Chapter 11 Answers. Practice Examples

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1 hapter Answers Practice Examples a. There are three half-filled p orbitals on, and one half-filled 5p orbital on I. Each halffilled p orbital from will overlap with one half-filled 5p orbital of an I. Thus, there will be three I bonds. The I atoms will be oriented in the same direction as the three p orbitals of : toward the x, y, and z-directions of a artesian coordinate system. Thus, the I I angles will be approximately 90 (probably larger because the I atoms will repel each other). The three I atoms will lie in the same plane at the points of a triangle, with the atom centered above them. The molecule is trigonal pyramidal. b. There are three half-filled orbitals on and one half-filled orbital on each. There will be three bonds, with bond angles of approximately90. The molecule is trigonal pyramid. (We obtain the same molecular shape if is assumed to be sp 3 hybridized, but bond angles are closer to09.5, the tetrahedral bond angle.) VSEPR theory begins with the Lewis structure and notes that there are three bond pairs and one lone pair attached to. This produces a tetrahedral electron pair geometry and a trigonal pyramidal molecular shape with bond angles a bit less than the tetrahedral angle of 09.5 because of the lone pair. Since VSEPR theory makes a prediction closer to the experimental bond angle of 07, it seems more appropriate in this case. a. Bent molecular geometry, sp 3 hybridization. b. See-saw molecular geometry, sp 3 d hybridization. 3a. Each central atom is surrounded by four electron pairs, requiring sp 3 hybridization. The carbon atoms have tetrahedral geometry, the oxygen atom is bent. 3b. The left-most and the right-most are surrounded by four electron pairs, and thus require sp 3 hybridization. The central carbon is surrounded by three electron groups and is sp hybridized. 4a. There are four bond pairs around the left-hand, requiring sp 3 hybridization. Three of the bonds that form are sigma bonds resulting from the overlap of a half-filled sp 3 hybrid orbital on with a half-filled s orbital on. The other has two attached electron groups, utilizing sp hybridization. The two atoms join with a sigma bond: overlap of sp on the lefthand with sp on the right-hand. The three bonds between and consist of a sigma bond (sp on with sp on ), and two pi bonds (p on with p on ). a b a b 4b. structure() structure(). In both structures, the central is attached to two other atoms, and possesses no lone pairs. The geometry of the molecule thus is linear and the hybridization on this central is sp. In structure () the bond results from the overlap of three pairs of half-filled orbitals: () sp x on b with p x on a forming a bond, () p y on b with p y on a forming a bond, and (3) p z on b opyright 0 Pearson anada Inc.

2 with p z on a also forming a bond. The bond is a coordinate covalent bond, and forms by the overlap of the full sp x orbital on b with the empty p orbital on. In structure () the == bond results from the overlap of two pairs of half-filled orbitals: () sp y on b with p y on forming a bond and () p z on b with p z on forming a bond. The == bond is a coordinate covalent bond, and is formed by two overlaps: () the overlap of the full sp y orbital on b with the empty p y orbital on to form a bond, and () the overlap of the half-filled p x orbital on b with the half-filled p x orbital on a to form a bond. Based on formal charge arguments, structure () is preferred, because the negative formal charge is on the more electronegative atom,. 5a. 53 kj/mol + Li 5b. The bond order in is ½. We would expect the ion to be stable, with a bond strength about half that of a hydrogen molecule. a 6a. + s s * s s * p p p * p * + e s s * s s * p p p * p * s s * s s * p p p * p *, bond order =.5., bond order = 0.5., bond order = b. The bond length increases as the bond order decreases. Longer bonds are weaker bonds. 7a. + s s p p p p, bond order =.0. p p s s p p 7b. B, bond order =.0. opyright 0 Pearson anada Inc.

3 F.. Atom Label sp hybrids np orbitals + A(S) (3p) 0 B() (p) (-) () (p) (-) D() (p) S 0 o non-bonding -orbitals 8a. -framework 3 x sp - sp -bonds -molecular orbitals F.. Atom Label sp hybrids p orbitals 0 A() 0 B() (-) () ~0 o non-bonding -orbitals 8b. -framework x sp - sp -bonds -molecular orbitals opyright 0 Pearson anada Inc. 3

4 Integrative Example A. (a) s s -system sp 3 -system s sp3 sp sp sp sp sp sp sp 3 s (b) s s (c) Antibonding moleclar orbitals Bonding moleclar orbitals ne p electron from each of the atoms is placed in the ring a a B. (a) 3 b b 3 (b) b - : (s) b (sp 3 ) a =: a (sp ) (sp ), : a (p) (p) -: (sp ) (sp 3 ) a - a : a (sp ) a (sp ) a - b : b (sp 3 ) a (sp ) - : (s) (sp 3 ) opyright 0 Pearson anada Inc. 4

5 Exercises. First,valence-bond theory clearly distinguishes between sigma and pi bonds. In valence-bond theory, it is clear that a sigma bond must be stronger than a pi bond, for the orbitals overlap more effectively in a sigma bond (end-to-end) than they do in a pi bond (side-to-side). Second, molecular geometries are more directly obtained in valence-bond theory than in Lewis theory. Third, Lewis theory does not explain hindered rotation about double bonds. 3a. Lewis theory does not describe the shape of the water molecule. 3b. In valence-bond theory using simple atomic orbitals, each bond results from the overlap of a s orbital on with a p orbital on. The angle between p orbitals is 90 so this method initially predicts a 90 bond angle. 3c. In VSEPR theory the molecule is categorized as being of the AX E type. The lone pairs repel each other more than do the bond pairs, explaining the smaller than 09.5º tetrahedral bond angle. 3d. In valence-bond theory using hybrid orbitals, each bond results from the overlap of a s orbital on with an sp 3 orbital on. The angle between sp 3 orbitals is 09.5º. 5. The central atom is sp hybridized in S, 3 -, and -. 7a. is the central atom. The molecule is linear and is sp hybridized. 7b. is the central atom. The molecule is trigonal planar around which is sp hybridized. 7c. l is the central atom. The electron-group geometry around l is tetrahedral, indicating that l is sp 3 hybridized. 7d. B is the central atom. The electron-group geometry is tetrahedral, indicating that B is sp 3 hybridized. 9. The hybridization is sp 3 d. Each of the three sigma bonds are formed by the overlap of a p orbital on F with one of the half-filled dsp 3 orbitals on l. The two lone pairs occupy dsp 3 orbitals. a. b. sp c. d. 3 sp d. sp 3. sp. e. 3 sp d. opyright 0 Pearson anada Inc. 5

6 3a. Planar molecule. The hybridization on is sp. 3b. Linear molecule. The hybridization for each is sp. 3c. either linear nor planar. The shape around the left-hand is tetrahedral and that has sp 3 hybridization. 3d. Linear molecule. The hybridization for is sp. 5a.. The bond is a bond, and the bond is composed of and bonds. 5b. : The bond is a bond, and each bond is composed of and bonds. l l 5c. l. All bonds are bonds except one of the bonds that comprise the == bond. The == bond is composed of one and one bond. 5d.. All single bonds in this structure are bonds. The double bond is composed of one and one bond. 7a. The geometry at is tetrahedral; is sp 3 hybridized. l l bond angles are09.5. Each l bond is represented by : c h lb3pg 3 sp 7b. The e - group geometry around is triangular planar, and is sp hybridized. The bond angle is about0. The bonds are: : ( p y ) ( sp ) : (sp ) l(3p z ) :( ) ( p ). p z z 7c. a =b. The geometry of a is tetrahedral, a is sp 3 hybridized and the a bond angle is (at least close to)09.5. The e - group geometry at is trigonal, is sp hybridized, and the bond angle is0. The four bonds are represented as 3 follows. : s a a sp b 3 sp sp sp py pz pz : : :. a b b 7d. The e - group geometry around is trigonal planar; all bond angles around are 0º, and the hybridization of is sp. The four bonds in the molecules are: p sp p sp p p :l 3 : :. x y z z opyright 0 Pearson anada Inc. 6

7 ((sp 3 ) - (s)) ((sp 3 ) - (s)) 3 ((sp ) - (sp 3 )) 4 ((sp ) - (p)) + ((p) - (p)) 5 ((sp 3 ) - (sp )) 6 ((sp 3 ) - (sp 3 )) 7 ((sp 3 ) - (sp 3 )) a. e n t r a l S i s s p h y b r i d i z e d a n d the t e r r m i n a l a n d S a t o m s a r e u n hybridized S S (S(sp )- S ( 3 p x )) (S(3p z )- S ( 3 p z )) (S(s p )- ( p z )) b. 3. a a b c b sp s sp sp sp sp : : : a a a b b c sp s sp p py p : : : c b c y a b y p p p p a z b z c z z : : σ: 3(sp ) 4(sp 3 ) σ: 3(sp ) (s) σ: (sp ) 3(sp ) π. (p) 3(p) σ: (sp ) 3(sp ) 5 3 σ: (sp) (p) σ: (sp ) (p) σ: (sp 5. ) 5(sp 3 ) π. (p) (p). There are 8 atoms that are on the same plane (,, -5, att. To 3). Furthermore, depending on the angle of rotation of the 3 groups (4 and 5), two atoms can also be added to this total. opyright 0 Pearson anada Inc. 7

8 7. The valence-bond method describes a covalent bond as the result of the overlap of atomic orbitals. Molecular orbital theory describes a bond as a region of space in a molecule where there is a good chance of finding the bonding electrons. The molecular orbital bond does not have to be created from atomic orbitals (although it often is) and the orientations of atomic orbitals do not have to be manipulated to obtain the correct geometric shape. There is little concept of the relative energies of bonding in valence-bond theory. In molecular orbital theory, bonds are ordered energetically. 9. bond order =.5 stable, bond order = stable. 3. In order to have a bond order higher than three, there would have to be a region in a molecular orbital diagram where four bonding orbitals occur together in order of increasing energy, with no intervening antibonding orbitals. o such region exists in any of the molecular orbital diagrams in Figure a. s 33b. s * 33c. s 33d. p 35a. : * * 4 * s s s s p p p 35b. + * * 4 : s s s s p p * * 4 35c. : s s s s p p * * 4 35d. : s s s s p p 35e. - * * 4 : s s s s p p 35f. + * * 4 : s s s s p * * 4 35g. B : s s s s p opyright 0 Pearson anada Inc. 8

9 σ* p σ* p p π* p p π* p p π p p σ p 37a. σ p π p 37b. Bond order is 3 for +,.5 for +. 37c. + is diamagnetic (all paired electrons), + is paramagnetic. 37d. + has the greater bond length, because there is less electron density between the two nuclei. F σ* p p π* p π p p σ p s σ* s 39. σ s s. The bond length in F + would be shorter. 4. The bonding requires that all six bonds must be equivalent. This can be achieved by creating six molecular orbitals three bonding and three antibonding into which the 6 electrons are placed. This creates a single delocalized structure for the 6 6 molecule. 43a. Delocalized molecular orbitals required. 43b. Delocalized molecular orbitals required. 43c. Delocalized molecular orbitals not required. opyright 0 Pearson anada Inc. 9

10 45a. Atomic number, which provides the location of the element in the periodic table, has some minimal predictive value in determining metallic character. 45b. The answer for this part is much the same as the answer to part (a), since atomic mass generally parallels atomic number for the elements. 45c. The number of valence electrons has no bearing on the metallic character of an element. 45d. Because metals occur in every period of the periodic table, there is no particular relationship between the number of electron shells and the metallic behavior of an element energy levels, electrons. 49a. Electrical conductor 49b. Insulator 49c. Insulator 49d. Semiconductor 49e. Electrical conductor 49f. Insulator 5a. o 5b. o 5c. o 5d. Yes 5e. o 5f. Yes 53. In ultra pure crystalline silicon, there are no extra electrons in the lattice that can conduct an electric current. If, however, the silicon becomes contaminated with arsenic atoms, then there will be one additional electron added to the silicon crystal lattice for each arsenic atom that is introduced. The arsenic atoms increase the conductivity of the solid by providing additional electrons that can carry a current after they are promoted into the conduction band by thermal excitation nm. This is IR-radiation. Integrative and Advanced Exercises 58. [e] 3s The half-filled 3s orbital on each a overlaps with another to form a covalent bond. There are electrons in a. These electrons are distributed in the molecular orbitals as opyright 0 Pearson anada Inc. 0

11 follows. s s * s s * p p p * p * 3s Again, we predict a single bond for a. A Lewis-theory picture of the bonding would have the two Lewis symbols for two a atoms uniting to form a bond: in. a a a a Thus, the bonding in a is very much like that 60a. s s * s s * p p p * p *, bond order =.5. s s * s s * p p p * p *, bond order =.0. 60b. [ ], [ ] 64. F a F a y a a z 3 3 F( p) ( sp ) ( sp ) ( sp ) sp ) ( p ) ( p ) ( p ) F 66a. For lf 3 the electron group geometry is trigonal bipyramidal and the molecule is T-shaped. For AsF 5 the electron group geometry and molecular shape are trigonal bipyramidal. For lf + the electron group geometry is tetrahedral and the ion is bent in shape. For AsF 6 the electron group geometry and the shape of this ion are octahedral. 66b. Trigonal bipyramidal electron group geometry is associated with sp 3 d hybridization. The l in lf 3 and As in AsF 5 both have sp 3 d hybridization. Tetrahedral electron group geometry is associated with sp 3 hybridization. Thus l in lf + has sp 3 hybridization. ctahedral electron group geometry is associated with sp 3 d hybridization, which is the hybridization adopted by As in AsF Suppose two e atoms in the excited state s s unite to form an e molecule. ne *0 0* possible configuration is. The bond order would be (4-0)/ =. s s s s z 70. The orbital diagrams for and are as follows. [e] sp p, [e] sp p antibonding molecular orbitals bonding molecular orbitals umber of -bonds = 3 This -bonding scheme produces three -bonds, which is identical to the number predicted by Lewis theory. opyright 0 Pearson anada Inc.

12 73. a b b c d a e 3 b -, d -, e - : (s) (sp 3 ) (all tetrahedral carbon uses sp 3 hybrid orbitals) c = b : c (sp ) b (p or sp ), : c (p) b (p) c - a : c (sp ) a (p or sp 3 ) d - a : d (sp 3 ) a (p or sp 3 ) a : a (sp) (sp) Two mutually perpendicular π -bonds: (p) (p) a - b : b (sp 3 ) a (sp) d - e : d (sp 3 ) e (sp 3 ) c - b : b (sp 3 ) c (sp ) 75a watts 75b. 8.9 amps 79. The wavelength for both will be the same, because they both have a conjugated π system. Feature Problems 80a kj 80b. 7.9 kj 80c kj 80d. atomization = 5358 kj (per mole of 6 6 ), Resonance energy = 68 kj. opyright 0 Pearson anada Inc.

13 σ* p s π* p π p p σ p 87. s s Self-Assessment Exercises 9. The answer is (c). 9. The answer is (c). 93. The answer is (a). 94. The answer is (b). 95. The answer is (c). 96. The answer is (d). 97. The answer is (a). 98. The answer is (c). 99. The valence-bond method using pure s and p orbitals incorrectly predicts a trigonal pyramidal shape with 90 F B F bond angles. 00. BrF 5 has six constituents around it; five are fluorine atoms, and the sixth is a lone pair. Therefore, the hybridization is sp 3 d, but the geometry is square pyramidal. 0. There are (a) 6 σ and (b) π bonds. 0. All three are paramagnetic. B has 3 bonding electrons, so the B B bond is the strongest. 03. The answer is (c) has the greater bond energy. opyright 0 Pearson anada Inc. 3

Chapter 9. Chemical Bonding II: Molecular Geometry and Bonding Theories

Chapter 9. Chemical Bonding II: Molecular Geometry and Bonding Theories Chapter 9 Chemical Bonding II: Molecular Geometry and Bonding Theories Topics Molecular Geometry Molecular Geometry and Polarity Valence Bond Theory Hybridization of Atomic Orbitals Hybridization in Molecules