Chemistry part 13, Julia Burdge,2e (2009) pptx

• The N atom has five valence eJectrons and four associated electrons one from each single bond and two from the double bond.. Each singly bonded 0 atom has six valence electrons and se

2j2 CHAPTER 8 Chemical Bonding I: Basic Concepts

r

Think About It Counting the total

number of valence electrons should

be relatively simple to do, but it is

often done hastily and is therefore

a potential source of error in this

type of problem Remember that

the number of valence electrons for

each element is equal to the group

number of that element

,

Multimedia Chemical Bonding - formal c h arge

ca l c ulation s

Draw the Lewis structure for carbon disulfide (CS2)

Strategy Use the procedure described in steps 1 through 6 on page 291 for drawing Lewis structures

Setup

Step 1: C and S have identical electronegativities We will draw the skeletal structure with the

unique atom, C, at the center

There are no electrons remaining after step 4, so step 5 does not apply

To complete carbon's octet, use one lone pair from each S atom to make a double bond to the C atom

Solution

:S=C=S: • •

Practice Problem A Draw the Lewis structure for NF3

Practice Problem B Draw the Lewis structure for CI0 3

Checkpoint 8.5 Drawing Lewis Structures

8 5 1 Identify the correct Lewis structure for

formic acid (HCOOH)

So far you have learned two different meth ods of electron "bookkeeping." In Chapter 4, you learned

about oxidation number s [ ~~ Sectio n 4 .4 ] , and in Section 8.4, you learned how to calculate tial charge s There is one additional commonly used method of electron bookkeeping namely ,

par-formal charge, which can be u se d to determine the most plausible Lewis structures when more

than one possibility exists for a compound Formal charge is determined by comparing the number

of electrons associated with an atom in a Lewis st ructure with the number of electron s that would

be associated with the i so lated atom In an i so lated atom, the number of electrons associated with

the atom is s imply the number of valence electrons (As usual , we need not be concerned with the

core electrons.)

To determine the number of electrons associated with an atom in a Lewis s tructure, keep in

mind the following:

SECTION 8.6 Lewis Structures and Formal Charge 293

o All the atom's nonbonding electrons are associated with the atom

o Half of the atom ' s bonding electrons are a ss ociated with the atom

formal charge = valence electron s - a ss ociated electron s Equa t ion 8. 2

An atom's formal charge is calculated as follows: We can illu s trate the concept of formal charge

using the ozone molecule (03) , U se the s tep-by- s tep method for dra w ing L ew i s s tructure s to draw

the Lewis structure for ozone , and then determine the formal charge on each 0 atom b y s

ubtract-ing the number of associated electron s from the number of v alence el e ctron s

ions, the formal charge s mu s t s um to the o v erall charge on the ion

Formal charge s do not repre s ent actual c harge s on atom s in a mole c ule In th e 0 3 molecule ,

for example , there is no evidence that the central atom bear s a net + 1 charge or that one of the

terminal atom s bear s a - 1 charge A ss ignin g formal char g e s to the atom s in th e L ew i s s tructure

merely help s us keep track of the electron s in v ol v ed in bonding in th e mol e cule

Sample Problem 8.7 let s you practice determinin g f o rmal charge s

The widespread use of fertilizers has resulted in the contamination of some groundwater with

nitrates, which are potentially harmful Nitrate toxicity is due primarily to its conversion in the body

to nitrite (N0 2 ), which interferes with the ability of hemoglobin to transport oxygen Determine the

formal charges on each atom in the nitrate ion (NO))

Strategy Use steps 1 through 6 on page 291 for drawing Lewis stuctures to draw the Lewis structure

of NO ) For each atom, subtract the associated electrons from the valence electrons

•

The N atom has five valence eJectrons and four associated electrons (one from each single bond and

two from the double bond) Each singly bonded 0 atom has six valence electrons and seven associated

electrons (six in three lone pairs and one from the single bond) The doubly bonded 0 atom has six

valence electrons and six associated electrons (four in two lone pairs and two from the double bond)

Solution The formal charges are as follows: + 1 (N atom), -1 (singly bonded 0 atoms), and 0

(doubly bonded 0 atom)

Practice Problem A Determine the formal charges on each atom in the carbonate ion (CO ~-)

Practice Problem B Determine the formal charges and use them to determine the overall charge, if

any, on the species represented by the following Lewis structure:

I

:O-S-O:

~I ~ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ ~

While you a r e n e w at deter m ini ng fo rmal

charges, it may be helpful to d r aw L ewis

structures with all dots, rathe r than d ashes

T his can make it easier to see how ma n y electrons ar e associated with ea c h atom

Think About It The sum of formal charges (+ 1) + (-1 ) + (- 1) +

(0) = -1 is equal to the overall charge on the nitrate ion

294 CHAPTER 8 Chemical Bonding I: Basic Concepts

Think About It For a molecule,

formal charges of zero are

preferred When there are

nonzero formal charges, they

should be consistent with the

electronegativities of the atoms in

the molecule A positive formal

charge on oxygen, for example, is

inconsistent with oxygen's high

electronegativity

Sometimes, there is more than one possible skeletal arrangement of atoms for the Lewis structure for a given species In such cases, we often can select the best skeletal arrangement by using formal charges and the following guidelines:

• For molecules, a Lewis structure in which all formal charges are zero is preferred to one in which there are nonzero formal charges

• Lewis structures with small formal charges (0 and + 1) are preferred to those with large mal charges (+2, +3, and so on)

for-• The best skeletal arrangement of atoms will give rise to Lewis structures in which the formal charges are consistent with electronegativities For example, the more electronegative atoms should have the more negative formal charges

Sample Problem 8.8 shows how formal charge can be used to determine the best skeletal arrangement of atoms for the Lewis structure of a molecule or polyatomic ion

Sample Problem 8.8

Formaldehyde (CH20), which can be used to preserve biological specimens, is commonly sold as a

37% aqueous solution Use formal charges to detellTline which skeletal arrangement of atoms shown here is the best choice for the Lewis structure of CH20

o

I

H-C-H

formal charges on the atoms in each one

In the structure on the left, the formal charges are as follows:

Both H atoms: 1 valence e - - 1 associated e - (from single bond) = 0

C atom: 4 valence e - - 5 associated e - (two in the lone pair, one from the single bond, and two from the double bond) = - 1

o atom: 6 valence e - - 5 associated e - (two from the lone pair, one from the single bond, and two from the double bond) = + 1

• • • •

H-C=O-H Formal charges o - J + 1 0

In the structure on the right, the fOlmal charges are as follows:

Both H atoms: 1 valence e - - 1 associated e - (from single bond) = 0

C atom: 4 valence e - - 4 associated e - (one from each single bond, and two from the double bond) = 0

o atom: 6 valence e - - 6 associated e - (four from the two lone pairs and two from the double bond) = 0

• • '0'

II

H-C-H Formal charges all zero

formal charge, which is inconsistent with oxygen's high electronegativity Therefore, the structure

on the right, in which both H atoms are attached directly to the C atom and all atoms have a formal charge of zero, is the better choice for the Lewis structure of CH20

group, -COOH Use formal charges to determine which of the two arrangements is better

• • '0'

arrangement of atoms in NCI2

Checkpoint 8.6 Lewis Structures and Formal Charge

8 6 1 Detennine the formal charges on H, C,

and N, respectively, in HCN

a) 0, + 1, and - 1

b) - 1, + 1, and 0 c) 0, - 1, and + 1 d) O, + I,and+l e) 0,0, and 0

Resonance

8 6 2 Which of the Lewis structures shown is

most likely preferred for NCO - ?

Our drawing of the Lewis structure for ozone (03) sat isfied the octet rule for the central 0 atom

because we placed a double bond betwe e n it and one of the two terminal 0 atoms In fact, we can

put the double bond at either end of the molecule , as show n by the following two equ i va lent Lewis

A single bond between 0 atoms s hould be longer than a double bond between 0 atoms, but

experi-mental evidence indicates that both of the bond s in 0 3 are equal in length (128 pm) Because

neither one of the se two Lewi s st ructure s accounts for the known bond lengths in 0 3, we u se both

Lewis str uctures to represent th e ozone molecule

Each of the Lewis structures is called a resonance structure A resonance structure i s o ne of two or more Lewis s tructure s for a s ingle molecule that cannot be repr esented accurately by only

one Lewi s st ructure The double-headed arrow indicates that the s tru c ture s s ho wn are resonance

s tructure s Like th e medieval European traveler to Africa who described a rhinoceros as a cross

between a griffin and a unicorn (two familiar but imaginar y an imal s), we de scribe ozone, a real

molecule, in term s of two familiar but n o ne xis tent st ructure s

A common mi sco nception about resonance i s that a molecule s uch as ozone so mehow s hifts

quickly back and forth from one re so nance structure to the ot her Neither resonance s tru cture,

though, adequately represents the actual molecule, w hi c h ha s its own unique , stable structure

" Resonance" is a human invention, de s igned to address the limitation s of a s impl e bonding model

To extend the animal analogy, a rhinocero s i s a di s tinct , real creature, not so me oscillation between

the mythical griffin and unicorn!

The carbonate ion provide s another example of resonan ce :

Ac cording to experimental evidence, all three carbon-oxygen bond s in CO ~- are equi vale nt

Therefore, the properties of the carbonate ion are best explained by considering its resonance

s tructures together

The concept of resonance applies equally well to organic sys tem s A good example is the

be nzene molecule (C6H6):

< •

If one of these re so nance structures conesponded to the actual st ructure of ben ze ne , there would

be two different bond length s between adjacent C atoms, one with the properties of a single bond

and the other with the propertie s of a double bond In fact, the di s tan ce between all adjacent C

a toms in benzene is 140 pm , which is s horter than a C-C bond (lS4 pm ) and l onge r than a C=C

bo nd (133 pm)

SECTION 8.7 Resonance 295

296 CHAPTER 8 Chemica l Bonding I: Basic Concepts

Th e representati o n of organic c ompounds is

d is cussed in m o re de t a il in Chapter 1 0

Think About It Always make sure

that resonance structures differ

only in the positions of the

electrons, not in the positions of the

atoms ,

A simpler way of drawing the structure of the benzene molecule and other compounds

con-taining the benzene ring is to show only the skeleton and not the carbon and hydrogen atoms, By this convention, the resonance structures are represented by

Note that the C atoms at the corners of the hexagon and the H atoms are not shown, although they

Resonance structures differ only in the positions of their electrons not in the positions of

their atom ' s, Thu s , :N=N=O: • • and :N-N - O: are resonance •• structures of each other, whereas :N=N=O: and :'N=O=N: are not o • • •

Sample Prob l em 8,9 shows how to draw resonance str u ctures,

High oil and gasoline prices have renewed interest in alternative methods of producing energy,

including the "clean" burning of coal Part of what makes "dirty" coal dirty is its high sulfur content

Burning dirty coal produces sulfur dioxide (S02)' among other pollutants, Sulfur dipxide is oxidized

Strategy Draw two or more Lewis structures for S03 in which the atoms are arranged the same way but the electrons are arranged differently,

Setup Following the steps for drawing Lewis structures, we determine that a correct Lewis structure

Solution :O=S-O: , • • :O-S - O: • • :O-S=O: ,

Practice Problem A Draw all possible resonance structures for the nitrate ion (NO) ,

Practice Problem B Draw three resonance structures for the tillocyanate ion (NCS- ), and determine

Exceptions to the Octet Rule

be drawn for the nitrite ion (NO z)?

(N and 0 must obey the octet rule,)

The octet rule almost alway s holds for second - period elements, Exceptions to the octet rule fa ll

into three categories:

SECTION 8.8 Exceptions to the Octet Rule 297

2 The central atom ha s fewer than eight electrons due to an odd number of electrons

Incomplete Octets

is fewer than eight Beryllium, for example, which is the Group 2A element in the second

period, has the electron configuration [He]2i Thus, it has two va lence electrons in the 2s

H-Be-H

Only four electrons surround the Be atom, so there i s no way to sa tisfy the octet rule for beryllium

in this molecule

Elements in Group 3A also tend to form compounds in which they are s urrounded by fewer

than eight electrons Boron, for example, ha s the electron configuration [He]2 i2pi, so it ha s only

three valence electrons Boron reacts with the halogens to form a class of compounds ha v in g the

boron atom in boron trifluoride:

We actually can satisfy t he octet rule for boron in BF 3 by u s in g a l o ne pair on one of the F

ftuo-rine carries a positive formal charge, a situation that is incon s i s tent with the electronegativities of

s ingle bond (137.3 pm) The shorter bond length would appear to s upp o rt the idea b e hind the three

resonance structures

On the other hand, boron trifluoride combines with ammonia in a reaction that i s better

rep-res ented using the Lewis s tructure in which boron ha s only six valence electrons around it:

The B-N bond in F3B-NH 3 i s different from the covalent bond s discussed so far in the

e nse that both electrons are contributed by the N atom This type of bond is called a coordinate

co valent bond (also referred to a s a dative bond) , which is d efined as a covalent bond in which

no t differ from those of a normal covalent bond ( i.e , the electrons are s hared in b ot h cases), th e

d ist inction is useful for keeping track of valence electrons and assigning formal charges

Odd Numbers of Electrons

So me molecules, s uch as nitrogen dioxide (NO ? ), contain an odd number of electrons

:O=N-O: • • •

B eca use we need an even number of electrons for every atom in a molecule to ha ve a complete

~u mber of electrons are so metime s referred to as free radicals (or ju st radicals) Many radical s

:rr e highly reactive, becau se there is a tendency for the unpaired electron t o form a covalent bond

' ith an unpaired electron on another molecule When two nitrogen dioxide molecules collide, for

•

298 CHAPTER 8 Chemical Bonding I: Basic Concepts

T he Amer i can Media Inc building in

Boca Raton, Florida

Severe flooding in New Orleans after

Hurricane Katrina in 2005

Think About It C l02 is u sed

primarily to bl eac h wood pulp in

the manufactur e of paper, but it i s

also u se d to bleach flour, di s infect

drinking water, and deodorize

certain indu s trial facilities

Recently, it ha s been used to

eradicate the toxic mold in home s

in New Orlean s that were dama ge d

by the devastating floodwater s of

Hunicane Katrina in 2005

The OH species is a radical, not to be confused

with the hydroxide ion (OW)

example, they form dinitrogen tetroxide, a molecule in which the octet rule is satisfied for bot h the

Beginning about a week after the September 11, 2001 , attacks, letter s containing anthrax bacteria

were mailed to several news media offices and to two U.S senators Of the 22 people who

subse-quently contracted anthrax, five died Anthrax is a spo re-forming b ac terium (Bacillus anthracis)

and, like s mallpox, i s cla ss ified by the CDC as a Category A biot e rrorism agent Spore - forming bacteria are notoriou s ly difficult to kill, making the cleanup of the buildings contaminated by anthrax costly and time-con s uming The American Media Inc (AMI) building in Boca Raton, Florida , was not deemed sa fe to enter until July of 2004, after it had been treated with ch l orine

•

dioxide (Cl O ?), the only structural fumigant approved by the Environmenta l Protection Agency

(EPA) for anthrax decontamination The effectiveness of C I 0 2 in killing anthrax and other hardy biological agents stems in part from it s being a radical, meaning that it contains an odd number

of electrons

Draw the Lewis st m c ture of chlorine dioxide (C l02)

Strategy The ske letal st mcture i s

O-CI-O

Thi s puts the unique atom , Cl, in the center and put s the more electronegative 0 atoms in termina l

pO SItIO n s

Setup There are a t ota l of 19 v alenc e electrons (seve n from the Cl and s ix from each of the two 0

atoms) We s ubtract four electrons to account for the two bond s in the s keleton , leaving us with 15

electrons to distribute as follows: three lone pairs on each 0 atom, one lone pair on the Cl atom, and the la s t remaining electron also on the Cl atom

• • • • • •

Solution :O-CI-O:

• • • ••

,

Practice Problem A Draw the Lewis s tmcture for the OH species

Practice Problem B Dra w the Lewi s s tructure for the NS2 molecule

Expanded Octets

Atoms of the s econd-period elements cannot have more than eight valence e l ectrons around them, but atoms of element s in and beyond the third period of the periodic table can In addition to the 3s and 3p orbitals, elements in the third period also have 3d orbita l s that can be used in bo n d i ng

The se orbitals enable an atom to form an expanded octet One compound in which there is an expanded octet is sulfur hexafluoride , a very stable compound The electron configuration of sul-

fur i s [Ne]3i3p 4 In SF6, each of sulfur's 6 valence electrons forms a covalent bond with a fluorine atom, so there are 12 electrons around the central sulfur atom:

SECTION 8.8 Exceptions to the Octet Rule 299

In Chapter 9 we will see that these 12 electrons, or six bonding pairs, are accommodated in six orbitals that originate from the one 3s, the three 3p, and two of the five 3d orbitals Sulfur also

forms many compounds in which it does obey the octet rule In sulfur dichloride, for instance , S is

surrounded by only eight electrons:

Draw the Lewis structure of boron triiodide (El3)

Strategy Follow the step-by-step procedure for drawing Lewis structures The skeletal structure is

I

I

I-B-I

Setup There are a total of 24 valence electrons in BI3 (three from the B and seven from each of

the three I atoms) We subtract 6 electrons to account for the three bonds in the skeleton, leaving 18

electrons to distri bute as three lone pairs on each I atom

however, BI3 can be drawn with a double bond in order to satisfy the octet of boron This gives rise to a total of four resonance structures:

Practice Problem B Draw the Lewis structure of boron trichloride (BCI3)

~I _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ ~

Draw the Lewis structure of arsenic pentafluoride (AsFs)

Strategy Follow the steps for drawing Lewis structures The skeletal structure already has more than

an octet around the As atom

Setup There are 40 total valence electrons [five from As (Group SA ) and seven from each of the five

F atoms (Group 7 A)] We subtract 10 electrons to account for the five bonds in the skeleton, leaving

30 to be distributed Next, place three lone pairs on each F atom, thereby completing all their octets

and using up all the electrons

Practice Problem A Draw the Lewis structure of phosphorus pentachloride (PCIs)

Practice Problem B Draw the Lewis structure of antimony pentafluoride ( SbFs)'

8eF 2 is also known as beryllium difluoride, but

form s with fluor i ne, t he name beryllium fluoride

Think About It Always make

sure that the number of electrons represented in your final Lewis

structure matches the total number

of valence electrons you are supposed to have

300 CHAPTER 8 Chemical Bonding I: Basic Concepts

Think About It Atoms beyond the

more than an octet of electrons,

in bonds or reside on the atom as

lone pairs

When drawing Lewis structures of compounds containing a centra l atom from the third period and beyond, the octet rule may be satisfied for all the atoms before all the valence electrons

have been used up When this happen s, the extra electrons should be placed as lone pairs on the

central atom Sample Problem 8.13 illu st rates this approach

Sample Problem 8.13 ••

Draw the Lewis structure of xenon tetrafluoride (XeF4)

Strategy Follow the steps for drawing Lewis structures The skeletal structure is

We subtract eight electrons to account for the bonds in the skeleton, leaving 28 to distribute We first

lone pairs on the Xe atom

of the diatomic hydrogen molecule, for examp l e, is

H 2 (g) - -+ H(g) + H(g) i1W = 436.4 kJ/mol

SECTION 8.9 Bond Enthalpy 301

Measuring the strength of covalent bond s in poly atomic molecule s i s more complicated For

OH(g ) • H(g ) + O (g )

D.W = 502 kJ / mol D.H o = 427 kJ / mol

are different For polyatomic molecule s , therefore, we speak of the avera ge bond enthalpy of

a particular bond For example , we can measure the enthalpy of the 0- H bond in 10 different

e nthalpies by 10 Table 8.6 lists the average bond enthalpies of a number of diatomic and

A comparison of the thermochemical c hanges that take place during a number of reaction s

com-b ustion of hydrogen gas in oxygen ga s is fairly e xoth e rmi c :

The formation of glucose from carbon dioxide and water , on the other hand , be s t achie v ed by

r he thermochemical nature of the reactions that molecule s underg o

In many cases, it is po s sible to predict the approximate enthalp y of a reaction b y u s ing the

av erage bond enthalpies Because energy is always required to break chemical bond s and chemical

b ond formation is a lways accompanied by a relea s e of energy , we can e s timate the e nthalp y of a

re action by counting the total number of bonds broken and formed in the reacti o n and recording all

th e corresponding enthalpy changes The enthalpy of reaction in the ga s pha s e i s given by

= total energy input (to br e ak bond s) - total energy r e le a s e d ( b y bond f o rm a ti o n )

w here BE stands for average bond enthalpy and 2, is the s ummation sign A s writt e n , Equation 8.3

in the reactants is less than the total energy releas e d when bond s are formed in the product s, then

~ o is negative and the reaction is exothermic [Figure 8.8 ( a ) ] On the other hand , "ifle ss energ y i s

re leased (bond making) than ab s orbed (bond breaking ), D.H o i s po s iti v e and the reacrion i s end o

-th ermic [Figure 8.8(b)]

•

To many students, Equation 8 3 ap p ea rs to b e backward Ordinarily you calcu l ate a change as

final minus initial Her e we a re determining th e

differ e nce between the amount of heat w e ha v e

to add to break reactant bonds a n d t he a m o unt

of heat releas e d when product bon d s for m T h e sign of the final answ e r tells us if t he p ro ces s is endo t hermic ( + ) or exothermic (-) o ve ra l l

Bond e n t halp i es for diatomic mol e cules

have more signific a nt figur e s than those for

polyato m ic mo l ecules Those f or poly a tomic

molec ul es are av er age va l ues base d on the

bonds in mor e than one compound

(a) an exothermic reaction and (b) an

endothermic reaction The t:.H o values

are calculated using Equation 5.19 and

tabulated t:.H ? values f rom Appendix 2

* Bond enthalpies sho w n in red are for diatomic molecules

' The c=o bond enthalp y i n CO2 i s 799 kJ / mo !

Reactant molecules

Atoms

Product molecules

s-s s=s

Product molecules

(b)

SECTION 8.9 Bond Enthalpy 303

If all the reactant s and product s are diat o mi c mol ec ule s, t h e n the e qu at ion f or the en thalp y

of reaction will yield accurate re s ult s because the bond enthalpie s of diatomic molecule s are

accu-rately known If s ome or all of th e reactant s and pr oducts are p o l y atomic molecules, th e e quati on

will yield only approximate result s becau s e th e bond enthalpies u s ed w ill be averages

Bonds to break: 4 C-H and 2 0=0

Bonds to form: 2 C=O and 4 H-O

[4(414 kJ / mol ) + 2(498.7 kJ/mol)] - [2(799 kJ / mol) + 4(460 k J / mol ) ] = - 785 k J /mo !

Practice Problem Use bond enthalpies from Table 8.6 to estimate the enthalpy of reaction for the

combination of carbon monoxide and oxygen to produce carbon dioxide:

Bond Enthalpy

8 9 1 Use data from Table 8.6 to estimate 8 9 2 Use data from Table 8.6 to estimate

t1Hr x n for the reaction of ethylene with t1Hr xn for the reaction of fl uorine and

hydrogen to produce ethane chlorine to produce ClF

Think About It Use Equation 5.19 [ ~ Section 5.6] and data from Appendix 2 to calculate this enthalpy

of reaction again; then compare your results using the two approaches

The difference in this case is due

to two things: Most tabulated bond

enthalpies are averages and, by

convention, we show the product

of combustion as liquid water-but average bond enthalpies apply to

species in the gas phase, where there

is little or no influence exerted by molecules

Remember that heats of reaction are expressed

in kJ/mol, where the "per mole" refers to pe r

mole of reaction as written [H SectIOn 5 3]

304 CHAPTER 8 Chemical Bonding I: Basic Concepts

Applying What You've Learned

Researchers in the early 1990s made the sensational announcement that nitric oxide

play an important role in human physiology They found that NO serves as a signal

including in the cardiovascular, nervous, and immune systems

The discovery of the biological role of nitric oxide has shed light on how

the heart, although how it worked wa s not understood We now know that nitroglycerin produces nitric oxide in the body, which ca u ses muscles to relax and allows the arter ie s

Research continues to uncover the role nitric oxide plays in biological processes,

and medicine continues to find new uses for this molecule

Problems:

•

use bond enthalpies to estimate D.H o for the reaction [ ~~ Sample Problem 8.13 ]