Chapter 13
The Chemistry of Solids
Chapter 13
The Chemistry of Solids
INSTRUCTOR’S NOTES
New to the 7th edition was a chapter devoted to the chemistry of solids. The chapter combined content previously
found in a chapter on liquids and solids with the content on bonding in ionic solids. The number of homework
problems with this chapter has been significantly increased in the 8th edition.
We emphasize with our students that although many substances are solids, there are many different ways in which
the atoms and or ions can be arranged (Table 13.1). The steps involved in calculating atom radius or density of a
solid from crystal structure data can be confusing to students. It is helpful to work many examples in lecture.
The ionic and molecular solids pictured in this chapter can be found on the ChemistryNow website in the Molecular
Modeling Database. The ionic solids pictured have been manipulated (by changing the ionic radii) in order to make
the ions appear with the correct radius.
When possible, bring models of ionic solids and/or cubic unit cells to class (available from Klinger Educational
Products, http://www.klingereducational.com) or use the Solid State Model Kit available from the Institute for
Chemical Education (http://ice.chem.wisc.edu) to build smaller models of crystalline solids.
SUGGESTED DEMONSTRATIONS
1.
Solids

Herman, Z.S. “Ionic Crystals: A Simple and Safe Lecture Demonstration of the Preparation of NaI
from Its Elements,” Journal of Chemical Education 2000, 77, 619.

Cady, S. G. “Use of Pom Pons to Illustrate Cubic Crystal Structures,” Journal of Chemical Education
1997, 74, 794.

The close packing of spheres (such as the photos on page 595) is easy to do on an overhead projector.
We have also used oranges and grapefruits, as well as Styrofoam balls.
253
Chapter 13
The Chemistry of Solids
SOLUTIONS TO STUDY QUESTIONS
13.1
Two possible unit cells:
Each unit cell contains 1 A square and 8 B squares, so the
simplest formula is AB8
13.2
The area inside the box is a unit cell. Each unit cell contains 2 A squares and 2 B squares,
so the simplest formula is AB.
13.3
8 corner Ca  1/8 = 1 Ca
6 edge O  1/2 = 3 O
1 internal Ti = 1 Ti
The formula is CaTiO3
13.4
8 corner Ti  1/8
= 1 Ti
4 face O  1/2
=2O
1 internal Ti
= 1 Ti
2 internal O
=2O
= 2 Ti total
= 4 O total
There are two TiO2 units per unit cell.
13.5
(a) 8 corner O  1/8 = 1 O
1 internal O
=1O
4 internal Cu
= 2 O total
= 4 Cu
= 4 Cu total
There is a ratio of two copper ions to one oxide ion for a formula of Cu 2O.
(b) The oxidation number of copper is +1.
13.6
(a) face-centered cubic
(b) tetrahedral holes
(c) (8 corner Ca2+  1/8) + (6 face Ca2+  1/2) = 4 Ca2+
The formula is CaF2
254
8 internal F– = 8 F–
Chapter 13
13.7
The Chemistry of Solids
Unit cell volume:
40.08 g
1 cm3
1 mol Ca
4 Ca atoms



= 1.73  10–22 cm3
1 mol Ca 1.54 g
unit cell
6.022  1023 atoms
Unit cell edge length:
V = 1.73  10–22 cm3 = (edge length)3
edge length = 3 1.73  10–22 cm3 = 5.57  10–8 cm
face diagonal = 4 · radius = 2  edge length
radius =
13.8
2  (5.57  10–8 cm)
= 1.97  10–8 cm = 197 pm
4
Assume that copper has a face-centered cubic unit cell. Calculate the density of the unit cell based on this
assumption and compare it with the actual density of copper:
63.546 g
1 mol Cu

= 1.0552  10–22 g/atom
23
1 mol Cu
6.02214  10 atoms
1.0552  10–22 g
4 Cu atoms
mass of unit cell =

= 4.2208  10–22 g/unit cell
1 Cu atom
1 unit cell
face diagonal = 4  atom radius = 2  edge length
4  atom radius
4  127.8 pm
edge length =
=
= 361.5 pm = 3.615  10–8 cm
2
2
mass of one Cu atom =
unit cell volume = (edge length)3 = (3.615  10–8 cm)3 = 4.723  10–23 cm3
unit cell density =
4.2208  10–22 g
= 8.937 g/cm3
4.723  10–23 cm3
The calculated density closely matches the actual density (8.95 g/cm3) so the copper unit cell is most likely
face-centered cubic.
13.9
Use the cell density or the ion radii to calculate the edge length.
Using cell density,
edge length =
3
4 KI
1 mol KI
166.00 g
1 cm3



= 3.53  10–22 cm3
1 unit cell
3.12 g
6.022  1023 formula units 1 mol KI
3.53  10–22 cm3 = 7.07  10–8 cm = 707 pm
Using ionic radii, edge length = 2(K+ radius) + 2(I– radius) = 2(133 pm) + 2(220 pm) = 706 pm
Note that one cannot in this problem assume the I- touch along the cell diagonal and use I- radius to find the
edge length. Edge length = 4(I- radius) = 4(220 pm)/21/2 = 622 pm. The size of the ions are such that the Iions cannot touch along the diagonal.
255
Chapter 13
13.10
The Chemistry of Solids
(8 corner Cl–  1/8) = 1 Cl–
1 internal Cs+ = 1 Cs+
Unit cell volume:
1 CsCl
1 mol CsCl
168.4 g
1 cm3
·


= 7.01  10–23 cm3
unit cell 6.022  1023 formula units 1 mol CsCl
3.99 g
Unit cell edge length:
V = 7.01  10–23 cm3 = (edge length)3
edge length
edge length = 3 7.01  10–23 cm3 = 4.12  10–8 cm
face diagonal
(face diagonal)2 = (edge length)2 + (edge length)2
face diagonal =
2 · edge length
(cube diagonal)2 = (edge length)2 + (face diagonal)2
(cube diagonal)2 = (edge length)2 + ( 2 · edge length)2
cube diagonal =
3 · (edge length)
cube diagonal =
3 · 412 pm
edge length
face diagonal
cube diagonal
cube diagonal = 714 pm
(2  Cl– radius) + (2  Cs+ radius) = cube diagonal
(2  181 pm) + (2  Cs+ radius) = 714 pm
Cs+ radius = 176 pm
13.11
The 1000 2s orbitals will combine to form 1000 molecular orbitals. In the lowest energy state, 500 of these
will be populated by pairs of electrons, and 500 will be empty.
13.12
4 orbitals/Mg atom · 6.022 x 1023 Mg atoms/mol = 24.09 x 1023 orbitals/mol = 2.409 x 1024 orbitals/mol.
Each Mg atom has a total of two electrons occupying its four valence orbitals. The molecular orbitals will
consequently similarly be ¼ full.
13.13
In metals, thermal energy causes some electrons to occupy higher energy orbitals in the band of molecular
orbitals. For each electron promoted, two singly occupied levels result: a negative electron above the Fermi
level and a positive hole below the Fermi level. Electrical conductivity results because in the presence of an
electric field, these negative electrons will move toward the positive side of the field, and the positive holes
will move toward the negative side.
13.14
A metal can absorb energy of nearly any wavelength, causing an electron to move to a higher energy state.
The now-excited system can immediately emit a photon of the same energy as the electron returns to the
original energy level. This rapid and efficient absorption and reemission of light makes metals appear
shiny.
13.15
In carbon, the band gap is too large for electrons to up in energy from the valence band to the conductance
band, whereas in silicon the band gap is small enough to permit this.
13.16
256
C > Si > Ge > Sn. The band gap gets smaller with increased metallic character.
Chapter 13
13.17
The Chemistry of Solids
An intrinsic semiconductor is one that naturally can have electrons move across the band gap, whereas an
extrinsic semiconductor requires the addition of dopants in order for it to conduct.
13.18
The aluminum has one less valence electron than the silicon. The Si-Al bonds form a discrete but empty
energy band at a level higher in energy than the valence band but lower than the conduction band. The gap
between the valence and acceptor bands is same, so electrons can be promoted with concurrent generation
of positive holes in the valence band. This is a p-type semiconductor.
13.19
RbI < LiI < LiF < CaO
13.20
Lattice energy depends directly on ion charges and inversely on the distance between ions. The sizes of the
Cl–, Br–, and I– ions fall in a relatively narrow range (181, 196, and 220 pm, respectively), and the ion sizes
change by only 15–24 pm from one ion to the next. Therefore, their lattice energies are expected to
decrease in a narrow range. The F– ion (133 pm), however, is only 74% as large as the Cl– ion, so the lattice
energy of NaF is much more negative.
13.21
As the ion-ion distance decreases, the force of attraction between the ions increases. This should make the
lattice more stable, and more energy should be required to melt the compound.
13.22
(a) NaCl
13.23
Li(s)  Li(g)
(b) MgO
(c) MgS
Li(g)  Li+(g) + e–
1/
2
F2(g)  F(g)
F(g) + e–  F–(g)
Li+(g) + F–(g)  LiF(s)
Li(s) + 1/2 F2(g)  LiF(s)
∆fH° = +159.37 kJ/mol
IE = +520. kJ/mol
∆fH° = +78.99 kJ/mol
EA = –328.0 kJ/mol
∆latticeU = –1037 kJ/mol
∆fH° = –607 kJ/mol
13.24
∆latticeU = –435.4 kJ/mol – 80.9 kJ/mol – 403 kJ/mol – 131.3 kJ/mol – (–349.0 kJ/mol) = –701.6 kJ/mol
13.25
(a) (8 corner C  1/8) + (6 face C  1/2) + (4 internal C) = 8 C atoms/unit cell
(b) face-centered cubic arrangement with carbon atoms in tetrahedral holes
13.26
(a) induced dipole/induced dipole
(b) Since there are only weak intermolecular forces between the layers in graphite, the layers slide over
each other easily, giving graphite a slippery feel. Pushing a pencil lead against paper causes some of
the carbon layers to rub off, leaving a black mark.
13.27
Type of solid: particles, forces of attraction
(a) metallic: metal atoms, metallic bonding
(b) ionic: ions, ion-ion interactions
(c) molecular: molecules, covalent bonds within the molecules and intermolecular forces between the
molecules
257
Chapter 13
The Chemistry of Solids
(d) network: extended network of covalently bonded atoms, covalent bonds
(e) amorphous: covalently bonded networks with no long-range regularity, covalent bonds
13.28
Ionic: hard, brittle, high melting point, poor electric conductivity as solid while good as liquid, often water
soluble
Metallic: malleable, ductile, good electric conductivity in solid and liquid, good heat conductivity, wide
range of hardness and melting points
Molecular: low to moderate melting points and boiling points, soft, poor electric conductivity in solid and
liquid
Network: wide range of hardness and melting points, poor electric conductivity (with some exceptions)
Amorphous: noncrystalline, wide temperature range for melting, poor electric conductivity (with some
exceptions)
13.29
Type of solid, particles, forces, properties
(a) network, covalently bonded atoms, covalent bonds, semiconductor
(b) amorphous, covalent bonds within the polymer molecules and dispersion forces between the polymer
molecules, thermal insulator
(c) network, covalently bonded atoms, covalent bonds, very hard material
(d) ionic, Ca2+ and TiO32- ions, ion-ion interactions, high melting point
13.30
(a) network, covalently bonded atoms, covalent bonds, semiconductor
(b) network, covalently bonded atoms, covalent bonds, brittle
(c) molecular, molecules, covalent bonds within the molecules and intermolecular forces between the
molecules, poor conductor
(d) ionic, Na+ and SO42- ions, ion-ion interactions, high melting point
13.31
15.5 g ·
1 mol C6 H6 9.95 kJ
·
= 1.97 kJ heat evolved (–1.97 kJ)
78.11 g
1 mol
+1.97 kJ of heat must be absorbed to convert the solid to a liquid.
13.32
The total heat required is the sum of the heat required to (1) heat the solid from 25 °C to its melting point
and (2) liquefy the solid at its melting point.
qsolid = (5.00 g Ag)(0.235 J/g·K)(1235 K – 298 K) = 1.10  103 J
3
2
1 mol Ag 11.3 kJ 10 J
·
·
= 5.24  10 J
107.9 g 1 mol 1 kJ
3
2
3
= (1.10  10 J) + (5.24  10 J) = 1.62  10 J
qmelting = 5.00 g ·
qtotal
258
Chapter 13
13.33
The Chemistry of Solids
(a) The positive slope of the solid/liquid equilibrium line indicates that liquid CO 2 is less dense than
solid CO2.
(b) gas phase
(c) no
13.34
(a) Xenon is a gas at room temperature and 1.0 atm pressure.
(b) Xenon is a liquid at 0.75 atm pressure and –114 ºC.
(c) When the pressure on a sample of liquid xenon is 380 mm Hg (0.5 atm), the temperature is between
–117 and –119 ºC.
(d) At –122 ºC, the vapor pressure of solid xenon is 0.25 atm.
(e) The solid phase is more dense than the liquid phase because the solid–liquid equilibrium line has a
positive slope.
13.35
The total heat required is the sum of the heat required to (1) heat the liquid from –50.0 ºC to its boiling
point, (2) vaporize the gas, and (3) heat the vapor to 0.0 ºC.
3
2
10 g
1 kJ
)(4.7 J/g  K)(239.9 K – 223.2 K) ∙ 3 = 9.4  10 kJ
1 kg
10 J
3
4
10 g 1 mol NH3 23.33 kJ
qevaporation = 12 kg ∙
∙
∙
= 1.6  10 kJ
1 kg
17.0 g
1 mol
3
2
10 g
1 kJ
qvapor = (12 kg ∙
)(2.2 J/g  K)( 273.2 K – 239.9 K) ∙ 3 = 8.8  10 kJ
1 kg
10 J
qliquid = (12 kg ∙
qtotal = 9.4  102 kJ + 1.6  104 kJ + 8.8  102 kJ = 1.8  104 kJ
13.36
The total heat required is the sum of the heat required to (1) cool the gas from 40.0 °C to its boiling point,
(2) condense the gas, and (3) cool the liquid to –40.0 °C.
1 mol g CCl2 F2
 117.2 J/mol  K  (243.4 K – 313.2 K) = –1350 J
120.9 g
3
1 mol CCl2 F2 –20.11 kJ 10 J
qcondensation = 20.0 g ∙
∙
∙
= –3330 J
120.9 g
1 mol
1 kJ
1 mol CCl2 F2
qliquid = 20.0 g ∙
∙ 72.3 J/mol  K ∙ (233.2 K – 243.4 K) = –122 J
120.9 g
3
qtotal = (–1350 J) + (–3330 J) + (–122 J) = –4.80  10 J
qgas = 20.0 ∙
259
Chapter 13
The Chemistry of Solids
800
700
13.37
normal
melting
point
Pressure (mm Hg)
600
normal
boiling
point
500
400
300
triple point
200
100
0
50
60
70
80
90
100
Temperature (K)
The estimated vapor pressure at 77 K is approximately 200 mmHg. The solid-liquid equilibrium line has a
positive slope, so the density of liquid O2 is less than that of solid O2.
13.38
(a) body-centered cubic
(b) (8 corner W  1/8) + (1 internal W) = 2 W atoms/unit cell
(c) (cube diagonal)2 = (edge length)2 + (face diagonal)2
edge length
(cube diagonal)2 = (edge length)2 + ( 2 · edge length)2
cube diagonal =
3 · (edge length)
cube diagonal =
3 · 316.5 pm
face diagonal
cube diagonal
cube diagonal = 548 pm
radius =
13.39
548pm
= 137 pm
4
(face diagonal)2 = (edge length)2 + (edge length)2
face diagonal =
2 · edge length
face diagonal =
2 · 409 pm
edge length
face diagonal
face diagonal = 578 pm
radius =
13.40
578pm
= 145 pm
4
8 corner Ca  /8
= 1 Ca
8 edge C  1/4
=2C
1 internal Ca
= 1 Ca
2 internal C
=2C
= 2 Ca total
= 4 C total
There is a ratio of two calcium ions to four carbon ions for a formula of CaC 2.
260
Chapter 13
13.41
The Chemistry of Solids
Unit cell volume:
3
– 23
3
4 Ir atoms 192.22 g 1 cm
1 mol Ir
∙
∙
∙
= 5.659  10 cm
unit cell 1 mol Ir 22.56 g 6.0221  1023 atoms
Unit cell edge length:
– 23
V = 5.659  10
3
3
cm = (edge length)
–23
edge length = 5.659  10
3
–8
cm = 3.839  10 cm
3
face diagonal = 4  radius = 2  edge length
–8
–8
2  (3.839  10 cm)
radius =
= 1.357  10 cm = 135.7 pm
4
13.42
Assume that vanadium metal has a face–centered cubic unit cell. Calculate the density of the unit cell based
on this assumption and compare it with the actual density of vanadium:
50.94 g 
1 mol V
–23
= 8.459  10 g
1 mol V 6.022  1023 atoms
–23
8.459  10 g 4 V atoms
–22

mass of unit cell =
= 3.384  10 g/unit cell
1 V atom
1 unit cell
face diagonal = 4  atom radius = 2  edge length
4  atom radius 4  132 pm
edge length =
=
= 373 pm = 3.73  10–8 cm
2
2
unit cell volume = (edge length)3 = (3.73  10–8 cm)3 = 5.19  10–23 cm3
3.384  10–22 g
unit cell density =
= 6.52 g/cm3
5.19  10–23 cm3
The calculated density does not agree with the actual density (6.11 g/cm3). If the unit cell is assumed to be
mass of one V atom =
body–centered cubic, the calculated density is 5.97 g/cm3; simple cubic results in a calculated density of
4.60 g/cm3. The vanadium unit cell is most likely body–centered cubic.
13.43
Calculate the mass of one CaF2 ion pair and compare this to the mass of one mole of CaF2:
–8
(5.46295  10 cm) 
3
–22
3.1805 g 1 unit cell
= 1.2963  10 g/CaF2

3
4 CaF2
1 cm
23
78.077 g/1 mol CaF2
= 6.0230  10 CaF2 /mol
–22
1.2963  10 g/CaF2
13.44
Calculate the mass of one iron atom and compare this to the mass of one mole of iron.
3

1m
102 cm
–23
3
V =  286.65 pm ∙ 12
∙
 = 2.3554  10 cm
10 pm 1 m 

7.874 g 1 unit cell
2.3554  10– 23 cm3 ∙
∙
= 9.2730  10–23 g/Fe atom
1 cm3 2 Fe atoms
55.845 g/1 mol Fe
= 6.0223  1023 Fe atoms/mol
9.2730  10–23 g/Fe atom
13.45
Diagram A:
area covered by circles
area of one circle
 r2
=
=
= 0.785 = 78.5%
total area
(2 rcircle )2
4 r2
261
Chapter 13
The Chemistry of Solids
Diagram B:
1
1
 base  height = (2 r)( 3 r)
2
2
1
1
area covered by circles =
 area of circle =  r 2
2
2
1
 r2
area covered
2
=
= 0.907 = 90.7%
1
area of triangle
(2 r)( 3 r)
2
area of triangle =
13.46
4 3
1 sphere  r
volume occupied by spheres
3
∙ 100% =
∙ 100% = 52%
3
(2 r)
% occupied space = total volume of unit cell
% empty space = 100 – % occupied space = 48%
13.47
(a) fcc lattice with ½ of tetrahedral holes occupied
(b)
8 Si atoms 28.086 g
1 mol Ir
·
·
= 3.731  10–22 g/unit cell
unit cell
1 mol Si 6.0221  1023 atoms
volume of unit cell = (5.431  10–8 cm)3 = 1.602  10–22 cm3
density =
3.731  10–22 g
= 2.329 g/cm3
1.602  10–22 cm3
Use geometric relationship between Si at cell corner, Si in tetrahedral hole, and Si in center of cell face
along with the cell face diagonal to calculate Si radius.
2 · edge length =
face diagonal =
2 · 543.1 pm = 768.1 pm
½ face diagonal = 384.1 pm
1
/ 2 (384.1 pm)
sin(109.5º/2) =
Si-Si distance
109.5°/2
Si in center
of cell face Si
Si-Si distance = 235.1 pm
1
Si radius = 117.6 pm
13.48
Si
Si in
tetrahedral hole
Si
/2 face diagonal
Si at cell
corner
Use geometric relationship between C atoms at cell corners and in center of cell faces and Si atoms in
tetrahedral holes along with the Si–C distance to calculate face diagonal.
1
sin(109.5º/2) =
1
/ 4 (face diagonal)
/ 4 (face diagonal)
=
Si-C distance
188.8 pm
face diagonal = 616.7 pm
edge length =
face diagonal
volume = (4.361 
262
2
10–8
cm)3
=
616.7 pm
2
= 8.293 
109.5°/2
= 436.1 pm
10–23
cm3
C in center
of cell face
C
Si
Si in
tetrahedral hole
C
1
/2 face diagonal
C at cell
corner
Chapter 13
The Chemistry of Solids
4 SiC
40.096 g
1 mol
·
·
= 2.663  10–22 g
unit cell 1 mol SiC 6.0221  1023 atoms
density =
13.49
2.663  10–22 g
= 3.211 g/cm3
8.293  10–23 cm3
(a) Mg2+ occupies 1/8 of 8 possible tetrahedral holes and Al3+ occupies 1/2 of 4 available octahedral holes.
(b) Fe2+ occupies 1/8 of 8 possible tetrahedral holes and Cr3+ occupies 1/2 of 4 available octahedral holes.
13.50
EA(2) = –418.0 kJ/mol – 2(107.3 kJ/mol) – 2(495.9 kJ/mol) – 249.1 kJ/mol – (–141.0 kJ/mol) + 2481 kJ/mol
EA(2) = 748.5 kJ/mol
13.51
140 x 103 J/mol · 1 mol/6.022 x 1023 photons · 1/6.6260693 x 10-34 J · s = 3.5 x 1014 s-1
 = c
 = c/ = (2.998 x 108 m/s)/3.5 x 1014 s-1 = 8.5 x 10-7 m = 8.5 x 102 nm
13.52
Increasing T increases energy available for electrons to jump the band gap.
13.53
Germanium has more metallic character and, thus, a smaller band gap and higher conductivity.
13.54
(a) n-type
(b) n-type
(c) p-type
(d) n-type
13.55
Because boron is deficient in comparison to carbon, this is a p-type conductor.
13.56
Molecular solids are made of molecules held together by intermolecular forces. Networks are made of
atoms held together by covalent bonds. Amorphous solids are networks of atoms held together by covalent
bonds but lacking long range order.
13.57
Lead sulfide has the same structure as sodium chloride, not the same structure as ZnS. There are 4 Pb 2+
ions and 4 S2– ions per unit cell, a 1:1 ratio that matches the compound formula. Both the unit cell and the
crystals have a cubic shape.
13.58
(a) Volume =
1 CaTiO3
135.9 g
1 mol
1 cm3
·
·
·
= 5.51  10–23 cm3
unit cell 1 mol CaTiO3 6.0221  1023 4.10 g
edge length =
3
5.51  10–23 cm3 = 3.80  10–8 cm = 380. pm
(b) Assume that the O2– and Ti4+ ions touch along center of cell.
edge length = 2(O2– radius) + 2(Ti4+ radius)
380. pm = 2(140 pm) + 2(Ti4+ radius)
Ti4+ radius = 50. pm
The calculated radius is smaller than the literature value.
263
Chapter 13
13.59
The Chemistry of Solids
(a) BBr3(g) + PBr3(g) + 3 H2(g)  BP(s) + 6 HBr(g)
(b) If B atoms are in an fcc lattice, then the P atoms must be in ½ of the tetrahedral holes.
(c) volume = (4.78  10–8 cm)3 = 1.09  10–22 cm3
4 BP
41.78 g
1 mol
·
·
= 2.775  10–22 g
unit cell 1 mol BP 6.0221  1023
density =
2.775 10–22 g
= 2.55 g/cm3
1.09  10–22 cm3
(d) Use geometric relationship shown in 13.35 and 13.36 to calculate B–P distance.
109.5°/2
B in center
of cell face
B
P in
tetrahedral hole
P
B at cell
corner
B
1
/2 face diagonal
2 · edge length =
face diagonal =
2 · 478 pm = 676 pm
½ face diagonal = 338 pm
1
sin(109.5º/2) =
/ 2 (338 pm)
B–P distance
B–P distance = 207 pm
13.60
In a face-centered cubic lattice of anions (X) there are a total of four anions per unit cell.
Type of holes occupied by cation (M)
Number of cations per unit cell
Formula of salt
all tetrahedral holes
8
M2X
half of the tetrahedral holes
4
MX
all octahedral holes
4
MX
It is not possible to have a cation:anion ratio of 3:1.
13.61
Assuming the spheres are packed in an identical way, the water levels will be identical. A face–centered
cubic lattice, for example, uses 74% of the available volume, regardless of sphere size.
13.62
Co2+ and Al3+
CoAl2O4:
Co2+:
[Ar]   

3d
Al3+:
[He] 
2s
SnCo2O4:
264
  
2p
Sn2+ and Co3+

paramagnetic, 3 unpaired electrons
4s
diamagnetic
Chapter 13
Sn2+:
Co3+:
The Chemistry of Solids
[Kr]

    
5s
4d
[Ar]  


diamagnetic
5p

3d
paramagnetic, 4 unpaired electrons
4s
SOLUTION TO APPLYING CHEMICAL PRINCIPLES: TIN DISEASE
1.
-tin has four atoms in the unit cell. -tin has eight atoms in the unit cell.
2.
Volume = 583 pm · 583 pm · 318 pm = 1.08 x 108 pm
(4 atoms Sn/1.08 x 108 pm )(1 mol/6.022 x 1023 atoms)(118.7 g/ 1 mol)(1 x 1010 pm/1 cm)3 = 7.29 g/cm3
3.
8 stoms Sn(1 mol/6.022 x 10 23 atoms)(118.7 g/ 1 mol) (1 cm3/5.769 g) (1 x 1010 pm/1 cm)3 = 2.734 x 108 pm3
Length = (2.734 x 108 pm3)1/3 = 649.0 pm
4.
The unit cell of -tin (tetragonal lattice) contains 4 atoms. The volume of space occupied by the atom is:
VSn atoms =
4 3
r · 4 = 4.70 x 107 pm3
3
VSn atoms 4.70 x 107 pm3
=
x 100 % = 43.5%
V -tin
1.08 x 108 pm3
The unit cell of -tin (cubic lattice) contains 8 atoms. The volume of space occupied by the atom is:
VSn atoms =
4 3
r · 8 = 9.39 x 107 pm3
3
VSn atoms 9.39 x 107 pm3
=
x 100 % = 34.4%
V -tin
2.73 x 108 pm3
265