Kelter P., Mosher M., Scott A. Chemistry. The Practical Science

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LUM

ner

FIGURE 9.23

MO diagram for H

. This diagram graph-

ically illustrates the energies and the

resulting orbital shapes of the different

molecular orbitals in the molecule.

FIGURE 9.24

Dilithium (Li

) MO diagram. The MO con-

ﬁguration can also be written in short-

hand format as σ

illustrate the energy of each of the bonding and antibonding orbitals and assist in

determination of the bond order for a molecule. Graphically, the shapes of each

of the molecular orbitals can be added to the diagram to get the overall picture of

the molecule.

We can draw the blueprint—the MO diagram—for molecular hydrogen (H

)

to illustrate how this is done. We show in Figure 9.23 that as the two hydrogen

atoms approach each other, their atomic 1s orbitals mix to provide two new mol-

ecular orbitals. The overlap of these orbitals possesses symmetry, which means

that they are equal in size and shape about the axis between the atoms (σ). One of

the molecular orbitals resulting from this overlap is lower in energy (the bonding

orbital, σ

, read as “sigma one s”) and the other is higher in energy (the anti-

bonding orbital, σ

∗

, read as “sigma one s star”). Placement of the two electrons

in H

follows the Pauli exclusion principle (two electrons with opposite spin per

orbital) and Hund’s rule (ﬁll lowest-energy orbitals ﬁrst) to ﬁll the bonding

orbital. The antibonding molecular orbital is left empty. The

highest-energy occu-

pied molecular orbital (HOMO)

is the molecular orbital that contains electrons (in

this case, the bonding orbital). The antibonding orbital ends up as the

lowest-

energy unoccupied molecular orbital (LUMO)

. The bond order for hydrogen can be

calculated ((2 bonding electrons – 0 antibonding electrons)兾2) as 1. We therefore

predict that molecular hydrogen has a single bond between the two nuclei.

What would happen to the bond order for this molecule if one of the electrons

were promoted to the LUMO?

The bond order for such a molecule would be zero

((1 bonding electron – 1 antibonding electron)兾2). With no net bond between

the two atoms, the molecule would break into the individual hydrogen atoms.

EXERCISE 9.8 MO Theory: More Power, Scotty

We showed in Exercise 9.2 that dilithium (of Star Trek fame) can be explained by the

valence bond model. What does MO theory say about dilthium (Li

Solution

The MO diagram for dilithium is shown in Figure 9.24. The HOMO is the σ

orbital and the LUMO is the σ

∗

orbital. The bond order for this molecule is

1 ((4 – 2)兾2), so MO theory suggests that Li

should exist as a molecule with one

bond between the two lithium atoms. The molecule does exist, but it quickly reacts

with other lithium atoms to make lithium metal. Because of this reactivity, it is not

possible to have a bottle of Li

PRACTICE 9.8

Use MO theory to determine whether He

is a stable molecule. What is the bond

order in this molecule?

See Problems 67, 68, and 71–74.

More About MO Diagrams

Lewis dot structures and valence bond theory agree that there are six bonding

electrons (that is, a triple bond) between the nitrogen atoms in gaseous nitrogen

). The MO diagram for the molecule should verify that the theory also gives

rise to a bond order of 3. In molecules containing more than just the s orbitals,

the MO diagram becomes much more complicated. Figure 9.25 shows that the

overlap of the 2p

orbitals on each N and the 2p

orbitals on each N, produces

four orbitals: two π and two π

∗

. The overlap of the 2p

orbitals gives a sigma mol-

ecular orbital. If we now place the electrons in the diagram, we see that they ﬁll

four bonding orbitals and one antibonding orbital. From the MO diagram, we

378 Chapter 9 Advanced Models of Bonding

Visualization: Sigma Bonding

and Antibonding Orbitals

Video Lesson: Applications of

the Molecular Orbital Theory

σ∗

π∗

2p2p

σ∗

2s 2s

σ∗

2s 2s

σ∗

π∗

2p2p

can calculate a bond order of 3 for molecular nitrogen, in

which the HOMO is the σ

orbital and there are two

LUMOs (the degenerate π

∗

orbitals).

The energy of the molecular orbitals in N

results in

the ordering shown in Figure 9.25. This is the same order-

ing that is observed in most of the diatomic molecules of

the second period of the periodic table. However, a differ-

ent ordering of the molecular orbitals is seen in O

and Ne

, as shown in Figure 9.26. The change in the order

of the molecular orbitals arises because the difference in

energy between the 2s and 2p atomic orbitals on O, F, and

Ne is relatively greater than the other elements in the sec-

ond row (see Figure 9.7). Note that the two electrons in

the HOMO are not paired in O

. This allows us to ratio-

nalize a property of molecular oxygen that is unexplain-

able by other bonding theories. The other models of

molecular oxygen correctly assign two bonds between the oxygen atoms, but

oxygen possesses properties that we would not predict on the basis of those

models. Oxygen, shown in Figure 9.27, is an example of a paramagnetic mole-

cule. The term

paramagnetism refers to the ability of a substance to be attracted

into a magnetic ﬁeld. This attraction arises because of the presence of unpaired

electrons within the molecule. A special case of this is

ferromagnetism (so

named because the effect is especially strong in iron), in which the paramag-

netic atoms are close enough together that they reinforce their attraction to the

9.3 Molecular Orbital Theory 379

FIGURE 9.25

MO diagram for N

FIGURE 9.26

MO diagram for O

. Note that molecular oxygen

has a different order for its molecular orbitals.

There are also two unpaired electrons in the MO

diagram. What is the bond order for O

Visualization: Molecular Orbital

Diagram (N

)

Liquid oxygen Liquid nitrogen

NS NS

(a) (b)

Energy

F–F FF

FIGURE 9.28

MO diagram of F

. The promotion of an electron from a π

MO to the σ

∗

MO reduces the bond order from 1 to 0. When

this happens, there is no net bond remaining between the

atoms. Addition of a photon of light to the molecule with a

wavelength of 754 nm causes the molecule to break in two.

magnetic ﬁeld, so the whole is, in effect, greater than the sum of its parts. The

opposite of paramagnetism is

diamagnetism. A diamagnetic molecule’s electrons

are paired, resulting in the molecule being repelled from a magnetic ﬁeld. Ni-

trogen (N

) is an example of a diamagnetic molecule because all of its electrons

are paired (see Figure 9.25).

Constructing models of molecules made with more than two atoms or

molecules made from two different atoms (heteronuclear diatomic molecules)

complicates the molecular orbital diagram because similar atomic orbitals on

each atom have different energy levels. A further complication is added because

the orbitals encompass the entire molecule and therefore must be constructed

from the linear combination of the atomic orbitals on every atom in the mole-

cule. For a 10-atom molecule, that would mean that each molecular orbital

would arise from the combination of 10 different atomic orbitals. Such a diagram

would be very complex indeed!

EXERCISE 9.9 MO Theory: The Power of a Photon

The halogens (F

,Cl

,Br

, and so on) are extremely reactive compounds. In fact,

their reaction can be initiated by a single photon of light, as we note at the start of

this chapter when we discussed 11-cis-retinal in the eye. Given that the photon will

promote one electron from a bonding orbital to an antibonding orbital of similar

symmetry, illustrate the reaction of F

with that photon using MO diagrams. The

MO diagram for F

is similar to that for O

Solution

The MO diagram for F

is shown in Figure 9.28. Promotion of an electron from the

orbital to an MO of the same symmetry, the σ

∗

orbital, would cause the bond

order for F

to become zero. The photon would effectively break the bond between

the ﬂuorine atoms. The reaction of photons with chloroﬂuorocarbons in the ozone

layer similarly breaks COCl bonds. The resulting chlorine atoms are extremely

reactive toward ozone (O

380 Chapter 9 Advanced Models of Bonding

FIGURE 9.27

The results of paramagnetism and dia-

magnetism. Molecular oxygen is para-

magnetic. Because of this, liquid oxygen

is attracted to a magnetic ﬁeld (a). Liquid

nitrogen is diamagnetic and is repelled

from the magnetic ﬁeld (b).

Visualization: Magnetic

Properties of Liquid Nitrogen

and Oxygen

Video Lesson: CIA

Demonstration: The

Paramagnetism of Oxygen

FIGURE 9.29

The MOs of conjugated systems. As the length of the conjugated π system increases, the energy

required to promote an electron decreases. Evidence of this is observed in the wavelength of

light absorbed by the molecules. Longer

π systems require less energy and a larger wavelength

(energy and wavelength are inversely proportional, E

PRACTICE 9.9

Illustrate the interaction of H

and O

with a single photon of light. After the pro-

motion of an electron, which molecule has the larger bond order?

See Problems 65 and 66.

Looking back to 11-cis-retinal, we can produce an MO diagram that

illustrates the location of the electrons in the conjugated π system. Although the

diagram is unusually complex, the energy levels of the conjugated π system are

more closely spaced as the π system increases in length. Promotion of an elec-

tron requires less and less energy as the π system lengthens, as indicated by the in-

creasing wavelength of absorbed light in the molecules of Figure 9.29. This occurs

because it takes less energy to promote an electron to a closer molecular orbital.

We pointed out earlier that this phenomenon is used by the photographic indus-

try in the development of color print ﬁlm. Here’s how. By constructing a

molecule that has a conjugated π system of just the right length, photo manufac-

turers can make molecules that absorb exactly the color needed to make your

pictures really stand out. We develop this idea further in the accompanying

NanoWorld/MacroWorld feature.

9.3 Molecular Orbital Theory 381

1,3-Butadiene

λ = 217 nm

1,3,5,7-Octatetraene

λ = 304 nm

cis-Retinal

λ = 325 nm

1,3,5-Hexatriene

λ = 268 nm

382 Chapter 9 Advanced Models of Bonding

“Smile, you’re on candid camera!” Everyone hopes never

to be ambushed with those words. In the nineteenth cen-

tury, when photography was in its infancy, candid pho-

tographs were not possible. In order to create a visual

memory of an event, you had to visit the photographer

and sit for a picture.“Sitting” meant holding a pose for

up to 30 seconds. Any movement would show as a blur

in the ﬁnal picture. However, the idea that an inexpen-

sive, yet realistic, picture could be created in such a short

time brought people to the photographer’s studio in

droves.

Scientists in the early 1800s were intrigued by the

idea of painting portraits without the artist’s pen. In the

summer of 1827, Joseph Nicéphore Niépce, using mater-

ial that hardened on exposure to light exposed ﬁlm for

eight hours in order to get a picture. Because no one

wanted to sit still that long, most of his pictures were

landscapes. In 1839, however, two other photographic

processes were developed, the daguerreotype (perfected

by Louis Daguerre) and the calotype (Henry Fox Talbot).

Although these techniques allowed pictures to be created

with much shorter exposures, they still produced a

black-and-white image (see Figure 9.30). This didn’t

matter much to the general public, and photographic

portrait studios seemed to pop up on every corner across

the globe. Despite some marginal success in producing

faint color images during the 1850s and 1890s, the ad-

vent of the color photograph as an inexpensive, repro-

ducible and stable process didn’t occur until the 1920s.

Researchers at Kodak Research Laboratories in 1935

ﬁnally hit upon the process that would bring the color

photograph to the world. Kodachrome®, their color

photography process, introduces color to a picture

by separating the three primary colors into speciﬁc

emulsion layers. How does it work? The emulsion layers

contain a compound that reacts when light strikes it. A

quick look at the chemistry behind the black-and-white

photograph will give us some insight.

In black-and-white photography, a piece of transpar-

ent plastic is coated with silver bromide crystals (see Fig-

ure 9.31) and a sensitizer. The plastic sheet is then placed

in a camera, and when the shutter is opened, the silver

bromide absorbs photons of light. In the presence of the

sensitizer, the silver cation is reduced by the sensitizer

(gains an electron) and becomes silver metal:

(s) + sensitizer n Ag°(s) + sensitizer

Then the ﬁlm is wound back into its container and

sent to be developed. The technician rinses the plastic

ﬁlm to remove unreacted silver bromide. The silver

metal remains in place on the plastic. The result is an in-

verted image called a negative. Next the technician shines

light through the negative onto another silver bromide/

sensitizer-coated support (typically paper this time) and

rinses off the unreacted silver bromide. Violà! We have a

positive image called a photograph.

Why is the ﬁlm coated with silver bromide instead of

silver chloride? We can answer this by examining the

wavelength of light that is absorbed; see Figure 9.32. Com-

parison of the two shows that silver bromide absorbs a

large amount of the visible light that enters our camera.

To get the best picture, we want to absorb as much light

as we can.

NanoWorld / MacroWorld

Big effects of the very small:

Color photography and the tricolor process

FIGURE 9.30

A daguerreotype of Michael Faraday

(see Chapter 19) taken sometime

between 1844 and 1860 at Mathew

Brady’s studio in New York City.

Because the process required the

subject to remain completely

motionless for up to 10 seconds,

most daguerreotypes were taken

in well-lit studios. Even though

the early daguerreotypes cost one

month’s wages for the average

person, they were much cheaper

than a painted portrait. This made

them an instant success.

FIGURE 9.31

A magniﬁed image of silver bromide crystals used in the photographic

industry. Note that the crystals are mostly hexagonal in shape and ﬂat

so that more surface area is exposed to incoming photons.

9.3 Molecular Orbital Theory 383

Silver bromide, though, doesn’t absorb the entire

spectrum of visible light. What color of light does silver

bromide absorb? Figure 9.32 shows that it absorbs blue

light. In order for us to make a full-color photograph, we

need to make some of the silver bromide on the plastic

absorb yellow light and some absorb red light. By mixing

these primary colors (blue, yellow, and red), we can recre-

ate the entire spectrum of colors. This is what happens in

the tricolor ﬁlm process. How, then, can we make silver

bromide absorb a different color of light?

The process requires coupling the silver bromide

with a dye molecule. These molecules are designed to ab-

sorb different colors by varying the length of their conju-

gated pi bonds. When the photon is absorbed into this

system, an electron in a π MO is promoted to a π

∗

MO,

and the dye molecule becomes excited. As with all

excited molecules, many fates exist for the pro-

moted electron. First, the molecule could release

a photon and return the promoted electron to

the

π MO, which would result in the original dye

molecule. The released photon would have the

same energy as the absorbed photon. Second,

the electron could collapse back to the π MO

through thermal excitation. In this process, the

dye molecule would release the energy by vibrat-

ing. Third, the electron could transfer to another

MO on another molecule. This is the fate that

gives rise to the photographic reaction.

The excited electron transfers from the dye

molecule to an antibonding orbital on the silver

bromide. And as it does so, our silver cation is re-

duced to silver metal. Even though the silver bro-

mide was unable to absorb the non-blue photon

of light, it still ended up as silver metal. Therefore,

all we have to do to make a color photograph is

ﬁnd two dye molecules that can transfer electrons

to the silver bromide. One should absorb yellow

light and the other should absorb red light. The exact

structures of these molecules are a closely guarded secret

in the photography industry, but the compounds shown

in Figure 9.33 have been used to do this in the past. Each

has an extended pi system that allows the molecules to

absorb light in the visible region of the spectrum.

Current advances in the photography industry in-

clude the development of four-color processes. These

processes make possible better deﬁnition of objects and

a more brilliant picture. Development of better ways

to coat the silver bromide evenly on plastic and the

preparation of silver bromide crystals that give the best

interaction with light are at the forefront of current re-

search. One day these developments may be able to com-

pletely eliminate “red eye” and make it easier to tolerate

being the subject of a candid shot.

Wavelength (nm)

Absorption coefficient (cm

–1

)

–2

260220 300 340 380 420 460 500 540 580

2.1 eV2.52.93.33.7

–1

AgCl

AgBr

FIGURE 9.32

Spectrum of silver bromide and silver chloride. Silver bromide absorbs more

of the visible spectrum than silver chloride. What color is silver bromide?

Basic Red 9, a magenta dye

NNH





A cyan dye





FIGURE 9.33

Dyes used in the photographic industry. Note the extended pi

bond networks in these molecules. The length of the conjuga-

tion is related to the wavelength of light that these molecules

can absorb.

OOH

Purpurin, a magenta dye

9.4 Putting It All Together

The models of each of the bonding theories provide increasingly better cor-

relation with the observed properties of many covalent compounds. Valence

bond theory works well at showing how the electrons in sigma bonds come to-

gether to create the skeletal framework of a molecule. At the same time, the local-

ized bonding picture described by valence bond theory conveniently determines,

among other things, the lengths of bonds in a molecule. The rules of VSEPR

modeling help identify the three-dimensional shape of the molecule, and Lewis

dot structures rapidly reveal the connectivity of atoms in a molecule. However,

molecular orbital theory works best at describing the behavior of electrons lo-

cated in the bonds.An approach to drawing a molecule that uses all of these mod-

els is often followed. We can show this approach using benzene (C

Benzene is a carcinogenic compound that is widely used in industry as a non-

polar solvent and as a starting compound in the manufacture of other useful

products such as phenol (an antiseptic compound) and nylon. Experiments have

revealed that benzene is a hexagon of six carbon atoms, each bonded to one hy-

drogen atom. It has also been observed experimentally that the distances between

carbon atoms in the molecule are identical. In the laboratory, benzene reacts as

though it has CPC bonds at each position in the molecule.

No single Lewis dot structure of benzene can be drawn to illustrate the exper-

imentally determined bond lengths and reactivity in the molecule. Considering

that benzene is best shown as a hybrid of two resonance structures, we can ad-

dress the experimental observation that all COC bonds in benzene are the same

length as shown in Figure 9.34. Valence bond theory can

also be used to indicate that each carbon is sp

hy-

bridized. The sigma bond network that makes up the

molecule deﬁnes 120° bond angles at each ﬂat sp

hy-

bridized carbon. Bonds between carbon atoms result

from the overlap of a 2sp

hybrid orbital with another

2sp

hybrid orbital. Each COC bond is 140 pm in length.

Bonds between the carbon and hydrogen atoms are the

result of 2sp

–1s orbital overlap. In addition, each carbon

atom contains a half-ﬁlled 2p orbital perpendicular to the

plane of the molecule.

However, the six half-ﬁlled 2p orbitals in benzene are

best handled by molecular orbital theory. The σ bond

network is separated from the π bond network in this ap-

proach. Doing so makes preparing a MO diagram a more

manageable task. The six 2p orbitals on the carbon atoms

in benzene have the same symmetry and energy, so they

can be mixed to provide six new π MOs. Three bonding

MOs and three antibonding MOs result (see Figure 9.35). We place the six elec-

trons into the new MOs, ﬁlling the three bonding MOs. The electrons in the MO

diagram occupy a π bond at every carbon atom, implying that there isn’t any dif-

ference in the carbons of benzene. Each COC bond has a bond order of 1.5. The

approach we have taken to draw the best model of benzene has provided us with

the best (and quickest) way to represent the experimentally determined proper-

ties of benzene (such as the COC bond lengths and the reactivity of benzene). We

no longer have to imagine different structures (arrived at from different models)

of benzene to account for the observed properties.

If we more closely examine the molecular orbital model of benzene, we ﬁnd

that the electrons in the π orbitals are spread out over the six atoms in the struc-

ture. This conjugated π system is said to be

delocalized. In fact, delocalized sys-

tems occur any time the electron density in a molecule can be distributed among

384 Chapter 9 Advanced Models of Bonding

Application

HEMICAL

ENCOUNTERS:

Benzene, Stability,

and MO Theory

Resonance hybrid

FIGURE 9.34

Delocalized π bonding in benzene.

Lewis dot structures and valence bond

models do a good job of describing

the sigma bond network in benzene.

FIGURE 9.35

Delocalized π bonding in benzene accord-

ing to MO theory.

Visualization: Delocalized

Pi Bonding in the Nitrate Ion

more than two atoms. Delocalization also can be seen in the carbonate ion

(CO

2–

) and in 11-cis-retinal. However, in the case of benzene (C

) and related

compounds, the delocalization of the electron density imparts a particular stabil-

ity to the molecule (more so than the delocalization of electrons in the carbonate

ion and 11-cis-retinal). The stability is so important to the molecule that it is

often written as shown in Figure 9.36, with a circle to illustrate the delocalization

of three π bonds in the molecule. Benzene’s stability makes it a good choice for

use as a reaction solvent. Unfortunately, this stability also makes it difﬁcult to me-

tabolize if accidentally ingested. One of the major problems with using benzene

is that it causes cancer in humans. Table 9.6 lists certain properties of benzene

and of some alternative solvents that have been used in place of it. Their proper-

ties aren’t identical to those of benzene, but their substitution for benzene has

done a great deal to safeguard the health and lives of industrial chemists.

9.4 Putting It All Together 385

π∗

Energy

Molecular orbital diagram for benzene.

FIGURE 9.36

Shorthand notations for benzene.

Common Solvents

Solubility

Solvent Structure Boiling Point Dipole Moment in Water

Benzene C

80.1°C 0 Very slight

Hexane C

68.7°C 0 Insoluble

Dichloromethane CH

40°C 1.60 Slight

Te t r a h y d r o f u r a n C

O 66°C 1.63 Miscible

To l u e n e C

111°C 0.36 Very slight

TABLE 9.6

Application

386 Chapter 9 Advanced Models of Bonding

9.5 Molecular Models in the Chemist’s Toolbox

Just as the best models in civil engineering give the best information on the out-

come of a construction project, our happiness with a particular molecular model

is based on our satisfaction with the information that the model provides. The

more rigorous models provide better information, but at a cost in time and

calculations that we may not be willing to accept. Valence bond theory quickly

determines the skeletal framework of a molecule by considering separate parts of

the molecule. It does not explain the molecule as a whole, nor does it consider

how electrons can be delocalized between more than two atoms. Molecular or-

bital theory, on the other hand, considers the molecule as a whole. It explains the

bonding patterns in detail, providing information about the energy of the elec-

trons in a particular orbital and the properties of the molecule. However, detailed

molecular orbital theory is difﬁcult to solve mathematically. Using approxima-

tions of molecular orbital theory, such as the LCAO–MO model, considerably

reduces the amount of time involved in determining the model, but at the ex-

pense of some of the detail provided by the theory. All of the models we have

discussed—Lewis dot structures, VSEPR, valence bond theory, and molecular

orbital theory—have a place in the toolbox of the chemist.

The Bottom Line

■

The molecular models that chemists construct

increase in complexity from Lewis dot structures,

to the VSEPR model, to valence bond theory, to

MO theory. This increase in complexity is accom-

panied by an increase in satisfactory agreement

with observed properties for the molecule.

(Section 9.1)

■

Valence bond theory, originated by Linus Pauling,

deﬁnes bonds as the overlap of atomic orbitals.

Sigma bonds result from end-on overlap of orbitals;

pi bonds result from side-to-side overlap of orbi-

tals. (Section 9.1)

■

Hybridization of atomic orbitals gives rise to new

orbitals that help explain bonding. Hybridization

also gives structures consistent with the VSEPR

model. (Section 9.2)

■

Mixing n atomic orbitals gives n hybridized orbitals.

(Section 9.2)

■

Hybridization can be used to explain the existence

of sigma (σ) and pi (π) bonds in molecules. (Sec-

tion 9.2)

■

Molecular orbital theory deﬁnes bonding with

orbitals that are not conﬁned to a single atom.

Bonds result from molecular orbitals in a molecule

that encompass the atoms. (Section 9.3)

■

Mixing n atomic orbitals gives n molecular orbitals.

One-half n of these are bonding; the other half are

antibonding. Bonding MOs are lower in energy

than the atomic orbitals from which they are con-

structed. Antibonding MOs are higher in energy

than the atomic orbitals from which they are con-

structed. (Section 9.3)

■

The Pauli exclusion principle and Hund’s rule must

be obeyed when placing electrons in the new MOs.

(Section 9.3)

■

MO diagrams can be used to identify the number

of bonds between atoms. (Section 9.3)

■

Paramagnetism results from unpaired electrons in

a molecule. Diamagnetism results from complete

pairing of all electrons in a molecule. (Section 9.3)

■

Electrons in conjugated π orbitals are said to be

delocalized, because the electron density can be

distributed among more than two atoms.

(Section 9.4)

Focus Your Learning 387

antibonding orbital An orbital that indicates a lack of

electron density between adjacent nuclei (no bond

exists when the orbital is occupied). The antibonding

orbital arises from the subtraction of two overlapping

atomic orbitals. (p. 375)

bond order The number of electrons in bonding orbitals

minus the number of electrons in antibonding or-

bitals, divided by 2. The bond order indicates the de-

gree of constructive overlap between two atoms.

(p. 377)

bonding orbital An orbital that indicates the presence

of electron density between adjacent nuclei (a bond

exists when the orbital is occupied). The bonding or-

bital arises from the addition of two overlapping

atomic orbitals. (p. 375)

conjugated π bonds An extended series of alternating

single and double bonds. (p. 374)

conjugation The presence of conjugated π bonds—that

is, a series of at least two double bonds alternating

with single bonds. (p. 374)

delocalized Term used to describe a π system wherein

the electron density in a molecule can be distributed

among more than two atoms. (p. 384)

diamagnetism The ability of a substance to be repelled

from a magnetic ﬁeld. This property arises because all

of the electrons in the molecule are paired.

(p. 380)

ferromagnetism A property of a compound that occurs

when paramagnetic atoms are close enough to each

other (such as in iron) that they reinforce their at-

traction to the magnetic ﬁeld, such that the whole is,

in effect, greater than the sum of its parts. (p. 379)

highest-energy occupied molecular orbital (HOMO) The

most energetic molecular orbital that contains at least

one electron. (p. 378)

hybridization The mathematical combination of two or

more orbitals to provide new orbitals of equal energy.

(p. 363)

linear combination of atomic orbitals–molecular orbitals

(LCAO–MO) theory

An approximation of molecular

orbital theory wherein atomic orbitals are added

together (both constructively and destructively) to

make molecular orbitals. (p. 375)

lowest-energy unoccupied molecular orbital (LUMO) The

least energetic molecular orbital that contains no

electrons. (p. 378)

molecular orbital (MO) diagram A diagram used to illus-

trate the different molecular orbitals available on a

molecule. (p. 377)

molecular orbital (MO) theory A theory that mathemati-

cally describes the orbitals on a molecule by treating

each electron as a wave instead of as a particle.

(p. 375)

nodal planes Flat, imaginary planes passing through

bonded atoms where an orbital does not exist.

(p. 370)

paramagnetism The ability of a substance to be attracted

into a magnetic ﬁeld. This attraction arises because of

the presence of unpaired electrons within the mole-

cule. (p. 379)

pi bond (π bond) A covalent bond resulting from side-

to-side overlap of orbitals. The pi bond possesses

a single nodal plane along the axis of the bond.

(p. 370)

sigma bond (σ bond) A covalent bond resulting from

end-on overlap of orbitals. The sigma bond does not

possess a nodal plane along the axis of the bond.

(p. 369)

symmetry The property associated with orbitals that

have similar size and shape. (p. 376)

valence bond theory The theory that all covalent bonds

in a molecule arise from overlap of individual valence

atomic orbitals. A modiﬁcation of this theory allows

valence atomic orbitals to include hybridized orbitals.

(p. 359)

Focus Your Learning

The answers to the odd-numbered problems and some selected

problems appear at the back of the book, as represented by the

blue numbering.

Section 9.1 Valence Bond Theory

Skill Building

1. a. Professional scientiﬁc journal articles, textbooks, and

popular writing have all depicted chemical bonds as a

typed dash between two atomic symbols. Although this

does communicate that the atoms are associated with

each other, cite one reason why it does not accurately

describe a chemical bond.

b. Why do chemists seek to provide better models to describe

the makeup of a chemical bond?

2. The VSEPR model of chemical bonding has been very suc-

cessful for some descriptions of chemical bonding, but what

is the major weakness of this popular model?

3. Write out the ground-state conﬁguration of silicon. Accord-

ing to the VSEPR model, how many bonds should one atom

of Si be able to form? Is your answer consistent with the for-

mula of SiCl

(a compound used in the formulation of some

smoke screens)?

Key Words