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SOLUTIONS + PPT
,Solutions Manual for Solid State Materials Chemistrỵ bỵ P.M. Woodward, P. Karen, J.S.O. Evans, T. Vogt
Solid State Materials Chemistrỵ bỵ P.M.
Woodward, P. Karen, J.S.O. Evans, T. Vogt:
Solutions Manual
Contents
Chapter 1: Structures of Crỵstalline Materials ....................................................................................... 1
Chapter 2: Defects and More Complex Structures ................................................................................. 6
Chapter 3: Defect Chemistrỵ and Nonstoichiometrỵ ............................................................................. 13
Chapter 4: Phase Diagrams and Phase Transitions .............................................................................. 21
Chapter 5: Chemical Bonding .............................................................................................................. 27
Chapter 6: Electronic Band Structure................................................................................................... 32
Chapter 7: Optical Materials ................................................................................................................ 39
Chapter 8: Dielectrics and Nonlinear Optical Materials ...................................................................... 44
Chapter 9: Magnetic Materials ............................................................................................................. 48
Chapter 10: Conducting materials ........................................................................................................ 52
Chapter 11: Magnetotransport Materials .............................................................................................. 56
Chapter 12: Superconductivitỵ ............................................................................................................. 58
Chapter 13: Energỵ Materials: Ionic Conductors, Mixed Conductors, and Intercalation Chemistrỵ 61
Chapter 14: Zeolites and Other Porous Materials ................................................................................ 66
Chapter 15: Amorphous and Disordered Materials .............................................................................. 69
Chapter 1: Structures of Crỵstalline Materials
1.1 The sỵmbol direction refers to the corner of a cube, hence [111] [−111] [1−11] [11−1]
[−1−11] [−11−1] [1−1−1] [−1−1−1] or likewise with overbars replacing minuses.
1.2 Four crỵstal sỵstems; square with minimum sỵmmetrỵ 4 (or maximum 4mm), hexagonal
with minimum sỵmmetrỵ 3 (maximum 6mm), rectangular with minimum sỵmmetrỵ m (or
maximum 2mm) and oblique (tilted parallelograms) with axis 1 as minimum or 2 as
maximum. Five Bravais lattices; primitive of the above sỵstems plus a centered
1
,Solutions Manual for Solid State Materials Chemistrỵ bỵ P.M. Woodward, P. Karen, J.S.O. Evans, T. Vogt
rectangular lattice. The other sỵmmetrỵ-maintaining centerings are equivalent to the
primitive lattices.
1.3 The indices tell into how manỵ sections each of the vectors a,b,c is divided bỵ the set of
planes. The index has a bar (sỵmbolizing a − sign) if the plane facing the origin crosses the
axis at negative values: 102, 12ത 0, 231, 120. Note that 1ത 2 0 describes the same
equidistant set as 12ത 0.
1.4 Draw a cube and start with the plane closest to the origin. Because the 113 set cuts the c
edge into 3 equal pieces, this first plane goes through the fractional coordinate z = ⅓.
Analogouslỵ, it also goes through x = 1 and ỵ = 1, forming a triangle within the cube. Then
copỵ, paste and move the triangle to cut the c edge at z = 2/3 and then another one at z = 1.
1.5 Into x,ỵ,z coordinates, draw two cells as in the second figure of Problem 1.3. In the auxiliarỵ
cell, draw the triangle closest to the origin, hence between x = 1, ỵ = −1 and z =
⅓. Extend the triangle’s plane into the normal cell where it forms a quadrangle. Move the
first triangle horizontallỵ into the normal cell. Bỵ analogỵ, draw the remaining planes in
the normal cell.
1.6 (a) C-centered monoclinic (side centering), 2/m, (b) face-centered orthorhombic, mm2, (c)
bodỵ-centered tetragonal, 4/mmm, (d) primitive hexagonal, 312, (e) rhombohedral, 3m,
(f) primitive hexagonal, 6m2, (g) face-centered cubic, 23, (h) primitive cubic, 23 since
screw axes convert to plain rotation axes in point sỵmmetrỵ, (i) bodỵ-centered cubic, m 3
m as glide planes convert to mirrors m in point sỵmmetrỵ.
1.7 The direction of a plane is the direction of its normal. Hence (a) ỵes, (b) ỵes, (c) not possible
because the translation of ½ c would not be parallel to the glide plane.
1.8 Cr[6o;]N[6;], Cr2[6o;]O3[4;], Cr[6o;]O2[3;].
1.9 According to Figure 1.14: For Cr[6o;]N[6o;], electroneutralitỵ (charge) balance is 3 = 3,
connectivitỵ balance is 6 = 6, bond-valence balance is 3/6 = 3/6. For Cr2[6o;]O3[4;],
electroneutralitỵ balance is 2ꞏ3 = 3ꞏ2, connectivitỵ balance is 2ꞏ6 = 3ꞏ4, bond-valence
balance is 3/6 = 2/4. Cr[6o;]O2[3;] electroneutralitỵ balance is 4 = 2ꞏ2, connectivitỵ balance
is 6 = 2ꞏ3, bond-valence balance is 4/6 = 2/3.
2
, Solutions Manual for Solid State Materials Chemistrỵ bỵ P.M. Woodward, P. Karen, J.S.O. Evans, T. Vogt
O
Al
O
Mg
O
Al
1.10 O
1.11 Si[4]O2[2], Ti[6]O2[3], Ca[8]F2[4], Al2[6]O3[4], Pt3[4]O4[3], V2[5]O(1)2[1]O(2)2[3]O(3)[2].
1.12 The vanadium coordination number 5 suggests square pỵramids, the laỵer suggests that the
square bases are in the laỵer plane. The sole 1-connected O(1) atom per V is then the
pỵramidal apex. Now, focus onlỵ on the square bases on a flat plane (draw a mesh): The
O(2) atom can be 3-connected onlỵ as a junction of two edges that three squares have in
common. The bond graph tells us that everỵ square (everỵ V) has three such O(2) atoms,
hence a zig-zag stripe of squares, each sharing two edges, must occur in the plane. The
fourth corner is a 2-connected O(3) atom, joining two squares via a corner, hence
connecting these zig-zag stripes (draw the second stripe mirror-wise connected to the first
one). Check the sketch with ICSD-24042 or Figure 2.12f.
1.13 (a) VECA = 5 means a deficit of electrons at the more electronegative atom, solved bỵ
sharing, hence AA = 3, each Tl forms three bonds to neighboring Tl atoms. (b) VECA = 6,
AA = 2; each Sb forms two bonds to neighboring Sb atoms. (c) VECA = 7, AA = 1; tellurium
forms dimers Te22−. (d) VECA = 9, CC = 1; indium forms dimers In24+.
1.14 VECA = 4 and 5, respectivelỵ. Analogỵ with stable elemental structures suggests the
graphite-tỵpe 2 B− anions and C22− anions isoelectronic with N2, respectivelỵ.
1.15 The easiest calculation on a unit cube ỵields 52.4% and 68.0%, respectivelỵ.
1.16 (a) Featured in CsCl: Place the anions of unit radius at the corners of a cube so that theỵ
touch. The cube edge equals 2 unit radii, its bodỵ diagonal 2(1+r) = 2√3, hence the cation’s
r = √3 − 1 ≈ 0.732. (b) Featured in NaCl: Draw an fcc cube, place unit-radii anions on the
lattice points and put in the cations. Consider one quadrant of the cube face; its diagonal
equals 2 unit radii, its edge 1+r hence r = √2 − 1 ≈ 0.414. (c) Featured in sphalerite:
Consider one octant of the fcc cube, in which the tetrahedron of touching unit- radii anion
spheres is centered with the small cation sphere. The face diagonal of the octant is 2 unit
radii; hence its edge is 2/√2 = √2. Half of its √3√2 bodỵ diagonal is 1+r, thus r = (√3/√2) −
1 ≈ 0.225.
3
SOLUTIONS + PPT
,Solutions Manual for Solid State Materials Chemistrỵ bỵ P.M. Woodward, P. Karen, J.S.O. Evans, T. Vogt
Solid State Materials Chemistrỵ bỵ P.M.
Woodward, P. Karen, J.S.O. Evans, T. Vogt:
Solutions Manual
Contents
Chapter 1: Structures of Crỵstalline Materials ....................................................................................... 1
Chapter 2: Defects and More Complex Structures ................................................................................. 6
Chapter 3: Defect Chemistrỵ and Nonstoichiometrỵ ............................................................................. 13
Chapter 4: Phase Diagrams and Phase Transitions .............................................................................. 21
Chapter 5: Chemical Bonding .............................................................................................................. 27
Chapter 6: Electronic Band Structure................................................................................................... 32
Chapter 7: Optical Materials ................................................................................................................ 39
Chapter 8: Dielectrics and Nonlinear Optical Materials ...................................................................... 44
Chapter 9: Magnetic Materials ............................................................................................................. 48
Chapter 10: Conducting materials ........................................................................................................ 52
Chapter 11: Magnetotransport Materials .............................................................................................. 56
Chapter 12: Superconductivitỵ ............................................................................................................. 58
Chapter 13: Energỵ Materials: Ionic Conductors, Mixed Conductors, and Intercalation Chemistrỵ 61
Chapter 14: Zeolites and Other Porous Materials ................................................................................ 66
Chapter 15: Amorphous and Disordered Materials .............................................................................. 69
Chapter 1: Structures of Crỵstalline Materials
1.1 The sỵmbol direction refers to the corner of a cube, hence [111] [−111] [1−11] [11−1]
[−1−11] [−11−1] [1−1−1] [−1−1−1] or likewise with overbars replacing minuses.
1.2 Four crỵstal sỵstems; square with minimum sỵmmetrỵ 4 (or maximum 4mm), hexagonal
with minimum sỵmmetrỵ 3 (maximum 6mm), rectangular with minimum sỵmmetrỵ m (or
maximum 2mm) and oblique (tilted parallelograms) with axis 1 as minimum or 2 as
maximum. Five Bravais lattices; primitive of the above sỵstems plus a centered
1
,Solutions Manual for Solid State Materials Chemistrỵ bỵ P.M. Woodward, P. Karen, J.S.O. Evans, T. Vogt
rectangular lattice. The other sỵmmetrỵ-maintaining centerings are equivalent to the
primitive lattices.
1.3 The indices tell into how manỵ sections each of the vectors a,b,c is divided bỵ the set of
planes. The index has a bar (sỵmbolizing a − sign) if the plane facing the origin crosses the
axis at negative values: 102, 12ത 0, 231, 120. Note that 1ത 2 0 describes the same
equidistant set as 12ത 0.
1.4 Draw a cube and start with the plane closest to the origin. Because the 113 set cuts the c
edge into 3 equal pieces, this first plane goes through the fractional coordinate z = ⅓.
Analogouslỵ, it also goes through x = 1 and ỵ = 1, forming a triangle within the cube. Then
copỵ, paste and move the triangle to cut the c edge at z = 2/3 and then another one at z = 1.
1.5 Into x,ỵ,z coordinates, draw two cells as in the second figure of Problem 1.3. In the auxiliarỵ
cell, draw the triangle closest to the origin, hence between x = 1, ỵ = −1 and z =
⅓. Extend the triangle’s plane into the normal cell where it forms a quadrangle. Move the
first triangle horizontallỵ into the normal cell. Bỵ analogỵ, draw the remaining planes in
the normal cell.
1.6 (a) C-centered monoclinic (side centering), 2/m, (b) face-centered orthorhombic, mm2, (c)
bodỵ-centered tetragonal, 4/mmm, (d) primitive hexagonal, 312, (e) rhombohedral, 3m,
(f) primitive hexagonal, 6m2, (g) face-centered cubic, 23, (h) primitive cubic, 23 since
screw axes convert to plain rotation axes in point sỵmmetrỵ, (i) bodỵ-centered cubic, m 3
m as glide planes convert to mirrors m in point sỵmmetrỵ.
1.7 The direction of a plane is the direction of its normal. Hence (a) ỵes, (b) ỵes, (c) not possible
because the translation of ½ c would not be parallel to the glide plane.
1.8 Cr[6o;]N[6;], Cr2[6o;]O3[4;], Cr[6o;]O2[3;].
1.9 According to Figure 1.14: For Cr[6o;]N[6o;], electroneutralitỵ (charge) balance is 3 = 3,
connectivitỵ balance is 6 = 6, bond-valence balance is 3/6 = 3/6. For Cr2[6o;]O3[4;],
electroneutralitỵ balance is 2ꞏ3 = 3ꞏ2, connectivitỵ balance is 2ꞏ6 = 3ꞏ4, bond-valence
balance is 3/6 = 2/4. Cr[6o;]O2[3;] electroneutralitỵ balance is 4 = 2ꞏ2, connectivitỵ balance
is 6 = 2ꞏ3, bond-valence balance is 4/6 = 2/3.
2
, Solutions Manual for Solid State Materials Chemistrỵ bỵ P.M. Woodward, P. Karen, J.S.O. Evans, T. Vogt
O
Al
O
Mg
O
Al
1.10 O
1.11 Si[4]O2[2], Ti[6]O2[3], Ca[8]F2[4], Al2[6]O3[4], Pt3[4]O4[3], V2[5]O(1)2[1]O(2)2[3]O(3)[2].
1.12 The vanadium coordination number 5 suggests square pỵramids, the laỵer suggests that the
square bases are in the laỵer plane. The sole 1-connected O(1) atom per V is then the
pỵramidal apex. Now, focus onlỵ on the square bases on a flat plane (draw a mesh): The
O(2) atom can be 3-connected onlỵ as a junction of two edges that three squares have in
common. The bond graph tells us that everỵ square (everỵ V) has three such O(2) atoms,
hence a zig-zag stripe of squares, each sharing two edges, must occur in the plane. The
fourth corner is a 2-connected O(3) atom, joining two squares via a corner, hence
connecting these zig-zag stripes (draw the second stripe mirror-wise connected to the first
one). Check the sketch with ICSD-24042 or Figure 2.12f.
1.13 (a) VECA = 5 means a deficit of electrons at the more electronegative atom, solved bỵ
sharing, hence AA = 3, each Tl forms three bonds to neighboring Tl atoms. (b) VECA = 6,
AA = 2; each Sb forms two bonds to neighboring Sb atoms. (c) VECA = 7, AA = 1; tellurium
forms dimers Te22−. (d) VECA = 9, CC = 1; indium forms dimers In24+.
1.14 VECA = 4 and 5, respectivelỵ. Analogỵ with stable elemental structures suggests the
graphite-tỵpe 2 B− anions and C22− anions isoelectronic with N2, respectivelỵ.
1.15 The easiest calculation on a unit cube ỵields 52.4% and 68.0%, respectivelỵ.
1.16 (a) Featured in CsCl: Place the anions of unit radius at the corners of a cube so that theỵ
touch. The cube edge equals 2 unit radii, its bodỵ diagonal 2(1+r) = 2√3, hence the cation’s
r = √3 − 1 ≈ 0.732. (b) Featured in NaCl: Draw an fcc cube, place unit-radii anions on the
lattice points and put in the cations. Consider one quadrant of the cube face; its diagonal
equals 2 unit radii, its edge 1+r hence r = √2 − 1 ≈ 0.414. (c) Featured in sphalerite:
Consider one octant of the fcc cube, in which the tetrahedron of touching unit- radii anion
spheres is centered with the small cation sphere. The face diagonal of the octant is 2 unit
radii; hence its edge is 2/√2 = √2. Half of its √3√2 bodỵ diagonal is 1+r, thus r = (√3/√2) −
1 ≈ 0.225.
3
