SOLUTIONS MANUAL for Solid State Materials Chemistry New Edition by Patrick M. Woodward, Pavel Karen

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Solutions Manual for Solid State Materials Chemistry by P.M. Woodward, P. Karen, J.S.O. Evans, T. Vogt

Solid State Materials Chemistry by P.M. Woodward, P. Karen, J.S.O. Evans, T. Vogt: Solutions Manual Contents Chapter 1: Structures of Crystalline Materials.......................................................................................1 Chapter 2: Defects and More Complex Structures ................................................................................6 Chapter 3: Defect Chemistry and Nonstoichiometry ...........................................................................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: Superconductivity ............................................................................................................58 Chapter 13: Energy Materials: Ionic Conductors, Mixed Conductors, and Intercalation Chemistry ..61 Chapter 14: Zeolites and Other Porous Materials ................................................................................66 Chapter 15: Amorphous and Disordered Materials .............................................................................69

Chapter 1: Structures of Crystalline Materials 1.1

The symbol 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 crystal systems; square with minimum symmetry 4 (or maximum 4mm), hexagonal with minimum symmetry 3 (maximum 6mm), rectangular with minimum symmetry m (or maximum 2mm) and oblique (tilted parallelograms) with axis 1 as minimum or 2 as maximum. Five Bravais lattices; primitive of the above systems plus a centered

1

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Solutions Manual for Solid State Materials Chemistry by P.M. Woodward, P. Karen, J.S.O. Evans, T. Vogt

rectangular lattice. The other symmetry-maintaining centerings are equivalent to the primitive lattices.

1.3

The indices tell into how many sections each of the vectors a,b,c is divided by the set of planes. The index has a bar (symbolizing a − sign) if the plane facing the origin crosses the axis at negative values: 102, 120, 231, 120. Note that 120 describes the same equidistant set as 120.

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.

Analogously, it also goes through x = 1 and y = 1, forming a triangle within the cube.

Then copy, 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,y,z coordinates, draw two cells as in the second figure of Problem 1.3. In the auxiliary cell, draw the triangle closest to the origin, hence between x = 1, y = −1 and z = ⅓. Extend the triangle’s plane into the normal cell where it forms a quadrangle. Move the first triangle horizontally into the normal cell. By analogy, draw the remaining planes in the normal cell.

1.6

(a) C-centered monoclinic (side centering), 2/m, (b) face-centered orthorhombic, mm2, (c) body-centered tetragonal, 4/mmm, (d) primitive hexagonal, 312, (e) rhombohedral, 3 m, (f) primitive hexagonal, 6 m2, (g) face-centered cubic, 23, (h) primitive cubic, 23 since screw axes convert to plain rotation axes in point symmetry, (i) body-centered cubic, m 3 m as glide planes convert to mirrors m in point symmetry.

1.7

The direction of a plane is the direction of its normal. Hence (a) yes, (b) yes, (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;], electroneutrality (charge) balance is 3 = 3, connectivity balance is 6 = 6, bond-valence balance is 3/6 = 3/6. For Cr2[6o;]O3[4;], electroneutrality balance is 2ꞏ3 = 3ꞏ2, connectivity balance is 2ꞏ6 = 3ꞏ4, bond-valence balance is 3/6 = 2/4. Cr[6o;]O2[3;] electroneutrality balance is 4 = 2ꞏ2, connectivity balance is 6 = 2ꞏ3, bond-valence balance is 4/6 = 2/3.

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Solutions Manual for Solid State Materials Chemistry by P.M. Woodward, P. Karen, J.S.O. Evans, T. Vogt

O Al Mg

O O

Al

1.10 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 pyramids, the layer suggests that

O

the square bases are in the layer plane. The sole 1-connected O(1) atom per V is then the pyramidal apex. Now, focus only on the square bases on a flat plane (draw a mesh): The O(2) atom can be 3-connected only as a junction of two edges that three squares have in common. The bond graph tells us that every square (every 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 by 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, respectively. Analogy with stable elemental structures suggests the graphite-type 2B− anions and C22− anions isoelectronic with N2, respectively.

1.15

The easiest calculation on a unit cube yields 52.4% and 68.0%, respectively.

1.16

(a) Featured in CsCl: Place the anions of unit radius at the corners of a cube so that they touch. The cube edge equals 2 unit radii, its body 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 unitradii 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 body diagonal is 1+r, thus r = (√3/√2) − 1 ≈ 0.225.

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Solutions Manual for Solid State Materials Chemistry by P.M. Woodward, P. Karen, J.S.O. Evans, T. Vogt

1.17

(a) Na–Cl = a/2 = 2.81 and Na–Na = a/√2 = 3.974 Å, (b) Ni–Ni = c/2 = 2.55 Å, (c) Ca–F = (a√3)/4 = 2.364 Å, (d) C–C = (a√3)/4 = 1.544 Å, (e) Ti–Ti = c = 2.9589 Å, (f) Ti–O = a/2 = 1.95, Sr–O = O–O = a/√2 = 2.758 Å.

1.18

The unit cell is oriented in analogy with Figure 1.22. There are 6 densest packed layers of spherical atoms along c in the hexagonal cell, hence the Ramsdell symbol is 6H. The layer sequence is BABCAC, the stacking of layers with local hexagonal and cubic environment is (hcc)2, the Jagodzinski–Wyckoff notation.

1.19

With a structure-drawing program, perform the sequence of the layer top-on and side-on orientations of the hexagonal unit cell, as indicated on Figure 1.26 (left). Evaluation of the subsequent shifts of the densest-packed layers shows that the repeating sequence is ABC hence cubic closest packing (ccp).

1.20

The filling is ⅔. The shifts of the densest packed planes are parallel with a or b, and, from the layer sequence along one of these shift directions, hcp is identified.

1.21

La has coordination number 9, and connectivity balance then dictates that Br is 3coordinated. Connecting three such large polyhedra at one vertex cannot be done without penetrating each other, as the edge angles at their vertices are rather obtuse. Both edgeand face-sharing is therefore present in LaBr3.

1.22

CrO2/2+2/1 and 1[Cr[4t]O[2n]O2[1]] with nonlinear bonds at O (tetrahedral angle).

1.23

All four anions cannot be equivalent (neither CA4/1, CA4/2, CA4/3 nor CA4/4 becomes CA3/2). The best compliance with the rule of parsimony is for Niggli formula CA3/3+1/2 of three 3-connected and one 2-connected A.

1.24

(a) CA8/2 = CA4, (b) CA8/4 = CA2 hence the all-edges-shared variant yields the CaF2 structure, (c) CA8/8 = CA all faces shared gives the CsCl structure.

1.25

Rewrite the chemical formula per one central atom. Having a sum of numerators equal to four, CN4/3 already corresponds to a tetrahedron and is the Niggli formula sought. This means that there is only one type of N vertex, of coordination number 3.

1.26

Parsimonial connectivites would occur for a polyhedron of 9 or 12 vertices; NLi9/3 or NLi12/4, respectively. A mixture of connectivity 3 and 4 must therefore occur for the coordination number 11 in between, NLi8/4+3/3.

1.27

Two different nodes, 6-coordinated Re and 2-coordinated O, hence a 6,2-connected net.

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