe3�,(h) Co2�, (i) Ni2�, (j) Cu�, (k) Cu2�, (l) Ag�,(m) Au�, (n) Au3�, (o) Pt2�.

8.30 Name the ions with �3 charges that have the fol-lowing electron configurations: (a) [Ar]3d3, (b) [Ar],(c) [Kr]4d6, (d) [Xe]4f145d6.

8.31 Which of the following species are isoelectronic witheach other? C, Cl�, Mn2�, B�, Ar, Zn, Fe3�, Ge2�.

8.32 Group the species that are isoelectronic: Be2�, F�,Fe2�, N3�, He, S2�, Co3�, Ar.

PERIODIC VARIATION IN PHYSICAL PROPERTIES

Review Questions

8.33 Define atomic radius. Does the size of an atom havea precise meaning?

8.34 How does atomic radius change (a) from left to rightacross a period and (b) from top to bottom in a group?

8.35 Define ionic radius. How does the size of an atomchange when it is converted to (a) an anion and (b) acation?

8.36 Explain why, for isoelectronic ions, the anions arelarger than the cations.

Problems

8.37 On the basis of their positions in the periodic table,select the atom with the larger atomic radius in eachof the following pairs: (a) Na, Cs; (b) Be, Ba;(c) N, Sb; (d) F, Br; (e) Ne, Xe.

8.38 Arrange the following atoms in order of decreasingatomic radius: Na, Al, P, Cl, Mg.

8.39 Which is the largest atom in Group 4A?8.40 Which is the smallest atom in Group 7A?8.41 Why is the radius of the lithium atom considerably

larger than the radius of the hydrogen atom?8.42 Use the second period of the periodic table as an ex-

ample to show that the size of atoms decreases as wemove from left to right. Explain the trend.

8.43 Indicate which one of the two species in each of thefollowing pairs is smaller: (a) Cl or Cl�; (b) Na orNa�; (c) O2� or S2�; (d) Mg2� or Al3�; (e) Au� orAu3�.

8.44 List the following ions in order of increasing ionicradius: N3�, Na�, F�, Mg2�, O2�.

8.45 Explain which of the following cations is larger, andwhy: Cu� or Cu2�.

8.46 Explain which of the following anions is larger, andwhy: Se2� or Te2�.

322 PERIODIC RELATIONSHIPS AMONG THE ELEMENTS

8.47 Give the physical states (gas, liquid, or solid) of therepresentative elements in the fourth period (K, Ca,Ga, Ge, As, Se, Br) at 1 atm and 25°C.

8.48 The boiling points of neon and krypton are �245.9°Cand �152.9°C, respectively. Using these data, esti-mate the boiling point of argon.

IONIZATION ENERGY

Review Questions

8.49 Define ionization energy. Ionization energy measure-ments are usually made when atoms are in thegaseous state. Why? Why is the second ionization en-ergy always greater than the first ionization energyfor any element?

8.50 Sketch the outline of the periodic table and showgroup and period trends in the first ionization energyof the elements. What types of elements have thehighest ionization energies and what types the low-est ionization energies?

Problems

8.51 Use the third period of the periodic table as an ex-ample to illustrate the change in first ionization en-ergies of the elements as we move from left to right.Explain the trend.

8.52 In general, ionization energy increases from left toright across a given period. Aluminum, however, hasa lower ionization energy than magnesium. Explain.

8.53 The first and second ionization energies of K are419 kJ/mol and 3052 kJ/mol, and those of Ca are590 kJ/mol and 1145 kJ/mol, respectively. Comparetheir values and comment on the differences.

8.54 Two atoms have the electron configurations 1s22s22p6

and 1s22s22p63s1. The first ionization energy of oneis 2080 kJ/mol, and that of the other is 496 kJ/mol.Match each ionization energy with one of the givenelectron configurations. Justify your choice.

8.55 A hydrogenlike ion is an ion containing only one elec-tron. The energies of the electron in a hydrogenlikeion are given by

En � �(2.18 � 10�18 J)Z2 ��n12��

where n is the principal quantum number and Z is theatomic number of the element. Calculate the ioniza-tion energy (in kJ/mol) of the He� ion.

8.56 Plasma is a state of matter consisting of positivegaseous ions and electrons. In the plasma state, a mer-cury atom could be stripped of its 80 electrons andtherefore would exist as Hg80�. Use the equation inProblem 8.55 to calculate the energy required for thelast ionization step, that is,

Hg79�(g) 88n Hg80�(g) � e�

ELECTRON AFFINITY

Review Questions

8.57 (a) Define electron affinity. (b) Electron affinity mea-surements are made with gaseous atoms. Why?(c) Ionization energy is always a positive quantity,whereas electron affinity may be either positive ornegative. Explain.

8.58 Explain the trends in electron affinity from aluminumto chlorine (see Table 8.4).

Problems

8.59 Arrange the elements in each of the following groupsin increasing order of the most positive electron affin-ity: (a) Li, Na, K, (b) F, Cl, Br, I.

8.60 Specify which of the following elements you wouldexpect to have the greatest electron affinity: He, K,Co, S, Cl.

8.61 Considering their electron affinities, do you think itis possible for the alkali metals to form an anion likeM�, where M represents an alkali metal?

8.62 Explain why alkali metals have a greater affinity forelectrons than alkaline earth metals.

VARIATION IN CHEMICAL PROPERTIES OF THEREPRESENTATIVE ELEMENTS

Review Questions

8.63 What is meant by the diagonal relationship? Nametwo pairs of elements that show this relationship.

8.64 Which elements are more likely to form acidic ox-ides? Basic oxides? Amphoteric oxides?

Problems

8.65 Use the alkali metals and alkaline earth metals as ex-amples to show how we can predict the chemicalproperties of elements simply from their electron con-figurations.

8.66 Based on your knowledge of the chemistry of the al-kali metals, predict some of the chemical propertiesof francium, the last member of the group.

8.67 As a group, the noble bases are very stable chemi-cally (only Kr and Xe are known to form compounds).Why?

8.68 Why are Group 1B elements more stable than Group1A elements even though they seem to have the sameouter electron configuration, ns1, where n is the prin-cipal quantum number of the outermost shell?

8.69 How do the chemical properties of oxides changefrom left to right across a period? From top to bot-tom within a particular group?

QUESTIONS AND PROBLEMS 323

8.70 Write balanced equations for the reactions betweeneach of the following oxides and water: (a) Li2O,(b) CaO, (c) SO3.

8.71 Write formulas for and name the binary hydrogencompounds of the second-period elements (Li to F).Describe how the physical and chemical propertiesof these compounds change from left to right acrossthe period.

8.72 Which oxide is more basic, MgO or BaO? Why?

ADDITIONAL PROBLEMS

8.73 State whether each of the following properties of therepresentative elements generally increases or de-creases (a) from left to right across a period and(b) from top to bottom within a group: metallic char-acter, atomic size, ionization energy, acidity of ox-ides.

8.74 With reference to the periodic table, name (a) a halo-gen element in the fourth period, (b) an element sim-ilar to phosphorus in chemical properties, (c) the mostreactive metal in the fifth period, (d) an element thathas an atomic number smaller than 20 and is similarto strontium.

8.75 Write equations representing the following processes:(a) The electron affinity of S�.(b) The third ionization energy of titanium.(c) The electron affinity of Mg2�.(d) The ionization energy of O2�.

8.76 Arrange the following isoelectronic species in orderof (a) increasing ionic radius and (b) increasing ion-ization energy: O2�, F�, Na�, Mg2�.

8.77 Write the empirical (or molecular) formulas of com-pounds that the elements in the third period (sodiumto chlorine) should form with (a) molecular oxygenand (b) molecular chlorine. In each case indicatewhether you would expect the compound to be ionicor molecular in character.

8.78 Element M is a shiny and highly reactive metal (melt-ing point 63°C), and element X is a highly reactivenonmetal (melting point �7.2°C). They react to forma compound with the empirical formula MX, a col-orless, brittle white solid that melts at 734°C. Whendissolved in water or when in the molten state, thesubstance conducts electricity. When chlorine gas isbubbled through an aqueous solution containing MX,a reddish-brown liquid appears and Cl� ions areformed. From these observations, identify M and X.(You may need to consult a handbook of chemistryfor the melting-point values.)

8.79 Match each of the elements on the right with its de-scription on the left:(a) A dark-red liquid Calcium (Ca)(b) A colorless gas that burns Gold (Au)

in oxygen gas Hydrogen (H2)

(c) A reactive metal that attacks Argon (Ar)water Bromine (Br2)

(d) A shiny metal that is used injewelry

(e) A totally inert gas8.80 Arrange the following species in isoelectronic pairs:

O�, Ar, S2�, Ne, Zn, Cs�, N3�, As3�, N, Xe.8.81 In which of the following are the species written in

decreasing order by size of radius? (a) Be, Mg, Ba,(b) N3�, O2�, F�, (c) Tl3�, Tl2�, Tl�.

8.82 Which of the following properties show a clear peri-odic variation? (a) first ionization energy, (b) molarmass of the elements, (c) number of isotopes of anelement, (d) atomic radius.

8.83 When carbon dioxide is bubbled through a clear cal-cium hydroxide solution, the solution appears milky.Write an equation for the reaction and explain howthis reaction illustrates that CO2 is an acidic oxide.

8.84 You are given four substances: a fuming red liquid,a dark metallic-looking solid, a pale-yellow gas, anda yellow-green gas that attacks glass. You are told thatthese substances are the first four members of Group7A, the halogens. Name each one.

8.85 For each pair of elements listed below, give threeproperties that show their chemical similarity:(a) sodium and potassium and (b) chlorine andbromine.

8.86 Name the element that forms compounds, under ap-propriate conditions, with every other element in theperiodic table except He, Ne, and Ar.

8.87 Explain why the first electron affinity of sulfur is200 kJ/mol but the second electron affinity is�649 kJ/mol.

8.88 The H� ion and the He atom have two 1s electronseach. Which of the two species is larger? Explain.

8.89 Predict the products of the following oxides with wa-ter: Na2O, BaO, CO2, N2O5, P4O10, SO3. Write anequation for each of the reactions. Specify whetherthe oxides are acidic, basic, or amphoteric.

8.90 Write the formulas and names of the oxides of thesecond-period elements (Li to N). Identify the oxidesas acidic, basic, or amphoteric.

8.91 State whether each of the following elements is a gas,a liquid, or a solid under atmospheric conditions. Alsostate whether it exists in the elemental form as atoms,as molecules, or as a three-dimensional network: Mg,Cl, Si, Kr, O, I, Hg, Br.

8.92 What factors account for the unique nature of hy-drogen?

8.93 The air in a manned spacecraft or submarine needsto be purified of exhaled carbon dioxide. Write equa-tions for the reactions between carbon dioxide and(a) lithium oxide (Li2O), (b) sodium peroxide(Na2O2), and (c) potassium superoxide (KO2).

324 PERIODIC RELATIONSHIPS AMONG THE ELEMENTS

8.94 The formula for calculating the energies of an elec-tron in a hydrogenlike ion is given in Problem 8.55.This equation cannot be applied to many-electronatoms. One way to modify it for the more complexatoms is to replace Z with (Z � �), where Z is theatomic number and � is a positive dimensionlessquantity called the shielding constant. Consider thehelium atom as an example. The physical significanceof � is that it represents the extent of shielding thatthe two 1s electrons exert on each other. Thus thequantity (Z � �) is appropriately called the “effectivenuclear charge.” Calculate the value of � if the firstionization energy of helium is 3.94 � 10�18 J peratom. (Ignore the minus sign in the given equation inyour calculation.)

8.95 Why do noble gases have negative electron affinityvalues?

8.96 The atomic radius of K is 216 pm and that of K� is133 pm. Calculate the percent decrease in volume thatoccurs when K(g) is converted to K�(g). [The vol-ume of a sphere is (�

43

�) r3, where r is the radius of thesphere.]

8.97 The atomic radius of F is 72 pm and that of F� is136 pm. Calculate the percent increase in volume thatoccurs when F(g) is converted to F�(g). (See Problem8.96 for the volume of a sphere.)

8.98 A technique called photoelectron spectroscopy isused to measure the ionization energy of atoms. Asample is irradiated with UV light, and electrons areejected from the valence shell. The kinetic energiesof the ejected electrons are measured. Since the en-ergy of the UV photon and the kinetic energy of theejected electron are known, we can write

h� � IE � �12

� mu2

where � is the frequency of the UV light, and m andu are the mass and velocity of the electron, respec-tively. In one experiment the kinetic energy of theejected electron from potassium is found to be 5.34 �10�19 J using a UV source of wavelength 162 nm.Calculate the ionization energy of potassium. Howcan you be sure that this ionization energy corre-sponds to the electron in the valence shell (that is, themost loosely held electron)?

8.99 Referring to the Chemistry in Action essay on p. 319,answer the following questions. (a) Why did it takeso long to discover the first noble gas (argon) onEarth? (b) Once argon had been discovered, why didit take relatively little time to discover the rest of thenoble gases? (c) Why was helium not isolated by thefractional distillation of liquid air?

8.100 The energy needed for the following process is1.96 � 104 kJ/mol:

Li(g) 88n Li3�(g) � 3e�

If the first ionization energy of lithium is 520 kJ/mol,calculate the second ionization energy of lithium, thatis, the energy required for the process

Li�(g) 88n Li2�(g) � e�

(Hint: You need the equation in Problem 8.55.)8.101 An element X reacts with hydrogen gas at 200°C to

form compound Y. When Y is heated to a higher tem-perature, it decomposes to the element X and hydro-gen gas in the ratio of 559 mL of H2 (measured atSTP) for 1.00 g of X reacted. X also combines withchlorine to form a compound Z, which contains 63.89percent by mass of chlorine. Deduce the identity ofX.

8.102 A student is given samples of three elements, X, Y,and Z, which could be an alkali metal, a member ofGroup 4A, and a member of Group 5A. She makesthe following observations: Element X has a metal-lic luster and conducts electricity. It reacts slowlywith hydrochloric acid to produce hydrogen gas.Element Y is a light-yellow solid that does not con-duct electricity. Element Z has a metallic luster andconducts electricity. When exposed to air, it slowlyforms a white powder. A solution of the white pow-der in water is basic. What can you conclude aboutthe elements from these observations?

8.103 Using the following boiling-point data and the pro-cedure in the Chemistry in Action essay on p. 302,estimate the boiling point of francium:

8.104 What is the electron affinity of the Na� ion?8.105 The ionization energies of sodium (in kJ/mol), start-

ing with the first and ending with the eleventh, are495.9, 4560, 6900, 9540, 13400, 16600, 20120,25490, 28930, 141360, 170000. Plot the log of ion-ization energy (y-axis) versus the number of ioniza-tion (x-axis); for example, log 495.9 is plotted versus1 (labeled I1, the first ionization energy), log 4560 isplotted versus 2 (labeled I2, the second ionization en-ergy), and so on. (a) Label I1 through I11 with theelectrons in orbitals such as 1s, 2s, 2p, and 3s.(b) What can you deduce about electron shells fromthe breaks in the curve?

8.106 Experimentally, the electron affinity of an elementcan be determined by using a laser light to ionize theanion of the element in the gas phase:

X�(g) � h� 88n X(g) � e�

QUESTIONS AND PROBLEMS 325

metal Li Na K Rb Cs

boiling point (°C) 1347 882.9 774 688 678.4

Referring to Table 8.4, calculate the photon wave-length (in nanometers) corresponding to the electronaffinity for chlorine. In what region of the electro-magnetic spectrum does this wavelength fall?

8.107 Explain, in terms of their electron configurations,why Fe2� is more easily oxidized to Fe3� than Mn2�

to Mn3�.8.108 The standard enthalpy of atomization of an element

is the energy required to convert one mole of an el-ement in its most stable form at 25°C to one mole ofmonatomic gas. Given that the standard enthalpy ofatomization for sodium is 108.4 kJ/mol, calculate theenergy in kilojoules required to convert one mole ofsodium metal at 25°C to one mole of gaseous Na�

ions.8.109 Write the formulas and names of the hydrides of the

following second-period elements: Li, C, N, O, F.Predict their reactions with water.

8.110 Based on knowledge of the electronic configurationof titanium, state which of the following compoundsof titanium is unlikely to exist: K3TiF6, K2Ti2O5,TiCl3, K2TiO4, K2TiF6.

8.111 Name an element in Group 1A or Group 2A that isan important constituent of each of the following sub-stances: (a) remedy for acid indigestion, (b) coolantin nuclear reactors, (c) Epsom salt, (d) baking pow-der, (e) gunpowder, (f) a light alloy, (g) fertilizer thatalso neutralizes acid rain, (h) cement, and (i) grit foricy roads. You may need to ask your instructor aboutsome of the items.

8.112 In halogen displacement reactions a halogen elementcan be generated by oxidizing its anions with a halo-gen element that lies above it in the periodic table.This means that there is no way to prepare elemen-tal fluorine, since it is the first member of Group 7A.Indeed, for years the only way to prepare elementalfluorine was to oxidize F� ions by electrolytic means.Then, in 1986, a chemist reported that by reactingpotassium hexafluoromanganate(IV) (K2MnF6) withantimony pentafluoride (SbF5) at 150°C, he had gen-erated elemental fluorine. Balance the followingequation representing the reaction:

K2MnF6 � SbF5 88n KSbF6 � MnF3 � F2

8.113 Write a balanced equation for the preparation of(a) molecular oxygen, (b) ammonia, (c) carbon diox-ide, (d) molecular hydrogen, (e) calcium oxide.Indicate the physical state of the reactants and prod-ucts in each equation.

8.114 Write chemical formulas for oxides of nitrogen withthe following oxidation numbers: �1, �2, �3, �4,�5. (Hint: There are two oxides of nitrogen with �4oxidation number.)

8.115 Most transition metal ions are colored. For example,a solution of CuSO4 is blue. How would you showthat the blue color is due to the hydrated Cu2� ionsand not the SO2�

4 ions?8.116 In general, atomic radius and ionization energy have

opposite periodic trends. Why?8.117 Explain why the electron affinity of nitrogen is ap-

proximately zero, while the elements on either side,carbon and oxygen, have substantial positive electronaffinities.

8.118 Consider the halogens chlorine, bromine, and iodine.The melting point and boiling point of chlorine are�101.0°C and �34.6°C while those of iodine are113.5°C and 184.4°C, respectively. Thus chlorine isa gas and iodine is a solid under room conditions.Estimate the melting point and boiling point ofbromine. Compare your values with those from ahandbook of chemistry.

8.119 While it is possible to determine the second, third,and higher ionization energies of an element, thesame cannot usually be done with the electron affini-ties of an element. Explain.

8.120 The only confirmed compound of radon is radon flu-oride, RnF. One reason that it is difficult to study thechemistry of radon is that all isotopes of radon areradioactive so it is dangerous to handle the substance.Can you suggest another reason why there are so fewknown radon compounds? (Hint: Radioactive decaysare exothermic processes.)

8.121 Little is known of the chemistry of astatine, the lastmember of Group 7A. Describe the physical charac-teristics that you would expect this halogen to have.Predict the products of the reaction between sodiumastatide (NaAt) and sulfuric acid. (Hint: Sulfuric acidis an oxidizing agent.)

8.122 As discussed in the chapter, the atomic mass of ar-gon is greater than that of potassium. This observa-tion created a problem in the early development ofthe periodic table because it meant that argon shouldbe placed after potassium. (a) How was this difficultyresolved? (b) From the following data, calculate theaverage atomic masses of argon and potassium: Ar-36 (35.9675 amu; 0.337%), Ar-38 (37.9627 amu;0.063%), Ar-40 (39.9624 amu; 99.60%); K-39(38.9637 amu; 93.258%), K-40 (39.9640 amu;0.0117%), K-41 (40.9618 amu; 6.730%).

8.123 Calculate the maximum wavelength of light (innanometers) required to ionize a single sodium atom.

8.124 Predict the atomic number and ground-state electronconfiguration of the next member of the alkali met-als after francium.

8.125 Why do elements that have high ionization energiesalso have more positive electron affinities? Which

326 PERIODIC RELATIONSHIPS AMONG THE ELEMENTS

group of elements would be an exception to this gen-eralization?

8.126 The first four ionization energies of an element areapproximately 738 kJ/mol, 1450 kJ/mol, 7.7 �103 kJ/mol, and 1.1 � 104 kJ/mol. To which periodicgroup does this element belong? Why?

8.127 Some chemists think that helium should properly becalled “helon.” Why? What does the ending in helium(-ium) suggest?

8.128 (a) The formula of the simplest hydrocarbon is CH4(methane). Predict the formulas of the simplest com-pounds formed between hydrogen and the followingelements: silicon, germanium, tin, and lead.(b) Sodium hydride (NaH) is an ionic compound.Would you expect rubidium hydride (RbH) to bemore or less ionic than NaH? (c) Predict the reactionbetween radium (Ra) and water. (d) When exposedto air, aluminum forms a tenacious oxide (Al2O3)coating that protects the metal from corrosion. Whichmetal in Group 2A would you expect to exhibit sim-ilar properties? Why?

8.129 Match each of the elements on the right with its de-scription on the left:(a) A pale yellow gas Nitrogen (N2)

that reacts with water. Boron (B)(b) A soft metal that reacts Aluminum (Al)

with water to produce Fluorine (F2)hydrogen. Sodium (Na)

(c) A metalloid that is hardand has a high meltingpoint.

(d) A colorless, odorlessgas.

(e) A metal that is more reactive thaniron, but does not corrodein air.

Answers to Practice Exercises: 8.1 (a) 1s22s22p63s23p64s2,(b) it is a representative element, (c) diamagnetic. 8.2 Li � Be �C. 8.3 (a) Li�, (b) Au3�, (c) N3�. 8.4 (a) N, (b) Mg. 8.5 No.8.6 (a) amphoteric, (b) acidic, (c) basic.

QUESTIONS AND PROBLEMS 327