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SYSTEMS WITH TWOOR MORE ATOMS

Chapter 12 MoleculesChapter 13 Solids — TheoryChapter 14 Solids — ApplicationsChapter 15 Statistical Mechanics

In Part III (Chapters 12 to 15) we treat applications of quantum mechanics tosystems that are larger than a single atom. In Chapter 12 we discuss how atomscombine to form molecules. Traditionally, this topic is one of the principalconcerns of chemistry, but the boundary between physics and chemistry is now

quite blurred. Whether we regard the study of molecules as part of physics orchemistry, it is certainly an important application of quantum mechanics.

In Chapters 13 and 14 we describe how atoms or molecules combine toform solids. The solid is our first example of a macroscopic system for whoseunderstanding we require quantum mechanics. Solid-state physics has spawneda dazzling array of electronic “solid-state” devices — transistors, LEDs,photo-voltaic cells, integrated circuits — that have made possible the electronic revo-lution of the last few decades. Chapter 13 is mainly about the theory of solids,and Chapter 14 about some of these applications.

Finally, Chapter 15 is an introduction to statistical mechanics,the statisti-

cal treatment of systems such as gases and solids, usually containing very largenumbers of atoms or molecules.

As far as possible, the chapters of Parts III and IV are designed to beindependent of one another. For example, you could jump directly to Chap-ter 15 without reading Chapters 12 through 14. If you want to read the chap-ters on solids (13 and 14), you would need to read just the first four sections of Chapter 12. If you are interested only in the foundations of solid-state physics,and not in the technological applications, you could read Chapter 13 only andskip 14. However, if you wish to understand the applications, you should firstread at least the first 5 sections of Chapter 13 before proceeding to Chap-

ter 14. If your main interest is in nuclear or particle physics, you could skipPart III entirely and proceed straight to Part IV.

F

FP

PO

O

367

PARTI I I

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368

C h a p t e r 12Molecules

12.1 Introduction12.2 Overview of Molecular Properties12.3 The Ionic Bond12.4 The Covalent Bond12.5 Directional Properties of Covalent Bonds★

12.6 Excited States of Molecules★

12.7 Molecular Spectra★

Problems for Chapter 12

★These sections can be omitted without significant loss of continuity

12.1 Introduction

A molecule is a stable, or nearly stable, bound state of two or more atoms. Anoxygen molecule, is a bound state of two oxygen atoms; a water molecule,

is a bound state of two hydrogens and one oxygen. In this chapter wediscuss the nature of the forces that bind atoms to one another and some of the properties of the resulting molecules.

The number of atoms in a molecule ranges from two, in a diatomic

molecule such as or CO, to several billions in large biological molecules. InFig. 12.1 we have illustrated the arrangement of the atoms in five different,relatively simple molecules, and Fig. 12.2 shows one of the largest known mol-ecules, DNA, which carries genetic information in organisms. Even largernumbers of atoms can form stable bound states. For example, if we took anylarge number of sodium atoms and the same number of chlorine atoms, then atroom temperature these could bind together to form a stable salt crystal; butwe would usually describe such a crystal as a solid, not a molecule.* If, however,

O2

H2O,O2 ,

OH H

Water, H2O

H H

Hydrogen, H2

C OO

Carbon dioxide, CO2 H C H

HH C

HH C

H

H

Propane, C3H8

C

H

H

C

C

C

C

C

H

H

H

H

Benzene, C6H6

FIGURE 12.1

The atoms within molecules are

arranged in many different ways:straight lines, triangles, chains, rings,and many others. The linesconnecting atoms representmolecular bonds, and double linesindicate double bonds, as describedin Section 12.4.

*There is no clear unambiguous distinction between molecules and solids. In fact, onecould argue that a salt crystal is just an enormous molecule. Nonetheless, this is notnormal usage. In particular, it is generally considered that any given molecular speciesshould have a definite number of constituent atoms,whereas one can add any numberof Na–Cl pairs to a salt crystal and still have a salt crystal which is just a little larger. Bythis criterion, then, a salt crystal is not a molecule.

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Section 12.1 • Introduction 369

(a)

(c)

C N

C C

CN O

CH

HH

HH

C

H

HH

CC

CPO O

O

O

H

H

C

O

O

H

H

O H

(b)

FIGURE 12.2

(a) A segment, about 10 nm long, of the double helix of the DNA molecule. (b) Each circlein (a) represents a base, comprising about 30 atoms, and there are a ltogether just four

different kinds of bases. This picture shows the atomic composition of the base calledthymine. (c) Electron micrograph of a segment of DNA about long (coated inplatinum to make it show up). Although this segment is about a thousand times longerthan the segment shown in (a), it is still only a tiny fraction of an entire molecule. Theaverage DNA molecule in a human chromosome is about 4 cm long and contains some 10billion atoms.

10 mm

we heated the salt until it vaporized, we would find that the atoms would moveapart in bound pairs, one Na atom bound to each Cl,and these pairs are exact-

ly what we do describe as molecules of NaCl. In this chapter we discuss onlymolecules, such as the NaCl molecule, containing a definite and reasonablysmall number of atoms. Further, we will consider only individual, isolated mol-ecules. Since the molecules in a solid or liquid can never really be consideredto be isolated, this means that this chapter is actually about molecules in a gas.The properties of solids and (very briefly) liquids are discussed in Chapters 13and 14.

A feature common to all molecules is that the force which holds theiratoms together is ultimately the electrostatic attraction between oppositecharges. At first, one might think there could be no electrostatic force be-

tween two neutral atoms. However, when two or more atoms are sufficientlyclose, their charges can redistribute themselves with pairs of opposite chargesclose together, so that attraction predominates over repulsion, bonding theatoms together to form a molecule.

Although all interatomic bonds are basically electrostatic, we can distin-guish five different ways in which the necessary redistribution of chargeoccurs. The resulting kinds of bond are called:

the ionic bondthe covalent bondthe hydrogen bond

the van der Waals bondthe metallic bond

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Ionic and covalent bonds are important in simple molecules and are the main

topics of this chapter.The hydrogen bond involves the sharing of a proton be-tween two negative ions. Since its main role is in large organic molecules,which are beyond the scope of this book, we will not discuss the hydrogenbond here.The van der Waals bond is very weak and is usually important onlywhen other types of bond do not act; it is the bond that holds some liquids andsolids together and is discussed briefly in Chapter 13. As its name implies, themetallic bond is what binds metals together and is described in Chapter 13.

In this chapter we focus mostly on diatomic molecules since they are rel-atively simple but still exemplify many of the general principles of molecularbinding. We begin with a brief overview of molecular properties in Section

12.2. In Sections 12.3 and 12.4 we give more detailed descriptions of the ionicand covalent bonds. In Section 12.5 we discuss how some molecular bondshave pronounced directional properties. This gives many molecules distinctshapes, which can be predicted using the properties of atomic wave functionsdescribed in Chapters 8 and 10. In Section 12.6 we discuss the excited states of molecules.We will see that the distribution of the energy levels in a molecule isquite different from that in an atom. For this reason, molecular spectra arequite different from typical atomic spectra, as we describe in Section 12.7.

12.2 Overview of Molecular Properties

The attraction that binds atoms together to form molecules is the electrostaticattraction between the positive nucleus of each atom and the negative elec-trons in all the other atoms. Because ordinary matter is electrically neutral,we are usually unaware of the enormous magnitude of the internal electricforces. It is only when the positive and negative charges are rearranged intoseparate regions that electric forces become evident. Such rearrangement iswhat happens within molecules on the microscopic scale. To illustrate themagnitude of the resulting forces,we first consider a thought experiment on a

macroscopic scale.

Example 12.1

Imagine the charges in 1 gram of hydrogen separated so that all of the elec-trons are at one place and all of the protons at another, separated from theelectrons by a distance equal to the earth’s diameter. What would be theforce of attraction between the electrons and the protons?

Each atom of hydrogen contains one electron and one proton, and 1gram of hydrogen contains Avogadro’s number, of atoms.Thus, the mag-

nitude of the charge of either sign is

The enormity of these charges becomes apparent when we calculate theattraction between them when separated by an earth diameter:

This is a force of roughly 60 tons!

F = k

Q2

d2= ¢8.99 * 109

N # m2

C2≤ *

19.63 * 104 C22

11.27 * 107

m

22

= 5.2 * 105 N

Q = NAe = 16.02 * 10232 * 11.60 * 10-19 C2 = 9.63 * 104

C

NA ,

Gerhard Herzberg(1904–1999,German-Canadian)

One of the giants of molecular

physics,Herzberg — a chemist — is perhaps best known for his de-finitive four-volume work on mol-ecular spectra and structure.Bornin Hamburg and educated at Göt-tingen and Berlin, he fled NaziGermany in 1935 and moved toCanada. His painstaking analyses of molecular spectra made possiblenumerous applications,such as theidentification of many molecules ininterstellar space. He won the

Nobel Prize in chemistry in 1971.

370 Chapter 12 • Molecules

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Section 12.2 • Overview of Molecular Properties 371

TABLE 12.1Sampling of eight diatomic molecules, four of which are bound ionically and four cova-lently. (Many molecules are bound by a mixture of both kinds of bonds.) The secondcolumn gives the bond length and the third gives the binding energy (or dissocia-tion energy) B.

Molecule (nm) B (eV) Bond

KCl 0.27 4.1LiF 0.16 5.9

IonicNaBr 0.25 3.7

NaCl 0.24 4.20.074 4.5

HCl 0.13 4.4Covalent

0.11 9.80.12 5.1O2

N2

H2

R0

R0 ,

It is the strong electrostatic interaction between the many charges withinordinary matter that is responsible for the strength of solids, the contact forcewhen one solid presses against another, and the energy released in chemicalreactions such as combustion. Most important, for our present purposes, it isthe electrostatic interaction that binds the atoms together within a molecule.

Before we try to calculate the binding strengths of molecules, let usconsider some of the observed parameters for a few simple molecules. InTable 12.1 we give a sampling of eight diatomic molecules. The first columnshows the chemical symbol and the second the bond length defined as thedistance between the two nuclei. The third column gives the dissociationenergy, or binding energy, B, the energy needed to separate the molecule intotwo neutral atoms, and the last column gives the type of bond.

We see from Table 12.1 that typical bond lengths in diatomic moleculesare one or two tenths of a nanometer, with ionic molecules tending to be a lit-tle larger than covalent. The binding energies are all of order a few eV. To un-derstand the values of these parameters, we must examine the distribution of charge inside the molecule. This distribution depends on the behavior of theatomic electrons as the atoms approach one another, and this behavior is de-termined mainly by the atoms’ positions in the periodic table. We now sketchthree types of behavior: first, the almost complete failure of the noble gases toform bonds; second, the formation of ionic bonds between atoms at oppositesides of the periodic table; and finally, the formation of covalent bonds, oftenbetween identical or similar atoms.

The Noble Gases

Until recently, it was believed that the elements in group VIIIof the periodic table formed no stable molecules, and because of this reluc-tance to mingle, they were named “inert” or “noble.” Although we now knowthat the noble atoms do form some molecules, all such molecules are veryweakly bound, and the names are still perfectly appropriate.

To see why the noble atoms interact so weakly with other atoms, recallfrom Section 10.7 that they are closed-shell atoms and are therefore spherical-

ly symmetric. Gauss’s law for the electric field tells us that a spherically sym-metric charge distribution of zero total charge produces no external electric

He, Ne, Ar, Á

R0 ,

∂∂

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Li F

(a)

FLi

(b)

FLi

(c)

e

FIGURE 12.3(a) When widely separated, thelithium and fluorine atoms areneutral. (b) When they approachone another, an electron cantransfer from the lithium to thefluorine. (c) The resulting chargedions are strongly attracted andform a stable LiF molecule.


*This second claim follows from Newton’s third law: The spherical distribution exertsno force on any external charge; therefore, the external charge exerts no force on thedistribution.

field and experiences no net force when placed in an applied field.* Equally

important, the electron distribution in a noble atom cannot easily be distortedbecause the excited states are high above the ground state (about 20 eV inHe). Thus, a noble atom, even when close to another atom, remains nearlyspherical, and there is very little force between the two atoms.

This argument shows correctly why there is little force between a nobleatom and any other atom,provided that they do not overlap. One could imag-ine that if the atoms were so close that they overlapped, there might be anattractive force;but even this is not the case: Once the atoms overlap, we needto take account of the quantized energy levels of the electrons, and we will seelater that because the noble atom has all closed shells, there can be very little

attraction, even when the atoms overlap.

Ionic Bonding

The simplest example of a pair of atoms that do form a strong bond is an alkali–halide pair, consisting of one alkali and one halogen atom. As we discussed inSection 10.7, the alkali metals have one loosely bound valenceelectron outside a closed shell. The halogens have one vacancyin their outer shell and hence have large electron affinity. If, for example, thealkali Li approaches the halogen F, the F atom can capture the valence elec-

tron of the Li atom,as indicated schematically in Fig. 12.3.After the transfer of an electron from Li to F, the resulting ions are stable and spherical, since eachnow has a closed outer shell. The strong electrostatic attraction between thesetwo oppositely charged stable spheres forms an ionic bond, which binds themtogether in a molecule of LiF.

Since both ions are spherically symmetric,they behave like point chargesof opposite sign and form an electric dipole. The dipole moment p of such apair of equal but opposite charges is defined to be

(12.1)

where q is the magnitude of either charge and d is their separation. We haveseen that a typical separation distance is so an ionically bondedmolecule should have a dipole moment of order

(12.2)

Measurements show that the dipole moments of alkali halide molecules are of this order,confirming that their bonding does indeed involve the transfer of anelectron, of the kind illustrated in Fig. 12.3 (see Problem 12.3).

Covalent Bonding

When two identical atoms approach one another, there is no difference inelectron affinities to cause the transfer of an electron as required for an ionicbond. Nonetheless, stable molecules such as and are known toexist. Furthermore, many other diatomic molecules, such as CO, have dipolemoments that are much smaller than that predicted in (12.2) and thus cannotbe ionically bonded. The bond that forms in all these cases involves a sharing,

O2H2 , N2 ,

p = ed ' 11.6 * 10-19 C2 * 12 * 10-10

m2 = 3.2 * 10-29 C # m

d L 0.2 nm,

p = qd

(F, Cl, Br, Á )(Li, Na, K, Á )

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(a)

(b)

FIGURE 12.4

Schematic plot of the distribution of the outer electrons in two atomsthat bond covalently. (a) The twoseparate atoms. (b) When theatoms form a covalent molecule,the wave functions for the outerelectrons interfere constructivelyand produce a concentration of charge in the region between thetwo nuclei. The two dots show the

positions of the two nuclei, and forclarity the distribution of innerelectrons is omitted entirely.

Section 12.2 • Overview of Molecular Properties 373

or co-ownership, of the valence electrons from both atoms, and is called a

covalent bond.We will discuss the covalent bond in detail in Section 12.4, but the main

result of that discussion is easily summarized with the help of Fig. 12.4, whichis a sketch of the probability density for the valence electrons around twoatoms joined by a covalent bond. Part (a) shows the electron distributionsaround the two well separated atoms. In (b) the atoms are close together and alarge proportion of the electron distribution is concentrated in a region be-tween the two nuclei. The electrostatic attraction of the positive nuclei towardthis concentration of negative charge is what bonds the atoms together.

Mixed BondsIn a purely ionic bond an electron is completely transferred from one atom toanother, and the dipole moment is given by (12.2). In a purely covalent bondthe shared electrons are equally divided between the atoms, and the dipolemoment is zero. While a few molecules (such as and ) are purely cova-lent, there are no molecules that are 100% ionic, and in most molecules thebonding is a combination of ionic and covalent. That is, there is a partial trans-fer of charge from one atom to the other and a partial sharing.This is signaledby a dipole moment that is not as large as expected for an ionic bond, but isstill greater than zero.

Estimating Bond Strengths

In the next two sections we discuss in some detail the strengths of ionic and co-valent bonds. Here we show that it is easy to get a rough estimate of theirstrengths. In particular, we can see why both kinds of bond lead to bindingenergies that are of order a few eV, as we saw in Table 12.1.

To estimate the binding energy of an ionic molecule, let us suppose thatone electron is completely transferred from one atom to the other. The poten-tial energy of the two resulting ions, at separation is and if

we substitute the observed value we find that

(12.3)

We will see in the next section that this potential energy is the dominant con-tribution to the molecule’s total energy. If we simply ignore all other contri-butions, we get the rough estimate That this is negativeconfirms that the molecule is a stable bound state, and the binding energy of the ionic molecule is (in this approximation)

(12.4)

Comparing this value with the measured values in Table 12.1, we see that ourestimate is a bit large.As we will see in Section 12.3, the terms that we have ne-glected in (12.4) reduce our value of B and give a result that agrees well withexperiment. Nevertheless, the rough estimate (12.4) is definitely of the rightorder of magnitude.

One can get a similar estimate of the binding energy of a covalent mole-

cule by making reasonable assumptions about the distribution of charge in thistype of molecule (see Problem 12.11). Here, however, we will point out only

B = -E Lke2

R0

L 7 eV

E L U L -7 eV.

U = -

ke2

R0

L -

1.44 eV # nm

0.2 nmL -7 eV

R0L

0.2 nm,

U = -ke2

>R0 ,R0 ,

O2H2

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that, as is clear from Table 12.1, the binding energy of covalent molecules is of

the same order of magnitude as that of ionic molecules, and this similarity canbe explained. The potential energy of the covalent molecule is the energy thatresults from concentrating one or two electrons between two positive charges,separated by a distance (as shown in Fig. 12.4).This potential energy is nat-

urally of order (Problem 12.11) which is the same expression(12.3) that we obtained for ionic molecules,* and we conclude — just what isobserved — that covalent and ionic molecules would have binding energies of comparable magnitudes.

Energy Released in Chemical ReactionsThe binding energy of molecules manifests itself on a macroscopic scale inchemical reactions. A chemical reaction is simply a regrouping of the atoms insome initial set of molecules to form different molecules; for example,

Since the initial and final molecules generally have different binding energies,this regrouping results in the release or absorption of energy (usually in theform of heat or light). Binding energies are of order a few eV, so the energy re-

leased in an individual regrouping can itself be of this same order.On a macro-scopic scale a few eV is a very small energy, but macroscopic amounts of matter contain enormous numbers of molecules, and the total energy releasedwhen a substantial amount of matter reacts can be very large, as the followingexample illustrates.

Example 12.2

Gasoline engines use the heat produced in the combustion of the carbon and

hydrogen in gasoline. One of the important sources of energy in this combus-tion is the oxidation of carbon to form carbon dioxide:

(12.5)

where the 11.4 eV released comes from the increased binding energy of themolecule as compared to that of the separate C and Find the total

energy released when 1 kg of carbon is oxidized. If this energy were usedwith 10% efficiency (a reasonable practical value) to drive a 1500-kg car up ahill, what elevation gain would it produce?

Because a mole of carbon is 12 grams, the number of carbon atoms in1 kg is

N =1000 grams

12 grams>mole* a6.0 * 1023

atoms

moleb = 5.0 * 1025 atoms

O2 .CO2

C + O2: CO2 + 11.4 eV

HCl + NaOH:H2O + NaCl

U L -ke2>R0 ,

R0


*We should emphasize that this is only an order-of-magnitude argument. It may helpto draw a parallel with the hydrogen atom, whose energy could be estimated asthe potential energy of an electron and proton a distance apart. This gives

which is twice the correct answer but is certainly the right order of magnitude.E L -ke2>aB,

aB

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The oxidation of each carbon atom yields 11.4 eV, so the total energy

released is

(12.6)

This energy results from burning 1 kg of carbon in the reaction (12.5). Sincethis reaction is the main source of energy in the combustion of gasoline andcarbon is the main constituent of gasoline, we can take (12.6) as a rough esti-mate of the energy released in burning 1 kg of gasoline. (For a more carefulestimate, see Problem 12.9.)

If 10% of the energy (12.6) were used to raise a 1500-kg car through avertical height h, then, since the gain in potential energy would be mgh, wecan calculate h as

This is the elevation gain produced by 1 kg of gasoline. Since a car’s gas tankholds at least 40 kg, we see that a tankful of gasoline can lift a car through avertical gain of some 25,000 m (or 16 miles) — some three times the totalheight of Mount Everest! This startling conclusion emphasizes that gasoline

is an extremely concentrated source of energy.*

12.3 The Ionic Bond

We have seen that an ionic bond is formed when an electron is transferredfrom one atom to another, producing a pair of oppositely charged ions thatattract one another strongly. This occurs when the first atom has a low ioniza-tion energy and can lose an electron easily, while the second has high electronaffinity and can bind an extra electron relatively well. We start this section by

considering the specific case of NaCl and examine in detail the energy balancein the formation of an NaCl molecule.

To begin, let us imagine that our Na and Cl atoms are far apart and elec-trically neutral (in which case they exert no force on one another). If we wereto transfer an electron from the Na to the Cl while the atoms were still farapart, this transfer would actually cost us energy.The energy needed to removean electron from Na is its ionization energy, the energygained when Cl captures an electron is its electron affinity,Thus the net cost of energy is

(12.7)

Clearly, the spontaneous transfer of an electron between the well-separatedatoms is something that will not occur. However, this situation changes as theatoms approach one another, as we now show.

To understand what happens as the two atoms come closer together, con-sider the two graphs in Fig. 12.5. These show the energy, as a function of the

¢E = IE(Na) - EA(Cl) = 5.1 eV - 3.6 eV = 1.5 eV

EA(Cl) = 3.6 eV.IE(Na) = 5.1 eV;

h =gain in PE

mg=

9.1 * 106 J11500 kg2 * 19.8 m>s22 L 620 m

= 5.7 * 1026 eV = 9.1 * 107

J

E = N * 111.4 eV2 = 15.0 * 10252 * 111.4 eV2Section 12.3 • The Ionic Bond 375

*We have assumed here that the car drives slowly up a steep incline, so that the workagainst rolling and air resistance is small compared to the work, mgh, against gravity.In the more usual context of a level road, the mgh term is zero and only rolling and

air resistance remain; under these conditions a tankful of gasoline takes one severalhundred miles (see Problem 12.5).

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1 2 E ( R )

R (nm)

Na Cl

Na Cl

E 1.5 eVRc

FIGURE 12.5Energy of sodium and chlorine as afunction of their internuclearseparation R. The dashedhorizontal line is the energy of thetwo neutral atoms; the solid curveis that of the two ions and

When R is less than thetwo ions have lower energy thanthe two neutral atoms.

Rc ,Cl-.Na+

distance R between the two nuclei, both for the neutral pair(dashed), and for the ions (solid curve).We have chosen the zero of energy as the energy of the two neutral atoms when far apart (and at rest).Since the neutral atoms exert no force on one another, their energy does notchange as we move R inward, and the corresponding graph (shown as thedashed line) is a constant:

Suppose instead that while the atoms are far apart we transfer anelectron from one to the other, and then slowly move them together. Whilethey are still far apart, the energy of the ions is higher than that of the neutralatoms by the amount calculated in (12.7). However, the ions areoppositely charged, and as they move closer, their potential energy,comes into play, so their total energy is

(12.8)

This energy is shown as the solid curve in Fig. 12.5 and decreases steadily asthe ions get closer.

At the critical separation shown as in Fig. 12.5, the energy of the twoions and has dropped to that of the neutral pair, This sepa-ration is easily calculated by setting in (12.8) equal to zero, to give

(12.9)

For the case of NaCl, with this gives

Once R is less than the ions have less energy than the neutral atoms; thatis, the potential energy of the ions more than offsets the cost of transferring the electron, and the transfer is energetically favored. To predictthe probability that the transfer actually occurs would require a detailed quan-tum calculation, but for our purposes, it is sufficient to note that the commonoccurrence of ionic molecules shows that the transfer does have significantprobability.

Once the electron is transferred, the two ions are strongly attracted and

can bind together to form a molecule. Their energy does not continue to de-crease indefinitely as R approaches zero. Rather, once the two ions overlap

¢E-ke2>RRc ,

Rc =ke2

¢E=

1.44 eV # nm

1.5 eV= 0.96 nm

¢E = 1.5 eV,

Rc =ke2

¢E

E1R2 Na + Cl.Cl-Na+ Rc

E1R2 = ¢E -ke2

R [Na+

+ Cl-]

-ke2>R,¢E = 1.5 eV

1R = q2E1R2 = constant = 0 [neutral atoms]

Na++ Cl-

Na + Cl

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Section 12.3 • The Ionic Bond 377

0.5 1.0

Rc

R0

E(R)

R (nm)

E n e r g y ( e V )

5

0

(E ke2

R(

B

FIGURE 12.6The general behavior of the energy

of the two ions andis shown by the solid curve. As longas the two ions do not overlap

the energy is just

Once theions overlap, repulsive forcesquickly dominate and climbssteeply up again. The minimum of

occurs at andwhere B is the

molecule’s binding energy.

E

1R0

2= -B,

R0 ,E

1R

2

E1R2E1R2 = ¢E - ke2>R.

1R g 0.3 nm2Cl-Na+E1R2

*We are still ignoring any kinetic energy of the two atoms. As we discuss in Section

12.6, there is a small, zero-point kinetic energy, and the ground state is actually slightlyabove E1R02.

appreciably, the force between them becomes repulsive, and the energyrises as shown in Fig. 12.6. One obvious source of this repulsion is the increas-ing force between the two positive nuclei as they get closer. Another impor-tant contribution is due to the Pauli exclusion principle: Electrons in the sameregion cannot occupy the same quantum state, and once the atoms overlap,some of the electrons must move to higher levels, increasing further.

The energy of the two ions has a minimum at the separation shownas in Fig. 12.6, and the two ions are in stable equilibrium at this separation.Therefore, is the center-to-center distance of the two atoms in the loweststate of the stable molecule, and the corresponding value of the energy,

is the ground-state energy of the molecule. * That is, where B isthe binding energy of the molecule.The observed values of these two parame-ters for NaCl are

Without a detailed calculation of the repulsive contributions to wecannot actually predict these values of and B. However, one can see inFig. 12.6 that when R is close to the repulsive contribution to is stillquite small; that is, near the exact shown by the solid curve, is

close to the value shown by the dashed curve. This suggeststhat we would get good approximations for and B by ignoring thedifference between these two curves; that is,we can approximate

(12.10)

This estimate for B is easily understood: The term is the electrostaticbinding energy of the two ions and is reduced by the energy that wasneeded to transfer the electron and form the ions.

In Section 12.2 we made the rough estimate

(12.11)

Comparing this with the new estimate (12.10), we see that this earlier approx-imation simply ignored the transfer energy and was therefore an overesti-mate, as we noted in Section 12.2. In fact, the improved approximation (12.10)

¢E,

B Lke2

R0

¢E

ke2>R0

B = -E1R02 Lke2

R0

- ¢E

E1R021¢E

-ke

2

>R2E

1R

2,R = R0

E1R2R0 ,R0

E1R2,

R0 = 0.24 nm and B = 4.2 eV

E1R02= -

B,

E

1R0

2,

R0

R0

E1R2 E1R2E1R2

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is still a slight overestimate since it ignores the repulsive contribution to

and this contribution necessarily reduces B.

Example 12.3

Given the known values of and for NaCl, estimate the dissociationenergy B of the NaCl molecule, using both of the approximations (12.11) and(12.10).

Since the rough estimate (12.11) gives

while the improved estimate (12.10) gives

As expected, both approximations overestimate the observed value of 4.2 eV.The first is within 40%,and the second a respectable 7%.

Valence

The NaCl molecule is formed by the transfer of one electron from an Naatom to a Cl atom, to give the closed-shell ions and Similar ionicmolecules, involving transfer of one electron, can be formed by combiningany alkali atom (closed-shell-plus-one, or group I) with any halogen (closed-shell-minus-one, or group VII).

Ionic molecules that involve the transfer of more than one electron arealso possible. For example, an atom like Mg from group II, has two electronsoutside closed shells and can lose these two electrons to become the closed-shell ion meanwhile an atom like O from group VI is two electrons shortof a closed shell, so can bind two extra electrons to form the closed-shellThus, Mg and O can combine to form the ionic molecule MgO by transfer of two electrons. Similarly, an atom such as Al from group III can lose three elec-trons, and an atom such as N from group V can gain three; thus, one can formthe ionic molecule AlN by transfer of three electrons from Al to N.

There are also many possible ionic molecules involving more than twoatoms. For example, one can form the ionic molecule in which the Mgatom loses two electrons (to become ) while each of the two F atomsgains one electron (to become ). In Table 12.2 we show the nine possibilitiesF-

Mg2+

MgF2 ,

O2-.Mg2+;

Cl-.Na+

B Lke2

R0

- ¢E = 16.0 - 1.52 eV = 4.5 eV

B L

ke2

R0 =

1.44 eV # nm

0.24 nm = 6.0 eV

R0 = 0.24 nm,

¢ER0

E

1R


TABLE 12.2

The nine possible molecules formed by combining the positive ions orwith the negative ions or Numbers in parentheses show the valence of each element concerned.

(1) (2) (3)

(1) NaF(2) MgO

(3) AlNAl2O3AlF3Al3+

Mg3N2MgF2Mg2+

Na3NNa2ONa+

N3O2F

N3-.F-, O2-,Al3+Na+, Mg2+,

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Linus Pauling(1901–1994,American)

One of the founding fathers of

quantum chemistry,Pauling appliedthe principles of quantum mechan-ics to the bonding of atoms inmolecules, and his book, The Na-ture of the Chemical Bond (1939),was one of the most influentialtexts of the period. He was one of the first to suggest that proteinmolecules are arranged in a spiral.Pauling won the Nobel Prize inchemistry in 1954 and the NobelPeace Prize, for his fight against

nuclear weapons, in 1962.

Section 12.4 • The Covalent Bond 379

*The ion is readily formed in hydrogen gas by an electric discharge, which can stripan electron from a neutral molecule. In isolation, is perfectly stable. In practice,

it eventually picks up a free electron and reverts to neutral Nevertheless,it survivesquite long enough for an accurate determination of its properties.

H2 .

H2+H2

H2+

for the ionic molecules formed from Na, Mg, or Al, each in combination with

one of the elements F, O,or N.When an atom can form ionic molecules, we define its valence as the

number of electrons it gains or loses in forming such a molecule. Elementsfrom groups I and VII (closed-shell one) have valence 1; elements fromgroups II and VI (closed-shell two) have valence 2; and so on. Knowing thevalences of two elements that form an ionic molecule, one can immediatelypredict the proportions in which the atoms combine. For example, Mg has va-lence 2 (closed-shell ion ) and N has valence 3 (closed-shell ion ). Toretain overall neutrality, we must combine three ions with every twoions,and the resulting compound is as shown in Table 12.2. In general,

the proportion of constituent atoms in any ionic molecule is just the reciprocalof the ratio of their valences.

12.4 The Covalent Bond

Most diatomic molecules that are not bound ionically are bound instead bythe covalent bond. As already described, the covalent bond involves a concen-tration of one or more electrons from each atom in the region between the twonuclei. To understand how this concentration occurs, we begin by considering

the simplest of all molecules, the molecular ion, which consists of two 2protons and a single electron.* Just as an understanding of the one-electronatom (hydrogen) helped us to build a theory of multielectron atoms inChapter 10, so we will see here that an understanding of the one-electron mol-ecule will help us to understand the covalent bonding of multielectronmolecules such as and

The Molecular Ion

In the single electron moves in the field of the two protons. Because theprotons are so heavy, they move very little compared to the electron, and it isa good approximation to treat them as stationary. (This is the same approxi-mation used in Chapters 8 and 10 to discuss atoms.) Within this approxima-tion, we must solve the Schrödinger equation for the single electron in thefield of the two fixed protons and find its lowest allowed energy E. Just as withionic molecules, this energy will depend on the internuclear distanceR; that is,

If has a minimum at some separation a stable moleculeexists, with bond length and energy

Before trying to solve the Schrödinger equation, we must choose a sys-tem of coordinates.A convenient choice is to put the two protons on the x axisat positions as shown in Fig. 12.7.The electron’s position is denotedx = ;R

>2,

E1R02.R0

R0 ,E1R2E = E1R2.

H2+

H2

H2O.O2 ,H2 ,1H+

22H2+

Mg3N2 ,N3-Mg2+

N3-Mg2+

;

;

Electron

Proton 2Proton 1

r 1r 2

r

x R/2 x R/20 x

FIGURE 12.7

The molecule consists of twoprotons and a single electron. Wetreat the protons as fixed at

on the x axis. Theelectron’s position relative to theorigin is denoted by r. Itselectrostatic potential energy isdetermined by the distances to thetwo protons, shown as and r2 .r1

x = ;R>2

H2+

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0

Proton 1

Proton 2 1

2

FIGURE 12.8Wave functions for the oneelectron in the molecule withthe two protons far apart

The plots show thevalues of on the internuclear axis.The function describes a state inwhich the electron is bound in the

state around proton 1 and istherefore affected very little by thedistant proton 2; is thecorresponding state with theelectron bound to proton 2.

c2

1s

c1

c1R W aB2.

H2+

*If we want these functions to be normalized, we should include an overall normaliza-tion factor; since this factor does not affect our discussion here, we omit it (but seeProblems 12.31 and 12.56).† For simplicity, we focus here on the wave functions on the x axis [and for brevity write

them as ]. The corresponding symmetry for arbitrary points is this:and c-1r2 = -c-1-r2.c+1r2 = c+1-r2 r = 1x, y, z2c;1x2

by r, and we must find the wave function that gives the minimum total en-

ergy for the molecule with a given separation R.A direct analytic solution of the Schrödinger equation for the electron in

is possible, but it is rather complicated and not especially illuminating. Forour purposes, it is more convenient to seek an approximate solution; and tothis end, we begin by supposing that the two protons are far apart — specifi-cally, that R is much larger than the Bohr radius Under these conditionsthe solutions of the Schrödinger equation are easy to see. If the electron isclose to proton 1, the effect of the distant proton 2 is small and the lowest pos-sible state is just the ground state for an electron bound to proton 1.The cor-responding wave function, which we denote by is the familiar wave

function for a hydrogen atom centered on proton 1:

(12.12)

where is the distance from proton 1 to the electron. There is a second statewith exactly the same energy, in which the electron is bound to proton 2, withwave function

(12.13)

We have sketched these two wave functions in Fig. 12.8, which shows the val-ues of and along the x axis (that is,the l ine joining the two protons).

The two wave functions and satisfy the Schrödinger equation withthe same energy; that is,they are degenerate.Therefore, any linear combination,

(12.14)

(for any two constants B and C ) also solves the Schrödinger equation with thissame energy. This wave function is a superposition of a state where the elec-tron is bound to proton 1 and another where it is bound to proton 2, and — in

principle at least — any such state is a possible state of our system. Two partic-ular such states that we will find to be especially important are*

(12.15)

and

(12.16)

These two wave functions are sketched in Fig. 12.9(a). Their characteristic

property is that they represent states where the electron is distributed equallyaround both protons. To see this, note that, as is clear from Fig. 12.9(a),†

(12.17)c+1x2 = c+1-x2 and c-1x2 = -c-1-x2

c- = c1 - c2

c+ = c1 + c2

c = Bc1 + Cc2

c2c1

c21r2c11r2c2

1r

2= Ae-r

2>aB

r1

c11r2 = Ae-r1>aB

1sc1

1r

2,

aB .

H2+

c

1r

2

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(a)

1 2

1 2

(b)

2

2

FIGURE 12.9(a) The wave functions andfor the electron in when thetwo protons are far apart. The plotsshow values of along theinternuclear axis. (b) Correspondingplots of the electron’s probabilitydensity (which are identical aslong as the protons are far apart).

ƒc; ƒ 2

c;

H2+

c-c+

that is, is an even function of x and is an odd function.Therefore,for either wave function, the probability density is the same at any point x

as at the point

(12.18)

with equal to either or This is illustrated in Fig. 12.9(b) and meansthat the electron distribution at any point near proton 1 is the same as at thecorresponding point near proton 2.

The individual wave functions and are often called atomic orbitalssince they describe states where the electron forms an atom with one of the nu-clei and is unaffected by the other. [Recall that “orbital” is simply an alternative

name for the spatial wave function for any one electron; that is, an orbitalspecifies the orbital motion — but not the spin state — of one electron. In anatom an orbital is specified by the quantum numbers n, l , m.] The combinations

and are called molecular orbitals since they describe states where the elec-tron is associated with both nuclei and belongs to the molecule as a whole. Wewill see that it is the molecular orbitals and that become the stationarystates of when the two protons come close enough to form a stable molecule.

When the two protons approach one another, the wave functionoverlaps proton 2 and vice versa, and the electron’s motion is influenced byboth protons. The simple atomic wave functions and of Eqs. (12.12) and

(12.13) are no longer solutions of the complete Schrödinger equation (whichincludes the potential energy due to both protons). Nevertheless, we can showthat the combinations remain reasonable approximations. Anycorrect solution should reflect the symmetry of the system, which has twoidentical force centers at that is, the wave functions for the station-ary states of should have the symmetry property (12.18). Now, as we have

just seen, the wave functions do have this symmetry. Further-more, these functions, being combinations of and reflect the presence of both protons. This suggests that the functions and could be reasonableapproximations to correct solutions of the Schrödinger equation, even though

and separately are not. Using a more advanced argument (called thevariational method), one can show that this suggestion is correct. Here we willsimply accept that and are reasonable approximate solutions, and usethem to deduce the general behavior of the energy of the molecule for thetwo states concerned.

Once the two protons are within one or two Bohr radii, the behavior of is as sketched in Fig. 12.10(a).The corresponding probability densities

and are shown in Fig. 12.10(b), and the important difference between thetwo states is immediately clear:The probability density is nonzero at, andnear to, the origin; in fact, it is larger in this region than either or sep-

arately (see Problems 12.31 and 12.56). This means that for the state theelectron distribution is enhanced in the region between the two protons —

c+ ,

ƒc2 ƒ 2ƒc1 ƒ 2ƒc+ ƒ 2

ƒc- ƒ 2ƒc+ ƒ 2c;

H2+

c-c+

c2c1

c-c+

c2c1

c; = c1 ; c2

H2+

x = ;R>2;

c; = c1 ; c2

c2c1

c11r2H2+

c-c+

c-c+

c1r2c2c1

c- .c+c

ƒc1x2 ƒ 2 = ƒc1-x2 ƒ 2

-x:ƒc ƒ 2

c-

1x

2c+

1x

2

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1 2

(a)

1 2

(b)

FIGURE 12.11

Schematic representation of thebehavior of the molecular orbitalsof as the nuclei

approach one another. The plusand minus signs show the signs of the wave functions. When thenuclei are far apart, bothdistributions consist of twoseparate spherical clouds. Oncethe clouds overlap, the wavefunctions and interfereconstructively in the state and

produce an enhanced densitybetween the two nuclei; in the statethey interfere destructively and

give a reduced density.c-

c+

c2c1

H2+c; = c1 ; c2

2

0.50

2

E n e r g y ( e V ) R0

E(R)

E(R)

R (nm)

FIGURE 12.12

The energy of the molecule as afunction of the distance R betweenthe two protons. The curveis the energy of the “bonding state”

and is that of theantibonding state c- .

E-1R2c+

E+1R2H2+

1 2 B( )

1 2 C ( )

(a) (b)

2

2

FIGURE 12.10

(a) Sketch of the wave functionsand for the electron in themolecule when the distance Rbetween the two protons iscomparable to the size of an Hatom. At the origin, is largerthan either or whereas isexactly zero. (The factors B and C are normalization constants; C is alittle larger than B — see Problem12.31 — and this is why the peaksof are a little taller than those

of ) (b) The correspondingprobability densities. (As in Figs.12.8 and 12.9, all plots show valuesalong the internuclear axis.)

c+

c-

c-c2c1

c+

H2+c-

c+

*We should emphasize that these figures are highly schematic.The distribution of oris not a clearly defined sphere but, rather, drops continuously from a maximum at

the nucleus and reaches zero only at infinity.Thus the region of overlap is not really assharply defined as our cartoons suggest.

c2

c1

precisely where it is strongly attracted by both protons. This lowers the mole-

cule’s energy (as compared to the energy when R is large) and causes themolecule to bind. By contrast, the wave function is exactly zero at the ori-gin.Therefore, is small in the region between the two protons, and the en-ergy is not low enough to bind in this state.

In Fig. 12.11 we show schematic two-dimensional pictures of the statesThe left side of each picture represents the state while the

atoms are well separated and the distribution consists of two distinct sphericalclouds.* The right side of each picture represents the distributions when thewave functions and begin to overlap. In the state the two functionsand interfere constructively in the region of overlap and concentrate the

probability density there. In the state they interfere destructively andreduce the corresponding density.

In Fig. 12.12 we have plotted the energy of the molecule, as afunction of the distance R between the two protons, for both of the statesand The zero of energy has been chosen as the energy when the electron isbound to one of the protons and the other proton is far away As wehave just argued, the energy of the bonding orbital decreases as theprotons get closer, until it reaches a minimum at a separation Once R isless than the repulsion of the two protons is dominant and increas-es rapidly as R approaches zero. The important point is that, just as with the

ionic molecules discussed in Section 12.3, the energy has a minimum at a sepa-ration and a stable molecule can form, with and binding energy

The calculated values,

agree perfectly with experiment.

R0 = 0.11 nm and B = 2.7 eV

B = -E+1R02.R = R0R0 ,

E+

1R

2R0 ,

R0 .c+E+1R2 1R: q2.

c- .c+

H2+E1R2c- ,

c2

c1c+c2c1

c; = c1 ; c2 .

ƒc- ƒ 2c-

H2+

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The curve in Fig. 12.12 increases steadily as R decreases and has

no minimum; so the molecule has no stable bound state corresponding tothe antibonding orbital

The Neutral Molecule

Armed with a theory of the one-electron molecule, we can now under-stand many simple multielectron molecules. We consider first the neutralmolecule with its two electrons. Just as in our discussion of multielectronatoms in Chapter 10, we start by ignoring the repulsion between the electrons.In this approximation each electron moves independently in the field of twoprotons, and its lowest level is just that of the bonding orbital of themolecule, with energy The ground state of with its two electrons,is obtained by putting both electrons into this lowest level, to give a total ener-gy of and hence binding energy This value is an overes-timate since it ignores the electrostatic repulsion of the two electrons.Nevertheless, it is reasonably close to the observed value

We found the ground state of by giving both electrons the same wavefunction According to the Pauli exclusion principle, this is possible only if the two electrons occupy different spin states;that is,the ground state of themolecule must have the two electrons’ spins antiparallel and hence have

— a prediction that is fully confirmed by experiment. If we try to as-semble an molecule from two H atoms whose electron spins are parallel ,then one electron can go into the bonding orbital but the exclusion princi-ple requires that the other go into the antibonding orbital or some otherhigher level. With the second electron in it is found that the total energy

has no minimum, and there is no corresponding molecule; if the secondelectron goes to another, higher level, a molecule may form, but with muchhigher energy — that is,it forms an excited state (as we discuss in Section 12.6).

The ground state of the molecule typifies the main features of all co-valent bonds. Each atom contributes one electron to the bond. The two elec-trons have their spins antiparallel, which allows them both to occupy the samebonding orbital with a large density between the two nuclei. The resulting“electron pair” produces a strong bond between the two atoms, but we will seethat because of the exclusion principle, they cannot be joined by a third elec-tron. It is the strong bond produced by a shared pair of electrons with antipar-allel spins and their inability to accept a third electron that are the essentialcharacteristics of the covalent bond.

Saturation of the Covalent Bond

One of the most striking features of many molecular bonds is the phenome-

non of saturation: Molecules containing given elements usually contain somedefinite number of atoms and cannot bind any additional atoms. For example,hydrogen forms the stable molecule as we have just seen, but a stablemolecule does not exist.

To understand why this is true, we have only to recall part of our discus-sion of multielectron atoms in Chapter 10. In the two-electron atom, He, bothelectrons can occupy the lowest orbital and are very tightly bound. In thethree-electron atom,Li, two of the electrons can occupy the lowest orbital, butthe exclusion principle requires the third electron to occupy the nextorbital, which is much less tightly bound.

Almost exactly the same argument applies to the molecules andIn both electrons can occupy the lowest molecular orbital (the bondingH2

H 3

:H2

12s211s2H3H2 ,

H2

E1R2 c- ,c-

c+ ,H2

stot = 0

H2

c+ .H2

B = 4.5 eV.

B = 5.4 eV.-5.4 eV,

H2 ,-2.7 eV.H2

+c+

H2

H2+ ,

H2

c- .H2

+

E-

1R

2

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TABLE 12.3

Bond lengths and binding energies of thecovalent molecules LiH, and

Molecule (nm) B (eV)

0.074 4.5LiH 0.16 2.4

0.27 1.0Li2

H2

R0

Li2 .H2 ,1B21R02

orbital ), and this molecule is tightly bound. In there is again an orbital

of lowest energy, with the probability density concentrated inside the triangleformed by the three protons. However, as with the Li atom, only two of thethree electrons can occupy this lowest orbital, and the third must occupy ahigher orbital.

This qualitative argument shows clearly that an molecule, if it isbound at all, should be less tightly bound than To decide whether or notthe molecule (or indeed the Li atom) actually is bound requires a detailedcalculation of the energy levels concerned. In the case of the Li atom, we knowthat it is bound, although with much lower ionization energy than He. In thecase of the molecule, it turns out that the higher levels are so much higher

that is not bound.The big difference between the Li atom and the mol-ecule results from several effects, one of the most important differences con-cerns the three protons in either system: In the Li atom, the three protons aretightly bound inside the nucleus by the strong nuclear force; as far as the atomis concerned, the proton–proton repulsion is effectively switched off.In the hy-pothetical molecule, the proton–proton repulsion raises the energy by somuch (several tens of eV) that the molecule is unstable.

If we consider possible larger hydrogen molecules,such as this situa-tion only becomes worse. Therefore, there are no hydrogen molecules beyondthe familiar, diatomic and the bonding between n hydrogen atoms “satu-

rates”at

Covalent Bonding of Multielectron Atoms

The formation of molecules from multielectron atoms is not as complicated asone might at first fear. This is because it is only the outer, or valence, electronsof each atom that overlap enough to have an important role in the molecularbond. A closed-shell-plus-one atom such as Li (with configuration ) orNa can, for our purposes, be regarded as a single electron in an

s state outside an inert positive “core.” This suggests that these atoms shouldbehave very much like hydrogen, the main difference being that the one va-

lence electron is in a state concentrated at much larger radius (being). Thus, we would anticipate that a lithium atom could bond covalently

with hydrogen to form LiH, or with a second lithium to form The moleculeshould be bigger than LiH, which should be bigger than and because of

the greater separations, the binding energies should be correspondingly small-er. All of these predictions are confirmed by the experimental data shown inTable 12.3.

Atoms with more than one valence electron are naturally more compli-cated, but here again the situation is not as bad as one might fear.Consider, forexample, the fluorine atom, with its seven valence electrons and configuration

(12.19)F: 1s2 2s2

2p5

H2 ,Li2

Li2 .3, Á

n = 2,

11s2 2s2

2p6 3s12 1s2

2s1

n = 2.

H2 ,

H4 ,

H3

H3H3

H3

H3

H2 .H3

H3c+

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*Recall that with this arrangement, the two electrons are farther apart (than if both

occupied the same orbital) and have slightly lower energy because their Coulombrepulsion is reduced.

Recall that there are three independent wave functions, or orbitals,which

we can take to be the functions introduced in Section 8.8, labeledand and we can rewrite the electron configuration (12.19) in more

detail as

(It does not matter which of the three orbitals gets the single electron; to bedefinite, we suppose that it is the ) The important point is that the orbitals

and are occupied by two electrons each. The electrons in each or-bital must therefore have their spins antiparallel; that is, they form an electron

pair that cannot participate in a covalent bond.Thus,the fluorine atom has justone electron available to share in a covalent bond with another atom. In otherwords, fluorine is, in this respect, similar to hydrogen. We would expect it tobond covalently with H to form HF, or with another F atom to form — andboth of these molecules are indeed observed.

The oxygen atom has one less electron than fluorine,with the configuration

Two of the electrons in the level must occupy the same orbital ( say)

but as we argued in Section 10.7, the other two will go, one each, into the tworemaining orbitals,* and That is, the detailed configuration of the oxy-gen atom is

This leaves two unpaired electrons that are free to be shared in covalent bondswith other atoms.Thus, oxygen can form a molecule like in which the Oatom forms one covalent bond with each of the two H atoms. It can also formtwo covalent bonds, with a second O atom to make the doubly bonded mole-

cule Because is doubly bonded, we would expect it to be more strong-ly bound than the singly bonded and this prediction is confirmed by theobserved binding energies given in Table 12.4.

F2 ,O2O2 .

H2O,

O: 1s2 2s2

2p 2 x 2p 1

y 2p 1z

2pz .2py

2px ,2p

O: 1s2 2s2

2p4

F2

2py2s, 2px

2pz .2p

F: 1s2 2s2

2p 2 x 2p 2

y 2p 1z

2pz ,2py ,2px ,

2p

TABLE 12.4

Binding, or dissociation, energies of the singly, doubly,and triply bonded molecules and

Molecule B (eV)

1.6

5.19.8N2

O2

F2

N2 .F2 , O2 ,

The nitrogen atom has one less electron than oxygen and has theconfiguration

N: 1s2 2s2

2p 1 x 2p 1

y 2p 1z

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Since it has three unpaired electrons that are available for sharing with other

atoms, it can, for example, form one bond with each of three H atoms to makean ammonia molecule It can also form three bonds with a second N atomto make the triply bonded Since has three bonds, it is very well bound,as seen in Table 12.4;and for this reason, gas is chemically very inactive.

Valence

In the context of covalent bonding, we define the valence of an atom as thenumber of electrons that it can share with other atoms in covalent bonds; H,Li, and F all have valence 1, while O has valence 2, and N has valence 3. In thecase of ionic molecules such as LiF, we defined the valence of an atom as the

number of electrons the atom can gain or lose to form the ionic molecule.(F has valence 1 because it gains one electron to become whereas Li hasvalence 1 because it loses one electron to become ) Since many atoms canbond both covalently and ionically, it is important to note that in all the simpleexamples discussed here, our two definitions give the same value for the va-lence. For instance, a closed-shell-plus-one atom such as Li or Na can lose itsouter electron to form an ionic bond, and it can share the outer electron toform a covalent bond. Either way, its valence is 1. Similarly, a closed-shell-minus-one atom such as F or Cl can gain an electron (which fills the one va-cancy in its outer shell) to form an ionic bond; but as we have just seen, it can

share the one unpaired electron in its outer shell to form a covalent bond.Either way, its valence is also 1.

On the basis of what has been said here, it should be clear that a mole-cule like LiF could bond ionically or covalently and that (at the level of ourqualitative treatment) we have no way to predict which mechanism will pre-vail. By measuring the dipole moment, as discussed in Section 12.2, one canfind the extent to which a given molecule is ionically or covalently bonded. Inthe case of the alkali halide molecules (such as LiF) it is found that the bond ispredominantly ionic, but in many other molecules the bond is found to be acombination of both types.

We should emphasize that we have focused our discussion on the sim-plest molecules and that many molecules are much more complicated. Per-haps the most important example of the complications that can occur, evenwith fairly simple atoms, concerns carbon.The ground state of carbon has theconfiguration

which suggests that carbon could contribute two electrons to covalent bondsand should have valence 2, as it does in CO. However, carbon usually has

valence 4, as is testified by the existence of such molecules as (methane)and The explanation of this apparent contradiction is actually quitesimple: The and levels are close together, and only a small energy isneeded to raise one electron to give the configuration

(12.20)

In this configuration carbon has four electrons available for sharing in cova-lent bonds, and the energy gained by forming four bonds amply repays thesmall energy needed to achieve the configuration (12.20).

Carbon can use its four valence electrons to form bonds in many differ-ent ways. For example, it can form the molecule as shown schematicallyCH4 ,

C: 1s2 2s1

2p3

2p2sCO2 .

CH4

C: 1s2 2s2

2p2

Li+.F-,

N2

N2N2 .NH3 .

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Section 12.5 • Directional Properties of Covalent Bonds 387

CH H

HH

Methane, CH4

(a)

H

HC C

H

H

Ethylene, C2H4

(b)

C

H

H

HH

H H

C

CC

C

C

Benzene, C6H6

(c)

FIGURE 12.13

Schematic views of three

hydrocarbon molecules:(methane), (ethylene), and(benzene). Each short line

represents a covalent bond formedby a shared pair of electrons. Theactual molecules are, of course,three-dimensional; for example, thefour H atoms in are arrangedsymmetrically at the four comers of a tetrahedron with the carbon at itscenter. In benzene the six bondsbetween the carbon pairs are

perhaps better thought of as allbeing equal mixtures of single anddouble bonds.

CH4

C6H6

C2H4CH4

y

x

2 pzz

y

x

2 p xz

y

x

2 p yz

(a) (b) (c)

FIGURE 12.14

Each of the three wave functionshas two lobes where the

probability density is largest. (Thesepictures show the contours wherethe density is 75% of its maximumvalue, for the wave functions of hydrogen.) The plus and minus signsindicate that the wave function is

positive in one lobe and negative inthe other.

2p

px , py , pz

in Fig. 12.13(a). However, it can also form the molecule (ethylene), in

which each carbon is joined to two hydrogens but is also doubly bonded tothe other carbon, as shown in Fig. 12.13(b). Another example of the numer-ous hydrocarbon molecules is (benzene), in which the six carbons are

joined in a ring of single and double bonds; this leaves room for one hydrogento attach to each carbon, as shown in Fig. 12.13(c). Because carbon can bondin so many ways to other carbon atoms and to other elements, such as H, N, O,and many more, there is a vast array of different carbon compounds. Sincemany of them play an important role in organisms,all carbon-containing mol-ecules (with a few exceptions such as CO and ) have come to be calledorganic molecules.

12.5 Directional Properties of Covalent Bonds★

★This section can be omitted without serious loss of continuity.

In this section we describe how the properties of atomic wave functions deter-mine the shapes of molecules. To begin, we review briefly the formation of asimple diatomic molecule like

As we saw in Section 12.4, the two electrons in both occupy the bond-

ing orbital, which corresponds to the combination of thewave functions for each separate atom,

Since these functions are both positive, they interfere constructively in theirregion of overlap (for the orbital ), as was indicated schematically inFig. 12.11(a). This enhances the charge density between the two nuclei andlowers the energy of the molecule.

Similar considerations apply to any diatomic molecule in which the va-

lence electrons are all in s states, and Fig. 12.11(a) could equally have repre-sented such molecules as and LiH. If we consider instead such atoms asN, O, or F, whose valence electrons are in p states, the situation is more inter-esting. Recall that p states have three possible wave functions, which we cantake to be the functions and introduced in Section 8.8 and illustratedin Fig. 8.21 (which we have redrawn here as Fig. 12.14). The important pointabout each of these p states is that its probability density is concentrated intwo lobes along one of the three coordinate axes. For example, the wavefunction of hydrogen is Eq.(8.94)

(12.21)c2pz = Aze-r>2aB

2pz

pzpx , py ,

Li2

c+

c1 = Ae-r1>aB and c2 = Ae-r2>aB

1sc+ = c1 + c2

H2

H2 .

CO2

C6H6

C2H4

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(b) Antibonding

(a) Bonding

H1 s

F 2 pz

FIGURE 12.15

Wave functions for the HFmolecule. (a) We obtain the

bonding orbital by adding the wavefunctions for the electron in Hand the electron in F, provided that the H atom approaches thepositive lobe of the function.(The wave function in F isshown here as a figure eight toemphasize that both lobes belongto a single atomic wave function.)(b) If the H atom approaches fromthe opposite direction, adding thetwo wave functions gives an

antibonding orbital. For clarity, thetwo atoms are shown with theirwave functions not yet overlapping.

2pz

2pz

2pz

1s


H

H

O

(a) (b) (c)

2 p y

2 pz H

HO

90

FIGURE 12.16

(a) The two unpaired electrons inan O atom occupy the and

orbitals. (b) A water molecule canform if two H atoms approach thepositive lobes of these two orbitals.(c) The centers of the atoms in theresulting molecule form an “L”with angle between the

arms. (The experimental value isas explained in the text.)u L 105°,

u = 90°

2pz2py

This gives maximum probability on the z axis and zero probability in the

plane (where ). For our present purposes, it is important to note that thefunction (12.21) is positive in its upper lobe (where ) and negative in thelower one (where ), and we have indicated these signs by plus and minussigns in Fig. 12.14.

Let us consider first an atom with just one unpaired p electron; forexample, fluorine, with configuration

The first eight electrons in F are all paired and can, for our present purposes,

be ignored. The single unpaired electron, which is distributed as inFig. 12.14(a), can pair with the one electron in a hydrogen atom to form themolecule HF. To arrange this,however, we must bring up the H atom in such away that the wave functions of the shared electrons interfere constructively intheir region of overlap; this requires that the H atom approach the positivelobe of the F wave function as in Fig. 12.15(a). If the H atom approaches fromthe opposite direction, as shown in Fig. 12.15(b), we obtain an antibondingorbital, and no molecule forms.

We can summarize our findings so far: When valence electrons are in s

states, their distribution is spherical, with no preferred direction for bonding.

When the valence electrons are in p states, their distribution is concentratedalong a definite axis and this means that they form directed bonds.

We are now ready to discuss some molecules that are more interesting,such as the water molecule, The oxygen atom, with configuration

has two unpaired electrons, whose distributions are sketched in Fig. 12.16(a).Each of the unpaired electrons can form a bond with one H atom, but only if the H atoms approach along the y and z axes, as shown in Fig. 12.16(b). This

implies that the water molecule should be shaped like a letter “L,” as shownschematically in Fig. 12.16(c). It is found that the water molecule does have an“L” shape, but the measured angle between the two arms is about 105°, some15° larger than predicted. However, the reason for this discrepancy is not hardto find:The two H atoms,having shifted their electrons toward the oxygen,areleft somewhat positive and repel one another,pushing the two arms a little far-ther apart than the 90° predicted by our simple model.

If the bonds in were purely covalent, the shared electrons would con-centrate at the midpoint between the O and each H. It is found, however, thatthey concentrate somewhat closer to the oxygen. In other words, the bonds are

partly ionic and there is a small net negative charge near the O, leaving a smallexcess of positive charge near the two hydrogens, as shown schematically in

H2O

O: 1s2 2s2

2p 2 x 2p 1

y 2p 1z

H2O.

F: 1s2 2s2

2p 2 x 2p 2

y 2pz1

z 6 0z 7 0

z = 0

xy

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Section 12.6 • Excited States of Molecules 389

H

H

O

FIGURE 12.17

Schematic view of the chargedistribution in a water molecule.The plus and minus signs hereindicate positive and negativecharges whose observed magnitudeis about (Problem 12.32).e

>3

90

H

z

y

x

H

H

N

FIGURE 12.19

The ammonia molecule, is a

pyramid with the N atom at onecomer. The H atoms at the otherthree comers form an equilateraltriangle. The angle between anytwo bonds is actually a littlemore than 90° because of theCoulomb repulsion between theH nuclei.

N ¬ Hu

NH3 ,

(a) (b) (c)

FIGURE 12.18

(a) and (b) A water molecule canattach itself to a positive ornegative charge. (c) Watermolecules tend to align and attractone another.

*This arrangement occurs in diamond and methane for example. The valenceelectrons occupy four orbitals called hybrids. Each of these is a combination of theusual s and p wave functions, and each concentrates the electron in a single lobe.The

electrostatic repulsion of the electrons pushes these as far apart as possible, so that theypoint symmetrically outward as described.

s-p(CH4),

Fig. 12.17. For this reason the molecule has an electric dipole moment,

which is responsible for several familiar properties of water: A water moleculecan easily pick up a charge of either sign, as shown in Fig. 12.18(a) and (b), andcan carry away charges from a body that is electrostatically charged. This ex-plains why walking across a nylon carpet on a moist day does not cause the an-noying electric shocks that we experience when the humidity is low.

For much the same reason,water molecules can dissolve ionic moleculessuch as NaCl by attaching themselves to the separate and ions. Final-ly, when properly oriented, two water molecules attract each other, as inFig. 12.18(c). This dipole–dipole attraction is what holds water together insnow and ice crystals.

Cl-Na+

H2O

Just as with water, we can predict the shape of the ammonia mole-cule, The nitrogen atom,with configuration

has three unpaired p electrons, each concentrated near one of three perpen-dicular axes.Thus N can bond with three atoms to form with one H atomon each of the three axes, as shown in Fig. 12.19. According to our simplemodel, the angle between any two of the bonds should bebut, just as with water, the repulsion between the positive protons increasessomewhat ( in this case).

Similar considerations apply to many other molecules. For example, itcan be shown that in certain carbon compounds the four valence electrons of

the carbon atom have lobes that point symmetrically outward toward the cor-ners of a regular tetrahedron centered on the nucleus.* This has importantimplications for the shapes of the many hydrocarbon molecules mentioned atthe end of the preceding section. For example,the methane molecule, isa regular tetrahedron with the four H atoms at its corners and the C atom at itscenter. Since the basic ideas remain the same,while the details get much morecomplicated, we will not pursue this topic further.

12.6 Excited States of Molecules★

★

This section can be omitted without serious loss of continuity,though you should readit if you want to study Section 12.7.

We have so far discussed only the ground states of molecules. However, mostinformation about molecules comes from the study of molecular spectra; andsince spectra result from transitions, which necessarily involve at least oneexcited state, we now extend our discussion to include excited states as well.

CH4 ,

u = 107°u

u = 90°,N ¬ Hu

NH3 ,

N: 1s2 2s2

2p 1 x 2p 1

y 2p 1z

NH3 .H2O,

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E n e r g y

Energylevels of molecule

Possible totalenergies of the two atoms

E2(R)E2

el

E1el

E0el

E1(R)

E0(R)

R

FIGURE 12.20

When two atoms are far apart,their total energy can have variousdifferent values

As R decreases, each of the curvesmay have a minimum, and if thishappens, there is a correspondingbound state of the molecule. Theresulting energy levels of themolecule are denoted by asexplained in the text.

Eiel ,

Ei1R21i = 0, 1, 2, Á 2E21R2, Á .E01R2, E11R2,

*Our only concern here is with valence electrons, since inner-shell electrons play littlepart in molecular bonding.

Electronic Levels

Our analysis of molecular ground states started by considering two atoms, atlarge separation R, both in their ground states with total energy Wefound that as the atoms move closer, the curve may decrease and gothrough a minimum. If this happens (as it does with for example), the twoatoms can bond to form a stable molecule,whose ground-state energy is simplythe minimum value of .

Suppose now that we start instead with one or more of the atom’svalence electrons* in excited states, so that the total energy of the two atoms is

As the two atoms approach one another, it may happen thatthe curve goes through a minimum, and if this does occur, we concludethat this molecule has an excited state with energy equal to the minimumvalue of We can describe this excited state as “stable” in the sense thatthe atoms will not fly apart spontaneously; however, it is not perfectly stablesince it can drop to the ground state (or any other lower level) by emittingradiation. In Fig. 12.20 we have sketched three molecular energy levels (la-beled 0, 1, 2) that come about in this way.

The three curves in Fig. 12.20 give the possible energies of the moleculeas functions of the separation R. When R is large, the energies

are the allowed total energies of the two separated atoms.The min-imum values of the functions are the corresponding energylevels of the molecule in question. These levels are called electronic levelsbecause they are the energies calculated using the Schrödinger equation forthe electrons while ignoring possible motions of the two nuclei; for this reason,they are denoted by as shown on the left of Fig. 12.20. As the pic-ture suggests, one would expect the spacing of these electronic levels of themolecule to be of the same order as the spacing of the atomic levels fromwhich they evolved.Therefore, since the spacing of valence levels in atoms isusually a few eV, the same is generally true for the electronic levels in mole-cules. This means that transitions between electronic levels in a moleculeinvolve photons of a few eV — that is,photons in or near the visible range.

Motion of the Nuclei

If the electronic levels just described were the only levels of mol-ecules, molecular spectra would look very like atomic spectra. There is, how-ever, a very important complication that we have so far ignored. Up to now,

E0el, E1

el, Á

E0el , E1

el, Á ,

E0

1R

2, E1

1R

2, Á

E1

1R

2, Á

E01R2,

E¿1R2.

E¿

1R

2E¿

1R

27 E

1R

2.

E1R2.

H2 ,E1R2 E1R2.

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(a)

350

(b)

400(nm)

FIGURE 12.21

Emission spectrum of gas in thenear ultraviolet. (a) Several broadlines, or bands, between 340 and410 nm. (b) High-resolutionenlargement of the three bands

near 380 nm shows that eachconsists of many separate lines.

N2

we have assumed that the two nuclei of our diatomic molecule are at rest;

that is, we have considered only the motion of the electrons and have com-pletely ignored the nuclear motion. (Note that by “nuclear motion”we donot

mean the internal motion of the nuclear constituents, but rather the motionof each nucleus as a whole.) We now consider the energy associated withthis motion and will find that the values of are quantized, with allowedvalues Thus, the allowed energies of the entire molecule,

are

When the molecule makes a transition, both the states of the electrons(labeled by i) and the nuclei (labeled by j ) can change. If we consider a transi-tion (up or down) between a level E and a higher level

then the photon absorbed or emitted in the transition has energy

(12.22)

The energies are, as we just argued, typically a few eV. On the otherhand, the energies are generally small fractions of an eV and are smallcorrections to However, there are many possible different values for

Therefore, (12.22) implies that the spectrum of photons absorbed oremitted, instead of being a single line for each value of will consist of many different lines, all close to corresponding to the many possiblesmall values of Unless a spectrometer has high resolution, it may fail toresolve some of these lines,which will then appear instead as a wideband. Sev-eral such bands in the emission spectrum of gas are illustrated in Fig. 12.21.In the low-resolution picture one sees several broad lines or bands; in the high-resolution picture three of these bands are seen clearly to be made of manynarrowly spaced lines.The occurrence of such band spectra at visible, or near-ly visible, frequencies is one of the chief characteristics of molecular spectra,and the measurement of their many constituent frequencies is a rich source of information about molecules.

N2

¢Enuc.¢Eel,

¢Eel,¢Enuc.

¢Eel.¢Enuc

¢Eel

= ¢Eel + ¢Enuc

Eg = E¿ - E =

1Ei¿

el- Ei

el

2+

1E j¿

nuc- E j

nuc

2

E = Eiel

+ E jnuc 4 E¿ = Ei¿

el+ E j¿

nuc

E¿,

E = Eiel

+ E jnuc

1i = 0, 1, 2, Á ; j = 0, 1, 2, Á

2

E = Eel+ Enuc,

E0nuc, E1

nuc, Á .Enuc

Enuc

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U (R) E0(R)

U (R0) E0el

0

E0(R0) k(R R0)212

R0 R

E n e r g y

FIGURE 12.23

If the electrons in a moleculeoccupy their ground state withenergy the potential-energyfunction that controls the slowvibrations of the two atoms is

Provided that thenuclei remain close to theequilibrium separation, canbe approximated by a parabola asshown by the dashed curve.

U1R2R0

U1R2 = E01R2.

E01R2,

correspond to different states of the electrons, and we see that for

each electronic state, there is a different potential-energy func-tion, to govern the vibrational motion.

Let us focus attention first on the electronic ground state, withpotential-energy function To simplify our discussion, we con-sider a molecule, such as HCl, in which one atom is much lighter than theother. In this case we can assume, to a good approximation, that the heavyatom (mass ) is stationary, and we have only to consider the motion of thelight atom (mass ).* As a further simplification, we note that thevibrational motion occurs along the line joining the two nuclei.Thus our prob-lem is reduced to solving the Schrödinger equation for a single body (the

lighter atom) moving in one dimension, with potential energyas sketched in Fig. 12.23.

As long as the atoms remain reasonably close to their equilibrium sep-aration the well of Fig. 12.23 can be closely approximated by a parabolaof the form

(12.25)

for some suitably chosen constant k, as shown by the dashed curve in thepicture (see Problems 12.37 and 12.43). This potential energy is the simpleharmonic oscillator, or SHO, potential, discussed in Section 7.9, wherek was

called the force constant. We found there that the allowed energies are [seeEq. (7.98)]†

(12.26)

where is the angular frequency of the corresponding classical oscillator(same mass m and force constant k):

vc = A k

m

vc

Envib

= 1n +122 Uvc 1n = 0, 1, 2, Á 2

E01R2 L E01R02 +12 k1R - R022

R0 ,

U1R2 = E01R2,

m V m¿

m¿

U1R2 = E01R2.i = 0,

U1R2 = Ei1R2,i = 0, 1, Á ,

E1

1R

2, Á

*Actually, the case that m and are of the same order is not much more difficult.Asdescribed briefly in Section 5.8 and Problem 5.21, the two-body problem with bothmasses moving is equivalent to the problem with one of the bodies fixed, except thatone must use the reduced mass, in place of m. Thus all of the for-mulas of this section are correct, whether or not provided that one replacesm by Note that if m is much less than then (see Problems 12.39 and12.40).† Equation (12.26) gives the allowed energies measured up from the bottom of the well

(12.25), and the total energy is therefore The first term is

the electronic energy and the term is the energy of vibration, which we

denote by Envib. A

n +1

2 B Uv

c

E0

el ,

E0

1R0

2E0

1R0

2+ An +

12 B Uvc .

m L mm¿,m.m V m¿,

m = mm¿>1m + m¿2m¿

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The lowest vibrational level has the zero-point energy and all the levelsgiven by (12.26) are spaced at equal intervals However, at high energiesthe approximation (12.25) breaks down and, as seen in Fig. 12.23, the true po-tential well rapidly becomes wider than its parabolic approximation. Thismeans that the higher levels are somewhat closer together, and there are nolevels above This is illustrated in Fig. 12.24.

We see in Fig. 12.24 that since the lowest possible energy isthe ground state of the molecule is higher by than the value we used inSections 12.3 and 12.4 (where we ignored motion of the nuclei). However, this

zero-point energy is usually quite small compared to electronic energies. Forexample,in LiBr, it is about 0.035 eV, as we see in the following example. Com-pared to typical electronic energies of order 1 eV, this is indeed fairly small.

Example 12.4

The parabola that best approximates the potential well for LiBr has aforce constant (see Problem 12.47). Use this to find the

zero-point energy for LiBr.

The mass of Br (80 u) is much greater than that of Li (7 u), so we canmake the approximation that the Br is fixed, and we get

Note that the spacing between adjacent vibrational levels is just twicethis, which is also small compared to typical electronicenergies.

We can apply similar considerations to any other electronic state. Foreach electronic state, labeled i, the curve defines the oscillator well thatcontrols the vibrational motion of the atoms, and for each such well there willbe a ladder of vibrational levels The levels of the mole-cule, including electronic and vibrational motion, are therefore as sketched inFig. 12.25.

Envib

= An +12 B Uvc .

Ei1R2 Uvc L 0.07 eV,

=200 eV # nm

2 A

800 eV>nm2

7 * 931.5 * 106 eV

L 0.035 eV

12 Uvc =

U

2 A

k

m=

Uc

2 A

k

mc2

12

Uvc

k L 800 eV

>nm2

E01R2

E0el1

2 Uvc

E0el

+12 Uvc ,

E = 0.

Uvc .

12

Uvc ,


E0el

0

12

h cn 0

12

34

5 6R

E n e r g y

FIGURE 12.24Vibrational levels of a diatomicmolecule in its lowest electronicstate. The numbers on the rightidentify the vibrational quantumnumber, n. For clarity, we haveshown a molecule with only sevenlevels; in most real molecules thereare more vibrational levels, moreclosely spaced.

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m

R0

m

CM

FIGURE 12.26

Rotational motion of a diatomic

molecule.


6h2/ I l 3

3h2/ I l 2

h2/ I l 1

0l 0

E r o t

FIGURE 12.27

Quantized energies of rotation,for a diatomic

molecule.l1l + 12 U2>2I,

E2el n 0

12

3 4

E1el

E0el

n 01

23

45 6

n 01

23

45 6

E n e r g y

FIGURE 12.25

Energy levels of a typical diatomicmolecule, including electronic andvibrational energies. The threeelectronic levels are shown as

For each electroniclevel, there is a ladder of vibrationalenergies,

with these are drawninside the appropriate well, withthe quantum number n shown atthe right.

n = 0, 1, Á ;

Envib

L An +12 B Uvc

E0el , E1

el , E2el .

*Notice that this result is the rotational analog of the more familiar result for thetranslational energy p2>2m.

Rotational Energy The rotational energies are easily calculated. We can think of the molecule asa dumbbell, which can rotate about its center of mass as shown in Fig. 12.26.The classical energy of rotation is where I is the moment of iner-tia and is the angular velocity. Since the angular momentum L is it fol-lows that and the rotational energy is*

This formula is also valid in quantum mechanics, provided that we use thecorrect quantized values for Hence, the allowed values of therotational energy of our diatomic molecule are

(12.27)

These values of are shown in Fig. 12.27, where we see — what is also clearfrom Eq. (12.27) — that the energies are all positive and that the spacingbetween the levels increases steadily with increasing l .

To estimate the magnitude of the rotational energies (12.27), we againconsider a molecule like HCl or LiBr in which one atom is much lighter thanthe other. In this case the rotational motion is — to a good approximation —

just an orbiting of the light atom (mass m) around the fixed heavy atom, andthe moment of inertia is simply

I = mR02

Erot

Elrot

= l1l + 12

U2

2I 1l = 0, 1, 2, Á 2

L2.l1l + 12 U2

Erot=

L2

2I

v = L>I Iv,v

Erot=

12

Iv2,

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(a) (b)

E2el

E1el

E0el

E n e r g y

n 01

2 3 4

n 0

n 0

n 0

n 1

n 2

1

1

2

2

3

3

4

4

5

5

6

6 4

5

6

74

5

6

3210

3210

43210

FIGURE 12.28

Energy levels of a diatomicmolecule have the form

(a) For eachelectronic level, there is a ladder of vibrational levels, labeled

(On the scale of thispicture the rotational levels are tooclosely spaced to be shown.)(b) Above each vibrational level is a“subladder” of very closely spacedrotational levels, as illustrated bythis enlargement of the threelowest vibrational levels. Therotational levels are labeled by thequantum numbershown on their left. Note that inmost molecules the rotationallevels are much more closelyspaced than shown here.

l = 0, 1, 2, Á ,

n = 0, 1, Á .

Enel

+ Envib

+ Elrot

In the case of HCl the observed value of is 0.13 nm and the light atom (H)

has mass Thus, for example, the level has energy

(12.28)

Since the electronic energies are of order 1 eV while the energies of vibrationare of order 0.1 eV, we see that rotational energies are much smaller than bothelectronic and vibrational energies.

Total Energy of the Molecule

We can now put our various results together. The total energy of a diatomicmolecule is

(12.29)

All three of these energies are quantized. The allowed values of we havedenoted by those for are denoted

and, at least for the lower levels, have the approximate formfinally, the allowed values of are Representative values forthe spacings of the various levels are

If we ignore the small rotational energies entirely, the energy levels of the molecule appear as shown in Fig. 12.25 [which we have reproduced as

Fig. 12.28(a)].Each electronic state, defines a potential well thati = 0, 1, Á ,

¢Erot ' 0.001 eV

¢Evib ' 0.1 eV

¢Eel ' 1 eV

l1l + 12 U2>2I.Erot An + 12 B Uvc ;

Envib

1n = 0, 1, Á

2EvibEi

el

1i = 0, 1, Á

2;

Eel

E = Eel+ Evib

+ Erot

L1200 eV # nm221938 * 106

eV2 * 10.13 nm22L 2 * 10-3

eV.

U

2

I=

U2

mR02

= 1 Uc22

mc2 R0

2

l=

1m=

938 MeV>c

2

.

R0

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Section 12.7 • Molecular Spectra 397

*We know from the kinetic theory of gases (Sec. 3.7) that the average translational kineticenergy is where denotes Boltzmann’s constant, Atroom temperature this energy is about 0.04 eV.† The main exception is that in homonuclear molecules (molecules such as with two

identical nuclei) a selection rule forbids these transitions and there is no rotationalspectrum.

H2

1T L 300 K2 kB = 8.62 * 10-5 eV>K.kB3kB T>2,

determines the vibrational motion of the atoms, and inside each well we

have drawn the ladder whose rungs represent the different vibrational ener-gies withIf we wish to take into account the rotational energies, then above each

rung of every ladder in Fig. 12.28(a) we must draw a “subladder” whose rungsrepresent the rotational energies On the scale of Fig. 12.28(a) the rungs of these subladders are too closely spaced to be distin-guished, but Fig. 12.28(b) shows an enlargement of the lowest three levels of part (a), each with its subladder of rotational levels. It is the level structure inFig. 12.28 — with rotational levels built on vibrational levels built on electron-ic levels — that determines the main features of molecular spectra, as we

describe in the next section.

12.7 Molecular Spectra★

★This section can be omitted without significant loss of continuity.

Knowing the structure of molecular energy levels — with their hierarchy of electronic, vibrational, and rotational energies — we can now discuss the spec-trum of radiation emitted or absorbed as a molecule changes its energy. Themany possible different spectral lines can be classified according to which of

the terms in the total energy,

changes in the corresponding transitions.

Rotational Transitions

We begin by discussing the rotational transitions, in which only the rotationalstate changes. Since the separation of rotational levels is of order thesame is true of the photons emitted or absorbed in rotational transitions. Pho-

tons with energy have wavelength of order 1 mm and are classified asmicrowave photons. Thus, the rotational spectrum of most molecules lies inthe microwave region. Microwave ovens exploit this fact; the frequency of theradiation in these ovens is designed to match the rotational modes of thewater molecule.

At room temperature and below, the average kinetic energy of the mole-cules in a gas is of order 0.04 eV or less.* This is so low that collisions betweenmolecules will seldom excite either the electronic or vibrational levels. On theother hand, it is quite enough to excite several rotational levels. Thus the mol-ecules of a gas at room temperature remain locked in their lowest electronic

and vibrational levels,but are continually being excited to several different ro-tational levels. In many molecules transitions between different rotational lev-els can occur rapidly† and the gas will emit microwave radiation as theseexcited molecules drop back to lower levels. Since this radiation is ratherweak, the most convenient way to study the rotational spectrum is usually to

10-

3 eV

10-3 eV,

E = Eel+ Evib

+ Erot

l1l + 12 U2>2I1l = 0, 1, Á 2.

n = 0, 1, Á .Envib

L An +12 B Uvc ,

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Transmittedintensity

100%

h2/ I h2/ I

2h2/ I

3h2/ I

Erot 0l 0

(b)(a)

3h2/ I 4h2/ I 1

2

6h2/ I 3

10h2/ I 4

E

FIGURE 12.29

The rotational spectrum of amolecule is produced by transitionsbetween different rotational levels,while the vibrational and electroniclevels remain unchanged. (a) Theallowed transitions satisfy

(b) General appearance

of the rotational spectrum inabsorption. A microwave beam of adjustable frequency is directedthrough the gas and is absorbedonly when its photons have energy

equal to one of the

transition energies U2>I, 2 U

2>I, Á .

1E = hf2

¢l = ;1.

*This is the same selection rule encountered in (11.46).As mentioned there, it follows

because the photon has spin 1 and must carry away exactly one unit of angularmomentum.

monitor the absorption of a microwave beam that is directed through the gas

and whose frequency can be adjusted. The different transitions then revealthemselves as sharp reductions in the transmitted beam whenever the mi-crowave photons have the right energy to excite the molecule from one levelto another.

Not all transitions that are energetically possible occur with appreciableprobability. As we discussed in Section 11.8, those transitions that have appre-ciable probability are called allowed and are determined by selection rules .Theselection rule for transitions between different rotational levels is*

(12.30)

That is, only those transitions in which l changes by one unit occur rapidlyenough to be easily observed. Therefore, the only allowed rotational transi-tions, up or down, have the form

(12.31)

as shown in Fig. 12.29(a). Since the energy of the rotational levels isthe energies of the photons emitted or absorbed in these transi-

tions have the form

(12.32)

This gives a series of equally spaced lines separated by the energy asillustrated in Fig. 12.29(b). By measuring the separation of any two adjacentlines in the rotational spectrum of a molecule, one can determine the mole-cule’s moment of inertia, I , and its bond lengthR0 .

U2>I,

=1l + 12 U2

I 1l = 0, 1, 2, 3, Á 2

Eg1l4 l + 12 = El+ 1rot

- Elrot

= 31l + 121l + 22 - l1l + 124

U2

2I

l1l + 12 U2>2I,El

rot

l4 l + 1 1l = 0, 1, 2, 3, Á 2

¢l = ;1

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Example 12.5

The four absorption lines of longest wavelength in lithium iodide gas arefound at 0.564,0.376,and 0.282 cm. Deduce the bond length of theLiI molecule.

Because the lines with longest wavelength correspond to photons withlowest energy, the four given lines should correspond to the lowest four tran-sitions of Fig. 12.29. The four energies are easily calculated.

(cm) (cm) Transition

1.130.5640.3760.282

These are seen to be in the ratios as required by (12.32), with

Since lithium has this gives

Note that in this calculation we treated the iodine nucleus as

fixed, so that with the mass of the lithium nucleus. If one

takes account of the small motion of the iodine, one gets (seeProblem 12.49).

Rotational spectra of molecules are useful in astronomy, where theymake possible the identification of many molecules in the large “molecularclouds” of our galaxy. These clouds are often so cool (a few tens of degreeskelvin) that intermolecular collisions almost never raise the molecules to excit-ed vibrational or electronic levels; nevertheless, they can excite several rota-tional levels,which then emit radiation that can be detected by radio telescopeson earth. For example, carbon monoxide is frequently detected in molecular

clouds by the microwaves emitted in the transitions and

Vibrational–Rotational Transitions

In many molecules one can observe transitions in which the vibrational levelchanges but the electronic level does not. However, in any such transition theselection rule (12.30) still applies, and the rotational level must also change.For this reason, these transitions are called vibrational–rotational transitions.We have seen that the spacing of vibrational levels is of order 0.1 eV, while thatof rotational levels is much smaller. Therefore, the photons emitted or ab-sorbed in vibrational–rotational transitions have energies of order 0.1 eV andare in the infrared.

l = 2: 1.l = 1: 0

R0 = 0.239 nm

m = 7 u,I = mR02

1mass = 127 u2 =

197 eV # nm

4 17 * 931.5 * 106 eV2 * 11.10 * 10-4 eV2 = 0.233 nm

R0 =U

4 m11.10 * 10-4 eV2 =

Uc

4 mc211.10 * 10-4 eV2

m=

7u=

7*

931.5 MeV>c2

,

U2

I=

U2

mR02

= 1.10 * 10-4 eV

1: 2 : 3 : 4 ,

3: 44.40 * 10-42: 33.30 * 10-41: 22.20 * 10-40: 11.10 * 10-4

EGL

E = hc>ll = 1.13,


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n 0 1l 3 2

n 0

n 1

E0el

FIGURE 12.30

In a vibrational–rotational transitionthe electronic state is unchanged,but both n and l change by oneunit. In the transition shown, nincreases and l decreases Since thetotal energy increases, thistransition entails the absorption of a photon. (Spacing of the rotationallevels is greatly exaggerated.)

1l = 3: 22.1n = 0: 12

0.30 0.31 0.32 0.33

E (eV)

7 6

T r a n s m i s s i o n

6 55 4

4 3 3 2

2 1

1 0 0 1

1 22 3

3 44 5

5 6

6 7

FIGURE 12.31

Vibrational–rotational absorptionspectrum of HBr. All linescorrespond to the vibrationaltransition with variousdifferent rotational transitions

as indicated.l4 l - 1,

n = 0: 1,

In addition to (12.30) there is also a selection rule for the vibrationalquantum number.

(12.33)

That is, among the many conceivable vibrational–rotational transitions, onlythose in which both n and l change by one unit are observed with appreciableprobability. A typical such transition is illustrated in Fig. 12.30. The energiesinvolved have the form

(12.34)

with the vibrational spacing much greater than the rotational.Therefore,if n decreases by one the total energy decreases whichever way l

changes and a photon is emitted with energy

(12.35)

If n increases by one, the molecule’s energy increases and a photon has to be

absorbed, with energy again given by (12.35). Either way, (12.35) implies thatthe vibrational–rotational spectrum consists of many lines, equally spaced oneither side of with spacing Since there are no transitions withthere is no line with actually equal to

A typical vibrational–rotational absorption spectrum, for HBr, is shownin Fig. 12.31. One sees clearly the equally spaced lines as predicted by (12.35),

Uvc .Eg

¢l = 0, U2>I. Uvc ,

= Uvc ;1l + 12 U2

I 1l = 0, 1, 2, Á 2Eg = 1En

vib- En- 1

vib 2 ; 1El+ 1rot

- Elrot21l4 l + 12,

1n: n - 12,1 Uvc2En

vib= An +

12 B Uvc and El

rot=

l

1l + 1

2 U

2

2I

¢n = ;1

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with a gap at the center corresponding to the absence of transitions withFrom the position of the central gap, one can measure the vibra-

tional parameter and from the spacing of the lines, one can read off therotational parameter as the following example shows.

Example 12.6

Use the graph in Fig. 12.31 to determine the parameters and for theHBr molecule. Hence, find the force constantk and bond length for HBr.

The central gap in a vibrational–rotational spectrum comes atand from Fig.12.31, we can read off as

(12.36)

Since where m is the mass of the hydrogen atom (as usual, weignore the motion of the heavier atom), a simple calculation gives

The spacing between adjacent lines in a vibrational–rotational spec-trum is and from Fig. 12.31, we find

(12.37)

Comparing (12.36) and (12.37), we see that the rotational energies aresmaller than the vibrational by two orders of magnitude, as stated earlier.Given that (where m is the mass of hydrogen), we find to be

(12.38)

The relative strengths of the absorption or emission lines in any spectrumdepend on how many of the molecules occupy the initial levels concerned, andthis occupancy depends on the temperature of the gas. (The higher the temper-ature, the greater the number of levels occupied.) Measurement of the relativeintensities of the lines in a spectrum like those of Fig. 12.31 or Fig. 12.29(b)allows one to find the temperature of the gas. This is especially useful in as-tronomy, where the gases of interest are inaccessible to direct measurement.

In fact, rotational and vibrational–rotational spectra are useful to as-tronomers for several other reasons as well. They can be used to identify themolecules in interstellar clouds that are too cool to emit at higher frequencies.Furthermore, microwave and IR radiations can pass through dust clouds thatare completely opaque to visible light. Thus, microwaves and IR can give in-formation about regions of our galaxy that are inaccessible to conventionaloptical astronomy. For example, they allow direct observations of the core of

=197 eV # nm

4 1938 * 106 eV2 * 10.0020 eV2 = 0.14 nm

R0 = A I

m= A

U2

m1 U2>I2 =Uc

4 mc21 U2>I2R0I = mR0

2

U2

I= 0.0020 eV

U2>I

= 2.4 * 103 eV

>nm2

k = mvc2

=mc21 Uvc221 Uc22

=1938 * 106

eV2 * 10.317 eV221197 eV # nm22

vc = 2 k>m ,

Uvc = 0.317 eV

Uvc

E = Uvc ,R0

U2>I Uvc

U2>I,

Uvc ;Eg

= Uvc.


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n 0

n 0

E iel

E0el

1

1

2

2

3

3

4

FIGURE 12.32

Typical electronic transitions startingfrom the lowest electronic andvibrational levels. Since there is noselection rule on n in electronictransitions, many different transitions,

arepossible for any one final electroniclevel Ei

el .

0: n, 1n = 0, 1, 2, 3, Á2,

n 0 0 0 1 0 2 0 3 0 4

(a)

(b)

FIGURE 12.33

(a) The electronic absorptionspectrum corresponding to thetransitions shown in Fig. 12.32.(b) At high resolution, each of thebands in part (a) is seen to be aseries of lines, corresponding to themany possible changes of rotational

energy.

the galaxy, where measurements of Doppler shifted spectra let us determinethe speed of rotation of the galaxy’s inner spiral arms.

Electronic Transitions

Any transition in which the electronic state of a molecule changes is called anelectronic transition. In electronic transitions both the vibrational and rota-tional states can also change.A typical electronic transition has the form

and the energy of the photon emitted or absorbed is

(12.39)

Since the largest of these three terms, is of order a few eV, the same istrue of and electronic transitions involve photons in the visible, or nearlyvisible, region.

Just like an atom, a molecule can make both upward and downward tran-sitions, producing absorption and emission spectra. If a gas sample is at roomtemperature,however, all the molecules will be in the lowest electronic and vi-brational states.In this case they can make no downward electronic transitions

and will produce no electronic emission spectrum.On the other hand, they canmake upward transitions — if exposed to light of the right frequency — andtheir electronic spectrum can be studied in absorption. Any observed transi-tion must start in the lowest electronic and vibrational levels since only theseare occupied. If we ignore for a moment the small rotational energies, theobserved transitions will have the form

(12.40)

The selection rule (12.33), which applies to vibrational transitions in

a single SHO well, does not apply to transitions between the levels of two dif-ferent wells.Therefore, the electronic transitions (12.40) can occur with manydifferent values of n:

as illustrated in Fig. 12.32. For any one final electronic level, these transi-tions will produce a series of equally spaced absorption lines, correspondingto the several possible final vibrational levels as shown inFig. 12.33(a).

n = 0, 1, 2, Á ,

0: n 1n = 0, 1, 2, Á 2¢n = ;1,

1E0 = E0el

+ E0vib2: 1E = Ei

el+ En

vib2

Eg ,¢Eel,

Eg = E¿ - E = ¢Eel+ ¢Evib

+ ¢Erot

1E = Eiel

+ Envib

+ Elrot24 1E¿ = Ei¿

el+ En¿

vib+ El¿

rot2

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Problems for Chapter 12 403

If we take account of the rotational energies, every one of the levelsshown in Fig. 12.32 has a ladder of many very closely spaced rotational levelsabove it. Thus, for each of the transitions (12.40), there are in reality manytransitions in which the rotational energy changes by small but differentamounts. Thus each of the lines in Fig. 12.33(a) is actually a band and is found,at high enough resolution,to be a series of closely spaced lines,as illustrated inFig. 12.33(b).

In summary, molecular spectra range from rotational spectra in the mi-crowave region, through vibrational–rotational spectra in the IR, to electronicspectra in the visible region. These provide three different windows, often re-quiring quite different experimental techniques, through which we study theproperties of molecules.They are also the basis of many practical applications,including lasers and masers, as discussed in Chapter 11.

CHECK LIST FOR CHAPTER 12CONCEPT DETAILS

Bond length (of diatomic molecule) Distance between nuclei (Sec. 12.2)

Binding, or dissociation, energy Energy to pull atoms apart (Sec. 12.2)

Ionic bonds Atoms exchange one or more electrons and are then bound

by the Coulomb attraction (Sec. 12.3).Dipole moments, (12.1)

Covalent bonds Atoms held together by shared electrons (Sec. 12.4)and molecules

Bonding and antibonding orbitalsDirectional properties★ (examples of and )

Valence Number of electrons an atom can gain or lose in an ionicbond.Number of electrons an atom can share in a covalent bond.

Excited states★ (12.29)

Electronic levels,

Vibrational levels, (12.26)

Rotational levels, (12.27)

Molecular spectra★ Rotational transitions (microwave) (12.32)Vibrational–rotational transitions (IR) (12.35)Electronic transitions (visible or nearly visible) (12.39)

Elrot

= l1l + 12 U2>2I ' 0.001 eV

Envib

= An +12 B Uvc

' 0.1 eV

Eiel ' 1 eV

E = Eel+ Evib

+ Erot

CH4H2O, NH3 ,

H2H2+

p = qd

PROBLEMS FOR CHAPTER 12the earth), what would be the resulting electrostaticforce of the sun on the earth, and what accelerationwould it produce? Compare this force with the corre-sponding gravitational force. (Make reasonable esti-mates.The mass of the earth is about andthat of the sun is The distance from earth

to sun is )

12.3 • If an ionic molecule results from the transfer of exact-ly one electron from one atom to the other, it shouldhave a dipole moment where is the bondR0p = eR0 ,

1.5 * 1011 m.

2 * 1030 kg.

6 * 1024 kg,

SECTION 12.2 (Overview of MolecularProperties)

12.1 • (a) If all the electrons in 1 gram of hydrogen could bedetached from all the protons and separated fromthem by 1 cm, what would be the total electrostaticforce between the electrons and protons? (b) Whatwould be the electrons’ acceleration in g’s (assumingthat they are somehow held together as a single body)?

12.2 • If all the electrons in the earth could be removedand attached to the sun (leaving all the protons on

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s

R0

FIGURE 12.34

(Problem 12.11)

length. Use the data in Table 12.1 (Section 12.2) topredict the dipole moments of KCl,LiF,NaBr,and NaCl

in The observed values are, respec-tively, and

express these as percentages of your predicted values. (These percentages measure theextent to which the molecules concerned are ionic.)

12.4 • If a diatomic molecule is ionically bonded by thecomplete transfer of one electron, its dipole momentshould be Given the data in the table thatfollows, discuss the extent to which the moleculesconcerned are ionically bonded.

Molecule: NaF HF CO

Bond length,(nm): 0.193 0.0917 0.113

Dipole moment,

12.5 • A 1984 Ford Escort traveling at experiences adrag force,because of rolling friction and air resistance,of about 360 N (roughly, 80 lb).* The energy releasedby 1 liter of gasoline is about and is usedwith about 18% efficiency.How far can the car travel on1 liter of gasoline at Convert your answer tomiles per gallon.

12.6 • In Example 12.2 (Section 12.2) we found that theenergy released in the burning of 1 kg of carbon togive is In that calculation we as-sumed that the original carbon atoms were separatefree atoms. [Specifically, the energy in Eq. (12.5) is forthe case that the C atom is free and unattached.] If the C atoms are bound in a liquid or solid, some ener-gy is needed to detach them, and the energy yield issmaller.Given that graphite (one of the solid forms of

pure carbon) has a binding energy of 7.4 eV peratom, find the energy released when 1 kg of graphiteburns in gas to form Compare your answerwith the result in Example 12.2.

12.7 • Calculate the energy released if 1 kg of graphite(one of the solid forms of pure carbon) burns in oxy-gen gas to give carbon monoxide, in the reaction

Don’t forget that energy is needed to dissociate thecarbon atoms from the graphite and the oxygen

atoms from the molecules.The binding energy of graphite is 7.4 eV per atom;the dissociation energy of the molecule is 5.1 eV, and that of CO is 11.1 eV.

12.8 • (a) The energy needed to dissociate the mole-cule into three separate atoms is 9.50 eV. Given thatthe dissociation energies of and are 4.48 and5.12 eV, calculate the energy released in the reaction

2H2 + O2: 2H2O

O2H2

H2O

O2

O2

2C + O2: 2CO

CO2 .O2

9.1 * 107 J.CO2

1 gal = 3.8 liters.)(1 mi = 1.6 km,50 mi>h?

3.2 * 107 J

50 mi>h

3.66 * 10-316.07 * 10-302.72 * 10-29p 1C # m2:

R0

p = eR0 .

3.00 * 10-29 C # m;3.04 * 10-29,3.42 * 10-29, 2.11 * 10-29,

coulomb # meters.

(b) The airship Hindenburg was held up by about 200million liters of hydrogen gas.Assuming that this was

at STP, estimate the energy released when the Hin-denburg exploded. Give your answer in “tons of TNT.” (One ton of TNT is the energyreleased by 1 ton of the explosive TNT.)

12.9 •• Gasoline is a variable mixture of various hydrocar-bons, such as octane To be definite, considerpure octane and calculate the energy released byburning 1 kg of octane in the reaction

using the following dissociation energies:

Molecule:

Dissociation energy (eV): 101.6 5.1 16.5 9.5

Note that these dissociation energies are the energiesneeded to separate each molecule completely into itsindividual constituent atoms. (Your answer hereshould be appreciably less than the value found inExample 12.2, which treated the oxidation of carbonatoms that were already separated.)

12.10 •• Consider a point charge q outside a spherically sym-metric, rigid charge distribution of zero total charge.(a) Use Gauss’s law to prove that the distribution ex-erts no force on q. (b) Next, show that no assembly of external charges can exert a force on a spherically sym-metric, rigid distribution of zero total charge.

12.11 •• As a simple classical model of the covalent bond,suppose that an molecule is arranged symmetrical-ly as shown in Fig. 12.34.Write down the total poten-tial energy U of the four charges and, treating theprotons as fixed, find the value of the electrons’ sepa-

ration s for which U is a minimum.Show that the min-imum value is (Note that youhave here treated the two protons as fixed at separa-tion Classically, they could not remain fixed in thisway; it is one of the triumphs of quantum mechanicsthat it explains how the protons have a stable equilib-rium separation, as we describe in Section 12.4. Notealso that you should not take the number 4.2 in ouranswer too seriously — especially in estimating thedissociation energy. First, the electrons have apprecia-ble kinetic energy, which we have not considered, sec-ond, even after the molecule has been dissociated into

R0 .

Umin L -4.2ke2>R0 .

H2

H2OCO2O2C8H18

2C8H18 + 25O2: 16CO2 + 18H2O

(C8H18).

4.3 * 109 J,

*E. J. Horton and W. D. Compton, Science, vol. 225, p. 587

(1984).

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two atoms, each atom still has an energy Both of these effects lower the dissociation energy, and a more

realistic estimate would be )

SECTION 12.3 (The Ionic Bond)

12.12 • (a) Use the graphs in Fig. 10.11 to find the energyneeded to transfer an electron between the fol-

lowing pairs of atoms: from Li to Br, from C to O,from Rb to F, from O to O. (b) Two of these pairsbond ionically and two do not;which are which?

12.13 • (a) Use the graphs in Fig. 10.11 to find the energyneeded to transfer an electron between the fol-

lowing pairs of atoms: from Cs to I,from N to O,fromN to N, from Na to Br. (b) Two of these pairs bondionically and two do not;which are which?

12.14 • The ionization energy of K is 4.34 eV and the elec-tron affinity of Cl is 3.62 eV.What is the critical radius

inside which the transfer of an electron from a Katom to a Cl is energetically favored?

12.15 • Use the ionization energies and electron affinitieslisted in Problem 12.18 to find the critical radius in-side which the transfer of an electron is energeticallyfavored, for each of the ionic molecules KI,NaBr, LiF.

12.16 • The ionization energy of potassium is 4.34 eV,while the electron affinity of chlorine is 3.62 eV.Given that the equilibrium separation of KCl is0.267 nm, estimate the dissociation energy of KClusing Eq. (12.10). Compare your answer with theobserved value

12.17 • The ionization energy of Li is 5.4 eV, and the elec-tron affinity of chlorine is 3.6 eV. Given that LiCl isionically bonded with dissociation energy 4.8 eV, esti-mate the equilibrium separation of LiCl.Will your

answer be an overestimate or underestimate?12.18 • Use the following data to estimate the dissociation

energies of the three ionic molecules KI, NaBr, andLiF. (IE denotes ionization energy and EA electronaffinity.) Compare your answers with the observedvalues given in the last column.Your answers shouldall be overestimates; explain why.

IE (eV) EA (eV) (nm) B (eV)

K: 4.34 I: 3.06 KI: 0.305 3.31

Na: 5.14 Br: 3.36 NaBr: 0.250 3.74

Li: 5.39 F: 3.40 LiF: 0.156 5.91

12.19 • Assuming that the following pairs of elements com-bine ionically, write down their chemical formulas: Caand S, Ca and Cl, Li and O, Na and P. [Hint: Use theperiodic table.]

12.20 • Assuming that the following pairs of elements com-bine ionically, write down their chemical formulas: Baand S, Ca and I, Sr and Br, K and N. [Hint: Use theperiodic table.]

R0

R0

B = 4.34 eV.

Rc

Rc

¢E

¢E

B ' ke2>R0 .

-ER . 12.21 •• (a) Given that Ba and S bond ionically, what wouldyou expect their valence to be? (b) Assuming that the

bond is purely ionic, predict the electric dipolemoment of BaS. (The measured bond length is

) (c) Express the observed dipole

moment, as a percentage of youranswer in part (b).

SECTION 12.4 (The Covalent Bond)

12.22 • A crucial fact for the discussion of Section 12.4 iscontained in Eq. (12.14), namely that if and sat-isfy the Schrödinger equation,then so does any linearcombination Starting with the standard

three-dimensional Schrödinger equation (8.2), provethis result.

12.23 • Assuming that they bond covalently as described inSection 12.4, predict the chemical formulas of oxygenfluoride and nitrogen fluoride.

12.24 • Hydrogen can bond covalently with phosphorus,tellurium, and bromine. Predict the chemical formu-las of the three resulting molecules.

12.25 • (a) Octane, is called a straight chain hydro-

carbon because its carbon atoms are arranged in astraight line. Draw a picture of the octane moleculeshowing all bonds, as in Fig. 12.13. (b) Do the samefor the straight-chain propane, (In both casesall carbon atoms have valence 4.)

12.26 • Make a sketch similar to those in Fig. 12.13 of ethane (in which both carbons have valence 4).

12.27 • Make a sketch similar to those in Fig.12.13 of acety-lene (in which both carbons have valence 4).[Hint: Remember that more than one bond can jointhe two carbon atoms.)

12.28 • The observed electric dipole moments of CO,ClF, and NaF are and

respectively. The correspondingbond lengths are 0.11, 0.16, and 0.19 nm. Express theobserved dipole moments as percentages of the val-ues you would predict if the molecules were purelyionic. Which of the molecules are predominantlyionic, and which are predominantly covalent?

12.29 •• (a) Assuming that the following pairs of elementscombine covalently, predict the formulas of the re-

sulting molecules.

(nm)

Cl and F 0.16

Br and Cl 0.21

I and Cl 0.23

(b) Use the observed dipole moments and bondlengths to confirm that these molecules are predomi-nantly covalent.

4.1 * 10-30

1.9 * 10-30

3.0 * 10-30

R0 p(C # m)

2.7 * 10-29 C # m,

3.0 * 10-30,3.7 * 10-31,

C2H2

C2H6

C3H8 .

C8H18 ,

Bc1 + Cc2 .

c2c1

3.62 * 10-29 C # m,

R0 = 0.251 nm.

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2q

q

q

O

H

H

p2

p1

d

FIGURE 12.35

(Problem 12.32)

12.30•• (a) What should be the valence of Be in itsground state (b) As discussed in Section

10.6, the and levels are close together, andwhen another atom is nearby, it is often energeti-cally favorable to promote one of the electronsto a state, so that a bond can form. What is thevalence of Be in the configuration(c) Predict the chemical formulas for the com-pounds of Be with fluorine, with oxygen, and withnitrogen.

12.31 ••• Consider the wave functions discussed inSection 12.4. In that discussion we did not worry aboutnormalization,but should strictly have been defined

as and where Band C are normalization constants needed to ensurethat (a) If and do not overlap (or,more precisely, if their overlap is negligible), show that

(Assume that and are them-selves normalized.) (b) If and overlap a little,

argue that B is a little less than and hence that atthe midpoint between the two protons, is just a lit-tle less than This proves our claim that con-centrates the probability density between the twoprotons. (c)Argue similarly that C must be a little larger

than

SECTION 12.5 (Directional Properties of Covalent Bonds★)

12.32 •• The water molecule is partially ionic, in that anelectron is partially transferred from each hydrogento the oxygen, as indicated in Fig. 12.35, where q de-notes the magnitude of each of the two chargestransferred. (a) Write down the electric dipole mo-ment p of the molecule in terms of the chargeq, the bond length d, and the angle (Note:The dipole moment of two charges q and a dis-

tance d apart is a vector of magnitude pointingfrom the negative to the positive charge. Thus,has two dipoles and as shown, and the totalmoment is ) (b) The measured values are

andFind the magnitude q of the charge transferred andexpress it as a fraction of the electron charge e.

u = 104.5°p = 6.46 * 10-30 C # m, d = 0.0956 nm,

p = p1 + p2 .p2p1

H2Oqd

-qu.H ¬ O

H2O

1

>1 2.

c+2 ƒc1 ƒ 2.ƒc+ ƒ 2

1>1 2c2c1

c2c1B = C = 1>1 2.

c2c11

ƒc ƒ 2 dV = 1.

c-=

C1c1-c22,c+

=B1c1

+c22c;

c;H2+

1s2 2s1

2p1 ?

2p2s

2p2s 11s2

2s2

2?

12.33 ••• We mentioned in Section 12.5 that the four bondsof a carbon atom (in a molecule like ) point sym-

metrically to the four comers of a regular tetrahedroncentered on the carbon nucleus. Prove that the anglebetween any two bonds is 109.5°.

SECTION 12.6 (Excited States of Molecules★)

12.34 • The force constant for the HI molecule isCalculate the vibrational en-

ergy of the lowest three vibrational levels of HI.(Treat the iodine atom as fixed.)

12.35 • The separation of the lowest two vibrational levels

in HBr is 0.33 eV. Find the force constant k of HBr.(Treat the bromine atom as fixed.)

12.36 • Calculate the lowest four rotational energies of theHI molecule in eV, and sketch them in an energy leveldiagram. (The bond length of HI is 0.16 nm, and youcan treat the iodine atom as stationary.)

12.37 •• Figure 12.6 shows the energy for theand ions that make up the NaCl molecule. Nearits minimum at can be approximated by aparabola: Use theinformation on the graph to make a rough estimate of the force constant k. [Hint: To the left of thegraph retains a reasonably parabolic shape as far asthe point where it crosses the horizontal axis.]

12.38 •• Consider an molecule rotating about its centerof mass (as in Fig. 12.26). (a) Derive an expression forits moment of inertia I in terms of m, the mass of an Oatom, and the bond length. (b) Use Eq. (12.27) tofind the lowest three rotational energy levels in eV.(The bond length of is 0.121 nm.)

12.39 •• Consider two atoms of masses m and bound a

distance apart and rotating about their center of mass (as in Fig. 12.26). (a) Calculate their moment of inertia, I , and prove that it can be written aswhere is the reduced mass

(12.41)

This shows that one can treat the rotational motion of any diatomic molecule as if only one of the atomswere moving,provided that one takes its mass to be(b) Prove that if then

12.40 •• Consider two classical atoms of mass m andseparated by a distance R, with their center of massfixed at the origin. (a) Write down expressions forthe atoms’ distances, r and from the origin, interms of and R. (b) Suppose that the twoatoms vibrate in and out along the internuclear axis.Show that their total kinetic energy can be expressedas where is the reduced mass de-fined in Eq. (12.41) (Problem 12.39). This resultproves for a classical molecule (what is true in quan-tum mechanics as well) that the vibrational motionof both atoms can be treated as if only one atomwere moving, provided that we take its mass to be m.

mK =12 m1dR>dt 22,

m, m¿,r¿,

m¿,

m L m.m V m¿,m.

m =mm¿

m + m¿

mI = mR0

2 ,

R0

m¿,

O2

R0 ,

O2

R0 ,

E

1R

2L E

1R0

2+

12

k

1R - R0

22.

R0 , E1R2Cl-

Na+E1R2

k = 2.0 * 103 eV>nm2.

CH4

407

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0 1 2 3

(cm)

I n t e n s i t y

FIGURE 12.36

(Problem 12.51)

0.27 0.28

E (eV)

FIGURE 12.37

(Problem 12.52)

12.41 •• Consider the molecules HCl and DCl, where D de-notes the deuterium, or heavy hydrogen, atom,

(a) Explain why the force constant k andbond length should be about the same for DCl asfor HCl. (b) The spacing of adjacent vibrational levelsin HCl is What is the correspondingspacing in DCl? (You may treat the Cl atom as fixed.)

12.42 •• The potential energy of two atoms that form a sta-ble molecule has a minimum at a separation asshown in Fig. 12.23. Consider two atoms with total ki-netic energy approaching one another from farapart. (a) What will be their KE when they reach theequilibrium separation (b) At what point in

the picture will the atoms come to rest? Will they re-main there? (c) Assuming that the total energy re-mains entirely in the form of the twoatoms, can the two atoms come to rest at Whatvalue must R eventually approach? (d) We see that if the atoms are to form a stable molecule, some of theirenergy must be dissipated while they are nearSuggest a mechanism to do this.

12.43 ••• (a) Use the curve in Fig. 12.12 to get a rough esti-mate of the force constant k for the molecule.(See the hint for Problem 12.37.) (b) Use your valueof k to estimate the zero-point energy of vibration in

(Since the two nuclei have equal mass, you willneed to use the reduced mass as in Problem 12.40.)

12.44 ••• A simple analytic function that can be used to ap-proximate the potential energy of the two atoms in adiatomic molecule is the Morse potential

(12.42)

where and S are positive constants, with(a) Sketch this function for

(b) In terms of and S, write down the bondlength and binding energy of a molecule whose po-

tential energy is given by (12.42). (c) Write down thefirst three terms of the Taylor series for about(see Appendix B) and show that they give the par-

abolic approximation (12.25). (d) Show that the forceconstant k is

SECTION 12.7 (Molecular Spectra★)

12.45 • The microwave spectrum of HCl consists of a seriesof equally spaced lines, apart. Find thebond length of HCl (treating the Cl atom as fixed).

12.46 • The microwave spectrum of HBr consists of a series

of equally spaced lines, apart. Find thebond length of HBr (treating the Br atom as fixed).

12.47 • The vibrational–rotational spectrum (12.35) of a di-atomic molecule is in the infrared region. At low res-olution it is seen as a single band centered on

The molecule LiBr shows a bright band inthe IR region, centered at Use this tofind the force constant k for LiBr (treating the Bratom as fixed).

12.48 •• A spectroscopist finds three lines in the far infraredspectrum of HI at 259, 195, and (and no other

lines in between). (a) Find the photon energies, and

156 mm

l = 17.8 mm.Eg = Uvc .

5.1 * 1011 Hz

6.4 * 1011 Hz

2A>S2.

R0

U1R2A, R0 ,

0 … R 6 q .R0 W S.A, R0 ,

U1R2 = A31e1R0 -R2>S- 122

- 14H2+.

H2+

R0 .

R0 ?KE + PE

R = R0 ?

K0

R0 ,

Uvc L 0.37 eV.

R0D = 2H.

identify the rotational transitions involved. (b) Hence,find the bond length of HI.

12.49 •• Repeat the calculation in Example 12.5 (Section12.7) of the bond length of LiI, but do not make theapproximation that the I atom is fixed. (Use the re-duced mass introduced in Problem 12.39.) Compareyour answer with that of Example 12.5, where we ig-nored the motion of the iodine atom.

12.50 •• A spectroscopist finds three lines in the micro-wave spectrum of CO at 1.29, and 0.86 mm(and no other lines in between). (a) Find the corre-sponding photon energies,and identify the transitions

involved. (b) Using the reduced mass of Problem12.39, find the bond length of CO.

12.51 •• The first four lines in the rotational spectrum of CaCl are shown schematically in Fig. 12.36. (Therelative strengths of the four lines depend on thetemperature and need not concern you here.) Usethis information to find the bond length of CaCl.(Since the two masses are comparable, you willneed to use the reduced mass introduced in Prob-lem 12.39.Assume that the Cl is chlorine 35.)

l = 2.59,

12.52 •• The vibrational–rotational spectrum of HI issketched in Fig. 12.37. Use the information in that fig-ure to find the force constant and bond length of HI.

12.53 ••• (a) Consider an HCl molecule in which the Clatom is Allowing for the motion of both atoms,find the wavelengths of the first two lines in the rota-tional spectrum. (Use the reduced mass introduced inProblem 12.39. Take and keep foursignificant figures.) (b) Do the same for the case thatthe Cl atom is (Use the same value for )(c) Make a sketch similar to Fig. 12.36 of these twolines in the emission spectrum of natural HCl, which

contains 3 parts to 1 part 37Cl.35Cl

R0 .37Cl.

R0 = 0.1270 nm,

35Cl.

408 Ch 12 M l l

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COMPUTER PROBLEMS

12.54 •• (Section 12.2) If you haven’t already done Problem12.11,do so now,and then make a plot of the potentialenergy as a function of s. (For the purposes of theplot,you may as well choose units such that andare equal to 1. Choose the ranges of U and s to showthe interesting behavior.) Does your graph confirm theresults of Problem 12.11?

12.55 •• (Section 12.6) The potential energy of the twoatoms in HCl can be approximated by the Morsepotential (12.42) (Problem 12.44) with

(a) Plot the potential energy for(Restrict the vertical axis to the

range ) (b) What are the bondlength and binding energy for HCl? (You may ignorethe zero-point vibration.) (c) What is the force con-stant k for HCl? [See Problem 12.44(d).] (d) Plot theSHO well (12.25) that approximates nearSuperpose this plot on your previous one.

R0 .U1R2-5 … U … 5 eV.

0 … R … 0.3 nm.U1R2

A = 4.57 eV, R0

= 0.127 nm, S = 0.0549 nm

R0ke2U1s2

12.56 ••• (Section 12.4) As a simple model of the ion,consider the following one-dimensional system: An

electron moves on the x axis, in the field of two iden-tical nuclei at Suppose that the wave func-tions for the electron bound to one nucleus, with the

other nucleus removed, are and

(a) Show that the normalization

constant A is and plot the two wave func-tions for The needed integrals are inAppendix B. (b) Now consider the bonding and anti-bonding wave functions and

and show that the normalization

constants and are equal to

respectively. (c) Plot the wave functions and forthe case that (with the nuclei reasonably far

apart). Repeat for the probability densities and

(d) Repeat all of part (c) but with the nucleicloser together at Comment on your results,with particular reference to Figs. 12.9 and 12.10.

d = 1.ƒc- ƒ 2.

ƒc+ ƒ 2d = 3

c-c+

1

>4 2 ; 2e-2d2

,B-B+

c- = B-

1c1 - c2

2,

c+ = B+1c1 + c22d = 3.12>p21>4,c2 = Ae-1x-d22.

c1 = Ae-1x+d22

x = ;d.

H2+

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