intermolecular forces: liquids and solids

Intermolecular Forces: Liquids and Solids

● Phases and Phase Diagrams● Liquids and Liquid Properties● Intermolecular Forces● Heating Curves● Introduction to Solids● Cubic Packing Arrangements● Closest-Packed Structures● Density of a Crystalline Solid● Ionic Solids and Interstitial Sites● The Born-Haber Cycle

Phases and Phase Diagrams

condensation

vaporization

depositionsublimation

fusion

freezing

incompressible

d ≈ 1 – 10 g mL−1

very compressible

d ≈ 1 – 10 g L−1 at SATP


A few definitions:

STP vs SATP

STP = Standard Temperature and Pressure (0 oC, 100 kPa)

SATP = Standard Ambient Temperature and Pressure (25 oC, 100 kPa)

100 kPa = 1 bar

Phases and Phase DiagramsDifferences between the different states of Matter:

Gas: Molecules move randomly and the intermolecular separations are large (i.e. most of a gas is empty space).

Liquids and Solids: the molecular motions are quite restricted and the intermolecular separations are small.

Solids: The molecules are often, but not always, arranged in regular, repeating patterns.

Substances exist in different phases and phase changes occur because molecules exert forces on each other. (Without intermolecular forces, all substances would behave as ideal gases!!) It takes energy to overcome the attractive intermolecular forces that cause molecules to aggregate. Therefore, sublimation, fusion and evaporation are all endothermic processes

A given substance will exist as a solid, liquid or gas depending on the temperature and pressure of the sample. A phase diagram shows the stable phases at each temperature and pressure.

Phases and Phase DiagramsPhase diagram of I2

Phases and Phase DiagramsTake note of the following points:1. Solid is the most stable phase at low T and high P. Gas is the stable phase at high T

and low P.

Phases and Phase DiagramsTake note of the following points:2. The S-L line shows the T’s and P’s at which both solid and liquid are stable and can coexist. It

also shows us how the melting temperature changes with pressure.

Phases and Phase DiagramsTake note of the following points:3. The L-G line curve shows the T’s and P’s at which both liquid and gas are stable and can

coexist. It also shows us how the boiling temperature changes with pressure.

Phases and Phase DiagramsTake note of the following points:4. For most substances, the S-L line has a positive slope, but for a few substances (most notably,

water but also bismuth and antimony), it has a negative slope!

For most substances



For water

Phases and Phase DiagramsTake note of the following points:5. At the triple point, all three phases are stable and coexist.

Phases and Phase DiagramsTake note of the following points:6. The G-L line ends abruptly at the critical point (Tc, Pc)

Phases and Phase DiagramsPhase diagram of I2

What is the phase of I2 at 25 oC and 1 atm? We are dealing with a solid.

25 oC

25 oC and 1 atm

Phases and Phase DiagramsPhase diagram of CO2

What is the phase of CO2 at 25 oC and 1 atm? We are dealing with a gas.

25 oC

25 oC and 1 atm1 atm

Phases and Phase DiagramsPhase diagram of H2O

What is the phase of H2O at 25 oC and 1 atm? We are dealing with a liquid.

25 oC

25 oC and 1 atm

1 atm



The slope of the S-L line is negative!

Phases and Phase DiagramsTake note of the following points:

1 m

1 m

10 m

A column of water 1 m ×1 m × 10 m occupies a volume of 10 m3 or 10,000 L. 1 L of water weighs 1 kg. 10,000 L of water weigh 10,000 kg.

The pressure exerted by 10,000 kg of water equals:

9.8 m2/s×10,000 kg/(1 m×1 m) 105 Pa 1 atm

10 m of water generates a 1 atm additional pressure.

Phases and Phase DiagramsTake note of the following points:We find liquid water at the bottom of the ocean.

The slope of the S-L line is negative!

Phases and Phase DiagramsTake note of the following points:Polymorphism: The existence of a solid substance in more than one form.

Other forms of ice obtained at several

thousands of atmospheres

Phases and Phase DiagramsTo summarize, a typical phase diagram looks like this:

T

supercriticalfluid

PLiquid

Gas

Solid

Tc

Pc

TvapTfus

1 atmA

B

solid-liquid coexistence line

liquid-vapour coexistence line

triple point

“normal” boiling point

“normal” melting point

critical point

Phases and Phase DiagramsSupercritical fluid:

T

supercritical

fluidP

(L)

(G)

(S)

Tc

Pc

A

B

solid-liquid coexistence

line


critical point

As one moves from A to B, the pressure increases and the density of the gas increases until it equals the density of the liquid. At this point, gas and liquid are indistinguishable, the interface between liquid and gas vanishes and we have a supercritical fluid.

If a gas is at T > TC (Point A in diagram), increasing the pressure of the gas does not yield a liquid but rather a supercritical fluid (Point B).

To take a gas at T > TC and transform it into a liquid, the temperature must first be reduced below TC. Then the pressure is increased to pass the liquid-vapor coexistence curve.


T

supercritical

fluidP

(L)

(G)

(S)

Tc

Pc

A

B


line


critical point

The phase boundary between liquid benzene and its vapour disappears at Tc.

Below Tc, the phase boundary is clearly visible.

Just below Tc, the phase boundary is barely visible.

At Tc, the phase boundary disappears.

T in

crea

ses

from

bel

ow T

C to

abo

ve T

C


Did you know? Supercritical CO2 is used to extract caffeine from coffee beans. The extracted caffeine can be sold to pharmaceutical or beverage companies. The critical point for CO2 is fairly low (Tc = 31 oC) and so, supercritical CO2 can be used at ambient temperatures without causing decomposition or “denaturing” of other compounds. Because it has low toxicity, a low critical temperature and is nonflammable, supercritical CO2 is becoming an increasingly important industrial and commercial solvent.

T

supercritical

fluidP

(L)

(G)

(S)

Tc

Pc



critical point

Phases and Phase DiagramsExamples: For a particular substance, the S-L coexistence curve has a negative slope.

a) What phase changes are possible if the pressure is increased at constant temperature T? Assume that T is less than Ttp, where Ttp is the triple point temperature.

Gas deposition solid melting liquid

T

supercritical

fluidP

(L)

(G)

(S)

Tc

Pc



critical point

TT

PT


b) What phase changes are possible if the pressure is increased at constant temperature T, assuming Ttp < T < Tc, where Ttp and Tc are the triple point and critical point temperatures, respectively.

Gas condensation liquid

T

supercritical

fluidP

(L)

(G)

(S)

Tc

Pc



critical point

TT

PT


c) True or False? The melting temperature increases as the pressure increases.

False

T

supercritical

fluidP

(L)

(G)

(S)

Tc

Pc



critical point

TT

PT

Tm


d) True or False? The solid is more dense than the liquid.

At a given temperature, when we increase the pressure, the density increases and the solid becomes a liquid. False

T

supercritical

fluidP

(L)

(G)

(S)

Tc

Pc



critical point

TT

PT


a) What is the phase of this substance at 25oC and 73 atm?

We are dealing with a liquid.

25 oC

25 oC and 73 atm73 atm

Examples: For a particular substance, the triple point is at 57 ºC and 5.1 atm, and the critical point is at 31oC and 73 atm.


b) What phase changes occur if the pressure is decreased from 75 atm to 0.001 atm at −60 oC? Assume that the solid-liquid line has a positive slope.

Solid sublimation gas

-60 oC

-60 oC and 75 atm

Examples: For a particular substance, the triple point is at 57 ºC and 5.1 atm, and the critical point is at 31oC and 73 atm.

-60 oC and 0.001 atm


condensation

vaporization

depositionsublimation

fusion

freezing

incompressible

d ≈ 1 – 10 g mL−1

very compressible

d ≈ 1 – 10 g L−1 at SATP

REVIEW


T

supercriticalfluid

P

Liquid

Gas

Solid

REVIEW


T

supercriticalfluid

PLiquid

Gas

Solid

Tc

Pc

TvapTfus

1 atmA

B



triple point

“normal” boiling point

“normal” melting point

critical point

REVIEW

Phases and Phase DiagramsExamples 12-45: Which substances listed in the table can exist as liquids at room temperature (~ 20 oC)?

Substance Tc, K Pc, atm

H2 33.3 12.8

N2 126.2 33.5

O2 154.8 50.1

CH4 191.1 45.8

CO2 304.2 72.9

HCl 324.6 82.1

NH3 405.7 112.5

SO2 431.0 77.7

H2O 647.3 218.3 T

supercritical

fluidP

(L)

(G)

(S)

Tc

Pc


line


critical point

T = 20 oC



H2 33.3 12.8

N2 126.2 33.5

O2 154.8 50.1

CH4 191.1 45.8

CO2 304.2 72.9

HCl 324.6 82.1

NH3 405.7 112.5

SO2 431.0 77.7

H2O 647.3 218.3

supercritical

fluidP

(L)

(G)

(S)

Tc = 240 oC

Pc


line


critical point

T = 20 oC



H2 33.3 12.8

N2 126.2 33.5

O2 154.8 50.1

CH4 191.1 45.8

CO2 304.2 72.9

HCl 324.6 82.1

NH3 405.7 112.5

SO2 431.0 77.7

H2O 647.3 218.3 T

supercritical

fluidP

(L)

(G)

(S)

Tc

Pc


line


critical point

T = 20 oC

20 oC < TC or TC > 293.15 K

Gases that can be liquified at room temperature are said to be “non-permanent gases”.Gases that cannot be liquified at room temperature are said to be “permanent gases”.

Phases and Phase DiagramsExamples 12-51: Phase diagram of phosphorousa) Indicate the phases present in the regions labeled with a question mark?

T

P

(L)

(G)

(S)

590 oC

43 atm

?

?

Phases and Phase DiagramsExamples 12-51: Phase diagram of phosphorousb) A sample of solid red phosphorous cannot be melted by heating in a container open to the atmosphere. Explain why this is so?

T

P

(L)

(G)

(S)

590 oC

43 atm

1 atm

Solid phosphorous can only be sublimed (S G) if it is heated at P = 1 atm.

Phases and Phase DiagramsExamples 12-51: Phase diagram of phosphorousc) Trace the phase changes that occur when the pressure on a sample is reduced from Point A to B, at constant temperature.

T

P

(L)

(G)

(S)

590 oC

43 atm

Solid condensation Liquid vaporization Gas

A

B

Liquids and Liquid Properties

T

supercritical

fluidP

(L)

(G)

(S)



critical point

We know that if the temperature of a gas is lowered sufficiently, the gas will condense to a liquid. Why is this? As T is lowered, the average kinetic energy of the molecules decreases. At some point, the molecules will no longer have enough kinetic energy to overcome the attractive forces that draw the molecules together. Consequently, the molecules cluster together to form a liquid.

Condensation


T

supercritical

fluidP

(L)

(G)

(S)



critical point

The freezing of a liquid can be explained in the same way: If the temperature of a liquid is lowered sufficiently, the molecules will not have enough kinetic energy to overcome attractive forces that draw the molecules closer together

the liquid freezes.

Freezing

Liquids and Liquid PropertiesThe physical properties of a liquid depend on the strength and nature of the intermolecular forces. We shall examine why the following physical properties are different from substance to substance.

vapour pressure = equilibrium pressure of vapour that forms above a liquid in a closed container

normal boiling point (Tvap) = temperature at which the vapour pressure of the liquid equals 1 atm

surface tension (g) = energy required to increase the surface area of a liquid

viscosity (η) provides a measure of a fluid’s resistance to flow; the speed of flow through a tube is inversely proportional to the viscosity

In general, the stronger the intermolecular attractions, the higher the boiling point, the greater the surface tension, the higher the viscosity and the lower the vapour pressure.

Liquids and Liquid Propertiesvapour pressure = equilibrium pressure of vapour that

forms above a liquid in a closed container

to vacuum

liquid

air



to vacuum

liquid

Liquid N2

air



to vacuum

solid

Liquid N2

air



to vacuum

solid

Liquid N2

vacuum



to vacuum

solid

vacuum



to vacuum

liquid

vacuum



to vacuum

liquid


T

supercritical

fluidP

(L)

(G)

(S)



critical point

The L-G line shows us how

1. the vapour pressure of a liquid changes with temperature

2. the boiling temperature of a liquid changes with pressure

Clausius-Clapeyron equation


T

supercritical

fluidP

(L)

(G)

(S)



critical point

Along the L-G line, both liquid and gas co-exist.At equilibrium, the rate of evaporation equals the rate of condensation


Liquid X(l)

Vapour X(g)

= standard enthalpy of vaporizationR = 8.3145 J K−1 mol−1

Liquids and Liquid PropertiesClausius-Clapeyron equation

HP

P R T T

ovap2

1 2 1

1 1ln

Hovap

The variation of vapour pressure with temperature is modeled reasonably well by the Clausius-Clapeyron equation:

The quantities appearing in this equation are described below.

P2 = vapour pressure at temperature T2

P1 = vapour pressure at temperature T1

Extremely important: Pay attention to the units!


T

supercritical

fluidP

(L)

(G)

(S)



critical point


(P1, T1)

(P2, T2)

HP

P R T T

ovap2

1 2 1

1 1ln


T

supercritical

fluidP

(L)

(G)

(S)



critical point


(P1, T1)

(P2, T2)

Example: a) If the vapour pressure of P4(l) is 10 Torr at 128 oC and 400 Torr at 251 oC, then what

is vapHo?

121

2 11TTR

HPPLn

ovap

12

1

2

11TT

PPRLn

H ovap

molkJLn

H ovap /4.52

15.2732511

15.2731281

400103145.8

Liquids and Liquid PropertiesClausius-Clapeyron equationExample: b) What is the normal boiling point of P4(l)?

121

2 11TTR

HPPLn

ovap

121

2 11TTP

PLnHR

ovap

1

2

12

11PPLn

HR

TT ovap

1

2

1

2 11

PPLn

HR

T

T

ovap

KLn

T 8.553

10760

400,523145.8

15.2731281

12

P1 = 10 Torr T1 = 128 oC P2 = 400 Torr T2 = 251 oCvapHo = 52.4 kJ/mol

Liquids and Liquid PropertiesClausius-Clapeyron equationExample: c) What is the vapour pressure at 200 oC?

121

2 11TTR

HPPLn

ovap

121

2 11expTTR

HPP o

vap

1212

11expTTR

HPP

ovap

torrP 10915.273128

115.273200

13145.8400,52exp102

P1 = 10 Torr T1 = 128 oC P2 = 400 Torr T2 = 251 oCvapHo = 52.4 kJ/mol







REVIEW

Intermolecular Forces

Dipole moments and polarizabilities of a few molecules. (Based on data from Physical Chemistry, 6th Edition, by P. Atkins, published by Freeman, 1998).

μ (in

debye)α/αHe

H2 0 4.1

HF 1.91 2.6

HCl 1.08 13

HBr 0.80 18

CO 0.12 10

CO2 0 13

H2O 1.85 7.5

NH3 1.47 11

He 0 1

Ar 0 8.4

CH4 0 13

CCl4 0 53

The purpose of this section is to understand how molecules interact with each other and how these interactions help us to understand trends in the physical properties of compounds (e.g. boiling points, vapour pressures, viscosity and surface tensions of liquids; densities and melting points of solids; deviations from ideal gas behaviour, etc.)

Generally speaking, differences in the physical properties of substances can often be rationalized by considering how molecules interact with each other at the molecular level. We’ll look at some of the types of intermolecular forces that act between pairs of molecules.

The molecules are not organized into perfect straight line because the molecules are in motion (i.e. each one possesses kinetic energy) and they “jiggle” out of perfect alignment.

Intermolecular Forces

μA μB

d

A. Dipole-dipole forces

Digging DeeperFor a pair of interacting polar molecules with dipole moments of μA and μB:

Some molecules possess a permanent dipole moment, μ, because the bond dipoles do not cancel out. Such molecules are said to be “polar” because one end of the molecule is slightly positive and the other end is slightly negative. The charge distribution of a polar molecule can be represented by an arrow that points from the positive end to the negative end. The dipole moments of a few molecules were given in the previous table.Polar molecules (i.e. dipoles) tend to orient themselves in a “head-to-tail” manner, as shown below:

6

22

dEnergyPotential BA

Intermolecular ForcesA. Dipole-dipole forcesExample: Iodine chloride, ICl, and bromine, Br2, have exactly the same number of electrons, and it is reasonable to assume that these molecules are essentially the same size. Yet the boiling points of ICl(l) and Br2(l) are quite different, 97 oC and 59 oC, respectively. Use your knowledge of dipole-dipole interactions to explain why ICl(l) has a higher boiling point than Br2(l).

Intermolecular ForcesA. Dipole-dipole forcesExample: Iodine chloride, ICl, and bromine, Br2, have exactly the same number of electrons, and it is reasonable to assume that these molecules are essentially the same size. Yet the boiling points of ICl(l) and Br2(l) are quite different, 97 oC and 59 oC, respectively. Use your knowledge of dipole-dipole interactions to explain why ICl(l) has a higher boiling point than Br2(l).

Chlorine is more electronegative than bromine. Consequently, I-Cl shows a dipole moment:

(d)ICl(d)

Br2 on the other hand cannot generate a dipole moment because the molecule is made of two identical bromine atoms.

Intermolecular ForcesB. London dispersion forces

The electrons in a molecule are in constant motion and at any particular instant, there may be an asymmetric distribution of electrons in the molecule (i.e. with a greater number of electrons at one end than at the other end). The asymmetric distribution of electrons gives rise to an instantaneous and temporary dipole moment (inst ). The formation of inst in a molecule causes (or induces) the formation of a dipole in neighbouring molecules. The induced dipole moment is ind . (Notice the head-to-tail arrangement of the instantaneous and induced dipole moments.)

----

-----

---

---

----

-----

---

---

Molecule A Molecule B

There is an attraction between inst and ind . The strength of the interaction increases as the “polarizabilities” of the molecules increase.

inst ind


The charge cloud of a large molecule is diffuse and easily polarized. The charge cloud of a small, compact molecule is not easily polarized. The polarizabilities of a few molecules were given earlier. Note that the larger the molecule, the larger the polarizability.

Polarizability(α)

provides a measure of the extent to which the charge cloud of a molecule can be distorted (i.e. polarized) by another molecule.

Digging DeeperFor a pair of molecules with polarizabilities of αA and αB:

αA αB

d 6dEnergyPotential BA


Remarks:London dispersion forces are most attractive when the molecules are large because large molecules have larger, more diffuse (i.e. more polarizable) charge clouds.London dispersion forces always contribute to the molecular interactions because all molecules have charge clouds and are therefore polarizable to some extent.

----

-----

---

---

----

-----

---

---


inst ind


The C-Cl bonds are very polar, but the bond dipoles cancel. The CCl4 molecule is nonpolar.

C

Cl Cl

Cl Cl

Example: Methane (CH4) and carbon tetrachloride (CCl4) are both nonpolar molecules. Use your knowledge of London dispersion forces to explain why the boiling point of CCl4(l) is much higher than that of CH4(l).

μ (in

debye)α/αHe

H2 0 4.1

HF 1.91 2.6

HCl 1.08 13

HBr 0.80 18

CO 0.12 10

CO2 0 13

H2O 1.85 7.5

NH3 1.47 11

He 0 1

Ar 0 8.4

CH4 0 13

CCl4 0 53

Carbon tetrachloride (CCl4) is a molecule that is much larger than methane (CH4). Consequently, CCl4 contains many more electrons which can generate an instantaneous dipole moment more readily than CH4. These dipole moments generate strong intermolecular forces between CCl4 molecules which are more difficult to break than in CH4. Consequently, Tb of CCl4 (78oC) is larger than Tb of CH4 (162 oC).

Intermolecular ForcesC. Hydrogen bonding forcesA special type of bond forms between molecules when the molecules contain a hydrogen atom bonded to N, O, or F. When H is bonded to N, O or F, the H atom carries a significant positive charge and it is strongly attracted to a lone pair on another molecule! When a hydrogen atom which is covalently bonded to one atom is simultaneously attracted to the lone pair on another atom, it is “bridging” two molecules, as shown below. Such a bond is called a hydrogen bond.

An intermolecular hydrogen bond “bridges” two molecules.

An intramolecular hydrogen bond bridges two parts of the same molecule.

Did you know? “inter” means “between” and “intra” means “within”.

XH

Y

intermolecular hydrogen bond

Intermolecular ForcesC. Hydrogen bonding forcesNote carefully:H is covalently bonded to X but is simultaneously attracted to a lone pair of electrons on Y. Both X and Y must be N, O or FHydrogen bonds are the strongest type of intermolecular force (but they are still weak in comparison to covalent and ionic bonding forces)

XH

Y

intermolecular hydrogen bond

dipole-dipole & LDFs

H bonds

covalent & ionic bonds

0.10-10 kJ mol−1

10−40kJ mol−1

100’s or 1000’s kJ mol−1

intermolecular forces chemical bonding forces

Intermolecular ForcesC. Hydrogen bonding forcesHydrogen bonds are important!H bonds between H2O’s in ice give the solid an open structureH bonds between H2O’s in water give the liquid a high BP, high surface tension and a large heat capacity.

Hydrogen bonding in water. This is Figure 12-7 of Petrucci 10e. Used with permission.

Intermolecular ForcesC. Hydrogen bonding forces

Hydrogen bonds are important!H bonds are especially important in biology (e.g. H bonds keep the two helices of DNA together; the structures and functions of proteins and enzymes are determined by H bonds)

The helical structures of proteins (above) and DNA (on the right) are stabilized by hydrogen bonds. These are Figures 28-12 and 28-26 of Petrucci 10e. Used with permission.

Intermolecular ForcesC. Hydrogen bonding forcesHydrogen bonds are important!H bonds are especially important in biology (e.g. H bonds keep the two helices of DNA together; the structures and functions of proteins and enzymes are determined by H bonds)

Intermolecular ForcesA. Dipole-dipole forces

Polar molecules with a dipole moment .

B. London dispersion forces

The formation of inst in a molecule causes (or induces) the formation of a dipole in neighbouring molecules.

C. Hydrogen bonding forces

A special type of bond forms between molecules when the molecules contain a hydrogen atom bonded to N, O, or F.

XH

Yintermolecular hydrogen bond

----

-----

---

---

----

-----

---

---


inst ind

REVIEW

Intermolecular ForcesC. Hydrogen bonding forcesA special type of bond forms between molecules when the molecules contain a hydrogen atom bonded to N, O, or F. When H is bonded to N, O or F, the H atom carries a significant positive charge and it is strongly attracted to a lone pair on another molecule!

XH

Yintermolecular hydrogen bond

Electronegativity Scale


C C

H

ClCl

H

C C

H

ClH

Clvs

(Z)-1,2-dichloroethene, also called cis-1,2-dichloroethene

(E)-1,2-dichloroethene, also called trans-1,2-dichloroethene

Example: Dichloroethene, C2H2Cl2, has several isomeric forms. Use your knowledge of intermolecular forces to predict whether (Z)-1,2-dichloroethene or (E)-1,2-dichloroethene has the higher boiling point. Lewis structures are given below.

The chlorine atom being more electronegative than carbon induces a dipole moment in the C-Cl bond. However, those dipole moments are opposite in (E)-1,2-dichloroethane and they cancel each other. The dipole moments of the C-Cl bond do not cancel each other in (Z)-1,2-dichloroethane. Thus the Z-isomer has a permanent dipole moment which induces strong intermolecular forces. The Z-isomer has a higher boiling point.

Tb = 60 oC Tb = 48 oC

A word of warning!Don’t over generalize the results of this example. You might end up making the wrong prediction!

Intermolecular ForcesC. Hydrogen bonding forcesExample: Consider the trans and cis isomers of C4H4O4. Which one has the higher melting point?

Fumaric acid generates intermolecular H-bonds which leads to the formation of a network where all the molecules are associated with another. Maleic acid forms intramolecular H-bonds and the intermolecular forces between molecules are weaker. Consequently, maleic acid has a lower melting point than fumaric acid.

Tm = 300 oC Tm = 140 oC

HOOH

O

O

Fumaric acid (trans)

O

OH

O

OH

Maleic acid (cis)

Intermolecular ForcesC. Hydrogen bonding forcesExample: Use your knowledge of intermolecular forces to predict whether CH3COCH3(l) or CH3CH2CH2OH(l) has the higher boiling point.

Acetone and 1-propanol have a similar number of C- and O-atoms. They both induce a dipole moment since O is more electronegative than C. However, 1-propanol can form H-bonds and acetone cannot. Consequently, acetone boils at a lower temperature than 1-propanol.

H3C CH3

OO

H

acetone 1-propanol

Tb = 56 oC Tb = 97 oC

Intermolecular ForcesC. Hydrogen bonding forcesExample: Which one of the liquids, HO-CH2CH2-OH or CH3CH2OH, has the highest vapour pressure at room temperature?

A single ethylene glycol (EG) can form more H-bonds with other EG molecules than ethanol can form with other ethanol molecules. Consequently, EG boils at a higher temperature. EG has a lower vapor pressure than ethanol.

Ethylene glycol Ethanol

Tb = 196 oC Tb = 78 oCOO

HH O

H


Example: Use your knowledge of intermolecular forces to predict the order of boiling points for H2O, H2S (hydrogen sulfide), H2Se (hydrogene selenide), and H2Te (hydrogen telluride).

All these molecules are bent, and as such, generate a dipole moment. HX

H


Example: Use your knowledge of intermolecular forces to predict the order of boiling points for H2O, H2S, H2Se and H2Te.

All these molecules are bent, and as such, generate a dipole moment.

Tb (oC) = 0.6055xM (g/mol) - 83.785

-80

-60

-40

-20

0

20

40

60

80

100

120

0 20 40 60 80 100 120 140 160

Tb,

o C

Mass, g/mol

H2STb = 60 oC

H2SeTb = 41 oC

H2TeTb = 2 oCH2O

Tb = 60 oC

Tb

-59.6 oC-41.3 oC

-2 oC

HX

H



All these molecules are bent, and as such, generate a dipole moment.

Tb (oC) = 0.6055xM (g/mol) - 83.785

-80

-60

-40

-20

0

20

40

60

80

100

120

0 20 40 60 80 100 120 140 160

Tb,

o C

Mass, g/mol

H2STb = 60 oC

H2SeTb = 41 oC

H2TeTb = 2 oC

H2OTb = 100 oC

HX

H

Tb

+100 oC-59.6 oC-41.3 oC

-2 oC


Tb (oC) = 0.6055xM (g/mol) - 83.785

-80

-60

-40

-20

0

20

40

60

80

100

120

0 20 40 60 80 100 120 140 160

Tb,

o C

Mass, g/mol

H2STb = 60 oC

H2SeTb = 41 oC

H2TeTb = 2 oC

H2OTb = 100 oC


By increasing the mass and the size of the molecule H2X, one increases the polarizability, and thus the strength of the interactions between the H2X molecules. The exception is H2O because water forms H-bonds: H2S < H2Se < H2Te < H2O

Tb

+100 oC-59.6 oC-41.3 oC

-2 oC

Intermolecular ForcesC. Hydrogen bonding forcesExample: Use your knowledge of intermolecular forces to predict the order of boiling points for H2O, H2S, H2Se and H2Te.

It should be noted that similar arguments can be applied to the hydrides of the group 15 and group 17 elements. Thus, in order of increasing BP, we have:

PH3 < AsH3 < SbH3 < NH3 and HCl < HBr < HI < HF

Heating CurvesA heating curve shows us how the temperature varies with the amount of heat added. Consider heating a sample of ice from ti = –10 oC to t = 150 oC at constant pressure.

Q

T

ti

tfus

tvap

melting solid

boiling liquid

Hfus Hvap

warming liquid

(slope2)warming solid

(slope1)

warming vapour(slope3)

qliqqsol

tf

qgas

Review this section on your own.

Introduction to SolidsSolids can be classified as crystalline or as amorphous, depending on whether we have regular or irregular packing of the atoms, molecules or ions that make up the solid. Crystalline and amorphous solids have rather different physical characteristics.

Crystalline solids

• regular repeating patterns• “sharp” melting points

Amorphous solids

• irregular packing• melt over a temperature range

Examples of amorphous solids include rubber, polystyrene, window glass, candle wax, and cotton candy. We can get irregular packing (amorphous solids) if there are impurities in the sample when the liquid freezes, or if the molecules are large or have flexible structures. For molecules that are large or very flexible (i.e. polymers!), it is statistically most probable that the molecules will be irregularly packed when the liquid freezes.

We shall focus exclusively on crystalline solids.

Introduction to SolidsFor crystalline solids, we use the following concepts to characterize the pattern of the packing arrangement.

three dimensional array of points that shows how the structural units are arranged in space.

Crystal lattice

Each structural unit could be a single atom, a molecule or an ion. In C(s), the structural unit is a carbon atom. In I2(s), the structural unit is an I2 molecule. In NH4NO3(s), the structural units are NH4

+ and NO3− ions.

the smallest building block that possesses the symmetry of the crystal lattice; a solid sample of any size can be “built” by stacking together unit cells.

Unit cell

Introduction to SolidsCrystalline solids are classified according to the nature of the bonding, or according to the geometry and symmetry of the packing arrangement. Let’s focus first on the nature of the bonding.

Crystalline solids

ionic Network covalent

molecular metallic

In an ionic solid, positive and

negative ions are held in their lattice

positions by (strong) ionic bonding

forces.

NaCl Tm = 800 oCLiF Tm = 848 oCZnO Tm = 1974 oC

In a network covalent solid, atoms are held

in their lattice positions by (strong)

covalent bonds.

SiO2 Tm= 1600 oCSiC Tm =2830 oCCdiamond Tm = 4440 oC

(at P = 12.4 GPa)

In a molecular solid, molecules are held

in their lattice positions by (weak)

intermolecular forces.

H2O Tm = 0 oC

S8 Tm = 113 oC

I2 Tm = 114 oC

In a metallic solid, metal cations are held in their lattice

positions by (strong) metallic

bonding forces.

Cu Tm = 1083 oCAg Tm = 962 oC

W Tm = 3422 oC

Introduction to SolidsThe Periodic Table can help understand the differences in Tm between NaCl, LiF, and ZnO. NaCl Tm = 800 oC

LiF Tm = 848 oCZnO Tm = 1974 oC

241

CA

CA

o rqqForceticElectrosta

Introduction to SolidsCrystalline solids are classified according to the nature of the bonding, or according to the geometry and symmetry of the packing arrangement. Let’s focus first on the nature of the bonding.

Crystalline solids

ionic Network covalent

molecular metallic

In an ionic solid, positive and

negative ions are held in their lattice

positions by (strong) ionic bonding

forces.

NaCl Tm = 800 oCLiF Tm = 848 oCZnO Tm = 1974 oC

In a network covalent solid, atoms are held

in their lattice positions by (strong)

covalent bonds.

SiO2 Tm= 1600 oCSiC Tm =2830 oCCdiamond Tm = 4440 oC

(at P = 12.4 GPa)

In a molecular solid, molecules are held

in their lattice positions by (weak)

intermolecular forces.

H2O Tm = 0 oC

S8 Tm = 113 oC

I2 Tm = 114 oC

In a metallic solid, metal cations are held in their lattice

positions by (strong) metallic

bonding forces.

Cu Tm = 1083 oCAg Tm = 962 oC

W Tm = 3422 oC

Introduction to SolidsMetallic solids can be viewed as an array of metal cations bathing in a sea of valence electrons.

+ + + + +

+ + + + +

+ + + + + A “sea” of delocalized

valence electrons

Electron sea model of metallic

bonding

Introduction to SolidsWhen we focus on the geometry of the packing arrangements, we find that there are seven basic shapes for unit cells. The shape of each unit cell is described in terms of three lengths (a, b and c) and three angles (, , and g). We will focus primarily on cubic unit cells. The others are mentioned only to emphasize that there are other types/shapes of unit cells besides cubic unit cells.

cubica = b = c = = g = 90otrigonal

a = b = c = = g ≠ 90otetragonal

a = b ≠ c = = g = 90o

hexagonala = b ≠ c = = 90o; and g = 120o

monoclinica ≠ b ≠ ca= g = 90o

and ≠ 90otriclinica ≠ b ≠ c ≠ ≠ g ≠ 90o orthorhombic

a ≠ b ≠ c = = g = 90o

Crystalline solids

a b

c

g




a = b ≠ c = = g = 90o




a ≠ b ≠ c = = g = 90o

Crystalline solids

a

a

a



Crystal lattice


+ and NO3− ions.


Unit cell

REVIEW

Cubic Packing ArrangementsTo investigate the cubic packing arrangements, we are going to imagine packing together identical, hard spheres each having a radius R.

A. Simple cubic packing Let’s consider ”building” a simple cubic packing arrangement on top of a table. We generate the first layer by arranging the spheres as shown (on the left) below. Subsequent layers are added by lining up spheres with those in the previous layer as shown (on the right).

Layer 1

Layer 1

Layer 2

Layer 3

Cubic Packing ArrangementsA. Simple cubic packing

(i) What is the relationship between the edge length of the unit cell (a) and the atom’s radius (R)?

a

a

a

R

a = 2×R


(ii) What is the number of spheres contained in one unit cell?

a

a

a

R

Each cell contains eight one eighth of a sphere. Thus there are:

1818 cellunitperspheresofNumber


(iii) What is the coordination number of a sphere?

a

a

a

R

Each sphere touches four spheres in the plane, one on top, and one underneath. Thus each sphere touches 6 other spheres. The coordination number equals 6.


(iv) What is the fraction of empty space in a simple cubic crystal?

a

a

a

R

The unit cell occupies a volume VUC = (a)3 = (2×R)3 = 8×R3.

The unit cell contains one sphere of volume:

The fraction of empty volume = (VUC – Vsphere)/VUC = 1 – /6 = 0.476

3

34 RVsphere

Cubic Packing ArrangementsB. Body-centred cubic packing(i) What is the relationship between the edge length of the unit cell (a) and the atom’s

radius (R)?

Since the atoms are touching along the cube’s diagonal (body diagonal), its length equals:

R + 2×R + R = 4×R

R

a


radius (R)?

aBody Diagonal = 4×R

4×Ra

a

What is the square diagonal?


radius (R)?


(Square Diagonal)2 = a2 + a2 = 2×a2

a

a a

a 2Square Diagonal

a2

4×R


radius (R)?


4×R


a 2Square Diagonal

(Body Diagonal)2 = a2 + (2×a)2 = 3×a2

Body Diagonal = 3×a = 4×R

aR 43 Ra

34

or

aa2

Cubic Packing ArrangementsB. Body-centred cubic packing(ii) What is the number of spheres contained in one unit cell?

Each corner is occupied by one eighth of an atom and there is a full atom at the center of the unit cell. Thus the cell contains:

R

a

2118181 cellunitperatomsofNumber

Cubic Packing ArrangementsB. Body-centred cubic packing(iii) What is the coordination number of a sphere?

The atom at the center of the unit cell touches 8 other atoms. The coordination number equals 8.

R

a

Cubic Packing ArrangementsB. Body-centred cubic packing(iv) What is the fraction of empty space in a body centered cubic crystal?

R

a

The unit cell occupies a volume VUC = (a)3 =

The unit cell contains two spheres, each sphere of volume:

The fraction of empty volume = (VUC – 2×Vsphere)/VUC =

3

34 RVsphere

33

3364

34 RR

319.0831

Cubic Packing ArrangementsC. Face-centred cubic packing(i) What is the relationship between the edge length of the unit cell (a) and the atom’s

radius (R)?

R

a

Cubic Packing ArrangementsC. Face-centred cubic packing(i) What is the relationship between the edge length of the unit cell (a) and the atom’s

radius (R)?

a

a

a


Square Diagonal = 2×a = 4×R

Ra 22 or aR 22

1

Cubic Packing ArrangementsC. Face-centred cubic packing(ii) What is the number of spheres contained in one unit cell?

R

a

Each one of the eight corners is occupied by one eighth of an atom and each one of the 6 faces is occupied by one half of an atom.

413818

216 cellunitperatomsofNumber

Cubic Packing ArrangementsC. Face-centred cubic packing(iii) What is the coordination number of a sphere?

The atom at the center of the face touches 4 atoms in the plane shown on the figure.


The atom at the center of the face touches 4 more atoms in the plane shown on the figure.


In total, the atom at the center of the face touches 12 atoms. The coordination number equals 12.

Cubic Packing ArrangementsC. Face-centred cubic packing(iv) What is the fraction of empty space in a face-centered cubic crystal?

R

a

The unit cell occupies a volume VUC = (a)3 =

The unit cell contains four spheres, each sphere of volume:

The fraction of empty volume = (VUC – 4×Vsphere)/VUC =

3

34 RVsphere

3321622 RR

260.023

1

Closest Packed Structures

Same layer as the layer with the grey balls.

Different layer from the layer with the grey balls.

Closest Packed Structures“x” type dimple (vertically

aligned with a sphere in layer 1)“y” type dimple (not

vertically aligned with a sphere in layer 1 or layer 2)

There are two types of “dimples” to choose from

when forming layer 3!!


choose “x” only choose “y” only

Spheres in layer 3 are vertically aligned with those in layer 1. That

is, layer 3 is a repeat of layer 1. ABAB ... closest-packing

Spheres in layer 3 are not aligned vertically with those in layer 1 or

layer 2. Layer 3 is a distinct layer. ABCABC ... closest-packing

“x” type dimple (vertically aligned with a sphere in layer 1)

“y” type dimple (not vertically aligned with a

sphere in layer 1 or layer 2)

Spheres in layer 3 are vertically aligned with those in layer 1. That

is, layer 3 is a repeat of layer 1. ABAB ... closest-packing

Spheres in layer 3 are not aligned vertically with those in layer 1 or

layer 2. Layer 3 is a distinct layer. ABCABC ... closest-packing


The unit cell for ABAB ... closest-packing is “hexagonal”. Thus, this type of closest-packing is also called “hexagonal closest-packing” or hcp for short.

The unit cell for ABCABC ... closest-packing is “face-centred cubic”. Thus, this type of closest-packing is also called “cubic closest-packing” or ccp for short.


The unit cell for ABAB ... closest-packing is “hexagonal”. Thus, this type of closest-packing is also called “hexagonal closest-packing” or hcp for short.

The unit cell for ABCABC ... closest-packing is “face-centred cubic”. Thus, this type of closest-packing is also called “cubic closest-packing” or ccp for short.

You are expected to remember that cubic closest-packing (ccp)

and face-centered cubic (fcc) are the same!!

Density of a Crystalline Solid

Keep in mind:

The formula d = ncell Matom /(NA a3) should be used only for identical hard spheres in a cubic lattice. If you have more than one type of sphere (e.g. of different sizes or masses) or unit cell that is not cubic, then you should start from d = mcell / Vcell.

The density of a solid depends on the microscopic details of the packing arrangement.

3aN

Mn

Vm

Vmd A

atomatoms

cell

cell

mass of one atom

Matom is molar mass, in g mol−1

NA = 6.022×1023 mol −1

Density of a Crystalline SolidExample: Tungsten, W, crystallizes in one of the three cubic arrangements. If the edge length of the unit cell is 321 pm and the density is 18.5 g cm-3, then what type of crystal lattice does tungsten have? What is the radius of a tungsten atom?

3aN

Mn

Vm

Vmd A

atomatoms

cell

cell

Matom = 184 g.mol1


3aN

Mn

Vm

Vmd A

atomatoms

cell

cell

Matom = 184 g.mol1

A

atomatoms

NM

adn3

2003.2

10022.6.184

)10321().(5.18

123

1

3103

molmolg

cmcmgnatoms


Simple cubic

Body-centered cubic

Face-centered cubic

1

2

4

# of atoms/unit cellRelationship

between a and R

R = a/2

aR 43

aR 22

1

pmR 13932143

Density of a Crystalline SolidThe edge length can be determined experimentally using x-ray diffraction. When x-rays are passed through a crystalline solid, the x-rays are deflected from their paths by the atoms of the solid and interfere with each other to produce an interference pattern – a “diffraction pattern” – that can be analyzed to determine the geometry of the crystal lattice.

Simple cubic

Body-centered cubic

Face-centered cubic

1

2

4


between a and R

R = a/2

aR 43

aR 22

1



Crystal lattice


+ and NO3− ions.


Unit cell

REVIEW




a = b ≠ c = = g = 90o




a ≠ b ≠ c = = g = 90o

Crystalline solids

a b

c

g

REVIEW




a = b ≠ c = = g = 90o




a ≠ b ≠ c = = g = 90o

Crystalline solids

a

a

a

REVIEW


Simple cubic

Body-centered cubic

Face-centered cubic

1

2

4


between a and R

R = a/2

aR 43

aR 22

1

REVIEW

R = 0.43×a

R = 0.35×a

R = 0.50×a

Ionic Solids and Interstitial SitesUp to this point, we’ve focused on packing together identical hard spheres. Now, we’ll extend that model so that we can describe the structures of binary ionic solids. For many ionic solids, it is often the case that one of the ions forms a cubic lattice and the other ion occupies “holes” in that lattice. In order to understand the structures of ionic crystals, we must first examine the types of holes we can have. “Holes” (or interstitial sites) are named according to the “coordination number” of a small sphere that just fits into that hole. The types of holes that interest us the most are cubic, tetrahedral and octahedral holes.

Ionic Solids and Interstitial Sites

The small sphere has a coordination number of three.

trigonal hole

Ionic Solids and Interstitial Sitestetrahedral hole

coordination # = 4

Ionic Solids and Interstitial Sitesoctahedral hole

coordination # = 6

side view

top view

Ionic Solids and Interstitial Sitescubic hole

coordination # = 8

Ionic Solids and Interstitial SitesExample: Within the face-centered cubic (fcc) unit cell, there are both octahedral and tetrahedral holes. Examine the fcc unit cell to identify the locations of the octahedral and tetrahedral holes. How many holes of each type are there per unit cell?

¼ of an octahedral hole

One octahedral hole


¼ of an octahedral hole

One octahedral hole

There is ¼ of an octahedral hole (per edge) and an octahedral hole at the centre of the cell.

For the fcc cell, there are 4 spheres per cell and 4 octahedral holes holes. The ratio of spheres-to-holes is:

Example: Within the face-centered cubic (fcc) unit cell, there are both octahedral and tetrahedral holes. Examine the fcc unit cell to identify the locations of the octahedral and tetrahedral holes. How many holes of each type are there per unit cell?

# octahedral holes = ¼ hole/edge × 12 edges + 1 = 4 (per fcc unit cell)

# spheres : # octahedral holes = 4 : 4 = 1 : 1


a

b

c d

tetrahedral hole


There is one tetrahedral hole associated with each corner of the fcc cell (that is, the sphere at each corner can be considered to be the “cap” of a tetrahedron). Thus,

For the fcc cell, there are 4 spheres per celland 8 tetrahedral holes. The ratios of spheres-to-holes are:

# spheres : # tetrahedral holes = 4 : 8 = 1 : 2

# tetrahedral holes = 8 (per fcc cell)


a

b

c d

tetrahedral hole


In general: In a closest-packed structure containing N spheres, where N is a very large number, there are N octahedral holes and 2N tetrahedral holes.

Ionic Solids and Interstitial SitesHow big are tetrahedral, octahedral and cubic holes?Octahedral holes. Let ro be the radius of a sphere that just fits into the octahedral hole.

2R

2R

2ro

Let ro be the radius of a sphere that just fits into the octahedral hole.

222 )2()2()2( RrRRR o 22 )(48 orRR

orRR 2

22 )(2 orRR

RRro 414.0)21(

Ionic Solids and Interstitial SitesHow big are tetrahedral, octahedral and cubic holes?Cubic holes: Let ro be the radius of a sphere that just fits into the cubic hole.

Let rc be the radius of a sphere that just fits into the cubic hole. This sphere has its centre at the midpoint of the body-diagonal. Therefore, we need to find the length of the body-diagonal.

)(22 oo rRRrRBD 2R


a

a

a2

)(22 oo rRRrRBD

222 2 aaa Diagonal2 =

a2Diagonal =


222222 32)2( aaaaaBD

a

a2

)(22 oo rRRrRBD

RRaBD 32)2(33

RrRBD o 32)(2

RRro 732.0)13(

Ionic Solids and Interstitial SitesHow big are tetrahedral, octahedral and cubic holes?Tetrahedral holes: Let ro be the radius of a sphere that just fits into the tetrahedral hole.

b

c d

a

b

c

d

a

2×ro


b

c d

a

b

c

d

a

2×ro

)(22 oo rRRrRBD

2×R

x ?


b

c d

a

b

c

d

a

2222 )2(2 Rxxx

x

x

2×R

Rx 2


b

c d

a

b

c

d

a

2×ro

)(22 oo rRRrRBD

2×R

x = R2

22222 6)2()2()(4 RRRrRBD o

RrRBD o 6)(2


b

c d

a

b

c

d

a

2×ro

2×R

x = R2

RRro 225.0)126(

Holes RelationshipTrigonal Too small to worry about

TetrahedralOctahedral

cubic

Ionic Solids and Interstitial SitesSummary of hole types and the radius ratio rules

R414.0R225.0

R732.0

Trigonal

Tetrahedral

Octahedral

cubic

Ionic Solids and Interstitial SitesRadius Ratio Rules for Ionic SolidsLet R+ be the radius of the positive ion in a binary ionic solid and let R− be the radius of the negative ion.

414.0225.0

RR

732.0414.0

RR

RR732.0

Positive ions occupy tetrahedral sites in the lattice of negative ions. The coordination number of each positive ion is 4.

Positive ions occupy octahedral sites in the lattice of negative ions. The coordination number of each positive ion is 6.

Positive ions occupy cubic sites in the lattice of negative ions. The coordination number of each positive ion is 8.

Note: The rules above are just guidelines. There are many exceptions.

Note: The positive ions are too big for the tetrahedral sites and thus the negative ions are forced apart a little this increases the attraction between the “+” and “−” ions and decreases the repulsion between the “−” ions.

Ionic Solids and Interstitial SitesSodium Chloride Structure (also called “rock salt structure”)

Cl− ions form a face-centered cubic (fcc) lattice and Na+ ions occupy 100% of the octahedral holes in the chloride lattice.

Cl

Na

Ionic Solids and Interstitial SitesCesium Chloride Structure

Cl

Cs

Cl− ions form a simple cubic lattice and Cs+ ions occupy cubic holes in the chloride lattice.

Ionic Solids and Interstitial SitesZinc Blende Structure

S2− ions form a face-centered cubic (fcc) lattice and Zn2+ ions occupy half (50%) of the tetrahedral hole.

Why are only half of the tetrahedral holes occupied?

How many spheres: 4 S2

How many tetrahedral holes: 8 for only 4 Zn2+

Ionic Solids and Interstitial SitesZinc Blende Structure

S2− ions form a face-centered cubic (fcc) lattice and Zn2+ ions occupy half (50%) of the tetrahedral hole.

Why are only half of the tetrahedral holes occupied? There are 4 S2− ions per cell. There must be four Zn2+ ions per cell because ZnS is a 1:1 salt. Therefore, only 4 of 8 tetrahedral holes are occupied by Zn2+ ions.



Crystal lattice


+ and NO3− ions.


Unit cell

REVIEW




a = b ≠ c = = g = 90o




a ≠ b ≠ c = = g = 90o

Crystalline solids

a b

c

g

REVIEW




a = b ≠ c = = g = 90o




a ≠ b ≠ c = = g = 90o

Crystalline solids

a

a

a

REVIEW


Simple cubic

Body-centered cubic

Face-centered cubic

1

2

4


between a and R

R = a/2

aR 43

aR 22

1

REVIEW

R = 0.43×a

R = 0.35×a

R = 0.50×a

Holes RelationshipTrigonal Too small to worry about

TetrahedralOctahedral

cubic

Ionic Solids and Interstitial SitesSummary of hole types and the radius ratio rules

R414.0R225.0

R732.0

Trigonal

Tetrahedral

Octahedral

cubic

REVIEW

Ionic Solids and Interstitial SitesFluorite Structure Ca2+ ions form a fcc lattice and F− ions

occupy all of the tetrahedral holes.

Thus, there are 4 Ca2+ ions per unit cell and 8 F− ions.

How many spheres: 4 Ca2+

How many tetrahedral holes: 8 F

Ionic Solids and Interstitial SitesAntifluorite Structure Sodium oxide, Na2O, adopts the so-called

antifluorite structure. The antifluorite structure is the “reverse” of the fluorite structure in the sense that the negative ions form a fcc lattice and the positive ions occupy all of the tetrahedral holes. In the case of Na2O, the O2− ions form a fcc lattice and the Na+ ions occupy 100% of the tetrahedral holes.

O2 Na

Ionic Solids and Interstitial SitesExample: In beryllium oxide, BeO, the oxide ions form a face-centred cubic lattice and the beryllium ions occupy tetrahedral sites in the lattice of O2- ions. What fraction of the holes is occupied by the Be2+ ions? (Ans: 50%)

How many spheres: 4 O2

How many tetrahedral holes: 8 tetrahedral holes for Be2

With 8 negative charges from O2 anions and 16 positive charges from Be2+ cations, the charge balance would not work. Only 4 Be2+ generating 8 positive charges can be present.

50% of the tetrahedral holes can be occupied by Be2+ cations.

Ionic Solids and Interstitial SitesExample: To “build” the sodium chloride structure, we would first arrange Cl ions in a face-centred cubic structure and then insert Na+ ions into the octahedral sites of the Cl lattice. Assume that Na+ and Cl are hard spheres with radii of 97 pm and 181 pm, respectively. Calculate the density of sodium chloride, in g cm3. (Ans: 2.26 g cm−3)

Cl

Na

a

4 spheres (Cl)

4 octahedral holes (Na+)

31233103 .26.2)10022.6()10556(

)/5.3523(4)(4 cmgmolcm

molgNaMMdA

ClNa

a = 2×(97+181) = 556 pm

Cl

Cs

Ionic Solids and Interstitial SitesExample: In cesium chloride, the Cl ions form a simple cubic lattice and the Cs+ ions occupy cubic holes in the chloride lattice. However, the Cs+ ions force the Cl ions apart so that none of the chloride ions are in direct contact with each other. If the density of cesium chloride, CsCl, is 3.988 g cm3, then what is the distance (R+ + R) ? (Ans: 357 pm)

a

1 spheres (Cl)

1 cubic hole (Cs+)

A

ClCs

NaMMd

3

)(Body Diagonal = 3a

A

ClCs

NdMMa

)(3

pmcmcmgNd

MMaA

ClCs 4121012.410022.6.988.3

5.359.132 83 2333

where aRRBD 3)(2

Cl

Cs

Ionic Solids and Interstitial SitesExample: In cesium chloride, the Cl ions form a simple cubic lattice and the Cs+ ions occupy cubic holes in the chloride lattice. However, the Cs+ ions force the Cl ions apart so that none of the chloride ions are in direct contact with each other. If the density of cesium chloride, CsCl, is 3.988 g cm3, then what is the distance (R+ + R) ? (Ans: 357 pm)

a

Body Diagonal = 3a

pma 412

pmaBD 71441233

)(2714 RRpmBD

pmpmRR 3572

714

The Born Haber CycleThe stability of an ionic solid is quantified in terms of its lattice energy.

Lattice Energy = energy change when gas phase ions combine to form an ionic solid

Na+(g) + Cl−(g) → NaCl(s)

Lattice energy (Hlattice) for NaCl(s)

Hlattice

The lattice energy cannot be measured directly. However, we can obtain it indirectly from other thermochemical data using the Born-Haber cycle.

The Born-Haber cycle involves the following steps:

• Write down the “formation reaction” for the solid from the elements under standard conditions (T = 25 oC and P = 105 bar).

• Convert the elements into gas-phase atoms.• Convert the atoms into gas-phase ions.• Combine the ions to form the solid.

The Born Haber CycleBorn-Haber cycle for NaCl(s)

• Write down the “formation reaction” for the solid.• Convert the elements into gas-phase atoms.• Convert the atoms into gas-phase ions.• Combine the ions to form the solid.

Na(s) + ½ Cl2(g) NaCl(s)Hf

o

Na(g) + Cl(g)

Hsubo ½ DCl-Cl

Na+(g) + Cl(g)Hlattice

o

(dissociation)(sublimation)

(first ionization energy)

IE(1)(first electron

affinity)

EA(1)

The Born Haber Cycle

Born-Haber cycle for NaCl(s)State function does not depend on the path:

University of

d(home – UW) = constant

path(UW home) varies

http://www.uwaterloo.ca/



http://www.feds.ca/files/2012/08/NEW-BOMBER-LOGO-copy.png



The Born Haber CycleBorn-Haber cycle for NaCl(s)


Na(s) + ½ Cl2(g) NaCl(s)Hf

o

Na(g) + Cl(g)

Hsubo ½ DCl-Cl

Na+(g) + Cl(g)Hlattice

o


(first ionization energy)

IE(1)(first electron

affinity)

EA(1)

Apply Hess’ Law: Hfo = Hsub

o + IE(1)Na(g) + ½ DCl-Cl + EA(1)Cl(g) + Hlattice Thus: Hlattice = (Hsub

o + IE(1)Na(g) + ½ DCl-Cl + EA(1)Cl(g)) Hfo

The Born Haber CycleBorn-Haber cycle for MgF2(s)


Mg(s) + F2(g) MgF2(s)Hf

o

Mg(g) + 2 F(g)

Hsubo DF-F

Mg2+(g) + 2 F(g)Hlattice

o


(first and second ionization energy)

IE(1) + IE(2)(first electron

affinity)

EA(1)

Apply Hess’ Law: Hfo = Hsub

o + IE(1)Mg(g) +IE(2)Mg(g) + DF-F + 2×EA(1)F(g) + Hlattice

Thus: Hlattice = (Hsubo + IE(1)Mg(g) +IE(2)Mg(g) + DF-F + 2×EA(1)F(g)) + Hf

o

The Born Haber Cycle

Born-Haber cycle for MgF2(s)

Hess’ Law: Hfo = Hsub

o + IE(1)Mg(g) +IE(2)Mg(g) + DF-F + 2×EA(1)F(g) + Hlattice

Born-Haber cycle for NaCl(s)Hess’ Law: Hf

o = Hsubo + IE(1)Na(g) + ½ DCl-Cl + EA(1)Cl(g) + Hlattice

By comparing the expressions we wrote for NaCl(s) and MgCl2(s), we can see that the expression we obtain for ΔHf

o will vary from salt to salt. Therefore, to obtain the correct expression, you must first construct the Born-Haber cycle.



All Done!

intermolecular forces: liquids and solids

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