Unit 1: Liquids and Solids

Kinetic-Molecular Description of Liquids and Solids

 Solids and liquids are

condensed states.

 The atoms, ions, or molecules in solids and liquids are much closer to one another than in gases.

 Solids and liquids are highly incompressible.

 Liquids and gases are

fluids.

 They easily flow.

 The

intermolecular attractions

in liquids and solids are strong.

2

Table 13-1, p. 448

Kinetic-Molecular Description of Liquids and Solids

 Schematic representation of the three common states of matter.

4

Kinetic-Molecular Description of Liquids and Solids

 If we compare the

strengths of interactions

among particles and the

degree of ordering

of particles, we see that Gases< Liquids < Solids 

Miscible liquids

are soluble in each other.  Examples of miscible liquids: • Water dissolves in alcohol.

• Gasoline dissolves in motor oil.

 The natural diffusion rate of liquids is slower than gases 5

Kinetic-Molecular Description of Liquids and Solids



Immiscible liquids

are insoluble in each other.

 Two examples of immiscible liquids: • Water does not dissolve in oil.

• Water does not dissolve in cyclohexane.

6

Kinetic-Molecular Description of Liquids and Solids

 Solid particles do not readily diffuse into other solids  However, analysis of 2 different blocks of solids e.g. Cu and Pb that have been pressed together for a period of years show that each block contains some atoms of the other element  solids do diffuse but very slowly and if pressure is applied.

Fig. 13-2, p. 449

Intermolecular Attractions and Phase Changes



Inter

molecular forces forces

between

(IMF) refer to the individual particles (atoms, molecules, ions) of a substance  These forces are quite weak relative to

intra

molecular forces i.e. covalent and ionic bonds

within

compounds  If it were not for IMF, condensed phases would not exist  IMF influence physical properties 8

Intermolecular Attractions and Phase Changes The important intermolecular attractions from strongest attraction to the weakest attraction are: 1. Ion-ion interactions (ionic bond)  The

force of attraction

between 2 oppositely charged ions is directly proportional to the charges on the ions (say q + and q )  F α q + x q • Thus ionic substances containing multiple charged ions e.g. Mg 2+ , Al 3+, etc. have stronger forces of attraction ( & thus higher m.p. and b.p.) than those with singly charged ions  The

force of attraction

between 2 oppositely charged ions is also inversely proportional to the square distance b/w the ions  F α 1/d 2 9

Intermolecular Attractions and Phase Changes 1.

Ion-ion interactions (ionic bond)  So for a series of similarly charged ions, the closer approach of smaller ions results in stronger interionic attractive forces  Higher m.p. and b.p

Smaller the ions



stronger the ionic bond





Intermolecular Attractions and Phase Changes

Example 1: Arrange the following ionic compounds in the expected order of increasing melting and boiling points.

NaF, CaO, CaF 2

You do it!

What important points must you consider?

+ -



2+ 2

 

2+ Ca O 2-

11

Intermolecular Attractions and Phase Changes

2. Dipole-dipole interactions  Occur between polar covalent molecules because of the attraction of the δ+ and δ- atoms of another molecule 12

Intermolecular Attractions and Phase Changes

2. Dipole-dipole interactions e.g. BrF (polar molecule).

Each polar molecule is shaded with regions of high e-s density ( red ) and regions of high positive charge ( blue ). Attractive forces are shown as blue arrows and repulsive forces as red arrows Molecules tend to arrange themselves to maximize attractions by bringing regions of opposite charge together while minimizing repulsions by separating regions of like charge 13

Intermolecular Attractions and Phase Changes 2. Dipole-dipole interactions  An increase in temp  increase in translational, rotational & vibrational motion of molecules  random orientation of molecules relative to each other  Consequently, dipole-dipole interactions become less important as temp   All these factors make compounds having only dipole-dipole interactions

more volatile

than ionic compounds 14

Intermolecular Attractions and Phase Changes

3. London Forces are very weak.

 • • • Also known as: Instantaneous dipole-induced dipole interactions Dispersion forces London dispersion forces  They are the weakest of the intermolecular forces.

 They exist in

ALL

molecules  This is the only attractive force in

nonpolar molecules.

15

Intermolecular Attractions and Phase Changes

3. London Forces are very weak  In a group of Ar atoms, the temporary dipole in one atom induces other atomic dipoles.

• Each atom’s e- cloud is attracted by the nucleus of the other atom or is repelled by the other atoms’ e-s cloud 16

Intermolecular Attractions and Phase Changes

 Similar effects occur in a group of I 2 molecules.

17

Intermolecular Attractions and Phase Changes Polarizability increases with increasing numbers of e and therefore with increasing sizes of molecules Therefore,

dispersion forces are generally stronger for larger molecules

For molecules that are large or quite polarizable the total effect of the dispersion forces can even be higher than dipole-dipole interactions or H-bonding 18

Intermolecular Attractions and Phase Changes  NOTE:  The term “Van der Waals forces” usually refers to: • Dipole - dipole interactions • London forces 19

Intermolecular Attractions and Phase Changes

4. Hydrogen bonding  They are NOT really chemical bonds  Are a special case of a very strong dipole-dipole interaction.

Criteria for strong H-bonding: 1. A

hydrogen bond donor

: polar covalent molecule containing H attached to either one of the three small, highly electronegative elements – F, O or N.

2. A

hydrogen bond acceptor

: highly electronegative elements – F, O or N.

20

Intermolecular Attractions and Phase Changes 4. Hydrogen bonding  Consider H 2 O molecules • Each O atom can form two H-bonds and • Each H atom can form one H-bond δ+ δ δ+ δ+ δ δ+ 21

Intermolecular Attractions and Phase Changes Hydrogen bonding in (b) methanol, CH 3 OH and (c) ammonia, NH 3 The H-bonds are due to electrostatic attraction between the δ+ charged H of one molecule to the δ- charge O or N of another.

22

Fig. 13-4, p. 452

Intermolecular Attractions and Phase Changes

4. Hydrogen bonding  • • • Typical H- bond energies range from 15 – 20kJ /mol This is four – five times greater than that of dipole-dipole interactions As a result, H-bonds exert a considerable influence on the properties of substances E.g. H-bonds are responsible for the unusually high b.p. and m.p. of water, methanol and ammonia 23

Intermolecular Attractions and Phase Changes

• The unusually high bps of NH 3 , H 2 O and HF are due to H-bonding • CH 4 is non-polar  weak London forces only • As molecular mass increases (e.g Grp 4) bp increases because of increased dispersion forces.

Boiling points of some hydrides as a function of molecular weight

24

So to summarize…..

Type of substance

ionic polar Non-polar

Intermolecular force present (strongest to weakest)

Ion- ion interaction Hydrogen bonding OR Permanent dipole- dipole London/ Dispersion forces

Note:

ALL molecules contain dispersion forces 25

The Liquid State - Properties

Viscosity

 Viscosity is the resistance to flow.

 For example, compare how water pours out of a glass (low viscosity) compared to molasses, syrup or honey (high viscosity).

 Oil for your car is bought based on this property.

 10W30 or 5W30 describes the viscosity of the oil at high and low temperatures.

26

The Liquid State - Properties

Viscosity

 For a liquid to flow, molecules must be able to slide past each other.

 The stronger the intermolecular forces (IM)  the more viscous the liquid Substances that have a great ability to form H-bonds usually have high viscosities 27

The Liquid State - Properties

Viscosity

 Increasing the size and surface area of molecules  increased viscosity due to increased dispersion forces Pentane, C 2 H 5 Viscosity = 0.215 centipoise at 25 o C dodecane, C 12 H 26 Viscosity = 1.38 centipoise at 25 o C 28

The Liquid State - Properties

Viscosity

 As temp  , molecules move more rapidly, their kinetic energies are able to overcome IM forces  a decrease in viscosity  An example of viscosity of two liquids. 29

The Liquid State - Properties

Surface Tension

 Molecules below the surface of a liquid are influenced by IM attractions from all directions  Those on the surface are attracted unevenly; are only attracted toward the interior  pulls the surface toward the center  Surface tension is a measure of the unequal attractions that occur at the surface of a liquid.

 It is a measure of the forces that must be overcome to expand the surface area of a liquid 30

The Liquid State - Properties

Surface Tension

31

Fig. 13-9, p. 457

The Liquid State - Properties

Surface Tension

 Coating glass with silicone polymer greatly reduces adhesion of water to the glass  The left side of each glass has been treated with Rain-X which contains a silicone polymer.  Water on the treated side forms droplets that are easily swept away.

32

p. 457

The Liquid State - Properties

Surface Tension

Droplets of mercury of glass  The small droplets are almost spherical, whereas larger ones are flattened due to the effects of gravity  This shows that surface tension has more influence on the shape of the small (lighter droplets) 33

p. 457

The Liquid State

 Floating paper clip demonstration of surface tension.

34

The Liquid State

Capillary Action

 Capillary action is the ability of a liquid to rise (or fall) in a glass tube or other container 35

The Liquid State

 Cohesive forces are the forces that hold liquids together.

 Adhesive forces are the forces between a liquid and another surface.

 Capillary rise implies that the: • Adhesive forces > cohesive forces  Capillary fall implies that the: • Cohesive forces > adhesive forces 36

The Liquid State

 Water exhibits a capillary rise.

 Mercury exhibits a capillary fall.

Mercury Water

37

The Liquid State

 Capillary action also affects the meniscus of liquids.

• Capillary action helps plant roots take up water and dissolved nutrients from soil • Roots, like glass, exhibit strong adhesive forces for water.

38

The Liquid State

 

Evaporation

Evaporation is the process in which molecules escape from the surface of a liquid and become a gas.

Evaporation is temperature dependent.

39

The Liquid State

Evaporation

Liquid continuously evapourates from an open vessel Equilibrium between liquid and vapour is established in a closed container in which molecules return to the liquid at the same rate as they leave it.

A bottle in which liquid-vapour equilibrium has been established. The droplets have condensed 40

Fig. 13-10ab, p. 458

The Liquid State

Distribution of kinetic energies of molecules in a liquid at different temperatures. At the lower temperature, a smaller fraction of the molecules have the same energy required to escape from the liquid, so evapouration is slower.

41

The Liquid State

Vapor Pressure



DEFINITION:

Vapor pressure is the pressure exerted by a liquid’s vapour on its surface at equilibrium.

 Because rate of evapouration increases in increasing temperature  Vapour pressure of liquids always increases with increasing temperature 42

The Liquid State

 Vapor Pressure (torr) and boiling point for three liquids at different temperatures.

0 o C

diethyl ether 85 ethanol water 12 5

20 o C

442 44 18

30 o C

647 74 32

normal boiling point

36 o C 78 o C 100 o C  Easily vapourized liquids are said to be volatile  They have relatively high vapour pressures 43

The Liquid State

Vapor Pressure as a function of temperature.

Notice that the plot is not linear Each substance exists as a liquid for temp & presure to the left of its curv e Each substance exists as a gas for temp & pres to the right of its curve The normal boiling point of a liquid is the temp at which its vapour pressure = 1 atm (760 torr) What are the intermolecular forces in each of these compounds?

You do it!

44

The Liquid State

Stronger attractive forces

Vapor Pressure

Lower vapour pressure Higher boiling point Increasing temperature Higher vapour pressure 45

The Liquid State

Boiling Points and Distillation

 The

boiling point

is the temperature at which the liquid’s vapor pressure is equal to the applied pressure.

 The

normal boiling point

is the boiling point when the pressure is exactly 1 atm (760 torr).

  E.g. water boils at 100 0 C at 1 atm If the applied pressure is lower than 1 atm, e.g. on a mountain  water boils at a lower temp • Takes longer to cook food on a mountain because the temp of boiling water is lower 46

The Liquid State

Distillation

 Distillation is a process in which a mixture or solution is separated into its components on the basis of the differences in boiling points of the components.

 Different liquids have different cohesive forces  different vapour pressures  boil at different temp • Distillation is another vapour pressure phenomenon. 47

The Liquid State

Lab setup for distillation  During distillation of an impure liquid, nonvolatile substances remain in the flask  The liquid is vapourized and condensed before being collected in the receiving flask.

48

The Liquid State

Heat Transfer Involving Liquids



The

specific heat heat capacity

(J/g

. o

C) or

molar

(J/mol

. o

C) of a liquid is the amount of heat that must be added to a stated mass of liquid to raise its temp by 1

o

C

with no change in phase

 It is given the symbol “C” 49

The Liquid State

Heat Transfer Involving Liquid • Example : How much heat is released by 200. g of H 2 O as it cools from 85.0

o C to 40.0

o C? The specific heat of water is 4.184 J/g o C.

?

J  2.00

 10 2 g(4.184

J g o C )( 85 .

0  40 .

0 o C ) ?

J  3 .

76  10 4 J  37.6

kJ 50

The Liquid State

• Molar heat capacity is the amount of heat required to raise the temperature of

one mole

of a substance 1.00 o C.

 Example : The molar heat capacity of ethyl alcohol, C 2 H 5 OH, is 113 J/mol o C. How much heat is required to raise the temperature of 125 g of ethyl alcohol from 20.0

o C to 30.0

o C?

1 mol C 2 H 5 OH = 46.0 g 51

The Liquid State

Step 1: Find out how many mols of liquid you have since you were given molar heat capacity (mass/molar mass = mols) ? mol = 125 g    ? mol = 125 g     113 J mol C 2      113 J o mol C       o C     kJ kJ 52

The Liquid State

 The calculations we have done up to now tell us the energy changes as long as the substance remains in

a single phase

.

 Next, we must address the energy associated with phase changes.

 For example, solid to liquid or liquid to gas and the reverse.

 Heat of Vaporization is the amount of heat required to change 1.00 g of a liquid substance to a gas at constant temperature.

 Heat of vaporization has units of J/g.

 Heat of Condensation is the reverse of heat of vaporization, phase change from gas to liquid.

1.00

g H 2 O (  ) at 100.0

o C   2260 J   2260 J 1 .

00 g H 2 O (g) at 100.0

o C 53

The Liquid State



Molar heat of vaporization or



H vap

The  H vap is the amount of heat required to change 1.00 mole of a liquid to a gas at constant temperature.  H vap has units of J/mol. 

Molar heat of condensation

The reverse of molar heat of vaporization is the heat of condensation.

1.00

mol H 2 O (  ) at 100.0

o C     40.7

kJ 40.7

kJ 1 .

00 mol H 2 O (g) at 100.0

o C 54

• Heats of vapourization reflect the strengths of IM forces • Heats of vapourization generally increase as bp and IM forces increase • Heats of vapourization generally increase as vapour pressure decrease

Table 13-5, p. 462

Table 13-6, p. 465

The Liquid State

 Predict the order of increasing boiling points for the following: H 2 S, H 2 O, CH 4 , H 2 , KBr 

At the molecular level what happens when a species boils?

57

The Liquid State

Compound MW(amu) B.P.( o C) CH 4 C 2 H 6 C 3 H 8 n-C 4 H 10 n-C 5 H 12 16 30 44 58 72 -161 -88 -42 -0.6

+36 Boiling point increases as molecular mass increases 58

The Liquid State

HF 20 19.5

HCl 37 - 85.0

HBr 81 - 67.0

HI 128 - 34.0

High bp of HF is due to the very polar H-F bond  strong IM forces. As we go down group effect of molecular mass has greater influence on bp 59

The Liquid State

Unusually high bp of water is due to H-bonding

VIA Hydrides 150 100 50 0 -50 -100 18 34 81 Molar Mass 130

60

The Liquid State

 Example: Arrange the following substances in order of increasing boiling points.

C 2 H 6 , NH 3 , Ar, NaCl, AsH 3

You do it!

Ar < C 2 H 6 < AsH 3 < NH 3 < NaCl nonpolar nonpolar polar very polar ionic London London dipole-dipole H-bonding ion-ion 61

The Liquid State

 Example : How many joules of energy must be absorbed by 500 g of H 2 O at 50.0

o C to convert it to steam at 120 o C ? The molar heat of vaporization of water is 40.7 kJ/mol and the molar heat capacities of liquid water and steam are 75.3 J/mol o C and 36.4 J/mol o C , respectively.

62

The Liquid State

 Problem-Solving Tip:  A problem such as this can be broken down into steps so that each involves

either

a temperature change

or

a phase change, but not both.

• A temperature change uses specific heat of the substance; remember that each different phase has its own specific heat.

• A phase change always takes place with no change in temperature, so that the calculation does not involve temperature.

63

The Liquid State

?

 18     J     o C    10 5 J Next, let’s calculate the energy required to boil the water.

    .

 3 mol J      5 J Finally, let’s calculate the heat required to heat steam from 100 to 120 o C.

   .

J       5 J 64

The Liquid State

  The total amount of energy for this process is the sum of the 3 pieces we have calculated.

5 J    5 J    5 3 J or 1.26 10 kJ 5 J   65

The Solid State

Normal Melting Point  The

normal melting point

(

freezing point

) is the temperature at which the solid melts (liquid and solid in equilibrium) at exactly 1.00 atm of pressure.

 The melting point increases as the strength of the intermolecular attractions increase.

Solid melting freezing Liquid 66

The Solid State

A typical heating curve at constant pressure

• As heat is added to a solid, the temperature rises.

• After enough heat is added to bring solid to its mp, additional heat is used to break IM forces to bring solid to liquid • During the melting process the temperature remains constant • After melting any additional heat increases the temperature of the liquid until boiling is 67 reached.

The Solid State

During any phase change, Solid Liquid Liquid Gas and the

temperature remains constant

68

Heat Transfer Involving Solids Heat of Fusion 

Heat of fusion

is the amount of heat required to melt

one gram

of a solid at its melting point at constant temperature.

1.00

g H 2 O (s) at 0 o C     334 J 334 J 1.00

g H 2 O (  ) at 0 o C •

Heat of crystallization

is the the heat of fusion.

reverse of 69

Heat Transfer Involving Solids

•

Molar heat of fusion or



H fusion

 The molar heat of fusion is the amount of heat required to melt

a mole

of a substance at its melting point.

The molar heat of crystallization is the reverse of molar heat of fusion 1.00

mole H 2 O (s) at 0 o C     6012 J 6012 J 1.00

mole H 2 O (  ) at 0 o C 70

The Solid State

A typical heating curve at constant pressure

• The length of the horizontal line for melting is proportional to the heat of fusion for the solid – higher the  H fus the longer the line • The horizontal line representing when a liquid boils is usually longer than the horizontal line for when a solid melts because the  H vap of a substance is usually higher that that of  H fus 71

Heat Transfer Involving Solids

  Heat ( or enthalpy) of solidification Heat ( or enthalpy) of solidification of a liquid is equal in magnitude to 

H fusion

It represents removal of sufficient heat from a given amount (1 mol or 1 g) of liquid to solidify the liquid at its freezing point  Here is a summary of the heats of transformation for water.

ice 1.00

mole H 2 O (s) at 0 o C     6012 J 6012 J 1.00

mole water H 2 O (  ) at 0 o C 1.00

mol water H 2 O (  ) at 100.0

o C     40.7

kJ 40.7

kJ steam 1 .

00 mol H 2 O (g) at 100.0

o C 72

Heat Transfer Involving Solids

 Example : Calculate the amount of heat required to convert 150.0 g of ice at -10.0

o C to water at 40.0

o C.

specific heat of ice is 2.09 J/g o C heat of fusion of ice = 334J/g specific heat of water = 4.18 J/g . o C

you do it

73

Heat Transfer Involving Solids

? J = (150.0 g)(2.09 o J g C o  3 ? J = (150.0 g)(334 J g  ? J = (150.0 g)(4.18 o J g C o  4  4 Note that most of the heat absorbed was to melt the ice.

74

Sublimation and the Vapor Pressure of Solids Sublimation  In the sublimation process the solid transforms directly to the vapor phase without passing through the liquid phase.

 Solid CO 2  or “dry” ice does this well.

Solids exhibit vapour pressures like liquids but generally much lower vapour pressures  Solids with very high vapour pressures sublime easily

solid

    condensati on 

gas

75

Sublimation and the Vapor Pressure of Solids Sublimation  Sublimation can be used to purify volatile solids  The high vapour pressure of the solid causes it to sublime when heated  Crystals of the purified substance are formed when the vapour is cooled.

 I 2 I 2 (Iodine) sublimes readily; vapour is purple 76

Phase Diagrams (P versus T)

 Phase diagrams are a convenient way to display all of the different phase transitions of a substance.

 Show the equilibrium pressure-temperature relationships among different phase in a closed system  This is the phase diagram for water.

77

Phase Diagrams (P versus T)



AC

: represent pressure-temp combinations for which liquid and gas coexist.

  Points above AC  liquid Points below AC  gas 

AB

: represent liquid-solid equilibrium conditions    Has a –ve slope because of unique property of water  solid is less dense than water Points to left of AB  ice Points to right of AB  liquid.

 AB is called a

melting curve

78

Phase Diagrams (P versus T)

 AD: sublimation curve  Solid and vapour in equilbrium  A: triple point  combinations of pressure & temp for which all 3 phases coexist.

 For water this is 4.58 torr and 0.01

o C 79

Phase Diagrams (P versus T)



P C

Compare water’s phase diagram to carbon dioxide’s phase diagram.

  Note the +ve slope of AB T C : critical temp  temp above which a gas cannot be liquified  gas and liquid do not coexist 

T C

 A substance above this temp is called a supercritical fluid P C : critical pressure pressure required to  liquefy a gas at it critical temp.

80

Amorphous Solids and Crystalline Solids  Amorphous solids

do not

molecular structure.

 have a well ordered Examples of amorphous solids include waxes, glasses, asphalt.

 Melting occurs over a range of temperatures for various portions of the sample as IM forces are overcome  Crystalline solids have well defined structures that consist of called extended array of repeating units

unit cells

.

 Exhibit sharp melting points The shattering of crystalline solids produces fragments having the same (or related) interfacial angles and structural characteristic of the original sample 81

Structure of Crystals

 Unit cells are the smallest repeating unit of a crystal.

 As an analogy, bricks are repeating units for buildings.

 There are seven basic crystal systems.

82

Fig. 13-23, p. 475

Structure of Crystals

 Different substances that crystallize in the same type of lattice with the same atomic arrangement are said to be isomorphous.

 A single substance that can crystallize in more than one arrangement is said to be polymorphous 84

Structure of Crystals

 We shall look at the three variations of the cubic crystal system.

 Simple cubic unit cells.

 The balls represent the positions of atoms, ions, or molecules in a simple cubic unit cell.

85

Structure of Crystals

 In a simple cubic unit cell each atom, ion, or molecule at a corner is shared by 8 unit cells  Thus 1 unit cell contains 8(1/8) = 1 atom, ion, or molecule.

86

Structure of Crystals

 Body centered cubic (bcc) has an additional atom, ion, or molecule in the center of the unit cell.

 On a body centered cubic unit cell there are 8 corners + 1 particle in center of cell.

 1 bcc unit cell • contains 8(1/8) + 1 = 2 particles.

87

Structure of Crystals

 A face centered cubic (fcc) unit cell has a cubic unit cell structure with an extra atom, ion, or molecule in each face.

 A face centered cubic unit cell has 8 corners and 6 faces.

 1 fcc unit cell contains • 8(1/8) + 6(1/2) = 4 particles.

88

Bonding in Solids

 We classify solids according to types of particles and bonding interactions.

 There are 4 categories:   Metallic solids Ionic solids   Simple Molecular solids Giant Covalent solids 89

Bonding in Solids

 Metallic Solids may be thought of as positively charged nuclei surrounded by a sea of electrons.

 The positive ions positions .

occupy the crystal lattice  Examples of metallic solids include:  Na, Li, Au, Ag, ……..

 Nearly all metals crystallize in one of three types of lattices:   Body-centered cubic (bcc) Face-centered cubic (fcc) also called cubic closed packed  Hexagonal close-packed 90

Bonding in Solids

(a) Spheres in same plane closely packed as possible. Each sphere touches 6 others (b) Spheres in 2 planes. Each sphere touches 6 others in its own layer and 3 in the layer below it, and 3 in the layer above it. Each sphere contacts 12 others therefore has a coordination number of 12.

91

(a) Expanded view of hexagonal close packed. 1 st and 3 rd layers are oriented in the same direction (b) Expanded view of cubic close packed. 1 st and 3 rd layers are oriented in opposite direction

Bonding in Solids

 Ionic Solids have ions that occupy the positions in the unit cell.

 Examples of ionic solids include:  CsCl, NaCl, ZnS 93

Bonding in Solids

(a) (b) (c) CsCl is a simple cubic. It is

NOT

body-centered, because the point at the center of the cell is not the same as the point at a corner NaCl is a face-centered cubic ZnS is a face-centered cubic 94

Bonding in Solids

 Representations of the crystal structure of NaCl  Na is grey and Cl is green   Ionic solids are usually poor thermal and electrical conductors In liquid (molten) state however, they are excellent electrical conductors because their ions are mobile 95

Fig. 13-28, p. 482

Bonding in Solids

 Molecular Solids have molecules of the positions of the unit cell .

in each  Molecular solids have low melting points, are volatile, and are electrical insulators.

 Examples of molecular solids include:  water, sugar, carbon dioxide, benzene 96

Bonding in Solids

The packing arrangement in molecular crystals depend on the shape of the molecule as well as on electrostatic attractions b/w +ve and –ve regions in the molecule.

97 (a) Solid carbon dioxide (b) benzene, C 6 H 6

Bonding in Solids

 Covalent Solids can be considered giant molecules that have covalently bonded atoms in an extended, rigid crystalline network  Some examples of covalent solids are: • Diamond, graphite, SiO 2 (sand), SiC 98

(a) In

diamond

sp 3 σ bonds each C atoms is tetrahedrally bonded to 4 other atoms through sp 3 (b) In

graphite

, C atoms are linked in planes by sp 2 -sp 2 σ and π bonds. The crystal is soft owing to the weak attractions b/w planes. Electrons move freely through the delocalized π bonding network in these planes but they do not jump b/w planes readily.

(c)

Quartz

(SiO 2 ): Each Si atom (gray) is bonded tetrahedrally to 4 O atoms (red).

Fig. 13-32, p. 485

Bonding in Solids

100

Bonding in Solids

Variations in Melting Points for Molecular Solids What are the intermolecular forces in each solid? Compound  ice Melting Point ( o C) 0.0

 ammonia -77.7

 benzene, C 6 H 6  napthalene, C 10 H 8 5.5

80.6

 benzoic acid, C 6 H 5 CO 2 H 122.4

101

Bonding in Solids

Variations in Melting Points for Covalent Solids Substance  sand, SiO 2  carborundum, SiC Melting Point ( o C) 1713 ~2700   diamond graphite >3550 3652-3697 102

Bonding in Solids

Variations in Melting Points for Ionic Solids Compound  LiF  LiCl  LiBr  LiI     CaF 2 CaCl 2 CaBr 2 CaI 2 Melting Point ( o C) 842 614 547 450 1360 772 730 740 103

Bonding in Solids

Variations in Melting Points for Metallic Solids Metal       Na Pb Al Cu Fe W Melting Point ( o C) 98 328 660 1083 1535 3410 104

Band Theory of Metals

 Sodium’s 3s orbitals can interact to produce overlapping orbitals  Electrical conductivity is due to the ability of any highest energy e-s in the “3s” band to jump to a slightly higher-energy vacant orbital in the same band 105

Band Theory of Metals

 The 3s orbitals can also overlap with unfilled 3p orbitals   This type of overlap becomes more important with Group II metals e.g. Mg’s 3s is filled with 2 e-s, thus without this overlap, the “3s” band in a Mg crystal would be filled 106

Band Theory of Metals

 Insulators have a large gap between the s and p bands.  Gap is called the

forbidden zone

.

 Semiconductors have a small gap between the bands.

107

Band Theory of Metals

 Metals are also good conductors of heat  They can absorb heat as e-s become thermally excited to low-lying vacant orbitals in the conduction band  The reverse process accompanies the release of heat.

 Metals have a lustrous appearance  Mobile e-s can absorb a wide range of wavelengths of radiant energy as they jump to higher energy levels  They emit photons of light and fall back to lower levels with the conduction band 108

 Metals are also malleable or ductile (a) In a metal, the +ve charged metal ions are immersed in a delocalized sea of e s. When the metal is distorted, the environment around the metal is essentially unchanged; there are no new repulsive forces (b) By contrast, when an ionic crystal is subjected to a force that causes it to slip along a plane, the increased repulsive forces b/w like charges causes the crystal to break along a plane.

Fig. 13-36, p. 490