Chemistry part 15, Julia Burdge,2e (2009) pdf

344 CHAPTER 9 Chemical Bonding II: Molecular Geometry and Bonding Theories Figure 9.14 Bond order determination for Li2 and Be 2' When we draw mo lecular orbital diagrams, we need only show the valence orbitals and electrons. D IT] [] [ill [ill 2s 2s 2s 2s Li atom Li atom Be atom Be atom Liz molecule Bez molecule 2-0 Bond order = = 1 2-2 Bond order = = 0 2 2 molecule is. The higher the bond order, the more stable the molecule. Bond order is calculated in the following way: Equation 9.1 bond order number of electrons in bonding molecular orbitals 2 number of electrons in antibonding molecular orbitals In the case of H z, where both electrons reside in the (TIs orbital, the bond order is [(2 - 0)/2] = 1. * In the case of He l> where the two additional electrons reside in the (Ti s orbital, the bond order is [(2 - 2)/2] = O. Molecular orbital theory predicts that a molecule with a bond order of zero will not exist and He 2, in fact, does not exist. We can do similar analyses of the molecules Li2 and Bez. (The Li and Be atoms have ground- state electron configurations of [Hel2sl and [Hel2i, respectively.) The 2s atomic orbitals also * combine to form the corresponding (T and (J molecular orbitals. Figure 9.14 shows the molecular . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . orbital diagrams and bond orders for Li z and Be ? As predicted by molecular orbital theory, Liz, with a bond order of 1, is a stable molecule, whereas Be z, with a bond order of 0, does not exist. 7T Molecular Orbitals In order to consider diatomic molecules beyond Bel> we must also consider the combination of p atomic orbitals. Like s orbitals, p orbitals combine both constructively, to give bonding molecular orbitals that are lower in energy than the original atomic orbitals, and destructively, to give anti- bonding molecular orbitals that are higher in energy than the original atomic orbitals. However, the orientations of Px, P Y' and pz orbitals give rise to two different types of molecular orbitals: (T molecular orbitals, in which the regions of electron density in the bonding and antibonding molecular orbitals lie along the internuclear axis, and '7T molecular orbitals, in which the regions of electron density affect both nuclei but do not lie along the internuclear axis. Orbitals that lie along the internuclear axis, as the 2px orbitals do in Figure 9.1S(a), point directly toward each other and combine to form (J molecular orbitals. Figure 9.1S(b) shows the combination of two 2px atomic orbitals to give two molecular orbitals designated (T zpx and (T;p x' Figure 9.1S(c) shows the relative energies of these molecular orbitals. Orbitals that are aligned parallel to each other, like the 2py and 2pz orbitals shown in Figure 9.1S(a), combine to form '7T molecular orbitals. These bonding molecular orbitals are des- ignated '7T z p , and '7Tz p,; the corresponding antibonding molecular orbitals are designated '7T ;p and >'< ) ~ • y . '7T 2Jl-' Often we refer to the molecular orbitals collectively using the designations '7Tzp and '7T;p , _. {, Y,z },", Figure 9.16(a) shows the constructive and destructive combination of parallel p orbitals. Figure 9 .16(b) shows the locations of the molecular orbitals resulting from the combination of PY' and pz orbitals relative to the two atomic nuclei. Again, electron density in the resulting bonding molecu- lar orbitals serves to hold the nuclei together, whereas electron density in the antibonding molecu- lar orbitals does not. Just as the p atomic orbitals within a particular shell are higher in energy than the s orbital in the same shell, all the molecular orbitals resulting from the combination of p atomic orbitals are higher in energy than the molecular orbitals resulting from the combination of s atomic orbitals. To understand better the relative energy levels of the molecular orbitals resulting from p-orbital combinations, consider the fluorine molecule (F z ). In general, molecular orbital theory predicts that the more effective the interaction or over- lap of the atomic orbitals, the lower in energy will be the resulting bonding molecular orbital and the higher in energy will be the resulting antibonding molecular orbital. Thus, the relative energy , y (a) ~ • + (b) (c) Figure 9.15 (a) Two sets of 2p orbitals. (b) The p atomic orbitals that point toward each other (P x) combine to give bonding and antibonding (T molecular orbitals. (c) The antibonding (T molecular orbital is higher in energy than the corresponding bonding (T molecular orbital. + • • + Bonding Antibonding (a) + • + Bonding Antibonding (b) Figure 9.16 Parallel p atomic orbitals combine to give 7f molecular orbitals. (a) Bonding and antibonding molecular orbitals shown separately. (b) Bonding and antibonding molecular orbitals shown together relative to the two nuclei. 345 346 CHAPTER 9 Chemical Bonding II: Molecular Geometry and Bonding Theories Figure 9.17 (a) Ordering of molecular orbital energies for O 2 and F 2 . (b) Ordering of molecular orbital energies for Li 2 , B }, C 2 , and N 2 . Bonding orbitals are darker; antibonding orbitals are lighter. • * (J2 s 0" 2s Bond order Bond length (pm) Bond enthalpy (klima!) D o D [ill 1 267 104.6 Magnetic properties Diamagnetic • (a) D o D D (b) D D D D levels of molecular orbitals in F2 can be represented by the diagram in Figure 9.17(a). The Px orbit- als, which lie along the internuclear axis, overlap most effectively, giving the lowest-energy bond- ing molecular orbital and the highest-energy antibonding molecular orbital. The order of orbital energies shown in Figure 9.17(a) assumes that p orbitals interact only with other p orbitals and s orbitals interact only with other s orbitals-that there is no significant interaction between sand p orbitals. In fact, the relatively smaller nuclear charges of boron, car- bon, and nitrogen atoms cause their atomic orbitals to be held less tightly than those of atoms with larger nuclear charges, and some s-p interaction does take place. This results in a change in the relative energies of the (J 2p and 1T2p , molecular orbitals. Although energies of several of the result- x },Z ing molecular orbitals change, the most important of these changes is the energy of the (J2p orbital, making it higher than the 1T2 p y" orbitals. The relative energy levels of molecular orbitals in the B 2 , C 2 , and N2 molecules can be represented by the diagram in Figure 9.l7(b), Molecular Orbital Diagrams Beginning with oxygen, the nuclear charge is sufficiently large to prevent the interaction of s and p orbitals. Thus, for O 2 and Ne 2, the order of molecular orbital energies is the same as that D D 1 1 [ill [ill 1 159 288.7 Paramagnetic D D H 1~ [ill [ill 2 131 627.6 Diamagnetic D 3 110 941.4 Diamagnetic D 1 1 H 1~ ill] [ill [ill 2 121 498.7 Paramagnetic D HH 1~ H ill] ill] ill] 1 142 156.9 Diamagnetic lliJ HH HH [0 lliJ [ill o (J2p x * 0"2s For simplicity the erls and er ts orbitals are omitted. The se two orbitals hold a total of four electrons. Remember that for 02 and F z, er2 px is lower in energy than 1T2p y' and 1T 2W Figure 9.18 Molecular orbital diagrams for second-period homonuclear diatomic molecules. , SECTION 9.6 Molecular Orbital Theory 347 for F 2 , which is shown in Figure 9.17(a). Figure 9.18 gives the molecular orbital diagrams, magnetic properties, bond orders, and bond enthalpies for Li 2 , B 2 , C 2 , N 2 , O 2 , F 2 , and Ne2' Note that the filling of molecular orbitals follows the same rules as the filling of atomic orbit- als [ ~~ Section 6.8]: • Lower energy orbitals fill first. • Each orbital can accommodate a maximum of two electrons with opposite spins . • Hund's rule is obeyed. There are several important predictions made by the molecular orbital diagrams in Figure 9.18. First, molecular orbital theory correctly predicts that Nez, with a bond order of 0, does not exist. Second, it correctly predicts the magnetic properties of the molecules that do exist. Both B2 and O 2 are known to be paramagnetic. Third, although bond order is only a qualitative measure of bond strength, the calculated bond orders of the molecules correlate well with the measured bond enthalpies. The N2 molecule, with a bond order of 3, has the largest bond enthalpy of the five mol- ecules. The B2 and F2 molecules, each with a bond order of 1, have the smallest bond enthalpies. Its ability to predict correctly the properties of molecules makes molecular orbital theory a power- ful tool in the study of chemical bonding. Sample Problem 9.8 shows how to use molecular orbital diagrams to determine the mag- netic properties and bond order of the superoxide ion . . . Sample Problem 9.8 The superoxide ion (0 2 ) has been implicated in a number of degenerative conditions, including aging and Alzheimer's disease. Using molecular orbital theory, determine whether O 2 is paramagnetic or diamagnetic, and then calculate its bond order. Strategy Start with the molecular orbital diagram for O 2 , add an electron, and then use the resulting diagram to determine the magnetic properties and bond order. Setup The molecular orbital diagram for O 2 is shown in Figure 9.18. The additional electron must be added to the lowest-energy molecular orbital available. • D or. CT2p x "- '" H 1 ',' , 1T2p y' 1T2 pz 1T2p y' 1T2 pz 1~ H CT2 px lliJ ;, lliJ " CT2 s CT2 s lliJ Molecular orbital diagram for O 2 Solution In this case, either of the two singly occupied 7T ;p orbitals can accommodate an additional electron. This gives a molecular orbital diagram in which there is one unpaired electron, making O 2 paramagnetic. The new diagram has six electrons in bonding molecular orbitals and three in • antibonding molecular orbitals. We can ignore the electrons in the (J 2s and (J2 s orbitals because their contributions to the bond order cancel each other, The bond order is (6 - 3)/2 = l.5. Practice Problem A Use molecular orbital theory to determine whether N ~ - is paramagnetic or diamagnetic, and then calculate its bond order. Practice Problem B Use molecular orbital theory to determine whether F ~+ is paramagnetic or diamagnetic, and then calculate its bond order. ~I " Think About It Experiments confirm that the superoxide ion is paramagnetic, Also, any time we add one or more electrons to an antibonding molecular orbital, as we did in this problem, we should expect the bond order to decrease, Electrons in anti bonding orbitals cause a bond to be less stable. 348 CHAPTER 9 Chemical Bonding II: Molecular Geometry and Bonding Theories Review the shapes of the d orbitals [ H~ Section 6. 7, Figure 6.20]. Bringing Chemistry to life Why Is Carbon Monoxide To x ic? . Hemoglobin is the substance in blood that is responsible for picking up O 2 from air in the lungs, and delivering it to the entire body. Hemoglobin is a protein molecule made up of four subunits, each of which contains a heme group. A heme group is a cyclic atTangement of atoms with an Fe 2 + ion anhe center: Carbon monoxide (CO) forms when carbon-containing substances are burned in a limited supply of oxygen. It is a colorless and odorless gas and is highly poisonous. The toxicity of CO lies in its unusual ability to bind very strongly to hemoglobin, thus preventing hemoglobin from binding oxyge n. When hemoglobin binds CO, the sigma bonds between C and 0, and between C and Fe, are best explained by considering the C atom to be sp-hybridized. The orbital diagram of the Fe 2 + ion is- [Ar]1 H 11 1111 11 I d xz d xy d yz d Z 2 d x 2 - y2 . . '" . . . . . . . . . . . . . . " . . A singly occupied sp orbital on the C atom overlaps with the singly occupied d z 2 orbital on the Fe 2+ ion: N N :-! o III C sp N N , The pi bonds are more easily understood by considering the molecular orbital diagram of CO. The two pi bonds between C and 0 are the result of there being a pair of electrons in each of the bond- ing 7T molecular orbitals, 7T2py and 7T2pz' Note that there are no electrons in the antibonding orbitals, 7Tt , ), and 7T3 p _. Because they are both empty, either of these 7T * orbitals can overlap with the filled d xz orbital o~ the Fe ?+ ion, forming a pi bond. This contributes to the strength of the carbon-iron bond, increasing hemoglobin's affinity for carbon monoxide. SECTION 9.6 Molecular Orbital Theory 349 co C 111 11 1 C 2p I H ~ ~ 2p N __ ~ N N Fe 2s N When hemoglobin binds Ob the sigma bonds between the two 0 atoms and between 0 and Fe can also be explained by sp hybridization. The sp hybrid orbital (on 0) overlaps with the d Z 2 orbital (on Fe 2 + ). . o II o N N N N According to the molecular orbital diagram of oxygen in Figure 9.18, however, the 1T ;p y and 1T ;p z orbitals each contain one electron. These singly occupied orbitals cannot overlap with the filled d xz orbital on iron, as this would result in more than two electrons in the region of overlap. There- fore, the bond between02 and the Fe 2+ ion, consisting only of a sigma bond, is weaker than that between CO and the Fe 2 + ion, which consists of a sigma bond and a pi bond. This makes hemo- globin's affinity for CO roughly 200 times greater than its affinity for O 2 , There is no simple rule for determining the order of the molecular orbitals of a heteronuclear diatomic molecule based on the order of the orbitals of the elements that make up the molecule. For CO, the molecular orbitals are ordered like those in C b rather than like those in O 2 , (See Fig- ure 9.18.) Sample Problem 9.9 lets you practice constructing molecular orbital diagrams for hetero- nuclear diatomics. ~ '~ . Sample Problem 9.9 ' . The importance of nitric oxide in biological systems was introduced in Chapter 8. Draw the molecular orbital diagram for nitric oxide. (Assume the ordering of molecular orbitals to be like that in 0 2') (Continued) • • 350 CHAPTER 9 Chemi cal Bonding II: Mo lecular Geometry and Bonding Theor i es 'N=O: • Think About It Often the bond order determined from a molecular orbital diagram corresponds to the number of bonds in the Lewis structure of the molecule. In the case of NO, though, the Lewis structure contains a double bond whereas the molecular orbital approach gives a bond order of 2.5. In fact, the molecular orbital approach gives a bond order that is more consistent with experimental data. The experimental bond enthalpy in NO is 631 kJ/mol, stronger than the tabulated value for a nitrogen-oxygen double bond [ H~ Section 8.9). Strategy Construct the diagram, ordering the molecular orbitals like those in O 2 (Figure 9.18). Include only the valence orbitals. Determine the total number of valence electrons and place them in the molecular orbitals starting with the lowest-energy orbital. Follow Hund's rule and the Pauli exclusion principle [ H~ Section 6.8). Setup The molecular orbitals in order of increasing energy are CF 25 < 0";5 < CF2p < 7T 2p , = 7T 2p_ < . '" * x ) (. 7T ;py = 7T 2p, < 0" 2p.( There are 11 valence electrons (five from N and six from 0). Solution NO o I H ~ ~ 2p Practice Problem A Draw the molecular orbital diagram for the cyanide ion (CN- ). (Assume the ordering of molecular orbitals to be like that in O 2 ,) Practice Problem B Given that BeO is diamagnetic, use a molecular orbital diagram to detenuine whether the orbitals are ordered like those of B~ or those of O 2 , Checkpoint 9.6 Molecular Orbital Theory 9. 6.1 Calculate the bond order of N ~ +, and 9.6.3 Calculate the bond order of He i . determine whether it is paramagnetic or diamagnetic. a) 0 b) 0.5 a) 2, paramagnetic c) 1.0 b) 2, diamagnetic d) 1.5 c) 3, paramagnetic e) 2 d) 3, paramagnetic e) 1, paramagnetic 9.6.4 Which if any of the following species has a bond order of O? (Select all that 9.6.2 Which of the following species is apply.) paramagnetic? (Select all that apply.) a) B ~+ a) ci- b) Ne ~+ b) O ~+ c) Fi- c) F ~+ d) He~ + d) F ~ - e) Hi- e) c i+ SECTION 9.7 Bonding Theories and Descriptions of Molecules with Delocalized Bonding 351 Bonding Theories and Descriptions of Molecules with Delocalized Bonding The progression of bonding theories in this chapter illustrates the importance of model develop- ment. Scientists use models to understand experimental results and to predict future observations. A model is useful as long as it agrees with observation. When it fails to do so, it must be replaced with a new model. What follows is a synopsis of the strengths and weaknesses of the bonding theories presented in Chapters 8 and 9: Lewis Theory Strength: The Lewis theory of bonding enables us to make qualitative predictions about bond strengths and bond lengths. Lewis structures are easy to draw and are widely used by chemists [ ~~ Section 8.9] . Weakness: Lewis structures are two dimensional, whereas molecules are three dimensional. In addition, Lewis theory fails to account for the differences in bonds in compounds such as H2o Flo and HF. It also fails to explain why bonds form. The Valence-Shell Electron-Pair Repulsion Model Strength: The VSEPR model enables us to predict the shapes of many molecules and polyatomic ions. Weakness: Because the VSEPR model is based on the Lewis theory of bonding, it also fails to explain why bonds form. Valence Bond Theory Strength: Valence bond theory describes the formation of covalent bonds as the overlap of atomic orbitals. Bonds form because the resulting molecule has a lower potential energy than the original, isolated atoms. Weakness: Valence bond theory alone fails to explain the bonding in many molecules such as BeCI 2 , BF 3 , and CH 4 , in which the central atom in its ground state does not have enough unpaired electrons to form the observed number of bonds. Hybridization of Atomic Orbitals Strength: The hybridization of atomic orbitals is not a separate bonding theory; rather, it is an extension of valence bond theory. Using hybrid orbitals, we can understand the bonding arid geometry of more molecules, including BeCl b BF 3 , and CH 4 . Weakness: Valence bond theory and hybrid orbitals fail to predict some of the important properties of molecules, such as the paramagnetism of 0 2' Molecular Orbital Theory Strength: Molecular orbital theory enables us to predict accurately the magnetic and other properties of molecules and ions. Weakness: Pictures of molecular orbitals can be very complex. Although molecular orbital theory is in many ways the most powerful of the bonding mod- els, it is also the most complex, so we continue to use the other models when they do an adequate job of explaining or predicting the properties of a molecule. For example, if you need to predict the three-dimensional shape of an AB n molecule on an exam, you should draw its Lewis structure and apply the VSEPR model. Don't try to draw its molecular orbital diagram. On the other hand, if you need to determine the bond order of a diatomic molecule or ion, you should draw a molecular orbital diagram. In general chemistry, it is best to use the simplest theory that can answer a par- ticular question. Because they remain useful, we don't discard the old models when we develop new ones. In fact, the bonding in some molecules, such as benzene (C 6 H 6 ), is best described using a combina- tion of models. Benzene can be represented with two resonance structures [ ~~ Section 8.7] . ill According to its Lewis structure and valence bond theory, the benzene molecule contains twelve (J bonds (six carbon-carbon, and six carbon-hydrogen) and three 'IT bonds. From experimental evi- dence, however, we know that benzene does not have three single bonds and three double bonds between carbon atoms. Rather, there are six equivalent carbon-carbon bonds. This is precisely the 352 CHAPTER 9 Chemical Bonding II: Molecular Geometry and Bonding Theories H reason that two different Lewis structures are necessary to represent the molecule. Neither one alone accurately depicts the nature of the carbon-carbon bonds. In fact, the 1T bonds in benzene are delocalized, meaning that they are spread out over the entire molecule, rather than confined between two specific atoms. (Bonds that are confined between two specific atoms are called local- ized bonds.) Valence bond theory does a good job of describing the localized IT bonds in benzene, but molecular orbital theory does a better job of using delocalized 1T bonds to describe the bonding scheme in benzene. To describe the IT bonds in benzene, begin with a Lewis structure and count the electron domains on the carbon atoms. (Either resonance structure will give the same result.) Each C atom has three electron domains around it (two single bonds and one double bond). Recall from Table 9.4 that an atom that has three electron domains is Sp2 hybridized. To obtain the three unpaired electrons necessary on each C atom, one electron from each C atom must be promoted from the doubly occupied 2s orbital to an empty 2p orbital: promotion. C* IT] 2s i 11 11 11 1 2 p 3 This actually creates four unpaired electrons. Next, the orbitals are Sp2 hybridized, leaving one singly occupied, unhybridized 2p orbital on each C atom: ,- - - - - - - - -I 1"- - - - - - - ] C* i [] 11 11 11 1 hybridizatio~ i 11 11 11 1 i IT] : 2s 2p 2p: 2p : sp2 sp2 sp2: 2p · r ~ 1 1- _________________________ , The sp ? hybrid orbitals adopt a trigonal planar arrangement and overlap with one another (and with ls orbitals on H atoms) to form the IT bonds in the molecule. D The remaining un hybridized 2p orbitals (o ne on each C atom) combine to form molecular orbitals. Because the p orbitals are all parall el to one another, only 7f2p and 7i-ij, molecular orbitals form. The combination of these six 2p atomic orbitals forms six molecular orbitals: three bonding and three antibonding. These molecular orbitals are delocalized over the entire benzene mole~ule: H H- H H (J" bonds in benzene 7f molecular orbitals in benzene In the ground state, the lower-energy bonding molecular orbitals contain all six electrons. The electron density in the delocalized 1T molecular orbitals lies above and below the plane that con- tains all the atoms and the IT bonds in the molecule: Sample Problem 9.10 shows how to combine valence bond theory and molecular orbital theory to explain the bonding in the carbonate ion. , SECTION 9.7 Bonding Theories and Descriptions of Molecules with Delocalized Bonding 353 Sample Problem 9.10 It takes three resonance structures to represent the carbonate ion (Co j- ): 2- 2- 2- ·b· II . /C, . :0 / ' 0·. o • • '. • • , • , . . . . . None of the three, though, is a completely accurate depiction. As with benzene, the bonds that are shown in the Lewis structure as one double and two si ngle are actually three equivalent bonds. Use a combination of valence bond theory and molecular orbital theory to explain the bonding in CO j- . Strategy Starting with the Lewis st ructure , use valence bond theor y and hybrid orbitals to describe the (T bond s. Then u se molecular orbital theory to describe the delocalized 7i bonding. Setup The Lewis structure of the carbo nate ion shows three electron domains around the central C atom, so the carbon must be Sp2 hybridized. Solution Each of the s/ hybrid orbitals on the C atom overlaps with a singly occupied p orbital on an 0 atom, forming the three (T bond s. Each 0 atom has an additional, singly occupied p orbital, perpendicular to the one involved in (T bonding. The unhybridi zed p orbital on C overlaps with the p orbitals on 0 to form 7i bond s, which have electron den sities above and below the plane of the molecule. Because the s pecie s ca n be represented with re sonance st ruct ur es, we know that the 7i bond s are delocalized. Practice Problem A Use a combination of valence bond theory and molecular orbital theory to de scr ibe the bonding in ozone (0 3 ) . Practice Problem B Use a combination of valence bond theory and molecular orbital theory to describe the bonding in the nitrite ion (N0 2 ). 9.7.1 9.7.2 Bonding Theories and Descriptions of Molecules with Delocalized Bonding Which of the following contain one or 9. 7.3 Which of the following can hybrid more delocalized 7i bonds? (Select all orbitals be used for? (Select all that that apply.) apply.) a) O 2 a) to explain the geometry of a b) CO 2 molecule c) NO z b) to explain how a central atom can form more bonds than the numb er d) CH 4 of unpaired electrons in its ground e) CH 2 Cl 2 state configuration Which of the atoms in BCl 3 need c) to predict the geometry of a molecule d) to explain the magnetic propertie s hybrid orbitals to describe the bonding of a mol ec ule in the molecule? a) all four atoms e) to predict the magnetic properties of a molecule b) only the B atom c) only the three Cl atoms 9.7.4 Which of the following enables us to d) only the B atom and one Cl atom explain the paramagnetism of O 2 ? e) only the B atom and two Cl atoms a) Lewis theory b) valence bond theory c) valence-shell electron-pair repulsion d) hybridization of atomic orbitals e) molecular orbital theory Doub le bonds that appear in different places in different resonance structures represent delo c alized 7T bonds. Think About It Although the Lewi s structure of CO~ - shows three electron domains on one of the 0 atoms, we generally do not treat terminal atoms ( those with s ingl e bonds to only one other atom) as though they are hybridized because it is unnecessary to do so. [...]... States than ks in large part to the vigilance of one doctor at the FDA She was troubled by inadequate research into the safety of the drug and steadfastly refused to approve the drug maker's application H ~N ;=0 ;=0 N j o o Lenalidomide Thalidomide CC-4047 Scientists who develop new drugs such as lenalidomide and CC-4047 must understand the principles and concepts of organic chemistry In This Chapter,... babies in other countries suffered terrible birth defects as the result of the drug thalidomide 363 '"' 364 CHAPTER 10 Organic Chemistry Why Carbon Is Different Carbon-contain ing compounds that are considered inorganic are generally those that are obtai ned from minerals Organic chemistry is usually defined as the study of compounds that contain carbon This definition is not , such things... and then to Hi+ 9.49 The formation of Hz from two H atoms is an energetically favorable process Yet statistically there is less than a 100 percent chance that any two H atoms will undergo the reaction Apart from energy considerations, how would you account for this observation based on the electron spins in the two H atoms? H3C - 9.50 Draw a molecular orbital energy level diagram for each of the following... asterisk denotes the excited molecule (c) Is N ; diamagnetic or paramagnetic? (d) When N ; loses its excess energy and converts to the ground state Nb it emits a photon of wavelength 470 nm, which makes up part of the auroras' lights Calculate the energy difference between these levels 9.105 The Lewis structure for O 2 is Use molecular orbital theory to explain the bonding in the azide ion (N :;-) (The alTangement... Stereoisomerism Organic Reactions Addition Reactions Substitution Reactions Other Types of Organic Reactions Organic Polymers Addition Polymers Condensation Polymers Biological Polymers • • emlstr • Organic Chemistry and Drugs Beginning in 1957, the drug thalidomide was marketed in 48 countries around the world as a sleeping pill and as an antinausea medicine for pregnant women suffering from morning sickness... You've Learned o Muscone Recall from the beginning of the chapter that one of the fundamental scents for which we have olfactory receptors is musky Musk is a familiar, provocative scent that has been part of the human experience for thousands of years The primary odorous molecule in musk is muscone, a large, cyclic, organic molecule The natural source of musk is a gland on the abdomen of the mature... Scientists who develop new drugs such as lenalidomide and CC-4047 must understand the principles and concepts of organic chemistry In This Chapter, You Will Learn some of the basic concepts of organic chemistry and how the principles of chemical bonding contribute to the understanding of organic compounds and reactions Before you begin, you should review • Lewis structures and formal charge [~~ Sections... double bond consists of one sigma bond and one pi bond A triple bond consists of one sigma bond and two pi bonds It is generally best to use the bonding theory that most easily describes the bonding in a particular molecule or polyatomic ion In species that can be represented by two or more resonance structures, the pi bonds are delocalized, meaning that they are spread out over the molecule and not constrained... and many do not contain hydrogen Examples of organic compounds include the following: CH4 CSH 7 0 4 COOH CH3NH2 CCl4 Methane Ascorbic acid Methylamine Carbon tetrachloride Early in the study of organic chemistry there was thought to be some fundamental difference between compounds that came from living things, such as plants and animals, and those that came from nonliving things, such as rocks Compounds... variety of different types of organic compounds, each with their own characteristic properties, result from the following: Carbon's formation of chains is called catenation 365 366 CHAPTER 10 Organic Chemistry 1 Carbon's ability to form chains by bonding with itself A functional group is a group of atoms that determines many of a molecule's properties [ ~ Section 2.6] 2 The presence of elements . understand the principles and concepts of organic chemistry. Thalidomide was no t a pp rove d for u se in the Uni t ed States than ks in large part to the vi g ila n ce of one doctor. Theories Review the shapes of the d orbitals [ H~ Section 6. 7, Figure 6.20]. Bringing Chemistry to life Why Is Carbon Monoxide To x ic? . Hemoglobin is the substance in blood. order of a diatomic molecule or ion, you should draw a molecular orbital diagram. In general chemistry, it is best to use the simplest theory that can answer a par- ticular question. Because