16. NO 3, 5 + 3(6) + 1 = 24 e. 22. HCN, = 10 valence electrons

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Lesley Baker
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1 Solution to Chapts 9 & 10 Problems: 16. N 3, 5 + 3(6) + 1 = 24 e 22. HCN, = 10 valence electrons Assuming N is hybridized, both C and N atoms are sp hybridized. The C H bond is formed from overlap of a carbon sp 3 hybrid orbital with a hydrogen 1s atomic orbital. The triple bond is composed of one bond and two bonds. The sigma bond is formed from head-to-head overlap of the sp hybrid orbitals from the C and N atoms. The two bonds in the triple bond are formed from overlap of the two unhybridized p atomic orbitals from each C and N atom. CCl 2, (7) = 24 valence electrons Assuming all atoms are hybridized, the carbon and oxygen atoms are sp 2 hybridized, and the two chlorine atoms are sp 3 hybridized. The two C Cl bonds are formed from overlap of sp 2 hybrids from C with sp 3 hybrid orbitals from Cl. The double bond between the carbon and oxygen atoms consists of one and one bond. The bond in the double bond is formed from head-to-head overlap of an sp 2 orbital from carbon with an sp 2 hybrid orbital from oxy-gen. The bond is formed from parallel overlap of the unhybridized p atomic orbitals from each atom of C and. 29. a. b. N C B tetrahedral sp 3 trigonal pyramid sp nonpolar <109.5 polar The angles in N 3 should be slightly less than because the lone pair requires more space than the bonding pairs. c. d. 304

2 CHAPTER 9 CVALENT BNDING: RBITALS 305 V-shaped sp 3 trigonal planar sp 2 <109.5 polar 120 nonpolar e. f. H Be H 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 I Se square planar d 2 sp 3 octahedral d 2 sp 3 90 nonpolar 90 nonpolar k. l. I

3 CHAPTER 9 CVALENT BNDING: RBITALS 306 square pyramid d 2 sp 3 T-shaped dsp 3 90 polar 90 polar 30. a. S V-shaped, sp 2, a. N 2 : (σ 2s) 2 (σ 2s*) 2 (π 2p) 4 (σ 2p) 2 (π 2p*) 2 B.. = bond order = (8 4)/2 = 2, 2 : (σ 2s) 2 (σ 2s*) 2 (σ 2p) 2 (π 2p) 4 (π 2p*) 4 B.. = (8 6)/2 = 1, 2 : (σ 2s) 2 (σ 2s*) 2 (σ 2p) 2 (π 2p) 4 (π 2p*) 4 (σ 2p*) 2 B.. = (8 8)/2 = 0, not b. Be 2: (σ 2s) 2 (σ 2s*) 2 B.. = (2)/2 = 0, not B 2: (σ 2s) 2 (σ 2s*) 2 (π 2p) 2 B.. = (4 2)/2 = 1, Ne 2: (σ 2s) 2 (σ 2s*) 2 (σ 2p) 2 (π 2p) 4 (π 2p*) 4 (σ 2p*) 2 B.. = (8 8)/2 = 0, not 45. The electron configurations are: a. Li 2: (σ 2s) 2 B.. = (0)/2 = 1, diamagnetic (0 unpaired e ) b. C 2: (σ 2s) 2 (σ 2s*) 2 (π 2p) 4 B.. = (6 2)/2 = 2, diamagnetic (0 unpaired e ) c. S 2: (σ 3s) 2 (σ 3s*) 2 (σ 3p) 2 (π 3p) 4 (π 3p*) 2 B.. = (8 4)/2 = 2, paramagnetic (2 unpaired e )

4 CHAPTER 9 CVALENT BNDING: RBITALS There are 14 valence electrons in the M electron configuration. Also, the valence shell is n = 3. Some possibilities from row 3 having 14 valence electrons are Cl 2, SCl, S 2, and Ar : (σ 2s) 2 (σ 2s*) 2 (σ 2p) 2 (π 2p) 4 (π 2p*) 2 B.. = bond order = (8 4)/2 = 2 N 2: (σ 2s) 2 (σ 2s*) 2 (π 2p) 4 (σ 2p) 2 B.. = (8 2)/2 = 3 In 2, an antibonding electron is removed, which will increase the bond order to 2.5 [= (8 3)/2]. The bond order increases as an electron is removed, so the bond strengthens. In N 2, a bonding electron is removed, which decreases the bond order to 2.5 = [(7 2)/2]. So the bond strength weakens as an electron is removed from N a. CS; CS is polar and has dipole-dipole forces in addition to London dispersion (LD) forces. All polar molecules have dipole forces. C 2 is nonpolar and only has LD forces. To predict polarity, draw the Lewis structure and deduce whether the individual bond dipoles cancel. b. Se 2; both Se 2 and S 2 are polar compounds, so they both have dipole forces as well as LD forces. However, Se 2 is a larger molecule, so it would have stronger LD forces. c. H 2NCH 2CH 2NH 2; more extensive hydrogen bonding (H-bonding) is possible because two NH 2 groups are present. d. H 2C; H 2C is polar, whereas CH 3CH 3 is nonpolar. H 2C has dipole forces in addition to LD forces. CH 3CH 3 only has LD forces. e. CH 3H; CH 3H can form relatively strong H-bonding interactions, unlike H 2C. 51. A cubic closest packed structure has a face-centered cubic unit cell. In a face-centered cubic unit, there are: 8 corners 1/8 atom corner 1/ 2 atom 6 faces = 4 atoms face The atoms in a face-centered cubic unit cell touch along the face diagonal of the cubic unit cell. Using the Pythagorean formula, where l = length of the face diagonal and r = radius of the atom: 4r l l 2 + l 2 = (4r) 2 2l 2 = 16r 2 l = r 8 l

5 CHAPTER 9 CVALENT BNDING: RBITALS 308 l = r 8 = m 8 = m = cm Volume of a unit cell = l 3 = (5.57 Mass of a unit cell = 4 Ca atoms 8 10 cm) 3 22 = cm 3 1 mol Ca atoms g Ca mol Ca = g Ca Density = mass volume g = 1.54 g/cm cm 103 A:. solid B: liquid C: vapor D: solid + vapor E: solid + liquid + vapor : liquid + vapor G: liquid + vapor H: vapor triple point: E critical point: G Normal freezing point: Temperature at which solid-liquid line is at 1.0 atm (see following plot). Normal boiling point: Temperature at which liquid-vapor line is at 1.0 atm (see following plot ). 1.0 atm nfp nbp Because the solid-liquid line equilibrium has a positive slope, the solid phase is denser than the liquid phase.

Molecular Geometry and Bonding Theories. Chapter 9

Molecular Geometry and Bonding Theories Chapter 9 Molecular Shapes CCl 4 Lewis structures give atomic connectivity; The shape of a molecule is determined by its bond angles VSEPR Model Valence Shell Electron