18 CHAPTER 2 Bonding and Molecular Structure 12p 上上一w上上上上 thell too ) Figure 2.5 Head Tail Figure 2.6 11 2.11 Ground state Hybridized states Figure 2.7 TABLE2.2 NUMBER OF TYPE OF TYPE BOND ANGLE GEOMETRY REMAININGP'S BOND FORMED p 109.5 Tetrahedral* 0 sp- 120° Trigonal planar 1 180 Linear 2 *Sce Fig.1.2. (180angle between o bonds) SD Figure 2.8
18 CHAPTER 2 Bonding and Molecular Structure Figure 2.5 Figure 2.6 Figure 2.7 NUMBER OF TYPE OF TYPE BOND ANGLE GEOMETRY REMAINING p’s BOND FORMED sp3 109.5º Tetrahedral* 0 σ sp2 120º Trigonal planar 1 σ sp 180º Linear 2 σ TABLE 2.2 * See Fig. 1.2. Figure 2.8
CHAPTER 2 Bonding and Molecular Structure 19 equnaeaatanwiHpbiwiyishsbndgels@l0gPeofAosdoesoueofomtem Problem 2.7 The H,O molecule has a bond angle of 105.(a)What (a) ↑U↓↑↑ o has two degenerate orbitals the p and p with which to form two equivalent bonds to h However if o used these AO's.the bond angle would bewhich is the angle between theyand zaxes.Since the angle is actually 105.which is close to 109.5is presumed to use spHO's. LLL↑1p09 g0= 1s 2sp (b)Unshared pairs of electrons exert a greater repulsive force than do shared pairs,which causes a contraction of bond angles.The more unshared pairs there are.the greater is the contraction. Problem 2.8 Each H-N-H bond angle in:NH,is 107.What type of AO's does N use? ,N=T1↑ 「1s2s2p.2p,2p -(ground state) angle SPHO's: Apparently.for atoms in the second period forming more than one covalent bond (Be.B.C.N.and O).a hybrid HobeoonddenNompenodo (a)The HO's used by the central atom.in this case B.determine the shape of the molecule: 58=L7 522n.242p（oae) .Henc,use s 2sp (sphybrid stae) s8、 2sp3 -(sp'hybrid state) used for bonding The empty sphybrid orbital overlaps with a filled orbital of F-which holds two electrons: :f:+BF3→BFa(coordinate covalent bonding） The shape is tetrahedral;the bond angles are 109.5
Problem 2.7 The H2 O molecule has a bond angle of 105º. (a) What type of AO’s does O use to form the two equivalent σ bonds with H? (b) Why is this bond angle less than 109.5º? (a) O has two degenerate orbitals, the py and pz , with which to form two equivalent bonds to H. However, if O used these AO’s, the bond angle would be 90º, which is the angle between the y- and z-axes. Since the angle is actually 105º, which is close to 109.5º, O is presumed to use sp3 HO’s. 8 1 22 2 2 O = ground state ↑↓ ↓ ↑↓ ↑ ↑ ssp p p xyz ( ) CHAPTER 2 Bonding and Molecular Structure 19 8 3 3 1 2 O H = O's ↑↓ ↑↓ ↑↓ ↑ ↑ s sp ( ) sp (b) Unshared pairs of electrons exert a greater repulsive force than do shared pairs, which causes a contraction of bond angles. The more unshared pairs there are, the greater is the contraction. Problem 2.8 Each H—N—H bond angle in :NH3 is 107º. What type of AO’s does N use? 7 122 2 2 N = ground state ↑↓↑↓ ↑ ↑ ↑ ssp p p xyz ( ) If the ground-state N atom were to use its three equal-energy p AO’s to form three equivalent N—H bonds, each H—N—H bond angle would be 90º. Since the actual bond angle is 107º rather than 90º, N, like O, uses sp3 HO’s: 1s 7N = (sp3 HO’s) 2sp3 Apparently, for atoms in the second period forming more than one covalent bond (Be, B, C, N, and O), a hybrid orbital must be provided for each σ bond and each unshared pair of electrons. Atoms in higher periods also often use HO’s. Problem 2.9 Predict the shape of (a) the boron trifluoride molecule (BF3) and (b) the boron tetrafluoride anion (BF 4). All bonds are equivalent. (a) The HO’s used by the central atom, in this case B, determine the shape of the molecule: 5 122 2 2 B = ground state ↑↓↑↓ ↑ ssp p p xyz ( ) There are three sigma bonds in BF3 and no unshared pairs; therefore, three HO’s are needed. Hence, B uses sp2 HO’s, and the shape is trigonal planar. Each F—B—F bond angle is 120º. 5 2 2 1 2 2 B = hybrid state ↑↓ ↑ ↑ ↑ s sp p sp z ( ) 5 3 3 1 2 B = hybrid state ↑↓ ↑ ↑ ↑ s sp ( ) sp used for bonding The empty sp3 hybrid orbital overlaps with a filled orbital of F, which holds two electrons: The empty pz orbital is at right angles to the plane of the molecule. (b) B in BF 4 has four σ bonds and needs four HO’s. B is now in an sp3 hybrid state: The shape is tetrahedral; the bond angles are 109.5º.
20 CHAPTER 2 Bonding and Molecular Structure Problem 2.10 Arrange the s,p,and the three sp-type HO's in order of decreasing energy. The morescharacter in the orbital,the lower the energy.Therefore.the order of decreasing energy is p>sp>sp2>sp>s Problem 2.11 What effect does hybridization have on the stability of bonds? Hybrid orbitals can (a)overlap better and(b)provide greater bond angles,thereby minimizing the repulsion between pairs of electrons and making for great stability. HON =(Number of o bonds)+(Number of unshared pairs of electrons) The hybridi rom Table2.3.If more than four HO's e HO's (Fig 29a uded to octahedral HO's [Fig. 909 (a)sp'd HO's (b)sp'd HO's Figure 2.9 TABLE 2.3 HON PREDICTED HYBRID STATE p 45 The p method can also be used for the multicovalent elements in the second period and,with few exceptions.in the higher periods of the periodic table. Problem 2.12 Use the HON method to determine the hybridized state of the underlined elements: (a)CHCl3 (b)H2C=CH2 (c)O=C=O (d)HC=N:(e)H" NUMBER OF BONDS +NUMBER OF UNSHARED ELECTRON PAIRS HON HYBRID STATE 00 (d)c 22 22 (d)N 1 (e) 4 sp
Problem 2.10 Arrange the s, p, and the three sp-type HO’s in order of decreasing energy. The more s character in the orbital, the lower the energy. Therefore, the order of decreasing energy is p > sp3 > sp2 > sp > s Problem 2.11 What effect does hybridization have on the stability of bonds? Hybrid orbitals can (a) overlap better and (b) provide greater bond angles, thereby minimizing the repulsion between pairs of electrons and making for great stability. By use of the generalization that each unshared and σ-bonded pair of electrons needs a hybrid orbital, but π bonds do not, the number of hybrid orbitals (HON) needed by C or any other central atom can be obtained as HON (Number of σ bonds) + (Number of unshared pairs of electrons) The hybridized state of the atom can then be predicted from Table 2.3. If more than four HO’s are needed, d orbitals are hybridized with the s and the three p’s. If five HO’s are needed, as in PCl5 , one d orbital is included to give trigonal-bipyramidal sp3 d HO’s [Fig. 2.9(a)]. For six HO’s, as in SF6 , two d orbitals are included to give octahedral sp3d2 HO’s [Fig. 2.9(b)]. 20 CHAPTER 2 Bonding and Molecular Structure Figure 2.9 HON PREDICTED HYBRID STATE 2 sp 3 sp2 4 sp3 5 sp3d 6 sp3d2 TABLE 2.3 The preceding method can also be used for the multicovalent elements in the second period and, with few exceptions, in the higher periods of the periodic table. Problem 2.12 Use the HON method to determine the hybridized state of the underlined elements: NUMBER OF σ BONDS NUMBER OF UNSHARED ELECTRON PAIRS HON HYBRID STATE 4 04 sp3 3 03 sp2 2 02 sp 2 02 sp 1 12 sp 3 14 sp3 (a) (b) (c) (d) C (d) N (e)
CHAPTER 2 Bonding and Molecular Structure 21 2.4 Electronegativity and Polarity for ed hy atoms of dissimilar ele ativities is called nolar a no olarcovalent bond exists betweer atoms having a very small or zero difference in electronegativity.A few relative electronegativities are F4.0)>03.5)>C1.N3.0)>Br2.8)>S.C.12.5)>H(2.1) electro ie elemenofnt bond is relaiivelyn hreeles partial charges should not be confused with ionic charges.Polar bonds are indicated bythe head points vector sum o ond moments gives the net dipole moment of the molecule Problem 2.13 What do the molecular dipole momentsu=0 for CO,and u=1.84 D for H,O tell you about the shapes of these molecules? InCO,: 6=&-8 ad each individul bond moments cancel: arges about the 00 H.O also has polar bonds.However.since there is a net dipole moment.the individual bond moments do not cancel,and the molecule must have a bent shape: H resultant moment 2.5 Oxidation Number The oxidation numbe when t ing electrons are assigned to the more equal harge on the specie Problem 2.14 Determine the oxidation number of each C.(ON).in:(a)CH.(b)CH,OH,(c)CH,NH (d)H,C=CH Use the data(ON)=-3;(ON)=1;(ON)=- All examples are molecules;therefore.the sum of all (ON)values is (a(ON0+4OND.=0: (ON0.+(4X1)=0:(ON)=-4 (b)(ONc+(ONo+4ONH=0:(ONc+(-2)+4=0:(ONc=-2 (C)(ONc+(ON、+5ONa=0:(ONc+(-3)+5=0:(ONc=-2 (d)Since both Catoms areequivalent. 2(ONe+4ON)4=0: 2(ONe+4=0 (ON)e=-2
2.4 Electronegativity and Polarity The relative tendency of a bonded atom in a molecule to attract electrons is expressed by the term electronegativity. The higher the electronegativity, the more effectively does the atom attract and hold electrons. A bond formed by atoms of dissimilar electronegativities is called polar. A nonpolar covalent bond exists between atoms having a very small or zero difference in electronegativity. A few relative electronegativities are F(4.0) > O(3.5) > Cl, N(3.0) > Br(2.8) > S, C, I(2.5) > H(2.1) The more electronegative element of a covalent bond is relatively negative in charge, while the less electronegative element is relatively positive. The symbols δ and δ represent partial charges (bond polarity). These partial charges should not be confused with ionic charges. Polar bonds are indicated by ; the head points toward the more electronegative atom. The vector sum of all individual bond moments gives the net dipole moment of the molecule. Problem 2.13 What do the molecular dipole moments μ 0 for CO2 and μ 1.84 D for H2 O tell you about the shapes of these molecules? In CO2: CHAPTER 2 Bonding and Molecular Structure 21 O is more electronegative than C, and each C—O bond is polar as shown. A zero dipole moment indicates a symmetrical distribution of δ charges about the δ carbon. The geometry must be linear; in this way, individual bond moments cancel: H2O also has polar bonds. However, since there is a net dipole moment, the individual bond moments do not cancel, and the molecule must have a bent shape: 2.5 Oxidation Number The oxidation number (ON) is a value assigned to an atom based on relative electronegativities. It equals the group number minus the number of assigned electrons, when the bonding electrons are assigned to the more electronegative atom. The sum of all (ON)’s equals the charge on the species. Problem 2.14 Determine the oxidation number of each C, (ON)C, in: (a) CH4, (b) CH3OH, (c) CH3NH2, (d) H2C=CH2. Use the data (ON)N 3; (ON)H 1; (ON)O 2. All examples are molecules; therefore, the sum of all (ON) values is 0. (a) (ON)C 4(ON)H 0; (ON)C (4 1) 0; (ON)C 4 (b) (ON)C (ON)O 4(ON)H 0; (ON)C (2) 4 0; (ON)C 2 (c) (ON)C (ON)N 5(ON)H 0; (ON)C (3) 5 0; (ON)C 2 (d) Since both C atoms are equivalent, 2(ON)C 4(ON)H 0; 2(ON)C 4 0; (ON)C 2
22 CHAPTER 2 Bonding and Molecular Structure 2.6 Intermolecular Forces (a)Diplole-dipole interaction results from the attraction of theend of one polar molecule for theend of another polar molecule (b)Hydrogen-bond.X-H and:Y may be bridged X-H-:Y if X and Y are small,highly electronegative aloauchasROndNvHhndkloocurintnolaka (c)L on (van der wa in molecuerrry dipole momenmanc s)lorces. lectons or a non ar mol stantly changing.these induced dip The greater the molecular weight of the molecule,the greater the number of electrons and the greater these forces. The order of attraction is H-bond>>dipole-dipole>London forces unt for the following pro (a)Only CH,Cl is polar.and it has the highest boiling point.CH,has a lower molecular weight(16g/mole)than eretore has the lowe dipole a nof CH.CHF.CH,CH.CH,ha nly Lo on forces, e we nof all. Problem 2.16 The boiling oints of n-pentane and its isomer ne tane are36.2℃and9.5℃，respectively. Account for this difference.(See Problem 1.4 for the structural formulas.) Thes cule,influen r品o山uToc可a.e ces the bo .Rods can Thus.point is higher 2.7 Solvents sitely d by electros ic forces.thereby accounting for solve in a solvent.Nonpo molecules that have an H that can form an H-bond.Aprotic solvents are highly polar molecules that do not have an H that can form an H-bond. Problem 2.17 Classify the follov Nonpolar:(b)Because of the symmetrical tetrahedral molecular shape,the individual C-Cl bond moments cancel.(c)With few exceptions,hydrocarbons are nonpolar.Protic:(e)and (f).Aprotic:(a)and (d).The S=0 and C-Ogroups are strongly polar,and the H's attached to C do not typically form H-bonds. Problem 2.18 Mineral oil a mixture of high- molecular-weight hydrocarbons,dissolves in n-hexane but not in water or ethyl alcohol,CH,CH,OH.Explain
2.6 Intermolecular Forces (a) Diplole-dipole interaction results from the attraction of the δ end of one polar molecule for the δ end of another polar molecule. (b) Hydrogen-bond. X—H and :Y may be bridged X—H---:Y if X and Y are small, highly electronegative atoms such as F, O, and N. H-bonds also occur intramolecularly. (c) London (van der Waals) forces. Electrons of a nonpolar molecule may momentarily cause an imbalance of charge distribution in neighboring molecules, thereby inducing a temporary dipole moment. Although constantly changing, these induced dipoles result in a weak net attractive force. The greater the molecular weight of the molecule, the greater the number of electrons and the greater these forces. The order of attraction is H-bond >> dipole-dipole > London forces Problem 2.15 Account for the following progressions in boiling point. (a) CH4, 161.5ºC; Cl2, 34ºC; CH3Cl, 24ºC. (b) CH3CH2OH, 78ºC; CH3 CH2F, 46ºC; CH3CH2 CH3, 42ºC. The greater the intermolecular force, the higher the boiling point. Polarity and molecular weight must be considered. (a) Only CH3 Cl is polar, and it has the highest boiling point. CH4 has a lower molecular weight (16 g/mole) than has Cl2 (71 g/mole) and therefore has the lowest boiling point. (b) Only CH3CH2OH has H-bonding, which is a stronger force of intermolecular attraction than the dipoledipole attraction of CH3 CH2F. CH3 CH2 CH3 has only London forces, the weakest attraction of all. Problem 2.16 The boiling points of n-pentane and its isomer neopentane are 36.2ºC and 9.5ºC, respectively. Account for this difference. (See Problem 1.4 for the structural formulas.) These isomers are both nonpolar. Therefore, another factor, the shape of the molecule, influences the boiling point. The shape of n-pentane is rodlike, whereas that of neopentane is spherelike. Rods can touch along their entire length; spheres touch only at a point. The more contact between molecules, the greater the London forces. Thus, the boiling point of n-pentane is higher. 2.7 Solvents The oppositely charged ions of salts are strongly attracted by electrostatic forces, thereby accounting for the high melting and boiling points of salts. These forces of attraction must be overcome in order for salts to dissolve in a solvent. Nonpolar solvents have a zero or very small dipole moment. Protic solvents are highly polar molecules that have an H that can form an H-bond. Aprotic solvents are highly polar molecules that do not have an H that can form an H-bond. Problem 2.17 Classify the following solvents: (a) (CH3)2SO, dimethyl sulfoxide; (b) CCl4, carbon tetrachloride; (c) C6H6, benzene; (d) Dimethylformamide; (e) CH3OH, methanol; (f) liquid NH3. Nonpolar: (b) Because of the symmetrical tetrahedral molecular shape, the individual C—Cl bond moments cancel. (c) With few exceptions, hydrocarbons are nonpolar. Protic: (e) and (f). Aprotic: (a) and (d). The SO and CO groups are strongly polar, and the H’s attached to C do not typically form H-bonds. Problem 2.18 Mineral oil, a mixture of high-molecular-weight hydrocarbons, dissolves in n-hexane but not in water or ethyl alcohol, CH3 CH2 OH. Explain. 22 CHAPTER 2 Bonding and Molecular Structure