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Unformatted text preview: Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 10 CHAPTER Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals INTRODUCTION IN CHAPTER ORY. HERE 9 WE DISCUSSED BONDING IN TERMS OF THE LEWIS THE- 10.1 MOLECULAR GEOMETRY WE WILL STUDY THE SHAPE, OR GEOMETRY, OF MOLECULES. 10.2 DIPOLE MOMENTS GEOMETRY HAS AN IMPORTANT INFLUENCE ON THE PHYSICAL AND CHEM- 10.3 VALENCE BOND THEORY ICAL PROPERTIES OF MOLECULES, SUCH AS MELTING POINT, BOILING POINT, 10.4 HYBRIDIZATION OF ATOMIC ORBITALS AND REACTIVITY. 10.5 HYBRIDIZATION IN MOLECULES CONTAINING DOUBLE AND TRIPLE BONDS WE WILL SEE THAT WE CAN PREDICT THE SHAPES OF MOLECULES WITH CONSIDERABLE ACCURACY USING A SIMPLE METHOD BASED ON LEWIS THE LEWIS STRUCTURES. 10.6 MOLECULAR ORBITAL THEORY THEORY OF CHEMICAL BONDING, ALTHOUGH USEFUL 10.7 MOLECULAR ORBITAL CONFIGURATIONS AND EASY TO APPLY, DOES NOT TELL US HOW AND WHY BONDS FORM. A 10.8 DELOCALIZED MOLECULAR ORBITALS PROPER UNDERSTANDING OF BONDING COMES FROM QUANTUM ME- CHANICS. THEREFORE, IN THE SECOND PART OF THIS CHAPTER WE WILL APPLY QUANTUM MECHANICS TO THE STUDY OF THE GEOMETRY AND STABILITY OF MOLECULES. 367 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 368 CHEMICAL BONDING II: MOLECULAR GEOMETRY AND HYBRIDIZATION OF ATOMIC ORBITALS 10.1 The term “central atom” means an atom that is not a terminal atom in a polyatomic molecule. VSEPR is pronounced “vesper.” MOLECULAR GEOMETRY Molecular geometry is the three-dimensional arrangement of atoms in a molecule. A molecule’s geometry affects its physical and chemical properties, such as melting point, boiling point, density, and the types of reactions it undergoes. In general, bond lengths and bond angles must be determined by experiment. However, there is a simple procedure that allows us to predict with considerable success the overall geometry of a molecule or ion if we know the number of electrons surrounding a central atom in its Lewis structure. The basis of this approach is the assumption that electron pairs in the valence shell of an atom repel one another. The valence shell is the outermost electron-occupied shell of an atom; it holds the electrons that are usually involved in bonding. In a covalent bond, a pair of electrons, often called the bonding pair, is responsible for holding two atoms together. However, in a polyatomic molecule, where there are two or more bonds between the central atom and the surrounding atoms, the repulsion between electrons in different bonding pairs causes them to remain as far apart as possible. The geometry that the molecule ultimately assumes (as defined by the positions of all the atoms) minimizes the repulsion. This approach to the study of molecular geometry is called the valence-shell electron-pair repulsion (VSEPR) model, because it accounts for the geometric arrangements of electron pairs around a central atom in terms of the electrostatic repulsion between electron pairs. Two general rules govern the use of the VSEPR model: As far as electron-pair repulsion is concerned, double bonds and triple bonds can be treated like single bonds. This approximation is good for qualitative purposes. However, you should realize that in reality multiple bonds are “larger” than single bonds; that is, because there are two or three bonds between two atoms, the electron density occupies more space. • If a molecule has two or more resonance structures, we can apply the VSEPR model to any one of them. Formal charges are usually not shown. • With this model in mind, we can predict the geometry of molecules (and ions) in a systematic way. For this purpose, it is convenient to divide molecules into two categories, according to whether or not the central atom has lone pairs. MOLECULES IN WHICH THE CENTRAL ATOM HAS NO LONE PAIRS For simplicity we will consider molecules that contain atoms of only two elements, A and B, of which A is the central atom. These molecules have the general formula ABx, where x is an integer 2, 3, . . . . (If x 1, we have the diatomic molecule AB, which is linear by definition.) In the vast majority of cases, x is between 2 and 6. Table 10.1 shows five possible arrangements of electron pairs around the central atom A. As a result of mutual repulsion, the electron pairs stay as far from one another as possible. Note that the table shows arrangements of the electron pairs but not the positions of the atoms that surround the central atom. Molecules in which the central atom has no lone pairs have one of these five arrangements of bonding pairs. Using Table 10.1 as a reference, let us take a close look at the geometry of molecules with the formulas AB2, AB3, AB4, AB5, and AB6. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 10.1 MOLECULAR GEOMETRY 369 TABLE 10.1 Arrangement of Electron Pairs about a Central Atom (A) in a Molecule and Geometry of Some Simple Molecules and Ions in which the Central Atom Has No Lone Pairs NUMBER OF ELECTRON PAIRS ARRANGEMENT OF ELECTRON PAIRS* MOLECULAR GEOMETRY* EXAMPLES 180 2 A ð BOAOB Linear ð Linear BeCl2, HgCl2 B 3 120 A BF3 A ½ Â B Trigonal planar B Trigonal planar B 109.5 4 A ð A B ð CH4, NH4 B B Tetrahedral Tetrahedral B 90 ð ð B B A 120 PCl5 A B ð 5 B Trigonal bipyramidal Trigonal bipyramidal B 90 ð 6 90 B ð B A ð ð SF6 A B B B Octahedral Octahedral *The colored lines are used only to show the overall shapes; they do not represent bonds. AB2: Beryllium Chloride (BeCl2) The Lewis structure of beryllium chloride in the gaseous state is O O SCl O Be O ClS Q Q Because the bonding pairs repel each other, they must be at opposite ends of a straight Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 370 CHEMICAL BONDING II: MOLECULAR GEOMETRY AND HYBRIDIZATION OF ATOMIC ORBITALS line in order for them to be as far apart as possible. Thus, the ClBeCl angle is predicted to be 180°, and the molecule is linear (see Table 10.1). The “ball-and-stick” model of BeCl2 is AB3: Boron Trifluoride (BF3) Boron trifluoride contains three covalent bonds, or bonding pairs. In the most stable arrangement, the three BF bonds point to the corners of an equilateral triangle with B in the center of the triangle: S OS F A B DG O OS F SQ F Q According to Table 10.1, the geometry of BF3 is trigonal planar because the three end atoms are at the corners of an equilateral triangle which is planar: Thus, each of the three FBF angles is 120°, and all four atoms lie in the same plane. AB4: Methane (CH4) The Lewis structure of methane is H A HOC OH A H Since there are four bonding pairs, the geometry of CH4 is tetrahedral (see Table 10.1). A tetrahedron has four sides (the prefix tetra means “four”), or faces, all of which are equilateral triangles. In a tetrahedral molecule, the central atom (C in this case) is located at the center of the tetrahedron and the other four atoms are at the corners. The bond angles are all 109.5°. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 10.1 MOLECULAR GEOMETRY 371 AB5: Phosphorus Pentachloride (PCl5) The Lewis structure of phosphorus pentachloride (in the gas phase) is O S ClS HA O P ClS E OQ O S Cl A Q S ClS Q O S Cl Q The only way to minimize the repulsive forces among the five bonding pairs is to arrange the PCl bonds in the form of a trigonal bipyramid (see Table 10.1). A trigonal bipyramid can be generated by joining two tetrahedrons along a common triangular base: The central atom (P in this case) is at the center of the common triangle with the surrounding atoms positioned at the five corners of the trigonal bipyramid. The atoms that are above and below the triangular plane are said to occupy axial positions, and those that are in the triangular plane are said to occupy equatorial positions. The angle between any two equatorial bonds is 120°; that between an axial bond and an equatorial bond is 90°, and that between the two axial bonds is 180°. AB6: Sulfur Hexafluoride (SF6) The Lewis structure of sulfur hexafluoride is F SOS O SO A QS F QH E F S EH F F SO A OS Q SFS Q Q The most stable arrangement of the six SF bonding pairs is in the shape of an octahedron, shown in Table 10.1. An octahedron has eight sides (the prefix octa means “eight”). It can be generated by joining two square pyramids on a common base. The central atom (S in this case) is at the center of the square base and the surrounding atoms are at the six corners. All bond angles are 90° except the one made by the bonds between the central atom and the pairs of atoms that are diametrically opposite each other. That angle is 180°. Since the six bonds are equivalent in an octahedral molecule, we cannot use the terms “axial” and “equatorial” as in a trigonal bipyramidal molecule. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 372 CHEMICAL BONDING II: MOLECULAR GEOMETRY AND HYBRIDIZATION OF ATOMIC ORBITALS MOLECULES IN WHICH THE CENTRAL ATOM HAS ONE OR MORE LONE PAIRS Determining the geometry of a molecule is more complicated if the central atom has both lone pairs and bonding pairs. In such molecules there are three types of repulsive forces — those between bonding pairs, those between lone pairs, and those between a bonding pair and a lone pair. In general, according to the VSEPR model, the repulsive forces decrease in the following order: lone-pair vs. lone-pair repulsion lone-pair vs. bondingpair repulsion bonding-pair vs. bondingpair repulsion Electrons in a bond are held by the attractive forces exerted by the nuclei of the two bonded atoms. These electrons have less “spatial distribution” than lone pairs; that is, they take up less space than lone-pair electrons, which are associated with only one particular atom. Because lone-pair electrons in a molecule occupy more space, they experience greater repulsion from neighboring lone pairs and bonding pairs. To keep track of the total number of bonding pairs and lone pairs, we designate molecules with lone pairs as ABxEy, where A is the central atom, B is a surrounding atom, and E is a lone pair on A. Both x and y are integers; x 2, 3, . . . , and y 1, 2, . . . . Thus the values of x and y indicate the number of surrounding atoms and number of lone pairs on the central atom, respectively. The simplest such molecule would be a triatomic molecule with one lone pair on the central atom and the formula is AB2E. As the following examples show, in most cases the presence of lone pairs on the central atom makes it difficult to predict the bond angles accurately. AB2E: Sulfur Dioxide (SO2) The Lewis structure of sulfur dioxide is OPOOO S O QSO Q Because VSEPR treats double bonds as though they were single, the SO2 molecule can be viewed as consisting of three electron pairs on the central S atom. Of these, two are bonding pairs and one is a lone pair. In Table 10.1 we see that the overall arrangement of three electron pairs is trigonal planar. But because one of the electron pairs is a lone pair, the SO2 molecule has a “bent” shape. S S S S S O S JG O O Since the lone-pair versus bonding-pair repulsion is greater than the bonding-pair versus bonding-pair repulsion, the two sulfur-to-oxygen bonds are pushed together slightly and the OSO angle is less than 120°. AB3E: Ammonia (NH3) The ammonia molecule contains three bonding pairs and one lone pair: HOOOH N A H As Table 10.1 shows, the overall arrangement of four electron pairs is tetrahedral. But in NH3 one of the electron pairs is a lone pair, so the geometry of NH3 is trigonal pyra- Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 10.1 FIGURE 10.1 (a) The relative sizes of bonding pairs and lone pairs in CH4, NH3, and H2O. (b) The bond angles in CH4, NH3, and H2O. Note that dashed lines represent bond axes behind the plane of the paper, wedged lines represent bond axes in front of the plane of the paper, and thin solid lines represent bonds in the plane of the paper. MOLECULAR GEOMETRY 373 H C N O H H H H H H H H (a) H C N O H 109.5° H H H H 107.3° H 104.5° H H (b) midal (so called because it looks like a pyramid, with the N atom at the apex). Because the lone pair repels the bonding pairs more strongly, the three NH bonding pairs are pushed closer together: O N DAG HHH Thus the HNH angle in ammonia is smaller than the ideal tetrahedral angle of 109.5° (Figure 10.1). AB2E2: Water (H2O) A water molecule contains two bonding pairs and two lone pairs: HOOOH O Q S S The overall arrangement of the four electron pairs in water is tetrahedral, the same as in ammonia. However, unlike ammonia, water has two lone pairs on the central O atom. These lone pairs tend to be as far from each other as possible. Consequently, the two OH bonding pairs are pushed toward each other, and we predict an even greater deviation from the tetrahedral angle than in NH3. As Figure 10.1 shows, the HOH angle is 104.5°. The geometry of H2O is bent: O DG H H AB4E: Sulfur Tetrafluoride (SF4) The Lewis structure of SF4 is Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 374 CHEMICAL BONDING II: MOLECULAR GEOMETRY AND HYBRIDIZATION OF ATOMIC ORBITALS O SQ F OS F G DQ O S DG OS SO F F Q Q The central sulfur atom has five electron pairs whose arrangement, according to Table 10.1, is trigonal bipyramidal. In the SF4 molecule, however, one of the electron pairs is a lone pair, so the molecule must have one of the following geometries: F F F S F F S S F F F (a) SF4 Q (b) In (a) the lone pair occupies an equatorial position, and in (b) it occupies an axial position. The axial position has three neighboring pairs at 90° and one at 180°, while the equatorial position has two neighboring pairs at 90° and two more at 120°. The repulsion is smaller for (a), and indeed (a) is the structure observed experimentally. This shape is sometimes described as a distorted tetrahedron (or a folded square, or seesaw shape). The angle between the axial F atoms and S is 186°, and that between the equatorial F atoms and S is 116°. Table 10.2 shows the geometries of simple molecules in which the central atom has one or more lone pairs, including some that we have not discussed. GEOMETRY OF MOLECULES WITH MORE THAN ONE CENTRAL ATOM So far we have discussed the geometry of molecules having only one central atom. The overall geometry of molecules with more than one central atom is difficult to define in most cases. Often we can only describe the shape around each of the central atoms. For example, consider methanol, CH3OH, whose Lewis structure is H A HOC O OOH O Q A H FIGURE 10.2 CH3OH. The geometry of The two central (nonterminal) atoms in methanol are C and O. We can say that the three CH and the CO bonding pairs are tetrahedrally arranged about the C atom. The HCH and OCH bond angles are approximately 109°. The O atom here is like the one in water in that it has two lone pairs and two bonding pairs. Therefore, the HOC portion of the molecule is bent, and the angle HOC is approximately equal to 105° (Figure 10.2). GUIDELINES FOR APPLYING THE VSEPR MODEL Having studied the geometries of molecules in two categories (central atoms with and without lone pairs), let us consider some rules for applying the VSEPR model to all types of molecules: • Back Forward Main Menu Write the Lewis structure of the molecule, considering only the electron pairs around the central atom (that is, the atom that is bonded to more than one other atom). TOC Study Guide TOC Textbook Website MHHE Website TABLE 10.2 Geometry of Simple Molecules and Ions in which the Central Atom Has One or More Lone Pairs CLASS OF TOTAL NUMBER NUMBER OF NUMBER OF ARRANGEMENT OF MOLECULE OF ELECTRON PAIRS BONDING PAIRS LONE PAIRS ELECTRON PAIRS* GEOMETRY EXAMPLES Bent SO2, O3 Trigonal pyramidal NH3 Bent H2O Distorted tetrahedron (or seesaw) IF4 , SF4, XeO2F2 T-shaped ClF3 Linear XeF2, I3 Square pyramidal BrF5, XeOF4 š AB2E 3 2 1 A B B Trigonal planar š AB3E 4 3 1 A B B B Tetrahedral š 4 2 2 A š AB2E2 B B Tetrahedral AB4E 5 4 B 1 B A ð B B Trigonal bipyramidal 5 3 2 B A ð AB3E2 ð B B Trigonal bipyramidal 5 2 3 A ð ð AB2E3 ð B B Trigonal bipyramidal B B AB5E 6 5 1 B A B B Octahedral š B AB4E2 6 4 2 B Square planar XeF4, ICl4 A B B Octahedral *The colored lines are used to show the overall shape, not bonds. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 376 CHEMICAL BONDING II: MOLECULAR GEOMETRY AND HYBRIDIZATION OF ATOMIC ORBITALS Count the number of electron pairs around the central atom (bonding pairs and lone pairs). Treat double and triple bonds as though they were single bonds. Refer to Table 10.1 to predict the overall arrangement of the electron pairs. • Use Tables 10.1 and 10.2 to predict the geometry of the molecule. • In predicting bond angles, note that a lone pair repels another lone pair or a bonding pair more strongly than a bonding pair repels another bonding pair. Remember that there is no easy way to predict bond angles accurately when the central atom possesses one or more lone pairs. • The VSEPR model generates reliable predictions of the geometries of a variety of molecular structures. Chemists use the VSEPR approach because of its simplicity. Although there are some theoretical concerns about whether “electron-pair repulsion” actually determines molecular shapes, the assumption that it does leads to useful (and generally reliable) predictions. We need not ask more of any model at this stage in the study of chemistry. We will use the VSEPR model to predict the geometry of some simple molecules and polyatomic ions in the following example. EXAMPLE 10.1 Use the VSEPR model to predict the geometry of the following molecules and ions: (a) AsH3, (b) OF2, (c) AlCl4 , (d) I3 , (e) C2H4. Answer (a) The Lewis structure of AsH3 is O HOAs OH A H This molecule has three bonding pairs and one lone pair, a combination similar to that in ammonia. Therefore, the geometry of AsH3 is trigonal pyramidal, like NH3. We cannot predict the HAsH angle accurately, but we know that it must be less than 109.5° because the repulsion of the bonding electron pairs by the lone pair on As is greater than the repulsion between bonding pairs. (b) The Lewis structure of OF2 is FOF SOOOOOS QQQ Since there are two lone pairs on the O atom, the OF2 molecule must have a bent shape, like that of H2O. Again, all we can say about angle size is that the FOF angle should be less than 109.5° due to the stronger repulsive force between the lone pairs and the bonding pairs. (c) The Lewis structure of AlCl4 is O S Cl S A O O S Cl OAl OCl S Q Q A S Cl S Q S Since the central Al atom has no lone pairs and all four AlOCl bonds are equivalent, the AlCl4 ion should be tetrahedral, and the ClAlCl angles should all be 109.5°. (d) The Lewis structure of I3 is S SOOQOOS I I QIQ Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 10.2 DIPOLE MOMENTS 377 The central I atom has two bonding pairs and three lone pairs. From Table 10.2 we see that the three lone pairs all lie in the triangular plane, and the I3 ion should be linear. (e) The Lewis structure is C2H4 is H H G D CP C D G H H The CPC bond is treated as though it were a single bond. Because there are no lone pairs present, the arrangement around each C atom has a trigonal planar shape like BF3, discussed earlier. Thus the predicted bond angles in C2H4 are all 120°. H 120 H G D CPC 120 D G H 120 H Similar problems: 10.7, 10.8, 10.9. Comment (a) The I3 ion is one of the few structures for which the bond angle (180°) can be predicted accurately even though the central atom contains lone pairs. (b) In C2H4, all six atoms lie in the same plane. The overall planar geometry is not predicted by the VSEPR model, but we will see why the molecule prefers to be planar later. In reality, the angles are close, but not equal, to 120° because the bonds are not all equivalent. PRACTICE EXERCISE Use the VSEPR model to predict the geometry of (a) SiBr4, (b) CS2, and (c) NO3 . 10.2 DIPOLE MOMENTS In Section 9.3 we learned that hydrogen fluoride is a covalent molecule with a polar bond. There is a shift of electron density from H to F because the F atom is more electronegative than the H atom. The shift of electron density is symbolized by placing a crossed arrow ( ) above the Lewis structure to indicate the direction of the shift. For example, F HOOS Q The consequent charge separation can be represented as F HOOS Q where (delta) denotes a partial charge. This separation of charges can be confirmed in an electric field (Figure 10.3). When the field is turned on, HF molecules orient their negative ends toward the positive plate and their positive ends toward the negative plate. This alignment of molecules can be detected experimentally. A quantitative measure of the polarity of a bond is its dipole moment ( ) which is the product of the charge Q and the distance r between the charges: Q r (10.1) To maintain electrical neutrality, the charges on both ends of an electrically neutral di- Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 378 CHEMICAL BONDING II: MOLECULAR GEOMETRY AND HYBRIDIZATION OF ATOMIC ORBITALS FIGURE 10.3 Behavior of polar molecules (a) in the absence of an external electric field and (b) when the electric field is turned on. Nonpolar molecules are not affected by an electric field. – + + – – + – + + – – – + + – – + + + – – – + – + (a) In a diatomic molecule like HF, and the charge Q is equal to . + – + +– – + – + – + + – + +– + – + + – + – – + – +– – + – (b) atomic molecule must be equal in magnitude and opposite in sign. However, in Equation (10.1), Q refers only to the magnitude of the charge and not to its sign, so is always positive. Dipole moments are usually expressed in debye units (D), named for Peter Debye.† The conversion factor is 1D 3.33 10 30 Cm where C is coulomb and m is meter. Diatomic molecules containing atoms of different elements (for example, HCl, CO, and NO) have dipole moments and are called polar molecules. Diatomic molecules containing atoms of the same element (for example, H2, O2, and F2) are examples of nonpolar molecules because they do not have dipole moments. For a molecule made up of three or more atoms both the polarity of the bonds and the molecular geometry determine whether there is a dipole moment. Even if polar bonds are present, the molecule will not necessarily have a dipole moment. Carbon dioxide (CO2), for example, is a triatomic molecule, so its geometry is either linear or bent: KC N O O OPC PO linear molecule (no dipole moment) resultant dipole moment bent molecule (has a dipole moment) The arrows show the shift of electron density from the less electronegative carbon atom to the more electronegative oxygen atom. In each case, the dipole moment of the entire molecule is made up of two bond moments, that is, individual dipole moments in the polar CPO bonds. The bond moment is a vector quantity, which means that it has both magnitude and direction. The measured dipole moment is equal to the vector sum of the bond moments. The two bond moments in CO2 are equal in magnitude. Since † Peter Joseph William Debye (1884 – 1966). American chemist and physicist of Dutch origin. Debye made many significant contributions in the study of molecular structure, polymer chemistry, X-ray analysis, and electrolyte solution. He was awarded the Nobel Prize in Chemistry in 1936. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 10.2 FIGURE 10.4 Bond moments and resultant dipole moments in NH3 and NF3. DIPOLE MOMENTS 379 Resultant dipole moment = 1.46 D N N H H F F H F Resultant dipole moment = 0.24 D The VSEPR model predicts that CO2 is a linear molecule. they point in opposite directions in a linear CO2 molecule, the sum or resultant dipole moment would be zero. On the other hand, if the CO2 molecule were bent, the two bond moments would partially reinforce each other, so that the molecule would have a dipole moment. Experimentally it is found that carbon dioxide has no dipole moment. Therefore, we conclude that the carbon dioxide molecule is linear. The linear nature of carbon dioxide has been confirmed through other experimental measurements. Next let us consider the NH3 and NF3 molecules shown in Figure 10.4. In both cases the central N atom has a lone pair, whose charge density is away from the N atom. From Figure 9.5 we know that N is more electronegative than H, and F is more electronegative than N. For this reason, the shift of electron density in NH3 is toward N and so contributes a larger dipole moment, whereas the NF bond moments are directed away from the N atom and so together they offset the contribution of the lone pair to the dipole moment. Thus the resultant dipole moment in NH3 is larger than that in NF3. Dipole moments can be used to distinguish between molecules that have the same formula but different structures. For example, the following molecules both exist; they have the same molecular formula (C2H2Cl2), the same number and type of bonds, but different molecular structures: resultant dipole moment Cl Cl G D CP C D G H H H Cl G D CP C D G Cl H cis-dichloroethylene 1.89 D trans-dichloroethylene 0 Since cis-dichloroethylene is a polar molecule but trans-dichloroethylene is not, they can readily be distinguished by a dipole moment measurement. Additionally, as we will see in the next chapter, the strength of intermolecular forces is partially determined by whether molecules possess a dipole moment. Table 10.3 lists the dipole moments of several polar molecules. Example 10.2 shows how we can predict whether a molecule possesses a dipole moment if we know its molecular geometry. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 380 CHEMICAL BONDING II: MOLECULAR GEOMETRY AND HYBRIDIZATION OF ATOMIC ORBITALS TABLE 10.3 Molecules Dipole Moments of Some Polar MOLECULE GEOMETRY HF HCl HBr HI H2O H2S NH3 SO2 Linear Linear Linear Linear Bent Bent Pyramidal Bent DIPOLE MOMENT (D) 1.92 1.08 0.78 0.38 1.87 1.10 1.46 1.60 EXAMPLE 10.2 Predict whether each of the following molecules has a dipole moment: (a) IBr, (b) BF3 (trigonal planar), (c) CH2Cl2 (tetrahedral). (a) Since IBr (iodine bromide) is diatomic, it has a linear geometry. Bromine is more electronegative than iodine (see Figure 9.5), so IBr is polar with bromine at the negative end. Answer I O Br An analogy is an object that is pulled in the directions shown by the three bond moments. If the forces are equal, the object will not move. Thus the molecule does have a dipole moment. (b) Since fluorine is more electronegative than boron, each BOF bond in BF3 (boron trifluoride) is polar and the three bond moments are equal. However, the symmetry of a trigonal planar shape means that the three bond moments exactly cancel one another: F A B F DG F Consequently BF3 has no dipole moment; it is a nonpolar molecule. (c) The Lewis structure of CH2Cl2 (methylene chloride) is Cl A HOC OH A Cl This molecule is similar to CH4 in that it has an overall tetrahedral shape. However, because not all the bonds are identical, there are three different bond angles: HCH, HCCl, and ClCCl. These bond angles are close to, but not equal to, 109.5°. Since chlorine is more electronegative than carbon, which is more electronegative than hydrogen, the bond moments do not cancel and the molecule possesses a dipole moment: Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 10.3 Similar problems: 10.19, 10.21, 10.22. Cl A C LG L Cl H& H VALENCE BOND THEORY 381 resultant dipole moment Thus CH2Cl2 is a polar molecule. PRACTICE EXERCISE Does the AlCl3 molecule have a dipole moment? 10.3 Recall that an object has potential energy by virtue of its position. Back Forward Main Menu VALENCE BOND THEORY The VSEPR model, based largely on Lewis structures, provides a relatively simple and straightforward method for predicting the geometry of molecules. But as we noted earlier, the Lewis theory of chemical bonding does not clearly explain why chemical bonds exist. Relating the formation of a covalent bond to the pairing of electrons was a step in the right direction, but it did not go far enough. For example, the Lewis theory describes the single bond between the H atoms in H2 and that between the F atoms in F2 in essentially the same way — as the pairing of two electrons. Yet these two molecules have quite different bond dissociation energies and bond lengths (436.4 kJ/mol and 74 pm for H2 and 150.6 kJ/mol and 142 pm for F2). These and many other facts cannot be explained by the Lewis theory. For a more complete explanation of chemical bond formation we look to quantum mechanics. In fact, the quantum mechanical study of chemical bonding also provides a means for understanding molecular geometry. At present, two quantum mechanical theories are used to describe covalent bond formation and the electronic structure of molecules. Valence bond (VB) theory assumes that the electrons in a molecule occupy atomic orbitals of the individual atoms. It permits us to retain a picture of individual atoms taking part in the bond formation. The second theory, called molecular orbital (MO) theory, assumes the formation of molecular orbitals from the atomic orbitals. Neither theory perfectly explains all aspects of bonding, but each has contributed something to our understanding of many observed molecular properties. Let us start our discussion of valence bond theory by considering the formation of a H2 molecule from two H atoms. The Lewis theory describes the HOH bond in terms of the pairing of the two electrons on the H atoms. In the framework of valence bond theory, the covalent HOH bond is formed by the overlap of the two 1s orbitals in the H atoms. By overlap, we mean that the two orbitals share a common region in space. What happens to two H atoms as they move toward each other and form a bond? Initially, when the two atoms are far apart, there is no interaction. We say that the potential energy of this system (that is, the two H atoms) is zero. As the atoms approach each other, each electron is attracted by the nucleus of the other atom; at the same time, the electrons repel each other, as do the nuclei. While the atoms are still separated, attraction is stronger than repulsion, so that the potential energy of the system decreases (that is, it becomes negative) as the atoms approach each other (Figure 10.5). This trend continues until the potential energy reaches a minimum value. At this point, when the system has the lowest potential energy, it is most stable. This condition corresponds to TOC Study Guide TOC Textbook Website MHHE Website 382 CHEMICAL BONDING II: MOLECULAR GEOMETRY AND HYBRIDIZATION OF ATOMIC ORBITALS Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry Back Microwave Ovens — Dipole Moments at Work In the last twenty years the microwave oven has become a ubiquitous appliance. Microwave technology enables us to thaw and cook food much more rapidly than conventional appliances do. How do microwaves heat food so quickly? In Chapter 7 we saw that microwaves are a form of electromagnetic radiation (see Figure 7.3). Microwaves are generated by a magnetron, which was invented during World War II when radar technology was being developed. The magnetron is a hollow cylinder encased in a horseshoe-shaped magnet. In the center of the cylinder is a cathode rod. The walls of the cylinder act as an anode. When heated, the cathode emits electrons that travel toward the anode. The magnetic field forces the electrons to move in a circular path. This motion of charged particles generates microwaves, which are adjusted to a frequency of 2.45 GHz (2.45 109 Hz) for cooking. A “waveguide” directs the microwaves into the cooking compartment. Rotating fan blades reflect the microwaves to all parts of the oven. The cooking action in a microwave oven results from the interaction between the electric field component of the radiation with the polar molecules — mostly water — in food. All molecules rotate at room temperature. If the frequency of the radiation and that of the molecular rotation are equal, energy can be transferred from microwave to the polar molecule. As a result, the molecule will rotate faster. The frequency 2.45 GHz is particularly suitable for increasing the rotational energy of water molecules. The friction from rapidly rotating water molecules eventually heats up the surrounding food molecules. The reason that a microwave oven can cook food so fast is that the radiation is not absorbed by nonpolar molecules and can therefore reach different parts of food at the same time. (Depending on the amount of water present, microwaves can penetrate food to a depth of several inches.) In a conventional oven, heat can affect the center of foods only by conduction (that is, by transfer of heat from hot air molecules to cooler molecules in food in a layer-by-layer fashion), which is a very slow process. The following points are relevant to the operation of a microwave oven. Plastics and Pyrex glasswares do not contain polar molecules and are therefore not affected by microwave radiation. (Styrofoam and certain plastics cannot be used in microwaves because they melt from the heat of the food.) Metals, however, reflect microwaves, thereby shielding the food and possibly returning enough energy to the microwave emitter to overload it. Because microwaves can induce Rotating blades Magnetron Anode Waveguide Cathode Magnet A microwave oven. The microwaves generated by the magnetron are reflected to all parts of the oven by the rotating fan blades. Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 10.3 383 is that at room temperature a thin film of liquid water quickly forms on the surface of frozen food and the mobile molecules in that film can absorb the radiation to start the thawing process. + Electric field of microwave – Direction of wave + – (a) + + Electric field of microwave Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry a current in the metal, this action can lead to sparks jumping between the container and the bottom or walls of the oven. Finally, although water molecules in ice are locked in position and therefore cannot rotate, we routinely thaw food in a microwave oven. The reason VALENCE BOND THEORY Direction of wave – – (b) Interaction between the electric field component of the microwave and a polar molecule. (a) The negative end of the dipole follows the propagation of the wave (the positive region) and rotates in a clockwise direction. (b) If, after the molecule has rotated to the new position the radiation has also moved along to its next cycle, the positive end of the dipole will move into the negative region of the wave while the negative end will be pushed up. Thus the molecule will rotate faster. No such interaction can occur with nonpolar molecules. substantial overlap of the 1s orbitals and the formation of a stable H2 molecule. If the distance between nuclei were to decrease further, the potential energy would rise steeply and finally become positive as a result of the increased electron-electron and nuclearnuclear repulsions. In accord with the law of conservation of energy, the decrease in potential energy as a result of H2 formation must be accompanied by a release of energy. Experiments Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 384 CHEMICAL BONDING II: MOLECULAR GEOMETRY AND HYBRIDIZATION OF ATOMIC ORBITALS Potential energy FIGURE 10.5 Change in potential energy of two H atoms with changes in their distance of separation. At the point of minimum potential energy, the H2 molecule is in its most stable state, and the bond length is 74 pm. The orbital diagram of the F atom is shown on page 272. 10.4 0 Distance of separation show that as a H2 molecule is formed from two H atoms, heat is given off. The converse is also true. To break a HOH bond, energy must be supplied to the molecule. Thus valence bond theory gives a clearer picture of chemical bond formation than the Lewis theory does. Valence bond theory states that a stable molecule forms from reacting atoms when the potential energy of the system has decreased to a minimum; the Lewis theory ignores energy changes in chemical bond formation. The concept of overlapping atomic orbitals applies equally well to diatomic molecules other than H2. Thus a stable F2 molecule forms when the 2p orbitals (containing the unpaired electrons) in the two F atoms overlap to form a covalent bond. Similarly, the formation of the HF molecule can be explained by the overlap of the 1s orbital in H with the 2p orbital in F. In each case, VB theory accounts for the changes in potential energy as the distance between the reacting atoms changes. Because the orbitals involved are not the same kind in all cases, we can see why the bond energies and bond lengths in H2, F2, and HF might be different. As we stated earlier, Lewis theory treats all covalent bonds the same way and offers no explanation for the differences among covalent bonds. HYBRIDIZATION OF ATOMIC ORBITALS The concept of atomic orbital overlap should apply also to polyatomic molecules. However, a satisfactory bonding scheme must account for molecular geometry. We will discuss three examples of VB treatment of bonding in polyatomic molecules. sp3 HYBRIDIZATION Consider the CH4 molecule. Focusing only on the valence electrons, we can represent the orbital diagram of C as Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 10.4 z z y z z y x 385 HYBRIDIZATION OF ATOMIC ORBITALS y x y x x Hybridization z z y z z y y x x y x x FIGURE 10.6 Formation of sp3 hybrid orbitals. hg hh 2s 2p Because the carbon atom has two unpaired electrons (one in each of the two 2p orbitals), it can form only two bonds with hydrogen in its ground state. Although the species CH2 is known, it is very unstable. To account for the four COH bonds in methane, we can try to promote (that is, energetically excite) an electron from the 2s orbital to the 2p orbital: h hhh 2s 2p Now there are four unpaired electrons on C that could form four COH bonds. However, the geometry is wrong, because three of the HCH bond angles would have to be 90° (remember that the three 2p orbitals on carbon are mutually perpendicular), and yet all HCH angles are 109.5°. To explain bonding in methane, VB theory uses hypothetical hybrid orbitals, which are atomic orbitals obtained when two or more nonequivalent orbitals of the same atom combine in preparation for covalent bond formation. Hybridization is the term applied to the mixing of atomic orbitals in an atom (usually a central atom) to generate a set of hybrid orbitals. We can generate four equivalent hybrid orbitals for carbon by mixing the 2s orbital and the three 2p orbitals: hhhh sp3 orbitals sp3 is pronounced “s-p three.” Back Forward Main Menu Because the new orbitals are formed from one s and three p orbitals, they are called sp3 hybrid orbitals. Figure 10.6 shows the shape and orientations of the sp3 orbitals. These four hybrid orbitals are directed toward the four corners of a regular tetrahe- TOC Study Guide TOC Textbook Website MHHE Website 386 CHEMICAL BONDING II: MOLECULAR GEOMETRY AND HYBRIDIZATION OF ATOMIC ORBITALS H C H H H FIGURE 10.7 Formation of four bonds between the carbon sp3 hybrid orbitals and the hydrogen 1s orbitals in CH4. dron. Figure 10.7 shows the formation of four covalent bonds between the carbon sp3 hybrid orbitals and the hydrogen 1s orbitals in CH4. Thus CH4 has a tetrahedral shape, and all the HCH angles are 109.5°. Note that although energy is required to bring about hybridization, this input is more than compensated for by the energy released upon the formation of COH bonds. (Recall that bond formation is an exothermic process.) The following analogy is useful for understanding hybridization. Suppose that we have a beaker of a red solution and three beakers of blue solutions and that the volume of each is 50 mL. The red solution corresponds to one 2s orbital, the blue solutions represent three 2p orbitals, and the four equal volumes symbolize four separate orbitals. By mixing the solutions we obtain 200 mL of a purple solution, which can be divided into four 50-mL portions (that is, the hybridization process generates four sp3 orbitals). Just as the purple color is made up of the red and blue components of the original solutions, the sp3 hybrid orbitals possess both s and p orbital characteristics. Another example of sp3 hybridization is ammonia (NH3). Table 10.1 shows that the arrangement of four electron pairs is tetrahedral, so that the bonding in NH3 can be explained by assuming that N, like C in CH4, is sp3-hybridized. The ground-state electron configuration of N is 1s22s22p3, so that the orbital diagram for the sp3-hybridized N atom is h h h hg sp3 orbitals N H H H FIGURE 10.8 The sp3hybridized N atom in NH3. Three sp3 hybrid orbitals form bonds with the H atoms. The fourth is occupied by nitrogen’s lone pair. Three of the four hybrid orbitals form covalent NOH bonds, and the fourth hybrid orbital accommodates the lone pair on nitrogen (Figure 10.8). Repulsion between the lone-pair electrons and electrons in the bonding orbitals decreases the HNH bond angles from 109.5° to 107.3°. It is important to understand the relationship between hybridization and the VSEPR model. We use hybridization to describe the bonding scheme only when the arrangement of electron pairs has been predicted using VSEPR. If the VSEPR model predicts a tetrahedral arrangement of electron pairs, then we assume that one s and three p orbitals are hybridized to form four sp3 hybrid orbitals. The following are examples of other types of hybridization. sp HYBRIDIZATION The beryllium chloride (BeCl2) molecule is predicted to be linear by VSEPR. The orbital diagram for the valence electrons in Be is hg 2s 2p We know that in its ground state Be does not form covalent bonds with Cl because its electrons are paired in the 2s orbital. So we turn to hybridization for an explanation of Be’s bonding behavior. First we promote a 2s electron to a 2p orbital, resulting in h h 2s 2p Now there are two Be orbitals available for bonding, the 2s and 2p. However, if two Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 10.4 z HYBRIDIZATION OF ATOMIC ORBITALS z z y y x z y Hybridization x 387 y x x FIGURE 10.9 Formation of sp hybrid orbitals. Cl atoms were to combine with Be in this excited state, one Cl atom would share a 2s electron and the other Cl would share a 2p electron, making two nonequivalent BeCl bonds. This scheme contradicts experimental evidence. In the actual BeCl2 molecule, the two BeCl bonds are identical in every respect. Thus the 2s and 2p orbitals must be mixed, or hybridized, to form two equivalent sp hybrid orbitals: hh sp orbitals Cl Be Cl FIGURE 10.10 The linear geometry of BeCl2 can be explained by assuming that Be is sp-hybridized. The two sp hybrid orbitals overlap with the two chlorine 3p orbitals to form two covalent bonds. empty 2p orbitals Figure 10.9 shows the shape and orientation of the sp orbitals. These two hybrid orbitals lie on the same line, the x-axis, so that the angle between them is 180°. Each of the BeCl bonds is then formed by the overlap of a Be sp hybrid orbital and a Cl 3p orbital, and the resulting BeCl2 molecule has a linear geometry (Figure 10.10). sp2 HYBRIDIZATION Next we will look at the BF3 (boron trifluoride) molecule, known to have planar geometry based on VSEPR. Considering only the valence electrons, the orbital diagram of B is hg 2s h 2p First, we promote a 2s electron to an empty 2p orbital: h 2s 2 sp is pronounced “s-p two.” hh 2p Mixing the 2s orbital with the two 2p orbitals generates three sp2 hybrid orbitals: hhh sp2 orbitals empty 2p orbital These three sp2 orbitals lie in the same plane, and the angle between any two of them is 120° (Figure 10.11). Each of the BF bonds is formed by the overlap of a boron sp2 hybrid orbital and a fluorine 2p orbital (Figure 10.12). The BF3 molecule is planar with all the FBF angles equal to 120°. This result conforms to experimental findings and also to VSEPR predictions. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 388 CHEMICAL BONDING II: MOLECULAR GEOMETRY AND HYBRIDIZATION OF ATOMIC ORBITALS FIGURE 10.11 Formation of sp2 hybrid orbitals. You may have noticed an interesting connection between hybridization and the octet rule. Regardless of the type of hybridization, an atom starting with one s and three p orbitals would still possess four orbitals, enough to accommodate a total of eight electrons in a compound. For elements in the second period of the periodic table, eight is the maximum number of electrons that an atom of any of these elements can accommodate in the valence shell. It is for this reason that the octet rule is usually obeyed by the second-period elements. The situation is different for an atom of a third-period element. If we use only the 3s and 3p orbitals of the atom to form hybrid orbitals in a molecule, then the octet rule applies. However, in some molecules the same atom may use one or more 3d orbitals, in addition to the 3s and 3p orbitals, to form hybrid orbitals. In these cases the octet rule does not hold. We will see specific examples of the participation of the 3d orbital in hybridization shortly. To summarize our discussion of hybridization, we note that • • F F • B • F • FIGURE 10.12 The sp2 hybrid orbitals of boron overlap with the 2p orbitals of fluorine. The BF3 molecule is planar, and all the FBF angles are 120°. Back Forward The concept of hybridization is not applied to isolated atoms. It is a theoretical model used only to explain covalent bonding. Hybridization is the mixing of at least two nonequivalent atomic orbitals, for example, s and p orbitals. Therefore, a hybrid orbital is not a pure atomic orbital. Hybrid orbitals and pure atomic orbitals have very different shapes. The number of hybrid orbitals generated is equal to the number of pure atomic orbitals that participate in the hybridization process. Hybridization requires an input of energy; however, the system more than recovers this energy during bond formation. Covalent bonds in polyatomic molecules and ions are formed by the overlap of hybrid orbitals, or of hybrid orbitals with unhybridized ones. Therefore, the hybridization bonding scheme is still within the framework of valence bond theory; electrons in a molecule are assumed to occupy hybrid orbitals of the individual atoms. Table 10.4 summarizes sp, sp2, and sp3 hybridization (as well as other types that we will discuss shortly). Main Menu TOC Study Guide TOC Textbook Website MHHE Website 10.4 TABLE 10.4 HYBRIDIZATION OF ATOMIC ORBITALS 389 Important Hybrid Orbitals and Their Shapes PURE ATOMIC ORBITALS OF THE CENTRAL ATOM HYBRIDIZATION OF THE CENTRAL ATOM NUMBER OF HYBRID ORBITALS SHAPE OF HYBRID ORBITALS EXAMPLES 180 s, p sp 2 BeCl2 Linear s, p, p sp2 3 BF3 120 Planar 109.5 s, p, p, p sp3 4 CH4, NH4 Tetrahedral 90 s, p, p, p, d sp3d 5 PCl5 120 Trigonal bipyramidal 90 s, p, p, p, d, d sp3d 2 6 SF6 90 Octahedral Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 390 CHEMICAL BONDING II: MOLECULAR GEOMETRY AND HYBRIDIZATION OF ATOMIC ORBITALS PROCEDURE FOR HYBRIDIZING ATOMIC ORBITALS Before going on to discuss the hybridization of d orbitals, let us specify what we need to know in order to apply hybridization to bonding in polyatomic molecules in general. In essence, hybridization simply extends Lewis theory and the VSEPR model. To assign a suitable state of hybridization to the central atom in a molecule, we must have some idea about the geometry of the molecule. The steps are as follows: Draw the Lewis structure of the molecule. Predict the overall arrangement of the electron pairs (both bonding pairs and lone pairs) using the VSEPR model (see Table 10.1). 3. Deduce the hybridization of the central atom by matching the arrangement of the electron pairs with those of the hybrid orbitals shown in Table 10.4. 1. 2. The following example illustrates this procedure. EXAMPLE 10.3 Determine the hybridization state of the central (underlined) atom in each of the following molecules: (a) HgCl2, (b) AlI3, and (c) PF3. Describe the hybridization process and determine the molecular geometry in each case. Table 7.3 gives the electron configurations of all the elements. (a) The ground-state electron configuration of Hg is [Xe]6s24f 145d10; therefore, the Hg atom has two valence electrons (the 6s electrons). The Lewis structure of HgCl2 is Answer O O SCl O HgO ClS Q Q There are no lone pairs on the Hg atom, so the arrangement of the two electron pairs is linear (see Table 10.1). From Table 10.4 we conclude that Hg is sp-hybridized because it has the geometry of the two sp hybrid orbitals. The hybridization process can be imagined to take place as follows. First we draw the orbital diagram for the ground state of Hg: hg 6s 6p By promoting a 6s electron to the 6p orbital, we get the excited state: h h 6s 6p The 6s and 6p orbitals then mix to form two sp hybrid orbitals: hh sp orbitals empty 6p orbitals The two HgOCl bonds are formed by the overlap of the Hg sp hybrid orbitals with the 3p orbitals of the Cl atoms. Thus HgCl2 is a linear molecule. (b) The ground-state electron configuration of Al is [Ne]3s23p1. Therefore, Al has three valence electrons. The Lewis structure of AlI3 is Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 10.4 HYBRIDIZATION OF ATOMIC ORBITALS 391 O SIS A O SI OAl Q A SIS Q There are three bonding pairs and no lone pair on the Al atom. From Table 10.1 we see that the shape of three electron pairs is trigonal planar, and from Table 10.4 we conclude that Al must be sp2-hybridized in AlI3. The orbital diagram of the groundstate Al atom is hg h 3s 3p By promoting a 3s electron into the 3p orbital we obtain the following excited state: h hh 3s 3p The 3s and two 3p orbitals then mix to form three sp2 hybrid orbitals: hhh sp2 orbitals empty 3p orbital The sp2 hybrid orbitals overlap with the 5p orbitals of I to form three covalent AlOI bonds. We predict that the AlI3 molecule is trigonal planar and all the IAlI angles are 120°. (c) The ground-state electron configuration of P is [Ne]3s23p3. Therefore, the P atom has five valence electrons. The Lewis structure of PF3 is F OO O F P A F There are three bonding pairs and one lone pair on the P atom. In Table 10.1 we see that the overall arrangement of four electron pairs is tetrahedral, and from Table 10.4 we conclude that P must be sp3-hybridized. The orbital diagram of the groundstate P atom is hg hhh 3s 3p By mixing the 3s and 3p orbitals, we obtain four sp3 hybrid orbitals. h h h hg sp3 orbitals Similar problems: 10.31, 10.32. Back Forward Main Menu As in the case of NH3, one of the sp3 hybrid orbitals is used to accommodate the lone pair on P. The other three sp3 hybrid orbitals form covalent POF bonds with the 2p orbitals of F. We predict the geometry of the molecule to be trigonal pyramidal; the FPF angle should be somewhat less than 109.5°. TOC Study Guide TOC Textbook Website MHHE Website 392 CHEMICAL BONDING II: MOLECULAR GEOMETRY AND HYBRIDIZATION OF ATOMIC ORBITALS PRACTICE EXERCISE Determine the hybridization state of the underlined atoms in the following compounds: (a) SiBr4 and (b) BCl3. HYBRIDIZATION OF s, p, AND d ORBITALS We have seen that hybridization neatly explains bonding that involves s and p orbitals. For elements in the third period and beyond, however, we cannot always account for molecular geometry by assuming that only s and p orbitals hybridize. To understand the formation of molecules with trigonal bipyramidal and octahedral geometries, for instance, we must include d orbitals in the hybridization concept. Consider the SF6 molecule as an example. In Section 10.1 we saw that this molecule has octahedral geometry, which is also the arrangement of the six electron pairs. Table 10.4 shows that the S atom is sp3d 2-hybridized in SF6. The ground-state electron configuration of S is [Ne]3s23p4: hg hg h h 3s 3p 3d Since the 3d level is quite close in energy to the 3s and 3p levels, we can promote 3s and 3p electrons to two of the 3d orbitals: h 3s 32 sp d is pronounced “s-p three d two.” hhh hh 3p 3d Mixing the 3s, three 3p, and two 3d orbitals generates six sp3d 2 hybrid orbitals: hhhhhh sp3d2 orbitals Recall that when n 2, 0 and 1. Thus we can only have 2s and 2p orbitals. empty 3d orbitals The six SOF bonds are formed by the overlap of the hybrid orbitals of the S atom and the 2p orbitals of the F atoms. Since there are 12 electrons around the S atom, the octet rule is violated. The use of d orbitals in addition to s and p orbitals to form an expanded octet (see Section 9.9) is an example of valence-shell expansion. Secondperiod elements, unlike third-period elements, do not have 2d energy levels, so they can never expand their valence shells. Hence atoms of second-period elements can never be surrounded by more than eight electrons in any of their compounds. Example 10.4 deals with valence-shell expansion in a third-period element. EXAMPLE 10.4 Describe the hybridization state of phosphorus in phosphorus pentabromide (PBr5). Answer The Lewis structure of PBr5 is O O S BrS S Br A QH O P BrS E OQ O S Br A Q S BrS Q Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 10.5 Ground state 2s 2s 2p hg sp 2 393 Table 10.1 indicates that the arrangement of five electron pairs is trigonal bipyramidal. Referring to Table 10.4, we find that this is the shape of five sp3d hybrid orbitals. Thus P must be sp3d-hybridized in PBr5. The ground-state electron configuration of P is [Ne]3s23p3. To describe the hybridization process, we start with the orbital diagram for the ground state of P: 2p Promotion of electron sp2Hybridized state HYBRIDIZATION IN MOLECULES CONTAINING DOUBLE AND TRIPLE BONDS hhh 3s 3p 2pz orbitals FIGURE 10.13 The sp2 hybridization of a carbon atom. The 2s orbital is mixed with only two 2p orbitals to form three equivalent sp2 hybrid orbitals. This process leaves an electron in the unhybridized orbital, the 2pz orbital. 3d Promoting a 3s electron into a 3d orbital results in the following excited state: h hhh 3s h 3p 3d Mixing the one 3s, three 3p, and one 3d orbitals generates five sp3d hybrid orbitals: hhhhh sp3d orbitals Similar problem: 10.40. empty 3d orbitals These hybrid orbitals overlap the 4p orbitals of Br to form five covalent POBr bonds. Since there are no lone pairs on the P atom, the geometry of PBr5 is trigonal bipyramidal. PRACTICE EXERCISE Describe the hybridization state of Se in SeF6. 10.5 90° 120° FIGURE 10.14 Each carbon atom in the C2H4 molecule has three sp2 hybrid orbitals (green) and one unhybridized 2pz orbital (gray), which is perpendicular to the plane of the hybrid orbitals. Back Forward Main Menu HYBRIDIZATION IN MOLECULES CONTAINING DOUBLE AND TRIPLE BONDS The concept of hybridization is useful also for molecules with double and triple bonds. Consider the ethylene molecule, C2H4, as an example. In Example 10.1 we saw that C2H4 contains a carbon-carbon double bond and has planar geometry. Both the geometry and the bonding can be understood if we assume that each carbon atom is sp2hybridized. Figure 10.13 shows orbital diagrams of this hybridization process. We assume that only the 2px and 2py orbitals combine with the 2s orbital, and that the 2pz orbital remains unchanged. Figure 10.14 shows that the 2pz orbital is perpendicular to the plane of the hybrid orbitals. Now how do we account for the bonding of the C atoms? As Figure 10.15(a) shows, each carbon atom uses the three sp2 hybrid orbitals to form two bonds with the two hydrogen 1s orbitals and one bond with the sp2 hybrid orbital of the adjacent C atom. In addition, the two unhybridized 2pz orbitals of the C atoms form another bond by overlapping sideways [Figure 10.15(b)]. A distinction is made between the two types of covalent bonds in C2H4. The three bonds formed by each C atom in Figure 10.15(a) are all sigma bonds ( bonds), covalent bonds formed by orbitals overlapping end-to-end, with the electron density concentrated between the nuclei of the bonding atoms. The second type is called a pi bond ( bond), which is defined as a covalent bond formed by sideways overlapping orbitals with electron density concentrated above and below the plane of the nuclei of the bonding atoms. The two C atoms form a pi bond as shown in Figure 10.15(b). It TOC Study Guide TOC Textbook Website MHHE Website 394 CHEMICAL BONDING II: MOLECULAR GEOMETRY AND HYBRIDIZATION OF ATOMIC ORBITALS FIGURE 10.15 Bonding in ethylene, C2H4. (a) Top view of the sigma bonds between carbon atoms and between carbon and hydrogen atoms. All the atoms lie in the same plane, making C2H4 a planar molecule. (b) Side view showing how the two 2pz orbitals on the two carbon atoms overlap, leading to the formation of a pi bond. (c) The interactions in (a) and (b) lead to the formation of the sigma bonds and the pi bond in ethylene. Note that the pi bond lies above and below the plane of the molecule. H 1s H 1s C C H 1s H 1s (a) 2 pz 2 pz π H C C H σ σ σ C 2 pz C H σ π (b) FIGURE 10.16 Another view of pi bond formation in the C2H4 molecule. Note that all six atoms are in the same plane. It is the overlap of the 2pz orbitals that causes the molecule to assume a planar structure. σ H (c) 2 pz π H H C H H C H C H C H H π Ground state 2s 2p 2s 2p spHybridized state Promotion of electron sp orbitals 2py 2pz FIGURE 10.17 The sp hybridization of a carbon atom. The 2s orbital is mixed with only one 2p orbital to form two sp hybrid orbitals. This process leaves an electron in each of the two unhybridized 2p orbitals, namely, the 2py and 2pz orbitals. Back Forward is this pi bond formation that gives ethylene its planar geometry. Figure 10.15(c) shows the orientation of the sigma and pi bonds. Figure 10.16 is yet another way of looking at the planar C2H4 molecule and the formation of the pi bond. Although we normally represent the carbon-carbon double bond as CPC (as in a Lewis structure), it is important to keep in mind that the two bonds are different types: One is a sigma bond and the other is a pi bond. The acetylene molecule (C2H2) contains a carbon-carbon triple bond. Since the molecule is linear, we can explain its geometry and bonding by assuming that each C atom is sp-hybridized by mixing the 2s with the 2px orbital (Figure 10.17). As Figure 10.18 shows, the two sp hybrid orbitals of each C atom form one sigma bond with a hydrogen 1s orbital and another sigma bond with the other C atom. In addition, two pi bonds are formed by the sideways overlap of the unhybridized 2py and 2pz orbitals. Thus the CqC bond is made up of one sigma bond and two pi bonds. The following rule helps us predict hybridization in molecules containing multiple bonds: If the central atom forms a double bond, it is sp2-hybridized; if it forms two double bonds or a triple bond, it is sp-hybridized. Note that this rule applies only to Main Menu TOC Study Guide TOC Textbook Website MHHE Website 10.5 395 HYBRIDIZATION IN MOLECULES CONTAINING DOUBLE AND TRIPLE BONDS 2 pz H 2 pz σ C H 1s π H 1s C C C π C π 2 py 2 py π (a) FIGURE 10.18 Bonding in acetylene, C2H2. (a) Top view showing the overlap of the sp orbitals between the C atoms and the overlap of the sp orbital with the 1s orbital between the C and H atoms. All the atoms lie along a straight line; therefore acetylene is a linear molecule. (b) Side view showing the overlap of the two 2py orbitals and of the two 2pz orbitals of the two carbon atoms, which leads to the formation of two pi bonds. (c) Formation of the sigma and pi bonds as a result of the interactions in (a) and (b). (b) C σ H (c) atoms of the second-period elements. Atoms of third-period elements and beyond that form multiple bonds present a more complicated picture and will not be dealt with here. EXAMPLE 10.5 Describe the bonding in the formaldehyde molecule whose Lewis structure is H H G O C PQ O D 2 Assume that the O atom is sp -hybridized. We note that the C atom (a second-period element) has a double bond; therefore, it is sp2-hybridized. The orbital diagram of the O atom is Answer h hg hg h sp2 2pz 2 Similar problems: 10.36, 10.90. Two of the sp orbitals of the C atom form two sigma bonds with the H atoms, and the third sp2 orbital forms a sigma bond with a sp2 orbital of the O atom. The 2pz orbital of the C atom overlaps with the 2pz orbital of the O atom to form a pi bond. The two lone pairs on the O atom are placed in its two remaining sp2 orbitals (Figure 10.19). H π σ σ C π O •• σ •• FIGURE 10.19 Bonding in the formaldehyde molecule. A sigma bond is formed by the overlap of the sp2 hybrid orbital of carbon and the sp2 hybrid orbital of oxygen; a pi bond is formed by the overlap of the 2pz orbitals of the carbon and oxygen atoms. The two lone pairs on oxygen are placed in the other two sp2 orbitals of oxygen. H PRACTICE EXERCISE Describe the bonding in the hydrogen cyanide molecule, HCN. Assume that N is sp-hybridized. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 396 CHEMICAL BONDING II: MOLECULAR GEOMETRY AND HYBRIDIZATION OF ATOMIC ORBITALS FIGURE 10.20 Liquid oxygen caught between the poles of a magnet, because the O2 molecules are paramagnetic. 10.6 MOLECULAR ORBITAL THEORY Valence bond theory is one of the two quantum mechanical approaches that explain bonding in molecules. It accounts, at least qualitatively, for the stability of the covalent bond in terms of overlapping atomic orbitals. Using the concept of hybridization, valence bond theory can explain molecular geometries predicted by the VSEPR model. However, the assumption that electrons in a molecule occupy atomic orbitals of the individual atoms can only be an approximation, since each bonding electron in a molecule must be in an orbital that is characteristic of the molecule as a whole. In some cases, valence bond theory cannot satisfactorily account for observed properties of molecules. Consider the oxygen molecule, whose Lewis structure is OPO O OQ Q According to this description, all the electrons in O2 are paired and the molecule should therefore be diamagnetic. But experiments have shown that the oxygen molecule is paramagnetic, with two unpaired electrons (Figure 10.20). This finding suggests a fundamental deficiency in valence bond theory, one that justifies searching for an alternative bonding approach that accounts for the properties of O2 and other molecules that do not match the predictions of valence bond theory. Magnetic and other properties of molecules are sometimes better explained by another quantum mechanical approach called molecular orbital (MO) theory. Molecular orbital theory describes covalent bonds in terms of molecular orbitals, which result from interaction of the atomic orbitals of the bonding atoms and are associated with the entire molecule. The difference between a molecular orbital and an atomic orbital is that an atomic orbital is associated with only one atom. BONDING AND ANTIBONDING MOLECULAR ORBITALS According to MO theory, the overlap of the 1s orbitals of two hydrogen atoms leads to the formation of two molecular orbitals: one bonding molecular orbital and one antibonding molecular orbital. A bonding molecular orbital has lower energy and greater stability than the atomic orbitals from which it was formed. An antibonding molecular orbital has higher energy and lower stability than the atomic orbitals from which it was formed. As the names “bonding” and “antibonding” suggest, placing electrons Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 10.6 Wave 1 Wave 1 Wave 2 Wave 2 397 in a bonding molecular orbital yields a stable covalent bond, whereas placing electrons in an antibonding molecular orbital results in an unstable bond. In the bonding molecular orbital the electron density is greatest between the nuclei of the bonding atoms. In the antibonding molecular orbital, on the other hand, the electron density decreases to zero between the nuclei. We can understand this distinction if we recall that electrons in orbitals have wave characteristics. A property unique to waves allows waves of the same type to interact in such a way that the resultant wave has either an enhanced amplitude or a diminished amplitude. In the former case, we call the interaction constructive interference; in the latter case, it is destructive interference (Figure 10.21). The formation of bonding molecular orbitals corresponds to constructive interference (the increase in amplitude is analogous to the buildup of electron density between the two nuclei). The formation of antibonding molecular orbitals corresponds to destructive interference (the decrease in amplitude is analogous to the decrease in electron density between the two nuclei). The constructive and destructive interactions between the two 1s orbitals in the H2 molecule, then, lead to the formation of a sigma bonding molecular orbital ( 1s) and a sigma antibonding molecular orbital ( #s): 1 a sigma bonding molecular orbital a sigma antibonding molecular orbital 1s Sum of 1 and 2 MOLECULAR ORBITAL THEORY 1s Sum of 1 and 2 formed from 1s orbitals (a) (b) FIGURE 10.21 Constructive interference (a) and destructive interference (b) of two waves of the same wavelength and amplitude. The two electrons in the sigma molecular orbital are paired. The Pauli exclusion principle applies to molecules as well as to atoms. formed from 1s orbitals where the star denotes an antibonding molecular orbital. In a sigma molecular orbital (bonding or antibonding) the electron density is concentrated symmetrically around a line between the two nuclei of the bonding atoms. Two electrons in a sigma molecular orbital form a sigma bond (see Section 10.5). Remember that a single covalent bond (such as HOH or FOF) is almost always a sigma bond. Figure 10.22 shows the molecular orbital energy level diagram — that is, the relative energy levels of the orbitals produced in the formation of the H2 molecule — and Molecule Destructive interaction # σ 1s Energy Atom Constructive interaction Bonding sigma molecular orbital Atom 1s Antibonding sigma molecular orbital 1s σ 1s (a) (b) FIGURE 10.22 (a) Energy levels of bonding and antibonding molecular orbitals in the H2 molecule. Note that the two electrons in the 1s orbital must have opposite spins in accord with the Pauli exclusion principle. Keep in mind that the higher the energy of the molecular orbital, the less stable the electrons in that molecular orbital. (b) Constructive and destructive interactions between the two hydrogen 1s orbitals lead to the formation of a bonding and an antibonding molecular orbital. In the bonding molecular orbital, there is a buildup between the nuclei of electron density, which acts as a negatively charged “glue” to hold the positively charged nuclei together. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 398 CHEMICAL BONDING II: MOLECULAR GEOMETRY AND HYBRIDIZATION OF ATOMIC ORBITALS the constructive and destructive interactions between the two 1s orbitals. Notice that in the antibonding molecular orbital there is a node between the nuclei that signifies zero electron density. The nuclei are repelled by each other ’s positive charges, rather than held together. Electrons in the antibonding molecular orbital have higher energy (and less stability) than they would have in the isolated atoms. On the other hand, electrons in the bonding molecular orbital have less energy (and hence greater stability) than they would have in the isolated atoms. Although we have used the hydrogen molecule to illustrate molecular orbital formation, the concept is equally applicable to other molecules. In the H2 molecule we consider only the interaction between 1s orbitals; with more complex molecules we need to consider additional atomic orbitals as well. Nevertheless, for all s orbitals, the process is the same as for 1s orbitals. Thus, the interaction between two 2s or 3s orbitals can be understood in terms of the molecular orbital energy level diagram and the formation of bonding and antibonding molecular orbitals shown in Figure 10.22. For p orbitals the process is more complex because they can interact with each other in two different ways. For example, two 2p orbitals can approach each other endto-end to produce a sigma bonding and a sigma antibonding molecular orbital, as shown in Figure 10.23(a). Alternatively, the two p orbitals can overlap sideways to generate a bonding and an antibonding pi molecular orbital [Figure 10.23(b)]. a pi bonding molecular orbital a pi antibonding molecular orbital 2p formed from 2p orbitals 2p formed from 2p orbitals In a pi molecular orbital (bonding or antibonding), the electron density is concentrated above and below an imaginery line joining the two nuclei of the bonding atoms. Two electrons in a pi molecular orbital form a pi bond (see Section 10.5). A double bond is almost always composed of a sigma bond and a pi bond; a triple bond is always a sigma bond plus two pi bonds. 10.7 MOLECULAR ORBITAL CONFIGURATIONS To understand properties of molecules, we must know how electrons are distributed among molecular orbitals. The procedure for determining the electron configuration of a molecule is analogous to the one we use to determine the electron configurations of atoms (see Section 7.8). RULES GOVERNING MOLECULAR ELECTRON CONFIGURATION AND STABILITY In order to write the electron configuration of a molecule, we must first arrange the molecular orbitals in order of increasing energy. Then we can use the following guidelines to fill the molecular orbitals with electrons. The rules also help us understand the stabilities of the molecular orbitals. • Back Forward Main Menu The number of molecular orbitals formed is always equal to the number of atomic orbitals combined. TOC Study Guide TOC Textbook Website MHHE Website 10.7 MOLECULAR ORBITAL CONFIGURATIONS 399 Molecule # σ 2p Destructive interaction Antibonding sigma molecular orbital + Energy Atom Atom 2p 2p Constructive interaction σ 2p Bonding sigma molecular orbital + (a) Destructive interaction Molecule Antibonding pi molecular orbital # π 2p Energy Atom 2p + Atom 2p π 2p + Constructive interaction Bonding pi molecular orbital (b) FIGURE 10.23 Two possible interactions between two equivalent p orbitals and the corresponding molecular orbitals. (a) When the p orbitals overlap end-to-end, a sigma bonding and a sigma antibonding molecular orbital form. (b) When the p orbitals overlap side-to-side, a pi bonding and a pi antibonding molecular orbital form. Normally, a sigma bonding molecular orbital is more stable than a pi bonding molecular orbital, since side-to-side interaction leads to a smaller overlap of the p orbitals than does end-to-end interaction. We assume that the 2px orbitals take part in the sigma molecular orbital formation. The 2py and 2pz orbitals can interact to form only molecular orbitals. The behavior shown in (b) represents the interaction between the 2py orbitals or the 2pz orbitals. • • • • • Back Forward Main Menu The more stable the bonding molecular orbital, the less stable the corresponding antibonding molecular orbital. The filling of molecular orbitals proceeds from low to high energies. In a stable molecule, the number of electrons in bonding molecular orbitals is always greater than that in antibonding molecular orbitals because we place electrons first in the lower-energy bonding molecular orbitals. Like an atomic orbital, each molecular orbital can accommodate up to two electrons with opposite spins in accordance with the Pauli exclusion principle. When electrons are added to molecular orbitals of the same energy, the most stable arrangement is predicted by Hund’s rule; that is, electrons enter these molecular orbitals with parallel spins. The number of electrons in the molecular orbitals is equal to the sum of all the electrons on the bonding atoms. TOC Study Guide TOC Textbook Website MHHE Website 400 CHEMICAL BONDING II: MOLECULAR GEOMETRY AND HYBRIDIZATION OF ATOMIC ORBITALS # σ 1s # σ 1s # σ 1s σ 1s σ 1s σ 1s H+ 2 H2 He + 2 He 2 Energy # σ 1s σ 1s FIGURE 10.24 Energy levels of the bonding and antibonding molecular orbitals in H2 , H2, He2 , and He2. In all these species, the molecular orbitals are formed by the interaction of two 1s orbitals. HYDROGEN AND HELIUM MOLECULES Later in this section we will study molecules formed by atoms of the second-period elements. Before we do, it will be instructive to predict the relative stabilities of the simple species H2 , H2, He 2 , and He2, using the energy-level diagrams shown in Figure 10.24. The 1s and #s orbitals can accommodate a maximum of four electrons. The 1 total number of electrons increases from one for H2 to four for He2. The Pauli exclusion principle stipulates that each molecular orbital can accommodate a maximum of two electrons with opposite spins. We are concerned only with the ground-state electron configurations in these cases. To evaluate the stabilities of these species we determine their bond order, defined as bond order The quantitative measure of the strength of a bond is bond dissociation energy, or bond energy (Section 9.10). The superscript in ( 1s)1 indicates that there is one electron in the sigma bonding molecular orbital. Back Forward ( 1 number of electrons 2 in bonding MOs number of electrons in antibonding MOs ) (10.2) The bond order indicates the strength of a bond. For example, if there are two electrons in the bonding molecular orbital and none in the antibonding molecular orbital, the bond order is one, which means that there is one covalent bond and that the molecule is stable. Note that the bond order can be a fraction, but a bond order of zero (or a negative value) means the bond has no stability and the molecule cannot exist. Bond order can be used only qualitatively for purposes of comparison. For example, a bonding sigma molecular orbital with two electrons and a bonding pi molecular orbital with two electrons would each have a bond order of one. Yet, these two bonds must differ in bond strength (and bond length) because of the differences in the extent of atomic orbital overlap. We are ready now to make predictions about the stability of H2 , H2, He2 , and He2 (see Figure 10.24). The H2 molecular ion has only one electron, in the 1s orbital. Since a covalent bond consists of two electrons in a bonding molecular orbital, H2 has only half of one bond, or a bond order of 1 . Thus, we predict that the H2 molecule 2 may be a stable species. The electron configuration of H2 is written as ( 1s)1. The H2 molecule has two electrons, both of which are in the 1s orbital. According to our scheme, two electrons equal one full bond; therefore, the H2 molecule has a bond order of one, or one full covalent bond. The electron configuration of H2 is ( 1s)2. As for the He2 molecular ion, we place the first two electrons in the 1s orbital and the third electron in the #s orbital. Because the antibonding molecular orbital is 1 destabilizing, we expect He2 to be less stable than H2. Roughly speaking, the instability resulting from the electron in the #s orbital is balanced by one of the 1s elec1 Main Menu TOC Study Guide TOC Textbook Website MHHE Website 10.7 401 Molecule # σ 2s Atom Atom 2s 2s Energy FIGURE 10.25 Molecular orbital energy level diagram for the Li2 molecule. The six electrons in Li2 (Li’s electron configuration is 1s22s1) are in the 1s, # , and 1s 2s orbitals. Since there are two electrons each in 1s and # 1s (just as in He2), there is no net bonding or antibonding effect. Therefore, the single covalent bond in Li2 is formed by the two electrons in the bonding molecular orbital 2s. MOLECULAR ORBITAL CONFIGURATIONS σ 2s # σ 1s 1s 1s σ 1s trons. The bond order is 1 (2 1) 1 and the overall stability of He2 is similar to that 2 2 of the H2 molecule. The electron configuration of He2 is ( 1s)2( #s)1. 1 In He2 there would be two electrons in the 1s orbital and two electrons in the # 1s orbital, so the molecule would have a bond order of zero and no net stability. The electron configuration of He2 would be ( 1s)2( #s)2. 1 To summarize, we can arrange our examples in order of decreasing stability: H2 H2 , He2 He2 We know that the hydrogen molecule is a stable species. Our simple molecular orbital method predicts that H2 and He2 also possess some stability, since both have bond orders of 1 . Indeed, their existence has been confirmed by experiment. It turns out that 2 H2 is somewhat more stable than He2 , since there is only one electron in the hydrogen molecular ion and therefore it has no electron-electron repulsion. Furthermore, H2 also has less nuclear repulsion than He2 . Our prediction about He2 is that it would have no stability, but in 1993 He2 gas was found to exist. The “molecule” is extremely unstable and has only a transient existence under specially created conditions. HOMONUCLEAR DIATOMIC MOLECULES OF SECOND-PERIOD ELEMENTS We are now ready to study the ground-state electron configuration of molecules containing second-period elements. We will consider only the simplest case, that of homonuclear diatomic molecules, or diatomic molecules containing atoms of the same elements. Figure 10.25 shows the molecular orbital energy level diagram for the first member of the second period, Li2. These molecular orbitals are formed by the overlap of Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 402 CHEMICAL BONDING II: MOLECULAR GEOMETRY AND HYBRIDIZATION OF ATOMIC ORBITALS 1s and 2s orbitals. We will use this diagram to build up all the diatomic molecules, as we will see shortly. The situation is more complex when the bonding also involves p orbitals. Two p orbitals can form either a sigma bond or a pi bond. Since there are three p orbitals for each atom of a second-period element, we know that one sigma and two pi molecular orbitals will result from the constructive interaction. The sigma molecular orbital is formed by the overlap of the 2px orbitals along the internuclear axis, that is, the x-axis. The 2py and 2pz orbitals are perpendicular to the x-axis, and they will overlap sideways to give two pi molecular orbitals. The molecular orbitals are called 2px, 2py, and 2pz orbitals, where the subscripts indicate which atomic orbitals take part in forming the molecular orbitals. As shown in Figure 10.23, overlap of the two p orbitals is normally greater in a molecular orbital than in a molecular orbital, so we would expect the former to be lower in energy. However, the energies of molecular orbitals actually increase as follows: 1s # 1s 2s # 2s 2py 2pz 2px # 2py # 2pz # 2px The inversion of the 2px orbital and the 2py and 2pz can be explained qualitatively. The electrons in the 1s, #s, 2s and #s orbitals tend to concentrate between the two 1 2 nuclei. Since the 2px orbital has a geometry similar to that of the s orbitals, its electrons will concentrate in the same region. Consequently, electrons in the 2px orbital will experience greater repulsion from electrons in the filled s and # orbitals than s those in the 2py and 2pz orbitals. Note that the repulsion between the electrons in the 2px and inner s orbitals decreases from left to right in the second period. As a result, the 2px orbital is higher in energy than the 2py and 2pz orbitals for B2, C2, and N2, but lower in energy than the 2py and 2pz orbitals for O2 and F2. With these concepts and Figure 10.26, which shows the order of increasing energies for 2p molecular orbitals, we can write the electron configurations and predict the magnetic properties and bond orders of second-period homonuclear diatomic molecules. We will consider a few examples. THE LITHIUM MOLECULE (Li2) The electron configuration of Li is 1s22s1, so Li2 has a total of six electrons. According to Figure 10.25, these electrons are placed (two each) in the 1s, #s, and 2s molec1 ular orbitals. The electrons of 1s and #s make no net contribution to the bonding in 1 Li2. Thus the electron configuration of the molecular orbitals in Li2 is ( 1s)2( #s)2( 2s)2. 1 Since there are two more electrons in the bonding molecular orbitals than in antibonding orbitals, the bond order is 1 [see Equation (10.2)]. We conclude that the Li2 molecule is stable, and since it has no unpaired electron spins, it should be diamagnetic. Indeed, diamagnetic Li2 molecules are known to exist in the vapor phase. THE CARBON MOLECULE (C2) The carbon atom has the electron configuration 1s22s22p2; thus there are 12 electrons in the C2 molecule. From the bonding scheme for Li2, we place four additional carbon electrons in the 2py and 2pz orbitals. Therefore, C2 has the electron configuration Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 10.7 MOLECULAR ORBITAL CONFIGURATIONS 403 Molecule σ #px 2 π #py π #pz 2 2 Energy Atom Atom 2px 2py 2pz σ 2px 2px 2py 2pz π 2py π 2pz FIGURE 10.26 General molecular orbital energy-level diagram for the second-period homonuclear diatomic molecules Li2, Be2, B2, C2, and N2. For simplicity, the 1s and 2s orbitals have been omitted. Note that in these molecules the 2px orbital is higher in energy than either the 2py or the 2pz orbitals. This means that electrons in the 2px orbitals are less stable than those in 2p y and 2p z. This aberration stems from the different interactions between the electrons in the 2px orbital, on one hand, and 2py and 2pz orbitals, on the other hand, with the electrons in the lower-energy s orbitals. For O2 and F2, the 2px orbital is lower in energy than 2py and 2pz. ( 2 #2 2 #2 2 2 1s) ( 1s) ( 2s) ( 2s) ( 2py) ( 2pz) Its bond order is 2, and the molecular should be diamagnetic. Again, diamagnetic C2 molecules have been detected in the vapor state. Note that the double bonds in C2 are both pi bonds because of the four electrons in the two pi molecular orbitals. In most other molecules, a double bond is made up of a sigma bond and a pi bond. THE OXYGEN MOLECULE (O2) As we stated earlier, valence bond theory does not account for the magnetic properties of the oxygen molecule. To show the two unpaired electrons on O2, we need to draw an alternative to the resonance structure present on p. 396: TOOOT OO QQ This structure is unsatisfactory on at least two counts. First, it implies the presence of a single covalent bond, but experimental evidence strongly suggests that there is a double bond in this molecule. Second, it places seven valence electrons around each oxygen atom, a violation of the octet rule. The ground-state electron configuration of O is 1s22s22p4; thus there are 16 electrons in O2. Using the order of increasing energies of the molecular orbitals discussed above, we write the ground-state electron configuration of O2 as ( Back Forward Main Menu TOC 2 #2 2 #2 2 2 2#1#1 1s) ( 1s) ( 2s) ( 2s) ( 2px) ( 2py) ( 2pz) ( 2py) ( 2pz) Study Guide TOC Textbook Website MHHE Website 404 CHEMICAL BONDING II: MOLECULAR GEOMETRY AND HYBRIDIZATION OF ATOMIC ORBITALS TABLE 10.5 Properties of Homonuclear Diatomic Molecules of the Second-period Elements* Li2 B2 C2 N2 O2 F2 # # 2px # 2py, 2px # hh hg hg hg hg 2py, 2pz 2px hh hg hg hg hg hg hg hg hg hg hg hg hg hg hg hg hg 2s 1 267 104.6 1 159 288.7 2 131 627.6 3 110 941.4 2 121 498.7 1 142 156.9 2s 2s Bond order Bond length (pm) Bond dissociation energy (kJ/mol) Magnetic properties 2pz # 2pz # and # 2py, hg 2py, 1s # hg 2px *For simplicity the and 2pz. hg hg 2pz 2s Diamagnetic Paramagnetic Diamagnetic Diamagnetic Paramagnetic Diamagnetic # 1s orbitals are omitted. These two orbitals hold a total of four electrons. Remember that for O2 and F2, 2px is lower in energy than 2py According to Hund’s rule, the last two electrons enter the #py and #pz orbitals with 2 2 parallel spins. Ignoring the 1s and 2s orbitals (since their net effects on bonding are zero), we calculate the bond order of O2 using Equation (10.2): 1 2 bond order (6 2) 2 Therefore, the O2 molecule has a bond order of 2 and is paramagnetic, a prediction that corresponds to experimental observations. Table 10.5 summarizes the general properties of the stable diatomic molecules of the second period. Example 10.6 shows how MO theory can help predict molecular properties of ions. EXAMPLE 10.6 The N2 ion can be prepared by bombarding the N2 molecule with fast-moving electrons. Predict the following properties of N2 : (a) electron configuration, (b) bond order, (c) magnetic character, and (d) bond length relative to the bond length of N2 (is it longer or shorter?). Answer (a) Since N2 has one fewer electron than N2, its electron configuration is (see Table 10.5) ( 2 #2 2 #2 2 2 1 1s) ( 1s) ( 2s) ( 2s) ( 2py) ( 2pz) ( 2px) (b) The bond order of N2 is found by using Equation (10.2): bond order 1 2 (9 4) 2.5 (c) N2 has one unpaired electron, so it is paramagnetic. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 10.8 Similar problems: 10.55, 10.56. DELOCALIZED MOLECULAR ORBITALS 405 (d) Since the electrons in the bonding molecular orbitals are responsible for holding the atoms together, N2 should have a weaker and, therefore, longer bond than N2. (In fact, the bond length of N2 is 112 pm, compared with 110 pm for N2.) PRACTICE EXERCISE Which of the following species has a longer bond length: F2 or F2 ? 10.8 DELOCALIZED MOLECULAR ORBITALS So far we have discussed chemical bonding only in terms of electron pairs. However, the properties of a molecule cannot always be explained accurately by a single structure. A case in point is the O3 molecule, discussed in Section 9.8. There we overcame the dilemma by introducing the concept of resonance. In this section we will tackle the problem in another way — by applying the molecular orbital approach. As in Section 9.8, we will use the benzene molecule and the carbonate ion as examples. Note that in discussing the bonding of polyatomic molecules or ions, it is convenient to determine first the hybridization state of the atoms present (a valence bond approach), followed by the formation of appropriate molecular orbitals. THE BENZENE MOLECULE H C H H C C C C H C H H FIGURE 10.27 The sigma bond framework in the benzene molecule. Each carbon atom is sp2-hybridized and forms sigma bonds with two adjacent carbon atoms and another sigma bond with a hydrogen atom. Benzene (C6H6) is a planar hexagonal molecule with carbon atoms situated at the six corners. All carbon-carbon bonds are equal in length and strength, as are all carbonhydrogen bonds, and the CCC and HCC angles are all 120°. Therefore, each carbon atom is sp2-hybridized; it forms three sigma bonds with two adjacent carbon atoms and a hydrogen atom (Figure 10.27). This arrangement leaves an unhybridized 2pz orbital on each carbon atom, perpendicular to the plane of the benzene molecule, or benzene ring, as it is often called. So far the description resembles the configuration of ethylene (C2H4), discussed in Section 10.5, except that in this case there are six unhybridized 2pz orbitals in a cyclic arrangement. Because of their similar shape and orientation, each 2pz orbital overlaps two others, one on each adjacent carbon atom. According to the rules listed on p. 398, the interaction of six 2pz orbitals leads to the formation of six pi molecular orbitals, of which three are bonding and three antibonding. A benzene molecule in the ground state therefore has six electrons in the three pi bonding molecular orbitals, two electrons with paired spins in each orbital (Figure 10.28). Unlike the pi bonding molecular orbitals in ethylene, those in benzene form delocalized molecular orbitals, which are not confined between two adjacent bonding atoms, but actually extend over three or more atoms. Therefore, electrons residing in any of these orbitals are free to move around the benzene ring. For this reason, the structure of benzene is sometimes represented as in which the circle indicates that the pi bonds between carbon atoms are not confined to individual pairs of atoms; rather, the pi electron densities are evenly distributed Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 406 CHEMICAL BONDING II: MOLECULAR GEOMETRY AND HYBRIDIZATION OF ATOMIC ORBITALS Top view Side view (a) (b) FIGURE 10.28 (a) The six 2pz orbitals on the carbon atoms in benzene. (b) The delocalized molecular orbital formed by the overlap of the 2pz orbitals. The delocalized molecular orbital possesses pi symmetry and lies above and below the plane of the benzene ring. Actually, these 2pz orbitals can combine in six different ways to yield three bonding molecular orbitals and three antibonding molecular orbitals. The one shown here is the most stable. throughout the benzene molecule. The carbon and hydrogen atoms are not shown in the simplified diagram. We can now state that each carbon-to-carbon linkage in benzene contains a sigma bond and a “partial” pi bond. The bond order between any two adjacent carbon atoms is therefore between 1 and 2. Thus molecular orbital theory offers an alternative to the resonance approach, which is based on valence bond theory. (The resonance structures of benzene are shown on p. 349.) THE CARBONATE ION Cyclic compounds like benzene are not the only ones with delocalized molecular orbitals. Let’s look at bonding in the carbonate ion (CO2 ). VSEPR predicts a trigonal 3 planar geometry for the carbonate ion, like that for BF3. The planar structure of the carbonate ion can be explained by assuming that the carbon atom is sp 2-hybridized. The C atom forms sigma bonds with three O atoms. Thus the unhybridized 2pz orbital of the C atom can simultaneously overlap the 2pz orbitals of all three O atoms (Figure 10.29). The result is a delocalized molecular orbital that extends over all four nuclei in such a way that the electron densities (and hence the bond orders) in the carbon-to- O O O C O O C O FIGURE 10.29 Bonding in the carbonate ion. The carbon atom forms three sigma bonds with the three oxygen atoms. In addition, the 2pz orbitals of the carbon and oxygen atoms overlap to form delocalized molecular orbitals, so that there is also a partial pi bond between the carbon atom and each of the three oxygen atoms. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website SUMMARY OF FACTS AND CONCEPTS 407 oxygen bonds are all the same. Molecular orbital theory therefore provides an acceptable alternative explanation of the properties of the carbonate ion as compared with the resonance structures of the ion shown on p. 349. We should note that molecules with delocalized molecular orbitals are generally more stable than those containing molecular orbitals extending over only two atoms. For example, the benzene molecule, which contains delocalized molecular orbitals, is chemically less reactive (and hence more stable) than molecules containing “localized” CPC bonds, such as ethylene. SUMMARY OF KEY EQUATIONS • Q r (10.1) • bond order SUMMARY OF FACTS AND CONCEPTS Back Forward Main Menu Expressing dipole moment in terms of charge (Q) and distance of separation (r) between charges. 1 number of electrons 2 in bonding MOs number of electrons in antibonding MOs (10.2) 1. The VSEPR model for predicting molecular geometry is based on the assumption that valence-shell electron pairs repel one another and tend to stay as far apart as possible. 2. According to the VSEPR model, molecular geometry can be predicted from the number of bonding electron pairs and lone pairs. Lone pairs repel other pairs more forcefully than bonding pairs do and thus distort bond angles from the ideal geometry. 3. Dipole moment is a measure of the charge separation in molecules containing atoms of different electronegativities. The dipole moment of a molecule is the resultant of whatever bond moments are present. Information about molecular geometry can be obtained from dipole moment measurements. 4. There are two quantum mechanical explanations for covalent bond formation: valence bond theory and molecular orbital theory. In valence bond theory, hybridized atomic orbitals are formed by the combination and rearrangement of orbitals from the same atom. The hybridized orbitals are all of equal energy and electron density, and the number of hybridized orbitals is equal to the number of pure atomic orbitals that combine. 5. Valence-shell expansion can be explained by assuming hybridization of s, p, and d orbitals. 6. In sp hybridization, the two hybrid orbitals lie in a straight line; in sp 2 hybridization, the three hybrid orbitals are directed toward the corners of a triangle; in sp3 hybridization, the four hybrid orbitals are directed toward the corners of a tetrahedron; in sp 3d hybridization, the five hybrid orbitals are directed toward the corners of a trigonal bipyramid; in sp3d 2 hybridization, the six hybrid orbitals are directed toward the corners of an octahedron. 7. In an sp2-hybridized atom (for example, carbon), the one unhybridized p orbital can form a pi bond with another p orbital. A carbon-carbon double bond consists of a sigma bond and a pi bond. In an sp-hybridized carbon atom, the two unhybridized p orbitals can form two pi bonds with two p orbitals on another atom (or atoms). A carbon-carbon triple bond consists of one sigma bond and two pi bonds. 8. Molecular orbital theory describes bonding in terms of the combination and rearrangement of atomic orbitals to form orbitals that are associated with the molecule as a whole. 9. Bonding molecular orbitals increase electron density between the nuclei and are lower in energy than individual atomic orbitals. Antibonding molecular orbitals have a region of zero electron density between the nuclei, and an energy level higher than that of the individual atomic orbitals. 10. We write electron configurations for molecular orbitals as we do for atomic orbitals, filling in electrons in the order of increasing energy levels. The number of molecular orbitals always equals the number of atomic orbitals that were combined. The Pauli exclusion principle and Hund’s rule govern the filling of molecular orbitals. TOC Study Guide TOC Textbook Website MHHE Website 408 CHEMICAL BONDING II: MOLECULAR GEOMETRY AND HYBRIDIZATION OF ATOMIC ORBITALS 11. Molecules are stable if the number of electrons in bonding molecular orbitals is greater than that in antibonding molecular orbitals. 12. Delocalized molecular orbitals, in which electrons are free to move around a whole molecule or group of atoms, are formed by electrons in p orbitals of adjacent atoms. Delocalized molecular orbitals are an alternative to resonance structures in explaining observed molecular properties. KEY WORDS Antibonding molecular orbital, p. 397 Bond order, p. 400 Bonding molecular orbital, p. 397 Delocalized molecular orbital, p. 405 Dipole moment ( ), p. 377 Homonuclear diatomic molecule, p. 401 Hybrid orbital, p. 385 Hybridization, p. 385 Molecular orbital, p. 396 Nonpolar molecule, p. 378 Pi bond ( bond), p. 393 Pi molecular orbital, p. 398 Polar molecule, p. 378 Sigma bond ( bond), p. 393 Sigma molecular orbital, p. 397 Valence shell, p. 368 Valence-shell electron-pair repulsion (VSEPR) model, p. 368 QUESTIONS AND PROBLEMS MOLECULAR GEOMETRY Review Questions 10.1 How is the geometry of a molecule defined and why is the study of molecular geometry important? 10.2 Sketch the shape of a linear triatomic molecule, a trigonal planar molecule containing four atoms, a tetrahedral molecule, a trigonal bipyramidal molecule, and an octahedral molecule. Give the bond angles in each case. 10.3 How many atoms are directly bonded to the central atom in a tetrahedral molecule, a trigonal bipyramidal molecule, and an octahedral molecule? 10.4 Discuss the basic features of the VSEPR model. Explain why the magnitude of repulsion decreases in the following order: lone pair-lone pair lone pairbonding pair bonding pair-bonding pair. 10.5 In the trigonal bipyramidal arrangement, why does a lone pair occupy an equatorial position rather than an axial position? 10.6 The geometry of CH4 could be square planar, with the four H atoms at the corners of a square and the C atom at the center of the square. Sketch this geometry and compare its stability with that of a tetrahedral CH4 molecule. Problems 10.7 Predict the geometries of the following species using the VSEPR method: (a) PCl3, (b) CHCl3, (c) SiH4, (d) TeCl4. 10.8 Predict the geometries of the following species: (a) AlCl3, (b) ZnCl2, (c) ZnCl2 . 4 Back Forward Main Menu TOC 10.9 Predict the geometry of the following molecules using the VSEPR method: (a) HgBr2, (b) N2O (arrangement of atoms is NNO), (c) SCN (arrangement of atoms is SCN). 10.10 Predict the geometries of the following ions: (a) NH4 , (b) NH2 , (c) CO2 (d) ICl2 , (e) ICl4 , 3 (f ) AlH4 , (g) SnCl5 , (h) H3O , (i) BeF2 . 4 10.11 Describe the geometry around each of the three central atoms in the CH3COOH molecule. 10.12 Which of the following species are tetrahedral? SiCl4, SeF4, XeF4, CI4, CdCl2 4 DIPOLE MOMENTS Review Questions 10.13 Define dipole moment. What are the units and symbol for dipole moment? 10.14 What is the relationship between the dipole moment and bond moment? How is it possible for a molecule to have bond moments and yet be nonpolar? 10.15 Explain why an atom cannot have a permanent dipole moment. 10.16 The bonds in beryllium hydride (BeH2) molecules are polar, and yet the dipole moment of the molecule is zero. Explain. Problems 10.17 Referring to Table 10.3, arrange the following mole- Study Guide TOC Textbook Website MHHE Website QUESTIONS AND PROBLEMS 10.18 10.19 10.20 10.21 cules in order of increasing dipole moment: H2O, H2S, H2Te, H2Se. The dipole moments of the hydrogen halides decrease from HF to HI (see Table 10.3). Explain this trend. List the following molecules in order of increasing dipole moment: H2O, CBr4, H2S, HF, NH3, CO2. Does the molecule OCS have a higher or lower dipole moment than CS2? Which of the following molecules has a higher dipole moment? 409 10.29 How would you distinguish between a sigma bond and a pi bond? 10.30 Which of the following pairs of atomic orbitals of adjacent nuclei can overlap to form a sigma bond? Which overlap to form a pi bond? Which cannot overlap (no bond)? Consider the x-axis to be the internuclear axis, that is, the line joining the nuclei of the two atoms. (a) 1s and 1s, (b) 1s and 2px, (c) 2px and 2py, (d) 3py and 3py, (e) 2px and 2px, (f) 1s and 2s Problems Br H G D C PC D G H Br Br Br G D C PC D G H H (a) (b) 10.22 Arrange the following compounds in order of increasing dipole moment: 10.31 Describe the bonding scheme of the AsH3 molecule in terms of hybridization. 10.32 What is the hybridization of Si in SiH4 and in H3SiOSiH3? 10.33 Describe the change in hybridization (if any) of the Al atom in the following reaction: AlCl3 Cl 88n AlCl4 10.34 Consider the reaction Cl Cl Cl Cl Cl Cl Cl BF3 Cl 10.35 Cl Cl (b) (a) Cl (c) (d) 10.36 VALENCE BOND THEORY Review Questions 10.23 What is valence bond theory? How does it differ from the Lewis concept of chemical bonding? 10.24 Use valence bond theory to explain the bonding in Cl2 and HCl. Show how the atomic orbitals overlap when a bond is formed. 10.25 Draw a potential energy curve for the bond formation in F2. 10.37 10.38 10.39 HYBRIDIZATION Review Questions 10.26 What is the hybridization of atomic orbitals? Why is it impossible for an isolated atom to exist in the hybridized state? 10.27 How does a hybrid orbital differ from a pure atomic orbital? Can two 2p orbitals of an atom hybridize to give two hybridized orbitals? 10.28 What is the angle between the following two hybrid orbitals on the same atom? (a) sp and sp hybrid orbitals, (b) sp2 and sp2 hybrid orbitals, (c) sp3 and sp3 hybrid orbitals Back Forward Main Menu TOC 10.40 10.41 NH3 88n F3BONH3 Describe the changes in hybridization (if any) of the B and N atoms as a result of this reaction. What hybrid orbitals are used by nitrogen atoms in the following species? (a) NH3, (b) H2NONH2, (c) NO3 What are the hybrid orbitals of the carbon atoms in the following molecules? (a) H3COCH3 (b) H3COCHPCH2 (c) CH3OCqCOCH2OH (d) CH3CHPO (e) CH3COOH Specify which hybrid orbitals are used by carbon atoms in the following species: (a) CO, (b) CO2, (c) CN . What is the hybridization state of the central N atom in the azide ion, N3 ? (Arrangement of atoms: NNN.) The allene molecule H2CPCPCH2 is linear (the three C atoms lie on a straight line). What are the hybridization states of the carbon atoms? Draw diagrams to show the formation of sigma bonds and pi bonds in allene. Describe the hybridization of phosphorus in PF5. How many sigma bonds and pi bonds are there in each of the following molecules? H H H Cl A A G D CP C ClO CO Cl H3COCP COCq COH D G A A H H H H Study Guide TOC (a) (b) Textbook Website (c) MHHE Website 410 CHEMICAL BONDING II: MOLECULAR GEOMETRY AND HYBRIDIZATION OF ATOMIC ORBITALS 10.42 How many pi bonds and sigma bonds are there in the tetracyanoethylene molecule? NqC NqC G D CP C D G CqN CqN MOLECULAR ORBITAL THEORY Review Questions 10.43 What is molecular orbital theory? How does it differ from valence bond theory? 10.44 Define the following terms: bonding molecular orbital, antibonding molecular orbital, pi molecular orbital, sigma molecular orbital. 10.45 Sketch the shapes of the following molecular orbitals: # # 1s, ls, 2p, and 2p. How do their energies compare? 10.46 Explain the significance of bond order. Can bond order be used for quantitative comparisons of the strengths of chemical bonds? Problems 10.47 Explain in molecular orbital terms the changes in HOH internuclear distance that occur as the molecular H2 is ionized first to H2 and then to H2 . 2 10.48 The formation of H2 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? 10.49 Draw a molecular orbital energy level diagram for each of the following species: He2, HHe, He2 . Compare their relative stabilities in terms of bond orders. (Treat HHe as a diatomic molecule with three electrons.) 10.50 Arrange the following species in order of increasing stability: Li2, Li2 , Li2 . Justify your choice with a molecular orbital energy level diagram. 10.51 Use molecular orbital theory to explain why the Be2 molecule does not exist. 10.52 Which of these species has a longer bond, B2 or B2 ? Explain in terms of molecular orbital theory. 10.53 Acetylene (C2H2) has a tendency to lose two protons (H ) and form the carbide ion (C2 ), which is pres2 ent in a number of ionic compounds, such as CaC2 and MgC2. Describe the bonding scheme in the C2 2 ion in terms of molecular orbital theory. Compare the bond order in C2 with that in C2. 2 Back Forward Main Menu TOC 10.54 Compare the Lewis and molecular orbital treatments of the oxygen molecule. 10.55 Explain why the bond order of N2 is greater than that of N2 , but the bond order of O2 is less than that of O2 . 10.56 Compare the relative stability of the following species and indicate their magnetic properties (that is, diamagnetic or paramagnetic): O2, O2 , O2 (superoxide ion), O2 (peroxide ion). 2 10.57 Use molecular orbital theory to compare the relative stabilities of F2 and F2 . 10.58 A single bond is almost always a sigma bond, and a double bond is almost always made up of a sigma bond and a pi bond. There are very few exceptions to this rule. Show that the B2 and C2 molecules are examples of the exceptions. DELOCALIZED MOLECULAR ORBITALS Review Questions 10.59 How does a delocalized molecular orbital differ from a molecular orbital such as that found in H2 or C2H4? What do you think are the minimum conditions (for example, number of atoms and types of orbitals) for forming a delocalized molecular orbital? 10.60 In Chapter 9 we saw that the resonance concept is useful for dealing with species such as the benzene molecule and the carbonate ion. How does molecular orbital theory deal with these species? Problems 10.61 Both ethylene (C2H4) and benzene (C6H6) contain the CPC bond. The reactivity of ethylene is greater than that of benzene. For example, ethylene readily reacts with molecular bromine, whereas benzene is normally quite inert toward molecular bromine and many other compounds. Explain this difference in reactivity. 10.62 Explain why the symbol on the left is a better representation of benzene molecules than that on the right. 10.63 Determine which of these molecules has a more delocalized orbital and justify your choice. Study Guide TOC Textbook Website MHHE Website QUESTIONS AND PROBLEMS (Hint: Both molecules contain two benzene rings. In naphthalene, the two rings are fused together. In biphenyl, the two rings are joined by a single bond, around which the two rings can rotate.) 10.64 Nitryl fluoride (FNO2) is very reactive chemically. The fluorine and oxygen atoms are bonded to the nitrogen atom. (a) Write a Lewis structure for FNO2. (b) Indicate the hybridization of the nitrogen atom. (c) Describe the bonding in terms of molecular orbital theory. Where would you expect delocalized molecular orbitals to form? 10.65 Describe the bonding in the nitrate ion NO3 in terms of delocalized molecular orbitals. 10.66 What is the state of hybridization of the central O atom in O3? Describe the bonding in O3 in terms of delocalized molecular orbitals. 411 10.75 Briefly compare the VSEPR and hybridization approaches to the study of molecular geometry. 10.76 Describe the hybridization state of arsenic in arsenic pentafluoride (AsF5). 10.77 Draw Lewis structures and give the other information requested for the following: (a) SO3. Polar or nonpolar molecule? (b) PF3. Polar or nonpolar molecule? (c) F3SiH. Show the direction of the resultant dipole moment. (d) SiH3 . Planar or pyramidal shape? (e) Br2CH2. Polar or nonpolar molecule? 10.78 Which of the following molecules are linear? ICl2 , IF2 , OF2, SnI2, CdBr2 10.79 Draw the Lewis structure for the BeCl2 ion. Predict 4 its geometry and describe the hybridization state of the Be atom. 10.80 The N2F2 molecule can exist in either of the following two forms: ADDITIONAL PROBLEMS 10.67 Which of the following species is not likely to have a tetrahedral shape? (a) SiBr4, (b) NF4 , (c) SF4, (d) BeCl2 , (e) BF4 , (f) AlCl4 4 10.68 Draw the Lewis structure of mercury(II) bromide. Is this molecule linear or bent? How would you establish its geometry? 10.69 Sketch the bond moments and resultant dipole moments for the following molecules: H2O, PCl3, XeF4, PCl5, SF6. 10.70 Although both carbon and silicon are in Group 4A, very few SiPSi bonds are known. Account for the instability of silicon-to-silicon double bonds in general. (Hint: Compare the atomic radii of C and Si in Figure 8.5. What effect would the larger size have on pi bond formation?) 10.71 Predict the geometry of sulfur dichloride (SCl2) and the hybridization of the sulfur atom. 10.72 Antimony pentafluoride, SbF5, reacts with XeF4 and XeF6 to form ionic compounds, XeF3 SbF6 and XeF5 SbF6 . Describe the geometries of the cations and anion in these two compounds. 10.73 Draw Lewis structures and give the other information requested for the following molecules: (a) BF3. Shape: planar or nonplanar? (b) ClO3 . Shape: planar or nonplanar? (c) H2O. Show the direction of the resultant dipole moment. (d) OF2. Polar or nonpolar molecule? (e) NO2. Estimate the ONO bond angle. 10.74 Predict the bond angles for the following molecules: (a) BeCl2, (b) BCl3, (c) CCl4, (d) CH3Cl, (e) Hg2Cl2 (arrangement of atoms: ClHgHgCl), (f) SnCl2, (g) H2O2, (i) SnH4. Back Forward Main Menu TOC F D D NPN F F G D NPN F (a) What is the hybridization of N in the molecule? (b) Which structure has a dipole moment? 10.81 Cyclopropane (C3H6) has the shape of a triangle in which a C atom is bonded to two H atoms and two other C atoms at each corner. Cubane (C8H8) has the shape of a cube in which a C atom is bonded to one H atom and three other C atoms at each corner. (a) Draw Lewis structures of these molecules. (b) Compare the CCC angles in these molecules with those predicted for an sp3-hybridized C atom. (c) Would you expect these molecules to be easy to make? 10.82 The compound 1,2-dichloroethane (C2H4Cl2) is nonpolar, while cis-dichloroethylene (C2H2Cl2) has a dipole moment: Cl Cl AA HOCO COH AA HH 1,2-dichloroethane Cl G D C PC D G H H cis-dichloroethylene Cl The reason for the difference is that groups connected by a single bond can rotate with respect to each other, but no rotation occurs when a double bond connects the groups. On the basis of bonding considerations, explain why rotation occurs in 1,2-dichloroethane but not in cis-dichloroethylene. Study Guide TOC Textbook Website MHHE Website 412 CHEMICAL BONDING II: MOLECULAR GEOMETRY AND HYBRIDIZATION OF ATOMIC ORBITALS 10.83 Does the following molecule have a dipole moment? Cl H G D C PC P C D G H Cl (Hint: See the answer to Problem 10.39.) 10.84 So-called greenhouse gases, which contribute to global warming, have a dipole moment or can be bent or distorted into shapes that have a dipole moment. Which of the following gases are greenhouse gases? N2, O2, O3, CO, CO2, NO2, N2O, CH4, CFCl3 10.85 The bond angle of SO2 is very close to 120°, even though there is a lone pair on S. Explain. 10.86 3′-azido-3′-deoxythymidine, commonly known as AZT, is one of the drugs used to treat acquired immune deficiency syndrome (AIDS). What are the hybridization states of the C and N atoms in this molecule? O H C N C C O CH3 NH2 C N H C C O Main Menu C N H H 10.87 The following molecules (AX4Y2) all have octahedral geometry. Group the molecules that are equivalent to each other. Forward H C N HOO CH2 O A CH HC AA AA HC CH A A N H B N B N Back 10.88 The compounds carbon tetrachloride (CCl4) and silicon tetrachloride (SiCl4) are similar in geometry and hybridization. However, CCl4 does not react with water but SiCl4 does. Explain the difference in their chemical reactivities. (Hint: The first step of the reaction is believed to be the addition of a water molecule to the Si atom in SiCl4.) 10.89 Write the ground-state electron configuration for B2. Is the molecule diamagnetic or paramagnetic? 10.90 What are the hybridization states of the C and N atoms in this molecule? TOC 10.91 Use molecular orbital theory to explain the difference between the bond energies of F2 and F2 (see Problem 9.108). 10.92 Referring to the Chemistry in Action on p. 382, answer the following questions: (a) If you wanted to cook a roast (beef or lamb), would you use a microwave oven or a conventional oven? (b) Radar is a means of locating an object by measuring the time for the echo of a microwave from the object to return to the source and the direction from which it returns. Would radar work if oxygen, nitrogen, and carbon dioxide were polar molecules? (c) In early tests of radar at the English Channel during World War II, the results were inconclusive even though there was no equipment malfunction. Why? (Hint: The weather is often foggy in that region.) 10.93 The geometries discussed in this chapter all lend Study Guide TOC Textbook Website MHHE Website QUESTIONS AND PROBLEMS themselves to fairly straightforward elucidation of bond angles. The exception is the tetrahedron, because its bond angles are hard to visualize. Consider the CCl4 molecule, which has a tetrahedral geometry and is nonpolar. By equating the bond moment of a particular COCl bond to the resultant bond moments of the other three COCl bonds in opposite directions, show that the bond angles are all equal to 109.5°. 10.94 Referring to Table 9.4, explain why the bond dissociation energy for Cl2 is greater than that for F2. (Hint: The bond lengths of F2 and Cl2 are 142 pm and 199 pm, respectively.) Back Forward Main Menu TOC 413 10.95 Use molecular orbital theory to explain the bonding in the azide ion (N3 ). Answers to Practice Exercises: 10.1 (a) Tetrahedral, (b) linear, (c) trigonal planar. 10.2 No. 10.3 (a) sp3, (b) sp 2. 10.4 sp3d2. 10.5 The C atom is sp-hybridized. It forms a sigma bond with the H atom and another sigma bond with the N atom. The two unhybridized p orbitals on the C atom are used to form two pi bonds with the N atom. The lone pair on the N atom is placed in the sp orbital. 10.6 F2 . Study Guide TOC Textbook Website MHHE Website C HEMISTRY IN T HREE D IMENSIONS Buckyball and Other Large Allotropes of Carbon 335 pm Graphite is made up of layers of six-membered rings of carbon. Geodesic domes were a favorite design element of the architect and philosopher R. Buckminster Fuller, for whom buckminsterfullerene was named because of its structural resemblance to Fuller’s domes. The geometry of a buckyball C60 (top) resembles a soccer ball (bottom). Scientists arrived at this structure by fitting together paper cutouts of enough hexagons and pentagons to accommodate 60 carbon atoms at the points where they intersect. In 1985 chemists at Rice University in Texas used a high-powered laser to vaporize graphite in an effort to create unusual molecules believed to exist in interstellar space. Mass spectrometry revealed that one of the products was an unknown species with the formula C60. Because of its size and the fact that it is pure carbon, this molecule has an exotic shape, which the researchers worked out using paper, scissors, and tape. Subsequent spectroscopic and X-ray measurements confirmed that C60 is shaped like a hollow sphere with a carbon atom at each of the 60 vertices. Geometrically, buckyball (short for “buckminsterfullerene”) is the most symmetrical molecule known. In spite of its unique features, however, its bonding scheme is straightforward. Each carbon is sp2-hybridized, and there are extensive delocalized molecular orbitals over the entire structure. The discovery of buckyball generated tremendous interest within the scientific community. Here was a new allotrope of carbon with an intriguing geometry and unknown properties to investigate. Since 1985 chemists have created a whole class of fullerenes, with 70, 76, and even larger numbers of carbon atoms. Moreover, buckyball has been found to be a natural component of soot, and the fullerenes C60 and C70 turned up in a rock sample from northwestern Russia. Buckyball and its heavier members represent a whole new concept in molecular architecture with far-reaching implications. Studies have shown that the fullerenes and their compounds can act as high-temperature superconductors and lubricants, as well as catalysts. One fascinating discovery, made in 1991 by Japanese scientists, was the identification of structural relatives of buckyball. These molecules are hundreds of 414 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website The structure of a buckytube that consists of a single layer of carbon atoms. Note that the truncated buckyball “cap,” which has been separated from the rest of the buckytube in this view, has a different structure than the graphite-like cylindrical portion of the tube. Chemists have devised ways to open the cap in order to place other molecules inside the tube. An electron micrograph of a multiwalled buckytube. The thickness of 14 adjacent graphite-like layers is about 5 nanometers. nanometers long with a tubular shape and an internal cavity about 15 nanometers in diameter. Dubbed “buckytubes” or “nanotubes” (because of their size), these molecules have two distinctly different structures. One is a single sheet of graphite that is capped at both ends with a kind of truncated buckyball. The other is a scroll-like tube having anywhere from 2 to 30 graphite-like layers. Buckytubes are several times stronger than steel wires of similar dimensions and one day they might be used to make ultralight bicycles, rocket motor casings, and tennis racquets. They could also serve as casting molds for very thin metal wires to be used in microscopic integrated circuits or as “bottles” for storing molecules. In the first biological application of buckyball, chemists at the University of California at San Francisco and Santa Barbara made a discovery in 1993 that could help in designing drugs to treat AIDS. The human immunodeficiency virus (HIV) that causes AIDS reproduces by synthesizing a long protein chain, which is cut into smaller segments by an enzyme called HIV-protease. One way to stop AIDS, then, might be to inactivate the enzyme. When the chemists reacted a water-soluble derivative of bucky-ball with HIV-protease, they found that it binds to the portion of the enzyme that would ordinarily cleave the reproductive protein, thereby preventing the HIV virus from reproducing. Consequently, the virus could no longer infect the human cells they had grown in the laboratory. The buckyball compound itself is not a suitable drug for use against AIDS because of potential side effects and delivery difficulties, but it does provide a model for the development of such drugs. Computer-generated model of the binding of a buckyball derivative to the site of HIV-protease that normally attaches to a protein needed for the reproduction of HIV. The buckyball structure fits tightly into the active site, thus preventing the enzyme from carrying out its function. 415 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website ...
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