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

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. , M ultimedia Chemical Bonding - formal ch arge ca lculation s. Draw the Lewis structure for carbon disulfide (CS 2 ). 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. S-C-S Step 2: The total number of valence electrons is 16, six from each S atom and four from the C atom [2(6) + 4 = 16]. Step 3: Subtract four electrons to account for the bonds in the skeletal structure, leaving us 12 electrons to distribute. Step 4: St ep 5: Step 6: Distribute the 12 remaining electrons as three lone pairs on each S atom. • • • • :S-C-S: •• •• 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 ato m. Solution • • :S=C=S: • • Practice Problem A Draw the Lewis structure for NF 3 . 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 (H COO H ). • • • • • • a) H-C-O-O-H • • • • • • • • • • b) H-O=C=O-H • • • • • • c) H -O-C-O-H • • • • • • '0' • • II d) H-C-O-H '0' • • I e) :O-C-H I H 8.5.2 Identify the correct Lewis structure for hydrogen peroxide (H 2 0 2 ) . " a) H-O=O-H b) H-O=O-H •• •• c) H-O-O-H • • • • d) H=O=O=H • • • • e) H-H-O-O: •• • • Lewis Structures and Formal Charge So far you have learned two different methods of electron "bookkeeping." In Chapter 4, you learned about oxidation numbers [ ~~ Sectio n 4.4], and in Section 8.4, you learned how to calculate par- tial charges. There is one additional commonly used method of electron bookkeeping namely, formal charge, which can be used 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 structure with the number of electrons that would be associated with the isolated atom. In an isolated atom, the number of electrons associated with the atom is simply 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 structure, 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 associated with the atom. formal charge = valence electrons - associated electrons Equation 8. 2 An atom's formal charge is calculated as follows: We can illustrate the concept of formal charge using the ozone molecule (0 3 ) , Use the step-by-step method for drawing Lewis structures to draw the Lewis structure for ozone, and then determine the formal charge on each 0 atom by subtract- ing the number of associated electrons from the number of valence electrons. 2 unshared + 6 s h~red = 5 e- 4 unshared + 4 s har ed = 6 e- 2 Valence e- e- associated with atom Difference (formal charge) 6 un shared + 2 s \ared = 7 e- 6 6 6 6 5 7 o + 1 -1 \ / :0=0-0: • • • As with oxidation numbers, the sum of the formal char ges mu st equal the overall charge on the species [I ~~ Section 4.4]. Because 0 3 is a molecule, its formal charges mu st s um to zero. For ions, the formal charges mu st s um to the overall charge on the ion. Formal charges do not repres ent actual charges on atoms in a molecule. In the 0 3 molecule, for example, there is no evidence that the central atom bears a net + 1 charge or that one of the terminal atoms bears a - 1 charge. Assigning formal charges to the atoms in the Lewis structure merely helps us keep track of the electrons inv ol ved in bonding in the molecule. Sample Problem 8.7 lets you practice determining formal charges. The widespread use of fertilizers has resulted in the contamination of some groundwater with nitrates, which are potentially harmful. Nitrate to xicity is due primarily to its conversion in the body to nitrite ( N0 2 ), which interferes with the ability of hemoglobin to transport oxyge n. 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 L ew is structure of NO) . For each atom, subtract the associated electrons from the valence electrons. Setup '0' • • I . :O-N=O: •• • The N atom has five valence eJ ectrons 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 ve n 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 structur e: I :O-S-O: • • • • • • ~I ~ ________________________________________________________ ~ While you are new at determ ini ng fo rmal charges, it may be helpful to draw L ewis . structures with all dots, rathe r than d ashes. T his can make it easier to see how ma ny electrons ar e associated with ea ch atom. • •• • • 0: O' 0: . Remember that f or the pu r pose of counting associated elect ron s, tho se s har ed by two at oms are evenly split between them. Think About It The sum of formal charges (+ 1) + ( -1 ) + (- 1) + (0) = -1 is equal to the ov er all 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 s pecies. In such cases, we often can select the best skeletal arrangement by using formal charges and the following guidelines: • For molecule s, 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 for- mal charges (+2, +3, and so on). • 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 (CH 2 0), 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 CH 2 0 . H-C - O-H o I H-C-H Strategy Complete the Lewis structures for each of the CH 2 0 skeletons shown and determine the formal charges on the atoms in each one. Setup The completed Lewis structures for the skeletons shown are: • • • • H-C=O-H • • '0' II H-C-H 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 ato m: 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 va lence 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 Solution Of the two possible arrangements, the structure on the left has an 0 atom with a positive 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 CH 2 0. Practice Problem A Two possible arrangements are shown for the Lewis structure of a carboxyl group, -COOH. Use formal charges to determine which of the two arrangements is better. • • '0' II • • • • • • -C-O-H -C-O=O-H Practice Problem B Use Lewis structures and formal charges to determine the best skeletal arrangement of atoms in NCI 2 . 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 -? a) [~=C=qj b) ~B-c=oJ c) ~N=C-QJ d) ~B-c =q J e) [~ =C-QJ Our drawing of the Lewis structure for ozone (0 3 ) satisfied the octet rule for the central 0 atom because we placed a double bond between it and one of the two terminal 0 atoms. In fact, we can put the double bond at either end of the molecule, as shown by the following two equivalent Lewis structures: '0=0-0: < -_. '. • • • • • :0-0=0: . A single bond between 0 atoms should be longer than a double bond between 0 atoms, but experi- mental evidence indicates that both of the bonds in 0 3 are equal in length (128 pm). Because neither one of these two Lewis structures accounts for the known bond lengths in 0 3 , we use both Lewis structures to represent the ozone molecule. Each of the Lewis structures is called a resonance structure. A resonance structure is one of two or more Lewis structures for a single molecule that cannot be represented accurately by only one Lewis structure. The double-headed arrow indicates that the structures shown are resonance structures. Like th e medieval European traveler to Africa who described a rhinoceros as a cross between a griffin and a unicorn (two familiar but imaginary animals), we describe ozone, a real molecule, in terms of two familiar but nonexistent structures. A common mi sconception about resonance is that a molecule such as ozone somehow shifts quickly back and forth from one resonance structure to the other. Neither resonance structure, though, adequately represents the actual molecule, which has its own unique, stable structure. "Resonance" is a human invention, designed to address the limitations of a simple bonding model. To extend the animal analogy, a rhinoceros is a distinct, real creature, not some oscillation between the mythical griffin and unicorn! The carbonate ion provides another example of resonance: 2- ?- - 2- '0' • • '0' • • '0' • • . I 'O=C-O: • • • • . . I . :O-C=O • • • • II :O-C -O : • • • • < < • • Ac cording to experimental evidence, all three carbon-oxygen bonds in CO ~- are equivale nt. Therefore, the properties of the carbonate ion are best explained by considering its resonance structures together. The concept of resonance applies equally well to organic systems. A good example is the benzene molecule (C 6 H 6 ): < • If one of these resonance structures conesponded to the actual structure of benzene, there would be two different bond lengths between adjacent C atoms, one with the properties of a single bond and the other with the properties of a double bond. In fact, the di stan ce between all adjacent C atoms in benzene is 140 pm , which is shorter than a C-C bond (lS4 pm) and longer than a C=C bond (133 pm). SECTION 8.7 Resonance 295 296 CHAPTER 8 Chemica l Bonding I: Basic Concepts Th e representati on of organic compounds is d is cussed in mo re de ta 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 , "" are' understood to 'be' th 'ere, Only the bonds between the C atoms are shown, Resonance structures differ only in the positions of their electrons not in the positions of •• •• their atom' s, Thus, :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 Problem 8,9 shows how to draw resonance structures, 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 (S0 2)' among other pollutants, Sulfur dipxide is oxidized in the atmosphere to form sulfur trioxide (S0 3), willch subsequently combines with water to produce sulfuric acid-a major component of acid rain, Draw a ll possible resonance structures of sulfur trioxide, Strategy Draw two or more Lewis structures for S0 3 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 for S0 3 contains two sulfur-oxygen single bonds and one sulfur-oxygen double bond, '0' , , , I :O=S-O: , But the double bond can be put in anyone of three positions in the molecule, , , • • '0' , . U '0' , . • I II I • 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 the formal charges in each resonance structure, . Checkpoint 8.7 Resonance 8.7.1 Indicate which of the following are 8.7.2 •• •• resonarlce structures of :Cl -Be-Cl: • • • • (s elect all that apply), , . a) :Cl=Be=Cr • • b) :Cl=Be-Cl : c) :Be=Cl - <:;:}: d) :CI-Be=Cl: . , e) :Be=C l =Cr , . Exceptions to the Octet Rule How many resonance structures can be drawn for the nitrite ion (NO z )? (N and 0 must obey the octet rule,) a) 1 b) 2 c) 3 d) 4 e) 5 The octet rule almost always holds for second-period elements, Exceptions to the octet rule fall into three categories: SECTION 8.8 Exceptions to the Octet Rule 297 I. The central atom has fewer than eight electrons due to a shortage of electrons. 2. The central atom has fewer than eight electrons due to an odd number of electrons. 3. The central atom has more than eight electrons. Incomplete Octets In so me compounds the number of electrons surrounding the central atom in a stable molecule 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 valence electrons in the 2s orbital. In the gas phase, beryllium hydride (BeH 2 ) exists as discrete molecules. The Lewis structure of BeH 2 is H-Be-H Only four electrons surround the Be atom, so there is no way to satisfy the octet rule for beryllium in this molecule. Elements in Group 3A also tend to form compounds in which they are surrounded by fewer than eight electrons. Boron, for example, has the electron configuration [He]2 i2pi, so it has only three valence electrons. Boron reacts with the halogens to form a class of compounds having the general formula BX 3 , where X is a halogen atom. Thus, there are only six electrons around the boron atom in boron trifluoride: • • 'F' • • · . I :F-B-F: • • • • We actually can satisfy the octet rule for boron in BF 3 by using a lone pair on one of the F atoms to form a double bond between the F atom and boron. This gives rise to three additional resonance structures: • • :F: . . I F=B-F: • • • • • • • • . F' · . II :F-B-F: • • • • •• 'F' • • . . I . • • :F-B=F: • • • Although these resonance structures result in boron carrying a negative formal charge while ftuo- rine carries a positive formal charge, a situation that is inconsistent with the electronegativities of the atoms involved, the experimentally determined bond length in BF 3 (130 .9 pm) is shorter than a single bond (137.3 pm). The shorter bond length would appear to support the idea behind the three resonance structures. On the other hand, boron trifluoride combines with ammonia in a reaction that is better rep- resented using the Lewis structure in which boron has only six valence electrons around it: • • ·F· • • . . I :F -B + . . I :F: • • H I :N-H I H •• :F: H . . I I ) :F-B-N-H . . I I :F: H • • It seems, then, that the properties of BF 3 are best explained by all four resonance structure s. The B-N bond in F 3 B-NH 3 is different from the covalent bonds discussed so far in the ense that both electrons are contributed by the N atom. This type of bond is called a coordinate co valent bond (also referred to as a dative bond), which is defined as a covalent bond in which one of the atoms donates both electrons. Although the properties of a coordinate covalent bond do not differ from those of a normal covalent bond (i.e., the electrons are shared in both cases), the distinction is useful for keeping track of valence electrons and assigning formal charges. Odd Numbers of Electrons Some molecules, such as nitrogen dioxide (NO?), contain an odd number of electrons. :O=N-O: • • • Because we need an even number of electrons for every atom in a molecule to have a complete octet, the octet rule cannot be obeyed for all the atoms in these molecules. Molecules with an odd ~u mber of electrons are sometimes referred to as free radicals (or ju st radicals). Many radicals :rr e highly reactive, because there is a tendency for the unpaired electron to 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 The American Media Inc. building in Boca Raton, Florida. Severe flooding in New Orleans after Hurricane Katrina in 2005. Think About It C l0 2 is used primarily to bleach wood pulp in the manufacture of paper, but it is also used to bleach flour, di sinfect drinking water, and deodorize certain industrial facilities. Recently, it has been used to eradicate the toxic mold in homes in New Orleans that were damaged by the devastating floodwaters 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 Nand 0 atoms. " 'l)". , ,~ \ I ' N, + ,N , j \ '0: :0, , ' •• Bringing Chemistry to Life T he Power of Radica ls " 'l>: :0 \\ I ' • N-N . 1 \\ :0. 0" • • •• 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, is cla ssified by the CDC as a Category A biot errorism 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 CI0 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. Sample Problem 8.10 . Draw the Lewis stmcture of chlorine dioxide (C l0 2 ) . Strategy The skeletal stmcture is O-CI-O This puts the unique atom, Cl, in the center and puts the more electronegative 0 atoms in terminal pO SItIO n s. Setup There are a total of 19 valence electrons (seven from the Cl and s ix from each of the two 0 atoms). We subtract four electrons to account for the two bonds in the skeleton, 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 last remaining electron also on the Cl atom. • • • • • • Solution :O-CI-O: • • • •• . . . . . . . . . . . , Practice Problem A Draw the Lewis stmcture for the OH species. Practice Problem B Draw the Lewis structure for the NS 2 molecule. Expanded Octets Atoms of the s econd-period elements cannot have more than eight valence el 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 ls that can be used in bo ndi 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 is [Ne]3i3p 4. In SF 6 , 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: :F: "p' I F: '. "S __ .0 :F I F: •• • • :F: •• 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: I • • • • • • :Cl -S -Cl: • • • • • • Sample Problems 8.11 and 8.12 involve compounds that do not obey the octet rule. . _. . . Sample Problem 8.11. . Draw the Lewis structure of boron triiodide (El 3 ). 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 BI 3 (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. Solution . I . • • I :I-B-I: •• • • . I . • • I :I-B-I: . I . • • • I • • :I=B-I: • • • Think About It Boron is one of the elements that does not always follow the octet rule. Like BF 3 , 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: • • T . I . • • II I • :I-B-I: • • :I-B=I: • • • •• •••••••••••••••••••••••••••••• •••• • Practice Problem A Draw the Lewis structure of beryllium fluoride ( BeF 2 ). Practice Problem B Draw the Lewis structure of boron trichloride (BCI 3 ). ~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. F I / F F-As I " F F 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. Solution :F: . I ;f: :F-As I " Ii: 'F' " . • • Practice Problem A Draw the Lewis structure of phosphorus pentachloride (PCI s ). Practice Problem B Draw the Lewis structure of antimony pentafluoride ( SbF s )' 8eF 2 is also known as beryllium difluoride, but because this is the only compound that beryllium forms with fluori ne, t he name beryllium fluoride is unambiguous. 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 second period can accommodate more than an octet of electron s, whether those electrons are used in bonds or reside on the atom as lone pairs. When drawing Lewis structures of compounds containing a central 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 happens, the extra electrons should be placed as lone pairs on the central atom. Sample Problem 8.13 illustrates this approach. Sample Problem 8.13 •• Draw the Lewis structure of xenon tetrafluoride (XeF 4 ). Strategy Follow the steps for drawing Lewis structures. The skeletal structure is F I F-Xe-F I F Setup There are 36 total valence electrons (eight from Xe and seven from each of the fOllr F atoms). We subtract eight electrons to account for the bonds in the skeleton, leaving 28 to distribute. We first complete the octets of all four F atoms. When this is done, four electrons remain, so we plilce two lone pairs on the Xe atom. Solution • • :F: . . . . I :F- Xe- F: . . I . . . . :F: • • Practice Problem A Draw the Lewis structure of the iodine tetrachloride ion (ICI 4 ). Practice Problem B Draw the Lewis structure of krypton difluoride (KrF 2)' Checkpoint 8.8 Exceptions to the Octet Rule 8.8.1 In which of the following species does 8.8.3 In which species does the central atom the central atom not obey the octet obey the octet rule? (Select a ll that rule? apply.) a) CI0 2 a) I) b) CO 2 b) BH 3 c) BrO) c) AsF6 d) HeN d) N0 2 e) ICl 4 e) CI0 2 8.8.2 Which elements cannot have more than 8.8.4 How many lone pairs are there on the an octet of electrons? (Select all that central atom in the Lewis structure of apply.) ICI 2 ? a) N a) 0 b) C b) I c) S c) 2 d) Br d) 3 e) 0 e) 4 Bond Enthalpy One measure of the stability of a molecule is its bond enthalpy, which is the enthalpy change asso- . ciated with breaking a particular bond in 1 mole of gaseous molecules. (Bond enthalpies in solids and liquids are affected by neighboring molecules.) The experimentally determined bond enthalpy of the diatomic hydrogen molecule, for example, is H2(g) - -+. H(g) + H(g) i1W = 436.4 kJ/mol SECTION 8.9 Bond Enthalpy 301 According to this equation, breaking the covalent bond s in 1 mole of ga seous H2 molecules requires 436.4 kJ of energy. For the less stable chlorine molecule, Cl i g) +. CI(g) + CI(g) D.W = 242.7 kJ/mol Bond enthalpies can also be directly mea sured for heteronuclear diatomic molecules, s uch as HCI, HCI(g) +. H(g) + Cl(g) as well as for molecules containing multiple bonds: 0 2 (g ) +. O(g) + O(g) N 2 (g) • N(g) + N( g) D.W = 431.9 kJ/ mol D.W = 498.7 kJ/ mol D.W = 941.4kJ / mol Measuring the strength of covalent bonds in poly atomic molecules is more complicated. For example, measurements show that the energy needed to break the first 0- H bond in H?O is differ- ent from that needed to break the second 0- H bond: H 2 0(g) +. H(g) + OH( g) OH(g ) • H(g) + O (g ) D.W = 502 kJ / mol D.H o = 427 kJ / mol In each case, an 0- H bond is broken, but the first step requires the input of more energy than the second. The difference between the two D.H o values suggests that the second 0- H bond itself undergoes change, because of the changes in its chemical environment. We can now understand why the bond enthalpy of the s ame 0- H bond in two different molecules, such as methanol (CH 3 0H) and water (H 2 0 ), will not be the same: their environments are different. For polyatomic molecules, therefore, we speak of the average bond enthalpy of a particular bond. For example, we can measure the enthalpy of the 0- H bond in 10 different polyatomic molecules and obtain the average 0- H bond enthalpy by dividing the s um of the bond enthalpies by 10. Table 8.6 lists the average bond enthalpies of a number of diatomic and poly- atomic molecules. As we noted earlier, triple bonds are stronger than double bonds, and double bonds are stronger than single bonds. A comparison of the thermochemical changes that take place during a number of reactions reveals a strikingly wide variation in the en thai pies of different reactions. For example, the com- bustion of hydrogen gas in oxygen ga s is fairly exothermic: D.H o = - 285.8 kJ/ mol The formation of glucose from carbon dioxide and water, on the other hand, best achiev ed by pho- tosynthesis, is highly endothermic: D.H 0 = 2801 kJ/mol We can account for such variations by looking at the stability of individual reactant and product molecules. After all, most chemical reactions involve the making and breaking of bond s. There- fore, knowing the bond enthalpies and hence the stability of molecules re veals something about rhe thermochemical nature of the reactions that molecules undergo. In many cases, it is possible to predict the approximate enthalpy of a reaction by using the av erage bond enthalpies. Because energy is always required to break chemical bond s and chemical bond formation is always accompanied by a release of energy, we can estimate the enthalpy of a reaction by counting the total number of bonds broken and formed in the reaction and recording all the corresponding enthalpy changes. The enthalpy of reaction in the ga s pha se is given by D.W = 2,BE(reactants) - 2,BE(products) Equation 8.3 = total energy input (to br eak bond s) - total energy released (by bond formation) . . where BE stands for average bond enthalpy and 2, is the summation sign. As written, Equation 8.3 rakes care of the sign convention for D.H o. Thus, if the total energy input needed to break bond s in the reactants is less than the total energy released when bonds are formed in the products, then ~ o is negative and the reaction is exothermic [Figure 8.8(a)]. On the other hand, "ifless energy is released (bond making) than absorbed (bond breaking), D.H o is po sitive and the reacrion is endo- thermic [Figure 8.8(b)]. • To many students, Equation 8 .3 ap p ea rs to be backward. Ordinarily you calcu l ate a change as final minus initial. Her e we a re determining the differe nce between the amount of heat we ha ve to add to break reactant bonds and t he amount of heat releas ed when product bon ds form. The sign of the final answ er tells us if t he p ro ces s is endo t hermic (+ ) or exothermic (-) o ve ra l l. [...]... F :F: 'N' g b) The difference in electronegativity is (3.5 - 3.0) = 0.5, making this a polar covalent bond c) The partial charges on Nand o 0 o in NO are +0.029 and - 0.029, respectively d) :N=Q: and l ' ' ' e) Formal charges are indicated on each - '.0,, -;:::-0: Lewis structure in part (d) f) /::"Ho for the N 1 1 + explosive decomposition of nitroglycerin H :0: H is -l.94 X 103 kJ/mo! a) • • •... fundamental scents: floral, pepperminty, musky, pungent, camphorlike, ethereal, and putrid Different parts of a larger molecule may have distinctly different shapes, enabling it to stimulate more than one type of olfactory receptor This results in the perception of a combination of odors For example, parts of the benzaldehyde molecule are shaped such that they stimulate the camphorlike, floral, and... they will arrange themselves to be as far apart as possible, thus minimizing the repulsive interactions between them We can visualize the arrangement of electron domains using balloons, as shown in Figure 9.1 Like the B atoms in our ABx molecules, the balloons are all connected to a central, fixed point, which represents the central atom (A) When they are as far apart as possible, they adopt the five geometries... enthalpies in Table 8.6 (b) Calculate the enthalpy of reaction from the standard enthalpies of formation (see Appendix 2) of the reactant and product molecules, and compare the result with your answer for part (a) 8.63 Classify the following substances as ionic compounds or covalent compounds containing discrete molecules: CH4 , KF, CO, SiCI4, BaCl z· 8.77 Which of the following are ionic compounds and... Draw resonance structures for the carboxylate group NHr,C H 6 • (a) (b) :O-C-O: •• :O=C-o.: •• •• (d) :O-C-O: •• •• 8.107 Draw Lewis structures for the following chlorofluorocarbons (CFCs), which are partly responsible for the depletion of ozone in the stratosphere: (a) CFCI 3 , (b) CF2Cl z, (c) CHF1CI, (d) CF3CHF2· 8.108 Draw Lewis structures for the following organic molecules: C ZH 3F, C 3H 6 ,... reaction I H-C-C-O 97 (c) • For each of the following organic molecules draw a Lewis structure in which the carbon atoms are bonded to each other by single bonds: (a) CZH 6 , (b) C 4H JQ, (c) C SH 12 For parts (b) and (c), show only structures in which each C atom is bonded to no more than two other C atoms 6, > N \ H-C-H H H :P=C=o.: •• 8.106 N I • •• The following is a simplified (skeletal) structure... t1Ho for the reaction Cl K (g) 8.129 Complete the Lewis structure and indicate the coordinate covalent bonds in the molecule 8.120 8.121 The hydroxyl radical (OH) plays an important role in atmospheric chemistry It is highly reactive and has a tendency to combine with an H atom from other compounds, causing them to break up Thus OH is so metimes called a "detergent" radical because it helps to clean... propellant It is used to displace air from potato chip bags in order to extend shelf life and as the propellant in whipped cream canisters In recent years N 20 has become popular as a recreational drug, due in part to its ready availability to consumers Although N 2 0 is legal, it is regulated by the FDA; its sale and distribution for the purpose of human consumption are not permitted Bond Use formal charges... Problem 8.1] [I~~ • b) Classify the bond in NO as nonpolar, polar, or ionic c) Given the experimentally determined dipole moment (0.16 D) and the bond length o (1.15 A), determine the magnitude of the partial charges in the NO molecule [ ~~ Sample Problem 8.5] d) Draw the Lewis structures for NO and for nitroglycerin (C3HsN30 9)' [ ~~ Sample Problem 8.6] e) Determine the formal charges on each atom... of the olfactory system Their work shed new light on how our sense of smell works Although it is somewhat more complicated than can be explained in detail here, the odor of a molecule depends in large part on which of the olfactory receptors it stimulates Olfactory receptors are specialized proteins, located on hairlike cilia in the back of the nose Stimulation of a receptor triggers an electrical signal . For parts (b) and (c), show only structures in which each C atom is bonded to no more than two other C atoms. Dr aw Lewis structures for the following chlorofluorocarbons (CFCs), which are partly. " 'l)". , ,~ I ' N, + ,N , j '0: :0, , ' •• Bringing Chemistry to Life T he Power of Radica ls " 'l>: :0 \ I ' • N-N effectiveness of CI0 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.