CHAPTER 2 Bonding and Molecular Structure 2.1 Atomic Orbitals e pr energ aeobae except for the s,each sublevel having some number of e al-en renerate)orbitals differing in their spatial orientation:and.(d)the electron spin.designatedor Table2.1 shows the distribution and desig- nation of orbitals. TABLE2.1 Principal energy level,n 3 Maximum no.of electrons,2m 2 8 18 32 Sublevels [n in number] 2s.2p 3x,3p,3d 4s.4p.4d.45 Maximum electrons per sublevel 26 2,6,10 2.6.10.14 Designations of filled orbitals 1s2 2x32p 3r23p3d0 4s2.4p4d,4f4 Orbitals per sublevel 1,3 1,3,5 1,35.7 13
CHAPTER 2 CHAPTER 2 13 Bonding and Molecular Structure 2.1 Atomic Orbitals An atomic orbital (AO) is a region of space about the nucleus in which there is a high probability of finding an electron. An electron has a given energy, as designated by (a) the principal energy level (quantum number) n, related to the size of the orbital; (b) the sublevel s, p, d, f, or g, related to the shape of the orbital; (c) except for the s, each sublevel having some number of equal-energy (degenerate) orbitals differing in their spatial orientation; and, (d) the electron spin, designated ↑ or ↓. Table 2.1 shows the distribution and designation of orbitals. TABLE 2.1 Principal energy level, n 12 3 4 Maximum no. of electrons, 2n2 2 8 18 32 Sublevels [n in number] 1s 2s, 2p 3s, 3p, 3d 4s, 4p, 4d, 4f Maximum electrons per sublevel 2 2, 6 2, 6, 10 2, 6, 10, 14 Designations of filled orbitals 1s2 2s2, 2p6 3s2, 3p6 , 3d10 4s2, 4p6, 4d10, 4f 14 Orbitals per sublevel 1 1, 3 1, 3, 5 1, 3, 5, 7
14 CHAPTER 2 Bonding and Molecular Structure The s orbital is a sphere around the nucleus,as shown in cross section in Fig.2.1(a).A p orbital is two donopposite sides of the nucThe teporbitals are ab they are oriente along t ex-,y-,and respectvely a n Reg al fan or node and is usually assigned a+ y-强Xi5 (a)s Orbital c8 (b)pOrbitals Figure 2.1 Three principles are used to distribute electrons in orbitals: L这0咖f:k么3尔“ 2.Pauli excusio principle.No more than two electrons can occupy an orbital and then only if they have 3.HRa℃ One elec olaced in each orbital so that the electrons ha etic-they are attracted to a magnetic field.) Problem 2.1 Show the distribution of electrons in the atomic orbitals of ()carbon and(b)oxygen. a dash re (a)Atomic number of Cis 6. 个！个个 1s 2s 2Px 2p,2P: The two 2p electrons are unpaired in each of two p orbitals(Hund's rule). (b)Atomic number ofis8. L↑个 1s 2s 2px 2p 2pa 2.2 Covalent Bond Formation-Molecular Orbital(MO)Method A covalent bond forms by overlap(fusion)of two AO's-one from each atom.This overlap produces a
The s orbital is a sphere around the nucleus, as shown in cross section in Fig. 2.1(a). A p orbital is two spherical lobes touching on opposite sides of the nucleus. The three p orbitals are labeled px , py , and pz because they are oriented along the x-, y-, and z-axes, respectively [Fig. 2.1(b)]. In a p orbital, there is no chance of finding an electron at the nucleus—the nucleus is called a node point. Regions of an orbital separated by a node are assigned and signs. These signs are not associated with electrical or ionic charges. The s orbital has no node and is usually assigned a . 14 CHAPTER 2 Bonding and Molecular Structure ↑↓↑↓ ↑ ↑ 122 2 2 ssp p p xyz ↑↓↑↓ ↑↓ ↑ ↑ 122 2 2 ssp p p xyz Figure 2.1 Three principles are used to distribute electrons in orbitals: 1. “Aufbau” or building-up principle. Orbitals are filled in order of increasing energy: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, and so on. 2. Pauli exclusion principle. No more than two electrons can occupy an orbital and then only if they have opposite spins. 3. Hund’s rule. One electron is placed in each equal-energy orbital so that the electrons have parallel spins, before pairing occurs. (Substances with unpaired electrons are paramagnetic—they are attracted to a magnetic field.) Problem 2.1 Show the distribution of electrons in the atomic orbitals of (a) carbon and (b) oxygen. A dash represents an orbital; a horizontal space between dashes indicates an energy difference. Energy increases from left to right. (a) Atomic number of C is 6. The two 2p electrons are unpaired in each of two p orbitals (Hund’s rule). (b) Atomic number of O is 8. 2.2 Covalent Bond Formation—Molecular Orbital (MO) Method A covalent bond forms by overlap (fusion) of two AO’s—one from each atom. This overlap produces a new orbital, called a molecular orbital (MO), which embraces both atoms. The interaction of two AO’s can produce two kinds of MO’s. If orbitals with like signs overlap, a bonding MO results which has a high
CHAPTER 2 Bonding and Molecular Structure 15 ensity hetween the atoms and therefore has a lower density)between the atoms and therefore has a higher energy than the individual AO's.An asterisk indicates antibonding. Head head over 一+ σ) wO-9 ③ CX⊙dX○一∞ aipp) (a)o Bonding d⊙一 ⊙⊙ ④ dCX⊙一 g⊙ CX⊙m⊙X⊙一0OX⊙ o(pp) ()Antibonding Figure 2.2 Two parallel p orbitals overlap side by side to form a pi bond IFig.2.3(a)1 or a bond iFig.2.3(b)1 The bond axis lies in a nodal plane (plane of zero electronic density)perpendicular to the cross-sectional plane of theπbond. Single bor oneand onebond.A triple bond isoneand twobond aaaoehM0 se the lecule.it is best to visualize most of them as being localize between pars of bonding atom.This description of bonding is called linear combination of (LCAO). 一 P y (a)πBonding and (b)n'Antibonding F月gure2.3
electron density between the atoms and therefore has a lower energy (greater stability) than the individual AO’s. If AO’s of unlike signs overlap, an antiboding MO* results which has a node (site of zero electron density) between the atoms and therefore has a higher energy than the individual AO’s. An asterisk indicates antibonding. Head-to-head overlap of AO’s gives a sigma (σ) MO—the bonds are called σ bonds, [Fig. 2.2(a)]. The corresponding antibonding MO* is designated σ* [Fig. 2.2(b)]. The imaginary line joining the nuclei of the bonding atoms is the bond axis, whose length is the bond length. CHAPTER 2 Bonding and Molecular Structure 15 Figure 2.2 Two parallel p orbitals overlap side by side to form a pi (π) bond [Fig. 2.3(a)] or a π* bond [Fig. 2.3(b)]. The bond axis lies in a nodal plane (plane of zero electronic density) perpendicular to the cross-sectional plane of the π bond. Single bonds are σ bonds. A double bond is one σ and one π bond. A triple bond is one σ and two π bonds (a πz and a πy , if the triple bond is taken along the x-axis). Although MO’s encompass the entire molecule, it is best to visualize most of them as being localized between pairs of bonding atoms. This description of bonding is called linear combination of atomic orbitals (LCAO). Figure 2.3
16 CHAPTER 2 Bonding and Molecular Structure Problem 2.2 What type of MO results from side-to-side overlap of ans and ap orbital? The overlap is depicted in Fig.2.4.The bonding strength generated from the overlap between the +s AO Figure 2.4 Problem 2.3 List the differences betweenabond and a bond. Bond 1.Formed by head-to-head overlap of AO's. 1.Formed by lateral overlap of p orbitals (or p and dorbitals). 2.Has eylindrical charge symmetry about bond axis.2.Has maximum charge density in the 3.Dross. onal plane of the orbitals 4.Has high er en 5.Only one bondist between two atoms. 5.One r bs ben atom Problem 2.4 Show the electron distribution in MO's of (a)H()H(c)H(d)He Predict which are unstable Fill the lower-energy MO first with no more than two electrons (a)H,has atotal of two electrons,therefore: ↑↓ GG 个 Stable (excess of one bonding electron).Has less bonding strength than H (c)Hformed theoretically from H:and H.,has three electrons: 00* Stable (has net bond strength of one bonding electron).The antibonding electron cancels the bonding strength of one of the bonding electrons. (d)He,has four electrons,two from each He atom.The electron distribution is 百0*
Problem 2.2 What type of MO results from side-to-side overlap of an s and a p orbital? The overlap is depicted in Fig. 2.4. The bonding strength generated from the overlap between the s AO and the portion of the p orbital is canceled by the antibonding effect generated from overlap between the s and the portion of the p. The MO is nonbonding (n); it is no better than two isolated AO’s. 16 CHAPTER 2 Bonding and Molecular Structure Figure 2.4 Problem 2.3 List the differences between a σ bond and a π bond. σ Bond π Bond 1. Formed by head-to-head overlap of AO’s. 1. Formed by lateral overlap of p orbitals (or p and d orbitals). 2. Has cylindrical charge symmetry about bond axis. 2. Has maximum charge density in the cross-sectional plane of the orbitals. 3. Has free rotation. 3. Does not have free rotation. 4. Has lower energy. 4. Has higher energy. 5. Only one bond can exist between two atoms. 5. One or two bonds can exist between two atoms. Problem 2.4 Show the electron distribution in MO’s of (a) H2 , (b) H+ 2 , (c) H 2 , (d) He2 . Predict which are unstable. Fill the lower-energy MO first with no more than two electrons. (a) H2 has a total of two electrons, therefore: ↑↓ σ σ * Stable (excess of two bonding electrons). (b) H 2, formed from H and H·, has one electron: ↑ σ σ * Stable (excess of one bonding electron). Has less bonding strength than H2. (c) H 2, formed theoretically from H: and H·, has three electrons: ↑↓ ↑ σ σ * Stable (has net bond strength of one bonding electron). The antibonding electron cancels the bonding strength of one of the bonding electrons. (d) He2 has four electrons, two from each He atom. The electron distribution is ↑↓ ↑↓ σ σ * Not stable (antibonding and bonding electrons cancel, and there is no net bonding). Two He atoms are more stable than a He2 molecule.
CHAPTER 2 Bonding and Molecular Structure 17 e the o MO fom ed from2s AO's has a higher energy than from IsAO's. predict whet Liz.(b)Bez can exist. The MO levels are as follows:with energy increasing from left to right. (a)Li,has six electrons.which fill the MO levels to give 仙！！ 0u0i,02x02 designated ((().Li,has an excess of two electrons in bonding MO's and therefore can exist:it is by no means the most stable form of lithium. (b)Be,would have eight electrons: 仙！仙付 01002,02 Bond orderNumber of valence eltro inMO)(Number of valence electrons inMO) The bond order is usually ual to the number of o and bonds between two atoms-in other words.I for a single bond,2 for a double bond,3 for a triple bond. Problem 2.6 The MO's formed when the two sets of the three 2p orbitals overlap are 2G2p2p.G2p )hw ry pis te()Was The valence sequence of MO's formed from overlap of then-2 AO's of diatomic molecules is 02.0i不n不2n02p不2nTin02n O has 12 elecrons to be placed in these MO's.giving ((pp)) 4 electrons.The bond order is 1/2 of 4.or 2:the two O's are joined by a net double bond. 2.3 Hybridization of Atomic Orbitals A carbon atom must provide four equal-energy orbitals in order to form four equivalent o bonds.as in methane CH It is assumed that the fourequivalent orbitals are formed by blending the 2s and the three 2p AO's give four new hybrid orbitals,called spHO's (Fig.2.5).The shape of an spHO is shown in Fig.2.6.The larger lobe ctron n orbital of g mate to form th plays an im ee Fig the a rtant role in an The AO's of carbon can hybridize in ways other than spas shown in Fig.2.7.Repulsion betwee pairs of elec
Problem 2.5 Since the σ MO formed from 2s AO’s has a higher energy than the σ * MO formed from 1s AO’s, predict whether (a) Li2, (b) Be2 can exist. The MO levels are as follows: σ1s σ*1sσ2sσ*2s , with energy increasing from left to right. (a) Li2 has six electrons, which fill the MO levels to give CHAPTER 2 Bonding and Molecular Structure 17 ↑↓ ↑↓ ↑↓ σ1s σ1s σ 2s σ 2s * * designated (σ1s )2(σ*1s )2(σ2s)2 . Li2 has an excess of two electrons in bonding MO’s and therefore can exist; it is by no means the most stable form of lithium. (b) Be2 would have eight electrons: ↑↓ ↑↓ ↑↓ ↑↓ σ1s σ1s σ 2s σ 2s * * There are no net bonding electrons, and Be2 does not exist. Stabilities of molecules can be qualitatively related to the bond order, defined as Bond order (Number of valence electrons in ≡ MO’s) – (Number of valence electrons in MO*’s) 2 The bond order is usually equal to the number of σ and π bonds between two atoms—in other words, 1 for a single bond, 2 for a double bond, 3 for a triple bond. Problem 2.6 The MO’s formed when the two sets of the three 2p orbitals overlap are ππσππσ 222222 pppppp yz xyz x *** (the π and π* pairs are degenerate). (a) Show how MO theory predicts the paramagnetism of O2. (b) What is the bond order in O2? The valence sequence of MO’s formed from overlap of the n 2 AO’s of diatomic molecules is σσπ π σ π π σ 222 2 2 2 2 2 ssp p p p p p yz xyz x * * * * O2 has 12 elecrons to be placed in these MO’s, giving ( )( )( )( )( )( )( 22 2 2 2 1 σσπ π σ π 222 2 2 2 ssp p p p yzxy * * π 2 pz * ) 1 (a) The electrons in the two, equal-energy, π* MO*’s are unpaired; therefore, O2 is paramagnetic. (b) Electrons in the first two molecular orbitals cancel each other’s effect. There are 6 electrons in the next 3 bonding orbitals and 2 electrons in the next 2 antibonding orbitals. There is a net bonding effect due to 4 electrons. The bond order is 1/2 of 4, or 2; the two O’s are joined by a net double bond. 2.3 Hybridization of Atomic Orbitals A carbon atom must provide four equal-energy orbitals in order to form four equivalent σ bonds, as in methane, CH4 . It is assumed that the four equivalent orbitals are formed by blending the 2s and the three 2p AO’s give four new hybrid orbitals, called sp3 HO’s (Fig. 2.5). The shape of an sp3 HO is shown in Fig. 2.6. The larger lobe, the “head,” having most of the electron density, overlaps with an orbital of its bonding mate to form the bond. The smaller lobe, the “tail,” is often omitted when HO’s are depicted (see Fig. 2.11). However, at times the tail plays an important role in an organic reaction. The AO’s of carbon can hybridize in ways other than sp3 , as shown in Fig. 2.7. Repulsion between pairs of electrons causes these HO’s to have the maximum bond angles and geometries summarized in Table 2.2. The sp2 and sp HO’s induce geometries about the C’s as shown in Fig. 2.8. Only σ bonds, not π bonds, determine molecular shapes.