Chemistry part 11, Julia Burdge,2e (2009) ppsx

SECTION 7.2 The Modern Periodic Table 241 Based on the type of s ubshell containing the outermost electron s, the elements can be divided into ategories the main group elements, the nob

240 CHAPTER 7 Electr on Configuration and the Periodic Table

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Essential elements in the human body

a ll biological molecule s Carbon i s perhaps the mo st versati le element because it has the ability to

fo rm various types of chemical bond s Carbon atoms can form bond s to each other, linking up to form an enormous variety of chains and ring structures

The metals play severa l different roles in living syste m s As cations (Na +, K +, Ca 2 + , and

Mg 2+), they serve to maintain the balance between intracellular and extracellular fluid s, nerve transmissions, and other activities.1'hey are also needed for protein functions For example, the Fe2 + ion binds oxygen in hemoglobin molecules and Cu2 + , Zn2 + , and Mg2+ ion s are essential for enzyme activ it y In addition, calcium in the form of Cas(P04MOH) and Ca3(P04)2 is an essential component of teeth and bones

These elements are necessary for biological functions s uch as growth, the transport of gen for metabolism, and defense against disease There is a delicate balance in the amounts of these elements in our bodie s Too much or too littl e over an extended period of time can lead to serious illness, retard ation, or even death

oxy-Checkpoint 7.1 Development of the Periodic Table

7 1 1 Which of the following elements would

you expect to have chemical properties most similar to tho se of S?

a) P b) C I c) Se d) Na e) Sr

7.1 2

The Modern Periodic Table

Which of the following elements would you expect to have properties simi l ar to those ofBa?

a) Sr b) Rb c) Na d) K e) B

Figure 7.2 shows the modern periodic table together with the outermost gro und-state electron figurations of the elements (The electron configurations of the elements are also given in Figure 6.25.) Starting with hydrogen, the electronic subshells are filled in the order shown in Figure 6.23 [ ~~ Section 6 8]

con-•

SECTION 7.2 The Modern Periodic Table 241

Based on the type of s ubshell containing the outermost electron s, the elements can be divided into

ategories the main group elements, the noble gases, the transition e l ements (or tran s ition

met-als), the lanthanid es, and the actinides The main group elements ( also called the r e pr esenta ti ve

e lements) are the elements in Groups lA through 7 A With th e exce ption of helium, each of the

n oble gases (the Group 8A elements) ha s a completely filled p s ubshell The electron configura

-t io n s are I i for helium and ninp6 for the other noble ga ses, where n i s the principal quantum

n um ber for the outermost shell

The transition metal s are the element s in Groups IB and 3B through 8B Transition metals

e ithe r have incompletely filled d s ub s hell s or readily produc e catio ns with incompletely filled d .,

ubs hells According to this defi nition, the elements of Group 2B a r e not transition metals They are

d -b lock elements, though, so they ge nerally are included in th e discussion of transitio n metals

The lanth anides and actinides are sometimes calledf-block transition elements because they

h av e incompletely filled f s ub shells Figure 7.3 distinguishe s the groups of elements discu sse d

he re

There is a distinct pattern to the electron configurations of the elements in a particular group

ee, for example , the electron configurations of Groups IA and 2A in Table 7.1 Each member of

G ro up 1 A ha s a noble gas core plus one additional electron , giving each alkali metal the general

el ectron configuration of [noble gas]nsl Similarly , the Group 2A alkaline earth metals have a

no ble gas core and an outer electron configuration of ni

In this context, outermost electrons refe rs to those that are placed in orbitals last using the

Aufbau principle [ ~ Section 6.8)

The met a ls of Group 2B typically form + 2 ions, although they can also form + 1 i ons In either case, the electron configuration includes

a completed dsubshell [ ~~ Section 7.6] Thus , strictly speaking, Zn, Cd, and Hg are not

transition metals

,i

"" ~

Why Are There Two Different Sets of Numbers

at the Top of the Periodic Table?

The numbering of the transition metal groups 3B through 7B

Group 3A) each have three outer electrons However, because

their outer electrons reside in different types of atomic orbitals

(s and d orbitals in the case of Sc; s and p orbitals in the case

IE and 2B have filled d subs hells [ ~~ Section 6.9] Unlike the

to the one and two electrons, respectively, that they have in s

Figure 7.3 Periodic table with

color-coding of main group elements,

noble gases, transition metals, group

2B metals, lanthanides, and actinides

elements that appear beneath them in the periodic table, cannot

be classified in this way and are all placed in Group SB

The designation of A and B groups is not universal In Europe, the practice is to use B for main group elements and

A for transition or d-block elements, which is just the opposite

of the American convention The International Union of Pure and Applied Chemistry (IUPAC) has recommended eliminating ambiguity by numbering the columns sequentially with Arabic

been accepted universally, and many modern periodic tables retain the traditional group designations Periodic tables in this text display both the IUPAC-recommended Arabic numerals and the traditional American numbering system Discussions in the

8A

18 3A 4A SA 6A 7A

13 14 15 16 17 He 1

B C N 0 F Ne 2 4B SB 6B 7B 1s8B~ IB 2B

SECTION 7.2 The Modern Periodic Table 243

The outermost electrons of an atom are called valence electrons, which are the ones involved

in the formation of chemical bonds between atoms The similarity of the valence electron

configu-rations (i.e., they have the same number and type of valence electrons) is what makes the elements

in the same group resemble one another chemically This observation holds true for the other main

group elements as well For instance, the halogens (Group 7 A) all have outer electron

configura-tions of ninp5, and they have similar properties

In predicting properties for Groups 3A through 7 A, we must take into account that each of these groups contains elements on both sides of the zigzag line that divides metals and nonmetals

For example, the elements in Group 4A all have the same outer electron configuration, ns 2np2, but

there is considerable variation in chemical properties among these elements because carbon is a

nonmetal, silicon and germanium are metalloids, and tin and lead are metals

As a group, the noble gases behave very similarly The noble gases are chemically inert because they all have completely filled outer ns and np subshells, a condition that imparts unusual

stability With the exception of krypton and xenon, they do not undergo chemical reactions

Although the outer electron configuration of the transition metals is not always the same within a group and there is often no regular pattern in the way the electron configuration changes

from one metal to the next in the same period, all transition metals share many characteristics

(multiple oxidation states, richly colored compounds, magnetic properties, and so on) that set

them apalt from other elements These properties are similar because all these metals have

incom-pletely filled d subshells Likewise, the lanthanide and actinide elements resemble one another

because they have incompletely filled f subshells

Sample Problem 7.2 shows how to determine the electron configuration from the number of electrons in an atom

Without using a periodic table, give the ground-state electron configuration and block designation

(s-, p-, d-, orf-block) of an atom with (a) 17 electrons, (b) 37 electrons, and (c) 22 electrons Classify

each atom as a main group element or transition metal

Strategy Use the Aufbau principle discussed in Section 6.8 Start writing each electron

configuration with principal quantum number n = 1, and then continue to assign electrons to orbitals

in the order presented in Figure 6.23 until all the electrons have been accounted for

Setup According to Figure 6.23, orbitals fill in the following order: Is , 2s, 2p, 3s, 3p, 4s, 3d, 4p,

Ss, 4d, Sp, 6s , and so on Recall that an s subshell contains one orbital, a p subshell contains three

orbitals, and a d subshell contains five orbitals Remember, too, that each orbital can accommodate a

maximum of two electrons The block designation of an element corresponds to the type of subshell

occupied by the last electrons added to the configuration according to the Aufbau principle

Solution

(a) li2s 2 2p63s23i, p-block, main group

(b) li2s22p 6 3s 2 3p 6 4i3d104iSsl, s-block, main group

(c) li2s22p 63 s 23 p 64i 3d 2 , d-block, transition metal

Practice Problem A Without using a periodic table, give the ground-state electron configuration and

block designation (s-, p-, d-, orf-block) of an atom with (a) 15 electrons, (b) 20 electrons, and (c) 35

electrons

Practice Problem B Identify the elements represented by (a) l S22s22i3s23pl, (b) l s 22i2p63i3i4s2

3d lO , and (c) lS 2 2s 2 2p63s23p64s23dl04isi

Representing Free Elements in Chemical Equations

Having classified the elements according to their ground-state electron configurations, we can now

learn how chemists represent elements in chemical equations

etals Because metals do not exist in discrete molecular units but rather in complex,

three-dimensional networks of atoms, we always use their empirical formulas in chemical equations

Think About It Consult Figure

6.25 to confirm your answers

244 CHAPTER 7 Electron Configuration and the Periodic Table

The empirical formulas are the same as the symbols that represent the elements For example, the empirical formula for iron is Fe, th e sa me as the symbo l for the element

equa- ,

Recall that allotropes are different forms of the tions Carbon, for examp le , exists in several allotropic forms R egar dle ss of the allotrope, we u se same elem ent [ H~ Section 2.6] it s emp iri ca l formula C to represent elemental carbon in chemical equations Often the symbol C

will be followed b y the spec ific allotrope in parenthe ses as in the equation representing the

conver-s ion of grap hite to diamond, two of carbon's allotropic forms:

Checkpoint 7.2

C(graphite) - _ C(d iamond )

For nonmetals that ex i s t as polyatomic molecules, we generally use the molecular formula in

equations: H2, N2, O ? , F2, C12, Br b 12, and P4, for example In the case of sulfur, however, we

usu-ally u se the empirical formula S rather than the molecular formula S 8 ' Thus, instead of writing the equation for the combustion of s ulfur a s

we u s u a ll y write

although, technically, both ways are correct

Kr, Xe, a nd Rn

also represent them with their empirical formulas that i s, their symbo l s: B , Si, Ge, and so on

7.2.1 Which e le ctro n configuration is correct fo r a gennan ium ( Ge )

7.2.2 Which of the following equations correctly repr ese nt the

c hemical r eactio n in which grap hite combines with s ulfur to

Shielding is also known as screening

Core electrons are tho se i n the completed inner

Effective Nuclear Charge

As we have seen, the e lectron configurations of th e elements s how a periodic variation with

incr easi ng atomic number In this and the next few sect ions , we will examine how electron figuration ex plain s the periodic variation of phy s ical and chemical properties of the elements We

con-begin by introducing the concept of effect ive nucl ear charge

Nuclear charge (Z) is simply the number of protons in the nucleus of an atom Effective nuclear charge (ZerrJ is the actual magnitude of po s itive charge that is "experienced" by an elec-

tron in the atom Th e o nly atom in which the nucl ear charge and effective nuclear charge are the

same is hydrogen , which has only one e lectron In all other atoms, the electrons are s

imultane-ously attracted to the nucleus and repelled by one another This re s ults in a phenomenon known as

shielding An electron in a many-electron atom is partially shielded from the positive charge of the nucleu s b y the other electrons in the atom

One way to illu strate how electrons in an atom s hi e ld one another i s to consider the amounts

of energy required to remove the two electrons from a helium atom, s hown in Figure 7.4

Experi-ment s s how that it take s 3.94 X 10 - 1 8 J to remove the first electron but 8.72 X 10 - 1 8 J to remove the second one There i s no s hielding once the first electron is removed, so the second electron feel s the full effect of the + 2 nuclear charge and i s more difficult to remove

Although all the electrons in an atom shield one another to some extent, tho se that are most

effective at s hielding are the core electrons As a result, the value of Z e ff increa ses s teadily from

What Causes the Periodic Trends in Properties?

Many of the periodic trends in properties of the elements can be

explained using Coulomb's* law, which states that the force ( F)

between two charged objects (QJ and Q 2 ) is dire ctly proportional

to the product of the two charges and inversel y proportional to

the distance (d) between the object s squared: Force is inver s ely

proportional to d 2 , whereas energy is inversely proportional to d

[I ~~ Section 5.1] The SI unit of force is the newton (N =

m kg/s 2 ) and the SI unit of energy is the joule (J = m2 • kg /s2)

When the charge s have opposite signs, F is negative indicating

the s ame s ign , F is po s itive - indicating a repulsive force Table

7.2 s how s how the magnitude of the attractive force between two oppo s itely charged objects at a fixed di s tance from each other varie s with change s in the magnitudes of the charge s

F et: QJ X Q 2

d 2

* Ch a rle s Au g u s tin d e C o ul o mb ( 17 3 6 - 1 8 06 ) Fr e nch ph ys ici s t Coulomb did

re se ar c h in el ec tri c i ty and m ag neti s m and a pplied N e w ton 's inver se s quar e law to

ele c tricit y H e al s o in ve nted a tor s i o n balanc e

le ft to right across a period of the periodic table becau s e the number of core electron s remains the

s ame (only the number of protons, Z, and the number of val e nce electrons increases)

As we move to the right across period 2 , the nuclear charge increase s b y 1 with each new

e lement, but the effective nuclear charge increa s es only by an average of 0.64 (If the valence

elec-tr ons did not shield one another , the effective nuclear charge would al s o incr e a s e by I each time a

p roton was added to the nucleus.)

Z

Z e ff (felt by valence electron s )

Li

3 1.28

Be

4 1.91

In general, the effective nuclear charge is given by

Z eff = Z - IT

B

5 2.42

F

9

5.10

Equation 7.1

w here IT is the shielding constant The shielding con s tant i s greater than z ero but smaller than Z

The change in Z eff as we move from the top of a group to the bottom i s generally le ss s

ignifi-c a nt than the change as we move acros s a period Although each step down a group repre s ent s a

l ar ge increase in the nuclear charge, there is al s o an additi o nal s hell of core e lectron s to s hield the

y a lence electrons from the nucleus Consequently , the eff ec tive nuclear charge changes le ss than

rb e nuclear charge as we move down a column of the periodic table

Fig u re 7.4 Removal of the first

electron in He requires less energy than removal of the second electron because

of shielding

245

246 CHAPTER 7 Electron Configuration and the Periodic Table

(a)

(b ) Figure 7.5 (a) Atomic radius in

metals is defined as half the distance

between adjacent metal atoms

(b) Atomic radius in nonmetals is

defined as half the distance between

bonded identical atoms in a molecule

~'!!-- Multimedia

Periodic Table atomic radius

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Periodic Trends in Properties of Elements

Several physical and chemical propertie s of the elements depend on effective nuclear charge To understand the trends in these properties, it is helpful to visualize the electrons of an atom in

shells Recall that the value of the principal quantum number (n) increases as the distance from

the nucleus increases [ ~~ Section 6.7] If we take this statement literally, and picture all the electron s in a shell at the s ame distance from the nucleus , the result is a sphere of uniformly dis- tributed negative charge , with its distance from the nucleus depending on the value of n With this

a s a s tarting point , we will examine the periodic trends in atomic radius , ionization energy, and electron affinity

Atomic Radius

Intuitively , we think of the atomic radius as the distance between the nucleus of an atom and its valence shell (i.e , the outermost shell that is occupied by one or more electrons), because we usu- ally envi s ion atom s a s spheres with discrete boundaries According to the quantum mechanical model of the atom, though , there is no specific distance from the nucleus beyond which an electron may not be found [ ~~ Section 6.7] Therefore , the atomic radius requires a specific definition

There are two ways in which the atomic radius is commonly defined One is the metallic radius, which is half the distance between the nuclei of two adjacent, identical metal atoms [Fig- ure 7.S(a ) ] The other is the covalent radius, which is halfthe distance between adjacent, identical nuclei in a molecule [Figure 7.S(b)]

Figure 7.6 shows the atomic radii of the main group elements according to their positions

in the periodic table There are two distinct trends The atomic radius decreases as we move from

left to right acros s a period and increases from top to bottom as we move down within a group

SECTION 7.4 Periodic Trends in Properties of Elements 247

The increase down a group is fairly easily explained As we step down a column, the outermost

occupied shell has an ever-increasing value of n, so it lies farther from the nucleus, making the

radius bigger

Now let's try to understand the decrease in radius from left to right across a period Although

this trend may at first seem counterintuitive, given that the number of valence electrons is increasing

with each new element, consider the shell model in which all the electrons in a shell form a uniform

sphere of negative charge around the nucleus at a distance specified by the value of n As we move

from left to right across a period, the effective nuclear charge increases and each step to the right

adds another electron to the valence shell Coulomb's law dictates that there will be a more powerful

attraction between the nucleus and the valence shell when the magnitudes of both charges increase

The result is that as we step across a period th e valence shell is drawn closer to the nucleus, making

the atomic radius smaller Figure 7.7 shows how the effective nuclear charge, charge on the valence

shell, and atomic radius vary across period 2 We can picture the valence shells in all the atoms as

being initially at the same distance (determined by n) from the nuclei, but being pulled closer by a

larger attractive force resulting from increases in both Zeff and the number of valence electrons

Sample Problem 7.3 shows how to use these trends to compare the atomic radii of different elements

Referring only to a periodic table, arrange the elements P, S, and 0 in order of increasing atomic

radius

Strategy Use the left-to-right (decreasing) and top-to-bottom (increasing) trends to compare the

atomic radii of two of the three elements at a time

Setup Sulfur is to the right of phosphorus in the third row, so sulfur should be smaller that

phosphorus Oxygen is above sulfur in Group 6A, so oxygen should be smaller than sulfur

Solution: 0 < S < P

Practice Problem A Referring only to a periodic table, arrange the elements Ge, Se, and F in order

of increasing atomic radius

Practice Problem B For which of the following pairs of elements can the atomic radii not be

compared using the periodic table alone: P and Se, Se and Cl, or P and O?

-~

Ionization Energy

Ionization energy (IE) is the minimum energy required to remove an electron from an atom in the

gas phase Typically, we express ionization energy in kJ/mol, the number of kilojoules required to

remove a mole of electrons from a mole of gaseous atoms Sodium, for example, has an ionization

energy of 495.8 kJ/mol, meaning that the energy input required to drive the process

Na(g) • Na + (g) +

e-is 495.8 kJ/mol Specifically, this is the first ionization energy of sodium, lE I (Na), which

corre-ponds to the removal of the most loosely held electron Figure 7.8(a) shows the first ionization

to right and increase from top to bottom, the

observed radii do not vary in as regular a way as

do the main group elements

to confirm the order Note that there are circumstances under which the trends alone will be insufficient to

compare the radii of two elements

Using only a periodic table, for example, it would not be possible to determine that chlorine (r = 99 pm) has a larger radius than oxygen

(r = 73 pm)

from left to right across a period because of the increased electrostatic

attraction between the effective nuclear

charge and the charge on the valence

shell The dark circle shows the atomic

size in each case

248 CHAPTER 7 Electron Configuration and the Periodic Table

Figure 7.8 (a) First ionization energies (in kJ / mol) of the main gro up elements (b) First ionization energy as a function of atomic number

As with atom ic radius, ionization energy

changes in a s imilar but somewhat less regular

way among the transition elements

ele-ment Both of the se interruptions of the upward trend in lE \ can be explained by u s ing electron configuration

-tion 6.8, Figure 6.23] Within a given shell, electrons with the higher value of.e have a higher

s how s the relative energies of an s s ubshell (.e = 0) and a p s ub s hell (.e = 1) Ionization of an

in the same orbital makes it ea s ier to remove one of them, making the ionization energy for the

The first ionization energy l Ei decreases as we move from top to bottom within a group due

to the increasing atomic radius Although the effective nuclear charge does not change s

Figure 7.9 (a) It i s harder to remove an e l ectron from an s orbital than it i s to remove an electron from a p orbital with the same principal quantum

number (b) Within a p subs hell , it i s easier to remove an e lectron from a doubly occup i ed orbital than from a si ngly occ upied orbital

SECTION 7.4 Periodic Trends in Properties of Elements 249

*Cells shaded with blue represent the removal of core electrons

s hell increases According to Coulomb 's law, the attractive force between a valence electron and

th e effective nuclear charge gets weaker as the distance between them increa ses This make s it

eas ier to remove an electron , and so lEI decrea s es

It is possible to remove additional electrons in subsequent ionization s, giving IE z, IE 3 , and so

o n The second and third ionizations of so dium, for example, can be repre se nted, respectively , as

it is harder to remove an electron from a cation than from an atom (a nd it gets even harder as the

ch arge on the cation increases) Table 7.3 li s t s the ionization energie s of the elements in period

2 and of sodium These data show that it take s much more energy to remove core electrons than

[0 remove valence electrons There are two reasons for this First , core electrons are closer to the

n ucleus, and second, core electrons experience a greater effective nuclear charge because there are

f e wer filled shells shielding them from the nucleu s Both of the se factors contribute to a greater

at tractive force between the electron s and the nucleus , which must be overcome in order to remove

m e electrons

Would you expect Na or Mg to have the greater first ionization energy (lEI ) ? Which s hould ha ve the

greater second ionization energy (IE 2 )?

Strategy Consider effective nuclear charge and electron configuration to compare the ionization

energies Effective nuclear charge increa ses from left to right in a period ( thu s increasing IE) , and it

is more difficult to remove a paired core electron than an unpaired valence electron

Setup Na is in Group lA, and Mg i s be s ide it in Group 2A Na ha s one vale nce e lectron , and Mg

has two valence electrons

Solution lEI (Mg) > lEI (Na) becau se Mg i s to the right of Na in the periodic table (i.e , Mg has the

greater effective nuclear charge, so it i s more difficult to remove it s electron) lE z(Na) > IE 2 (Mg)

because the s econd ionization of Mg remo ves a valence electron, whereas the seco nd ionization of

Na removes a core electron

Practice Problem A Which element , Mg or AI, will have the higher fir s t i o ni za tion energy and

which will have the higher third ionization energy?

Practice Problem B Explain why Rb ha s a lower fEI than Sr, but Sr ha s a lower fE z than Rb

energies of Na and Mg are 496

a nd 738 kJ / mol , re s pectively The second ionization energies of Na and Mg are 4562 and 1451 kJ / mol ,

re s pectivel y

250 CHAPTER 7 Electro n C onfigurat i on and the Period i c Table

Although IE and EA both increase f rom l e ft to

right across a period , an increase in IE means

that it is less likely th a t an electron will b e

r emoved from an ato m An i ncrease in fA, o n

the other hand, means that it is more likely that

an electron will be a cc epted by an atom

(in kllmol) of the main group elements

(b) Electron affinity as a function of

atomic number

Electron Affinity

Electron affinity (EA) is the energy released (the negative of the enthalpy change fJ.H) when an atom in the gas phase accepts an electron Consider the process in which a gaseous chlorine atom accepts an electron:

Cl(g) + e - - _ Cl - (g) fJ.H = -349.0 kJ/mol

A negative value of fJ.H indicates an exothermic process [ ~~ Section S 3 ], so 349.0 kJ/mol of energy are released (the definition of electron affinity) when a mole of gaseous chlorine atoms accepts a mole of electrons A positive electron affinity indicates a process that is energetically favorable In general, the larger and more positive the EA value, the more favorable the process and the more apt it is to occur Figure 7.10 s h ows electron affinities for the main group ele m ents

Like ionization energy, electron affinity increases from left to right across a period This trend

in EA is due to the increase in effective nuclear charge from left to right (i.e., it becomes

increases) There are also periodic interruptions of the upward trend of EA from left to right,

simi-lar to those observed for lEb although they do not occur for the same elements For examp l e, the

of a Group SA element is lower than that for the corresponding Group 4A element These tions to the trend are due to the electron configurations of the elements involved

Sr +5.03

Ba +13 95

J +295

At +270

8A

18

He (0 0)

Ne ( - 29)

Ar (-35)

Kr

(-39)

Xe ( - 41)

SECTION 7.4 Periodic Trends in Properties of Elements 251

subshell, it is easier to add an electron to an empty orbital than to add one to an orbital that already contains an electron

It is harder to add an electron to a Group 2A element (ni) than to the Group lA element

(ns1 ) in the same period because the electron added to the Group 2A element is placed in an

orbital of higher energy (a p orbital versus an s orbital) Likewi se, it is harder to add an electron

to a Group 5A element (ns 2 np 3) than to the corresponding Group 4A element (ninp2) because

the electron added to the Group 5A element must be placed in an orbital that already contains an

electron Figure 7.11 illustrates these points

Just as more than one electron can be removed from an atom, more than one electron can also be added to an atom While many first electron affinities are po s itive, subsequent electron

affinities are always negative Considerable energy is required to overcome the repulsive forces

between the electron and the negatively charged ion The addition of two electrons to a gaseous

oxygen atom can be represented as:

cantly endothermic process such as the addition of an electron to a gaseous 0- ion happens only

in concert with one or more exothermic processes that more than compensate for the required

e nergy input

Sample Problem 7.5 lets you practice using the periodic table to compare the electron ties of elements

Strategy Consider the effective nuclear charge and electron configuration to compare the electron

Setup (a) Al is in Group 3A and Si is beside it in Group 4A AI ha s three va lence electrons

([Ne]3i3p\ and Si has four valence electrons ([Ne]3i3p 2 )

Solution (a) EA 1 (Si) > EA 1 (AI) becau se Si is to the right of Al and therefore has a greater effective

(b) EA] (Si) > EA] (P) because although P is to the right of Si in the third period of the periodic table

(giving P the larger Zeff), adding an electron to a P atom require s placing it in a 3p orbital that is

Practice Problem A Would you expect Mg or Al to ha ve the higher EAl?

Pr ac tice Problem B Why is the EA] for Ge greater than the EA] for As ?

Valence orbital diagram for P

Think About It The first electron

affinities of AI, Si, and Pare 42.5,

134 , and 72.0 kJ / mol, respectively

252 CHAPTER 7 E lectron C onf i guration a n d t h e P eriodic T ab l e

Malleability i s t he p rop e rty t h at allows meta l s

to be po u n ded in to th in sheets Ductility, the

capacit y t o be dra wn out into wi r e s , is another

ch ara c teristic of meta ls

-• Be shiny, lustrous, and malleable

• Be good conductors of both heat and electricity

• Have low ionization energies (so they commonly form cations)

• Form basic, ionic compounds with oxygen (metal oxides) Nonmetals, on the other hand, tend to

• Vary in color and lack the shiny appearance associated with metals

• Be brittle, rather than malleable

• Be poor conductors of both heat and electricity

• Have high electron affinities (so they commonly form anions)

Metallic character increases from top to bottom in a group and decreases from left to right within

a period

Metalloids are elements with properties intermediate between those of metals and nonmet

-als Because the definition of metallic character depends on a combination of properties, there may

be some variation in the elements identified as metalloids in different sources Astatine (At), for

example, is listed as a metalloid in some sources and a nonmetal in others

7 4 1

7 4 2

An'ange the elements Ca, Sr, and Ba in

order of increasing IE ]

For each of the following pairs of

elements, indicate which will have the

Recall from Chapter 2 that we can use the periodic table to predict the charges on many

of the ions formed by main group elements Elements in Groups lA and 2A, for example, form

ions with charges of + 1 and + 2, respectively Elements in Groups 6A and 7 A form ions with

charges of - 2 and -1 , respectively Knowing something about electron configurations enables us

to explain these charges