Chemistry part 14, Julia Burdge,2e (2009) pdf

318 CHAPTER 9 Chemical Bonding II: Molecular Geometry and Bonding Theories Think About It Compare these results with the information in Figure 9.2 and Table 9.2. Make sure that you can draw Lewis structures cOITectly. Without a correct Le wis structure, you will be unable to determine the shape of a molecule. molecular geometries where there are one or more lone pairs on the central atom. Note the posi- tions occupied by the lone pairs in the trigonal bipyramidal electron-domain geometry. When there are lone pairs on the central atom in a trigonal bipyramid, the lone pairs preferentially occupy equatorial po sitions because repulsion is greater when the angle between electron domains is 90° or less. Placing a lone pair in an axial po sition would put it at 90° to three other electron domains. Placing it in an equatorial position put s it at 90° to only two other domains, thus minimizing the number of strong repulsive interactions. All po sitions are equivalent in the octahedral geometry, so one lone pair on the central atom can occupy any of the po sitions. If there is a second lone pair in this geometry, though, it must occupy the po sition opposite the first. This arrangement minimizes the repulsive forces between the two lone pairs (they are 180° apart instead of 90° apart). In summary, the steps to determine the electron-domain and molecular geometries are as follows: 1. Draw the Lewis structure of the molecule or polyatomic ion. 2. Count the number of electron domains on the central atom. 3. Determine the electron-domain geometry by applying the VSEPR model. 4. Determine the molecular geometry by considering the positions of the atoms only. Sample Problem 9.1 shows how to determine the shape of a molecule or poly atomic ion. Sample Problem 9.1 . Determine the s hape s of (a) 50 3 and (b) ICI 4 . Strategy Use Lewis structures and the VSEPR model to determine first the electron-domain geometry and then the molecular geometry (shape). Setup (a) The Lewis structure of 50 3 is: •• '0' • • • I •. :O=S-O: • •• There are three electron domains on the central atom: one double bond and two single bonds. (b) The Le wis structure ofICl 4 is: •• :CI: · . .1 :CI-·I CI: · . J . . . :CI: • • There are six electron domains on the central atom in ICl 4 : four single bonds and two lone pairs. Solution (a) According to the VSEPR model , three electron domains will be aITanged in a trigonal plane. Since there are no lone pairs on the central atom in 50 3 , the molecular geometry is the same as the electron-domain geometry. Therefore, the shape of S0 3 is trigonal planar. Electron-domain geometry: trigonal planar -_. Molecular geometry: trigonal planar (b) Six electron domains will be aITanged in an octahedron. Two lone pairs on an octahedron will be located on opposite s ide s of the centr'al atom, making the shape of ICl 4 square planar. Electron-domain geometry: octahedral • Molecular geometry: square planar SECTION 9.1 Molecular Geometry 3 19 Practice Problem A Determine the shapes of (a) CO 2 and (b) SCI 2 . Practice Problem B Determine the shapes of (a) SbF 5 and (b) BrF5' Deviation from Ideal Bond Angles Some electron domains are better than others at repelling neighboring domains. As a result, the bond angles may be slightly different from those shown in Figure 9.2. For example, the electron- domain geometry of ammonia (NH 3) is tetrahedral, so we might predict the H-N - H bond angles to be 109.5°. In fact, the bond angles are about 107°, slightly smaller than predicted. The lone pair on the nitrogen atom repels the N - H bond s more strongly than the bond s repel one another. It therefore "squeezes" them closer together than the ideal tetrahedral angle of 109.5°. In effect, a lone pair takes up more space than the bonding pairs. This can be understood by considering the attractive forces involved in determining the location of the electron pairs. A lone pair on a central atom is attracted only to the nucleus of that atom. A bonding pair of electrons, on the other hand, is simultaneously attracted by the nuclei of both of the bonding atoms. As a result, the lone pair ha s more freedom to spread out and greater capa city to repel other electron domains. Also, because they contain more electron den sity, multiple bond s repel more strongly than single bonds. Consider the bond angles in each of the following example s: 10 1. 5 ° -+ 1 84.8° BrFS Geometry of Molecules with More Than One Central Atom Thus far we have considered the geometries of molecule s having only one central atom. We can determine the overall geometry of more complex molecules by treating them as though they have multiple central atoms. Methanol (CH 3 0 H), for example, has a central C atom and a central 0 atom, as shown in the following Lewis structure: H I H-C-O-H I H Both the C and the 0 atoms are sUlTounded by four electron domains. In the case of C, they are three C-H bond s and one C-O bond. In the case of 0 , they are one O- C bond , one O-H bond, and two lone pair s. In each case, the electron-domain geometry is tetrahedral. However, the molecular geometry of the C part of the molecule is tetrahedral, whereas the mol ecular geometry of the 0 part of the molecule is bent. Note that although the Lewis structure makes it appear as though there is a 180° angle between the O-C and O-H bo nd s, the angle is actually approxi- mately 109.5°, the angle in a tetrahedral arrangement of electron domains. Sample Problem 9.2 shows how to determine when bond angles differ f rom ideal values. Acetic acid, the substance that gives vinegar its characteristic smell and sour taste, is sometimes used in combination with corticosteroids to treat certain types of ear infections. Its Lewis structure is • • H"O' I II H-C-C-O-H I H Determine the molecular geometry about each of the central atoms, and determine the approximate value of each of the bond angles in the molecule. Which if any of the bond angles wo uld you expect to be smaller than th e ideal values? (Continued) When we specify the geo metry of a particular portion of a molecule, we refer to it as the geometry" about" a particular atom. In met hanol, for ex ample, we say that the geometry is tetrahedral about the C atom and bent about the 0 atom. 320 CHAPTER 9 Chemi cal Bo nding II : Mo lecular Geomet ry and Bondi ng T heor i es Think About It Compare these answers with the information in Figure 9.2 and Table 9.2. Strategy Identify the central atoms and count the number of electron domains around each of them. Use the VSEPR model to determine each electron-domain geometry, and the information in Table 9.2 to determine the molecular geometry about each central atom. Setup The leftmost C atom is surrounded by four electron domains: one C-C bond and three C- H bonds. The middle C atom is surrounded by three electron domains: one C- C bond, one C-O bond, and one C=O (double) bond. The 0 atom is surrounded by four electron domains: one O- C bond, one O-H bond, and two lone pairs. Solution The electron-domain geometry of the leftmost C is tetrahedral. Because all four electron domains are bonds, the molecular geometry of this part of the molecule is also tetrahedra l. The electron- domain geometry of the middle C is trigonal planar. Again, because all the domains are bonds, the molecular geometry is also trigonal planar. The electron-domain geometry of the 0 atom is tetrahedral. Because two of the domains are lone pairs, the molecular geometry about the 0 atom is bent. Bond angles are determined using electron-domain geometry. Therefore, the approximate bond angles about the leftmost Care 109.5°, those about the middle Care 120°, and those about the 0 are 109.5°. The angle between the two single bonds on the middle carbon will be less than 120° because the double bond repels the single bonds more strongly than they repel each other. Likewise, the bond angle between the two bonds on the 0 will be less than 109.5° because the lone pairs on 0 repel the single bonds more strongly than they repel each other and push the two bond ing pa irs closer together. The angles are labeled as follows: - 109.5° > 120° H '0' H < 120° < 109.5° Practice Problem A Ethanolamine (HOCH 2 CH 2 NH 2 ) has a sme ll similar to ammonia and is commonly found in biological tissue s. Its Lewis structure is H H . . I I H- O-C-C - N- H . . I I I H H H Determine the molecular geometry about each central atom and label all the bond angles. Cite any expected deviations from ideal bond angles. , Practice Problem B Determine the molecular geometry about each central atom in H 2 S0 4 , Label any expected deviations from ideal bond angles. Its Lewis structure is '0' • • • • I •. H-O-S - O- H I :0: Molecular Geometry 9.1.1 What are the electron-domain 9.1.2 What are the electron-domain geometry and mol ec ular geometry geometry and mo lecular geometry of CO ~-? ofC IO;-? a) tetrahedral, trigonal planar a) tetrahedral, trigonal planar b) tetrahedral, trigonal pyramidal b) tetrahedral, trigonal pyramidal c) trigonal pyramidal, trigonal c) trigonal pyramidal, trigonal pyramidal pyramidal d) trigonal planar, trigonal planar d) trigonal planar, trigonal planar e) tetrahedral, tetrahedral e) tetrahedral, tetrahedral SECTION 9.2 Molecular Geometry and Polarity 321 9.1.3 What is the approximate value of the bond angle indicated? H H I I H-C=C~H a) < 90° b) < 109.5° c) > 109.5° d) > 120° e) < 120° 9.1.4 What is the approximate value of the bond angle indicated? H Ir;:, H-C-O-H I •• H a) < 180° b) > 180° c) < 109.5° d) > 109S e) < 90° Molecular Geometry and Polarity Molecular geometry is tremendously important in understanding the physical and chemical behav- ior of a substance. Molecular polarity, for example, is one of the most important consequences of molecular geometry, because molecular polarity influences physical, chemical, and biological properties. Recall from Section 8.4 that a bond between two atoms of different electronegativities is polar and that a diatomic molecule containing a polar bond is a polar molecule. Whether a mol- ecule made up of three or more atoms is polar depends not only on the polarity of the individual bonds, but also on its molecular geometry. Each of the CO 2 and H 2 0 molecules contains two identical atoms bonded to a central atom and two polar bonds. However, only one of these molecules is polar. To understand why, think of each individual bond dipole as a vector. The overall dipole moment of the molecule is determined by vector addition of the individual bond dipoles. In the case of CO 2 , we have two identical vectors pointing in opposite directions. When the vectors are placed on a Cartesian coordinate system, they have no y component and their x compo- nents are equal in magnitude but opposite in sign. The sum of these two vectors is zero in both the x and y directions. Thus, although the bonds in CO 2 are polar, the molecule is nonpolar. 2:.x = 0 x t ~- 2:.y = 0 overall 2:. = 0 y . . The vectors representing the bond dipoles in water, although equal in magnitude and opposite in the x direction, are not opposite in the y direction. Therefore, although their x components sum to zero, their y components do not. This means that there is a net resultant dipole and H 2 0 is polm: x- ' y 2:.x = 0 2:.y * 0 overall 2:. * 0 Dipole moments can be used to distinguish between molecules that have the same chemical formula but different arrangements of atoms. Such compounds are called structural isomers. For ex ample, there are two structural isomers of dichloroethylene (C 2 H 2 CI 2 ). Because the individual bond dipoles sum to zero in trans-dichloroethylene, the trans isomer is nonpolar: 2:.x = 0 2:.y = 0 overall: 2:. = 0 x ; - y Multimedia Chemical bonding-molecu lar geometry and polarity (interadi ve) . Recall that we can represent an individual bond dipole us ing a crossed arr ow that po i nts toward the more ele ctronegative atom [ ~~ Section 8.41 Can Bond Dipoles Cancel One Another in More Complex Molecules? In ABx molecules where x :> 3, it may be less obvious whether the individual bond dipoles cancel one another. Consider the mol- ecule BF 3 , for example, which has a trigonal planar geometry: \0 0 4X= 0 x +-7l-;:-+, ~30;:-; 0:- 4Y = 0 overall 4 = 0 60° y We will simplify the math in this analysis by assigning the vec- tors representing the three identical B - F bonds an arbitrary mag- nitude of 1.00. The x, y coordinates for the end of arrow 1 are (0, 1.00). Determining the coordinates for the ends of arrows 2 and 3 requires the use of trigonometric functions. You may have learned the mnemonic SOH CAH TOA, where the letters stand for: Sin = Opposite over Hypotenuse Cos = Adjacent over Hypotenuse Tangent = Opposite over Adjacent The x coordinate for the end of arrow 2 corresponds to the length of the line opposite the 60° angle. The hypotenuse of the triangle has a length of 1.00 (the arbitrarily assigned value). Therefore, using SOH, opposite sin 60° = 0.866 = = ~-  hypotenuse opposite 1 so the x coordinate for the end of arrow 2 is 0.866. The magnitude of the y coordinate corresponds to the length of the line adjacent to the 60 ° angle. Using TOA, tan 60° = 1.73 = opposite = 0.866 adjacent adjacent . 0.866 adjacent = = 0.500 1.73 so the y coordinate for the end of arrow 2 is -0.500. (The trigo- nometric formula gives us the length of the side. We know from the diagram that the sign of this y component is negative.) Arrow 3 is similar to arrow 2. Its x component is equal in magnitude but opposite in sign, and its y component is the same 322 magnitude and sign as that for arrow 2. Therefore, the x and y coordinates for all three vectors are x Y Arrow 1 0 1 Arrow 2 0.866 -0.500 Arrow 3 -0.866 - 0.500 Sum = 0 0 Because the individual bond dipoles (represented here as the vec- tors) sum to zero, the molecule is nonpolar overall. Although it is somewhat more complicated, a similar anal- ysis can be done to show that all x, y, and 2 coordinates sum to zero when there are four identical polar bonds arranged in a tet- rahedron about a central atom. In fact, any time there are identi- cal bonds symmetrically distributed around a central atom, with no lone pairs on the central atom, the molecule will be nonpolar overall, even if the bonds themselves are polar. In cases where the bonds are distributed symmetrically around the central atom, the nature of the atoms surrounding the central atom determines whether the molecule is polar overall. For example, CCl 4 and CHCl 3 have the same molecular geom- etry (tetrahedral), but CCl 4 is nonpolar because the bond dipoles cancel one another. In CHCI 3 , however, the bonds are not all identical, and therefore the bond dipoles do not sum to zero. The CHCl 3 molecule is polar. 2 x = 4X = 0 4Y = 0 42 = 0 overall 4 = 0 4X = 0 4Y * 0 42 = 0 overall: 4 * 0 SECTION 9.2 Molecular Geometry and Polarity 323 The bond dipoles in the cis isomer do not cancel one another, so cis-dichloroethylene is polar: 2,X = 0 2,y * 0 overall: 2, * 0 x ,k y Because of the difference in polarity, these two isomers can be distinguished experimentally by measuring the dipole moment. I Sample Problem 9.3 shows you how to determine whether a molecule is polar. . . . . h Samp'le Problem 9.3 .' Detemune whether (a) PCl s and (b) H 2 CO (C double bonded to 0) are polar. Strategy For each molecule, draw the Lewis structure, use the VSEPR model to detennine its molecular geometry, and then deternline whether the individual bond dipoles cancel. Setup (a) The Lewis structure of PCIs is :CI: tl~c.r :q-P~ . It .cr. :CI: . . With five identical electron domains around the central atom, the electron-domain and molecular geometries are trigonal bipyramidal. The equatorial bond dipoles will cancel one another, just as in the case of BF 3 , and the axial bond dipoles will also cancel each other. (b) The Lewis structure of H 1C O is '0' til C~ H-7 ""' H The bond dipoles, although symmetrically distributed around the C atom, are not identical and therefore will not sum to zero. Solution (a) PCIs is nonpolar. (b) H 2 CO is polar. Practice Problem A Deternline whether (a) CH1Cl l and (b) XeF4 are polar. Practice Problem B Deternline whether (a) NBr 3 and (b) BCl 3 are polar. ~ Checkpoint 9.2 Molecular Geometry and Polarity • 9.2.1 Identify the polar molecules in the 9.2.2 Identify the nonpolar molecules in the following group: HBr, CH 4 , CS 2 . following group: SO l, NH 3 , XeF 2 . a) HBronly a) S0 2, NH 3 , and XeF 2 b) HBr and CS l b) S0 2 only c) HBr , CH 4 , and CS 2 c) XeF 2 only d) CH 4 and CS 2 d) S0 2 andXeF 2 e) CH 4 only e) S02 and NH 3 Think About It Make sure that your Lewis structures are correct and that you count electron domains on the central atom carefully. This will give you the correct electron-domain and molecular geometries. Molecular polarity depends both on individual bond dipoles and molecular geometry. How r~ EI ctF r • • • In Chapter 6 we learned that although an electron is a particle with a known ma ss, it exhibits wavelike properties. The quantum mechan- ical model of the atom, which gives rise to the familiar shapes of s and p atomic orbitals, treats electrons in atoms as waves, rather than particles. Therefore, rather than use , llTOW S to denote the locations and spins of electrons, we will adopt a convention whereby a singly occupied orbital will appear as a light color and a doubly occupied orbital will appear as a darker version of the same color. In the rep- resentations of orbitals that follow, atomic s orbitals will be repre- sented as yellow, and atomic p orbitals will be represented as light blue if singly occupied and darker blue if doubly occupied. (Empty p orbitals will appear white.) When two electrons occupy the same atomic orbital, their spins are paire d. Paired electrons occupy the same orbital and have opposite spins [I ~~ Section 6.6J. 324 H-H • • • • :F-F: • • •• •• H-F: Lewis dot structures of H 2 , F 2 , and HF - F2 HF Ball-and-stick models Two si ngly occupied s orbitals each containing one electron Two singly occupied p orbita ls each containing one electron T wo singly occupi ed orbitals (one s, one p) each contai nin g one electron Overlapped s orbital s, sharing th e pair of electron s, both doubly occupied Overlapped p orbital s, sharing the pair of electron s, . both doubly occupied Overlapped orbitals (one s, one p), sharing the pair of electrons, both doubly occupied Valence Bond Theory The Lewis theory of chemical bonding provides a relatively simple way for us to visualize the arrangement of electrons in molecules. It is insufficient, however, to explain the differences between the covalent bonds in compounds such as H z, Fz, and HF. Although Lewis theory describes the bonds in these three molecules in exactly the s ame way, they really are quite different from one another, as evidenced by their bond lengths and bond enthalpies listed in Table 9.3. Understand- ing these differences and why covalent bonds form in th e first place requires a bonding model that combines Lewi s's notion of atoms sharing electron pairs and the quantum mechanical de scriptions of atomic orbitals. According to valence bond theory, atoms share electrons when an atomic orbital on one atom overlaps with an atomic orbital on the other. Each of the overlapping atomic orbitals must contain a single, unpaired electron. Furthermore, the two electrons shared by the bonded atoms must have opposite spins [ ~~ Section 6.6] . The nuclei of both atoms are attracted to the shared pair of electrons. It is this mutual attraction for the shared electrons that holds the atoms t oge th er . 0 Bond Length (A) Bond Enthalpy (kJlmol) H2 0.74 436.4 F2 1.42 150.6 HF 0.92 568.2 SECTION 9.3 Valence Bond Theory 325 The H - H bond in H2 forms when the singly occupied ls orbitals of the two H atoms . . . . . . . . . . . . . . . . . . . . . . . . . . .,. . • • • • overlap: H H H H Similarly, the F- F bond in F2 forms when the singly occupied 2p orbitals of the two F atoms overlap: Recall that the ground-state electron configuration of the F atom is [He]2i2 p 5 [ ~~ Section 6.8] . ( The ground-state orbital diagram of F is shown in the margin.) We can also depict the formation of an H - F bond using the valence bond model. In this case, the singly occupied ls orbital of the H atom overlaps with the singly occupied 2p orbital of the F atom: H H According to the quantum mechanical model, the sizes, shape s, and energies of the Is orbital of H and the 2p orbital of F are different. Therefore, it is not surprising that the bonds in H 2 , Fb and HF "ary in strength and length. Wh y do covalent bonds form? According to valence bond theory, a covalent bond will form between two atoms if the potential energy of the re sulting molecule is lower than that of the iso- lated atoms. Simply put , this means that the formation of covalent bonds is exothermic. While this fact may not seem intuitively obvious, you know that energy must be supplied to a molecule in order to break covalent bonds [ ~~ Section 8.9] . Because the formation of a bond is the reverse process, we should expect energy to be given off. , . . . . . . . . . . . . . . , . . . . . . Valence bond theory also introduces the concept of directionality to chemical bonds. For example, we expect the bond formed by the overlap of a p orbital to coincide with the axis along which the p orbital lies. Consider the molecule H 2 S. Unlike the other molecules that we have ~ n countered , H 2 S does not have the bond angle that Lewis theory and the VSEPR model would lead us to predict. (With four electron domains on the central atom, we would expect the bond angle to be on the order of 1 09 .5 ° .) In fact, the H - S - H bond angle is 92°. Lo oking at this in terms Kee p in min d that alt ho ugh th er e are still j ust two el ect ron s, eac h at om "thinks" it owns them bo t h, so w he n the si ng ly occ up ied or bita ls o ve rl ap, both o rbital s end up doub ly o cc upied. Orbital diagram for F Re call tha t the enthalpy cha n ge for a forward process an d th at for t he rev er se p ro ces s di ffer only in sign: !!.H forward = - !!. H r"""rse [ ~ Section 5.3] 326 CHAPTER 9 Chemical Bonding II: Molecular Geometry and Bonding Theories In order for you to understand the material in t his section a nd Sect i on 9.4, you must be able to draw orbital diagrams for ground- state ele c tron configurations [~ ~ Section 6.8]. Think About It Because the 4p orbitals on the Se atom are all mutually perpendicular, we should expect the angles between bonds formed by their overlap to be approximately 90 °. of valence bond theory, the central atom (S) has two unpaired electrons, each of which resides in a '3p orbitai. the ' orbhai' dIagram 'fo'[ · ti1 e ground-state electron configuration of the S atom is S [Ne] [TI] IHI1 11 I ? 3s- 3p4 Remember that p orbitals are mutually perpendicular, lying along the x, y, and z axes [ ~~ Section 6.7]. We can rationalize the observed bond angle by envisioning the overlap of each of the singly occupied 3p orbitals with the Is orbital of a hydrogen atom: H H H H In summary, the important features of valence bond theory are as follows: o A bond forms when singly occupied atomic orbitals on two atoms overlap. o The two electrons shared in the region of orbital overlap must be of opposite spin. o Formation of a bond results in a lower potential energy for the system. Sample Problem 9.4 shows how to use valence bond theory to explain the bonding in a molecule. Sample Problem 9.4 Hydrogen selenide (H 2 Se) is a foul-smelling ga s that can cause eye and respiratory tract infianunation. Th e H-Se-H bond angle in H 2 Se is approximately 92°. Use valence bond theory to describe the bonding in this molecule. Strategy Consider the central atom's ground-state electron configuration, and determine what orbitals are available for bond formation. Setup The ground-state electron configuration of Se is [Ar]4 i3d J 0 4l. Its orbital diagram (showing only the 4p orbitals) is IHI1 11 I 4p4 Solution Two of the 4p orbitals are singly occupied and therefore available for bonding. The bonds in H 2 Se form as the result of the overlap of a hydrogen Is orbital with each of these orbitals on the Se atom. Practice Problem A Use valence bond theory to describe the bonding in phosphine (PH 3 ), which has H - P-H bond angles of approximately 94°. Practice Problem B Use valence bond theory to describe the bonding in arsine (AsH 3 ), which has H-A s -H bond angles of approximately 92°. Checkpoint 9.3 Valence Bond Theory 9.3.1 Which of the following atoms, in its ground state, does not have unpaired electrons? (Select all that apply.) a)O b)Be c)B d) F e) Ne 9.3.2 According to valence bond theory, how many bonds would you expect a nitrogen atom (in its ground state) to form? a)2 b)3 c)4 d)5 e)6 SECTION 9.4 Hybridization of Atomic Orbitals 327 Hybridization of Atomic Orbitals Although valence bond theory is useful and can explain more of our experimental observations than Lewis bond theory, it fails to explain the bonding in many of the molecules that we encounter. According to valence bond theory, for example, an atom must have a singly occupied atomic orbital in order to form a bond with another atom. How then do we explain the bonding in BeC1 2 ? The cen- tral atom, Be, has a ground-state electron configuration of [He]2s 2, so it has no unpaired electrons. With no singly occupied atomic orbitals in its ground state, how does. Be form two bonds? Furthermore, in cases where the ground-state electron configuration of the central atom does have the required number of unpaired electrons, how do we explain the observed bond angles? Carbon, like sulfur, has two unpaired electrons in its ground state. Using valence bond theory as our guide, we might envision the formation of two covalent bonds with oxygen, as in CO 2 , If . . . . . . . . . . . . . . . . . . . the two unpaired electrons on C (each residing in a 2p orbital) were to form bonds, however, the O-C-O bond angle should be on the order of 90°, like the bond angle in H 2 S. In fact, the bond angle in CO 2 is 180°: Actual bond angle is 180°. Bond angle should be 90°. In order to explain these and other observations, we need to extend our discussion of orbital over- lap to include the concept of hybridization or mixing of atomic orbitals. The idea of hybridization of atomic orbitals begins with the molecular geometry and works backward to explain the bonds and the observed bond angles in a molecule. To extend our dis- cussion of orbital overlap and introduce the concept of hybridization of atomic orbitals, we first consider beryllium chloride (BeCI 2 ), which has two electron domains on the central atom. Using its Lewis structure (shown in the margin) and the VSEPR model, we predict that BeC1 2 will have a . . . . . . , . . . . . . . . . CI- Be -Cl bond angle of 180°. If this is true, though, how does Be form two bonds with no unpaired electrons, and why is the angle between the two bonds 180°? In order to answer the first part of the question, we envision the promotion of one of the elec- trons in the 2s orbital to an empty 2p orbital. Recall that electrons can be promoted from a lower atomic orbital to a higher one [~ . Se ction 6.3]. The ground-state electron configuration is the one in which all the electrons occupy orbitals of the lowest possible energy. A configuration in which one or more electrons occupy a higher energy orbital is called an excited state. An excited state generally is denoted with a star (e.g., Be* for an excited-state Be atom). Showing only the valence orbitals, we can represent the promotion of one of the valence electrons of beryllium as Be promotion • Be * 2p With one of its valence electrons promoted to the 2p subshell, the Be atom now has two unpaired electrons and therefore can form two bonds. However, the orbitals in which the two unpaired elec- trons reside are different from each other, so we would expect bonds formed as a result of the over- lap of these two orbitals (each with a 3p orbital on a CI atom) to be different: 2s 3p 2p 3p Experimentally, though, the bonds in BeCl 2 are identical in length and strength. •• •• :CI-Be-Cl: • • • • The ground-state orbital diagram for Cis: [ill [ill 11 11 1 • Media Player/ MPEG Content Hybrid orbitals-orbital hybrid i zation a nd valence bond theory. - _ _ Multimedia Hybrid or bi tal s-o rbi tal hy br id i zation (interactive). Remember that Be is one of the atoms that does not obey the octet rule [H~ Section 8. 11 . The Lewis structure of BeCl, is • • • • :CI-Be-CI: • • • • [TIJ IHIHI1 I 3s 2 3 p s Orbital diagram for CI [...]... (b) The 2s orbital and one of the 2p orbitals on Be combine to form two sp hybrid orbitals Unoccupied orbitals are shown in white (c) Like any two electron domains, the hybrid orbitals on Be are 180 apart (d) The hybrid orbitals on Be each overlap with a singly occupied 3p orbital on a CI atom, + I> 2s -I> 2p sp (,>0 c> x x z (b) (c) Cl _ (d) sp SECTION 9.4 Hybridization of Atomic... white.) (c) Hybrid orbitals on B overlap with 2p orbitals on F + + 2s 2p 2p (a) y y x x - -;; z (b) i (c) z 330 CHAPTER 9 Chemical Bonding II: Molecular Geometry and Bonding Theories they ' wiii' n6't be 'part 6f the 'dls'c usslon 'cit bondln'g' in this chapter As we will see in Section 9.5, though, unhybridlzed atomic orbitals that do contain electrons are important in our description of the bonding in . arrangement minimizes the repulsive forces between the two lone pairs (they are 180° apart instead of 90° apart). In summary, the steps to determine the electron-domain and molecular geometries. tetrahedral. However, the molecular geometry of the C part of the molecule is tetrahedral, whereas the mol ecular geometry of the 0 part of the molecule is bent. Note that although. (Continued) When we specify the geo metry of a particular portion of a molecule, we refer to it as the geometry" about" a particular atom. In met hanol, for ex ample,