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

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9.1 Valence Bond Theory

Just as light can be absorbed by the molecule known as retinal, it can also be

absorbed by the diatomic halogens. Chlorine (Cl

) and ﬂuorine (F

) both absorb

light according to the following reactions:

+ light (h

) n 2 Cl•

+ light (h

) n 2 F•

The resulting very reactive atoms are called radicals (remember these from Chap-

ter 8) because they contain one unpaired electron. It is this reactivity that ac-

counts for chlorine’s effects on the ozone layer and ﬂuorine’s ability to react with

normally unreactive atoms such as xenon.

The Lewis dot structures of the halogens indicate that they contain nonpolar

covalent bonds and similar bond lengths. However, on the basis of the different

wavelengths of light (remember that energy is inversely proportional to wave-

length!) required to break the bond between these atoms (Figure 9.4), we can rea-

son that chlorine and ﬂuorine must have different bond strengths. It appears that

they have some differences in their bonds that the Lewis dot structure model does

not identify.

Knowing the exact makeup of a bond can give us a good picture of its

strength. For instance, we discussed in Chapter 8 how bond strengths can be used

to arrive at a rough approximation of the enthalpy of a reaction. We also noted

that bond energies were averaged to get the values we saw in the table. The fact

that they were averaged implies that not all bonds between the same atoms have the

same energy. We can explore this idea by studying the combustion reactions of

both ethane (CH

) and acetylene (CHqCH). Experimentally, as shown in

Table 9.1, the combustion of acetylene is accompanied by an enthalpy change of

–1300 kJ/mol, compared to the –1559 kJ/mol associated with the combustion of

ethane. Using the bond energy values from Chapter 8, we end up calculating

enthalpy changes that are different than these values. Even though the COC

bond in ethane (376.1 kJ/mol) is much weaker than the CqC bond of acetylene

(962 kJ/mol), this difference doesn’t account for the experimentally determined

difference in the enthalpy of combustion between ethane and acetylene.

Why are

our numbers so different?

Some of the missing energy is accounted for by the dif-

ferences in COH bond energy. Experimentally, researchers have determined that

the energy of the COH bond in ethane (410 kJ/mol) is much smaller than that

of the COH bond in acetylene (536 kJ/mol). However, according to Lewis dot

structures and to the VSEPR models, there should be no difference in energy

between the bonds because they are both COH bonds. We need to construct a

better model of these compounds that takes this difference into account.

358 Chapter 9 Advanced Models of Bonding

F F F

λ  752 nm

Cl Cl Cl

λ  492 nm

FIGURE 9.4

The interaction of light with chlorine

and ﬂuorine imply that these two

molecules possess different bond

strengths.

Calculated Versus Experimental ∆

The experimental and calculated ∆

H values for the fuels listed do not agree. The

temperature of the combustion is listed only for reference.

Flame ∆

H (kJ/mol) ∆

H (kJ/mol)

Compound Temperature (K) Experimental Calculated

Hydrogen (H

) 2490 –242 –242

Methane (CH

) 2285 –890 –803

Ethane (CH

) 2338 –1559 –1429

Ethylene (CH

PCH

) 2643 –1411 –1324

Acetylene (CHqCH) 2859 –1300 –1257

TABLE 9.1

Video Lesson: Valence Bond

Theory

One such “better model” of a covalent bond is the valence bond model. In the

1930s, Linus Pauling (whom we remember from Chapters 7 and 8), devised va-

lence bond theory (VB theory) to address the inadequacy of the bonding models

of G. N. Lewis.

What Is a Valence Bond?

Pauling’s valence bond theory envisions a bond as the overlap of atomic orbitals

on adjacent atoms. Because two electrons interact in a bond, overlapping two

half-ﬁlled orbitals supplies these electrons, which are covalently shared. If all of

the bonds were treated this way, an atom would be able to satisfy the octet (or

duet, for hydrogen) rule.

Molecular hydrogen’s use in industry to make such products as

the hydrogenated vegetable oils used in processed foods and its

possible use as an alternative to gasoline in motor vehicles makes it

a meaningful compound to study. The electron conﬁguration for a

hydrogen atom is 1s

. When the 1s orbital on the hydrogen atoms

overlap, a bond results. Because the electrons are shared by both

orbitals, each of the hydrogen atoms possesses the 1s

electron

conﬁguration, as shown in Figure 9.5, which is a full 1s orbital (the

duet rule is satisﬁed for both hydrogen atoms).

EXERCISE 9.1 Modiﬁcations to the Structure of HF

We noted in Chapter 8 that hydrogen ﬂuoride is often used by master glassworkers

as they etch a design into a piece of art. Some of the world-famous Steuben art has

been made via this etching technique. Using valence bond theory, describe which

orbitals overlap to form a covalent bond between the hydrogen and ﬂuorine atoms

in HF.

First Thoughts

Our ﬁrst thoughts bring to mind the Lewis dot structure of HF. We note that the

structure includes three lone pairs around the ﬂuorine atom and the one bonding

pair of electrons between the two atoms. To construct the valence bond model for

HF, we need to consider the electron conﬁgurations of these atoms. How many

p orbitals are there in the second principal energy level of ﬂuorine? Which type of

orbital on ﬂuorine might overlap with the hydrogen 1s to share a needed electron?

Solution

The conﬁguration of the valence electrons in ﬂuorine is 2s

. Overlap of the

hydrogen 1s orbital with one of the 2p orbitals allows the electrons to be shared

between the two atoms. The valence bond model of HF in Figure 9.6 shows a 1s–2p

orbital overlap that constitutes the covalent bond.

9.1 Valence Bond Theory 359

Application

HH HH

FIGURE 9.5

Overlap of orbitals makes the bond in H

. As two hydrogen

atoms approach, their 1s orbitals overlap to form a bond. The

overlap allows both hydrogen atoms to satisfy the duet rule.

H F H-F

FIGURE 9.6

Orbital overlap in HF.

Further Insights

Does the valence bond model we constructed imply anything about the properties

of the bond? What is the strength of the bond? Does the s–s orbital overlap in H

result in a stronger bond than the s–p orbital overlap in HF? Does the overlap of the

s orbital in the ﬁrst principal energy level and the p orbital in the second principal

energy level indicate anything about the bond? We answer these questions next.

PRACTICE 9.1

Using valence bond theory, describe the orbital overlap in F

, a very reactive mole-

cule used to manufacture Teﬂon ((C

)

) and other ﬂuorinated compounds.

See Problems 2, 9, and 10.

Application of Valence Bond Theory

Because the degree of electron sharing is related to the strength of a bond, orbitals

that exhibit more overlap result in stronger covalent bonds. What determines the

amount of overlap? The keys are the relative energy, as shown in Figure 9.7, and

the size of the atomic orbitals. More speciﬁcally:

■

Smaller orbitals overlap more than larger orbitals.

■

Orbitals with similar sizes overlap more than orbitals with mismatched sizes.

■

Orbitals with similar energies overlap more than orbitals with very different

energies.

The hydrides LiH, NaH, and KH—used as

bases in chemical reactions, in the removal of oxide

coatings on metals, and in processes to make puri-

ﬁed hydrogen gas—make excellent case studies. In

each of these substances, the bonds between adja-

cent atoms result in covalent overlap of a 1s orbital

of the hydrogen atom and the ns valence orbital of

the metal atom. The valence bond in LiH results

from a 2s–1s orbital overlap. The valence bond in

NaH results from the overlap of a 3s (Na) and a 1s

(H) orbital. Similarly, overlap in KH results from a

4s orbital and a 1s orbital. Because the energy and

the size of the metal’s s orbital are dramatically

greater in potassium than in lithium, the overlap of

the potassium 4s and hydrogen 1s orbitals isn’t well

matched (see Figure 9.8). We would therefore pre-

dict the bond in LiH to be much stronger (better

overlap) than the bond in KH.

Table 9.2 lists the bond energies for LiH, NaH,

and KH as 238 kJ/mol, 185.7 kJ/mol, and 174.6 kJ/

mol, respectively. Note in the table the relatively low FOF bond energy. Although

the 2p–2p orbital overlap is expected to be quite good, the electronegativity of

each halogen atom competes with the orbital overlap. The electrons that partici-

pate in the bond between the two ﬂuorine atoms are held more tightly to the

atoms. This results in a decreased electron density between the atoms in F

—and

an unusually low bond energy.

Valence bond theory also addresses any misconceptions reﬂected in the Lewis

dot structure and VSEPR models about the lengths of bonds. There doesn’t

appear to be any difference in the bond length for H

compared to F

if we use

only Lewis dot structures as our model. Experimentally, however, we know that the

bond lengths are different. Which is longer, the bond in the hydrogen molecule or

that in ﬂuorine? The difference in bond lengths results from a difference in the

orbitals that overlap to form the covalent bond. Orbitals that extend farther from

the nucleus result in bonds that are longer. The key question then, is which orbital

360 Chapter 9 Advanced Models of Bonding

Atomic number

Potential energy (eV)

–50

2468101214161820

–45

–40

–35

–30

–25

–20

–15

–10

–5

FIGURE 9.7

Relative energy of the atomic orbitals. The energy level of the atomic

orbitals decreases with increasing nuclear charge, and the atomic

orbital becomes more stable because it has a lower potential energy.

FIGURE 9.8

Overlap of atomic s orbitals.

9.1 Valence Bond Theory 361

Bond Energies Resulting from Orbital Overlap

As the difference in size of the overlapping orbitals increases, the strength of the resulting bond decreases.

The data shown here are for diatomic molecules.

Bond ∆

diss

H (kJ/mol) Orbital Overlap Representation of Orbital Overlap

HOH 435.8 1s–1s

LiOH 238.0 1s–2s

NaOH 185.7 1s–3s

KOH 174.6 1s–4s

RbOH 167.0 1s–5s

HOF 569.9 1s–2p

HOCl 431.9 1s–3p

HOBr 366.3 1s–4p

HOI 298.4 1s–5p

FOF 158.3 2p–2p

TABLE 9.2

ula_09_06

reaches farther from the nucleus, an s or a p? The end-on overlap of two p orbitals

makes a longer bond than the overlap of two s orbitals, because the average elec-

tron density lies farther from the nucleus of the atom. For this reason, the bond

in F

is longer (141.7 pm) than the bond in H

(74.6 pm). Overlap of two 1s or-

bitals makes a shorter bond than the overlap of two 2s orbitals for the same rea-

son. We’ll discuss this in greater detail in Section 9.2.

EXERCISE 9.2 Using Valence Bond Theory: Beam Me Up, Scotty

According to Star Trek fans, dilithium crystals (Li

) power much of the universe of

the future. Judging on the basis of valence bond theory, does this compound actu-

ally exist? Does Na

exist? Which would have a longer bond?

Solution

Because each of the lithium atoms has a half-ﬁlled 2s orbital, we’d predict that their

overlap should provide a stable dilithium molecule. Dilithium does exist, but be-

cause the formation of metallic lithium is so favorable, dilithium can be observed

only as a gas at high temperatures. Two sodium atoms can have overlap of their

3s orbitals to make a bond. So, theoretically, disodium should also exist. Because

the disodium molecule is made up of bigger orbitals, we’d also predict it to have a

longer bond than dilithium.

PRACTICE 9.2

Which of these molecules has the longest bond: Cl

,HCl,or F

See Problems 13–15.

What’s Wrong with This Model?

Methane (CH

) is the major component of natural gas. Piped into our homes, it

undergoes combustion to heat our water, cook our food, and warm our rooms.

Let’s build a valence bond model of methane. We start by writing the valence elec-

tron conﬁgurations of the atoms in methane. Immediately, we note that the va-

lence electron conﬁgurations of carbon (2s

) and hydrogen (1s

) indicate a

problem. The valence shell of carbon contains a completely full 2s orbital, two

partially ﬁlled p orbitals, and one completely unﬁlled p orbital. Valence bond

theory indicates that we should be able to make only two bonds to the hydrogen

atoms resulting from the overlap of the two partially ﬁlled p orbitals with the

hydrogen 1s orbitals, as shown in Figure 9.9. Something is wrong here.

We know that the Lewis dot structure model of methane correctly shows

four bonds. VSEPR models represent the molecule as a tetrahedral structure

with four equal bonds. Therefore, each of the COH bonds in methane must be

made up of the same types of orbitals. But unless we modify the valence bond

theory to account for our experimental evidence, we’re sure to build a structure

that will fail. In the next section, we will describe a modiﬁcation that enables us

to correct this discrepancy.

HERE’S WHAT WE KNOW SO FAR

■

Lewis dot structure and the VSEPR model do not properly explain such prop-

erties of molecules as bond lengths and bond energies. Valence bond theory

(VB theory) was introduced by Linus Pauling to better explain molecular

properties.

■

A valence bond is seen as the overlap of atomic orbitals on adjacent atoms.

362 Chapter 9 Advanced Models of Bonding

■

In VB theory, smaller orbitals overlap more than larger orbitals.

■

In VB theory, orbitals similar in size overlap more than orbitals with mis-

matched sizes.

■

In VB theory, orbitals with similar energies overlap more than orbitals with

very different energies.

■

Bond length and bond energy in simple molecules can be explained by valence

bond theory.

■

VB theory does not properly explain bond angles in molecules, so we must

modify it to make a better model.

9.2 Hybridization

Linus Pauling (we have already discussed his contributions in the areas of elec-

tronegativity and valence bond theory) advanced the theory of hybridization to

address the problems associated with VB theory. The result ﬁxes the problems

with the valence bond approach to model construction. In 1954, Pauling was

awarded the Nobel Prize in chemistry for his years of research into the nature of

the chemical bond.

Hybridization Deﬁned

A hybrid is a mixture of two species. Hybrid roses, tulips, marigolds, and lilies add

beauty to the garden. Much as red roses and white roses can be interbred to give

pink hybrids, we can mathematically combine two or more orbitals to provide

new orbitals of equal energy in the process of

hybridization.

9.2 Hybridization 363

Atomic orbitals on four hydrogen atoms

Atomic orbitals on carbon atom

1s 1s 1s

2s1s 2p 2p 2p

FIGURE 9.9

Valence bond model of methane. Atomic orbitals that

contain only one electron can overlap to form a bond.

These are shown with a dotted line. The orbitals that

cannot overlap to form a bond (either too many elec-

trons or not enough electrons) have an “

X” on the

dotted line. The model needs some corrections because

there aren’t enough orbitals to make four bonds.

Just as gardeners combine red and white

roses to obtain pink hybrids, chemists

mathematically combine two or more

orbitals to represent new orbitals of

equal energy in the process of hybridiza-

tion. The Pinocchio rose and the Crimson

Glory rose can be combined to give the

Fashion rose.

Pinocchio rose Crimson Glory rose Fashion rose

Tutorial: Hybridization

Recall from Chapter 6 that orbitals with the same energy are known as degen-

erate. What types of orbitals can be mixed together? The degree to which an or-

bital mixes with another is directly related to the difference in energy of the two

orbitals. Typically, we hybridize only orbitals of the same subshell, such as a hy-

brid made using 2s and 2p orbitals, resulting in a set of new orbitals that have

the properties of all the orbitals from which they were mixed. The preparation

of pink paint offers an analogy to hybridization. To make a good pink paint, we

must mix red and white paint that have the same base (i.e., latex or oil). We

don’t get a good mixture by combining one can of latex paint and one can of oil

paint.

How many orbitals do we make when we hybridize atomic orbitals? Just as if

we were to mix one can of red paint and one can of white paint to get two cans of

pink paint, we should expect to get two hybridized orbitals if we mix two atomic

orbitals. The number of orbitals that are hybridized determines the number of

new orbitals that are made. The orbitals that result from this mixing will be de-

generate (have the same energy) and will have the same shape, but they will be

oriented in different directions.

What is the energy of the resulting hybridized

orbitals?

Consider our paint example again. We expect the color of the mixed

paint to be the weighted average of the colors that we added. Similarly, the energy

of the resulting hybridized orbitals should be the weighted average of the energies

of the atomic orbitals that were mixed.

sp, sp

, and sp

Orbitals

To determine the hybridized orbitals used by an atom, we follow a brief series of

steps that convert atomic orbitals into hybridized orbitals. These steps are out-

lined in Table 9.3, and we will show in detail how the process works for methane

(CH

). The Lewis dot structure of methane indicates that we should have one

bond from each hydrogen atom to the carbon atom.

Step 1: Write the electron conﬁguration of the carbon atom. The electron con-

ﬁguration of carbon (1s

) indicates that there are only two half-ﬁlled

orbitals.

Step 2: In order to place four hydrogen atoms in degenerate bonds around the

carbon atom, we need to make four orbitals on carbon that will be able to in-

teract with the four hydrogen 1s orbitals. How do we show that the carbon

makes four bonds?

Step 3: We hybridize the existing orbitals to make four orbitals. Knowing that we

need to mix four orbitals to make four orbitals, we take the available 2s orbital

and the three 2p orbitals and mix them to make four new degenerate orbitals, as

shown in Figure 9.10.

364 Chapter 9 Advanced Models of Bonding

To make pink paint, we can mix red

paint and white paint. The resulting

paint is a hybrid—a product that has

characteristics of both paints.

Algorithm for Hybridizing Atomic Orbitals

Step 1: Electron Conﬁguration Write the electron conﬁguration for the atom that will be hybridized.

Step 2: Bonds Needed Note the number of attached atoms and lone pairs, and select the orbitals to be

hybridized.

Step 3: Hybridize Mix the orbitals to make an identical number of new orbitals. Note the energy of

the hybridized orbitals.

Step 4: Rewrite Electron Conﬁguration Redraw the electron conﬁguration and include the hybridized orbitals. Place the

electrons in the new orbitals according to the Aufbau principle, Hund’s rule, and

the Pauli exclusion principle.

Step 5: Form Valence Bonds Identify which orbitals will overlap to form a bond.

Step 6: Examine Structure Describe the geometry and bond lengths for the new bond.

TABLE 9.3

Video Lesson: An Introduction to

Hybrid Orbitals

The new orbitals are given a name in order to distinguish them from the 2s and

2p orbitals, but to show their relationship to these orbitals, we call them the

2sp

orbitals. The name of the new orbitals illustrates that one 2s orbital and

three 2p orbitals are mixed to make four new orbitals.

Step 4: The next step in Table 9.3 is to show the electron conﬁguration of our

hybridized carbon atom. We can write 1s

(2sp

)

to show the hybrid orbitals.

Carefully observe how this is written because the notation is tricky. Each of the

four 2sp

orbitals on our carbon atom contains only one electron and is able

to participate in bonding.

Step 5: The new shape of the orbitals (large at one end and small at the other) is

shown in Figure 9.10. Does the geometry of this hybridized carbon atom—that

is, with what are now four equivalent pairs surrounding it—agree with that

obtained from the VSEPR model? In other words,

does our answer make sense?

Step 6: Let’s look at the energy changes that occurred with this hybridization.

Energy was added to two 2s electrons on carbon in order to promote them to

the new 2sp

orbitals. A small amount of energy was removed when two 2p

electrons moved to the new orbitals. The net result is that the carbon atom

did increase its energy in making the new hybrid orbitals. However,

when the hybridized carbon combines with the four hydrogen atoms,

a large amount of energy is released as the new C—H bonds are

formed. Although the hybridization process itself is energetically up-

hill, the end result (four equal bonds) is still energetically downhill

(and quite favorable). Why? What reduces the energy? In methane,

the s–sp

overlap between each hydrogen and the central carbon atom

forms the four bonds (Figure 9.10). The outcome of the resulting va-

lence bond model now agrees with the Lewis dot structure and the

VSEPR model. Moreover, we also have an idea of the relative bond

lengths and strengths in methane.

Boron triﬂuoride (BF

) was mentioned in Chapter 8. This compound is par-

ticularly useful in enhancing the reactivity of molecules during the synthesis of

new pharmaceutical agents. Using the ideas of hybridization and valence bond

theory, what is the structure of BF

? The electron conﬁguration of boron is

. The Lewis dot structure of boron triﬂuoride indicates that we should

have three single bonds, so we must hybridize the orbitals on boron to make three

9.2 Hybridization 365

Atomic orbitals on carbon Hybrid orbitals on carbon

Note: Only two bonds are possible. Note: Four bonds are possible

with the new hybridized orbitals.

2p 2p

Energy

2sp

Energy

FIGURE 9.10

Hybridization to give four orbitals.

(+Energy) + (–Energy) = net increase in energy

–Energy

+Energy

2p 2p

2sp

366 Chapter 9 Advanced Models of Bonding

Atomic orbitals on boron Hybrid orbitals on boron

2p 2p

Energy

2sp

Energy

Atomic orbitals on boron Hybrid orbitals on boron

Note: Only one bond is possible. Note: Three bonds are possible

with the new hybridized orbitals.

FIGURE 9.11

Hybridization to give three orbitals in BF

Atomic orbitals on beryllium Hybrid orbitals on beryllium

2p 2p

Energy

2sp

2p 2p

2sp

Energy

Atomic orbitals on beryllium Hybrid orbitals on beryllium

Note: No bonds are possible. Note: Two bonds are possible

with hybridized orbitals.

FIGURE 9.12

Hybridization to give two orbitals in BeH

orbitals. Following the steps in Table 9.3, we mix the 2s orbital and two of the 2p

orbitals (all from boron) to make the new orbitals. The three resulting orbitals

have the designation 2sp

, as shown in Figure 9.11. We didn’t hybridize one of the

2p orbitals because it wasn’t needed to make a bond (i.e., the empty unhybridized

2p orbital is still available on the boron atom). We rewrite the electron conﬁgura-

tion for boron triﬂuoride to end up with three orbitals that are available to bond

with the half-ﬁlled 2p orbital on each unhybridized ﬂuorine atom (p–sp

overlap).

We can build a model of beryllium hydride (BeH

) showing covalent bonds

between the beryllium (1s

) and hydrogen atoms. There are no electrons in

the atomic 2p orbitals. Because we know that there are two bonds in BeH

,we hy-

bridize two orbitals to make two new 2sp orbitals, as shown in Figure 9.12.

Rewriting the electron conﬁguration shows that we can now make bonds to the

hydrogen atoms, resulting in an overlap of the 1s and 2sp orbitals. In beryllium

hydride, there are two empty, unchanged 2p orbitals.

Selected Hybrid Orbitals

COH Bond COC Bond

Hybrid Example Length (pm) Angle Length (pm)

OCH

109 109.5° 154

Orbital

PCH

108 120° 134

Energy

sp CHqCH 106 180° 120

TABLE 9.4

Subtraction

Addition

FIGURE 9.13

Mixing an s and a p orbital. Addition and

subtraction of the s and p orbitals gives

two sp orbitals.

180°

—

90°

120°

109.5°

Increasing length of orbital

9.2 Hybridization

367

EXERCISE 9.3 Hybridization in Common Molecules

Describe the hybridization of the central atom in each of these molecules:

a. H

Ob.NH

c. AlH

Solution

a. Lewis dot structure models for water indicate that the central oxygen has two

bonds and two lone pairs. Mixing the four orbitals (one 2s and three 2p) on

the oxygen atom allows us to have two lone pairs of equal energy and two

equal bonds to the adjacent hydrogen atoms. The oxygen atom possesses 2sp

hybridized orbitals.

b. In a similar fashion, the nitrogen atom requires four things (three bonds and

one lone pair) in ammonia. Mixing the four orbitals in its valence shell gives

us three bonds to the hydrogen atoms and one lone pair (all degenerate in en-

ergy). The nitrogen atom is 2sp

hybridized.

c. Aluminum has only three electrons in its valence shell. On the basis of its

electron conﬁguration, only one bond would be allowed. Because three points

of attachment are needed (one to each hydrogen), we mix the 3s and two of

the 3p orbitals to make an 3sp

hybridized aluminum.

PRACTICE 9.3

What is the hybridization of the central atom in each of the following substances?

a. OF

b. H

Sc.NH

See Problems 23, 24, 33, 34, 43, and 44.

Shapes of the Hybrids

Hybrid orbitals have a shape that results from the mixing of the corresponding

atomic orbitals. If we add an s orbital and a p orbital together, the result is one of

the two hybrid sp orbitals. If we subtract the s orbital from the p orbital, we get the

other hybrid sp orbital as shown in Figure 9.13. The angles between the resulting

orbitals agree with what we expect from VSEPR rules. The sp hybridized atom has

bonds with 180° angles, the sp

hybridized atom forms bonds with 120° angles,

and the sp

hybridized atom contains bonding angles of 109.5° (see Figure 9.14;