Chemistry part 20, Julia Burdge,2e (2009) pot

The other types of cubic cells, shown in Figure 12.16, are the body-centered cubic cell bcc and the face-centered cubic cell fcc.. Unlike the simple cube, the second layer of atoms in

(a) (b)

Figure 12.15 Arrangement of identical spheres in a simple cubic cell (a) Top view of one layer of spheres (b) Definition of a simple cubic cell

Primiti v e cubic Body-centered cubic Face-centered cubic

Figure 12.16 Three types of cubic cells The top view makes it easier to see the locations of the lattice points, but the bottom view is more realistic, with the spheres touching one another

Figure 12.17 In the body-centered cubic arrangement, the spheres in each layer rest in the depressions between spheres in the previous layer

The other types of cubic cells, shown in Figure 12.16, are the body-centered cubic cell (bcc) and the face-centered cubic cell (fcc) Unlike the simple cube, the second layer of atoms in the body-centered cubic arrangement fits into the depressions of the first layer and the third layer fits into the depressions of the second layer (Figure 12.17)

The coordination number of each atom in the bcc structure is 8 (each sphere is in contact with four others in the layer above and four others in the layer below) In the face-centered cubic cell, there are atoms at the center of each of the six faces of the cube, in addition to the eight corner atoms The coordination number in the face-centered cubic cell is 12 (each sphere is in contact with four others in its own layer, four others in the layer above, and four others in the layer below)

Because every unit cell in a crystalline solid is adjacent to other unit cells, most of a cell's atoms are shared by neighboring cells (The atom at the center of the body-centered

SECTION 12.3 Crystal Structure 475

Figure 12.18 (a) A comer atom in any cell is shared by eight unit cells (b) An edge atom is shared by

four unit cells (c) A face-centered atom in a cubic cell is shared by two unit cells

cubic cell is an exception.) In all types of cubic cells, for example, each corner atom belongs to

eight unit cells whose corners all touch [Figure 12.18(a)] An atom that lies on an edge, on the

other hand, is shared by four unit cells [Figure 12.18(b)], and a face-centered atom is shared

by two unit cells [Figure 12.18(c)] Because a simple cubic cell has lattice points only at each

of the eight corners, and because each corner atom is shared by eight unit cells, there will be

the equivalent of only one complete atom contained within a s imple cubic unit cell ( Figure

12.19] A body-centered cubic cell contains the equivalent of two complete atoms, one in the

center and eight shared corner atoms A face-centered cubic cell contains the equivalent of four

complete atoms-three from the six face-centered atoms and one from the eight shared corner

atoms

Closest Packing

There is more empty space in the simple cubic and body-centered cubic cells than in the

face-centered cubic cell Closest packing, the most efficient arrangement of atoms, starts with the

structure shown in Figure 12.20(a), which we call layer A Focu s ing on the only atom that

is surrounded completely by other atoms, we see that it has six immediate neighbors in its

own layer In the second layer, which we call layer B, atoms are packed into the depressions

between the atoms in the first layer so that all the atoms are as close together as possible

[Fig-ure 12.20(b)]

There are two ways that a third layer of atoms can be arranged They may sit in the sions between second-layer atoms such that the third-layer atoms lie directly over atoms in the first

depres-layer [Figure 12.20(c)] In this case, the third depres-layer is also labeled A Alternatively , atoms in the

third layer may sit in a different set of depressions such that they do not lie directly over atoms in

the first layer [Figure 12.20(d)] In this case, we label the third layer C

Figure 12.20 (a) In a close-packed layer, each sphere is in contact with six others (b) Spheres in the

second layer fit into the depressions between the first-layer spheres (c) In the hexagonal close-packed

structure, each third-layer sphere is directly over a first-layer sphere (d) In the cubic close-packed structure,

each third-layer sphere fits into a depression that is directly over a depression in the first layer

Figure 12.19 Because each sphere

is shared by eight unit cells and there are eight comers in a cube, there is the equivalent of one complete sphere inside a simple cubic unit cell

•

Figure 12.21 An arrangement for

obtaining the X-ray diffraction pattern

of a crystal The shield prevents the

intense beam of undiffracted X-rays

from damaging the photographic plate

476

Virtually all we know about crystal structure has been learned from X-ray diffraction studies

X-ray diffraction is the scattering of X rays by the units of a crystalline solid The scattering,

or diffraction patterns, produced are used to deduce the arrangement of particles in the solid lattice

In Section 6.1 we discussed the interference phenomenon associated with waves (see Figure 6.4) Because X rays are a form of electromagnetic radiation (i.e., they are waves), they exhibit interference phenomena under suitable conditions In 1912 Max von Laue4 correctly suggested that, because the wavelength of X rays is comparable in magnitude to the distances between lattice points in a crystal, the lattice should be able to diffract X rays An X-ray dif- fraction pattern is the result of interference in the waves associated with X rays

Figure 12.21 shows a typical X-ray diffraction setup A beam of X rays is directed at

a mounted crystal When X-ray photons encounter the electrons in the atoms of a crystalline solid, some of the incoming radiation is reflected, much as visible light is reflected by a mirror; the process is called the scattering of X rays

To understand how a diffraction pattern arises, consider the scattering of X rays by the atoms in two parallel planes (Figure 12.22) Initially, the two incident rays are in phase with each other (their maxima and minima occur at the same positions) The upper wave is scat- tered, or reflected, by an atom in the first layer, while the lower wave is scattered by an atom in the second layer For these two scattered waves to be in phase again, the extra distance traveled

by the lower wave (the sum of the distance between points Band C and the distance between points C and D) must be an integral multiple of the wavelength ("-) of the X ray; that is,

Equation 12.5 BC + CD = 2d sin 8 = n"- n = 1, 2, 3,

where 8 is the angle between the X rays and the plane of the crystal and d is the distance between adjacent planes Equation 12.5 is known as the Bragg equation, after William H Bragg and Sir William L Bragg s The reinforced waves produce a dark spot on a photographic film for each value of that satisfies the Bragg equation

x tube

X-ray beam

Shield Crystal

Photographic

plate

4 Max Theodor Felix von Laue (1879 - 1960) German physicist Von Laue received the Nobel Prize in Physics in

1914 for his discovery of X-ray diffraction

5 William Henry Bragg (1862-1942) and Sir William Lawrence Bragg (1890-1972) English physicists, father and son Both worked on X-ray crystallography The younger Bragg formulated the fundamental equation for X-ray dif- fraction The two shared the Nobel Prize in Physics in 1915

Incident ray s Reflected rays

d sin e

Figure 12.22 Reflection of X rays from two layers of atoms The lower wave travels a distance 2d

sin 0 longer than the upper wave does For the two waves to be in phase again after reflection, it must be

true that 2d sin 0 = 11"-, where "- is the wavelength of the X ray and 11 = 1, 2, 3, The sharply defined

spots in Figure 12.21 are observed only if the crystal is large enough to consist of hundreds of parallel

layers

X rays of wavelength 0.154 nm strike an aluminum crystal; the rays are reflected at an angle of

19.3° Assuming that 11 = 1, calculate the spacing between the planes of aluminum atoms (in pm)

that is responsible for this angle of reflection

Strategy Use Equation 12.5 to solve for d

Practice Problem A X rays of wavelength 0.154 nm are diffracted from a crystal at an angle of

14.17° Assuming that 11 = 1, calculate the distance (in pm) between layers in the crystal

Practice Problem B At what angle will X rays of wavelength 0.154 nm be diffracted from a

crystal if the distance (in pm) between layers in the crystal is 188 pm? (Assume 11 = 1 )

The X-ray diffraction technique offers the most accurate method for determining bond lengths and bond angles in molecules in solids Because X rays are scattered by electrons,

chemists can construct an electron-density contour map from the diffraction patterns by using

a complex mathematical procedure Basically, an electron-density contour map tells u s the

relative electron densities at various locations in a molecule The densities reach a maximum

near the center of each atom In this manner, we can determine the positions of the nuclei and

hence the geometric parameters of the molecule

•

1 nm = 1000 pm

Think About It The distance between layers of atoms in a crystal should be similar in magnitude

to the wavelength of the X rays diffracted by the crystal (compare 0.154 nm with 0.233 nm)

477

The noble gases, which are monatomic,

crystallize in the ccp structure, with the

exception of he l ium, which crystallizes in the

Note that this arrangement is the same as the face-centered unit cell

Figure 12.23 shows the exploded views and the structures resulting from these two

arrange-ments The ABA arrangement [Figure 12.23(a)] is known as the hexagonal close-packed (hcp)

structure, and the ABC arrangement [Figure 12.23(b)] is the cubic close-packed (ccp) structure,

which corresponds to the face-centered cube already described In the hcp structure, the spheres in every other layer occupy the same vertical position (ABABAB ), while in the ccp structure, the spheres in every fourth layer occupy the same vertical position (ABCABCA ) In both struc- tures, each sphere has a coordination number of 12 (each sphere is in contact with six spheres in its own layer , three spheres in the layer above, and three spheres in the layer below) Both the hcp and ccp structures represent the most efficient way of packing identical spheres in a unit cell, and the coordination number cannot exceed 12

Many metals form crystals with hcp or ccp structures For example, magnesium, titanium, and zinc crystallize with their atoms in an hcp array, while aluminum, nickel, and silver crystal- lize in the ccp arrangement A substance will crystallize with the arrangement that maximizes the

" ,

stability of the solid

Figure 12.24 summarizes the relationship between the atomic radius r and the edge

length a of a simple cubic cell, a body-centered cubic cell, and a face-centered cubic cell This relationship can be used to determine the atomic radius of a sphere in which the density of the crystal is known

SECTION 12.3 C rysta l Str uctu re 47 9

Sample Problem 12.4 illustrates the relationships between the unit cell type, cell sions, and density of a metal

dimen-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 (A)

Strategy Using the given density and the mass of gold contained within a face-centered cubic unit

cell, determine the volume of the unit cell Then, use the volume to determine the value of a, and use

the equation supplied in Figure 12.24(c) to find r Be sure to use consistent units for mass, length, and

volume

Setup 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- Figure 12.24(c)] D = m/Vand

Using the calculated volume and the relationship V = a 3 (rearranged to solve for a ) , we determine

the length of a side of a unit cell:

a = 3.yv = 3~ 6.78 X 10-23 cm3 = 4.08 X 10 - 8 cm

Using the relationship provided in Figure 12.22(c) (rearranged to solve for r), we determine the

radius of a gold atom in centimeters:

Practice Problem A When silver crystallizes, it forms face-centered cubic cells The unit cell edge

length is 4.087 A Calculate the density of silver

Practice Problem B The density of sodium metal is 0.971 g/cm3 , and the unit cell edge length is

o 4.285 A Determine the unit cell (simple, body-centered, or face-centered cubic) of sodium metal

12.3.1 Nickel has a face-centered cubic unit

cell with an edge length of 352.4 pm

Calculate the density of nickel

a) 2.227 g/cm3

b) 4.455 g/cm3 c) 38.99 g/cm3

d) 8.908 g/cm3 e) 11.14 g/cm3

12.3.2 At what angle would you expect X rays

of wavelength 0.154 nm to be reflected

from a crystal in which the distance between layers is 312 pm? (Assume

to be on the order of 1 A, so thIS

answer is reasonable

It i s i mportant to realize that the lattice points

u sed to define a unit cell must all be identicaL

W e can d efine the unit cell of Na( 1 based on

t h e pos iti ons of the Na + ions or the positions of

t he (1 - i ons

It is a common mistak e to i dentify th e (5(1

struc ture a s b od y - c entered cubic Remember

t h at t h e l a tti c e po i nts used to define a unit ce ll

m ust a ll be id enticaL In this case, they are all (l

-i ons ( 5 ( 1 has a simp l e cubic unit cell

Figure 12.25 The unit c e ll of an

ionic compound can be defined ei t her

by (a ) the positions of anion s or ( b ) the

po s it i on s of cation s

Figure 12.26 Cr ys tal s tructur es of

(a) CsCl , ( b) ZnS, and ( c ) CaF2 In e a ch

ca s e , the s maller s phere repre se nt s the

,

unit cell defined b y the arrangement of anions Look closely at the unit cell s hown in Figure 12.25(a) It

i s defined a s fcc b y the po s ition s of th e Cl- ion s Recall that there is the equivalent of four sphere s tained in the fc c unit cell ( half a s phere at each of s ix fa c es and one-eighth of a sphere at each of eight comer s) In thi s ca s e the s phere s are Cl- ion s, s o the unit ceU of N aC1 contains four Cl - ions Now look

con-at the po s ition s of the N a + ion s Ther e are Na + ions centered on each edge of the cube, in addition to

o ne N a + at the cent e r Each s phere on the cube 's edge is s hared b y four unit cell s, and there are 12 such

ed g e s Thu s, the unit cell in Figure 12 2 5 ( a ) al s o contains four N a + ions ( one-quarter sphere at each of

1 2 edge s, gi v ing three s phere s, and one s phere at the center) The unit cell of an ionic compound always contain s the s ame ratio of cation s to anion s a s the empirical formula of the compound

Figure 12.26 s ho ws the cry s tal s tructure s of three ionic compound s : C s Cl, ZnS, and CaF2

." " - ." '

Ce s ium chloride [Figur e 12.26 ( a ) ] ha s the s imple cubic lattice Despite the apparent similarity of

SECTION 12.4 Types of Crystals 481

the fonnulas of CsCl and NaCl, CsCl adopts a different arrangement because the Cs + ion is much

larger than the Na+ ion Zinc sulfide [Figure l2.26(b)] has the zincblende structure, which is based

on the face-centered cubic lattice If the So- ion s occupy the lattice points, the smal ler Zn2+ ions are

arranged tetrahedrally about each So- ion Other ionic compounds that have the zincblende structure

include CuCl, BeS, CdS, and HgS Calcium fluoride [Figure 12.26(c)] has the fluorite structure The

unit cell in Figure 12.26(c) is defined based on the positions of the cations, rather than the position s

of the anions The Ca2+ ions occupy the lattice point s, and each F - ion is surrounded tetrahedrally by

four Ca2+ ion s The compounds SrF2, BaF z, BaClz, and PbF2 also have the fluorite s tructure

Sample Problems 12.5 and 12.6 show how to determine the number of ions in a unit cell and

the density of an ionic crystal , re spec tively

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

Strategy Determine the contribution of each ion in the unit cell based on its position

Setup Referring to Figure 12.26, 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 each contribute one-half

Solution The ZnS unit cell contains four Zn2+ ions (interior) and four S2- ions [8 X J (corners) and

6 X ~ (faces)]

Practice Problem A Referring to Figure 12.26, determine how many of each ion are contained

within a unit cell of CaF2

Practice Problem B Referring to Figure 12.26, determine how many of each ion are contained

within a unit cell of CsC!

Sample Problem 12.6

The edge length of the NaCl unjt cell is 564 pm Determine the density of NaCl in g/cm3

Strategy Use the number of Na + and CC ions in a unit cell (four of each) to determine the mass of

a unit cel! 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

+ Setup The masses of Na and CI- ions are 22.99 amu and 35.45 amu, respectively The conversion

factor from amu to grams is

1 g

6.022 X 1023 amu

so the masses of the Na + and CI - ions are 3.818 X 10- 23 g and 5.887 X 10-23 g, respectively The

unit cell length is

564ym X 1 X 1O- 12fi{ x ' 1 cm = 5.64 X 10-8 cm

lym 1 X 1O-2.-m

Solution The mass of a unit cell is 3.882 X 10- 22 g (4 X 3.818 X 10- 23 g + 4 X 5.887 X 10- 23 g)

The volume of a unit cell is 1.794 X 10- 22 cm3 [(5.64 X 10-8 cm)3] Therefore, the density is given by

3.882 X 10- 22 g

1.794 X 10 cm

Practice Problem A LiF has the same unit cell as NaCl (fcc) The edge length of the LiF unit cell is

402 pm Determine the density of LiF in g/cm3

Practice Problem B NiO also adopts the face-centered-cubic arrangement Given that the density of

NiO is 6.67 g/cm3, calculate the length of the edge of its unit cell (in pm)

Think About It Make sure that

the ratio of cations to anions

that you determine for a unit cell

matches the ratio expressed in the

compound's empirical formula

Note tha t the mass of an atomic ion is treated the same as the mass of th e parent atom I n

the se cases, the mass of an electron is not

significant [ ~ Section 2.2, Table 2.1]

· 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 especially with regard to length

conversion-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

getting the centimeter-meter

conversion upside down would

result in a calculated density of 2.16 X 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

Most ionic crystals have high melting points, an indication of the strong cohesive forces holding the ions together A measure of the stability of ionic crystals is the lattice energy

conduct electricity because the ions are fixed in position In the molten (melted) state or when solved in water, however, the compound's ions are free to move and the resulting liquid conducts electricity

dis-Covalent Crystals

In covalent crystals, atoms are held together in an extensive three-dimensional network entirely

by covalent bonds Well-known examples are two of carbon's allotropes: diamond and graphite

In diamond, each carbon atom is Sp3 -hybridized and bonded to four other carbon atoms [Figure 12.27 (a)] The strong covalent bonds in three dimensions contribute to diamond's unusual hard-

atoms are arranged in six-membered rings [Figure 12.27(b)] The atoms are all sp2-hybridized, and each atom is bonded to three other atoms The remaining unhybridized 2p orbital on each carbon atom is used in pi bonding In fact, each layer of graphite has the kind of delocalized molecular orbital that is present in benzene [ ~~ Sect i o n 9.7] Because electrons are free to move around in this extensively delocalized molecular orbital, graphite is a good conductor of electricity in direc- tions along the planes of the carbon atoms The layers are held together by weak van der Waals forces The covalent bonds in graphite account for its hardness; however, because the layers can slide past one another, graphite is slippery to the touch and is effective as a lubricant It is also used

as the "lead" in pencils

Another covalent crystal is quartz (Si02) The arrangement of silicon atoms in quartz is similar to that of carbon in diamond, but in quartz there is an oxygen atom between each pair of Si atoms Because Si and 0 have different electronegativities, the Si -0 bond is polar Nevertheless, Si02 is similar to diamond in many respects, such as being very hard and having a high melting point (161O°C)

335 pm

Figure 12.27 Structures of (a) diamond and (b) graphite Note that in diamond, each carbon atom is bonded in a tetrahedral arrangement to four other carbon atoms In graphite, each carbon atom is bonded in a trigonal planar arrangement to three other carbon atom s The distance between layers in graphite is 335 pm

SECTION 12.4 Types of Crystals 483

In a molecular crystal, the lattice points are occupied by molecules , so the attractive forces between

them are van der Waals forces and/or hydrogen bonding An example of a molecular crystal is

solid sulfur dioxide (S02), in which the predominant attractive force i s a dipole-dipole interaction

Intermolecular hyqrogen bonding is mainly respon s ible for maintaining the

three-dimensionallat-tice of ice (Figure 12.28) Other examples of molecular crystals are 12, P4, and S s

Except in ice, molecules in molecular crystals are generally packed together as closely as their size and shape allow B ec ause van der Waals forces and hydrogen bonding are usually quite

weak compared with covalent and ionic bonds , molecular crystals are more easily broken apart than

ionic and covalent crystals Indeed, mo st molecular crystals melt at temperature s below 100 ° C

Metallic Crystals

Every lattice point in a metallic crystal is occupied by an atom of the same metal Metallic crystals

are generally body-centered cubic, face-centered cubic, or hexagonal clo se -packed Consequently,

metallic elements are usually very den se

The bonding in metal s i s quite different from that in other types of crystals In a metal, the bonding electrons are delocalized over the entire crystal In fact, metal atom s in a crystal can be

imagined as an array of po s itive ion s immersed in a sea of delocalized valence electrons (F igure

12.29) The great cohesive force re s ulting from delocalization is re spo n s ible for a metal's s trength,

whereas the mobility of the delocalized electrons makes metals good conductors of heat and

elec-tricity Table 12.4 summarizes the properties of the four different type s of crystals di sc u ssed Note

that the data in Table 12.4 refer to the solid pha se of each s ub sta nce listed

Dispersion and dipole-dipole forces, hydrogen bond s

Metallic bonds

t lncluded in this category are crystals made up of individual atoms

three-by long dotted lines between 0 and

H The empty space in the structure accounts for the low density of ice, relati ve to liquid water

Figure 12.29 A cross section of a metallic c rystal Each circled po s itive

c harge represents the nucleu s and inner electrons of a metal atom The grey

area surrounding the positive metal ions indicates the mobile "sea" of electrons

Examples

NaCl, LiF, MgO, CaC03

C ( diamond ),* Si02 (quartz)

All metallic elements, such

as Na, Mg, Fe, Cu

Think About It Metals typically

have high densities, so common

sense can help you decide whether

or not your calculated answer is

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

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

1 g 6.022 X 1023 amu

so the mass of an Ir atom is 3.192 X 10-22 g The unit cell length is

Solids are mo s t stable in cr ys talline form However , if a solid is formed rapidly (e.g., when a liquid

is cooled quickly ), its atoms or molecules do not have time to align themselves and may become locked in positions other than tho s e of a regular crystal The resulting solid is said to be amor-

phous Amorphous solids, such as glas s , lack a regular three-dimensional arrangement of atoms

In this section , we will briefly discuss the properties of glass

Glas s i s one of civilization's most valuable and versatile materials It is also one of the olde s t gla s s articles date back as far as 1000 B.C Glass commonly refers to an optically trans-

parent fu s ion product of inorganic materials that has cooled to a rigid state without crystallizing

By fusion product we mean that the glass is formed by mixing molten silicon dioxide (Si02), its chief component , with compound s such as sodium oxide (NazO), boron oxide (B 2 0 3), and certain transition metal oxide s for color and other properties In some respects, glass behaves more like a liquid than a s olid

There are about 800 different types of glass in common use today Figure 12.30 shows dimensional schematic representations of crystalline quartz and amorphous quartz glass Table 12.5 list s the composition and properties of quartz, Pyrex, and soda-lime glass The color of glass

two-is due largely to the presence of oxides of metal ions mostly transition metal ions For ple, green gla s s contains iron(III) oxide (Fe 2 0 3 ) or copper(II) oxide (CuO), yellow glass contains uranium(IV) oxide (UO z ), blue glass contains cobalt(Il) and copper(II) oxides (CoO and CuO), and red glass contains small particles of gold and copper

exam-Phase Changes

A phase is a homogeneous part of a system that i s s eparated from the rest of the system by a defined boundary When an ice cube floats in a glass of water, for example, the liquid water is one phase and the s olid water (the ice cube) is another Although the chemical properties of water are the same in both phases, the physical properties of a solid are different from those of a liquid

well-SECTION 12.6 Phase Changes 485

Low thel llIal expan s ion, transparent to a wide

range of wavelength s U s ed in optical research

Low thermal expan s ion; transparent to visible and infrared , but not to ultraviolet light Used

thermal shocks Tran s mit s visible light but

bottles

phase change phas e ' change s in a " s y s tem are ' generaliy c au s ed by the addition ' or ' removal ' of

Example

Evaporation (or vaporization) of water

Condensation of water vapor

these pha s e changes: vaporization and c o nden s ation Figure 12.31 s ummari z e s the v arious type s

of phase changes

Liquid-Vapor Phase Transition

tempera-ture When the vapor pressure reaches the external pressure , the liquid boil s In fact , the boiling

point of a substance is defined as the temperature at which it s vapor pre ss ure equals the external,

atmospheric pressure As a result , the boiling point of a s ub s tance varie s with the external pre s

sea level, the vapor pressure of water ( or any liquid ) reache s th e external pre ss ure at a lower

point is related to the molar heat o/vaporization (t:lllvap), the amount of heat required to vapori z e

a mole of sub s tance at its boiling point Indeed, the data in Table 12.6 s how that the boiling point

by the strength of intermolecular forces For example , argon (Ar) and methane (ClI4) , which have

a-tion Diethyl ether (C 2 lIsOC 2 lI S ) ha s a dipole moment , and the dipole-dipole forces account for it s

moderately high boiling point and !1H v ap Both ethanol ( C 2 lI s OlI ) and water ha v e s trong hydrogen

Figure 12 30 Two-dimensional representation of (a) crystalline quartz

quartz glass The small spheres

P hase chan ge s a r e p hysic al cha ng e s

[ ~~ Section 1.4]

T he t em p e ra tu re at which the v a por pr essu r e

o f a l i quid is e q u a l t o 1 atm i s calle d the normal boiling point

Figure 12.31 The six possible

phase changes: melting (fusion),

vaporization, sublimation, deposition,

condensation, and freezing

To is the highest temperature at which a

substance c an exi s t as a liquid

•

Vaporization liquid_gas

Boiling Point (0C)

-186

80.1 78.3 34.6

357

-164

100

Condensation gas_liquid

Freezing

liquid • so lid

6.3 31.0 39.3

26.0

59.0

9.2 40.79

bonding, which accounts for th eir high boiling points and large !1H v ap values Strong metallic

bonding causes mercury to have the highest boiling point and !1Hv a p of the liquids in Table 12.6

Interestingly, benzene (C6H6), although nonpolar, has a high polarizability due to the distribution

of it s electrons in delocali ze d 'IT molecular orbitals The di spe rsion forces that result can be as strong as (or even stronger than) dipole-dipole forces and/or hydrogen bonds

The opposite of vaporization is condensation In principle, a gas can be liquefied (made to

condense) either by cooling or by applying pressure Cooling a sa mple of gas decreases the kinetic energy of it s molecule s, so eventually the molecule s aggregate to form small drops of liquid

Applying pressure to the gas (compress ion ), on the other hand , reduces the distance between ecules, so they can be pulled toget her by intermolecular attractions Many liquefication processes

mol-u se a combination of reduced temperature and increased pressure

Every substance ha s a critical temperature (Tc) above which its gas phase cannot be fied , no matter how great the applied pressure Critical pressure (Pc) is the minimum pressure that must be applied to liquefy a substance at its critical temperature At temperatures above the

lique-critical temperature , there i s no fundamental distinction between a liquid and a gas we simply

have a fluid A fluid at a temperature and pressure that exceed T c and P c> respectively, is called a

supercriticalfluid Supercritical fluids have some remarkable properties and are used as so lvents

in a wide variety of indu stria l applications The first such large- sca le industrial use was the

decaf-feination of coffee with s up erc ritical CO ?

Table 12.7 li sts the critical temperatures and critical pre ss ur es of a number of common

sub-sta nce s The critical temperature of a s ubstance reflects the s trength of its intermolecular forces Benzene, ethanol, mercury , and water, which have strong intermolecular forces, also have high

critical temperatures compared with the other substances listed in the table

Solid-Liquid Phase Transition

The transformation of liquid to sol id i s called freezing, and the reverse process is called melting,

or fusion The melting point of a solid or the freezing point of a liquid is the temperature at which