10 Liquids-Solids 6.pdf

CHAPTER 10:

INTERMOLECULAR FORCES AND

THE PHYSICAL PROPERTIES OF

LIQUIDS AND SOLIDS

The magnitude of

inter

molecular forces determines

whether a substance is a gas, liquid, or solid

•Intermolecular forces are attractive forces between particles. (just as interstates run between states and international means between countries)

•The state of a substance depends largely on the balance between the kinetic energies of the particles and the energies of attraction between particles.

Gas Liquid Solid

Intra- Versus Inter- Molecular Forces

•

Intra

molecular

forces



Ionic bonding

(bonding

within

an ionic compound) and

covalent bonding

(bonding

within

a molecule) are two of

the strongest forces that hold molecules together.

It requires 927 kJ to break the two intramolecular covalent O–H– bonds in 1 mol of water.

•

Intermolecular forces



Ion–Dipole



Dipole–Dipole

It requires 41 kJ to overcome the intermolecular attractions between water molecules and convert 1 mol of liquid water to water vapor at 100°C.



Induced Dipole–Induced Dipole (London Dispersion Forces)

Attractive forces that act between atoms or molecules in a pure

substance are collectively called

van der Waals forces

.

Dipole-dipole interactions

are attractive forces that act between

polar molecules.

The magnitude of the attractive forces depends on the magnitude of

the dipole.

Intermolecular Forces

Ion-Dipole Forces

:

Exist

between

an

ion

and

the partial charge

on the end of a

polar molecule

.

Ion–Dipole

Forces: important for solutions of

ionic substances in polar solvents

Dipole

–

Dipole

Forces: exist between

oppositely charged ends of polar molecules.

•

These forces are effective only when polar molecules

are very close together.

•

The magnitude of the attractive forces depends on

the magnitude of the dipole.

Boiling Point

Reflects

the Magnitude of

Intermolecular Forces

A substance in which the particles are held together by

larger intermolecular attractions will require more energy

to separate the particles and will therefore boil at a

higher temperature.

Why does this happen?

_Te, Se, S and O are in the same group. We expect their hydrogen compounds to be chemically similar. Extrapolating from the graph, we would expect H2O to boil at -64°C.

There is a 164°C difference between the predicted and measured boiling point.

Why?

100

-64 -60.3

-41.3 -2.2

-100 -50 0 50 100

T

em

p

(

°C

)

H2Te

H2Se _H

2S

H2O

H2O ??

Hydrogen bonding

is a special type of dipole-dipole interaction.

Hydrogen bonding only occurs in molecules that contain H bonded

to a small, highly electronegative atom such as N, O, or F.

Hydrogen Bonding

F

H

F

Hydrogen Bonding:

a very strong type of dipole–dipole interaction that

only occurs in molecules that contain H bonded to a

small, highly electronegative atom (N, O, or F)

O H d+ H d

-d+

Hydrogen bonding in liquid water Polar water molecule

-200 -100 0 100

B

oi

ling

Poi

nt

(

°

C)

H2O

H2S H2Se H2Te

HF

HCl HBr HI

SbH3

AsH3

PH3

NH3

CH4 SiH4

GeH4 SnH4

Snowflakes and Ice Crystals Reflect the

Arrangement of Water Molecules at the

Molecular Level!

Hydrogen Bonding is Expected in Molecules

that have –OH or –NH Groups

H

O

H

H

O

CH

3

d

+

d

-d

+

Polar

Water

molecule

Polar

Methanol

molecule

Polar

Ammonia

molecule

d

+

d

+

N

H

H

H

d

+

d

-d

+

d

+

d

-Hydrogen Bonding in Alcohols is Similar to

Hydrogen Bonding in Water

Hydrogen bonding

in liquid water

Hydrogen bonding

in an alcohol

Having

more than one –OH Group

increases

the number of

hydrogen bonds

and therefore,

increases

the

boiling point

97.2 290 187 0 100 200 300 400

1-propanol 1,2-propanediol 1,2,3-propanetriol

T e m p ( °C )

CH CH2

OH

CH2

OH OH

CH2 CH2

OH

CH3

CH CH2

OH

CH3

OH

Hydrogen Bonding in DNA

Thymine – Adenine

2 hydrogen bonds

Cytosine – Guanine

3 hydrogen bonds

In which of the following substances, in the

liquid state, would hydrogen bonding occur?

a)

c)

b)

d)

C

H

H

H

C

H

H

H

Cl

N

London Dispersion Forces

: result from the

Coulombic attractions between instantaneous

dipoles of nonpolar molecules

You might think that two nearby nonpolar molecules

A and B would be unaffected by each other.

A

B



The electric field of a molecule fluctuates rapidly. For

the briefest of moments, the electrons may be on one

side of the molecule.



When that happens, molecule B can be considered

to have a temporary separation of charge.

A

B

The neighboring molecules A and C “feel” the electric

field of B and undergo a spontaneous adjustment in

their electrons’ positions, giving them a

complementary temporary separation of charge

(remember “like charges repel”, “opposites attract”).

A

B

C

London Dispersion Forces are Weak

•

Dispersion forces

or

London dispersion forces

result from

the Coulombic attractions between

instantaneous dipoles

of non-polar molecules.

The Number and Kind of London Dispersion

Forces Affect the Boiling Point

1.

Relative atomic weight of the atoms involved

1.

Number of atoms in the molecule

1.

Shape of the molecule

Heavier Molecules Have Higher Boiling Points

-100 0 100 200 300 400 500 600

B

oili

n

g

p

oin

t (

K

)

F2 Cl2

Br2 I2

He

Ar

Ne

Intermolecular Forces

Boiling Points of Alkanes

Increases with the number of carbon atoms due to the

induced dipole–induced dipole attractions

(London Dispersion forces)

-200 -100 0 100 200 300 400

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Number of carbons

B o il in g p o in t (° C ) Room temperature

Shape Affects Boiling Point

The more compact the molecule,

the less surface area there is for interactions,

and therefore the lower the boiling point.

pentane

bp 36.1°C 2-methylbutane bp 28°C 2,2-dimethylpropane bp 9.5°C

C H C H H H H C H H C H H C H H H C H C H H H C C H H C H H H H H H C H C H H C C C H H H H H H H H H

A Comparison of Intermolecular Forces

What kinds of intermolecular forces exist in the

following molecules?

a)

Ar

b)

HCl

c)

HF

d)

CH

4

e)

CaCl

2

f)

CCl

4

g)

CH

3

COOH

h)

CH3COCH3

i)

H

2

S

Worked Example 12.1

Strategy Draw Lewis dot structures and apply VSEPR theory to determine whether each molecule is polar or nonpolar. Nonpolar molecules exhibit dispersion forces only. Polar molecules exhibit dipole-dipole interactions and dispersion forces. Polar molecules with N–H , F–H, or O–H bonds exhibit dipole-dipole interactions (including hydrogen bonding) and dispersion forces.

(a) (b) (c) (d) What kind(s) of intermolecular forces exist in (a) CCl4(l), (b) CH3COOH(l),

Worked Example 12.1 (cont.)

(a) (b) (c) (d)

Solution (a) CCl4 is nonpolar, so the only intermolecular forces are dispersion

forces.

(b) CH3COOH is polar and contains an O–H bond, so it exhibits dipole-dipole

interactions (including hydrogen bonding) and dispersion forces. (c) CH3COCH3 is polar but does not contain N–H , F–H, or O–H bonds, so it

exhibits dipole-dipole interactions and dispersion forces.

(d) H2S is polar but does not contain N–H , F–H, or O–H bonds, so it exhibits

dipole-dipole interactions and dispersion forces.

Think About It Being able to draw correct Lewis structures is, once again, vitally important. Review, if you need to, the procedure for drawing them.

Properties of Liquids

Surface tension

Viscosity

Vapor Pressure

Surface tension: the amount of energy required to

stretch or increase the surface of a liquid

The molecules on the surface feel a net inward pull.

Molecules with large IMF, like polar molecules, tend to

have high surface tension.

Capillary action:

movement of a liquid up a narrow tube



Two types of forces bring about capillary action:



Cohesion

is the attraction between like molecules



Adhesion

is the attraction between

unlike

molecules

Adhesive forces are greaterthancohesive

forces

Cohesive forces are greaterthanadhesive

forces

Viscosity:

a measure of a fluid’s resistance to flow



The higher the viscosity the more slowly a liquid flows.



Liquids that have high IMFs have higher viscosities.

Factors that Affect the Viscosity of a Liquid

0 500 1000 1500 2000 2500 3000

0 10 20 30 40 50 60 70 80 90 100

Temp (°C)

V

is

cosi

ty

(

cS

t)

0 500 1000 1500 2000 2500 3000

0 10 20 30 40 50 60 70 80 90 100

Temp (°C)

V

is

cosi

ty

(

cS

t)

Viscosity

increases

with

increasing molecular weight.

Glycerol is a thick, syrupy liquid. In

terms of IMFs, explain why glycerol

has a higher viscosity than water.



Since it is a bigger molecule, glycerol has greater

London dispersion forces that water.



Glycerol

can make

three

hydrogen

bonds

whereas

water

can make only

one

hydrogen

bond

.



Therefore, glycerol molecules have more attraction for

one another than water molecules have for one

another and so have a higher viscosity.

Phase Changes

If a molecule at the surface of the liquid has enough

kinetic energy, it can escape to the gas phase in a

process called

evaporation

or

vaporization

.

If the container is open, evaporation will continue until

the liquid evaporates away.

Low temp

N

um

be

r of

m

ole

cule

s

High temp

Kinetic energy Energy needed to escape liquid

In a sealed vessel, vapor pressure increases until the

rate of evaporation equals the rate of condensation

H

2

O(

l

)

⇌

H

2

O(

g

)

Evaporation:

H

2

O(

l

) → H

2

O(

g

)

Condensation:

H

2

O(

l

) ← H

2

O(

g

)

When the forward process and reverse process are occurring at the same rate, the system is in

dynamic equilibrium.

In a sealed vessel, vapor pressure increases until the

rate of evaporation equals the rate of condensation

exerted by its vapor when the liquid and vapor

The vapor pressure of a liquid is the pressure

states are in equilibrium.

The exponential rise in vapor pressure with increasing

temperature allows us to use natural logarithms to express

the nonlinear relationship as a linear relationship

Clausius–Clapeyron Equation

ln P = natural log of vapor pressure

ΔH

vap

= the molar heat of vaporization

R = the gas constant (8.314 J/K•mol)

T = the kelvin temperature

C is an experimentally determined constant

vap

ln

P

H

C

RT



 -

+

Clausius–Clapeyron Equation

: A Plot of

ln

P

Versus

1/T(K)

Gives a Straight Line with a

Slope of –

H

_vap

/R

R

=

8.314

J

mol×K

y =

m

x

+ b

y

=

mx

+

b

Clausius–Clapeyron Equation: Can be used to

calculate the



H

_vap

of a liquid from its measured

vapor pressure at two temperatures

Strategy Given the vapor pressure at one temperature, P1, use the equation

below to calculate the vapor pressure at a second temperature, P2.

Temperature must be expressed in kelvins, so T1 = 291.15 K and T2 = 305.15 K.

Because the molar heat of vaporization is given in kJ/mol, we will have to convert it to J/mol for the units of R to cancel properly: ΔHvap = 2.6×104 J/mol. The

inverse function of ln x is ex.

Diethyl ether is a volatile, highly flammable organic liquid that today is used mainly as a solvent. (It was used as an anesthetic during the nineteenth century and as a recreational intoxicant early in the twentieth century during prohibition, when ethanol was difficult to obtain.) The vapor pressure of diethyl ether is 401 mmHg at 18°C, and its molar heat of vaporization is 26 kJ/mol. Calculate its vapor pressure at 32°C.

ln = − P1

P2

ΔHvap

R 1 T2

1 T1

Solution

Think About It It is easy to switch P1 and P2 or T1 and T2 accidentally and get

the wrong answer to a problem such as this. One way to help safeguard against this common error is to verify that the vapor pressure is higher at the higher

temperature.

ln = P1

P2

2.6×104 _J/mol

8.314 J/K∙mol 305.15 K −1 291.15 K 1 = −0.4928

P1

P2 = e

−0.4928 _{= 0.6109}

P1

0.6109= P2

P2 = _mmHg401

0.6109

= 6.6×102

What is the boiling point of water at a pressure of 0.750 atm?

You need two points: (P1,T1) and (P2,T2)

P1 = 1.00 atm (the normal boiling point)

T1 = 100°C = 373 K (the normal boiling point) P2 = 0.750 atm

Hvap = 40,670 J/mol

R = 8.314 J/mol•K

T

2

=

1

ln

P

1

P

2

æ

è

ç

ö

ø

÷

D

H

vap

R

æ

è

ç

ö

ø

÷

+

1

T

1

æ

è

ç

ç

ç

ç

ö

ø

÷

÷

÷

÷

=

D

H

vap

T

1

D

H

vap

+

RT

1

ln

P

P

1

2

æ

è

ç

ö

ø

÷

=

365

K

=

92

°

C

Using the following data, determine



H

vap

for hexafluorobenzene (C

6

F

6

)

T(K)

P(torr)

280.

32.42

300.

92.47

320.

25.1

330.

334.4

340.

482.9

y =

m

x

+ b

1/T ln P

0.00357 3.48 0.00333 4.53 0.00313 5.42 0.00303 5.81 0.00294 6.18

3.00 3.50 4.00 4.50 5.00 5.50 6.00 6.50

0.00280 0.00290 0.00300 0.00310 0.00320 0.00330 0.00340 0.00350 0.00360 0.00370

ln

P

1/T

=

35,700

J

mol

=

35.7

kJmol

Use Excel’s slope function to determine the

slope of the line:

1/T ln P

0.00357 3.48 0.00333 4.53 0.00313 5.42 0.00303 5.81 0.00294 6.18

DH

vap

=

-slope

´

R

=SLOPE(

known_y’s

,

known_x’s

)

known_y’s known_x’s

=

35,700

J

mol

=

35.7

kJmol

Determine the normal boiling point for hexachlorobenzene

given that



H

_vap

= 35,700 J/mol

T

2

=

1

ln

P

1

P

₂

æ

è

ç

ö

ø

÷

D

H

_vap

R

æ

è

ç

ö

ø

÷

+

1

T

₁

æ

è

ç

ç

ç

ç

ö

ø

÷

÷

÷

÷

T (K) P (torr)

280. 32.42 300. 92.47 320. 225.1 330. 334.4 340. 482.9

1/T ln P

0.00357 3.48 0.00333 4.53 0.00313 5.42 0.00303 5.81 0.00294 6.18

3.00 3.50 4.00 4.50 5.00 5.50 6.00 6.50

0.00280 0.00290 0.00300 0.00310 0.00320 0.00330 0.00340 0.00350 0.00360 0.00370

ln

P

1/T

y

=

mx

+

b

x

=

y

-

b

m

1

T

=

ln

P

-

intercept

slope

Use Excel’s slope and intercept functions:

x

=

y

-

b

m

1

T

=

ln

P

-

intercept

slope

T

=

slope

ln(760)

-

intercept

=SLOPE(

known_y’s

,

known_x’s

)

=INTERCEPT(

known_y’s

,

known_x’s

)

Normal boiling point = 352.3 K = 79.1°C

Consider the following vapor pressure versus

temperature plot for three different substances A, B,

and C. If the three substances are CH

4

, SiH

4

, and

NH

3

, match each curve to the correct substance.

Va

por

pre

ss

ure

(t

orr)

Temperature (°C)

A B C

When a substance goes from one phase to another

phase, it has undergone a

phase change

.

Example

Phase Change

Freezing

H

2

O(l)



H

2

O(s)

Evaporation (vaporization)

H

2

O(l)



H

2

O(g)

Melting (fusion)

H

2

O(s)



H

2

O(l)

Condensation

H

2

O(g)



H

2

O(l)

Sublimation of dry ice

CO

2

(s)



CO

2

(g)

Deposition of iodine

I

2

(g)



I

2

(s)

Phase changes are generally caused by the addition

or removal of energy, usually in the form of heat.

Vapor Pressure Increases

as Temperature Increases

The

boiling point

of a

liquid is the

temperature at which

the vapor pressure

becomes equal to the

external atmospheric

pressure

exerted on

the liquid.

The

normal boiling

point

of a liquid is the

temperature at which a

liquid boils at a

pressure of 1 atm

.

Frozen Fruit Juice Concentrate: At a Reduced

Pressure, the Boiling Point is Lower

Some of the water in the fruit juice is boiled away at a

reduced pressure, thus concentrating the juice without

heating it to a high temperature (which spoils the taste

of the juice and reduces its nutritional value).

Pressure Cooker: At

Increased Pressure, the

Boiling Point is Higher



For every 10°C increase in temperature, food cooks

twice as fast (takes half the amount of time).



With an internal pressure of 2.02 atm in the pressure

cooker, water boils at 121°C.



Food cooks four times faster in the pressure cooker.

The Molar Heat of Vaporization (

H

vap

)is the

amount of heat required to vaporize a mole of

substance at its boiling point.

The boiling point increases as

H

_vap

increases.































The Molar Heat of Fusion (

H

vap

) is the energy

required to melt one mole of a solid.

A Typical Heating Curve

Solid

Boiling point

Vapor

Liquid

Solid and liquid in

equilibrium

Liquid and vapor in

equilibrium

Time

Te

m

per

at

ure

Melting point

The Molar Enthalpy of Sublimation is the

energy required to sublime one mole of a solid.



Sublimation is the process by which molecules go

directly from the solid phase to the vapor phase.



Deposition is the reverse process of sublimation.

Δ

H

_sub

= Δ

H

_fus

+ Δ

H

_vap

(a) Calculate the amount of heat deposited on the skin of a person burned by 1.00 g of liquid water at 100.0°C and (b) the amount of heat deposited by 1.00 g of steam at 100.0°C. (c) Calculate the amount of energy necessary to warm 100.0 g of water from 0.0°C to body temperature and (d) the amount of heat required to melt 100.0 g of ice 0.0°C and then warm it to body temperature. (Assume that body temperature is 37.0°C.)

Strategy For the purpose of following the sign conventions, we can designate the water as the system and the body as the surroundings. (a) Heat is transferred from hot water to the skin in a single step: a temperature change. (b) The transfer of heat from steam to the skin takes place in two steps: a phase change and a temperature change. (c) Cold water is warmed to body temperature in a single step: a temperature change. (d) The melting of ice and the subsequent warming of the resulting liquid water takes place in two steps: a phase change and a temperature change. In each case, the heat transferred during a temperature change depends on the mass of the water, the specific heat of the water, and the change in temperature. For the phase changes, the heat transferred depends on the amount of water (in moles) and the molar heat of vaporization (ΔHvap) or molar

heat of fusion (ΔHfus). In each case, the total energy transferred or required is the

sum of the energy changes for the individual steps.

The specific heat is 4.184 J/g∙°C for water and 1.99 J/g∙°C for steam. ΔHvap is

40.79 kJ/mol and ΔHfus is 6.01 kJ/mol. Note: The ΔHvap of water is the amount of

heat required to vaporize a mole of water. However, we want to know how much heat is deposited when water vapor condenses, so we use −40.79 kJ/mol.

Solution (a) ΔT = 37.0°C – 100.0°C = –63.0°C

q = msΔT = 1.00 g × ×–63.0°C

Thus, 1.00 g of water at 100.0°C deposits 0.264 kJ of heat on the skin. (The negative sign indicates that heat is given off by the system and absorbed by the surroundings.)

(b)

q1 = nΔHvap = 0.0555 mol ×

q2 = msΔT = 1.00 g × ×–63.0°C

The overall energy deposited on the skin by 1.00 g of steam is the sum of q1 and q2:

–2.26 kJ + (–0.264 kJ) = –2.53 kJ 4.184 J

g∙°C = –2.64×102 J = –0.264 kJ

1.00 g

18.02 g/mol = 0.0555 mol water −40.79 kJ

mol = –2.26 kJ 4.184 J

g∙°C = –2.64×102 J = –0.264 kJ

Solution (c) ΔT = 37.0°C – 0.0°C = 37.0°C

q = msΔT = 1.00 g × ×37.0°C

The energy required to warm 100.0 g of water from 0.0°C to 37.0°C is 15.5 kJ.

(d)

q1 = nΔHfus = 5.55 mol ×

q2 = msΔT = 100.0 g × ×37.0°C

The energy rquired to melt 100.0 g of ice at 0.0°C and warm it to 37.0°C is the sum of q1 and q2:

33.4 kJ + 15.5 kJ = 48.9 kJ 4.184 J

g∙°C = 1.55×104 J = 15.5 kJ

100.0 g

18.02 g/mol = 5.55 mol water 6.01 kJ

mol = 33.4 kJ 4.184 J

g∙°C = 1.55×104 J = 15.5 kJ

Think About It In problems that include phase changes, the q

values corresponding to the phase-change steps will be the largest contributions to the total. If you find that this is not the case in your solution, check to see if you have made the common error of neglecting to convert the q values corresponding to temperature

changes from J to kJ.

A

phase diagram

summarizes the conditions

at which a substance exist as a

s

,

l

, or

g

.

Phase

Diagram

Phase

Diagram for

Water

Strategy Each point on the phase diagram corresponds to a pressure-temperature combination. The normal boiling and melting points are the temperatures at which the substance undergoes phase changes. These points fall on the phase boundary lines. The triple point is where the three phase boundaries meet.

Using the following phase diagram, (a) determine the normal boiling point and the normal melting point of the substance, (b) determine the physical state of the substance at 2 atm and 110°C, and (c) determine the pressure and temperature that correspond to the triple point of the substance.

Solution By drawing lines corresponding to a given pressure and/or temperature, we can determine the temperature at which a phase change occurs, or the physical state of the substance under specified conditions.

(a)

The normal boiling and melting points are ~140°C and ~205°C, respectively.

Solution (b)

At 2 atm and 110°C the substance is a solid.

Solution (c)

The triple point occurs at ~0.8 atm and ~115°C.

Think About It The triple point of this substance occurs at a pressure below atmospheric pressure. Therefore, it will melt rather than sublime when it is heated under ordinary conditions.

Use the phase diagram for carbon to answer the following questions:

1. How many triple points are in the phase diagram?

2. What phases coexist at each triple point?

3. What happens if graphite is subjected to very high pressures at room temperature?

Crystal Structure

A

crystalline solid

possess rigid and long-range order; its atoms,

molecules, or ions occupy specific positions.

A

unit cell

is the basic repeating structural unit of a crystalline solid.

There are seven types of unit cells.

Crystal Structure

The coordination number is defined as the number of atoms surrounding an atom in a crystal lattice.

The value of the coordination number indicates how tightly the atoms are packed together. The basic repeating unit in the array of atoms is called a simple cubic cell.

Crystal Structure

There are three types of cubic cells.

Crystal Structure

In a

body-centered cubic cell (bcc)

the

spheres in each layer rest in the

depressions between spheres in the

previous layer.

The coordination number is 8.

Crystal Structure

Most of a cell’s atoms are shared by neighboring cells.

Crystal Structure

A corner atom is shared

by eight unit cells. An edge atom is shared by four unit cells. A face-centered atom is shared by two unit cells.

A simple cubic cell has the equivalent of only one complete atom

contained within the cell.

Crystal Structure

8 atoms at corners 1 equivlent atom 1  8

A body-centered cubic cell has two equivalent atoms:

A face-centered cubic cell contains four complete atoms:

Crystal Structure

8 atoms at corners 1 equivalent atom 1 atom in the center

2 equivalent atoms total

 

+



1 8

8 atoms at corners 1 equivalent atom

6 atoms on faces 3 equivalent atoms 4 equivalent atoms total

 

+  



1 8 1 2

Hexagonal close-packed (hcp)

structure:

Crystal Structure

Close packing starts with a layer of atoms (A)

Atoms in the second layer (B) fit into the depressions of the first layer

Hexagonal close-packed structure. Site directly over an

atom in layer A

Cubic close-packed (ccp)

structure:

Crystal Structure

Site directly over an atom in layer A (hcp)

Site NOT directly over an atom in layer A

(ccp)

Cubic close-packed structure

Closest packing:

Crystal Structure

Hexagonal close-packing

Edge length (

a

) and radius (

r

) are related:

Crystal Structure

Simple cubic Body-centered cubic Face-centered cubic

Strategy Using the given density and the mass of gold continued within a face-centered cubic unit cell, determine the volumes of the unit cell. Then, use the volume to determine the value of a, and use the equation a = √8r to find r. Be

sure to use consistent units for mass, length, and volume. The face-centered cubic unit cell contains a total of four atoms of gold [six faces, each shared by two unit cells, and eight corners, each shared by eight unit cells]. d = m/V and V = a3_.

Gold crystallizes in a cubic close-packed structure (face-centered cubic unit cell) and has a density of 19.3 g/cm3_{. Calculate the atomic radius of an Au atom in}

angstroms (Å).

Solution First, we determine the mass of gold (in grams) contained within a unit cell:

m = 4 atoms× × = 1.31×10-21 _{g/unit cell} unit cell 6.022×1 mol1023 _atoms 197.0 g Au_{1 mol Au}

Solution

Then we calculate the volume of the unit cell in cm3_:

V = = = 6.78×10-23 _cm3

Using the calculated volume and the relationship V = a3 (rearranged to solve for a), we determine the length of a side of a unit cell:

a = = √6.78×10-23 _cm3 _{= 4.08}×10-8 _cm

Using the relationship provided a = √8r (rearranged to solve for r), we determine

the radius of a gold atom in centimeters.

r = = = 1.44×10-8 _cm

Finally, we convert centimeters to angstroms:

1.44×10-8 _{cm × × = 1.44 Å}

m d

1.31×10-21

g 19.3 g/cm3

3_V 3

a

√8 4.08×10

-8

cm

√8

1×10-2

m 1 cm

1 Å 1×10-10

m

Think About It Atomic radii tend to be on the order of 1 Å, so this answer is reasonable.

Ionic crystals are composed of charged ions that are held together by

Coulombic attraction.

The unit cell of an ionic compound can be defined be either the

positions of the anions or the positions of the cations.

Types of Crystals

Crystal structures of three ionic compounds:

Types of Crystals

CsCl

Simple cubic lattice ZincblendeZnS structure (based on FCC)

CaF2

fluorite structure (based on FCC)

Strategy Determine the contribution of each ion in the unit cell based on its position. Referring to the figure, the unit cell has four Zn2+ _{ions completely}

contained within the unit cell, and S2- _{ions at each of the eight corners and at each}

of the six faces. Interior ions (those completely contained within the unit cell) contribute one, those at the corners each contribute one-eighth, and those on the faces contribute one-half.

How many of each ion are contained within a unit cell of ZnS?

Solution The ZnS unit cell contains four Zn2+ _{ions (interior) and four S}2- _ions

[8 × (corners) and 6 1 × (faces)] 8 1 2

Strategy Use the number of Na+ _{and Cl}- _{ions in a unit cell (four of each) to}

determine the mass of a unit cell. Calculate volume using the edge length given in the problem statement. Density is mass divided by volume (d = m/V). Be careful

to use units consistently.

The masses of Na+ _{and Cl}- _{ions are 22.99 amu and 35.45 amu, respectively. The}

conversion factor from amu to grams is

so the masses of the Na+ _{and Cl}- _{ions are 3.818}×10-23 _{g and 5.887}×10-23 _g,

respectively. The unit cell length is

564 pm × × = 5.64×10-8 _cm

The edge length of the NaCl unit cell is 564 pm. Determine the density of NaCl in g/cm3_.

1 g 6.022×1023

amu

1×10-12

m 1 pm

1 cm 1×10-2

m

Solution The mass of the unit cell is 3.882×10-22 _{g (4} × 3.818×10-23 _{g + 4} ×

5.887×10-23 _{g). The volume of a unit cell is 1.794}×10-22 _cm3 _[(5.64×10-8 _cm)3_].

Therefore, the density is given by

d = = 2.16 g/cm_1.7943.882_××₁₀10_-22-22 _cm g ₃ 3

Think About It If you were to hold a cubic centimeter (1 cm3_{) of salt in your}

hand, how heavy would you expect it to be? Common errors in this type of problem include errors of unit conversion–especially with regard to length and volume. Such errors can lead to results that are off by many orders of magnitude. Often you can use common sense to gauge whether or not a calculated answer is reasonable. For instance, simply getting the centimeter-meter conversion upside down would result in a calculated density of 2.16×1012 _g/cm3_{! You} know that a

cubic centimeter of salt doesn’t have a mass that large. (That’s billions of kilograms!) If the magnitude of a result is not reasonable, go back and check your work.

In covalent crystals, atoms are held together in an extensive

three-dimensional network entirely by covalent bonds.

Types of Crystals

In molecular crystals, the lattice points are occupied by molecules;

the attractive forces between them are van der Waals forces and/or

hydrogen bonding.

Types of Crystals

Strategy A face-centered metallic crystal contains four atoms per unit cell [8 × (corners) and 6 × (faces)]. Use the number of atoms per cell and the atomic mass to determine the mass of a unit cell. Calculate volume using the edge length given in the problem statement. Density is then mass divided by volume (d = m/V). Be sure to make all necessary unit conversions.

The mass of an Ir atom is 192.2 amu. The conversion factor from amu to grams is

so the mass of an Ir atom is 3.192×10-22 _{g. The unit cell length is}

383 pm × × = 3.83×10-8 _cm

The metal iridium (Ir) crystallizes with a face-centered cubic unit cell. Given that the length of the edge of a unit cell is 383 pm, determine the density of iridium in g/cm3_.

1 g 6.022×1023

amu

1×10-12

m 1 pm

1 cm 1×10-2

m

1

8 1 2

Solution The mass of the unit cell is 1.277×10-21 _{g (4} × 3.192×10-22 _{g). The}

volume of a unit cell is 5.618×10-23 _cm3 _[(3.83×10-8 _cm)3_{]. Therefore, the}

density is given by

d = = 22.7 g/cm1.277×10-21 g 3

5.62×10-23 _cm3

In metallic crystals, every lattice point is occupied by an atom of the

same metal.

Electrons are delocalized over

the entire crystal.

Delocalized electrons make metals

good conductors.

Large cohesive force resulting from

delocalization makes metals strong.

Types of Crystals

Types of Crystals

Amorphous Solids

Amorphous solids

lack a regular three-dimensional arrangement of

atoms.

Glass

is an amorphous solid.

Glass is a fusion product.

SiO

2

is the chief component.

Na

2

O and B

2

O

3

are typically fused with molten SiO

2

and allowed to

cool without crystallizing.

Amorphous Solids

Amorphous Solids