CHAPTER 8 BONDING: GENERAL CONCEPTS Ionic solids are held together by strong electrostatic forces that are omnidirectional.

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1 CAPTER 8 BDIG: GEERAL CCEPTS 1 CAPTER 8 BDIG: GEERAL CCEPTS Questions 15. a. This diagram represents a polar covalent bond as in. In a polar covalent bond, there is an electron rich region (indicated by the red color) and an electron poor region (indicated by the blue color). In, the more electronegative atom (on the red side of the diagram) has a slightly greater ability to attract the bonding electrons than does (on the blue side of the diagram), which in turn produces a dipole moment. b. This diagram represents an ionic bond as in a. ere, the electronegativity differences between the a and are so great that the valence electron of sodium is transferred to the chlorine atom. This results in the formation of a cation, an anion, and an ionic bond. c. This diagram represents a pure covalent bond as in 2. Both atoms attract the bonding electrons equally, so there is no bond dipole formed. This is illustrated in the electrostatic potential diagram as the various red and blue colors are equally distributed about the molecule. The diagram shows no one region that is red nor one region that is blue (there is no specific partial negative end and no specific partial positive end), so the molecule is nonpolar. 18. Ionic solids are held together by strong electrostatic forces that are omnidirectional. i. or electrical conductivity, charged species must be free to move. In ionic solids, the charged ions are held rigidly in place. nce the forces are disrupted (melting or dissolution), the ions can move about (conduct). ii. Melting and boiling disrupts the attractions of the ions for each other. Because these electrostatic forces are strong, it will take a lot of energy (high temperature) to accomplish this. iii. If we try to bend a piece of material, the ions must slide across each other. or an ionic solid the following might happen: strong attraction strong repulsion Just as the layers begin to slide, there will be very strong repulsions causing the solid to snap across a fairly clean plane. iv. Polar molecules are attracted to ions and can break up the lattice.

2 2 CAPTER 8 BDIG: GEERAL CCEPTS These properties and their correlation to chemical forces will be discussed in detail in Chapters 10 and a. Rb < K < a b. Ga < B < c. Br < < d. S < < 30. a. Sn b. Tl Br c. Si d. 32. The order of E from igure 8.3 is: a. Rb (0.8) = K (0.8) < a (0.9), different b. Ga (1.6) < B (2.0) < (3.5), same c. Br (2.8) < (3.0) < (4.0), same d. S (2.5) < (3.5) < (4.0), same Most polar bonds using actual E values: a. C most polar (Sn predicted) b. Al Br most polar (Tl Br predicted). c. Si (same as predicted). d. Each bond has the same polarity, but the bond dipoles point in opposite directions. xygen is the positive end in the bond dipole, and oxygen is the negative end in the bond dipole ( predicted). 36. The possible ionic bonds that can form are between the metal Cs and the nonmetals P,, and. These ionic compounds are Cs 3P, Cs 2, and Cs. The bonding between the various nonmetals will be covalent. P 4, 2, and 2 are all pure covalent (or just covalent) with equal sharing of the bonding electrons. P will also be a covalent bond because P and have identical electronegativities. The other possible covalent bonds that can form will all be polar covalent because the nonmetals involved in the bonds all have intermediate differences in electronegativities. The possible polar covalent bonds are P and. ote: The bonding among cesium atoms is called metallic. This type of bonding between metals will be discussed in Chapter a. 2; both 2 and 3 have permanent dipole moments in part due to the polar and bonds. But because oxygen is more electronegative than nitrogen, one would expect 2 to have a slightly greater dipole moment. This diagram has the more intense red color on one end and the more intense blue color at the other end indicating a larger dipole moment. b. 3; this diagram is for a polar molecule, but the colors are not as intense as the diagram in part a. ence, this diagram is for a molecule which is not as polar as 2. Since is less electronegative than, 3 will not be as polar as 2. c. C 4; this diagram has no one specific red region and has four blue regions arranged symmetrically about the molecule. This diagram is for a molecule which has no dipole moment. This is only true for C 4. The C bonds are at best, slightly polar because carbon and hydrogen have similar electronegativity values. In addition, the slightly polar

3 CAPTER 8 BDIG: GEERAL CCEPTS 3 C bond dipoles are arranged about carbon so that they cancel each other out, making C 4 a nonpolar molecule. See Example a. Al 3+ and ; Al 3, aluminum chloride b. a + and 2 ; a 2, sodium oxide c. Sr 2+ and ; Sr 2, strontium fluoride d. Ca 2+ and S 2 ; CaS, calcium sulfide 46. a. Sr 2+ : [Ar]4s 2 3d 10 4p 6 ; Cs + : [Kr]5s 2 4d 10 5p 6 ; In + : [Kr]5s 2 4d 10 ; Pb 2+ : [Xe]6s 2 4f 14 5d 10 b. P 3 and S 2 : [e]3s 2 3p 6 ; Br : [Ar]4s 2 3d 10 4p a. e has 10 electrons. Al, Mg 2, and a 2 are some possible ionic compounds where each ion has 10 electrons. b. CaS, K 3P, and K are some examples where each ion is isoelectronic with Ar; i.e., each ion has 18 electrons. c. Each ion in Sr 3As 2, SrBr 2, and Rb 2Se is isoelectronic with Kr. d. Each ion in BaTe and CsI is isoelectronic with Xe. 52. All these ions have 18 e ; the smallest ion (Sc 3+ ) has the most protons attracting the 18 e, and the largest ion has the fewest protons (S 2 ). The order in terms of increasing size is Sc 3+ < Ca 2+ < K + < < S 2. In terms of the atom size indicated in the question: 54. a. V > V 2+ > V 3+ > V 5+ b. Cs + > Rb + > K + > a + c. Te 2 > I > Cs + > Ba 2+ d. P 3 > P 2 > P > P e. Te 2 > Se 2 > S 2 > Mg(s) Mg(g) Δ = 150. kj (sublimation) Mg(g) Mg + (g) + e Δ = 735 kj (IE 1) Mg + (g) Mg 2+ (g) + e Δ = 1445 kj (IE 2) 2(g) 2 (g) Δ = 154 kj (BE) 2 (g) + 2 e 2 (g) Δ = 2( 328) kj (EA) Mg 2+ (g) + 2 (g) Mg 2(s) Δ = 2913 kj (LE) Mg(s) + 2(g) Mg 2(s) o Δ f = 1085 kj/mol 66. Sometimes some of the bonds remain the same between reactants and products. To save time, only break and form bonds that are involved in the reaction. C + 2 C

4 4 CAPTER 8 BDIG: GEERAL CCEPTS a. Bonds broken: Bonds formed: 1 C (891 kj/mol) 1 C (305 kj/mol) 2 (432 kj/mol) 2 C (413 kj/mol) 2 (391 kj/mol) Δ = 891 kj + 2(432 kj) [305 kj + 2(413 kj) + 2(391 kj)] = 158 kj b Bonds broken: Bonds formed: 1 (160. kj/mol) 4 (565 kj/mol) 4 (391 kj/mol) 1 (941 kj/mol) 2 (154 kj/mol) Δ = 160. kj + 4(391 kj) + 2(154 kj) [4(565 kj) kj] = 1169 kj 68. C + C C C Bonds broken: Bonds formed: 1 C (1072 kj/mol) 1 C C (347 kj/mol) 1 C (358 kj/mol) 1 C= (745 kj/mol) 1 C (358 kj/mol) Δ = [ ] = 20. kj 84. a. P 3 has (7) = 32 valence electrons.

5 CAPTER 8 BDIG: GEERAL CCEPTS 5 P Skeletal structure P Lewis structure ote: This structure uses all 32 e while satisfying the octet rule for all atoms. This is a valid Lewis structure. S 4 2 has 6 + 4(6) + 2 = 32 valence electrons. S 2- ote: A negatively charged ion will have additional electrons to those that come from the valence shell of the atoms. The magnitude of the negative charge indicates the number of extra electrons to add in. Xe 4, 8 + 4(6) = 32 e P 4 3, 5 + 4(6) + 3 = 32 e Xe 4 has 7 + 4(6) + 1 = 32 valence electrons - ote: All of these species have the same number of atoms and the same number of valence electrons. They also have the same Lewis structure. b. 3 has 5 + 3(7) = 26 valence electrons. S 3 2, 6 + 3(6) + 2 = 26 e Skeletal structure Lewis structure

6 6 CAPTER 8 BDIG: GEERAL CCEPTS S 2- P 3 3, 5 + 3(6) + 3 = 26 e 3, 7 + 3(6) + 1 = 26 e ote: Species with the same number of atoms and valence electrons have similar Lewis structures. c. 2 has 7 + 2(6) + 1 = 20 valence - Skeletal structure Lewis structure S 2, 6 + 2(7) = 20 e P 2, 5 + 2(7) + 1 = 20 e ote: Species with the same number of atoms and valence electrons have similar Lewis structures. d. Molecules ions that have the same number of valence electrons and the same number of atoms will have similar Lewis structures. 86. a. 2, 5 + 2(6) = 17 e 2 4, 2(5) + 4(6) = 34 e Plus others Plus other resonance structures b. B 3, 3 + 3(1) = 6 e 3, 5 + 3(1) = 8 e B

7 CAPTER 8 BDIG: GEERAL CCEPTS 7 B 3 3, = 14 e B In reaction a, 2 has an odd number of electrons, so it is impossible to satisfy the octet rule. By dimerizing to form 2 4, the odd electron on two 2 molecules can pair up, giving a species whose Lewis structure can satisfy the octet rule. In general, odd-electron species are very reactive. In reaction b, B 3 is electron-deficient. Boron has only six electrons around it. By forming B 3 3, the boron atom satisfies the octet rule by accepting a lone pair of electrons from 3 to form a fourth bond See Exercise 84 for the Lewis structures of P 3, S 4 2, 4 and P 4 3. All these compounds/ions have similar Lewis structures to those of S 2 2 and Xe 4 shown below. ormal charge = [number of valence electrons on free atom] [number of lone pair electrons on atom + 1/2(number of shared electrons of atom)]. a. P 3: P, C = 5 1/2(8) = +1 b. S 4 2 : S, C = 6 1/2(8) = +2 c. 4 :, C = 7 1/2(8) = +3 d. P 4 3 : P, C = 5 1/2(8) = +1 e. S 2 2, 6 + 2(6) + 2(7) = 32 e f. Xe 4, 8 + 4(6) = 32 e - S Xe S, C = 6 1/2(8) = +2 Xe, C = 8 1/2(8) = +4 g. 3, 7 + 3(6) + 1 = 26 e h. 4 3, 5 + 4(6) + 3 = 32 e 3-, C = 7 2 1/2(6) = +2, C = 5 1/2(8) = a. P 3 has 5 + 3(7) = 26 valence electrons. b. S 2 has 6 + 2(7) = 20 valence electrons. P S

8 8 CAPTER 8 BDIG: GEERAL CCEPTS Trigonal pyramid; all angles are < V-shaped; angle is < c. Si 4 has 4 + 4(7) = 32 valence electrons. Si Tetrahedral; all angles are ote: In P 3, S 2, and Si 4, there are four pairs of electrons about the central atom in each case in this exercise. All of the structures are based on a tetrahedral geometry, but only Si 4 has a tetrahedral structure. We consider only the relative positions of the atoms when describing the molecular structure All have polar bonds; in Si 4, the individual bond dipoles cancel when summed together, and in P 3 and S 2, the individual bond dipoles do not cancel. Therefore, Si 4 has no net dipole moment (is nonpolar), and P 3 and S 2 have net dipole moments (are polar). or P 3, the negative end of the dipole moment is between the more electronegative chlorine atoms, and the positive end is around P. or S 2, the negative end is between the more electronegative atoms, and the positive end of the dipole moment is around S Li + (g) + 2 (g) 2 Li(s) = 2( 829 kj) 2 Li(g) 2 Li + (g) + 2 e = 2(520. kj) 2 Li(s) 2 Li(g) = 2(166 kj) 2 (g) 2 (g) + 2 (g) = 2(427 kj) 2 (g) + 2 e 2 (g) = 2( 349 kj) 2 (g) 2(g) = (432 kj) 2 Li(s) + 2 (g) 2 Li(s) + 2(g) = 562 kj

9 CAPTER 8 BDIG: GEERAL CCEPTS 9 CAPTER 9 CVALET BDIG: RBITALS 18. C 4 has 4 + 4(7) = 32 valence electrons. C 4 has a tetrahedral arrangement of the electron pairs about the carbon atom that requires sp 3 hybridization. The four sp 3 hybrid orbitals from carbon are used to form the four bonds to chlorine. The chlorine atoms also have a tetrahedral arrangement of electron pairs, and we will assume that they are also sp 3 hybridized. The C sigma bonds are all formed from overlap of sp 3 hybrid orbitals from carbon with sp 3 hybrid orbitals from each chlorine atom. 29. a. b. C tetrahedral sp 3 trigonal pyramid sp nonpolar <109.5 polar The angles in 3 should be slightly less than because the lone pair requires more space than the bonding pairs. c. d. V-shaped sp 3 trigonal planar sp 2 <109.5 polar 120 nonpolar B

10 10 CAPTER 8 BDIG: GEERAL CCEPTS e. f. Be a b Te linear sp see-saw dsp nonpolar a) 120, b) 90 polar g. h. a b As Kr trigonal bipyramid dsp 3 linear dsp 3 a) 90, b) 120 nonpolar 180 nonpolar i. j. Kr 90 o Se square planar d 2 sp 3 octahedral d 2 sp 3 90 nonpolar 90 nonpolar k. l. I I square pyramid d 2 sp 3 T-shaped dsp 3 90 polar 90 polar 37. a. Add lone pairs to complete octets for each and.

11 CAPTER 8 BDIG: GEERAL CCEPTS 11 a c C C b e d 2 C C 2 C f C g C 3 h Azodicarbonamide methyl cyanoacrylate ote: 2, C 2 ( 2C), and C 3 are shorthand for nitrogen or carbon atoms singly bonded to hydrogen atoms. b. In azodicarbonamide, the two carbon atoms are sp 2 hybridized, the two nitrogen atoms with hydrogens attached are sp 3 hybridized, and the other two nitrogens are sp 2 hybridized. In methyl cyanoacrylate, the C 3 carbon is sp 3 hybridized, the carbon with the triple bond is sp hybridized, and the other three carbons are sp 2 hybridized. c. Azodicarbonamide contains three π bonds and methyl cyanoacrylate contains four π bonds. d. a) b) 120 c) 120 d) 120 e) 180 f) 120 g) h) a. To complete the Lewis structure, add two lone pairs to each sulfur atom. C C sp3 sp sp sp sp sp sp sp2 sp2 3 C C C C C C C C C C C 2 S S b. See the Lewis structure. The four carbon atoms in the ring are all sp 2 hybridized, and the two sulfur atoms are sp 3 hybridized. c. 23 σ and 9 π bonds. ote: C 3 ( 3C), C 2, and C are shorthand for carbon atoms singly bonded to hydrogen atoms.