9 Bond Properties: Order, Length, and Energy

8.9 Bond Properties: Order, Length, and Energy







␦+



Overall

displacement of

bonding electrons



␦+



H



H

C



Cl−



C



␦−



Displacement of

bonding electrons



Cl



␦+



Cl



H

C



H+



␦

␦−

A, polar, diplacement of

bonding electrons to one

side of the molecule



381



C



Displacement of

bonding electrons



Cl



␦

␦−

B, not polar, no net

displacement of bonding

electrons to one side

of the molecule



Think about Your Answer The electrostatic potential surfaces reflect the fact that molecule A is

polar because the electron density is shifted to one side of the molecule. Molecule B is not polar

because the electron density is distributed symmetrically.



A = cis-1,2-dichloroethylene



B = trans-1,2-dichloroethylene



Check Your Understanding

The electrostatic potential surface for OSCl2 is pictured here.



(a)



Draw a Lewis electron dot picture for the molecule, and give the formal charge of each atom.



(b) What is the molecular geometry of OSCl2? Is it polar?



revIeW & cHecK FOr SectIOn 8.8

1.



Which of the hydrogen halides is the most polar?

(a)



2.



HF



(b) HCl



(c)



HBr



(d) HI



POCl3



(d) SO2Cl2



Which of the molecules listed below is nonpolar?

(a)



BCl3



(b) PCl3



(c)



8.9 BondProperties:Order,Length,andenergy

Bond Order

The order of a bond is the number of bonding electron pairs shared by two atoms

in a molecule (Figure 8.16). You will encounter bond orders of 1, 2, and 3, as well as

fractional bond orders.

When the bond order is 1, there is only a single covalent bond between a pair

of atoms. Examples are the bonds in molecules such as H2, NH3, and CH4. The



kotz_48288_08_0344-0399.indd 381



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382



c h a p t er 8   Bonding and Molecular Structure



Figure 8.16   Bond order. The

four C—H bonds in methane each

have a bond order of 1. The two CPO

bonds of CO2 each have a bond order

of two, whereas the nitrogen–

nitrogen bond in N2 has an order of 3.



H

C



H



H

H



O



Bond order 1



C



N



O



Bond order 2



N



Bond order 3



bond order is 2 when two electron pairs are shared between atoms, such as the

CPO bonds in CO2 and the CPC bond in ethylene, H2CPCH2. The bond order is

3 when two atoms are connected by three bonds. Examples include the carbon–

oxygen bond in carbon monoxide, CO, and the nitrogen–nitrogen bond in N2.

Fractional bond orders occur in molecules and ions having resonance structures. For example, what is the bond order for each oxygen–oxygen bond in O3?

Each resonance structure of O3 has one OOO single bond and one OPO double

bond, for a total of three shared bonding pairs accounting for two oxygen–oxygen

links.

Bond order = 1

Bond order = 2



O

O



O



Bond order for each

oxygen–oxygen bond

= 32 , or 1.5



One resonance structure



We can define the bond order between any bonded pair of atoms X and Y as







Bond order ϭ



number of shared pairs in all X— Y bonds

number of X—Y links in the molecule or ion







(8.2)



For ozone, there are three bond pairs involved in two oxygen–oxygen links, so the

bond order for each oxygen–oxygen bond is 3⁄2, or 1.5.



Bond Length



4A



5A



6A



C



N



O



Si



P



S



Relative sizes of some atoms of Groups 4A,

5A, and 6A.



Bond lengths are related to atom sizes.

C—H

N—H

O—H

110

98

94 pm

Si—H

P—H

S—H

145

138

132 pm



kotz_48288_08_0344-0399.indd 382



Bond length, the distance between the nuclei of two bonded atoms, is clearly related

to the sizes of the atoms (Section 7.5). In addition, for a given pair of atoms, the

order of the bond plays a role.

Table 8.8 lists average bond lengths for a number of common chemical bonds.

It is important to recognize that these are average values. Neighboring parts of a

molecule can affect the length of a particular bond, so there can be a range of values for a particular bond type. For example, Table 8.8 gives the average COH bond

as 110 pm. However, in methane, CH4, the measured bond length is 109.4 pm,

whereas the COH bond is only 105.9 pm long in acetylene, H—CqCOH. Variations as great as 10% from the average values listed in Table 8.8 are possible.

Because atom sizes vary in a regular fashion with the position of the element in

the periodic table (Figure 7.6), we can predict trends in bond lengths. For example,

the HOX distance in the hydrogen halides increases in the order predicted by the

relative sizes of the halogens: HOF < HOCl < HOBr < HOI. Likewise, bonds

between carbon and another element in a given period decrease going from left to

right, in a predictable fashion; for example, COC > CON > COO. Trends for

multiple bonds are similar. A CPO bond is shorter than a CPS bond, and a CPN

bond is shorter than a CPC bond.

The effect of bond order is evident when bonds between the same two atoms

are compared. For example, the bonds become shorter as the bond order increases

in the series COO, CPO, and CqO:

Bond



COO



CPO



CqO



Bond order

Average bond length (pm)



1

143



2

122



3

113



11/19/10 9:36 AM



8.9  Bond Properties: Order, Length, and Energy







383



Table 8.8  Some Average Single- and Multiple-Bond Lengths in Picometers (pm)*

Single Bond Lengths

Group



H



1A



4A



5A



6A



7A



4A



5A



6A



7A



7A



7A



H



C



N



O



F



Si



P



S



Cl



Br



I



74



110



98



94



92



145



138



132



127



142



161



154



147



143



141



194



187



181



176



191



210



140



136



134



187



180



174



169



184



203



132



130



183



176



170



165



180



199



128



181



174



168



163



178



197



234



227



221



216



231



250



220



214



209



224



243



208



203



218



237



200



213



232



228



247



C

N

O

F

Si

P

S

Cl

Br

I



266



Multiple Bond Lengths



−12



*1 pm = 10



CPC



134



CqC



121



CPN



127



CqN



115



CPO



122



CqO



113



NP0



115



NqO



108



m.



The carbonate ion, CO32−, has three equivalent resonance structures. Each CO

bond has a bond order of 1.33 (or 4⁄3) because four electron pairs are used to form

three carbon–oxygen links. The CO bond distance (129 pm) is intermediate between a COO single bond (143 pm) and a CPO double bond (122 pm).

2−



O



Bond order = 2

Bond order = 1



C

O



O

Bond order = 1



Average bond order



= 43 , or 1.33

Bond length



= 129 pm



Bond Dissociation Enthalpy

The bond dissociation enthalpy is the enthalpy change for breaking a bond in a

molecule with the reactants and products in the gas phase. The process of breaking bonds

in a molecule is always endothermic, so ∆rH for bond breaking is always positive.

Molecule (g)



Energy supplied = ∆H > O

Energy released = ∆H < O



Molecular fragments (g)



Suppose you wish to break the carbon–carbon bonds in ethane (H3COCH3),

ethylene (H2CPCH2), and acetylene (HCqCH). The carbon–carbon bond orders

in these molecules are 1, 2, and 3, respectively, and these bond orders are reflected

in the bond dissociation enthalpies. Breaking the single C—C bond in ethane



kotz_48288_08_0344-0399.indd 383



11/19/10 9:36 AM



384



c h a p t er 8   Bonding and Molecular Structure



requires the least energy in this group, whereas breaking the C—C triple bond in

acetylene requires the most energy.

H3C—CH3(g) → H3C(g) + ​CH3(g)







ΔrH ​= ​+368 kJ/mol-rxn



H2CPCH2(g) → H2C(g) + ​CH2(g)     ∆rH ​= ​+682 kJ/mol-rxn

HCqCH(g) → HC(g) + ​CH(g)







ΔrH = ​+962 kJ/mol-rxn



The energy supplied to break carbon–carbon bonds must be the same as the

energy released when the same bonds form. The formation of bonds from atoms or radicals in the gas phase is always exothermic. This means, for example, that ΔrH for the

formation of H3COCH3 from two CH3(g) radicals is −368 kJ/mol-rxn.

H3C⋅ (g) ​+ ​⋅CH3(g) → H3C−CH3(g)    ∆rH ​= ​−368 kJ/mol-rxn



Generally, the bond energy for a given type of bond (a C—C bond, for example)

varies somewhat, depending on the compound, just as bond lengths vary from one

molecule to another. Thus, data provided in tables must be viewed as average bond

dissociation enthalpies (Table 8.9). The values in such tables may be used to estimate

the enthalpy change for a reaction, as described below.

In reactions between molecules, bonds in reactants are broken, and new

bonds are formed as products form. If the total energy released when new bonds

form exceeds the energy required to break the original bonds, the overall



• Variability in Bond Dissociation

Enthalpies  The values of ∆r H for ethane, ethylene, and acetylene in the text

are for those molecules in particular.

The bond dissociation enthalpies in

Table 8.9 are average values for a range

of molecules containing the indicated

bond.



Table 8.9  Some Average Bond Dissociation Enthalpies (kJ/mol)*

Single Bonds

H

C

N



H



C



N



O



F



Si



P



S



Cl



Br



I



436



413



391



463



565



328



322



347



432



366



299



346



305



358



485



—



—



272



339



285



213



163



201



283



—



—



—



192



—



—



146



—



452



335



—



218



201



201



155



565



490



284



253



249



278



222



—



293



381



310



234



201



—



326



—



184



226



255



—



—



242



216



208



193



175



O

F

Si

P

S

Cl

Br

I



151



Multiple Bonds

NPN



418



CPC



610



NqN



945



CqC



835



CPN



615



CPO



745



CqN



887



CqO



1046



OPO (in O2)



498



*Sources of dissociation enthalpies: I. Klotz and R. M. Rosenberg, Chemical Thermodynamics, 4th Ed., p. 55,

New York, John Wiley, 1994; and J. E. Huheey, E. A. Keiter, and R. L. Keiter, Inorganic Chemistry 4th Ed.,

Table E. 1, New York, HarperCollins, 1993. See also Lange’s Handbook of Chemistry, J. A. Dean (ed.), McGrawHill Inc., New York.



kotz_48288_08_0344-0399.indd 384



11/19/10 9:36 AM



385



8.9 Bond Properties: Order, Length, and Energy







reaction is exothermic. If the opposite is true, then the overall reaction is

endothermic.

Let us use bond dissociation enthalpies to estimate the enthalpy change for the

hydrogenation of propene to propane:



H



H



H H



C



C



C



H(g) + H



H(g)



H



H



H



H H



C



C



H

propene



C



H(g)



H H

propane



The first step is to identify what bonds are broken and what bonds are formed. In

this case, the CPC bond in propene and the HOH bond in hydrogen are broken.

A COC bond and two COH bonds are formed.

Bonds broken:



1 mol of CPC bonds and 1 mol of HOH bonds



H



H



H H



C



C



C



H(g) + H



H(g)



H

Energy required = 610 kJ for CPC bonds + 436 kJ for H—H bonds = 1046 kJ/mol-rxn



CASE STUDY



Ibuprofen, A Study in Green Chemistry



Ibuprofen, C13H18O2, is now

one of the world’s most common, over-the-counter drugs. Itt is an effective anti-inflammatory drug and is used to

treat arthritis and similar conditions. Unlike

aspirin it does not decompose in solution, so

ibuprofen can be applied to the skin as a

topical gel, thus avoiding the gastrointestinal

problems sometimes associated with aspirin.



drug did not end up in the drug itself, that is,

the method had a very poor atom economy

(page 168). Recognizing this, chemists

looked for new approaches for the synthesis

of ibuprofen. A method was soon found that

uses only three steps, severely reduces

waste, and uses reagents that can be recovered and reused. A plant in Texas now makes

more than 7 million pounds of ibuprofen

annually by the new method, enough to

make over 6 billion tablets.



Questions:



FigureA



Ibuprofen, C13H18O2



Ibuprofen was first synthesized in the

1960s by the Boots Company in England,

and it soon became an important product.

Millions of pounds are sold every year.

Chemists realized, however, that the

method of making ibuprofen took six steps,

wasted chemicals, and produced byproducts that required disposal. Many of

the atoms of the reagents used to make the



kotz_48288_08_0344-0399.indd 385



1. The third and final step in the synthesis

involves the transformation of an OH

group with CO to a carboxylic acid group

(−CO2H) (Figure B). Using bond enthalpy

data, is this an endothermic or exothermic step?

2. Do any of the atoms in an ibuprofen

molecule have a formal charge of zero?

3. What is the most polar bond in the

molecule?

4. Is the molecule polar?

5. What is the shortest bond in the

molecule?



H3C



OH



H3C



CH

HC

HC



C

C



CH



CO



HC



CH



catalyst



HC



CH2

H3C



C

H



CO2H

CH

C

C



CH

CH



CH2

CH3



H3C



C

H



CH3



FigureB The final step in the synthesis of

ibuprofen.



6. What bond or bonds have the highest

bond order?

7. Are there any 120˚ bond angles in ibuprofen? Any 180˚ angles?

8. If you were to titrate 200. mg of ibuprofen to the equivalence point with 0.0259

M NaOH, what volume of NaOH would

be required?

Answers to these questions are available in

Appendix N.



11/19/10 9:36 AM



386



c h a p t er 8   Bonding and Molecular Structure



Bonds formed:



1 mol of COC bonds and 2 mol of COH bonds



H



H



H H



C



C



H



H H



C



H(g)



Energy evolved ​= ​346 kJ for C—C bonds + ​2 mol × 413 kJ/mol for COH bonds ​= ​

1172 kJ/mol-rxn



By combining the enthalpy changes for breaking bonds and for making bonds,

we can estimate ∆rH for the hydrogenation of propene and predict that the reaction

is exothermic.



• ∆rH from Enthalpies of Formation 



Using ∆f H° values for propane and propene, we calculate ∆r H° = −125.1 kJ/

mol–rxn. The bond dissociation enthalpy calculation is in excellent agreement with that from enthalpies of

formation in this case.



ΔrH ​= ​1046 kJ/mol-rxn ​− ​1172 kJ/mol-rxn ​= ​−126 kJ/mol-rxn



In general, the enthalpy change for any reaction can be estimated using the

equation

 ∆rH ​= ​Σ∆H(bonds broken) − Σ∆H(bonds formed) 







(8.3)



Such calculations can give acceptable results in many cases.



EXAMPLE 8.14 ​Using Bond Dissociation Enthalpies

Problem  Acetone, a common industrial solvent, can be converted to 2-propanol, rubbing alcohol, by hydrogenation. Calculate the enthalpy change for this reaction using bond enthalpies.



H

O

H3C



C



O

CH3(g) + H



H(g)



H3C



C



CH3(g)



H

acetone



2-propanol



 

What Do You Know?  You know the molecular structures of the reactants and the products.

Strategy  Determine which bonds are broken and which are formed. Add up the enthalpy

changes for breaking bonds in the reactants and for forming bonds in the product. The difference in the sums of bond dissociation enthalpies can be used as an estimate of the

enthalpy change of the reaction (Equation 8.3).

Solution

Bonds broken: 1 mol of CPO bonds and 1 mol of H—H bonds



O

H3C



C



CH3(g) + H



H(g)



ΣΔH(bonds broken) ​= ​745 kJ for CPO bonds + ​436 kJ for H—H bonds ​= ​1181 kJ/mol-rxn

Bonds formed: 1 mol of C—H bonds, 1 mol of C—O bonds, and 1 mol of O—H bonds



H

O

H3C



C



CH3(g)



H



kotz_48288_08_0344-0399.indd 386



11/19/10 9:36 AM



8.9 Bond Properties: Order, Length, and Energy







387



ΣΔH(bonds formed) = 413 kJ for C—H + 358 kJ for C—O + 463 kJ for O—H = 1234 kJ/mol-rxn

ΔrH = ΣΔH(bonds broken) − ΣΔH(bonds formed)

ΔrH = 1181 kJ − 1234 kJ = −53 kJ/mol-rxn

Think about Your Answer The overall reaction is predicted to be exothermic by 53 kJ per mol of

product formed. This is in good agreement with the value calculated from ΔfH° values (−55.8 kJ/

mol-rxn).

Check Your Understanding

Using the bond dissociation enthalpies in Table 8.9, estimate the enthalpy of combustion of gaseous methane, CH4 to give water vapor and carbon dioxide gas.



revIeW & cHecK FOr SectIOn 8.9

Which of the following species has the largest N—O bond?

(a)

2.



NO3−



(b) NO+



(c)



NO2−



(d) H2NOH



Which of the following species has the largest C—N bond order?

(a)



CN−



A CLOSER LOOK



(b) OCN−



(c)



(d) N(CH3)3



DNA—Watson, Crick, and Franklin



A. Barrington Brown/Science Source/Photo Researchers, Inc.



DNA is the substance in

every plant and animal that

carries the exact blueprint of that plant or

animal. The structure of this molecule, the

cornerstone of life, was uncovered in 1953,

and James D. Watson, Francis Crick, and

Maurice Wilkins shared the 1962 Nobel Prize

in Medicine and Physiology for the work. It

was one of the most important scientific

discoveries of the 20th century. The story of

this discovery has been told by Watson in his

book The Double Helix.



James D. Watson and Francis Crick. In a photo

taken in 1953, Watson (left) and Crick (right) stand by

their model of the DNA double helix (at The University of Cambridge, England). Together with Maurice

Wilkins, Watson and Crick received the Nobel Prize

in Medicine and Physiology in 1962.



kotz_48288_08_0344-0399.indd 387



CH3NH2



When Watson was a graduate student at

Indiana University, he had an interest in the

gene and said he hoped that its biological

role might be solved “without my learning

any chemistry.” Later, however, he and Crick

found out just how useful chemistry can be

when they began to unravel the structure of

DNA.

Solving important problems requires

teamwork among scientists of many kinds,

so Watson went to Cambridge in 1951. There

he met Crick, who, Watson said, talked

louder and faster than anyone else. Crick

shared Watson’s belief in the fundamental

importance of DNA, and the pair soon

learned that Maurice Wilkins and Rosalind

Franklin at King’s College in London were

using a technique called x-ray crystallography to learn more about DNA’s structure.

Watson and Crick believed that understanding this structure was crucial to understanding genetics. To solve the structural problem, however, they needed experimental

data of the type that could come from the

experiments at King’s College.

The King’s College group was initially

reluctant to share their data, and, what is

more, they did not seem to share Watson

and Crick’s sense of urgency. There was also

an ethical dilemma: Could Watson and Crick

work on a problem that others had claimed

as theirs? “The English sense of fair play



Henry Grant Collection/Museum of London



1.



Rosalind Franklin of King’s College, London. She

died in 1958 at the age of 37. Because Nobel Prizes

are never awarded posthumously, she did not share

in this honor with Watson, Crick, and Wilkins. For

more on Rosalind Franklin, read Rosalind Franklin:

The Dark Lady of DNA, by Brenda Maddox.



would not allow Francis to move in on

Maurice’s problem,” said Watson.

Watson and Crick approached the problem through a technique chemists now use

frequently—model building. They built models of the pieces of the DNA chain, and they

tried various chemically reasonable ways of

fitting them together. Finally, they discovered that one arrangement was “too pretty

not to be true.” Ultimately, the experimental

evidence of Wilkins and Franklin confirmed

the “pretty structure” to be the real DNA

structure.



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388



c h a p t er 8 Bonding and Molecular Structure



3.



Use bond dissociation enthalpies to estimate the enthalpy change for the decomposition of

hydrogen peroxide.

2 H2O2(g) n 2 H2O(g) + O2(g)

(a)



+352 kJ/mol-rxn



(c)



(b) −206 kJ/mol-rxn



+1037 kJ/mol-rxn



(d) −2204 kJ/mol-rxn



8.10 DNA,revisited

This chapter opened with some questions about the structure of DNA, one of the

key molecules in all biological organisms. Now that you have studied this chapter,

we can say more about the structure of this important molecule and why it looks the

way it does.

As shown in Figure 8.17, each strand of the double-stranded DNA molecule

consists of three repeating parts: a phosphate, a deoxyribose molecule (a sugar molecule with a five-member ring), and a nitrogen-containing base. (The bases in DNA

can be one of four molecules: adenine, guanine, cytosine, and thymine; in Figure

8.17, the base is adenine.) Two units of the backbone (without the adenine on the

deoxyribose ring) are also illustrated in Figure 8.17a.

One important point here is that the repeating unit in the backbone of DNA

consists of the atoms OOPOOOCOCOC. Each atom has a tetrahedral electronpair geometry. Therefore, the chain cannot be linear. In fact, the chain twists as one

moves along the backbone. This twisting gives DNA its helical shape.

Five-member deoxyribose ring

is slightly puckered owing to

tetrahedral geometry around

each C and O atom.



Angles in this ring are all about 120°. In each major

resonance structure for this ring, each C is surrounded

by one double bond and two single bonds, and each

N is surrounded by one double bond, one single bond,

Pho

and one lone pair.

Sug

Pho

Sug

Pho

Sug

G



C

Phosphate group, PO43−

The electron pair and

molecular geometry are

both tetrahedral.



O

P



Adenine



A



T



A



T

C

Pho



Pho



Sugar

(deoxyribose portion)



Pho

Sug

Sug A T

Pho



A



Pho

Sug



P—O—C geometry is bent.

The O atom is surrounded by two bond

pairs and two lone pairs resulting in a

bent molecular geometry around this O.



T

G



Pho



P O C



Sug



C Sug

Sug

Pho

Pho

Sug

Sug

C G

Pho

Pho

Sug



Sug

Pho

Sug

Pho



Repeating unit

of DNA backbone:

1 P atom

2 O atoms

3 C atoms



C



Base



C



Sug

Pho

Sug



C



P



O



C



O



Pho

A Sug

Sug T

Pho

Pho

Sug C Sug

Pho

Sug

T A Sug

Pho

Pho

A

T



Base



(a) A repeating unit consists of a phosphate portion, a deoxyribose portion (a sugar molecule with a five-member ring),

and a nitrogen-containing base (here adenine) attached to the deoxyribose ring.

(b) Two of the four bases in DNA, adenine and thymine.

The electrostatic potential surfaces help to

visualize where the partially charged

atoms are in these molecules

and how they can interact.



Figure8.17 Bonding in the DNA molecule.



kotz_48288_08_0344-0399.indd 388



11/19/10 9:36 AM



8.10  DNA, Revisited







In DNA this N atom is

attached to a sugar.



389



FIGURE 8.18   Adenine, one of

the four bases in the DNA molecule.  Notice that all of the C and N

atoms of the five- and six-member

rings have trigonal-planar electronpair geometry. The N atom in the

five-member ring is attached to a

sugar in DNA.



Why are there two strands in DNA with the OOPOOOCOCOC backbone on

the outside and the nitrogen-containing bases on the inside? This structure arises

from the polarity of the bonds in the base molecules attached to the backbone. For

example, the N–H bonds in the adenine molecule are very polar, which

leads to a special type of intermolecular force—hydrogen bonding—binding the

adenine to thymine in the neighboring chain (Figure 8.17b). You will learn more

about this in Chapter 12 when we explore intermolecular forces and also in The

Chemistry of Life: Biochemistry (pages 491–507).

The rings in the nitrogen-containing bases are all flat with trigonal-planar

electron-pair geometry around each atom in the rings. Let us examine the electronpair geometries in one of the bases, adenine (Figure 8.18). There are two major

resonance structures for this ring system. In these, each carbon atom is surrounded

by one double bond and two single bonds, leading to a trigonal-planar electron-pair

geometry. Each nitrogen atom in the rings, except the one attached to the sugar, is

surrounded by a double bond, a single bond, and a lone pair, likewise leading to

trigonal-planar electron-pair geometries around these atoms. The nitrogen attached to the sugar, however, is different from what we would normally predict. It is

surrounded by three single bonds and one lone pair. We would normally expect it

to have a tetrahedral electron-pair geometry (and trigonal-pyramidal molecular geometry), but it does not. Instead, the bonding groups assume a trigonal-planar geometry, and the lone pair is in a plane perpendicular to the bonds. After studying

Chapter 9, you will understand how this allows the electrons in this lone pair to interact with the electrons in the rings’ double bonds in a favorable way.



chapter goals revisited

Now that you have studied this chapter, you should ask whether you have met the chapter

goals. In particular, you should be able to:

Understand the difference between ionic and covalent bonds



a. Describe the basic forms of chemical bonding—ionic and covalent—and the

differences between them, and predict from the formula whether a compound

has ionic or covalent bonding, based on whether a metal is part of the formula

(Section 8.1).

b. Write Lewis symbols for atoms (Section 8.2).

Draw Lewis electron dot structures for small molecules and ions



a. Draw Lewis structures for molecular compounds and ions (Section 8.2).

Study Questions: 5–12.

b. Understand and apply the octet rule; recognize exceptions to the octet rule

(Sections 8.2–8.5). Study Questions: 5–12, 60.



kotz_48288_08_0344-0399.indd 389



  and 

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11/19/10 9:36 AM



390



c h a p t er 8   Bonding and Molecular Structure



c. Write resonance structures, understand what resonance means, and know

how and when to use this means of representing bonding (Section 8.4). Study

Questions: 9, 10.

Use the valence shell electron-pair repulsion (VSEPR) theory to predict the shapes of

simple molecules and ions and to understand the structures of more complex molecules.



a. Predict the shape or geometry of molecules and ions of main group elements

using VSEPR theory (Section 8.6). Table 8.10 shows a summary of the relation

between valence electron pairs, electron-pair and molecular geometry, and

molecular polarity. Study Questions: 17–24, and Go Chemistry Module 12.

Use electronegativity and formal charge to predict the charge distribution in molecules

and ions, to define the polarity of bonds, and to predict the polarity of molecules.



a. Calculate formal charges for atoms in a molecule based on the Lewis structure

(Section 8.3). Study Questions: 13–16, 38.

b. Define electronegativity and understand how it is used to describe the unequal

sharing of electrons between atoms in a bond (Section 8.7).

c. Combine formal charge and electronegativity to gain a perspective on the

charge distribution in covalent molecules and ions (Section 8.7). Study

Questions: 27–38, 79.

d. Understand why some molecules are polar whereas others are nonpolar

(Section 8.8). (See Table 8.7.)

e. Predict the polarity of a molecule (Section 8.8). Study Questions: 41, 42, 82,

83, 85, 90, and Go Chemistry Module 13.

Understand the properties of covalent bonds and their influence on molecular structure



a. Define and predict trends in bond order, bond length, and bond dissociation

enthalpy (Section 8.9). Study Questions: 43–50, 62, 85.

b. Use bond dissociation enthalpies in calculations (Section 8.9 and

Example 8.14). Study Questions: 51–56, 73.



Table 8.10  Summary of Molecular Shapes and Molecular Polarity

Pairs of

Valence Electrons



Electron-Pair

Geometry



Number of

Bond Pairs



Number of

Lone Pairs



Molecular

Geometry



Molecular

Dipole?*



Examples



2



Linear



2



0



Linear



No



BeCl2



3



Trigonal planar



3

2



0

1



Trigonal planar

Bent



No

Yes



BF3, BCl3

SnCl2(g)



4



Tetrahedral



4

3

2



0

1

2



Tetrahedral

Trigonal pyramidal

Bent



No

Yes

Yes



CH4, BF4−

NH3, PF3

H2O, SCl2



5



Trigonal bipyramidal



5

4

3

2



0

1

2

3



Trigonal bipyramidal

Seesaw

T-shaped

Linear



No

Yes

Yes

No



PF5

SF4

ClF3

XeF2, I3−



6



Octahedral



6

5

4



0

1

2



Octahedral

Square pyramidal

Square planar



No

Yes

No



SF6, PF6−

ClF5

XeF4



*For molecules of AXn, where the X atoms are identical.



kotz_48288_08_0344-0399.indd 390



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▲ more challenging  blue-numbered questions answered in Appendix R







391



Key Equations

Equation 8.1 (page 354)  ​Used to calculate the formal charge on an atom in a

molecule.

Formal charge of an atom in a molecule or ion = NVE ​− ​[LPE + ​1⁄2(BE)]



Equation 8.2 (page 382)  ​Used to calculate bond order.

Bond order ϭ



number of shared pairs in all X— Y bonds

number of X—Y links in the molecule or ion



Equation 8.3 (page 386)  ​Used to estimate the enthalpy change for a reaction using

bond dissociation enthalpies.

∆rH ​= ​Σ∆H(bonds broken) − Σ∆H(bonds formed)



Study Questions

  Interactive versions of these questions are

assignable in OWL.

▲ denotes challenging questions.

Blue-numbered questions have answers in Appendix R and

fully worked solutions in the Student Solutions Manual.



Practicing Skills

Valence Electrons and the Octet Rule

(See Section 8.1.)

1. Give the periodic group number and number of valence

electrons for each of the following atoms.

(a)O

(d)Mg

(b)B

(e)F

(c)Na

(f) S

2. Give the periodic group number and number of valence

electrons for each of the following atoms.

(a)C

(d)Si

(b)Cl

(e)Se

(c)Ne

(f) Al

3. For elements in Groups 4A–7A of the periodic table,

give the number of bonds an element is expected to

form if it obeys the octet rule.

4. Which of the following elements are capable of forming

compounds in which the indicated atom has more than

four valence electron pairs?

(a)C

(d)F

(g)Se

(b)P

(e)Cl

(h)Sn

(c)O

(f) B

Lewis Electron Dot Structures

(See Sections 8.2, 8.4 and 8.5 and Examples 8.1, 8.2, 8.4, and 8.5.)

5. Draw a Lewis structure for each of the following molecules or ions.

(a)NF3

(c)HOBr

(b)ClO3−

(d)SO32−



kotz_48288_08_0344-0399.indd 391



6. Draw a Lewis structure for each of the following molecules or ions:

(a)CS2

(b)BF4−

(c)HNO2 (where the arrangement of atoms is HONO)

(d)OSCl2 (where S is the central atom)

7. Draw a Lewis structure for each of the following molecules:

(a) chlorodifluoromethane, CHClF2 (C is the central

atom)

(b)acetic acid, CH3CO2H (basic structure pictured below)

H



H



O



C



C



O



H



H

(c)acetonitrile, CH3CN (the framework is H3COCON)

(d)allene, H2CCCH2

8. Draw a Lewis structure for each of the following

mole­cules:

(a)methanol, CH3OH

(b)vinyl chloride, H2CPCHCl, the molecule from

which PVC plastics are made

(c)acrylonitrile, H2CPCHCN, the molecule from which

materials such as Orlon are made

H



H



H



C



C



C



N



9. Show all possible resonance structures for each of the

following molecules or ions:

(a)sulfur dioxide, SO2

(b)nitrous acid, HNO2

(c)thiocyanate ion, SCN−

10. Show all possible resonance structures for each of the

following molecules or ions:

(a)nitrate ion, NO3−

(b)nitric acid, HNO3

(c)nitrous oxide (laughing gas), N2O (where the bonding is in the order N—N—O)

11. Draw a Lewis structure for each of the following molecules or ions:

(a)BrF3

(c)XeO2F2

(b)I3−

(d)XeF3+



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