chemical bonding i: basic concepts -...

10-1 Lewis Theory:An Overview

10-2 Covalent Bonding:An Introduction

10-3 Polar Covalent Bondsand ElectrostaticPotential Maps

10-4 Writing Lewis Structures

10-5 Resonance10-6 Exceptions to the

Octet Rule10-7 Shapes of Molecules10-8 Bond Order and

Bond Lengths10-9 Bond Energies➣ FOCUS ON Molecules in

Space: Measuring BondLengths

372

C O N T E N T S

CHEMICALBONDING I:

BASIC CONCEPTS

Consider all that we already know about chemical compounds.We can determine their compositions and write their formu-las. We can represent the reactions of compounds by chemical

equations and perform stoichiometric and thermochemical calcula-tions based on these equations. And we can do all this without reallyhaving to consider the ultimate structure of matter—the structure ofatoms and molecules. Yet the shape of a molecule—that is, thearrangement of its atoms in space—often defines its chemistry. Ifwater had a different shape, its properties would be significantlydifferent, and life as we know it would not be possible.

In this chapter, we will describe the interactions between atomscalled chemical bonds. Most of the discussion centers on the Lewis the-ory, one of the simplest methods of representing chemical bonding.We will also explore another relatively simple theory, one for predict-ing probable molecular shapes. Throughout the chapter, we will tryto relate these theories to what is known about molecular structuresfrom experimental measurements. In Chapter 11 we will examine the

Computer-generated electron charge-density maps (gray) and electrostaticpotential maps (colored) of methanol and ethanol The gray figures show the extent of electron charge density while thecolored figures show the distribution of charge in the molecule. In thischapter, we study ideas that enable us to predict the geometric shapesand polarity of molecules.

(CH3CH2OH).(CH3OH)

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10-1 Lewis Theory: An Overview 373

Since 1962, a number ofcompounds of Xe and Krhave been synthesized. As wewill see in this chapter, afocus on noble-gas electronconfigurations can still beuseful, even if the idea thatthey confer completeinertness is invalid.

▲

The term covalent wasintroduced by IrvingLangmuir.

▲

Gilbert Newton Lewis(1875–1946)Lewis’s contribution to thestudy of chemical bonding isevident throughout this text.Equally important, however,was his pioneering introduc-tion of thermodynamics intochemistry.

▲

subject of chemical bonding in greater depth, and in Chapter 12 we will describeintermolecular forces—forces between molecules—and explore further therelationship between molecular shape and the properties of substances.

LEWIS THEORY: AN OVERVIEW

In the period from 1916–1919, two Americans, G. N. Lewis and Irving Lang-muir, and a German, Walther Kossel, advanced an important proposal aboutchemical bonding: Something unique in the electron configurations of noblegas atoms accounts for their inertness, and atoms of other elements combinewith one another to acquire electron configurations like noble gas atoms. Thetheory that grew out of this model has been most closely associated with G. N.Lewis and is called the Lewis theory. Some fundamental ideas associatedwith Lewis’s theory are

1. Electrons, especially those of the outermost (valence) electronic shell, playa fundamental role in chemical bonding.

2. In some cases, electrons are transferred from one atom to another. Positiveand negative ions are formed and attract each other through electrostaticforces called ionic bonds.

3. In other cases, one or more pairs of electrons are shared between atoms. Abond formed by the sharing of electrons between atoms is called acovalent bond.

4. Electrons are transferred or shared in such a way that each atom acquiresan especially stable electron configuration. Usually this is a noble gas con-figuration, one with eight outer-shell electrons, or an octet.

LEWIS SYMBOLS AND LEWIS STRUCTURES

Lewis developed a special set of symbols for his theory. A Lewis symbol con-sists of a chemical symbol to represent the nucleus and core (inner-shell)electrons of an atom, together with dots placed around the symbol to representthe valence (outer-shell) electrons. Thus, the Lewis symbol for silicon, which hasthe electron configuration , is

Electron spin had not yet been proposed when Lewis framed his theory, andso he did not show that two of the valence electrons are paired and two

are unpaired. We will write Lewis symbols in the way Lewis did. We willplace single dots on the sides of the symbol, up to a maximum of four. Then wewill pair up dots until we reach an octet. Lewis symbols are commonly writtenfor main-group elements but much less often for transition elements. Lewissymbols for several main-group elements are written in Example 10-1.

E X A M P L E 1 0 - 1Writing Lewis Symbols. Write Lewis symbols for the following elements: (a) N, P,As, Sb, Bi; (b) Al, I, Se, Ar.

Solution(a) These are group 15 elements, and their atoms all have five valence elec-

trons The Lewis symbols all have five dots.

(b) Al is in group 13; I, in group 17 ; Se, in group 16; Ar, in group 18.

Note that for main-group elements, the number of valence electrons, and hence thenumber of dots appearing in a Lewis symbol, is equal to the group number forthe s-block elements and to the “group number minus 10” for the p-block elements.

I Se ArAl

P As Sb BiN

1ns2np32.

13p2213s22

Si

[Ne]3s23p2

10-1

Stress to the student theimportance of countingvalence electrons for many

aspects of bonding.

Periodic Trends:Lewis Structures activity

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374 Chapter 10 Chemical Bonding I: Basic Concepts

Practice Example A: Write Lewis symbols for Mg, Ge, K, and Ne.

Practice Example B: Write the Lewis symbols expected for Sn, and

A Lewis structure is a combination of Lewis symbols that represents eitherthe transfer or the sharing of electrons in a chemical bond.

(10.1)

(10.2)

In these two examples, we designated the electrons involved in bond forma-tion differently— from one atom and from the other. This helps to em-phasize that an electron is transferred in ionic bonding and that a pair ofelectrons is shared in covalent bonding. Of course, it is impossible to distin-guish between electrons, and henceforth we will use only dots to repre-sent electrons in Lewis structures.

Lewis’s work dealt mostly with covalent bonding, which we will emphasizethroughout this chapter. However, Lewis’s ideas also apply to ionic bonding,and we briefly describe this application next.

LEWIS STRUCTURES FOR IONIC COMPOUNDS

In Section 3-2, we learned that the formula unit of an ionic compound is the sim-plest electrically neutral collection of cations and anions from which the chemi-cal formula of the compound can be established. The Lewis structure of sodiumchloride (structure 10.1) represents its formula unit. For an ionic compound of amain-group element, (1) the Lewis symbol of the metal ion has no dots if all thevalence electrons are lost, and (2) the ionic charges of both cations and anionsare shown. These ideas are further illustrated through Example 10-2.

E X A M P L E 1 0 - 2Writing Lewis Structures of Ionic Compounds. Write Lewis structures for the fol-lowing compounds: (a) BaO; (b) (c) aluminum oxide.

Solution(a) Write the Lewis symbol, and determine how many electrons each atom

must gain or lose to acquire a noble-gas electron configuration. Ba losestwo electrons, and O gains two.

(b) A Cl atom can accept only one electron because it already has seven va-lence electrons. One more will give it a complete octet. On the other hand,a Mg atom must lose two electrons to have the electron configuration ofthe noble gas neon. So two Cl atoms are required for each Mg atom.

Mg � [Mg]2�2[ Cl ]�

Cl

Cl Lewis structure

OBa � [Ba]2�[ O ]2�

Lewis structure

MgCl2 ;

1 # 2

1 # 21*2

ClH� � H�ClCovalent bonding(sharing ofelectrons):

Lewis symbols Lewis structure

ClNa� � [Na]�[�Cl ]�Ionic bonding(transfer ofelectrons):

Lewis symbols Lewis structure

S2-.Tl+,Br-,

No bond is 100% ionic.All ionic bonds have somecovalent character.

Bacskay, G. G., Reimers, J. R.,Nordholm, S. “The

Mechanism of Covalent Bonding.”J. Chem. Educ. 1997: 74, 1494(December 1997).

Reimers, J. R., Bacskay, G. G.,Nordholm, S. “The Basics of

Covalent Bonding.” J. Chem. Educ.1997: 74, 1503 (December 1997).

PETRMC10_372-422-hr 12/15/05 6:28 PM 74

10-2 Covalent Bonding: An Introduction 375

(c) The formula of aluminum oxide follows directly from the Lewis struc-ture. The combination of one Al atom, which loses three electrons, andone O atom, which gains two, leaves an excess of one lost electron. Tomatch the numbers of electrons lost and gained, the formula unit mustbe based on two Al atoms and three O atoms.

Practice Example A: Write plausible Lewis structures for (a) and (b)

Practice Example B: Write plausible Lewis structures for (a) calcium iodide; (b)barium sulfide; (c) lithium oxide.

The compounds described in Example 10-2 are binary ionic compounds con-sisting of monatomic cations and monatomic anions. Commonly encounteredternary ionic compounds consist of monatomic and polyatomic ions. Bondingbetween atoms within the polyatomic ions is covalent. Some ternary ioniccompounds are considered later in the chapter.

With the exception of ion pairs such as that may be found in thegaseous state, formula units of solid ionic compounds do not exist as separate en-tities. Instead, each cation is surrounded by anions and each anion by cations.These very large numbers of ions are arranged in an orderly network called anionic crystal (Fig. 10-1). Ionic crystal structures and the energy changes accompa-nying the formation of ionic crystals are described in Chapter 12.

CONCEPT ASSESSMENT ✓How many valence electrons do the Lewis symbols for the elements in group 16have? Which of the following are correct Lewis symbols for sulfur?

COVALENT BONDING: AN INTRODUCTION

A chlorine atom shows a tendency to gain an electron, as indicated by its elec-tron affinity From which atom, sodium or hydrogen, can theelectron most readily be extracted? Neither atom gives up an electron freely,but the energy required to extract an electron from ismuch smaller than that for In Chapter 9 we learned thatthe lower its ionization energy, the more metallic an element is; sodium ismuch more metallic than hydrogen (recall Figure 9-11). In fact, hydrogen isconsidered to be a nonmetal. A hydrogen atom in the gaseous state does notgive up an electron to another nonmetal atom. Bonding between a hydrogenatom and a chlorine atom involves the sharing of electrons, which leads to acovalent bond.

To emphasize the sharing of electrons, let us think of the Lewis structure ofHCl in this manner.

H Cl

H (I1 = 1312 kJ>mol).Na (I1 = 496 kJ>mol)

(-349 kJ>mol).

10-2

(Na+Cl-)

Mg3N2.Na2S

Al

� 2[Al]3�3[ O ]2�

O

O

Lewis structureAl O

SS SS

FIGURE 10-1Portion of an ionic crystalThis structure of alternating

and ions extends inall directions and involvescountless numbers of ions.

Cl-Na+

▲

The Lewis structure of HClshows that the H atom doesnot obey the octet rule.

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The Lewis structures forand suggest that

these molecules have a linearshape. They do not. Lewistheory by itself does notaddress the question ofmolecular shape(see Section 10-7).

Cl2OH2O

▲

The broken circles represent the outermost electron shells of the bondedatoms. The number of dots lying on or within each circle represents the effec-tive number of electrons in each valence shell. The H atom has two dots, asin the electron configuration of He. The Cl atom has eight dots, correspondingto the outer-shell configuration of Ar. Note that we counted the two electronsbetween H and Cl twice. These two electrons are shared by the H and Clatoms. This shared pair of electrons constitutes the covalent bond. Writtenbelow are two additional Lewis structures of simple molecules.

As was the case for Cl in HCl, the O atom in the Lewis structure of andin is surrounded by eight electrons (when the bond-pair electrons aredouble counted). In attaining these eight electrons, the O atom conforms to theoctet rule—a requirement of eight valence-shell electrons for the atoms in aLewis structure. Note, however, that the H atom is an exception to this rule.The H atom can accommodate only two valence-shell electrons.

Lewis theory helps us to understand why elemental hydrogen and chlorineexist as diatomic molecules, and In each case, a pair of electrons isshared between the two atoms. The sharing of a single pair of electrons be-tween bonded atoms produces a single covalent bond. To underscore the im-portance of electron pairs in the Lewis theory the term bond pair applies to apair of electrons in a covalent bond, while lone pair applies to electron pairsthat are not involved in bonding. Also, in writing Lewis structures it is cus-tomary to replace some electron pairs with dashes (—), especially for bondpairs. These features are shown in the following Lewis structures.

Cl2 .H2

Cl2OH2O

� � �H andH HHO O Cl Cl ClCl�O O

water dichlorine oxide

(≠)

(10.3)

(10.4)

�

�

Bond pair

Lone pairs

or

Bond pair

H H H HH

or

H

Cl Cl ClCl ClCl

E X A M P L E 1 0 - 3Writing Simple Lewis Structures. Write a Lewis structure for the ammonia molecule.

SolutionTo write a Lewis structure you must know the formula of the molecule to be repre-sented. The formula of ammonia, shows how many and what types of atomsare in the structure. Another requirement for writing a Lewis structure is knowl-edge of the number of valence electrons associated with each of the atoms present.The valence electrons can then be represented in the Lewis symbols, as shown here.

Now we can assemble one N and three H atoms into a structure that gives the Natom a valence-shell octet and each of the H atoms two valence electrons (produc-ing the electron configuration of He).

Practice Example A: Write Lewis structures for and HOCl.

Practice Example B: Write Lewis structures for and C2H6.N2H4,NI3,

CH4,Br2 ,

HH

HN

H H H N

NH3,

Purser, Gordon H.“Lewis Structures Are

Models for Predicting MolecularStructure, Not ElectronicStructure.” J. Chem. Educ. 1999: 76,1013 (July 1999).

H2 Bond Formationanimation

Octet Rule activity

The use of the term “lonepair” can become confusingwhen discussing the Lewis

structures of molecules that containodd numbers of electrons as in thestructure of nitrogen monoxide.

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10-2 Covalent Bonding: An Introduction 377

COORDINATE COVALENT BONDS

The Lewis theory of bonding describes a covalent bond as the sharing of a pairof electrons, but this does not necessarily mean that each atom contributes anelectron to the bond. A covalent bond in which a single atom contributes bothof the electrons to a shared pair is called a coordinate covalent bond.

If we attempt to attach a fourth H atom to the Lewis structure of shownin Example 10-3, we encounter a difficulty. The electron brought by the fourth Hatom would raise the total number of valence electrons around the N atom tonine, so there would no longer be an octet. The molecule does not form, butthe ammonium ion, does, as suggested in Figure 10-2. That is, the lone pairof electrons on a molecule extracts an H atom from a HCl molecule, andthe electrons in the bond remain on the Cl atom. The result is equivalentto a ion joining with the molecule to form the ion,

(10.5)

while the abandoned electron pair remaining on the Cl atom converts it to aion.

The bond formed between the N atom of and the ion in structure(10.5) is a coordinate covalent bond. It is important to note, however, that oncethe bond has formed it is impossible to say which of the four bonds isthe coordinate covalent bond. Thus, a coordinate covalent bond is indistin-guishable from a regular covalent bond.

Another example of coordinate covalent bonding is found in the familiarhydronium ion.

(10.6)H HHO

�

N ¬ H

H+NH3

Cl-

H HH

HN

�

NH4

+NH3H+H ¬ Cl

NH3

NH4

+,NH4

NH3

HH

HN H H

H

HNH Cl �

�

Cl�

FIGURE 10-2Formation of the ammonium ion, The H atom of HCl leaves its electron with theCl atom and, as attaches itself to the molecule through the lone-pair electrons on theN atom. The ions and are formed.Cl-NH4

+

NH3H+,

NH4

�

▲

MULTIPLE COVALENT BONDS

In the preceding description of the Lewis model for covalent chemical bonding,we have used a single pair of electrons between two atoms to describe a singlecovalent bond. Often, however, more than one pair of electrons must be sharedif an atom is to attain an octet (noble gas electron configuration). and are two molecules in which atoms share more than one pair of electrons.

First, let’s apply the ideas about Lewis structures to From the Lewissymbols, we see that the C atom can share a valence electron with each Oatom, thus forming two carbon-to-oxygen single bonds.

But this leaves the C atom and both O atoms still shy of an octet. The problemis solved by shifting the unpaired electrons into the region of the bond, asindicated by the red arrows.

(10.7)

In Lewis structure (10.7), the bonded atoms are seen to share two pairs of elec-trons (a total of four electrons) between them—a double covalent bond ( “ ).

O O CO OCO OC

O OC O OC

CO2.

N2CO2

Coordinate covalent bondsoccur in the formation oftransition metal complexes,

discussed in Chapter 25.

Formation ofcoordinate covalent bondsis an example of Lewis

acid–base reactions.

It is easy to show theformation of coordinate

complexes. Dissolve whiteanhydrous CuCl2 in water; it turnslight blue due to the complex[Cu(H2O)6]2+. Add ammonia to thisand it turns deep blue as some ofthe water is replaced by ammoniato form complexes such as[Cu(NH3)4(H2O)2]2+.

Point out that theadditional bonds arealways bonds formed from

the p orbitals.

A common misconception,which is perpetuated by theway Lewis structures are

drawn, is that a double bond iscomposed of two single bonds anda triple bond is composed to threesingle bonds.

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Now let’s try our hand at writing a Lewis structure for the molecule.Our first attempt might again involve a single covalent bond and the incorrectstructure shown below.

Each N atom appears to have only six outer-shell electrons, not the expectedeight. The situation can be corrected by bringing the four unpaired electronsinto the region between the N atoms and using them for additional bondpairs. In all, we now show the sharing of three pairs of electrons between the Natoms. The bond between the N atoms in is a triple covalent bondDouble and triple covalent bonds are known as multiple covalent bonds.

(10.8)

The triple covalent bond in is a very strong bond that is difficult to breakin a chemical reaction. The unusual strength of this bond makes quiteinert. As a result, coexists with in the atmosphere and forms ox-ides of nitrogen only in trace amounts at high temperatures. The lack of reac-tivity of with is an essential condition for life on Earth. The inertness of

also makes it difficult to synthesize nitrogen compounds.Another molecule whose Lewis structure features a multiple bond is

which has a double bond.

(10.9)

The blue question mark suggests that there is some doubt about the validity ofstructure (10.9), and the source of the doubt is illustrated in Figure 10-3.The structure fails to account for the paramagnetism of oxygen—the mole-cule must have unpaired electrons. Unfortunately, no completely satisfactoryLewis structure is possible for but in Chapter 11, bonding in the mole-cule is described in a way that accounts for both the double bond and theobserved paramagnetism.

We could continue applying ideas introduced in this section, but our abilityto write plausible Lewis structures will be greatly aided by a couple of newideas that we introduce in Section 10-3.

CONCEPT ASSESSMENT ✓What types of bonds can be used to describe the chemical bonds in

CONCEPT ASSESSMENT ✓In which groups of the periodic table are the elements most likely to use multiple bonds?

POLAR COVALENT BONDS AND ELECTROSTATIC

POTENTIAL MAPS

We have introduced ionic and covalent bonds as though they are of two dis-tinctly different types: ionic bonds involving a complete transfer of electrons andcovalent bonds involving an equal sharing of electron pairs. Such is not the case,however, and most chemical bonds fall between the two extremes of 100%ionic and 100% covalent. A covalent bond in which electrons are not sharedequally between two atoms is called a polar covalent bond. In such a bond,

10-3

[BF4

-]?

O2O2,

O2

O O�O O O ?O O O

O2,N2(g)

O2N2

O2(g)N2(g)N2(g)

N2

N N N NN N or

( ‚ ).N2

N NN N (Incorrect)�

N2

FIGURE 10-3Paramagnetism of oxygenLiquid oxygen is attractedinto the magnetic field of alarge magnet.

▲

KEEP IN MINDthat merely being able towrite a plausible Lewisstructure does not provethat it is the correct electronicstructure. Proof can comeonly through confirmingexperimental evidence.

▲

Paramagnetism here is aresult of the presence ofunpaired electrons. Recall

that there are four types ofmagnetism: ferro (and anti)magnetism, diamagnetism,electron paramagnetism (treatedhere), and nuclear paramagnetism(responsible for NMR and MRI).

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10-3 Polar Covalent Bonds and Electrostatic Potential Maps 379

electrons are displaced toward the more nonmetallic element. The unequalsharing of the electrons leads to a partial negative charge on the more non-metallic element, signified by and a corresponding partial positive chargeon the more metallic element, designated by Thus we can represent thepolar bond in HCl by a Lewis structure in which the partial charges and indicate that the bond pair of electrons lies closer to the Cl than to the H.

The advent of inexpensive fast computers has allowed chemists to developmethods for displaying the electron distribution within molecules. This distri-bution is obtained, in principle, by solving the Schrödinger equation for a mol-ecule. Although the solution can only be obtained by using approximatemethods, these methods provide an electrostatic potential map, a way to vi-sualize the charge distribution within a molecule.

Before discussing these maps let us first review the notion of electron den-sity, or charge density, introduced in Chapter 8. There we saw that the behav-ior of electrons in atoms can be described by mathematical functions calledorbitals. The probability of finding an electron at some point in the three-dimensional region associated with an orbital is related to the square of anatomic orbital function. Typically, we refer to the region encompassing 95% ofthe probability of finding the electron as the shape of the orbital. In a similarway we can map the total electron density throughout a molecule, that is, notjust the density of a single orbital. The electron density surface that encom-passes 95% of the charge density in ammonia is depicted in Figure 10-4.

The electrostatic potential is the work done in moving a unit of positive chargeat a constant speed from one region of a molecule to another. The electrostaticpotential map is obtained by hypothetically probing an electron density sur-face with a positive point charge. The positive point charge will be attracted toan electron-rich region—a region of excess negative charge when all thecharges of the nuclei and electrons have been taken into account—and the elec-trostatic potential will be negative. On the other hand, if the point charge isplaced in an electron-poor region, a region of excess positive charge, the posi-tive point charge will be repelled, and the electrostatic potential will be positive.

ClHd� d�

d-d+d+ .

d- ,

�

SolidTransparent

Move probe over moleculeto measure potential

Probe at infinite distance

Move probe to surface of electron density FIGURE 10-4Determination of theelectrostatic potential mapfor ammoniaThe electrostatic potential atany point on the chargedensity surface of a moleculeis defined as the change inenergy that occurs when aunit positive charge isbrought to this point, startingfrom another point that isinfinitely far removed fromthe molecule. The surfaceencompassing the ammoniamolecule is analogous tothe 95% surface of electroncharge density for atomicorbitals discussed in Chapter8. The electrostatic potentialmap gives information aboutthe distribution of electroncharge within this surface.

▲

A permanent dipolemoment exists if there is aseparation of positive and

negative charge.

The following articlediscusses using electrondensity to describe the

bonding in a molecule: Gillespie,Ronald J. J. Chem. Educ. 2001:78, 1688.

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The procedure for making an electrostatic potential map is illustrated in Figure10-4, which shows the distribution of electron density in ammonia.

An electrostatic potential map gives information about the distribution ofelectron charge in a molecule. For example, in a neutral molecule, if the poten-tial at a point is positive, it is likely that an atom at this point carries a net pos-itive charge. An arbitrary “rainbow” color scheme is adopted in the display ofan electrostatic potential map. Red, the low-energy end of the spectrum, isused for regions of the most negative electrostatic potential, and blue is usedto color regions of the most positive electrostatic potential. Intermediate colorsrepresent intermediate values of the electrostatic potential. Thus, the potentialincreases from red through yellow to blue, as seen in the scale in Figure 10-5.For example, the blue-green color surrounding the hydrogen atoms in Figure10-4 suggests that they carry a slight positive charge. The nitrogen atom, beingclosest to the red region, carries a net negative charge.

Let us now look at the computed electrostatic potential maps for NaCl, and HCl (Fig. 10-5). We see that has a uniform distribution of electroncharge density as depicted by the uniform color distribution in the electrostat-ic potential map. This is typical for a nonpolar covalent bond and occurs in alldiatomic molecules containing identical atoms. The sodium chloride mole-cule, on the other hand, exhibits a highly nonuniform distribution of electroncharge density. The sodium atom is almost exclusively in the blue extreme ofpositive charge and the chlorine in the red extreme of negative charge. Thiselectrostatic potential map is typical of an ionic bond. Yet, it is clear from themap that the transfer of electron density from the sodium atom to the chlorineatom is not complete. That is, the NaCl bond is not completely ionic. Experi-ments show that the bond is only about 80% ionic. The molecule HCl also hasan unsymmetrical distribution of electron charge density, as indicated by thegradation of color in the electrostatic potential map. Note, however, that inthis case the chlorine atom is not completely in the extreme dark red corre-sponding to a large negative charge. Instead, it is in the orange-red region in-dicating a partial negative charge. Correspondingly, the hydrogen atom has apartial positive charge, as indicated by the pale blue. The electrostatic poten-tial map clearly depicts the polar nature of the bond in HCl.

Cl2

Cl2 ,

750 kJ mol�1

500 kJ mol�1

250 kJ mol�1

0 kJ mol�1

�250 kJ mol�1

Positiveextreme

Negativeextreme

Cl2

HCl

NaCl

Electrostatic potential isthe work done in moving aunit of positive charge ata constant speed from oneregion of a molecule toanother.

▲

FIGURE 10-5The electrostaticpotential maps forsodium chloride, hydrogenchloride, and chlorineThe dark red and dark blueon the electrostatic potentialmap correspond to theextremes of the electrostaticpotential, negative topositive, for the particularmolecule for which the mapis calculated. To get a reliablecomparison of differentmolecules, the values of theextremes in electrostaticpotential (in ) mustbe the same for all of themolecules compared. In themaps shown here the rangeis to 750 kJ mol-1.-250

kJ mol-1

▲

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Periodic Trends:Electronegativityanimation

ELECTRONEGATIVITY

We expect the bond to be polar because the Cl atom has a greateraffinity for electrons than does the H atom. Electron affinity is an atomicproperty, however, and more meaningful predictions about bond polaritiesare those based on a molecular property, one that relates to the ability ofatoms to lose or gain electrons when they are part of a molecule rather thanisolated from other atoms.

Electronegativity (EN) describes an atom’s ability to compete for electronswith other atoms to which it is bonded. As such, electronegativity is related toionization energy (I) and electron affinity (EA). To see how they are related con-sider the reaction between two hypothetical elements A and B, which could givethe products or We represent these two reactions by the expressions

(10.10)

(10.11)

If the bonding electrons are shared approximately equally in these hypotheticstructures, we would expect that because neither extreme (or ) is favored. If we make the assumption that the resultant bond isnonpolar, then

which gives, after collecting terms for each atom,

(10.12)

Equation (10.12) tells us that a nonpolar bond will result when the differencebetween the ionization energy and the electron affinity is the same for bothatoms involved in the bond. The quantity provides a measure of theability of an atom to attract electrons (or electron charge density) to itself rela-tive to some other atom. Thus it is related to the electronegativity of the atom.

An element with a high ionization energy and an electron affinity that islarge and negative, such as fluorine, will have a large electronegativity rela-tive to an atom with a low ionization energy and a small electron affinity,such as sodium.

There are several methods for converting qualitative comparisons to actualnumerical values of the electronegativities of the elements. One widely usedelectronegativity scale, with values given in Figure 10-6, is that devised byLinus Pauling (1901–1994). Pauling’s EN values range from about 0.7 to 4.0. Ingeneral, the lower its EN, the more metallic the element is, and the higher theEN, the more nonmetallic it is. From Figure 10-6 we also see that electronega-tivity decreases from top to bottom in a group and increases from left to rightin a period of the periodic table. These are the expected trends when we inter-pret electronegativity in terms of the quantity That is, as the ioniza-tion energy (I) increases across the period we expect the electronegativity toincrease. The distinction between electron affinity and electronegativity isclearly seen when we consider the electron affinities of F andchlorine Although the electron affinity of Cl issomewhat more negative than that of F the EN of Cl (3.0) is sig-nificantly lower than that of F (4.0) because of the decreased ionization energyof chlorine relative to fluorine

CONCEPT ASSESSMENT ✓With the aid of only a periodic table, decide which is the most electronegative atom ofthe following sets of elements (a) As, Se, Br, I; (b) Li, Be, Rb, Sr; (c) Ge, As, P, Sn.

(1681 kJ>mol).(1251 kJ>mol)

(-328 kJ>mol),(-349 kJ>mol)(-349 kJ mol-1):

(-328 kJ mol-1)

1I - EA2.

ENA r 1IA - EAA2

1I - EA2

1IA - EAA2 = 1IB - EAB2

1IA + EAB2 = 1IB + EAA2

A-B+A+B-¢E1 = ¢E2

A + B ¡ A-B+ ¢E2 = 1IB + EAA2

A + B ¡ A+B- ¢E1 = 1IA + EAB2

A-B+.A+B-

H ¬ Cl

Students are frequentlymixed up aboutelectronegativity and

electron affinity. The former refersto bonded atoms in molecules only,while the latter refers to the abilityof an atom to attract an electronto itself.

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Perc

ent i

onic

cha

ract

er

100

75

50

25

010 2

IBr HBr

HCl

HF

LiBrKI CsIKBr

KCl

CsClNaCl

LiF

KF

CaFLiClLiI

H IICl

3Electronegativity difference

FIGURE 10-7Percent ionic character of a chemical bond as afunction of electronegativity difference

▲

Although Figure 10-7suggests that the bondbetween two identical metalatoms should be covalent[as it is in forexample], in solid metals,where bonding extendsthroughout a network ofmany, many atoms, thebonding is of a type calledmetallic (explored in thenext chapter).

Li2(g),

▲

Electronegativity values allow an insight into the amount of polar characterin a covalent bond based on electronegativity difference, — the absolutevalue of the difference in EN values of the bonded atoms. If for twoatoms is very small, the bond between them is essentially covalent. If islarge, the bond is essentially ionic. For intermediate values of the bondis described as polar covalent. A useful rough relationship between andpercent ionic character of a bond is presented in Figure 10-7.

Large EN differences are found between the more metallic and the morenonmetallic elements. Combinations of these elements are expected to pro-duce bonds that are essentially ionic. Small EN differences are expected fortwo nonmetal atoms, and the bond between them should be essentially cova-lent. Thus, even without a compilation of EN values at hand, you shouldbe able to predict the essential character of a bond between two atoms. Simplyassess the metallic/nonmetallic characters of the bonded elements from theperiodic table (recall Figure 9-11).

¢EN¢EN,

¢EN¢EN

≤EN

Li

Na

1.5

H

1.2

K1.0

Rb

Cs

Sr

Ba

Fr0.9

1.3 1.5

1.2

1.1

1.4

1.3

Ac

V1.6

1.6

1.5

1.8

2.4

1.5 1.8

1.9

1.9

2.2

2.2

1.8 1.8

2.2

2.2

2.2

2.2

1.9Zn1.6

1.9

2.4

1.7

1.9

Ga1.6

Ge1.8

1.7

1.8

1.8

1.8

As2.0

Se2.4

1.9

1.9

2.1

2.0

Br2.8

2.5

2.2

B2.0

C2.5

Al1.5

Si1.8

N3.0

O3.5

P2.1

S2.5

F4.0

Cl3.0

1

2

3 4 5 6 7 98 10 11 12

13 14 15 16 17

† *†

Lanthanides: 1.1–1.3Actinides: 1.3 –1.5

below 1.0

1.0–1.4

1.5–1.9

2.0–2.4

2.5–2.9

3.0–4.0

2.1

1.0

0.9

0.8

0.8

0.8

0.7

Be

Mg

Ca

1.0

0.9

Ra1.1

La*

Y

Sc Ti1.6

Zr

Hf Ta

Nb

Cr Mn Fe Co Ni Cu

Mo Tc Ru Rh Pd Ag

W Re Os Ir Pt Au Hg Tl Pb Bi Po At

Cd In Sn Sb Te I

FIGURE 10-6Electronegativities of the elementsAs a general rule, electronegativities decrease from top to bottom in a group and increasefrom left to right in a period of elements. The values are from L. Pauling, The Nature ofthe Chemical Bond, 3rd ed, Cornell University, Ithaca, NY, 1960, page 93. Values may besomewhat different when based on other electronegativity scales.

▲

An alternative way to explore the bondingcharacter and

electronegativity relationship isto use a method described inElectronegativity and the BondTriangle. Terry L. Meek and LeahD. Graner J. Chem. Educ. 2005:82, 325.

Many periodicproperties can berationalized by using the

variations in the magnitude ofthe atomic number, Z.

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E X A M P L E 1 0 - 4Assessing Electronegativity Differences and the Polarity of Bonds. (a) Whichbond is more polar, or (b) What is the percent ionic character ofeach of these bonds?

Solution(a) Look up EN values of H, Cl, and O in Figure 10-6, and compute

For the bond, For the bond, Because

its is somewhat greater, we expect the bond to be the morepolar bond.

(b) Determine the percent ionic character from Figure 10-7.

Practice Example A: Which of the following bonds are the most polar, that is,have the greatest ionic character:

Practice Example B: Which is the most polar bond: or

To illustrate the variation of bond polarity with electronegativity using elec-trostatic potential maps, consider the electrostatic potential maps for HCl, HBrand HI displayed below.

The gradation of the charge on the H atom ranges from quite positive inHCl to less positive in HBr and HI, as is seen in the gradation of color on the Hatom from dark blue to pale blue. This trend of decreasing charge separationin these three molecules corresponds to the decrease in electronegativity of thehalogen atoms from Cl to Br to I. Correspondingly, the halogen atom becomesless red—signifying a decreasing negative charge in going from chlorine to io-dine. Electrostatic potential maps are a powerful way of displaying the varia-tion of polarity within a group of related molecules. We will use computedelectrostatic potential maps later in this chapter and in subsequent chapters,whenever charge separation within a molecule contributes significantly tounderstanding the topic at hand.

E X A M P L E 1 0 - 5Identifying a Molecular Structure Using Electronegativity and ElectrostaticPotential Maps. Two electrostatic potential maps are shown below. One corre-sponds to NaF and the other to NaH. Which map corresponds to which molecule?

HCl HIHBr

O ¬ F?P ¬ O,C ¬ P,C ¬ S,

P ¬ Cl?N ¬ O,N ¬ H,H ¬ Br,

H ¬ O bond: ¢EN = 1.4; L35% ionic

H ¬ Cl bond: ¢EN = 0.9; L20% ionic

H ¬ O¢EN¢EN = 3.5 - 2.1 = 1.4.H ¬ O3.0 - 2.1 = 0.9.

¢EN =H ¬ ClENO = 3.5.ENCl = 3.0;ENH = 2.1;¢EN.

H ¬ O?H ¬ Cl

The following articlediscusses usingelectronegativity values

to predict bond type: Sproul,Gordon J. Chem. Educ. 2001: 78, 387.

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SolutionFirst look up the EN values of H, Na, and F in Figure 10-6, and compute

For the bond, For the bond, Because for NaF is greater, weexpect the bond to be the more polar bond. Consequently, we expect itselectrostatic potential map to show the greater range of colors when the electrostat-ic potential maps are based on the same scale. We conclude that the electrostatic po-tential map on the left represents NaF.

Practice Example A: Which of the following electrostatic potential maps corre-sponds to IF, and which to IBr?

Practice Example B: Which of the following electrostatic potential maps corre-sponds to and which to

WRITING LEWIS STRUCTURES

In this section we combine the ideas introduced in the preceding three sectionswith a few new concepts to write a variety of Lewis structures.

FUNDAMENTAL REQUIREMENTS

Let us begin with a reminder of some of the essential features of Lewis struc-tures that we have already encountered.

• All the valence electrons of the atoms in a Lewis structure must appear inthe structure.

• Usually, all the electrons in a Lewis structure are paired.• Usually, each atom acquires an outer-shell octet of electrons. Hydrogen,

however, is limited to two outer-shell electrons.• Sometimes, multiple covalent bonds (double or triple bonds) are needed.

Multiple covalent bonds are formed most readily by C, N, O, P, and S atoms.

SKELETAL STRUCTURES

The usual starting point in writing a Lewis structure is to designate theskeletal structure—all the atoms in the structure arranged in the order inwhich they are bonded to one another. In a skeletal structure with more thantwo atoms, we generally need to distinguish between central and terminalatoms. A central atom is bonded to two or more atoms, and a terminal atom isbonded to just one other atom. As an example, consider ethanol, CH3CH2OH.

10-4

CH3SH?CH3OH,

F ¬ Na¢EN¢EN = 4.0 - 0.9 = 3.1.F ¬ Na

¢EN = 2.1 - 0.9 = 1.2.H ¬ NaENF = 4.0.ENNa = 0.9;ENH = 2.1;¢EN.

Skeletal structures aresigma bonds.Additional multiple bonds

and resonance are described bypi bonds.

The teacher should pointout that, although Lewisstructures are useful, they

have a number of shortcomings:electron-deficient species and thosewith expanded valence shellsviolate the octet rule. Lewis theoryonly works well for the first- andsecond-row elements; it fails toproperly account for resonance; itdoes not predict structure; it doesnot account for molecules with anodd number of electrons; it doesnot predict the color ofcompounds; it does not accountfor paramagnetism; and it doesnot account for bond length andbond energy.

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10-4 Writing Lewis Structures 385

Its skeletal structure is the same as the following structural formula. In thisstructure, the central atoms—both C atoms and the O atom—are printed in red.The terminal atoms—all six H atoms—are printed in blue.

(10.13)

Here are a few additional facts about central atoms, terminal atoms, and skele-tal structures.

• Hydrogen atoms are always terminal atoms. This is because a H atom canaccommodate only two electrons in its valence shell, so it can form onlyone bond to another atom. (An interesting and rare exception occurs insome unusual boron–hydrogen compounds.)

• Central atoms are generally those with the lowest electronegativity. In the skele-tal structure (10.13), the atoms of lowest electronegativity happen to be H atoms, but as noted above, H atoms can only be terminalatoms. Next lowest in electronegativity are the C atoms, andthese are central atoms. The O atom has the highest electronegativity (3.5)but nevertheless is also a central atom. For O to be a terminal atom in struc-ture (10.13) would require it to exchange places with a H atom, but thiswould make the H atom a central atom and that is not possible. The chiefcases where O atoms are central atoms are in structures with a peroxo link-age or a hydroxy group Otherwise, expect an Oatom to be a terminal atom.

• Carbon atoms are always central atoms. This is a useful fact to keep in mindwhen writing Lewis structures of organic molecules.

• Except for the very large number of chain-like organic molecules, moleculesand polyatomic ions generally have compact, symmetrical structures. Thus, ofthe two skeletal structures below, the more compact structure on the rightis the one actually observed for phosphoric acid,

A STRATEGY FOR WRITING LEWIS STRUCTURES

At this point, let us incorporate a number of the ideas that we have consideredso far into a specific approach to writing Lewis structures. This strategy is de-signed to give you a place to begin as well as consecutive steps to follow toachieve a plausible Lewis structure.

1. Determine the total number of valence electrons that must appear in thestructure.Examples: In the molecule there are 4 valence electrons foreach C atom, or 8 for the two C atoms; 1 for each H atom, or 6 for the six Hatoms; and 6 for the lone O atom. The total number of valence electrons inthe Lewis structure of is

In the polyatomic ion there are 5 valence electrons for the P atom and6 for each O atom, or 24 for all four O atoms. To produce the charge of an additional 3 valence electrons must be brought into the structure. Thetotal number of valence electrons in the Lewis structure of is

5 + 24 + 3 = 32

PO4

3-

3- ,PO4

3-,

8 + 6 + 6 = 20

CH3CH2OH

CH3CH2OH,

H O

H

O P

(Incorrect)

O O H H O

O

O

P

(Correct)

O H

H

H3PO4.

( ¬ O ¬ H).( ¬ O ¬ O ¬ )

1EN = 2.52

1EN = 2.12

H

H

H

C

H

H

C O H

Stress to students thatthey will need to memorizethe rules and steps, and

practice a lot, to becomeproficient at writing Lewisstructures. In addition, studentsshould begin by writingstructures of molecules that haveonly one central atom before tryingto write the structures of morecomplicated molecules.

It is a good idea to tellstudents that before writingLewis structures for more

complicated molecules, thearrangement of the atoms (skeletalstructure) needs to be knownbeforehand.

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In the polyatomic ion there are 5 valence electrons for the N atomand 1 for each H atom, or 4 for all four H atoms. To account for the chargeof one of the electrons must be lost. The total number of valence elec-trons in is

2. Identify the central atoms(s) and terminal atoms.3. Write a plausible skeletal structure. Join the atoms in the skeletal structure

by single covalent bonds (single dashes, representing two electrons each).4. For each bond in the skeletal structure, subtract two from the total number

of valence electrons.5. With the valence electrons remaining, first complete the octets of the ter-

minal atoms. Then, to the extent possible, complete the octets of the centralatom(s). If there are just enough valence electrons to complete octets for allthe atoms, the structure at this point is a satisfactory Lewis structure.

6. If one or more central atoms is left with an incomplete octet after step 5,move lone-pair electrons from one or more terminal atoms to formmultiple covalent bonds to central atoms. Do this to the extent neces-sary to give all atoms complete octets, thereby producing a satisfactoryLewis structure.

Figure 10-8 summarizes this procedure for writing Lewis structures.

5 + 4 - 1 = 8NH4

+1+ ,

NH4

+,

Count the total number ofelectrons in the structure.

Identify the terminal atoms.

Complete the octets ofterminal atoms.

(H atoms require a duet.)

A satisfactoryLewis structure

is obtained.

Subtract the number of electronsused to this point from the

total number of valence electrons.Do any electrons remain?

Draw a skeletal structure.

Place two electrons in eachbond in the skeletal structure.

Yes

Yes

No

No

Do all atoms haveoctets (duets for H)?

Form multiple bondsas needed to

complete octets.

Place remainingelectrons on

the central atom(s).

FIGURE 10-8Summary scheme fordrawing Lewis structures

▲

Point out to studentsthat the charges associatedwith polyatomic ions give

the structures the electrons theyneed to obey the octet rule. So forexample, show students whathappens when you try to drawthe Lewis structures of NH4 or PO4without their ionic charges.

Writing Lewis Structuresactivity

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E X A M P L E 1 0 - 6Applying the General Strategy for Writing Lewis Structures. Write a plausibleLewis structure for cyanogen, a poisonous gas used as a fumigant and rocketpropellant.

Solution

Step 1. Determine the total number of valence electrons. Each of the two C atoms(group 14) has four valence electrons, and each of the two N atoms (group 15)has five. The total number of valence electrons is

Step 2. Identify the central atom(s) and terminal atoms. The C atoms have a lowerelectronegativity (2.5) than do the N atoms (3.0). C atoms are central atoms,and N atoms are terminal atoms.

Step 3. Write a plausible skeletal structure by joining atoms through single cova-lent bonds.

Step 4. Subtract two electrons for each bond in the skeletal structure. The threebonds in this structure account for 6 of the 18 valence electrons. This leaves12 valence electrons to be assigned.

Step 5. Complete octets for the terminal N atoms, and to the extent possible, thecentral C atoms. The remaining 12 valence electrons are sufficient only tocomplete the octets of the N atoms.

Step 6. Move lone pairs of electrons from the terminal N atoms to form multiplebonds to the central C atoms. Each C atom has only four electrons in its va-lence shell and needs four more to complete an octet. Thus, each C atom re-quires two additional pairs of electrons, which it acquires if we move twolone pairs from each N atom into its bond with a C atom, as shown below.

Practice Example A: Write plausible Lewis structures for (a) (b) HCN,and (c)

Practice Example B: Write plausible Lewis structures for (a) formic acid,HCOOH, and (b) acetaldehyde,

E X A M P L E 1 0 - 7Writing a Lewis Structure for a Polyatomic Ion. Write the Lewis structure for thenitronium ion,

Solution

Step 1. Determine the total number of valence electrons. The N atom (group 15) hasfive valence electrons, and each of the two O atoms (group 16) has six. How-ever, one valence electron must be removed to produce the charge of

The total number of valence electrons is Step 2. Identify the central atom(s) and terminal atoms. The N atom has a lower

electronegativity (3.0) than the O atoms (3.5). N is the central atom, and theO atoms are the terminal atoms.

Step 3. Write a plausible skeletal structure by joining atoms through single cova-lent bonds.

Step 4. Subtract two electrons for each bond in the skeletal structure. The two bondsin this structure account for 4 of the 16 valence electrons. This leaves 12 va-lence electrons to be assigned.

O ¬ N ¬ O

5 + 6 + 6 - 1 = 161+ .

NO2

+.

CH3CHO.

COCl2.CS2,

N C C N N C C N

N C C N

N ¬ C ¬ C ¬ N

4 + 4 + 5 + 5 = 18.

C2N2,

Writing Lewis Structures IIactivity

PETRMC10_372-422-hr 12/15/05 6:28 PM Page 387


Step 5. Complete octets for the terminal O atoms, and to the extent possible, thecentral N atom. The remaining 12 valence electrons are sufficient only tocomplete the octets of the O atoms.

Step 6. Move lone pairs of electrons from the terminal O atoms to form multiplebonds to the central N atom. The N atom has only four electrons in itsvalence shell and needs four more to complete an octet. Thus, the N atomrequires two additional pairs of electrons, which it acquires if wemove one lone pair from each O atom into its bond with the N atom, asshown below.

(10.14)

Practice Example A: Write plausible Lewis structures for the following ions: (a) (b) (c)

Practice Example B: Write plausible Lewis structures for the following ions: (a) (b) (c)

FORMAL CHARGE

Instead of writing Lewis structure (10.14) for the nitronium ion in Example10-6, we might have written the following structure.

(improbable) (10.15)

Despite the fact that this structure satisfies the usual requirements—the cor-rect number of valence electrons and an octet for each atom—we have markedit improbable because it fails in one additional requirement. Have you noticedthat in our strategy for writing Lewis structures, once the total number of va-lence electrons has been determined, there is no need to keep track of whichelectrons came from which atoms? Nevertheless, after we have a plausibleLewis structure, we can go back and assess where each electron apparentlycame from, and in this way we can evaluate formal charges. Formal charges(FC) are apparent charges on certain atoms in a Lewis structure that arisewhen atoms have not contributed equal numbers of electrons to the covalentbonds joining them. In cases where more than one Lewis structure seems pos-sible, formal charges are used to ascertain which sequence of atoms andarrangement of bonds is most satisfactory.

The formal charge on an atom in a Lewis structure is the number of valenceelectrons in the free (uncombined) atom minus the number of electrons as-signed to that atom in the Lewis structure, with the electrons assigned in thefollowing way.

• Count lone-pair electrons as belonging entirely to the atom on which theyare found.

• Divide bond-pair electrons equally between the bonded atoms.

Assigning electrons in this way is equivalent to writing that

= number lone-pair e- +12

number bond-pair e-

e- assigned to a bonded atom in a Lewis structure

(e-)

O N O�

NCO-.NH3OH+;BF4

-;

O2-.N2H5

+;NO+;

O N O�

O N O�

O N O�

Formal Chargesanimation

Students frequently confusethe rules for countingelectrons on atoms when

determining formal charge andwhen determining whether anatom is obeying the octet rule.When counting electrons forformal charge the bond-pairelectrons are divided equallybetween the bonded atoms,whereas all the bond-pairelectrons are assigned to an atomin a bond when counting electronsfor the octet rule.

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We will see someexceptions to the idea thatformal charges should bekept to a minimum inSection 10-6.

▲

KEEP IN MINDthat formal charges are notactual charges but simply“balances” in a form of“electron bookkeeping” thathelp us choose a probableLewis structure.

▲

Because formal charge is the difference between the assignment of valenceelectrons to a free (uncombined) atom and to the atom in a Lewis structure, it canbe expressed as

(10.16)

Now, let us assign formal charges to the atoms in structure (10.15), proceed-ing from left to right.

Formal charges in a Lewis structure can be shown using small, encirclednumbers.

(10.17)

The following are general rules that can help to determine the plausibilityof a Lewis structure based on its formal charges.

• The sum of the formal charges in a Lewis structure must equal zero for aneutral molecule and must equal the magnitude of the charge for a poly-atomic ion. [Thus for structure (10.17), this sum is ]

• Where formal charges are required, they should be as small as possible.

• Negative formal charges usually appear on the most electronegativeatoms; positive formal charges, on the least electronegative atoms.

• Structures having formal charges of the same sign on adjacent atoms areunlikely.

Lewis structure (10.17) conforms to the first two rules, but is not in good ac-cordance with the third rule. Despite the fact that O is the most electronegativeelement in the structure, one of the O atoms has a positive formal charge. Thegreatest failing, though, is in the fourth rule. Both the O atom on the left andthe N atom adjacent to it have positive formal charges. Structure (10.17) is nota satisfactory Lewis structure. By contrast, the Lewis structure of de-rived in Example 10-7 has only one formal charge, on the central N atom.It conforms to the rules completely and is the accepted Lewis structure.

E X A M P L E 1 0 - 8Using Formal Charges in Writing Lewis Structures. Write the most plausibleLewis structure of nitrosyl chloride, NOCl, one of the oxidizing agents present inaqua regia, a mixture of concentrated nitric and hydrochloric acids capable of dis-solving gold.

SolutionAlthough the formula is written as NOCl, we can pretty much reject the skeletalstructure It places the most electronegative atom as the central atom(see also Practice Example A). This then leaves the following as possible skeletalstructures.

O ¬ Cl ¬ N and O ¬ N ¬ Cl

N ¬ O ¬ Cl.

+1,NO2

+

+1 + 1 - 1 = +1.

O N O

�

�1 �1 �1

O FC � 6 valence e� in O � 2 lone-pair e� � (6 bond-pair e�) � 6 � 2 � 3 � �1

O

N FC � 5 valence e� in N � 0 lone-pair e� � (8 bond-pair e�) � 5 � 0 � 4 � �1

FC � 6 valence e� in O � 6 lone-pair e� � (2 bond-pair e�) � 6 � 6 � 1 � �1

121212

-12

number bond-pair e-

FC = number valence e- in free atom - number lone-pair e-

PETRMC10_372-422-hr 12/15/05 6:28 PM Page 389


Regardless of the skeletal structure chosen, the number of valence electrons(dots) that must appear in the final Lewis structure is

When we apply the four steps listed below to the two possible skeletal structures, weobtain a total of four Lewis structures—two for each skeletal structure. This dou-bling occurs because in step 4, there are two ways to complete the octets of the cen-tral atoms. The final Lewis structures obtained are labeled and (b2).(b1),(a2),(a1),

5 from N + 6 from O + 7 from Cl = 18

Based on structure ONCl is a better wayto write the formula ofnitrosyl chloride than NOCl.

(b1)

▲

O 1. Assign four electrons

2. Assign twelve more electrons

3. Assign the last two electrons

4. Complete the octet on the central atom

Cl N

O Cl N

O Cl N

O N Cl

O N Cl

ClO N

OO ClNO ClNNClO NCl

(a) (b)

(a1) (a2) (b1) (b2)

Evaluate formal charges using equation (10.16). In structure for the N atom,

for the O atom,

for the Cl atom,

Proceed in a similar manner for the other three structures. Summarize the formalcharges for the four structures.

(a1) (a2) (b1) (b2)N: 0 0O: 0 0Cl: 0

Select the best Lewis structure in terms of the formal-charge rules. First, note thatall four structures obey the requirement that formal charges of a neutral moleculeadd up to zero. In structure the formal charges are large ( on Cl and onN) and the negative formal charge is not on the most electronegative atom. Struc-ture has formal charges on all atoms, one of them large ( on Cl). Structure

is the ideal we seek—no formal charges. In structure we again have formalcharges. The best Lewis structure of nitrosyl chloride is

Practice Example A: Write a Lewis structure for nitrosyl chloride based on theskeletal structure and show that this structure is not as plausible as theone obtained in Example 10-8.

Practice Example B: Write two Lewis structures for cyanamide, animportant chemical of the fertilizer and plastics industries. Use the formal chargeconcept to choose the more plausible structure.

CONCEPT ASSESSMENT ✓For molecules, the most satisfactory Lewis structure may have no formal charges

in some cases and formal charges in others. For polyatomic ions,minimally the most satisfactory Lewis structure has a formal charge on at least oneatom. Explain the basis of these observations.

1FC = 02

NH2CN,

N ¬ O ¬ Cl,

O ClN

(b2),(b1)+2(a2)

-2+2(a1),

+1+2+2-1-1

-1-2

FC = 7 - 2 -12

162 = +2

FC = 6 - 4 -12

142 = 0

FC = 5 - 6 -12

122 = -2

(a1),

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10-5 Resonance 391

Bond lengths are discussedmore fully in Section 10-8.

▲

RESONANCE

The ideas presented in the previous section allow us to write many Lewisstructures, but some structures still present problems. We describe these prob-lems in the next two sections.

Although we usually think of the formula of oxygen as there are actual-ly two different oxygen molecules. Familiar oxygen is dioxygen, the othermolecule is trioxygen—ozone, The term used to describe the existence oftwo or more forms of an element that differ in their bonding and molecularstructure is allotropy— and are allotropes of oxygen. Ozone is found nat-urally in the stratosphere and is also produced in the lower atmosphere as aconstituent of smog.

When we apply the usual rules for Lewis structures for ozone, we come upwith these two possibilities.

But there is something wrong with both structures. Each suggests that oneoxygen-to-oxygen bond is single and the other is double. Yet, experimental ev-idence indicates that the two oxygen-to-oxygen bonds are the same; each hasa length of 127.8 pm. This bond length is shorter than the single-bondlength of 147.5 pm in hydrogen peroxide, , but it is longer thanthe double-bond length of 120.74 pm in diatomic oxygen, . The bondsin ozone are intermediate between a single and a double bond. The difficultyis resolved if we say that the true Lewis structure of is neither of the previ-ously proposed structures, but a composite, or hybrid, of the two, a fact that wecan represent as

(10.18)

The situation in which two or more plausible Lewis structures can be writ-ten but the “correct” structure cannot be written at all is called resonance. Thetrue structure is a resonance hybrid of plausible contributing structures. Accept-able contributing structures to a resonance hybrid must all have the sameskeletal structure (the atomic positions cannot change); they can differ only inhow electrons are distributed within the structure. In expression (10.18), thetwo contributing structures are joined by a double-headed arrow. The arrowdoes not mean that the molecule has one structure part of the time and theother structure the rest of the time. It has the same structure all the time. By aver-aging the single bond in one structure with the double bond in the other, wemight say that the oxygen-to-oxygen bonds in ozone are halfway between asingle and double bond, that is, 1.5 bonds. The fact that the electrons in ozoneare distributed over the whole molecule so as to produce two equivalentbonds is readily seen in the electrostatic potential map of ozone.

O OO OO O

O3

O O

OH HO

O ¬ O

O OO OO O

O3O2

O3.O2 ;

O2,

10-5

Electrostatic potential map ofozone

The two resonance structures in expression (10.18) are equivalent; that is,they contribute equally to the structure of the resonance hybrid. In manycases, there are several contributing resonance structures that do not con-tribute equally. For example, consider the azide anion, for which threeresonance structures are given below.

N3

-,

Electrostatic potential map ofthe azine anion

N N N N�

N N N�

N�

N�2 �1 0 0�1 �1 �1 �1 �2

We can decide which resonance structure likely contributes most to the hybridby applying the general rules for formal charges (page 389). The central reso-nance structure avoids the unlikely large formal charge of found on an Natom in the other two structures. Consequently, we expect that structure tocontribute most to the resonance hybrid of the azide anion.

-2

These two articles discussuseful analogies to helpexplain the concept of

resonance: Silverstein, Todd P.J. Chem. Educ. 1999: 76, 206 andStarkey, Ronald. J. Chem. Educ.1995: 72, 542.

PETRMC10_372-422-hr 12/15/05 6:28 PM Page 391


E X A M P L E 1 0 - 9Representing the Lewis Structure of a Resonance Hybrid. Write the Lewis struc-ture of the acetate ion,

SolutionThe skeletal structure has the three H atoms as terminal atoms bonded to a C atomas a central atom. The second C atom is also a central atom bonded to the first. Thetwo O atoms are terminal atoms bonded to the second C atom.

The number of valence electrons (dots) that must appear in the Lewis structure is

from H from C from O to establish charge of

Twelve of the valence electrons are used in the bonds in the skeletal structure, andthe remaining twelve are distributed as lone-pair electrons on the two O atoms.

In completing the octet of the C atom on the right, we discover that we can writetwo completely equivalent Lewis structures, depending on which of the two Oatoms furnishes the lone pair of electrons to form a carbon-to-oxygen double bond.The true Lewis structure is a resonance hybrid of the following two contributingstructures.

(10.19)H

H

H

C

O

C O H

H

H

C

O

C O

��

H

H

H

C

O

C O

�

1-c1 = 3 + 8 + 12 + 1 = 24+12 * 62+12 * 42+13 * 12

H

H

H

C

O

C O

CH3COO-.

Acetate anion Practice Example A: Draw Lewis structures to represent the resonance hybridfor the molecule.

Practice Example B: Draw Lewis structures to represent the resonance hybridfor the nitrate ion.

EXCEPTIONS TO THE OCTET RULE

The octet rule has been our mainstay in writing Lewis structures, and it willcontinue to be one. Yet at times, we must depart from the octet rule, as we willsee in this section.

ODD-ELECTRON SPECIES

The molecule NO has 11 valence electrons, an odd number. If the number ofvalence electrons in a Lewis structure is odd, there must be an unpaired elec-tron somewhere in the structure. Lewis theory deals with electron pairs anddoes not tell us where to put the unpaired electron; it could be on either the Nor the O atom. To obtain a structure free of formal charges, however, we willput the unpaired electron on the N atom.

N O

10-6

SO2

PETRMC10_372-422-hr 12/15/05 6:28 PM Page 392

10-6 Exceptions to the Octet Rule 393

The presence of unpaired electrons causes odd-electron species to be para-magnetic. NO is paramagnetic. Molecules with an even number of electronsare expected to have all electrons paired and to be diamagnetic. An importantexception is seen in the case of which is paramagnetic despite having 12valence electrons. Lewis theory does not provide a good electronic structurefor but the molecular orbital theory that we will consider in the next chap-ter is much more successful.

The number of stable odd-electron molecules is quite limited. More com-mon are free radicals, or simply radicals, highly reactive molecular fragmentswith one or more unpaired electrons. The formulas of free radicals are usuallywritten with a dot to emphasize the presence of an unpaired electron, such asin the methyl radical, , and the hydroxyl radical, . The Lewis structuresof these two free radicals are

Both of these free radicals are commonly encountered as transitory species inflames. In addition, is formed in the atmosphere in trace amounts as a re-sult of photochemical reactions. Many important atmospheric reactions involvefree radicals as reactants, such as in the oxidation of CO to Free radicals,because of their unpaired electron, are highly reactive species. The hydroxylradical, for example, is implicated in DNA damage that can lead to cancer.

INCOMPLETE OCTETS

Our initial attempt to write the Lewis structure of boron trifluoride leads to astructure in which the B atom has only six electrons in its valence shell—anincomplete octet.

(10.20)

We have learned to complete the octets of central atoms by shifting lone-pairelectrons from terminal atoms to form multiple bonds. One of three equivalentstructures with a boron-to-fluorine double bond is shown below.

(10.21)

An observation in support of structure (10.21) is that the bond lengthin (130 pm) is less than expected for a single bond. A shorter bond sug-gests that more than two electrons are present, that is, that there is multiple-bond character in the bond. On the other hand, the placement of formalcharges in structure (10.18) breaks an important rule—negative formal chargeshould be found on the more electronegative atom in the bond. In this struc-ture, the positive formal charge is on the most electronegative of all atoms—F.

The high electronegativity of fluorine (4.0) and the much lower one ofboron (2.0) suggest an appreciable ionic character to the boron-to-fluorinebond (see Figure 10-7). This suggests the possibility of ionic structures such asthe following.

(10.22)F

F

B� F

F

�

BF3

B ¬ F

F B F

F

�1�1

F

F

B F

F

OH CO CO2� H�

CO2.

#OH

H

H

C H HO

OHCH3

O2,

O2, Experimental evidencefor the paramagnetism of is shown in Figure 10-3.

O2

▲

The primary industrialuses of are not to producechemicals containing theelements boron or fluorine.Rather, is used becauseof properties stemming fromits electronic structure.In most cases the isrecovered and recycled.

BF3

BF3

BF3

▲

PETRMC10_372-422-hr 12/15/05 6:28 PM Page 393


In view of its molecular properties and chemical behavior, the best representa-tion of appears to be a resonance hybrid of structures (10.20, 10.21, and10.22), with perhaps the most important contribution made by the structurewith an incomplete octet (10.20). Whichever structure we choose to em-phasize, an important characteristic of is its strong tendency to form acoordinate covalent bond with a species capable of donating an electron pairto the B atom. This can be seen in the formation of the ion.

In the bonds are single bonds and the bond length is 145 pm.The number of species with incomplete octets is limited to some beryllium,

boron, and aluminum compounds. Perhaps the best examples are the boronhydrides. Bonding in the boron hydrides will be discussed in Chapter 22.

EXPANDED VALENCE SHELLS

We have consistently tried to write Lewis structures in which all atoms exceptH have a complete octet, that is, in which each atom has eight valence elec-trons. There are a few Lewis structures that break this rule by having 10 oreven 12 valence electrons around the central atom, creating what is called anexpanded valence shell. Describing bonding in these structures is an area ofactive interest among chemists.

Molecules with expanded valence shells typically involve nonmetal atomsof the third period and beyond that are bonded to highly electronegativeatoms. For example, phosphorus forms two chlorides, and We canwrite a Lewis structure for with the octet rule. In with five Cl atomsbonded directly to the central P atom, the outer shell of the P atom appears tohave ten electrons. We might say that the valence shell has expanded to 10electrons. In the molecule, the valence shell appears to expand to 12.

Expanded valence shells have also been used in cases where they appear togive a better Lewis structure than strict adherence to the octet rule, as suggest-ed by the two Lewis structures for the sulfate ion that follow.

The argument for including the expanded valence-shell structure is that it re-duces formal charges. Also, the experimentally determined sulfur-to-oxygenbond lengths in and are in agreement with this idea. The experi-mental results for summarized in structure (10.23), indicate that the

bond with O as a central atom and with an attached H atom is longerthan the bond with O as a terminal atom.S ¬ OS ¬ O

H2SO4,H2SO4SO4

2-

OSO

O

O

Normal octet

2�

�2

�1

�1

�1�1OSO

O

OExpanded valence

shell

2�

�1�1

ClPCl

Cl

Octet

P

Cl

Expanded valenceshell

Cl

Cl

Cl

ClS

F

Expanded valenceshell

F

F

F

F

F

SF6

PCl5 ,PCl3

PCl5 .PCl3

BF4

-,

F

F

BF�

�

F

FF

F

F

B

F

F

FF�

BF4

-

BF3

BF3

BF3

PETRMC10_372-422-hr 12/15/05 6:28 PM Page 394

10-7 Shapes of Molecules 395

(10.23)

Experimental evidence appears to support using an expanded valence shellin the Lewis structure of sulfuric acid. The experimentally determined bond length in the sulfate anion—149 pm—lies between the two bondlengths found in sulfuric acid, suggesting a partial double-bond character. Theexpanded valence-shell structure is suggestive of this partial double-bond char-acter, whereas the octet structure is not. For the sulfate anion, best agreementwith the observed bond lengths is found in a resonance hybrid havingstrong contributions from a series of resonance structures (10.24) based on ex-panded valence shells.

(10.24)

The problem with expanded valence-shell structures is, of course, to explainwhere the “extra” electrons go. This expansion has been rationalized by as-suming that after the 3s and 3p subshells of the central atom fill to capacity(eight electrons), extra electrons go into the empty 3d subshell. If we assumethat the energy difference between the 3p and 3d levels is not very large, the va-lence-shell expansion scheme seems reasonable. But is this a valid assumption?The use of the 3d orbitals for valence-shell expansion is a matter of scientificdispute.*† Although unresolved questions about the expanded valence-shellconcept may be unsettling, the point to keep in mind is that the unmodifiedoctet rule works perfectly well for most uses of Lewis structures. We will returnto this topic, together with several other unsettled issues, in the concludingsection of Chapter 11.

SHAPES OF MOLECULES

The Lewis structure for water gives the impression that the constituent atomsare arranged in a straight line.

However, the experimentally determined shape of the molecule is not linear.The molecule is bent, as shown in Figure 10-9. Does it really matter that the

molecule is bent rather than linear? The answer is, decidedly yes. As wewill learn in Chapter 12, the bent shape of water molecules helps to accountfor the fact that water is a liquid rather than a gas at room temperature. InChapter 13, we will find that it also accounts for the ability of liquid water todissolve so many different substances.

What we seek in this section is a simple model for predicting the approxi-mate shape of a molecule. Unfortunately, Lewis theory tells us nothing aboutthe shapes of molecules. On the other hand, it is an excellent place to begin.The next step is to use an idea based on repulsions between valence-shell elec-tron pairs. We will discuss this idea after defining a few terms.

H2O

OH H

10-7

O

O S O

O

O

O S O

2� 2�

O

O

O S O

O

2�

S ¬ O

S ¬ OS ¬ O

H

O

O

S

H

154 pm 143 pm

OO

*L. Suidan et al., J. Chem. Educ.: 72, 583 (1995).†G. H. Purser, J. Chem. Educ.: 78, 981 (2001).

O

H

d2

d1a

H

FIGURE 10-9Geometric shape of a moleculeTo establish the shape ofthe triatomic moleculeshown here, we need todetermine the distancesbetween the nuclei of thebonded atoms and the anglebetween adjacent bonds.In the bond lengths

and thebond angle a = 104.45°.d1 = d2 = 95.8 pm

H2O,

H2O

▲

PETRMC10_372-422-hr 12/15/05 6:28 PM Page 395


By molecular shape, we mean the geometric figure we get when joiningthe nuclei of bonded atoms by straight lines. Figure 10-9 depicts thetriatomic (three-atom) water molecule using a ball-and-stick model. Theballs represent the three atoms in the molecule, and the straight lines(sticks), the bonds between atoms. In reality, the atoms in the molecule arein close contact, but for clarity we show only the centers of the atoms.To have a complete description of the shape of a molecule, we need to knowtwo quantities.

• bond lengths, the distances between the nuclei of bonded atoms

• bond angles, the angles between adjacent lines representing bonds

We will concentrate on bond angles in this section and bond lengths inSection 10-8.

A diatomic molecule has only one bond and no bond angle. Because thegeometric shape determined by two points is a straight line, all diatomicmolecules are linear. A triatomic molecule has two bonds and one bond angle.If the bond angle is 180°, the three atoms lie on a straight line, and the mol-ecule is linear. For any other bond angle, a triatomic molecule is said to beangular, bent, or V-shaped. Some polyatomic molecules with more thanthree atoms have planar or even linear shapes. More commonly, however,the centers of the atoms in these molecules define a three-dimensionalgeometric figure.

VALENCE-SHELL ELECTRON-PAIR REPULSION (VSEPR) THEORY

The shape of a molecule is established by experiment or by a quantummechanical calculation confirmed by experiment. The results of theseexperiments and calculations are generally in good agreement with thevalence-shell electron-pair repulsion theory (VSEPR). In VSEPR theory,we focus on pairs of electrons in the valence electron shell of a central atomin a structure.

Electron pairs repel each other, whether they are in chemical bonds (bond pairs)or unshared (lone pairs). Electron pairs assume orientations about an atom to mini-mize repulsions.

This, in turn, results in particular geometric shapes for molecules.Another aspect of VSEPR theory is a focus not just on electron pairs but on

electron groups. A group of electrons can be a pair, either a lone pair or a bondpair, or it can be a single unpaired electron on an atom with an incompleteoctet, as in NO. A group can also be a double or triple bond between twoatoms. Thus in the molecule, the central C atom has two electrongroups in its valence shell. Each of the double bonds with its two electronpairs is treated as one electron group.

Consider the methane molecule, in which the central C atom hasacquired the electron configuration of Ne by forming covalent bonds withfour H atoms.

What orientation will the four electron groups (bond pairs) assume? The bal-loon analogy of Figure 10-10 suggests that electron-group repulsions will forcethe groups as far apart as possible—to the corners of a tetrahedron having theC atom at its center. The VSEPR method predicts, correctly, that is atetrahedral molecule.

CH4

CH

HH H

CH4,

OO C

FIGURE 10-10Balloon analogy to valence-shell electron-pair repulsionWhen two elongatedballoons are twisted together,they separate into four lobes.To minimize interferences,the lobes spread out into atetrahedral pattern. (Aregular tetrahedron hasfour faces, each an equilateraltriangle.) The lobes areanalogous to valence-shellelectron pairs.

▲

VSEPR can only rationalizemolecular structure; itcannot predict it. Counting

of electrons is important here. TheVSEPR model does not account forodd number of electrons, bondlength, bond strength, color, andparamagnetism.

Pfennig, B. W., Frock, R. L.“The Use of Molecular

Modeling and VSEPR Theory inthe Undergraduate Curriculum toPredict the Three-DimensionalStructure of Molecules.” J. Chem.Educ. 1999: 76, 1018 (July 1999).

VSEPR animation

PETRMC10_372-422-hr 12/15/05 6:28 PM Page 396


In and the central atom is also surrounded by four groups ofelectrons, but these molecules do not have a tetrahedral shape.

Here is the situation: VSEPR theory predicts the distribution of electrongroups, and in these molecules, electron groups are arranged tetrahedrallyabout the central atom. The shape of a molecule, however, is determined bythe location of the atomic nuclei. To avoid confusion, we will call the geomet-ric distribution of electron groups the electron-group geometry and the geo-metric arrangement of the atomic nuclei—the actual determinant of themolecular shape—the molecular geometry.

In the molecule, only three of the electron groups are bond pairs; thefourth is a lone pair. Joining the N nucleus to the H nuclei by straight lines out-lines a pyramid with the N atom at the apex and the three H atoms at the base;it is called a trigonal pyramid. We say that the electron-group geometry is tetra-hedral and the molecular geometry is trigonal-pyramidal.

In the molecule, two of the four electron groups are bond pairs andtwo are lone pairs. The molecular shape is obtained by joining the two H nu-clei to the O nucleus with straight lines. For the electron-group geometryis tetrahedral and the molecular geometry is V-shaped, or bent. The geometricshapes of and are shown in Figure 10-11, together with com-puter generated space-filling molecular models, electron charge density maps,and electrostatic potential maps.

H2ONH3,CH4,

H2O,

H2O

NH3

NH

H H and OH

H

H2O,NH3

VSEPRnotation:

H

H

H

HH

H

H

H

HO

C

AX4(a) (b)

N

AX3E(c)

AX2E2

FIGURE 10-11Molecular shapes based on tetrahedralelectron-group geometry ofand Molecular shapes are established bythe blue lines. Lone-pair electrons areshown as red dots along broken linesoriginating at the central atom. (a) Allelectron groups around the central atomare bond pairs. The blue lines that outlinethe molecule are different from the blacklines representing the carbon-to-hydrogenbonds. (b) The lone pair of electrons isdirected to the “missing” corner of thetetrahedron. The nitrogen-to-hydrogenbonds form three of the edges of atrigonal pyramid. (c) The moleculeis a bent molecule outlined by the twooxygen-to-hydrogen bonds.

H2O

H2ONH3 ,CH4 ,

▲

This article discussesthe two main methodsthat have been used to

determine the tetrahedralbond angle: ten Hoor, Marten. J. Chem. Educ. 2002: 79, 956.

The use of computer-generated models. Rasmol

(PC Mac) or Chime (Web) cangreatly enhance studentunderstanding. On the Internet,thousands of molecular structuresexist with the extension .pdb(Protein Data Bank). For example,http://www.eou.edu/chemweb/molmodel/mmp1.html and manyothers. Check CW for updates.

PETRMC10_372-422-hr 12/15/05 6:28 PM Page 397


In the VSEPR notation used in Figure 10-11, A is the central atom, X is a ter-minal atom or group of atoms bonded to the central atom, and E is a lone pairof electrons. Thus, the symbol signifies that two atoms or groups (X) arebonded to the central atom (A). The central atom also has two lone pairs ofelectrons (E). is an example of a molecule of the type.

For tetrahedral electron-group geometry, we expect bond angles of 109.5°,known as the tetrahedral bond angle. In the molecule, the measured bondangles are, in fact, 109.5°. The bond angles in and are slightly small-er: 107° for the bond angle and 104.5° for the bondangle. We can explain these less-than-tetrahedral bond angles by the fact thatthe charge cloud of the lone-pair electrons spreads out. This forces the bond-pair electrons closer together and reduces the bond angles.

POSSIBILITIES FOR ELECTRON-GROUP DISTRIBUTIONS

The most common situations are those in which central atoms have two, three,four, five, or six electron groups distributed around them.

Electron-group geometries

• two electron groups: linear• three electron groups: trigonal-planar• four electron groups: tetrahedral• five electron groups: trigonal-bipyramidal• six electron groups: octahedral

Figure 10-12 extends the balloon analogy to these cases. The cases for five- andsix-electron groups are typified by and molecules with expandedvalence shells.

The molecular geometry is the same as the electron-group geometry onlywhen all electron groups are bond pairs. These are for the VSEPR notation

(that is, and so on). In Table 10.1, the cases are illus-trated by photographs of ball-and-stick models. If one or more electron groupsare lone pairs, the molecular geometry is different from the electron-groupgeometry, although still derived from it. The relationship between electron-group geometry and molecular geometry is summarized in Table 10.1. To un-derstand all the cases in Table 10.1, we need two more ideas.

• The closer together two groups of electrons are forced, the stronger the repulsionbetween them. The repulsion between two electron groups is much strongerat an angle of 90° than at 120° or 180°.

AXnAX4,AX2, AX3,AXn

SF6,PCl5

H ¬ O ¬ HH ¬ N ¬ HH2ONH3

CH4

AX2E2H2O

AX2E2

FIGURE 10-12Several electron-groupgeometries illustratedThe electron-group geometriespictured are trigonal-planar(orange), tetrahedral (gray),trigonal-bipyramidal (pink), andoctahedral (yellow). The atomsat the ends of the balloons arenot shown and are not importantin this model.

▲

VSEPR theory works bestfor second-period elements.The predicted bond angleof 109.5° for is close tothe measured angle of 104.5°.For however, thepredicted value of 109.5° isnot in good agreement withthe observed 92°.

H2S,

H2O

▲

PETRMC10_372-422-hr 12/15/05 6:28 PM Page 398


TABLE 10.1 Molecular Geometry as a Function of Electron-Group Geometry

Number Numberof Electron- of IdealElectron Group Lone VSEPR Molecular BondGroups Geometry Pairs Notation Geometry Angles Example

2 linear 0 180°

3 trigonal- 0 120°planar

trigonal- 1 120°planar

4 tetrahedral 0 109.5°

tetrahedral 1 109.5°

tetrahedral 2 109.5°

5 trigonal- 0 90°, 120°bipyramidal

PC15AX5

OH2AX2E2

NH3AX3E

CH4AX4

aSO2AX2E

BF3AX3

BeC12AX2

(continues)

VSEPR: Basic MolecularConfigurations models

(linear)

(trigonal planar)

(bent)

(tetrahedral)

(trigonal-pyramidal)

(bent)

(trigonal-bipyramidal)

A

X

X

XXX

A X

X

AXX

X

A

X

XXX

A

X

X

A

X

X

X

XAX

PETRMC10_372-422-hr 12/15/05 6:28 PM Page 399


TABLE 10.1 (Continued)

Number Numberof Electron- of IdealElectron Group Lone VSEPR Molecular BondGroups Geometry Pairs Notation Geometry Angles Example

(seesaw)

(T-shaped)

(linear)

(octahedral)

(square-pyramidal))

(square-planar)

AX X

XX

A

X

X XXX

A

X

X XXX

X

A

X

X

A

X

X

X

A

X

X

X

Xtrigonal- 1 90°, 120°bipyramidal

trigonal- 2 90°bipyramidal

trigonal- 3 180°bipyramidal

6 octahedral 0 90°

octahedral 1 90°

octahedral 2 90°

For a discussion of the structure of see page 402.For a discussion of the placement of the lone-pair electrons in this structure, see page 401.b

SO2,a

XeF4AX4E2

BrF5AX5E

SF6AX6

XeF2AX2E3

C1F3AX3E2

SF4bAX4E

PETRMC10_372-422-hr 12/15/05 6:28 PM Page 400


• Lone-pair electrons spread out more than do bond-pair electrons. As a result,the repulsion of one lone pair of electrons for another lone pair is greaterthan, say, between two bond pairs. The order of repulsive forces, fromstrongest to weakest, is:

Consider with the VSEPR notation Two possibilities for itsstructure are presented in the margin, but only one is correct. The correctstructure (top) places a lone pair of electrons in the central plane of thebipyramid. As a result, two lone pair–bond pair interactions are 90°. In the in-correct structure (bottom), the lone pair of electrons is at the bottom of thebipyramid and results in three lone pair–bond pair interactions of 90°. This isa less favorable arrangement.

APPLYING VSEPR THEORY

The following four-step strategy can be used for predicting the shapes ofmolecules.

1. Draw a plausible Lewis structure of the species (molecule or polyatomic ion).2. Determine the number of electron groups around the central atom, and iden-

tify them as being either bond-pair electron groups or lone pairs of electrons.3. Establish the electron-group geometry around the central atom—linear,

trigonal-planar, tetrahedral, trigonal-bipyramidal, or octahedral.4. Determine the molecular geometry from the positions around the central

atom occupied by the other atomic nuclei—that is, from data in Table 10.1.

E X A M P L E 1 0 - 1 0Using VSEPR Theory to Predict a Geometric Shape. Predict the molecular geom-etry of the polyatomic anion

SolutionApply the four steps outlined above.

Step 1. Write the Lewis structure. The number of valence electrons isFrom I From Cl To establish ionic charge of

1

To join 4 Cl atoms to the central I atom and to provide octets for all the atoms, weneed 32 electrons. In order to account for all 36 valence electrons, we need to placean additional four electrons around the I atom as lone pairs. That is, we are forced toexpand the valence shell of the I atom to accommodate all the electrons required inthe Lewis structure.

Step 2. There are six electron groups around the I atom, four bond pairs and twolone pairs.

Step 3. The electron-group geometry (the orientation of six electron groups) isoctahedral.

Step 4. The anion is of the type which according to Table 10.1, leads toa molecular geometry that is square-planar.

Figure 10-13 suggests two possibilities for distributing bond pairs and lone pairs inThe square-planar structure is correct because the lone pair–lone pair interac-

tion is kept at 180°. In the incorrect structure, this interaction is at 90°, which resultsin a strong repulsion.

ICl4

-.

AX4E2,ICl4

-

Cl

I

Cl

ClCl

�

= 3614 * 72 +11 * 72�1

ICl4

-.

AX4E.SF4,

lone pair–lone pair 7 lone pair–bond pair 7 bond pair–bond pair

KEEP IN MINDthat we need to expand avalence shell only if morethan eight electrons must beaccommodated by the centralatom in a Lewis structure.This leads to structures basedon five or six electron groups.Otherwise, octet-based Lewisstructures are perfectlysatisfactory when applyingthe VSEPR theory.

▲

Cl

Cl

Cl

Cl

Cl

Cl

Cl

Cl

(Incorrect)

(Correct)

�

�

I

I

FIGURE 10-13Example 10-10 illustratedThe observed structure of

is square-planar.ICl4

-

▲

SF

F

FF

SF

F

FF

PETRMC10_372-422-hr 12/15/05 6:28 PM Page 401


SOO

SOO

O

H H

C

Practice Example A: Predict the molecular geometry of nitrogen trichloride.

Practice Example B: Predict the molecular geometry of phosphoryl chloride,an important chemical in the manufacture of gasoline additives, hydraulic

fluids, and fire retardants.

CONCEPT ASSESSMENT ✓The ions and differ by only two electrons. Would you expect them to havethe same geometric shape? Explain.

STRUCTURES WITH MULTIPLE COVALENT BONDS

In a multiple covalent bond, all electrons in the bond are confined to the re-gion between the bonded atoms, and together constitute one group of elec-trons. Let us test this idea by predicting the molecular geometry of sulfurdioxide. S is the central atom, and the total number of valence electrons is

The Lewis structure is the resonance hybrid of the two contribut-ing structures in the margin.

It is immaterial which structure we use. In either case, we count the electronsin the double covalent bond as one group. This bond and the sulfur-to-oxygensingle bond account for two electron groups. The third electron group aroundthe central S atom is a lone pair of electrons. The electron-group geometryaround the central S atom is that of three electron groups—trigonal-planar. Ofthe three electron groups, two are bonding groups and one is a lone pair. This isthe case of (see Table 10.1). The molecular shape is angular, or bent, withan expected bond angle of 120°. (The measured bond angle in is 119°.)

E X A M P L E 1 0 - 1 1Using VSEPR Theory to Predict the Shape of a Molecule with a Multiple Cova-lent Bond. Predict the molecular geometry of formaldehyde, used to makea number of polymers, such as melamine resins.

SolutionThe Lewis structure in the margin has a total of 12 valence electrons and C as thecentral atom. If all the bonds to the carbon atom were single bonds, C would lack anoctet. This deficiency is corrected by moving a lone pair of electrons from the Oatom into the carbon-to-oxygen bond, making it a double bond.

There are three electron groups around the C atom, two groups in the carbon-to-hydrogen single bonds and the third group in the carbon-to-oxygen double bond.The electron-group geometry for three electron groups is trigonal-planar. Because allthe electron groups are involved in bonding, the VSEPR notation for this moleculeis The molecular geometry is also trigonal-planar.

Practice Example A: Predict the shape of the COS molecule.

Practice Example B: Nitrous oxide, is the familiar laughing gas used asan anesthetic in dentistry. Predict the shape of the molecule.

MOLECULES WITH MORE THAN ONE CENTRAL ATOM

Although many of the structures of interest to us have only one central atom,VSEPR theory can also be applied to molecules or polyatomic anions withmore than one central atom. In such cases, the geometric distribution of termi-nal atoms around each central atom must be determined and the results thencombined into a single description of the molecular shape. We use this idea inExample 10-12.

N2ON2O,

AX3.

H2CO,

SO2

AX2E

3 * 6 = 18.

ICl2

+ICl2

-

POCl3,

The need for resonancestructures is an obviousfailing of the Lewis

structures. Note that the more validresonance structures that exist, themore stable the actual structure is.

PETRMC10_372-422-hr 12/15/05 6:28 PM Page 402


E X A M P L E 1 0 - 1 2Applying VSEPR Theory to a Molecule with More Than One Central Atom.Methyl isocyanate, is used in the manufacture of insecticides, such as car-baryl (Sevin). In the molecule, the three H atoms and the O atom are ter-minal atoms and the two C and one N atom are central atoms. Sketch this molecule.

SolutionTo apply the VSEPR method, let us begin with a plausible Lewis structure. Thenumber of valence electrons in the structure is

From C From N From O From H

In drawing the skeletal structure and assigning valence electrons, we first obtain astructure with incomplete octets. By shifting the indicated electrons, we can giveeach atom an octet.

The C atom on the left has four electron groups around it—all bond pairs. Theshape of this end of the molecule is tetrahedral. The C atom to the right, by formingtwo double bonds, is treated as having two groups of electrons around it. This dis-tribution is linear. For the N atom, three groups of electrons are distributed in atrigonal-planar manner. The bond angle should be about 120°.

Practice Example A: Sketch the methanol molecule, Indicate the bondangles in this molecule.

Practice Example B: Glycine, an amino acid, has the formula Sketch the glycine molecule, and indicate the various bond angles.

MOLECULAR SHAPES AND DIPOLE MOMENTS

Let us recall some facts that we learned about polar covalent bonds in Section10-3. In the HCl molecule, the Cl atom is more electronegative than the Hatom. Electrons are displaced toward the Cl atom, as clearly shown in the elec-trostatic potential map in the margin. The HCl molecule is a polar molecule.In the representation below, we use a cross-base arrow that points to theatom that attracts electrons more strongly.

The extent of the charge displacement in a polar covalent bond is given bythe dipole moment, The dipole moment is the product of a partial charge

and distance (d).(10.25)

If the product, has a value of the dipole moment, has a value called 1 debye, D. One experimental methodof determining dipole moments is based on the behavior of polar molecules inan electric field, suggested in Figure 10-14.

m,(C # m),3.34 * 10-30 coulomb # meterd * d,

m = d * d

1d2m.

d+H 6 Cld-

(6)

H2NCH2COOH.

CH3OH.

O

N

C

C

HH H

180�

120�

109�

C ¬ N ¬ C

ON CH

H

H

C ON CH

H

H

C

13 * 12 = 2211 * 6211 * 5212 * 42

CH3NCOCH3NCO,

Molecular Polarity activity

The students may havestudied vectors. Use ofvectors is a good way to

show that a molecule has or doesnot have a dipole moment.

PETRMC10_372-422-hr 12/15/05 6:28 PM Page 403


The polarity of the bond, as demonstrated on page 379, involvesa shift of the electron charge density toward the Cl atom, and this produces aseparation of the centers of positive and negative charge. Suppose, instead ofa shift in electron charge density, we think of an equivalent situation—thetransfer of a fraction of the charge of an electron from the H atom to the Clatom through the entire internuclear distance. Let us determine the magni-tude of this partial charge, To do this, we need the measured dipolemoment, 1.03 D; the bond length, 127.4 pm; and equation (10.25) re-arranged to

This charge is about 17% of the charge on an electron andsuggests that HCl is about 17% ionic. This assessment of the percent ioniccharacter of the bond agrees well with the 20% we made based onelectronegativity differences (recall Example 10-4).

Carbon dioxide molecules are nonpolar. To understand this observa-tion, we need to distinguish between the displacement of electron chargedensity in a particular bond and in the molecule as a whole. The elec-tronegativity difference between C and O causes a displacement of elec-tron charge density toward the O atom in each carbon-to-oxygen bondand gives rise to a bond dipole. However, because the two bond dipoles areequal in magnitude and point in opposite directions, they cancel eachother and lead to a resultant dipole moment of zero for the molecule. Thesymmetrical nature of the electron charge density is clear in the electrosta-tic potential map for in the margin.

O O � 0C �

CO2

CO2 .

H ¬ Cl

(1.602 * 10-19 C)

d =m

d=

1.03 D * 3.34 * 10-30 C # m>D

127.4 * 10-12 m= 2.70 * 10-20 C

H ¬ Cld.

H ¬ Cl

��

Electrode Electrode

d�

d�d�

d�

d�

d�

d�

d�

d�

d�

d�

d�

d�

d�

d�

d�

d�

d�

d�

d�

d�

d�

d�

d�

Field on

d�

d�

d�

d�

d�

d�

d�

d�

d�

d�

d�

d�d�

d�

d�

d�

d�

d�

d�

d�

d�d�

d� d�

Field off

FIGURE 10-14Polar molecules in an electric fieldThe device pictured is called an electrical condenser (or capacitor). It consists of a pairof electrodes separated by a medium that does not conduct electricity but consists ofpolar molecules. (a) When the field is off, the molecules orient randomly. (b) Whenthe electric field is turned on, the polar molecules orient in the field between thecharged plates so that the negative ends of the molecules are toward the positiveplate and vice versa.

▲

Electrostatic potential map ofcarbon dioxide

The following articlediscusses the commondifficulties and

misconceptions students haveabout determining the geometriesand polarities of molecules andposes some strategies foralleviating these problems:Furio, Carlos and Calatayud, MaLuisa. J. Chem. Educ. 1996: 73, 36.

PETRMC10_372-422-hr 12/15/05 6:29 PM Page 404


KEEP IN MINDthat the lack of a moleculardipole moment cannotdistinguish between thetwo possible moleculargeometries: tetrahedral andsquare-planar. To do this,other experimentalevidence, such as X-raydiffraction, is required.

▲

Cl

C

ClCl

C

(a) CCl4: a nonpolar molecule

(b) CHCl3: a polar molecule

H�

Cl

ClCl

Cl

FIGURE 10-15Molecular shapes anddipole moments(a) The resultant of two of the

bond dipoles is shownas a red arrow, and that of theother two, as a blue arrow.The red and blue arrowspoint in opposite directionsand cancel. The molecule is nonpolar. Thebalance of the chargedistribution in is clearlyseen in the electrostaticpotential map. (b) Theindividual bond dipoles docombine to yield a resultantdipole moment (red arrow)of 1.04 D. The electrostaticpotential map indicates thatthe hydrogen atom has apartial positive charge.

CCl4

CCl4

C ¬ Cl

▲

The fact that is nonpolar is experimental proof that the three atomsin the molecule lie along a straight line in the order Ofcourse, we can also predict that is a linear molecule with theVSEPR theory, based on the Lewis structure

Water molecules are polar. They have bond dipoles because ofthe electronegativity difference between H and O, and the bonddipoles combine to produce a resultant dipole moment of 1.84 D. Theelectrostatic potential map for water provides visual evidence of a netdipole moment on the water molecule. The molecule cannot be linear,for this would lead to a cancellation of bond dipoles, just as with We have predicted with the VSEPR theory that the molecule isbent, and the observation that it is a polar molecule simply confirmsthe prediction.

Carbon tetrachloride molecules are nonpolar. Based on the elec-tronegativity difference between Cl and C, we expect a bond dipole forthe bond. The fact that the resultant dipole moment is zeromeans that the bond dipoles must be oriented in such a way that theycancel. The tetrahedral molecular geometry of provides the sym-metrical distribution of bond dipoles that leads to this cancellation, asshown in Figure 10-15(a). Can you see that the molecule will be polar ifone of the Cl atoms is replaced by an atom with a different electro-negativity, say H? In the molecule, there is a resultant dipolemoment (Fig. 10-15b).

CHCl3,

CCl4

C ¬ Cl

CCl4 .

O

H

H104�

H2OCO2.

H2O.

C OO

CO2

O ¬ C ¬ O.CO2

Electrostatic potential map of water

PETRMC10_372-422-hr 12/15/05 6:29 PM Page 405


E X A M P L E 1 0 - 1 3Determining the Relationship Between Geometric Shapes and the Resultant Di-pole Moments of Molecules. Which of these molecules would you expect to bepolar:

SolutionPolar: ICl and NO are diatomic molecules with an electronegativitydifference between the bonded atoms. is a bent molecule with an electronega-tivity difference between the S and O atoms.Nonpolar: and is a diatomic molecule of identical atoms; hence noelectronegativity difference. For refer to Table 10.1. is a symmetrical pla-nar molecule (120° bond angles). The bond dipoles cancel each other.

Practice Example A: Only one of the following molecules is polar. Which is it,and why?

Practice Example B: Only one of the following molecules is nonpolar. Which isit, and why?

CONCEPT ASSESSMENT ✓The molecule has the dipole moment whereas for the similarmolecule Why do you suppose there is such a large differencein these two values?[Hint: What is the effect of the lone-pair electrons on the N atom?]

BOND ORDER AND BOND LENGTHS

The term bond order describes whether a covalent bond is singledouble or triple Think

of electrons as the “glue” that binds atoms together in covalent bonds. Thehigher the bond order—that is, the more electrons present—the more glue andthe more tightly the atoms are held together.

Bond length is the distance between the centers of two atoms joined by acovalent bond. A double bond between atoms is shorter than a single bond,and a triple bond is shorter still. You can see this relationship clearly in Table10.2 by comparing the three different bond lengths for the nitrogen-to-nitro-gen bond. For example, the measured length of the nitrogen-to-nitrogen triplebond in is 109.8 pm, whereas the nitrogen-to-nitrogen single bond in hy-drazine, is 147 pm.

Perhaps you can now also better understand the meaning of covalent radiusthat we introduced in Section 9-3. The single covalent radius is one-half the dis-tance between the centers of identical atoms joined by a single covalent bond.Thus, the single covalent radius of chlorine in Figure 9-8 (99 pm) is one-half

the bond length given in Table 10.2, that is, Furthermore, as arough generalization,

the length of the covalent bond between two atoms can be approximated as the sumof the covalent radii of the two atoms.

Some of these ideas about bond length are applied in Example 10-14.

12 * 199 pm.

H2N ¬ NH2,N2

(bond order = 3).(bond order = 2),(bond order = 1),

10-8

0.24 D.=NF3, mm = 1.47 D,NH3

NH3.CH2Cl2 ,PCl5 ,Cl3CCH3,

C2H4.H2O2,SF6,

B ¬ FBF3BF3,

Cl2BF3.Cl2

SO2

ICl, NO, SO2 .

Cl2 , ICl, BF3, NO, SO2 ?

It is a good idea to relatebond order and bondenergies to

thermodynamics. It is alsoimportant to remember that thesebond orders depend upon thelocalized bond model andresonance cannot be accounted forwith the energies in Table 10.3.

PETRMC10_372-422-hr 12/15/05 6:29 PM Page 406

10-8 Bond Order and Bond Lengths 407

TABLE 10.2 Some Average Bond Lengthsa

Bond Bond BondLength, Length, Length,

Bond pm Bond pm Bond pm

74.14 154 145110 134 123100 120 109.897 147 136

132 128 12091.7 116 145

127.4 143 121141.4 120 143160.9 178 199

228266

aMost values ( and so on) are averaged over a number of speciescontaining the indicated bond and may vary by a few picometers. Where a diatomicmolecule exists, the value given is the actual bond length in that molecule (and so on) and is known more precisely.

HF,N2,H2,

C ¬ H,N ¬ H,C ¬ H,

I ¬ IBr ¬ BrC1 ¬ C1C ¬ C1H ¬ IF ¬ FC “ OH ¬ BrO “ OC ¬ OH ¬ C1O ¬ OC ‚ NH ¬ FN “ OC “ NH ¬ SN ¬ OC ¬ NH ¬ ON ‚ NC ‚ CH ¬ NN “ NC “ CH ¬ CN ¬ NC ¬ CH ¬ H

E X A M P L E 1 0 - 1 4Estimating Bond Lengths. Provide the best estimate you can of these bond lengthsfor the (a) the nitrogen-to-hydrogen bonds in (b) the bromine-to-chlorinebond in BrCl.

Solution(a) The Lewis structure of ammonia (page 376) shows the nitrogen-to-

hydrogen bonds as single bonds. The value listed in Table 10.2 for thebond is 100 pm, so this is the value we would predict. (The mea-

sured bond length in is 101.7 pm.)(b) There is no bromine-to-chlorine bond length in Table 10.2 , so we need

to calculate an approximate bond length using the relationship betweenbond length and covalent radii. BrCl contains a single bond[imagine substituting one Br atom for one Cl atom in structure (10.4)].The length of the bond is one-half the bond length plus

one-half the bond length:

(The measured bond length is 213.8 pm.)

Practice Example A: Estimate the bond lengths of the carbon-to-hydrogenbonds and the carbon-to-bromine bond in

Practice Example B: In the thiocyanate ion, the length of the carbon-to-nitrogen bond is 115 pm. Write a plausible Lewis structure for this ion and describeits geometric shape.

CONCEPT ASSESSMENT ✓and are made up of the same atoms. How would you expect the nitro-

gen-to-oxygen bond lengths in these two ions to compare?NO2

+NO2

-

SCN-,

CH3Br.

214 pm.

=A12 * 288 pm B+A12 * 199 pm BBr ¬ Br

Cl ¬ ClBr ¬ Cl

Br ¬ Cl

NH3N ¬ HN ¬ H

NH3 ;

PETRMC10_372-422-hr 12/15/05 6:29 PM Page 407


BOND ENERGIES

Together with bond lengths, bond energies can be used to assess the suitabili-ty of a proposed Lewis structure. Bond energy, bond length, and bond orderare interrelated properties in this sense: the higher the bond order, the shorterthe bond between two atoms and the greater the bond energy.

Energy is released when isolated atoms join to form a covalent bond, and ener-gy must be absorbed to break apart covalently bonded atoms. Bond-dissociationenergy, D, is the quantity of energy required to break one mole of covalentbonds in a gaseous species. The SI units are kilojoules per mole of bonds

In the manner of Chapter 7, we can think of bond-dissociation energy as anenthalpy change or a heat of reaction. For example,

It is not hard to picture the meaning of bond energy for a diatomic mole-cule, because there is only one bond in the molecule. It is also not difficult tosee that the bond-dissociation energy of a diatomic molecule can be expressedrather precisely, as is that of With a polyatomic molecule such as the situation is different (Fig. 10-16). The energy needed to dissociate one moleof H atoms by breaking one bond per molecule

is different from the energy required to dissociate one mole of H atoms bybreaking the bonds in OH(g).

The two bonds in are identical; therefore, they should haveidentical energies. This energy, which we can call the bond energy in

is the average of the two values listed above: The bond energy in other molecules containing the OH group will be somewhatdifferent from that in For example, in methanol, the

bond-dissociation energy, which we can represent as isThe usual method of tabulating bond energies (Table 10.3) is as

averages. An average bond energy is the average of bond-dissociation energiesfor a number of different species containing the particular bond. Understand-ably, average bond energies cannot be stated as precisely as specific bond-dis-sociation energies.

As you can see from Table 10.3, double bonds have higher bond energiesthan do single bonds between the same atoms, but they are not twice as large.Triple bonds are stronger still, but their bond energies are not three times aslarge as single bonds between the same atoms. This observation about bondorder and bond energy will seem quite reasonable after multiple bonds aremore fully described in the next chapter.

Bond energies also have some interesting uses in thermochemistry. For a re-action involving gases, visualize the process

In this hypothetical process, we first break all the bonds in reactant moleculesand form gaseous atoms. For this step, the enthalpy change is

(reactants), where BE stands for bond energy. Next,we allow the gaseous atoms to recombine into product molecules. In this step,bonds are formed and The en-thalpy change of the reaction, then, is

(10.26) L ©BE (reactants) - ©BE (products)

¢Hrxn = ¢H (bond breakage) + ¢H (bond formation)

(products).- ©BE=(bond formation)¢H

©BE=(bond breakage)¢H

gaseous reactants ¡ gaseous atoms ¡ gaseous products

436.8 kJ>mol.D(H ¬ OCH3),O ¬ H

CH3OH,H ¬ O ¬ H.

O ¬ H463.4 kJ>mol.H2O,O ¬ H

H2OO ¬ H

O ¬ H(g) ¡ H(g) + O(g) ¢H = D(O ¬ H) = +428.0 kJ>mol

H ¬ OH(g) ¡ H(g) + OH(g) ¢H = D(H ¬ OH) = +498.7 kJ>mol

H2OO ¬ H

H2O,H2(g).

Bond formation: 2 H(g) ¡ H2(g) ¢H = -D(H ¬ H) = -435.93 kJ>mol

Bond breakage: H2(g) ¡ 2 H(g) ¢H = D(H ¬ H) = +435.93 kJ>mol

(kJ>mol).

10-9

428.0 kJ/mol

498.7 kJ/mol

435.93 kJ/mol

O H

O HH

H H

FIGURE 10-16Some bond energiescomparedThe same quantity of energy,

is requiredto break all bonds.In more energy isrequired to break the firstbond thanto break the second

The secondbond broken is that in theOH radical. The bondenergy in is the averageof the two values:463.4 kJ>mol.

H2OO ¬ H

(428.0 kJ>mol).

(498.7 kJ>mol)

H2O,H ¬ H

435.93 kJ>mol,

▲

KEEP IN MINDthat tabulated bond energiesare for isolated molecules inthe gaseous state. They donot apply to molecules inclose contact in liquidsand solids.

▲

PETRMC10_372-422-hr 12/15/05 6:29 PM Page 408

10-9 Bond Energies 409

TABLE 10.3 Some Average Bond Energiesa

Bond Bond BondEnergy, Energy Energy

Bond kJ mol Bond kJ mol Bond kJ mol

436 347 163414 611 418389 837 946464 305 222368 615 590565 891 142431 360 498364 736 159297 339 243

193151

Although all data are listed with about the same precision (three significant figures), somevalues are actually known more precisely. Specifically, the values for the diatomic molecules

HF, HCl, HBr, HI, and are actually bond-dissociation energies, rather than average bond energies.The value for the bonds in is 799 kJ>mol.CO2C “ Ob

I2Br2 ,Cl2 ,F2 ,(O “ O),O2(N ‚ N),N2H2,

a

I ¬ IBr ¬ BrC1 ¬ C1C ¬ C1H ¬ IF ¬ FbC “ OH ¬ BrO “ OC ¬ OH ¬ C1O ¬ OC ‚ NH ¬ FN “ OC “ NH ¬ SN ¬ OC ¬ NH ¬ ON ‚ NC ‚ CH ¬ NN “ NC “ CH ¬ CN ¬ NC ¬ CH ¬ H

///

The approximately equal sign in expression (10.26) signifies that someof the bond energies used are likely to be average bond energies rather thantrue bond-dissociation energies. Also, a number of terms often cancel outbecause some of the same types of bonds appear in the products as in the re-actants. We can base the calculation of just on the net number and typesof bonds broken and formed, as illustrated in Example 10-15.

E X A M P L E 1 0 - 1 5Calculating an Enthalpy of Reaction from Bond Energies. The reaction of methane

and chlorine produces a mixture of products called chloromethanes. One ofthese is monochloromethane, used in the preparation of silicones. Calculate

for the reaction

SolutionTo assess which bonds are broken and formed, it helps to draw structural formulas(or Lewis structures), as in Figure 10-17. To apply expression (10.26) literally, wewould break four bonds and one bond and form three bonds,one bond, and one bond. The net change, however, is the breaking ofone bond and one bond, followed by the formation of onebond and one bond.

Practice Example A: Use bond energies to estimate the enthalpy change for thereaction

Practice Example B: Use bond energies to estimate the enthalpy of formationof NH3(g).

2 H2(g) + O2(g) ¡ 2 H2O(g)

¢H for net bond breakage: 1 mol C ¬ H bonds +414 kJ1 mol Cl ¬ Cl bonds +243 kJ sum: +657 kJ

¢H for net bond formation: 1 mol C ¬ Cl bonds -339 kJ1 mol H ¬ Cl bonds -431 kJ sum: -770 kJ

Enthalpy of reaction: ¢Hrxn = 657 - 770 = -113 kJ

H ¬ ClC ¬ ClCl ¬ ClC ¬ H

H ¬ ClC ¬ ClC ¬ HCl ¬ ClC ¬ H

CH4(g) + Cl2(g) ¡ CH3Cl(g) + HCl(g)¢H

CH3Cl,(CH4)

¢Hrxn

1L2

Point out this only worksfor the localized bondmodel, and predicting if a

reaction is exothermic orendothermic for resonancestructures cannot be done usingbond energies.

Calculate the bond energy ofbenzene based upon a

localized bond model of six H ¬ Cbonds, three C¬C bonds, andthree C “C bonds. Compare thevalue with the experimental heat offormation and find that thelocalized bond model calculation,using Table 10.3, gives a value thatis about 150 kJ mol-1 less than theexperimental value. The differenceis due to resonance stability.

Enthalpy of Reactionactivity

PETRMC10_372-422-hr 12/15/05 6:29 PM Page 409

�

414 kJ/mol 243 kJ/mol

H

H

H

C H Cl Cl �

�339 kJ/mol �431 kJ/mol

H

H

H

C Cl H Cl

FIGURE 10-17Net bond breakage and formation in a chemical reaction—Example 10-15illustratedBonds that are broken are shown in red and bonds that are formed, in blue. Bondsthat remain unchanged are black. The net change is that one and onebond break and one and one bond form.H ¬ ClC ¬ Cl

Cl ¬ ClC ¬ H

▲

We have described how bond lengths contribute to ourknowledge of molecular structure (Section 10-8), but howcan bond lengths be measured? One method makes use ofthe interaction of electromagnetic radiation with the oscil-lating electric field produced by a rotating polar moleculein the gas phase.

Consider polar diatomic molecules such as HF. In thegas phase the molecules rotate about (1) an axis coincidingwith the internuclear axis and (2) an axis perpendicular tothe internuclear axis. It is this second rotation that createsthe oscillating electric field that interacts with incomingradiation; some energy is absorbed from the radiation andthis energy increases the rotational energy of the molecules.The rotational motion of a diatomic molecule is quantized,and the quantum mechanical rotational energy levels forthe molecule are given by

(10.27)Erot =12I

b AJ1J + 12Bh2

4p2 r

MOLECULES IN SPACE: MEASURING BOND LENGTHSFOCUS ON

where is the rotational energy; J is the rotational quan-tum number, which can have the values 1, 2, 3, andso on; and I is the moment of inertia for rotation about theaxis interacting with the electromagnetic radiation. Therole of the moment of inertia in rotational motion is equiv-alent to that of simple mass in translational motion; itsvalue is given by

(10.28)

where and are the masses of the atoms in the di-atomic molecule and is the internuclear separation be-tween the atoms, that is, the bond length.

The rotational energy given by equation (10.27) has theunit joule. However, rotational spectra are usually dis-played in the unit favored by spectroscopists— Atfirst glance, it is not obvious that the unit is related toan energy unit. Let us briefly show that it is. ConsiderPlanck’s equation, The unit of Planck’s constant,h, is J s; that of radiation frequency, is Thus the unitof E is Now suppose we replace the value of by its equivalent, Then we have and

So, we see that the reciprocal of the wave-length is proportional to the energy in joules, with the pro-portionality constant being If we express in we use but if is in then

Thus, to convert equation (10.27)in the unit joule to an equation in the unit we mustdivide both sides of the equation by hc, as shown below.

(10.29)

The various constants and the moment of inertia arecombined into a single constant, the rotational constant B.We can therefore write

(10.30)

where B =h

8p2 cI

(in cm-1)

eJ = B AJ1J + 12B (in cm-1),

Erot

hc= eJ =

h

8p2 cI

AJ1J + 12B (in cm-1)

Erot =12I

b AJ1J + 12Bh2

4p2 r (in J)

cm-1,c = 2.998 * 1010 cm s-1.

cm-1,1>lc = 2.998 * 108 m s-1;m-1,1>l1>hc.

E>hc = 1>l.E = hc>l,c>l.

nJ s s-1 = J.s-1.n,

E = hn.

cm-1cm-1.

r0

m2m1

I = ¢ m1 m2

m1 + m2≤r0

2

J = 0,Erot

Radiotelescopes detect the emissionspectra of rotationally excited molecules inclouds in interstellar space.

▲

410

PETRMC10_372-422-hr 12/15/05 6:29 PM Page 410

There is no advantage to using bond energies over enthalpy-of-formationdata. Enthalpies of formation are generally known rather precisely, whereasbond energies are only average values. But when enthalpy-of-formation dataare lacking, bond energies can prove particularly useful.

Another way to use bond energies is in predicting whether a reaction willbe endothermic or exothermic. In general, if

(exothermic)(reactants) (products)

and(endothermic)

(reactants) (products)

¢H 7 0weak bonds¡strong bonds

¢H 6 0strong bonds¡weak bonds

cm�1

6

4

3

2

1

0

5

0 12B8B4B

2B6B 10B

FIGURE 10-18Rotational energy levelsof a diatomic molecule

▲

With equation (10.30) we can calculate an energy leveldiagram (Fig. 10-18), in which is also displayed the expect-ed rotational spectrum for a molecule in the gas phase.A molecule can increase its rotational energy if the inci-dent radiation exactly fits the change in rotational energythat accompanies a change of rotational quantum numberof one, that is This gives

¢e = B * 21J i + 12 J i = 0, 1, 2, 3, Á

¢J = 1.

where is the initial rotational energy level. This relation-ship means that transitions from successive energy levelsto the levels above them are associated with energychanges that have steadily increasing values with an in-crement of 2B. Thus, the spacing between lines in therotational spectrum can be measured and the internuclearseparation calculated, yielding an experimental determi-nation of the bond length from equation (10.31).

(10.31)

with the masses of atoms expressed in kilograms.We have just shown how rotational spectra can be

used to determine bond lengths. However, rotationalspectra also play an important role in molecular astro-physics and radio astronomy. The wavelength of theelectromagnetic radiation absorbed in rotational spec-troscopy is in the 3 mm to 30 cm range, which is the mi-crowave region of the electromagnetic spectrum. In radioastronomy the observed emission lines are from the de-excitation of excited molecules in interstellar clouds.Molecules such as HCl, NaCl, PN, and CS have beendetected by combining an analysis of the observed emis-sion lines with a knowledge of pure rotational spectraobserved in the laboratory on Earth.

I = ¢ m1 m2

m1 + m2≤r0

2 =h

8p2 cB

(in J s2 = kg m2)

r0

J i

Sagittarius is probably the densest cloud in the Milky Way andtherefore contains many molecules. Some of the many species thathave been detected in space are OH, SO, SiO, SiS, NO, NS,HCl, PN, NH, CN, CO, CS, and C2 .CH+,

H2 ,

▲

411

PETRMC10_372-422-hr 12/15/05 6:29 PM Page 411


Summary10-1 Lewis Theory: An Overview—A Lewis symbol repre-sents the valence electrons of an atom using dots placedaround the chemical symbol. A Lewis structure is a combi-nation of Lewis symbols used to represent chemical bond-ing. Normally, all the electrons in a Lewis structure arepaired, and each atom in the structure acquires an octet—that is, there are eight electrons in the valence shell. Thepresence of eight electrons in the valence shell is known asthe octet rule. In Lewis theory, chemical bonds are classi-fied as ionic bonds, which are formed by electron transferbetween atoms, or covalent bonds, which are formed byelectrons shared between atoms. Most bonds, however,have partial ionic and partial covalent characteristics.

10-2 Covalent Bonding: An Introduction—In covalentbonds, pairs of electrons shared between two atoms toform the bonds are called bond pairs, while pairs of elec-trons not shared in a chemical bond are called lone pairs.A single pair of electrons shared between two atoms con-stitutes a single covalent bond. When both electrons in acovalent bond between two atoms are provided by onlyone of the atoms, the bond is called a coordinate covalentbond. To construct a Lewis structure for a molecule inwhich all atoms obey the octet rule it is often necessary foratoms to share more than one pair of electrons, thus form-ing multiple covalent bonds. Two shared electron pairsconstitute a double covalent bond and three shared pairsa triple covalent bond.

10-3 Polar Covalent Bonds and Electrostatic PotentialMaps—A covalent bond in which the electron pair isnot shared equally by the bonded atoms is called a polar co-valent bond. Whether a bond is polar or not can be predict-ed by comparing the electronegativity (EN) of the atomsinvolved (Fig. 10-6). The greater the electronegativity

difference between two atoms in a chemical bond,the more polar the bond and the more ionic its character.The electron charge distribution in a molecule can bevisualized by computing an electrostatic potential map(Figs. 10-4 and 10-5). The variation of charge in the moleculeis represented by a color spectrum in which red is the mostnegative and blue the most positive. Electrostatic potentialmaps are a powerful way of representing electron chargedistribution in both polar and nonpolar molecules.10-4 Writing Lewis Structures—To draw the Lewis struc-ture of a covalent molecule, one needs to know theskeletal structure—that is, which is the central atom andwhat atoms are bonded to it. Atoms that are bonded tojust one other atom are called terminal atoms (structure10.13). Typically, the atom with the lowest electronegativi-ty is a central atom. At times, the concept of formal charge(expression 10.16) is useful in selecting a skeletal structureand assessing the plausibility of a Lewis structure.10-5 Resonance—Often, more than one plausible Lewisstructure can be written for a species; this situation iscalled resonance. In these cases the true structure is a res-onance hybrid of two or more contributing structures.10-6 Exceptions to the Octet Rule—There are oftenexceptions to the octet rule. (1) Odd-electron species, suchas NO, have an unpaired electron and are paramagnetic.Many of these species are reactive molecular fragments,such as OH, called free radicals. (2) A few molecules haveincomplete octets in their Lewis structures, that is, notenough electrons to provide an octet for every atom. (3)Expanded valence shells occur in some compounds ofnonmetals of the third period and beyond. In these, thevalence shell of the central atom must be expanded to 10or 12 electrons in order to write a Lewis structure.

(≤EN)

Example 10-16 applies this idea to a reaction involving highly reactive,unstable species for which enthalpies of formation are not normally listed.

E X A M P L E 1 0 - 1 6Using Bond Energies to Predict Exothermic and Endothermic Reactions. One of thesteps in the formation of monochloromethane (Example 10-15) is the reaction of agaseous chlorine atom (a chlorine radical) with a molecule of methane. The productsare an unstable methyl radical and HCl(g). Is this reaction endothermic or exothermic?

SolutionFor every molecule of that reacts, one bond breaks, requiring 414 kJ permole of bonds; and one bond forms, releasing 431 kJ per mole of bonds.Because more energy is released in forming new bonds than is absorbed in breakingold ones, we predict that the reaction is exothermic.

Practice Example A: One naturally occurring reaction involved in thesequence of reactions leading to the destruction of ozone is

Is this reaction endothermic or exothermic?

Practice Example B: Predict whether the following reaction should be exother-

mic or endothermic: 2 HCl(g).+Cl2(g) ¡12

O2(g)+H2O(g)

O2(g).+NO(g)NO2(g) + O(g) ¡

H ¬ ClC ¬ HCH4

CH4(g) Cl(g)� CH3(g) HCl(g)�

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Integrative Example 413

10-7 Shapes of Molecules—A powerful method for pre-dicting the molecular geometry, or molecular shape, of aspecies is the valence-shell electron-pair repulsion theo-ry (VSEPR). The shape of a molecule or polyatomic iondepends on the geometric distribution of valence-shellelectron groups—the electron-group geometry—andwhether these groups contain bonding electrons or lonepairs (Fig. 10-11, Table 10.1). The angles between the elec-tron groups provide a method for predicting bond anglesin a molecule. An important use of information about theshapes of molecules is in establishing whether bonds in amolecule combine to produce a resultant dipole moment(Fig. 10-15). Molecules with a resultant dipole moment are

polar molecules; those with no resultant dipole momentare nonpolar.

10-8 Bond Order and Bond Lengths—Single, double, andtriple covalent bonds are said to have a bond order of 1, 2and 3 respectively. Bond length is the distance between thecenters of two atoms joined by a covalent bond. The greaterthe bond order, the shorter the bond length (Table 10.2).

10-9 Bond Energies—Bond-dissociation energy, D, is thequantity of energy required to break one mole of covalentbonds in a gaseous molecule. Average bond energies(Table 10.3) can be used to estimate enthalpy changes forreactions involving gases.

Integrative ExampleNitryl fluoride is a reactive gas useful in rocket propellants. Its mass percent composition is 21.55% N, 49.23% O, and29.23% F. Its density is at 20 °C and 1.00 atm pressure. Describe the nitryl fluoride molecule as completely aspossible—that is, its formula, Lewis structure, molecular shape, and polarity.

StrategyFirst, determine the empirical formula of nitryl fluoride from the composition data and the molar mass based on that for-mula. Next, determine the true molar mass from the vapor density data. Now the two results can be compared to estab-lish the molecular formula. Then, write a plausible Lewis structure based on the molecular formula, and apply VSEPRtheory to the Lewis structure to predict the molecular shape. Finally, assess the polarity of the molecule from the molecu-lar shape and electronegativity values.

Solution

2.7 g>L

To determine the empirical formula, use the method ofExample 3-5 (page 78). In 100.0 g of the compound,

mol F = 29.23 g F *1 mol F

18.998 g F= 1.539 mol F

mol O = 49.23 g O *1 mol O

15.999 g O= 3.077 mol O

mol N = 21.55 g N *1 mol N

14.007 g N= 1.539 mol N

The empirical formula is N1.539O3.077F1.539 = NO2F

The molar mass based on this formula is 14 + 32 + 19 = 65 g>mol

Use the method of Example 6-10 (page 194) to determinethe molar mass of the gas.

= 65 g>mol

=2.7 g>L * 0.0821 L atm mol-1 K-1 * 293 K

1.00 atm

molar mass =mRTPV

=dRT

P

Because the two molar mass results are the same, the mol-ecular and empirical formulas are the same: NO2F

Because N has the lowest electronegativity, it should bethe lone central atom in the Lewis structure. There are twoequivalent contributing structures to a resonance hybrid.

FO

O

N FO

O

N

�10 0 �1�1 0

0�1

Three electron groups around the N atom produce atrigonal-planar electron-group geometry. All the electrongroups participate in bonding, so the molecular geometryis also trigonal-planar. The predicted bond angles are120°. (The experimentally determined bondangle is 118°.)

F ¬ N ¬ O

N

F

OO

120�

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Exercises

Lewis Theory

1. Write Lewis symbols for the following atoms. (a) Kr;(b) Ge; (c) N; (d) Ga; (e) As; (f) Rb.

2. Write Lewis symbols for the following ions. (a)(b) (c) (d) (e) (f)

3. Write plausible Lewis structures for the followingmolecules that contain only single covalent bonds.(a) ICl; (b) (c) (d) (e)

4. Each of the following molecules contains at least onemultiple (double or triple) covalent bond. Give aplausible Lewis structure for (a) (b)(c) (d) FNO.

5. By means of Lewis structures, represent bonding be-tween the following pairs of elements. (a) Cs and Br;(b) H and Sb; (c) B and Cl; (d) Cs and Cl; (e) Li and O;(f) Cl and I. Your structures should show whether thebonding is essentially ionic or covalent.

6. Which of the following have Lewis structures that donot obey the octet rule:

Explain.7. Give several examples for which the following state-

ment proves to be incorrect. “All atoms in a Lewis struc-ture have an octet of electrons in their valence shells.”

8. Suggest reasons why the following do not exist asstable molecules: (a) (b) HHe; (c) (d)

9. Describe what is wrong with each of the followingLewis structures.

(a) NH H H

Ca

O

O

H3O.He2 ;H3 ;

NO2 ?SO4

2-,NH4

+,SO3,SF6,BF3,NH3,

Cl2CO;(CH3)2CO;CS2 ;

H2Se.NI3 ;OF2 ;Br2 ;

Sc3+.Se2-;Br-;K+;Sn2+;H-;

The molecule has a symmetrical shape, but because theelectronegativity of F is different from that of O, weshould expect the electron charge distribution in the mole-cule to be nonsymmetrical, leading to a small resultant di-pole moment. is a polar molecule.

AssessmentIn the contributing structures to the resonance hybridLewis structure, the F atom has a formal charge of zeroand the two O atoms have an average formal charge of

As a result, we might expect the oxygen-atom re-gion of the molecule to be the most negative in electroncharge density and the fluorine region to be more neutral.This conclusion is confirmed by the electrostatic potentialmap for Notice also the positive charge densityaround the nitrogen nucleus, which is in accord withthe fact that the nitrogen is the least electronegative of thethree elements.

NO2F.

-1>2.

NO2F

10. Describe what is wrong with each of the followingLewis structures.

(a) OO Cl

C[ ]N �(b)

11. Only one of the following Lewis structures is correct.Select that one and indicate the errors in the others.

(a) cyanate ion

N O

O

O C

C[

[

[

]C

N

2�

Cl

]�

]�

(b) carbide ion

(c) hypochlorite ion

(d) nitrogen(II) oxide

12. Indicate what is wrong with each of the followingLewis structures. Replace each one with a more ac-ceptable structure.

(a)

Cl[

OMg

]N

O

�

2�] O[

O N[

S C[

]� Cl[ ]�

]�(b)

(c)

(d)

(b)

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Exercises 415

Ionic Bonding

13. Write Lewis structures for the following ioniccompounds: (a) calcium chloride; (b) barium sulfide;(c) lithium oxide; (d) sodium fluoride.

14. Under appropriate conditions, both hydrogen and ni-trogen can form monatomic anions. What are theLewis symbols for these ions? What are the Lewisstructures of the compounds (a) lithium hydride;(b) calcium hydride; (c) magnesium nitride?

15. Derive the correct formulas for the following ioniccompounds by writing Lewis structures. (a) lithiumsulfide; (b) sodium fluoride; (c) calcium iodide;(d) scandium chloride.

16. Each of the following ionic compounds consists of acombination of monatomic and polyatomic ions. Rep-resent these compounds with Lewis structures.(a) (b) (c) Ca(OCl)2 .NH4Br;Ca(OH)2 ;

Formal Charge

17. Assign formal charges to each of the atoms in thefollowing structures.

(a)

C

[CH3 CH CH3]

O O

O2�

�

[H C C ]�

20. Although the notion that a Lewis structure in whichformal charges are zero or held to a minimum seemsto apply in most instances, describe several significantsituations in which this appears not to be the case.

21. What is the formal charge of the indicated atom ineach of the following structures?(a) the central O atom in (b) B in (c) N in (d) P in (e) I in

22. Assign formal charges to the atoms in the followingspecies, and then select the more likely skeletalstructure.(a) or (b) SCS or CSS(c) NFO or FNO(d) or or

23. The concept of formal charge helped us to choose themore plausible of the Lewis structures for givenin expressions (10.14) and (10.15). Can it similarlyhelp us to choose a single Lewis structure as mostplausible for Explain.

24. Show that the idea of minimizing the formal chargesin a structure is at times in conflict with the observa-tion that compact, symmetrical structures are morecommonly observed than elongated ones with manycentral atoms. Use as an illustrative example.ClO4

-

CO2H+?

NO2

+

OCl2SOSCl2SOCl2

H2ONHH2NOH

ICl4

-PCl5

NO3

-BF4

-O3

Lewis Structures

25. Write acceptable Lewis structures for the followingmolecules: (a) (b) (c) HONO;(d)

26. Two molecules that have the same formulas but dif-ferent structures are said to be isomers. (In isomers,the same atoms are present but linked together in dif-ferent ways.) Draw acceptable Lewis structures fortwo isomers of

27. The following polyatomic anions involve covalentbonds between O atoms and the central nonmetalatom. Propose an acceptable Lewis structure for each.(a) (b) (c) (d) HO2

-.CO3

2-;NO2

-;SO3

2-;

S2F2.

O2SCl2 .HOClO2 ;H2NOH;

28. Represent the following ionic compounds by Lewisstructures. (a) barium hydroxide; (b) sodium nitrite;(c) magnesium iodate; (d) aluminum sulfate.

29. Write a plausible Lewis structure for crotonaldehyde,a substance used in tear gas and

insecticides.30. Write a plausible Lewis structure for a substance

known as carbon suboxide.C3O2,

CH3CHCHCHO,

(b)

(c)

18. Assign formal charges to each of the atoms in the fol-lowing structures.

(a)

S

OO

I I

N

OO

(b)

(c)

19. Both oxidation state and formal charge involveconventions for assigning valence electrons to bondedatoms in compounds, but clearly they are notthe same. Describe several ways in which theseconcepts differ.

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31. Write Lewis structures for the molecules representedby the following molecular models.

32. Write Lewis structures for the molecules representedby the following molecular models.

(a)

(b)

(a)

(b)

33. Write Lewis structures for the molecules representedby the following line-angle formulas.[Hint: Recall page 69 and Figure 3-2.]

(a)

(b)

34. Write Lewis structures for the molecules representedby the following line-angle formulas.[Hint: Recall page 69 and Figure 3-2.]

(a)

(b)

O

NH2Cl

O

Cl O H

O

OH

HO

Polar Covalent Bonds and Electrostatic Potential Maps

35. Use your knowledge of electronegativities, but do notrefer to tables or figures in the text, to arrange the fol-lowing bonds in terms of increasing ionic character:

36. Which of the following molecules would you expectto have a resultant dipole moment (a)(b) (c) (d) HBr, (e) (f)(g) OCS? Explain.

37. What is the percent ionic character of each of the fol-lowing bonds: (a) (b) (c)(d)

38. Plot the data of Figure 10-6 as a function of atomicnumber. Does the property of electronegativity con-form to the periodic law? Do you think it should?

As ¬ O?Al ¬ O;O ¬ Cl;S ¬ H;

SiF4,H2CCl2,BF3,NO2,F2,(m);

K ¬ F.Br ¬ H,Na ¬ Cl,F ¬ H,C ¬ H,

39. Which electrostatic potential map corresponds toand which to H2C “ O?F2C “ O,

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Exercises 417

40. Match the correct electrostatic potential map corre-sponding to HOCl, FOCl, and HOF.

42. Two electrostatic potential maps are shown, one cor-responding to a molecule containing only Cl and F,the other P and F. Match them. What are the molecu-lar formulas of the compounds?

Resonance

43. Through appropriate Lewis structures, show that thephenomenon of resonance is involved in the nitrite ion.

44. Which of the following species requires a resonancehybrid for its Lewis structure, (a) (b)(c) or (d) Explain.

45. Dinitrogen oxide (nitrous oxide, or “laughing gas”) issometimes used as an anesthetic. Here are some dataabout the molecule: bond

bond Use these dataand other information from the chapter to comment onthe plausibility of each of the following Lewis struc-tures shown. Are they all valid? Which ones do youthink contribute most to the resonance hybrid?

length = 119 pm.N ¬ O113 pm;length =N ¬ NN2O

OH-?CO3

2-,OCl-,CO2,

46. The Lewis structure of nitric acid, is a reso-nance hybrid. How important do you think the contri-bution of the following structure is to the resonancehybrid? Explain.

NH O

O

O

HONO2,

ON N(a)

ON N(b)

ON N(c)

NN O(d)

Odd-Electron Species

47. Write plausible Lewis structures for the followingodd-electron species. (a) (b) (c)

48. Write plausible Lewis structures for the following freeradicals. (a) (b) (c)

49. Which of the following species would you expect tobe diamagnetic and which paramagnetic: (a)(b) OH; (c) (d) (e) (f) HO2 ?SO3

2-;SO3 ;NO3 ;OH-;

ClO–.HO2–;–C2H5 ;

NO3.ClO2 ;CH3 ;50. Write a plausible Lewis structure for and indi-

cate whether the molecule is diamagnetic or para-magnetic. Two molecules can join together(dimerize) to form Write a plausible Lewisstructure for and comment on the magneticproperties of the molecule.

N2O4,N2O4.

NO2

NO2,

41. Two electrostatic potential maps are shown, one cor-responding to a molecule containing only S and F, theother Si and F. Match them. What are the molecularformulas of the compounds?

Expanded Valence Shells

51. In which of the following species is it necessary to em-ploy an expanded valence shell to represent the Lewisstructure; Explain your choices.

ClO4

-?SF4,OSCl2,ICl3 ,PI3 ,PO4

3-,

52. Describe the carbon-to-sulfur bond in Thatis, is it most likely a single, double, or triple bond?

H2CSF4.

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Molecular Shapes

53. Use VSEPR theory to predict the geometric shapes ofthe following molecules and ions. (a) (b) HCN; (c)

(d) (e) NSF.54. Use VSEPR theory to predict the geometric shapes of

the following molecules and ions. (a) (b)(c) (d) (e)

55. Each of the following is either linear, angular (bent),planar, tetrahedral, or octahedral. Indicate the correctshape of (a) (b) (c) HCN; (d)(e)

56. Predict the geometric shapes of (a) CO; (b)(c) (d) (e) (f) (g)

57. One of the following ions has a trigonal-planar shape:Which ion is it? Explain.

58. Two of the following have the same shape. Whichtwo, and what is their shape? What are the shapes ofthe other two? HCN,

59. Each of the following molecules contains one or moremultiple covalent bonds. Draw plausible Lewis struc-tures to represent this fact, and predict the shape ofeach molecule. (a) (b) (c) ClNO2.Cl2CO;CO2 ;

NO3

-.SO3

2-,NI3 ,

CO3

2-.PF6

-;PO4

3-;SO3

2-;

AlF6

3-.SO2 ;SbCl5 ;ICl3 ;PH3 ;SiCl4 ;

BF4

-.SbCl6

-;N2O4 ;H2S;

BrF4

+.SO3 ;SOCl2 ;SO4

2-;PCl3 ;

NO3

-;NH4

+;N2 ;

60. Sketch the probable geometric shape of a molecule of(a) (b) (c)

(d)61. Use the VSEPR theory to predict the shapes of the an-

ions (a) (b) (that is, ); (c)(d)

62. Use the VSEPR theory to predict the shape of (a) themolecule (b) the molecule (c) the ion

(d) the ion (e) the ion 63. The molecular shape of is planar (see Table 10.1).

If a fluoride ion is attached to the B atom of through a coordinate covalent bond, the ion results. What is the shape of this ion?

64. Explain why it is not necessary to find the Lewisstructure with the smallest formal charges to make asuccessful prediction of molecular geometry in theVSEPR theory. For example, write Lewis structuresfor having different formal charges, and predictthe molecular geometry based on these structures.

SO2

BF4

-BF3

BF3

ClO3

-.ClO4

-;SF5

-;O2SF2 ;OSF2 ;

I3

-.PF6

-;SSO3

2-S2O3

2-ClO4

-;

(H3COCH3).C2H6O(H3CCH3);C2H6(NCCN);C2N2(O2NNO2);N2O4

Shapes of Molecules with More Than One Central Atom

65. Sketch the propyne molecule, Indicatethe bond angles in this molecule. What is the maxi-mum number of atoms that can be in the same plane?

66. Sketch the propene molecule, Indicatethe bond angles in this molecule. What is the maxi-mum number of atoms that can be in the same plane?

CH3CH “ CH2.

CH3C ‚ CH. 67. Lactic acid has the formula COOH.Sketch the lactic acid molecule, and indicate the vari-ous bond angles.

68. Levulinic acid has the formula Sketch the levulinic acid molecule, and indi-

cate the various bond angles.COOH.

CH3(CO)CH2CH2

CH3CH(OH)

Polar Molecules

69. Predict the shapes of the following molecules, andthen predict which would have resultant dipole mo-ments. (a) (b) (c) (d) (e) (f)

70. Which of the following molecules would you expectto be polar: (a) HCN; (b) (c) (d) OCS;(e) (f) (g) Give reasons for yourconclusions.

POF3 ?SiF4 ;SOCl2 ;CS2 ;SO3 ;

CH2Cl2.SF6 ;C2H4 ;H2S;NH3 ;SO2 ;

71. The molecule has a resultant dipole moment of2.2 D. Can this molecule be linear? If not, describe ashape that might account for this dipole moment.

72. Refer to the Integrative Example. A compound relatedto nitryl fluoride is nitrosyl fluoride, FNO. For thismolecule, indicate (a) a plausible Lewis structure and(b) the geometric shape. (c) Explain why the mea-sured resultant dipole moment for FNO is larger thanthe value for FNO2.

H2O2

Bond Lengths

73. Without referring to tables in the text, indicate whichof the following bonds you would expect to have thegreatest bond length, and give your reasons. (a)(b) (c) (d) BrCl.

74. Estimate the lengths of the following bonds and indi-cate whether your estimate is likely to be too high ortoo low: (a) (b)

75. A relationship between bond lengths and single cova-lent radii of atoms is given on page 406. Use this rela-tionship together with appropriate data from Table10.2 to estimate these single-bond lengths. (a)(b) (c) (d) C ¬ Br.C ¬ F;O ¬ Cl;

I ¬ Cl;

C ¬ F.I ¬ Cl;

Br2 ;N2 ;O2 ;

76. In which of the following molecules would you ex-pect the oxygen-to-oxygen bond to be the shortest: (a)

(b) (c) Explain.77. Refer to the Integrative Example. Use data from the

chapter to estimate the length of the bond in

78. Write a Lewis structure of the hydroxylamine mole-cule, Then, with data from Table 10.2, deter-mine all the bond lengths.

H2NOH.

FNO2.N ¬ F

O3 ?O2,H2O2,

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Integrative and Advanced Exercises 419

Bond Energies

79. A reaction involved in the formation of ozone in theupper atmosphere is Without referringto Table 10.3, indicate whether this reaction is en-dothermic or exothermic. Explain.

80. Use data from Table 10.3, but without performing de-tailed calculations, determine whether each of the fol-lowing reactions is exothermic or endothermic.(a)(b)

81. Use data from Table 10.3 to estimate the enthalpychange for the following reaction.

82. One of the chemical reactions that occurs in the forma-tion of photochemical smog is

Estimate the enthalpy change of this reaction byusing appropriate Lewis structures and data fromTable 10.3.

O2.+NO ¡ NO2+O3

¢H = ?C2H6(g) + Cl2(g) ¡ C2H5Cl(g) + HCl(g)

1¢H2

H2(g) + I2(g) ¡ 2 HI(g)CH4(g) + I(g) ¡ –CH3(g) + HI(g)

O2 ¡ 2 O.83. Estimate the standard enthalpies of formation at 25 °C

and 1 atm of (a) OH(g); (b) Write Lewisstructures and use data from Table 10.3, as necessary.

84. Use for the reaction in Example 10-15 and otherdata from Appendix D to estimate

85. Use bond energies from Table 10.3 to estimate theenthalpy change for the following reaction.

86. Equations (1) and (2) can be combined to yield theequation for the formation of from its elements.CH4(g)

C2H2(g) + H2(g) ¡ C2H4(g) ¢H = ?

1¢H2

¢H°f [CH3Cl(g)].¢H

N2H4(g).

Integrative and Advanced Exercises87. Given the bond-dissociation energies: nitrogen-to-

oxygen bond in NO, in in in

calculate for the reaction.

88. The following statements are not made as carefully asthey might be. Criticize each one.(a) Lewis structures with formal charges are incorrect.(b) Triatomic molecules have a planar shape.(c) Molecules in which there is an electronegativitydifference between the bonded atoms are polar.

89. A compound consists of 47.5% S and 52.5% Cl, bymass. Write a Lewis structure based on the empiricalformula of this compound, and comment on its defi-ciencies. Write a more plausible structure with thesame ratio of S to Cl.

90. A 0.325-g sample of a gaseous hydrocarbon occupiesa volume of 193 mL at 749 mmHg and 26.1 °C. Deter-mine the molecular mass, and write a plausible con-densed structural formula for this hydrocarbon.

91. A 1.24-g sample of a hydrocarbon, when completelyburned in an excess of yields 4.04 g and1.24 g Draw a plausible structural formula forthe hydrocarbon molecule.[Hint: There is more than one possible arrangement ofthe C and H atoms.]

92. Draw Lewis structures for two different moleculeswith the formula Is either of these moleculeslinear? Explain.

93. Sodium azide, is the nitrogen gas-forming sub-stance used in automobile air-bag systems. It is an ioniccompound containing the azide ion, In this ion, thetwo nitrogen-to-nitrogen bond lengths are 116 pm. De-scribe the resonance hybrid Lewis structure of this ion.

N3

-.

NaN3,

C3H4.

H2O.CO2O2(g),

2 NO(g) + 5 H2(g) ¡ 2 NH3(g) + 2 H2O(g)

¢H463 kJ>mol;H2O,O ¬ H389 kJ>mol;NH3,N ¬ H436 kJ>mol;

H2,H ¬ H631 kJ>mol;94. Use the bond-dissociation energies of and

in Table 10.3, together with data from Appendix D, toestimate the bond-dissociation energy of NO(g).

95. Hydrogen azide, is a liquid that explodes vio-lently when subjected to physical shock. In the molecule, one nitrogen-to-nitrogen bond length is 113pm, and the other is 124 pm. The bondangle is 112°. Draw Lewis structures and a sketch ofthe molecule consistent with these facts.

96. A few years ago the synthesis of a salt containing theion was reported. What is the likely shape of this

ion—linear, bent, zigzag, tetrahedral, seesaw, or square-planar? Explain your choice.

97. Carbon suboxide has the formula The carbon-to-carbon bond lengths are 130 pm and carbon-to-oxygen, 120 pm. Propose a plausible Lewis structureto account for these bond lengths, and predict theshape of the molecule.

98. In certain polar solvents, undergoes an ionizationreaction in which a ion leaves one moleculeand attaches itself to another. The products of the ion-ization are and Draw a sketch showing thechanges in geometric shapes that occur in this ionization(that is, give the shapes of and ).

99. Estimate the enthalpy of formation of HCN usingbond energies from Table 10.3, data from elsewhere inthe text, and the reaction scheme outlined as follows.

2 PCl5 ∆ PCl4

+ + PCl6

-

PCl6

-PCl4

+,PCl5 ,

PCl6

-.PCl4

+

PCl5Cl-PCl5

C3O2.

N5

+

H ¬ N ¬ N

HN3

HN3,

O2(g)N2(g)

Overall: C(s) + 2 H2(g) ¡ CH4(g) ¢Hf° = -75 kJ>mol

(2) C(g) + 2 H2(g) ¡ CH4(g) ¢H = ?

(1) C(s) ¡ C(g) ¢H = 717 kJ

Use the preceding data and a bond energy offor to estimate the bond energy.

Compare your result with the value listed in Table 10.3.C ¬ HH2436 kJ>mol

Overall: C(s) +12

N2(g) +12

H2(g) ¡ HCN(g) ¢H°f = ?

(2) C(g) +12

N2(g) +12

H2(g) ¡ HCN(g) ¢H° = ?

(1) C(s) ¡ C(g) ¢H° = ?

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106. R. S. Mulliken proposed that the electronegativityof an atom is given by

where I and EA are the ionization energy and elec-tron affinity of the atom, respectively. Using the elec-tron affinities and ionization energy values for thehalogen atoms up to iodine, estimate the value of kby employing the electronegativity values in Figure10-6. Estimate the electron affinity of At.

107. When molten sulfur reacts with chlorine gas, a vile-smelling orange liquid forms. When analyzed, theliquid compound has the empirical formula SCl.Several possible Lewis structures are shown below.Criticize these structures and choose the best one.

108. Hydrogen azide, can exist in two forms. Oneform has the three nitrogen atoms connected in aline; and the nitrogen atoms form a triangle in theother. Construct Lewis structures for these isomersand describe their shapes. Other interesting deriva-tives are nitrosyl azide and trifluoromethylazide Describe the shapes of these mole-cules based on a line of nitrogen atoms.

109. A pair of isoelectronic species for C and N exist withthe formula in which there is an bond.A corresponding fluoride of boron also exists.Draw Lewis structures for these species and describetheir shapes.

110. Acetone a ketone, will react with astrong base to produce the enolate anion,

Draw the Lewis structure of theenolate anion, and describe the relative contribu-tions of any resonance structures.

111. The species has been synthesized and hasbeen described as a tetrahedral anion. Comment onthis description.

PBr4

-

CH3(C “ O)CH2

-.(A-)

(CH3)C “ O,

X ¬ XX2O4

(CF3N3).(N4O)

HN3,

(a)

(b)

(c)

(d)

(e)

Cl S S Cl

Cl S S Cl

S Cl Cl S

Cl S S Cl

Cl S S Cl

x = k * 1I - EA2

(x)100. The enthalpy of formation of is

Use this value, with other appropriatedata from the text, to estimate the oxygen-to-oxygensingle-bond energy. Compare your result with thevalue listed in Table 10.3.

101. Use the VSEPR theory to predict a probable shape ofthe molecule and explain the source of anyambiguities in your prediction.

102. The enthalpy of formation of methanethiol,is Methanethiol can be syn-

thesized by the reaction of gaseous methanol andWater vapor is another product. Use this infor-

mation and data from elsewhere in the text to estimatethe carbon-to-sulfur bond energy in methanethiol.

103. For LiBr, the dipole moment (measured in the gasphase) and the bond length (measured in the solidstate) are 7.268 D and 217 pm, respectively. For NaCl,the corresponding values are 9.001 D and 236.1 pm.(a) Calculate the percent ionic character for eachbond. (b) Compare these values with the expectedionic character based on differences in electronegativ-ity (see Figure 10-7). (c) Account for any differences inthe values obtained in these two different ways.

104. One possibility for the electron-group geometry forseven electron groups is pentagonal-bipyramidal, asfound in the ion Write the VSEPR notationfor this ion. Sketch the structure of the ion, labelingall the bond angles.

105. The extent to which an acid (HA) dissociates in waterdepends upon the stability of the anion the morestable the anion, the more extensive is the dissociationof the acid. The anion is most stable when the negativecharge is distributed over the whole anion rather thanlocalized at one particular atom. Consider the follow-ing acids: acetic acid, fluoroacetic acid, cyanoaceticacid, and nitroacetic acid. Draw Lewis structures fortheir anions, including contributing resonance struc-tures. Rank the acids in order of increasing extent ofdissociation. Electrostatic potential maps for the fouranions are provided below. Identify which map corre-sponds to which anion, and discuss whether the mapsconfirm conclusions based on Lewis structures.

(A-);

[ZrF7]3-

H2S(g).

-22.9 kJ>mol.CH3SH(g),

F4SCH2,

-136 kJ>mol.H2O2(g)

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Self-Assessment Exercises 421

Feature Problems112. Pauling’s reasoning in establishing his original elec-

tronegativity scale went something like this: If weassume that the bond is nonpolar, its bond en-ergy is the average of the bond energies of and The difference between the calculated andmeasured bond energies of the bond is attrib-utable to the partial ionic character of the bond andis called the ionic resonance energy (IRE). If the IRE isexpressed in kilojoules per mole, the relationship be-tween IRE and the electronegativity difference is

To test this basis for an electroneg-ativity scale,(a) Use data from Table 10.3 to determine IRE for the

bond.(b) Determine for the bond.(c) Establish the approximate percent ionic charac-ter in the bond by using the result of part (b)and Figure 10-7. Compare this result with thatobtained in Example 10-4.

113. On page 405, the bond angle in the molecule isgiven as 104° and the resultant dipole moment as

(a) By an appropriate geometric calculation, deter-mine the value of the bond dipole in (b) Use the same method as in part (a) to estimatethe bond angle in given that the bond di-pole is 0.67 D and that the resultant dipole momentis m = 0.93 D.

H ¬ SH2S,

H2O.H ¬ O

m = 1.84 D.

H2O

H ¬ Cl

H ¬ Cl¢ENH ¬ Cl

IRE>96.=(¢EN)2

A ¬ BB ¬ B.

A ¬ AA ¬ B

(c) Refer to Figure 10-15. Given the bond dipoles1.87 D for the bond and 0.30 D for the bond, together with estimate the

bond angle in 114. Alternative strategies to the one used in this chapter

have been proposed for applying the VSEPR theoryto molecules or ions with a single central atom. Ingeneral, these strategies do not require writingLewis structures. In one strategy, we write

(1) the total number of electron pairs [(number ofvalence electrons) (electrons required for ioniccharge)] 2(2) the number of bonding electron pairs (numberof atoms) 1(3) the number of electron pairs around central atom

total number of electron pairs 3 [number ofterminal atoms (excluding H)](4) the number of lone-pair ofcentral atom pairs number of bonding pairs

After evaluating items 2, 3, and 4, establish theVSEPR notation and determine the molecular shape.Use this method to predict the geometrical shapes ofthe following: (a) (b) (c) (d)(e) (f) Justify each of the steps in thestrategy, and explain why it yields the same results asthe VSEPR method based on Lewis structures. Howdoes the strategy deal with multiple bonds?

PCl4

+.ClF4

-;SO2 ;ClF3 ;NH3 ;PCl5 ;

-electrons = number

*-=

-=

>;

=

CHCl3.H ¬ C ¬ Clm = 1.04 D,

C ¬ HC ¬ Cl

Self-Assessment Exercises115. In your own words, define the following terms:

(a) valence electrons; (b) electronegativity; (c) bond-dissociation energy; (d) double covalent bond;(e) coordinate covalent bond.

116. Briefly describe each of the following ideas: (a) for-mal charge; (b) resonance; (c) expanded valenceshell; (d) bond energy.

117. Explain the important distinctions between (a) ionicand covalent bonds; (b) lone-pair and bond-pairelectrons; (c) molecular geometry and electron-group geometry; (d) bond dipole and resultant di-pole moment; (e) polar molecule and nonpolarmolecule.

118. Of the following species, the one with a triple cova-lent bond is (a) (b) (c) (d)

119. The formal charges on the O atoms in the ionis (a) (b) (c) 0; (d)

120. Which molecule is nonlinear? (a) (b)(c) HCN; (d) NO.

121. Which molecule is nonpolar? (a) (b)(c) (d) FNO.

122. The highest bond-dissociation energy is found in(a) (b) (c) (d) I2 .Cl2 ;N2 ;O2 ;

NH3 ;CH2Cl2 ;SO3 ;

CO2 ;SO2 ;+1.-1;-2;[ONO]+

AlCl3 .CO2 ;CN-;NO3-;

123. The greatest bond length is found in (a) (b)(c) (d) BrCl.

124. Draw plausible Lewis structures for the followingspecies; use expanded valence shells where neces-sary. (a) (b) (c) (d)

125. Predict the shapes of the following sulfur-containingspecies. (a) (b) (c)

126. Without referring to tables or figures in the text otherthan the periodic table, indicate which of the follow-ing atoms, Bi, S, Ba, As, or Mg, has the intermediatevalue when they are arranged in order of increasingelectronegativity.

127. Use data from Tables 10.2 and 10.3 to determine foreach bond in this following structure (a) the bondlength and (b) the bond energy.

128. Construct a concept map illustrating the connectionsbetween Lewis dot structures, the shapes of mole-cules, and polarity.

C C Cl

O H

H

H

SO4

2-.SO3

2-;SO2 ;

BrF5.CO3

2-;PF3 ;Cl2O;

Br2 ;N2 ;O2 ;

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eMedia Exercises

129. In the Periodic Trends: Lewis Structures activity(Activebook 10-1), the similarity between the Lewisstructures of elements found within the same groupis illustrated. Use this concept to describe the simi-larity of the Lewis structures of ionic compounds in-volving atoms of a similar group (that is, NaCl,NaBr, and NaI).

130. After viewing the Bond Formation animation(Activebook 10-2), describe the forces involved incovalent bond formation in the simplest covalentcompound. Which of the forces are attractive andwhich are repulsive?

131. Using the principles described in the Writing LewisStructures activity and the Formal Charges anima-tion (Activebook 10-4), generate the possible Lewisdot structures of the molecules and Whatare the similarities and differences between the pre-dicted structures of the two molecules?

SO2.O3

H2

132. In the VSEPR animation (Activebook 10-7), a sequen-tial pattern in electron-group geometry is observedwith the addition of each pair of electrons. Althoughthe sequential addition of electron pairs (bonding ornonbonding) is not a realistic mechanism by whichmolecules are formed, it does illustrate the key prin-ciple behind the valence-shell electron-pair repulsiontheory. (a) Use this principle to predict the structure ifthe central atom of a molecule were bonded to eightidentical neighboring atoms. (b) Can you predict anapproximate bond angle?

133. Use the Molecular Polarity activity (Activebook 10-7)to predict which of the following molecules possess(i) polar covalent bonds, and (ii) a net molecular di-pole moment. (a) (b) (c) (d)(e) H2S.

NH3 ;CH4 ;CF2Cl2 ;CF4 ;

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chemical bonding i: basic concepts -...

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