Molecular Structure: Solids and Liquids - College of -...

93

10Molecular Structure: Solids and Liquids

10.1 a.b.c.d.

10.2 a.b.c.d.

10.3 a. Arb. Xec. Ard. none

10.4 a. noneb. Nec. noned. Kr

10.5 a.b.c.d.

10.6 a.b.c.d.

10.7 a.

b.

c.

d. orBH H

H

H

B

H

H

HH

– –

BH4� (8 e�)

Br N BrorN BrBr

BrBr

NBr3 (26 e�)

F S ForS F FSF2 (20 e�)

H F H ForHF (8 e�)

1N � 4H � 1(5 e) � 4(1 e) � 1 e (positive charge) � 8 valence electrons1C � 4H � 1O � 1(4 e) � 4(1 e) � 1(6 e) � 14 valence electrons3Br � 1N � 3(7 e) � 1(5 e) � 26 valence electrons2Br � 1S � 2(7 e) � 1(6 e) � 20 valence electrons

1O � 1H � 1(6 e) � 1(1 e) � 1 e(negative charge) � 8 valence electrons1C � 4C; � 1(4 e) � 4(7 e) � 4 � 28 � 32 valence electrons2I � 2(7 e) � 14 valence electrons2H � 1S � 2(1 e) � 1(6 e) � 8 valence electrons

1s22s22p63s23p63d104p6Sr2�

1s22s22p63s23p63d6Co3�:1s22s22p6Al3�

1s22s22p63s23p63d3Mn4�:

1s22s22p63s23p63d3V2�

1s22s22p63s23p6K�

1s22s22p63s23p64s23d104p65s24d105p6Ba2�

1s22s22p63s23p6Sc3�

: O2�:1s22s22p6O:1s22s22p4: Co3�:1s22s22p63s23p63d6Co:1s22s22p63s23p64s23d7: Al3�:1s22s22p6Al:1s22s22p63s23p1: Mn2�:1s22s22p63s23p63d5Mn:1s22s22p63s23p64s23d5

: Ti2�:1s22s22p63s23p63d2Ti:1s22s22p63s23p64s23d2 : Ni2�:1s22s22p63s23p63d8Ni:1s22s22p63s23p64s23d8 : S2�:1s22s22p63s23p6S:1s22s22p63s23p4 : Ca2�:1s22s22p63s23p6Ca:1s22s22p63s23p64s2

47374_10_p93-112.qxd 2/9/07 9:26 AM Page 93

94

Chapter 10

e.

f.

10.8 a.

b.

c.

d.

e.

f.

10.9 When using all the valence electrons does not give complete octets, it is necessary to writemultiple bonds.

10.10 If using all the available electrons for a molecule or polyatomic ion does not complete octets, thenmultiple bonds must be written.

10.11 Resonance occurs when we can write two or more formulas for the same molecule or ion.

10.12 A covalent compound has resonance when there is more than one way to write the electron-dotsymbol.

CC

H

or H

H

C H

H

H

CH H

H H

HHC2H6 (14 e�)

C FC

F

orF

ClFClCl Cl

CF2Cl2 (32 e�)

Si FSi

F

orF

FFFF F

SiF4 (32 e�)

O HO

H

or

H

HHH

++

H3O� (8 e�)

C ClCl

Cl

orClClCl

Cl

Cl

Cl

CCl4 (32 e�)

O HO H

H

or

HH2O (8 e�)

NHH

N NHN H H

H H

orH

N2H4 (14 e�)

orCH

H

H

C

H

H

OHO H HCH3OH (14 e�)

47374_10_p93-112.qxd 2/9/07 9:26 AM Page 94

10.13 a.

b.

c.

10.14 a.

b.

c.

10.15 a.

b.

10.16 a.

b.

10.17 a. linearb. Four electron groups around a central atom have a tetrahedral electron structure. When three

electron groups are bonded to the central atom, the structure is trigonal pyramidal.

10.18 a. Four groups of electrons bonded to a central atom with no lone pairs give a tetrahedral shape.b. Four electron pairs are arranged as a tetrahedron with angles. When only two pairs are

bonded to an atom, it has a bent shape.

10.19 The four electron pairs in have a tetrahedral arrangement, but three bonded atoms and onelone pair around a central atom give a trigonal pyramidal shape.

10.20 In the electron-dot formula of the central atom S has four electrons pairs arranged as a tetra-hedron. Because there are two bonded atoms and two lone pairs, has a bent shape. Thearrangement of electron pairs determines the angles between the pairs, whereas the number ofbonded atoms determines the shape of the molecule.

10.21 In the central atom B has three bonded atoms and no lone pairs, which gives , a trigonalplanar shape. In the central atom N has three bonded atoms and one lone pair, which give

a trigonal pyramidal shape.

10.22 In and the central atoms of C and O have four electron pairs arranged as tetrahedrons,which have angles. The shapes are different because has four bonded atoms (tetrahedralshape), whereas has two bonded atoms and two lone pairs (bent shape).

10.23 a. Three electron groups and three bonded atoms have a trigonal planar shape.b. The central oxygen atom has four electron pairs, but only two are bonded to fluorine atoms. Its

shape is bent with angles.109�

H2OCH4109�

H2O,CH4

NF3

NF3,BF3BF3,

H2SH2S,

PCl3

109�

N ON O NN N N ON2O (16e)

O CH O–

O CH O–

HCO2

� (18 e)

C NO–

N–

O–

O C C NOCN�

O

N OCl

O

N OClClNO2

orON +N O +

NO� (10 e�)

Clor S Cl

O

ClCl S

O

SOCl2 (26 e�)

orCH C H C CH HHCCH (10 e�)

Hor C H

O

CH H

O

H2CO (12 e�)

CC

H

or H C H

H

CH HH H

H2CCH2 (12 e�)

O orC C OCO (10 e�)

Molecular Structure: Solids and Liquids

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96

Chapter 10

c. The central atom C has two electron groups bonded to two atoms; HCN is linear.d. The central atom C has four electron pairs bonded to four chlorine atoms; has a

tetrahedral shape.e. The central atom Se has three electron groups with two bonded atoms and a lone pair, which

gives a bent shape with angles.

10.24 a. tetrahedral b. trigonal pyramidal c. bentd. linear e. trigonal planar

10.25 To find the total valence electrons for an ion, add the total valence electrons for each atom and addthe number of electrons indicated by a negative charge.a.

total for the electron-dot formula.

trigonal planar

b.total for the electron-dot formula

4 electron pairs around S; tetrahedral shape

c. tetrahedral

d. linear

10.26 To find the total valence electrons for an ion, add the total valence electrons for each atom and addthe number of electrons indicated by a negative charge.a.

total for the electron-dot formula

3 electron groups around N; 2 bonded atoms and 1 lone pair give a bent mol-ecular shape with angles.

b.total for the electron-dot formula

4 electron pairs around P and 4 bonded atoms give a tetrahedral shape.P OO

3–O

O

5 e� � 24 e� � 3 e� � 32 e�

P (5 valence electrons) � 4O (4 � 6 valence electrons) � charge (3 electrons) �120�

N OO–

5 e� � 12 e� � 1 e� � 18 e�

N (5 valence electrons) � 2O (2 � 6 valence electrons) � charge (1 electron) �

N OO+

NO2� (16 e�)

H

B HH

H

–

BH4� (8 e�)

O

S OO

2–

O

6 e� � 24 e� � 2 e� � 32 e�

S (6 valence electrons) � 4O (4 � 6 valence electrons) � charge (2 electrons) �

O

C OO

2–4 e� � 18 e� � 2 e� � 24 e�

C (4 valence electrons) � 3O (3 � 6 valence electrons) � charge (2 electrons) �

120�SeO2

CCl4

47374_10_p93-112.qxd 2/9/07 9:26 AM Page 96

c. 4 electron pairs around Cl; tetrahedral shape

d. 4 electron pairs around S and 3 bonded atoms give a trigonal pyramidalshape.

10.27 The electronegativity values increase going across a period.

10.28 The electronegativity values decrease going down a group.

10.29 A nonpolar covalent bond would have an electronegativity difference of 0.0 to 0.4.

10.30 A polar covalent bond would have an electronegativity difference of 0.4 to 1.8.

10.31 a. Electronegativity increases going up a group: K, Na, Lib. Electronegativity increases going across a period: Na, P, Clc. Electronegativity increases going across a period and at the top of a group: Ca, Br, O

10.32 a. Electronegativity increases going up a group: Br, Cl, Fb. Electronegativity increases going across a period: B, N, Oc. Electronegativity increases going across a period: Mg, S, F

10.33 a.

b.

c.

d.

e.

10.34 a.

b.

c. Br—Fd+ d–

Se—Fd+ d–

Si—Brd+ d–

B—Cld + d –

P—Brd + d –

C—Od + d –

Si—Pd + d –

N—Fd + d –

S FF

+F

O

Cl OO

O

–


97

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98

Chapter 10

d.

e.

10.35 a. Si Br electronegavity difference 1.0, polar covalentb. Li F electronegavity difference 3.0, ionicc. Br F electronegavity difference 1.2, polar covalentd. Br Br electronegavity difference 0, nonpolar covalente. N P electronegavity difference 0.9, polar covalentf. C P electronegavity difference 0.4, nonpolar covalent

10.36 a. Si O electronegavity difference 1.7, polar covalentb. K Cl electronegavity difference 2.2, ionicc. S F electronegavity difference 1.5, polar covalentd. P Br electronegavity difference 0.7, polar covalente. Li O electronegavity difference 2.5, ionicf. N P electronegavity difference 0.9, polar covalent

10.37 Electrons are shared equally between two identical F atoms, but unequally between nonidenticalatoms H and F.

10.38 CBr4 has four equal dipoles that cancel to give a nonpolar molecule. In NBr3, the three dipoles donot cancel in a tetrahedral arrangement with one lone electron pair. Therefore, NBr3 is a polarmolecule.

10.39 a. Two dipoles cancel; nonpolar. b. Dipoles do not cancel; polar.c. Four dipoles cancel; nonpolar. d. Three dipoles cancel; nonpolar.

10.40 a. Dipoles do not cancel; polar. b. Dipoles do not cancel; polar.c. Four dipoles cancel; nonpolar. d. Dipoles do not cancel; polar.

10.41 a. An attraction between the positive end of one polar molecule and the negative end of anotherpolar molecule is called dipole–dipole attraction.

b. An ionic bond is an attraction between a positive and negative ion.c. The weak attractions that occur between temporary dipoles in nonpolar molecules are

dispersion forces.d. The attraction between H and F, O, or N in a dipole is a hydrogen bond.e. The weak attractions that occur between temporary dipoles in nonpolar molecules are

dispersion forces.

10.42 a. dipole–dipole b. ionic c. dipole–dipoled. dispersion forces e. hydrogen bonds

10.43 a. Hydrogen bonding occurs between dipoles containing H and F, O, or N.b. Dispersion forces occur between temporary dipoles in nonpolar molecules.c. Dipole–dipole interactions occur between dipoles in polar molecules.d. Dispersion forces occur between temporary dipoles in nonpolar molecules.e. Dispersion forces occur between temporary dipoles in nonpolar molecules.

10.44 a. Dispersion forces occur between temporary dipoles in nonpolar molecules.b. Hydrogen bonds form between dipoles containing H and F, O, or N.c. Dipole–dipole interactions occur between dipoles in polar molecules.

Cl2

CCl4

99

9999

999999

N—Pd+ d–

N—Hd+ d–

47374_10_p93-112.qxd 2/9/07 9:26 AM Page 98

d. Hydrogen bonds form between dipoles containing H and F, O, or N.e. Hydrogen bonds form between dipoles containing H and F, O, or N.

10.45 a. A gas takes the shape and volume of its container. Thus, a gas has no definite volume or shape.b. The particles in a gas have little attraction between them and do not interact.c. A solid has a definite volume or shape.

10.46 a. liquid b. gas c. gas

10.47 a.

b.

c.

d.

10.48 a.

b.

c.

d.

10.49 a.

b.

c.

d.

10.50 a. b. 172 kJ releasedc. d.

10.51 a.

b. Two calculations are needed:

(1)

(2)

Total: c. Two calculations are needed:

(1)

(2)

Total: 33.9 kJ � 6.28 kJ � 40.2 kJ

15.0 g water � 100 �C �4.184 J

g �C�

1 kJ

1000 J � 6.28 kJ

15.0 g steam �2260 J

1 g steam�

1 kJ

1000 J � 33.9 kJ

16 700 J � 13 600 J � 30 300 J

50.0 g ice � 65�C �4.184 J

g �C� 13 600 J

50.0 g ice �334 J

1 g ice� 16 700 J

20.0 g � 57�C �4.184 J

g �C� 4800 J

1.1 � 104 kJ absorbed9.94 � 104 J absorbed2.26 � 104 J released

175 g steam �2260 J

1 g water�

1 kJ

1000 J � 396 kJ; heat is released

8.0 kg steam �2260 J

1 kg steam � 1.81 � 107 J; heat is released

50.0 g water �2260 J

1 g water�

1 kJ

1000 J � 113 kJ; heat is absorbed

10.0 g water �2260 J

1 g water � 22 600 J; heat is absorbed

5.00 kg ice �1000 g

1 kg�

334 J

1 g ice�

1 kJ

1000 J � 1670 kJ (absorbed)

145 g ice �334 J

1 g ice�

1 kJ

1000 J � 48.4 kJ (absorbed)

275 g water �334 J

1 g water � 9.19 � 104 J (released)

35.2 g water �334 J

1 g water � 1.18 � 104 J (released)

50.0 g water �334 J

1 g water�

1 kJ

1000 J � 16.7 kJ; released

225 g water �334 J

1 g water�

1 kJ

1000 J � 75.2 kJ; released

17.0 g ice �334 J

1 g ice � 5680 J; absorbed

65.0 g ice �334 J

1 g ice � 21 700 J; absorbed


99

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100

Chapter 10

d. Three calculations are needed:

(1)

(2)

(3)

Total:

10.52 a.

Total:

b.

Total:

c.

Total:

d.

Total:

10.53 Two calculations are needed:

Total:

10.54

Total:

10.55 a. dispersion forces b. hydrogen bondsc. dipole–dipole interactions d. dispersion forces

10.56 a. dispersion forces b. hydrogen bondsc. dispersion forces d. dipole–dipole interactions

260 kJ � 48.1 kJ � 38.4 kJ � 347 kJ

115 g �334 J

g�

1 kJ

1000 J� 38.4 kJ

115 g � 100.0�C � 4.184 J/g �C �1 kJ

1000 J� 48.1 kJ

115 g �2260 J

g�

1 kJ

1000 J� 260 kJ

91.9 kJ � 27.6 kJ � 119.5 kJ

275 g water � 24.0�C �4.184 J

g �C�

1 kJ

1000 J� 27.6 kJ

� 91.9 kJ275 g ice �334 J

1 g ice�

1 kJ

1000 J

20 700 J � 124 000 J � 145 000 J � 1.45 � 105 J

55.0 g �2260 J

g� 124 000 J

55.0 g � 90.0�C � 4.184 J/g �C � 20 700 J

55.0 mL �1.00 g

1 mL� 55.0 g

192 kJ � 35.6 kJ � 28.4 kJ � 256 kJ

85.0 g �334 J

g�

1 kJ

1000 J� 28.4 kJ

85.0 g � 100.0�C � 4.184 J/g �C �1 kJ

1000 J� 35.6 kJ

85.0 g �2260 J

g�

1 kJ

1000 J� 192 kJ

175 000 J � 32 900 J � 2.08 � 105J525 g � 15.0�C � 4.184 J/g �C � 32 900 J

525 g �334 J

1 g� 175 000 J

284 000 J � 44 500 J � 329 000 J � 3.29 � 105J125 g � 85.0�C � 4.184 J/g �C � 44 500 J

125 g �2260 J

g� 284 000 J

8.016 kJ � 10.04 kJ � 54.24 � 72.3 kJ (rounded to 0.1)

24.0 g steam �2260 J

1 g steam�

1 kJ

1000 J� 54.24 kJ (not rounded)

24.0 g water � 100�C �4.184 J

g �C�

1 kJ


24.0 g ice �334 J

1 g ice�

1 kJ


47374_10_p93-112.qxd 2/9/07 9:26 AM Page 100

10.57 is trigonal planar; all dipoles cancel and is a nonpolar molecule. is trigonal pyra-midal; the dipoles do not cancel and is a polar molecule.

10.58 is a linear molecule with two polar bonds at which makes the opposing dipoles cancel.is a bent molecule with two polar bonds at that do not cancel.

10.59 a. The liquid water in perspiration absorbs heat and changes to vapor. The heat needed for thechange is removed from the skin.

b. On a hot day, there are more liquid water molecules in the damp clothing that have sufficientenergy to become water vapor. Thus, water evaporates from the clothes more readily on a hotday.

c. Some water molecules evaporate, but they cannot escape from the sealed bag. The highhumidity in the bag allows some of the gaseous water to condense back to liquid, so the clotheswill not dry.

10.60 a. A liquid that evaporates quickly cools the injured area and numbs the pain.b. In a flat, shallow dish there are more molecules on the surface of the water and the water will

evaporate more quickly.c. The sandwich that is not wrapped will dry out more quickly. In a wrapped sandwich, the water

molecules cannot escape and water will not be lost as quickly.

10.61

10.62

100110

�10

50

0

Heat removed

melting point

boiling (condensation) point

liquid

gas

solid

T (°C)

150

100

50

0

�50Heat added

boiling point

melting point

liquid

gas

solid

T (°C)

120�SO2

180�,CO2

PCl3

PCl3BCl3BCl3


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102

Chapter 10

10.63 a. b.c. A represents the solid state. B represents the change from solid to liquid, or melting of the

substance. C represents the liquid state as temperature increases. D represents the change fromliquid to gas, or boiling of the liquid.

d. at solid; at liquid; at liquid; at gas

10.64 a. 3 b. 4 c. 2 d. 1 e. 5

10.65 To find the total valence electrons for an ion, add the total valence electrons for each atom and addthe number of valence electrons indicated by a negative charge. If charge is positive, remove oneor more valence electrons.a.b.c.d.e.

10.66 To find the total valence electrons for an ion, add the total valence electrons for each atom and addthe number of valence electrons indicated by a negative charge. If charge is positive, remove oneor more valence electrons.a.b.c.d.e.

10.67 a.

b.

c.

d.

e.

10.68 a. O CC orH CH O

H

CH

H H

H HH

H

H

H

H3COCH3 (20 e�)

C ClC orHClH

CH C

H

Cl

Cl

H2CCCl2 (24 e�)

N OO or O N O+ +

NO2� (16 e�)

O HN N OHorH H

HH

H2NOH (14 e�)

O ClCl O ClClorCl2O (20 e�)

B FF

–F

F

–

B

F

F

F

ForBF4

� (32 e�)

1P(5 e) � 3Cl(7 e) � 5 � 21 � 26 valence electrons1Se(6 e) � 2Cl(7 e) � 6 � 14 � 20 valence electrons1Cl(7 e) � 2O(6 e) � 1 e (negative charge) � 7 � 12 � 1 � 20 valence electrons2N(5 e) � 1O(6 e) � 10 � 6 � 16 valence electrons1C(4 e) � 1O(6 e) � 2Cl(7) � 4 � 6 � 14 � 24 valence electrons

1S(6 e) � 3O(6 e) � 2 e (negative charge) � 6 � 18 � 2 � 26 valence electrons1B(3 e) � 3Cl(7 e) � 3 � 21 � 24 valence electrons1P(5 e) � 4H(1 e) � 1 e (positive charge) � 5 � 4 � 1 � 8 valence electrons2C(4 e) � 4H(1 e) � 1O(6 e) � 8 � 4 � 6 � 18 valence electrons1C(4 e) � 2S(6 e) � 4 � 12 � 16 valence electrons

80�C,25�C,�40�C,�80�C,

60�C�60�C

47374_10_p93-112.qxd 2/9/07 9:26 AM 2

b.

c.

d.

e.

10.69 a. 16 valence electrons

b. 16 valence electrons

c. 16 valence electrons

10.70 a. 24 valence electrons

b. 24 valence electrons

c. 16 valence electrons

10.71 a. trigonal pyramidal

b. tetrahedral

c. linear

d. bent (120�)O S O O S OSO2

Cl Be ClBeCl2

BrSiBr

Br

Br

SiBr4

FNF

F

NF3

N NCO O– – –

O C C N

OCO–

O

OCO–

O

OCO–

O

ONO

O

ONO

O

ONO–

O––

S–

SNC–

C–

C N N S

ONO+

OO+ +

O N N O

NN O ON N

CH orH

CH C

H

C C NH H

C N(20 e�)

S FF orF

SF F

F+ +

(26 e�)

Cl OO or ClO O– –

ClO2� (20 e�)

C SS or CS SCS2 (16 e�)


103

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Chapter 10

10.72 a. tetrahedral

b. linear

c. trigonal planar

d. trigonal planar

10.73 a. Electronegativity is higher at the top of a group: I, Cl, Fb. Electronegativity increases left to right across a period: K, Li, S, Clc. Electronegativity is higher at the top of a group: Ba, Sr, Mg, Be

10.74 a. Se, Br, Clb. K, Li, S, Clc. Li, B, O, F

10.75 Determine the difference in electronegativity values:a. C—N is less (0.5)b. N—Br is less (0.2)c. Br—Cl is less (0.2)d. Br—Cl is less (0.2)e. C—S is less (0)

10.76 a.b.c.d.e.

10.77 A dipole arrow points from the atom with the lower electronegativity value (more positive) to theatom in the bond that has the higher electronegativity value (more negative).a. Si has the lower electronegativity value of 1.8, making Si the positive end of the dipole. The Cl

atom has a higher electronegativity value of 3.0.

b. C has the lower electronegativity value of 2.5, making C the positive end of the dipole. The Natom has a higher electronegativity value of 3.0.

c. Cl has the lower electronegativity value of 3.0, making Cl the positive end of the dipole. The Fatom has a higher electronegativity value of 4.0.

ClF

C N

Si Cl

P9OF9BrSi9ClP9ClC9O

C9O (1.0)Br9 I (0.3)S9Cl (0.5)N9F (1.0)C9O (1.0)

Cl B Cl

Cl

BCl3 (24 e�)

O C Cl

Cl

COCl2 (24 e�)

O or2–

O O2–

OO22� (14 e�)

NH

HHH or N

H

H

HH

+ +

NH4� (8 e�)

47374_10_p93-112.qxd 2/9/07 9:26 AM Page 104

d. C has the lower electronegativity value of 2.5, making C the positive end of the dipole. The Fatom has a higher electronegativity value of 4.0.

e. N has the lower electronegativity value of 3.0, making N the positive end of the dipole. The Oatom has a higher electronegativity value of 3.5.

10.78 A dipole arrow points from the atom with the lower electronegativity value (more positive) to theatom in the bond that has the higher electronegativity value (more negative).a. C has the lower electronegativity value of 2.5, making C the positive end of the dipole. The O

atom has a higher electronegativity value of 3.5.

b. N has the lower electronegativity value of 3.0, making N the positive end of the dipole. The Fatom has a higher electronegativity value of 4.0.

c. Cl has the lower electronegativity value of 3.0, making Cl the positive end of the dipole. The Oatom has a higher electronegativity value of 3.5.

d. S has the lower electronegativity value of 2.5, making S the positive end of the dipole. The Clatom has a higher electronegativity value of 3.0.

e. P has the lower electronegativity value of 2.1, making P the positive end of the dipole. The Fatom has a higher electronegativity value of 4.0.

10.79 a. polar covalent b. nonpolar covalent c. ionic d. nonpolar covalent e. nonpolar covalent

10.80 a. polar covalentb. nonpolar covalentc. ionicd. nonpolar covalente. polar covalent

(F 4.0 � F 4.0 � 0.0)(C 2.5 � H 2.1 � 0.4)

(Cl 3.0 � Na 0.9 � 2.1)(C 2.5 � C 2.5 � 0.0)

(Cl 3.0 � Si 1.8 � 1.2)

P F

S Cl

O Cl

N F

C O

N O

C F


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Chapter 10

10.81 a. 4 electron groups, 2 bonded atoms, bent

b. 4 electron groups, 2 bonded atoms, bent

c. 3 electron groups, 3 bonded atoms, trigonal planar

d. 4 electron groups, 4 bonded atoms, tetrahedral

e. 2 electron groups, 2 bonded atoms, linear

f. 4 electron groups, 3 bonded atoms, trigonal pyramidal

g. 4 electron groups, 4 bonded atoms, tetrahedral

10.82 a. Nonpolar. There are three nonpolar P—H bonds (no dipoles) in a trigonal pyramidal shape.b. Nonpolar. In a linear molecular, two equal dipoles directed away from each other at will

cancel.c. Nonpolar. There are three nonpolar N—Cl bonds (no dipoles) in a trigonal pyramidal shape.d. Polar. In a tetrahedral, a Cl provides a polar C—Cl bond that gives a negative end (Cl) to the

molecules.

10.83 a. Polar. A single polar bond makes the Br end more negative than the H end.

b. Nonpolar. In a linear molecule, two equal dipoles directed away from each other at willcancel.

c. Nonpolar. There are three nonpolar N—Cl bonds (no dipoles) in a trigonal pyramidal shape.

N

Cl

Cl Cl

O Si O Si OO Si

180�

H Br H Br

CH3Cl

180�

H N

H

H

H

+

NH4�

O P

O

O

3–

PO33�

S C SCS2

F C

F

F

F

CF4

O C

O

O

2–

CO32�

(109�)O

H

HH2O

(109�)O Br O–

BrO2�

47374_10_p93-112.qxd 2/9/07 1:23 PM Page 106

d. Polar. In a tetrahedral molecule, a Cl provides a polar C—Cl bond that gives a negative end(Cl) to the molecules.

e. Polar, trigonal pyramidal with one lone pair; dipoles do not cancel.

f. Polar. Two polar bonds do not oppose each other in the bent molecule. Both O—H bondsmake the O end of the molecule more negative than the H end.

10.84 a. nonpolarb. nonpolarc. polard. polare. nonpolarf. polar

10.85 a. In a molecule that has a trigonal planar shape, the dipoles cancel, making the molecule nonpolar.b. In a molecule with a bent shape and one lone pair, the dipoles do not cancel. The molecule is

polar.c. In a linear molecule with identical atoms, any dipoles cancel, making the molecule nonpolar.

10.86 a. tetrahedral; dipoles cancel; nonpolarb. trigonal pyramidal; dipoles do not cancel; polarc. tetrahedral; dipoles do not cancel; polar

10.87 a. hydrogen bond (3) b. hydrogen bond (3) c. dispersion (4)d. dipole–dipole (2) e. hydrogen bond (3) f. ionic (1)

10.88 a. Dipole–dipole attractions occur between the positive end of one polar molecule and the nega-tive end of another polar molecule.

b. Hydrogen bonds are strong dipole–dipole bonds that occur between a partially positive hydro-gen atom and one of the strongly electronegative atoms of F, O, or N.

c. Dispersion forces occur between temporary dipoles that form within nonpolar molecules.Larger nonpolar compounds containing a greater number of electrons produce more temporarydipoles within the molecule.

10.89 Heat is released (exothermic) as water vapor turns to rain or liquid water turns to snow. The heat re-leased in either of these processes warms the surrounding air, so the air temperature is in fact raised.

10.90 a. As the liquid water freezes, heat must be released into the air. The temperature in the orchardwill not go below until all the water has frozen.

b. Two calculations are required:

For a total of: 310 kJ � 1700 kJ � 2000 kJ � 2.0 � 103 kJ

5.0 kg �1000 g

1 kg�

334 J

1 g�

1 kJ

1000 J� 1700 kJ

5.0 kg �1000 g

1 kg�

4.184 J

g �C� 15�C �

1 kJ

1000 J� 310 kJ

0�C

O H

H

H2O

F N F

F

Cl C H

Cl

Cl

CH3Cl


107

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108

Chapter 10

10.91

water

Total:

10.92

Total heat absorbed:

10.93 a. b.

c. d.

10.94 a. Error: There should be 20 electrons.

Correct

b. Error: Carbon does not have 8 electrons.

Correct

c. Error: Nitrogen cannot have 10 electrons.

Correct

10.95 a. trigonal pyramidal b. tetrahedralc. linear d. trigonal planar

10.96 a. nonpolar b. nonpolarc. nonpolar d. polar

10.97 350 g of ice will be melted.

H N O H

H

H N O H

H

H C H

O

H C H

O

Cl O Cl

Cl O Cl

Cl C O C

H

H

H

O

H N N H

Cl C

H

C N

H

N C

H

H

O

H

Masswater ��49 900 J

(�40.0�C)�

1 g �C

4.184 J� 298 g water initially at 64.0�C

11 500 J � 38 400 J � 49 900 J

115 g �334 J

1 g� 38 400 J

115 g � 24.0�C �4.184 J

g �C� 11 500 J (heat absorbed)

34 kJ � 109 kJ � 143 kJ

325 g �334 J

1 g�

1 kJ

1000 J� 109 kJ0�C : ice 0�C:

325 g � 25�C �4.184 J

g �C�

1 kJ

1000 J� 34 kJ25�C : 0�C:

47374_10_p93-112.qxd 2/9/07 9:26 AM Page 108

10.98

a. liquid b. benzene melts c. liquidd. gas e. boiling temperature of

10.99

10.100 a. dispersion forces b. hydrogen bonds; highest boiling pointc. dispersion forces; lowest boiling point d. dipole–dipole interactionse. dipole–dipole interactions

Answers to Combining Ideas Chapters 9–10

CI 15 a. X is a metal. Y is a nonmetal.b. Y has the higher electronegativity.c.d. 1.

2.3. has the same electron configuration as Ne.

has the same electron configuration as Ar.4. magnesium chloride

e. 1.2.3. calcium bromide

CI 16

a. b. c. d. e.

CI 17 a. With two bonding atoms and two lone pairs, the shape is bent.H N H–

C Sid + d –

Al Cld + d –

I Fd + d –

N Fd + d –

C Fd + d –

CaBr2,Y� � 1s22s22p63s23p64s23d104p6X2� � 1s22s22p63s23p6Y � 1s22s22p63s23p64s23d104p5X � 1s22s22p63s23p64s2

MgCl2,Y�

X2�

Y� � 1s22s22p63s23p6X2� � 1s22s22p6Y � 1s22s22p63s23p5X � 1s22s22p63s2

X2�, Y�

�15 000 J

(�8.0�C)(4.184 J/g � C)� 450 g water

45.0 g �334 J

g� 15 000 J

80.1�C

100

80

60

40

20

0

T (°C)

boiling

melting

liquid

gas

solid


109

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110

Chapter 10

b. With four bonding atoms and no lone pairs, the shape is tetrahedral.

c. With two bonding atoms and two lone pairs, the shape is bent.

d. With four bonding atoms and no lone pairs, the shape is tetrahedral.

CI 18 a.

b.

c.

d.

e.

�2 mol CO2

1 mol EtOH�

44.01 g

1 mol CO2

�1 kg

1000 g� 10. kg

5.7 L �1000 mL

1 L�

0.796 g

1 mL�

1 mol EtOH

40.07 g

15 gal gasoline �10 gal ethanol

100 gal gasoline�

4 qt

1 gal�

1 L

1.057 qt� 5.7 L

C2H6O(g) � 3O2(g) : 2CO2(g) � 3H2O(g)

20.0 g �1 mol

40.07 g�

38.6 kJ

1 mol�

19.3 kJ

26.2 kJ

20.0 g � 140.�C �2.46 J

g �C�

1 kJ

1000 J� 6.89 kJ

100

50

0

�50

�100

�150

boiling

melting

Heat added

liquid

gas

solid

F

B FF

F

–

H O Br

Cl

P ClCl

Cl

+

47374_10_p93-112.qxd 2/9/07 9:26 AM Page 110

CI 19 a.

b. chloral hydrate: chloral: c.

CI 20 a. Ethylene glycol empirical formula is .

b.

c.d.

e. 11 lb �453.6 g

1 lb�

1 kg

1000 g�

4700 mg

1 kg�

1 g

1000 mg�

1 mL

1.11 g� 21 mL

C2H6O2(l) � 2O2(g) : H2C2O4(aq) � 2H2O(l)

O

C CO O

O

H H

H

C CO O

H

H

H H

H

CH3O

Mass Cl/Molar mass � 100 � 106.35 g Cl/165.39 g � 100 � 64.30% (by mass)C2H3O2Cl3 � 2(12.01) � 3(1.008) � 2(16.00) � 3(35.45) � 165.39 g/mol

C2HOCl3C2H3O2Cl3

Cl

C CCl H

O

Cl

Cl

C CCl

Cl

O H

O

H

H


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