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

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

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

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

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, 12̄ 0, 231, 120. Note that 1̄ 2 0 describes the same equidistant set as 12̄ 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.

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, 3m, (f) primitive hexagonal, 6m2, (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.

2

Solutions Manual for Solid State Materials Chemistry by 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 pyramids, the layer suggests that 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.

3

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; NLi 9/3 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.

4

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

1.28

(a) In the unit cell of CaO, connect Ca atoms at the bottom and top face to obtain the pseudotetragonal cell that corresponds to CaO2. In CaO2 the peroxide anion replaces the oxide anion; vertex decoration. (b) Sr2FeMoO6 is a site-ordered variant of SrMoO3 in which the single Mo site is replaced by an ordered arrangement of Fe and Mo sites.

1.29

The Laves phase is a spinel type without oxygens. It contains a network of vertex-sharing tetrahedral clusters of copper.

1.30

Using the chemical information about typical oxidation states: [Li+][4t][Mn2+Mn3+][6o]O4, [Zn2+][4t][Fe3+]2[6o]O4,

[Fe2+][4t][Fe2+Ti4+][6o]O4,

[Fe3+][4t][Fe2+Fe3+][6o]O4,

[Si4+][4t][Ni2+]2[6o]O4. [Nb5+][4t][Li+Zn2+][6o]O4, [W6+][4t][Na+]2[6o]O4.

1.31

(a) ⅛ of all tetrahedral and ½ of octahedral holes in ccp of O atoms are filled. (b) Orient the ccp structure as shown on Figure 1.26 (right). You see a plane occupied with octahedral holes alternating with a sandwich plane of both octahedral and tetrahedral occupancies. (c) Separating the individual layers in the drawing program shows that the octahedral hole plane has ¾ filling. The sandwich has a ¼ filling of octahedral holes together with a combined ¼ filling of tetrahedral holes positioned above the bottom-layer oxygens and ¼ filling of tetrahedral holes below the top-layer oxygens.

1.32

Draw the unit cell with Ti coordination octahedra and view it subsequently against three directions: [001], [110] and [−110]. You see the same tilt as in Figure 1.53, a+b−b−.

1.33

(SiO2+2/2)36−.

1.34

The atomic coordinates in Figure 1.57 suggest Al6Si 6O2412−, simplified AlSiO4−.

1.35

VECA for Cu is 12; the six electrons missing to 18 are gained by sharing via formation of six bonds per Cu. This suggests that the Cu tetrahedral clusters share corners.

1.36

Li+, VECA = 6 hence 2 electrons must be shared by each As. This suggests 1As− anionic chains.

1.37

K+, VECA = 4 hence 4 electrons must be shared by each In as in graphite or diamond.

Analogy of In with the heavier elements of the carbon group that exhibit covalent bonding, Si, Ge or gray Sn, suggest the diamond net, hence the

3 

In− anion. That is also

the reality.

1.38

Ca2+, VECA = 4 hence 4 electrons must be shared by each In. Analogy suggests a diamond net, 3In−. A search for CaIn2 and visualization in a structure-drawing program shows that the tetrahedral network is actually of the hexagonal-diamond type, lonsdaleite.

5

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

Chapter 2: Defects and More Complex Structures 2.1

The figure of 1019 can be derived from the given data and the values in Table 2.1. For a 1 cm3 crystal, the number of unit cells would be (0.01/3.61510−10)3 = 2.121022. Each unit cell contains 4 Cu sites. Using values from Table 2.1, the fractional concentration of neq  S   − H  = exp(21/8.3145)  vacancies is x = = exp  vib  exp   v N 0 + neq  k   kT  exp[(−123103)/(8.3145  1273)] = 1.1210−4. The total number of defects is then 4  2.121022  1.1210−4 = 9.51018.

2.2

Using values for H and ∆Svib from Table 2.1 gives xv = 4.8010−21, 1.1610−7 and 2.3010−4 at 300, 800, and 1357 K, respectively. The number of defects rises rapidly with temperature.

2.3

As stated in text, NaCl favors Schottky defects. We can use values in Table 2.1 to calculate xS as 4.5410−19 and 2.2310−7 at 300 and 700 K respectively. Each grain will contain [0.001/(5.6210−10)]3 = 5.6341018 unit cells hence 4  5.6341018 NaCl units. It has 10 defects at 300 K and 51012 defects at 700 K.

2.4

For Schottky defects, the derivation follows that outlined in the text but there is a configurational entropy term for both cations and anions. The configurational entropy term is given by:  ( N0 + n)! ( N0 + n)!   ( N0 + n)!  S = k ln  N !n!  N !n!  = 2k ln  N !n!   0 0   0 

The free energy change on formation of n defects is therefore given by: G = nH − nT S S

On

− 2kT ln vib

applying

 ( N0 + n )!     N 0 !n! 

Stirling’s

approximation,

this

can

be

G = nH S − nT Svib − 2kT ( N 0 + n ) ln ( N 0 + n ) − N 0 ln ( N 0 ) − n ln ( n )

Applying d ( G ) = 0 at equilibrium gives: dn

d

(G) = 0 = H −TS

dn

S

− 2kT ln vib

 N0 + neq  

neq



which can be rearranged to give: 6

expanded

to:

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

neq N +n 0

 Svib   −HS   Svib   −HS  or n = exp  2k  exp   2k exp  2kT  eq,Schottky  N 0 exp 2kT         eq

For Frenkel defects, the configurational entropy for n interstitials in a crystal of N0 sites with NI possible interstitial sites is given by: N !  !

S = k ln  N N ( − nI )!n! ( N − 0n)!n! 

I

0



On applying Stirling’s approximation, we get the free energy change on introducing n defects as: G = nH − nT S F

NI ln ( NI ) + N0 ln ( N0 ) − ( NI − n)ln ( NI − n) − kT   vib  −( N0 − n)ln ( N0 − n) − 2n ln (n) 

On differentiation and gathering terms we get:  (NI − neq )(N0 − neq )  d   neq 2 dn (G ) = 0 = H F − T Svib − kT ln    Assuming that NI and N0 are >> neq and rearranging gives: neq 2  exp  S vib   −H F   ,  k  exp    NI N0  kT 

which leads to the expression in the text, of 12  Svib   −HF  n  ( N N ) exp exp  .

2kT  2k  eq,Frenkel 0 I     2.5

This Problem shows how relatively simple experimental observations can tell you something about the defect structure of a material. (a) From weight percentages, the Ni:O molar ratio is 1:1.053. One can therefore express two limiting empirical formulae Ni0.950O or NiO1.053; the former implies nickel vacancies and the latter oxygen interstitials. Note that a range of compositions (e.g. Ni0.978O1.030) between these two extremes would also be possible. (b) Density can be calculated as the mass of a single unit cell divided by its volume. If the material were stoichiometric NiO, the mass of one mole of unit cells would be 474.6928/1000 kg, as there are four formula units per unit cell for NaCl-type structures. The density is then (474.6928/1000)/(NA  4.180310−30) or 6793 kg/m3. For Ni0.950O, the corresponding density would be 471.7581/(NA  4.180310−30) = 6526 kg/m3. For NiO1.05, the density would be 475.4928/(NA  4.180310−30) = 6866 kg/m3.

7

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

The experimental density is significantly less than that for a stoichiometric material implying that cation vacancies are the defect present. The density calculation for Ni 0.950O confirms this. Note that experimental densities change significantly for relatively small changes in composition. (c) One would normally use X-ray diffraction to determine a unit-cell parameter. We can think of the effect of a vacancy using a Vegard’s law type argument. The vacancy is larger than a Ni2+ on the same site as the non-compensated local anions will repel, and the nonstoichiometric material would have a larger cell parameter. Experimentally, the value is around 4.19 Å for ~5% Ni3+.

2.6

From the weight percentage, the Zn:O molar ratio is 1.03:1, and the limiting empirical formulae are Zn1.03O or ZnO0.97. According to the data in Chapter 1.4.2, the wurtzite-type unit cell contains 2 formula units and its volume is V=abcsin(). Densities are calculated as in Problem 2.5. Zn1.03O has a density 5813 kg/m3, and ZnO0.97 5644 kg/m3.

Experimental density is then consistent with Zn1.03O that suggests Zn interstitials behind the nonstoichiometry.

2.7

Oxidation states: (a) Fe2+; Fe2.29+ or Fe2+0.616Fe3+0.256; Fe2.66+ of 1 FeII per 2 FeIII in a spinel Fe3+Fe2.5+2O4 (in Chapter 11 we describe how Fe3O4 contains Fe3+ on tetrahedral sites and a 50:50 mixture of Fe2+ and Fe3+ on octahedral sites); Fe2+ as it contains S22− anions. Fe4+ and S2− is unlikely as S2− would be oxidized by tetravalent iron. (b) Fe2+; Fe1.82+. (c) La3+ in both. Fe2+; Fe1.9+. (d) Y3+, Fe2.5+ but can charge order to Fe2+ and Fe3+ as discussed in Chapter 11; Nd3+, Ba2+, Fe3+; Nd3+, Ba2+, Co3.5+.

2.8

Oxidation states: (a) Ti4+ (note difference from FeS2 in previous Problem); Li+, Ti3.3+. (b) La3+, Ca2+; Mn in three cases: Mn3+; Mn3.2+; formally Mn3.5+ but could charge order to Mn3+ and Mn4+. (c) La3+, Cu in three cases: Cu2+; Cu2.15+; Cu2.15+. Note different ways to achieve Cu optimal doping as explored in superconductivity Chapter 12. (d) Ba2+, Pb4+; Ba2+, Bi4+ but expect charge ordering to the 6s2-stabilized Bi3+ and Bi5+; Ba2+, K+, Bi4.4+.

2.9

If it were an ideal NaCl-type TiO, the density would be 5797 kg/m3. This suggests that the synthesized stoichiometric sample contains significant defects, with the experimentalto-theoretical density ratio suggesting Ti0.85O0.85.

2.10

This Problem is a little bit involved, so it is best set up in a spreadsheet. Create columns for z in TiOz, density, cell parameter, mass of cell, relative molecular mass (RMM) of cell from the density and expected RMM for a formula TiOz. Then take the ratio of the 8

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

expected RMM to RMM from density values to give x and y: TiO1.32 ≡ Ti0.74O0.98; TiO1.12 ≡ Ti0.81O0.91; TiO0.69 ≡ Ti0.96O0.66. If you plot graphs of Ti and O occupancies against z and fit straight lines through each, they have values of x = 0.83 and y = 0.85 at z = 1, i.e. the 1:1 material is about Ti0.85O0.85 from these data.

2.11

z in TiOz

Cell (Å)

Density (g/cm3)

1.32

4.1608

4.713

1.12

4.1755

4.867

0.69

4.2212

4.992

For ideal cell contents of Nb4O4, the theoretical density is 9.695 g/cm3. This implies the real material is Nb0.75O0.75 with cell contents Nb3O3. It remains cubic by removing one Nb (the site at cell corners) and one O (the cell center) leaving square-planar coordinations of both cations and anions. This was described as a 4,4 net in Chapter 1.

The structure can be described in space group Pm3̄ m with Nb on 3c and O on 3d sites.

2.12

If

we

assume

Vegard’s

law,

we

can

rearrange

Equation

(2.6)

to

give

x =a(x)−aAY (aBY −aAY), which yields x = 0.3. Assuming a similar relationship for the band gap gives Eg = 0.3  2.24 + (1−0.3)  1.42 = 1.66 eV. 2.13

Using a(x, y) = aInP and introducing appropriate cell parameters for other phases into Equation (2.8) from the text will confirm this expression.

2.14

This material is In1−xGaxP1−yAsy taken as a solid solution between n- and p-type InP-based structures. Equation (2.8) and the x and y values suggest a = 0.1272  5.65 + 0.1128  5.45 + 0.4028  6.06 + 0.3572  5.87 = 5.87 Å; the material is lattice-matched with InP.

2.15

Na1−2xCaxxCl vNa′ and CaNa•; Ca1−xYxF2+x YCa• and Fi′; LixSi1−xAlxO2 Lii• and AlSi′; Zr1−xCaxO2−xx CaZr′′ and vO••; (Ca1−xNax)(Al2−xSi2+x)O8 NaCa′ and SiAl•.

2.16

Twinning types expected are: (a) An orthorhombic cell with a  b will mimic a tetragonal cell. We can imagine making a mistake when “building a wall” by adding bricks that are rotated around the c-axis by 90 degrees or flipped around the diagonal of the square face, see Figure (a) below. More formally, the different domains are related by either a 4-fold axis along c or 2-fold axes along the direction family ⟨1 1 0⟩ that would be present in a tetragonal cell. (b) This might mimic an orthorhombic cell with  = 90, and the twins 9

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

might be related by symmetry elements of this cell; 2-fold axes along a and c or mirror planes perpendicular to a and c. (c) Here we have a unit cell where c  2a. We could therefore imagine a twin boundary as in Figure (c) where we have two “vertical cells” placed alongside each “horizontal cell”. The domains would be related by a 4-fold rotation about the b axis. (d) This one is a little harder to spot. From Figure (d), a monoclinic cell with a = c ( any angle) will mimic a C-centered orthorhombic cell. Here it is a rectangle drawn around the lozenge of the monoclinic ac projection and shifted along the monoclinic [101] direction such that the corner (lattice point) of the monoclinic cell repeats at the center of the rectangle. One possibility is that the a  c structure has twin domains related by a 2-fold axis of this orthorhombic cell, which is the [101] or [−101] direction of the monoclinic cell.

2.17

This Problem is the 3D analogue of the 2D sketch of Figure 2.9c, and we can tackle it in a number of ways. In cubic BaTiO3 the Ti lies on average at the center of the TiO6 octahedron (being disordered over 8 orientations). If the symmetry is reduced to P4mm, we can imagine the Ti moving towards any corner of the octahedron (at least in the average structure where Ti is disordered over 4 orientations), which retains the 4 fold symmetry. We could choose the direction of displacement to be along any one of +a, −a, +b, −b, +c, −c, and we would expect 6 domain variants to form. Alternatively, we can use International Tables Volume A (see Appendix B) to find the translationengleiche subgroups of Pm3̄ m (#221), which are listed as maximal non-isomorphic subgroups of type I: We see that the subgroup P4mm is not among them, but P4/mmm is, of index 3.

The term “index” means that the symmetry operations are diluted by a factor of 3. If we go to the entry for P4/mmm (#123) it has P4mm as a maximal subgroup of type I and of index 2. In total, the symmetry operations are diluted by a factor of 3×2=6. We expect 6 variants to form. Another alternative, if you have a background in molecular symmetry, is to use the ratio of the order (number of symmetry operations) of each point group as read 10

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

from character tables. For m 3̄ m (Oh) the order is 48, and for 4mm (C4v) the order is 8. The number of variants is 48/8 = 6. A final method is to use a web-based software package such as ISODISTORT that has a number of very useful tools for studying symmetrylowering phase transitions. In this case we could use ISODISTORT’s “method 2” to perform the symmetry reduction and obtain a list of the number of domains and their relative orientations. This is often a convenient method for exploring multiple phase transitions.

2.18

(a) At high temperatures, it is an fcc cube occupied by one “mixed” atom at each lattice point. In the low-temperature structure, the Au is at the corner of the cell and the Cu on the face centers. (b) Both structures have point group m 3̄ m so this is a klassengleiche transition involving the loss of translational symmetry (here the centering). (c) As it’s a klassengleiche transition, antiphase domains will form on cooling. We can imagine selecting any one of the four metal sites in the high T fcc cell as the Au site. Four antiphase domain variants will then form as ordered regions grow to meet each other from each of these nucleation choices in a manner equivalent to Figure 2.10.

Equivalently, International Tables Volume A (see Appendix B) lists in its Fm3̄ m (#225) entry the space group Pm3̄ m (#221) as a maximal non-isomorphic subgroup of type IIa (klassengleiche) of index 4.

2.19

One way of working out composition is to consider the Niggli formula (Chapter 1.4.4) for the unit cell depicted. You may want to sketch cells surrounding that shown to help you.

The octahedral connectivities sum 4×MoO2/3+4/2 + 4×MoO1/3+5/2 + 8×MoO6/2 = Mo16O46 can be verified by counting sites in the drawing. It suggests a crystallographic shear structure of Mo8O23, with an average oxidation state of Mo5.75+.

2.20

The (Mo7W7)O41 member of the MnO3n−1 shear-structure type.

2.21

Each unit cell of NaCl-type FeO has cell contents Fe4O4. The superstructure has 250 times the volume of a single cell and contents Fe1000O1000. If it contains 16 V12T4, the composition would be Fe808Fe64O1000 or Fe0.872O.

2.22

5869 kg/m3 and 5660 kg/m3.

2.23

Fe0.875O.

2.24

The defect can be sketched based on one octant of the NaCl structure—i.e. a cube of length a/2 with Fe3+ at its center and O2− on 4 diagonal corners; the other corners 11

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

represent Fe2+ vacancies. The nearest Fe2+ sites to an Fe2+ vacancy are arranged in a cuboctahedron around the vacancy, three vertices of which are unoccupied due to the vacancy’s own V4T defect, hence there are 9 occupied Fe sites nearest to a Fe vacancy.

The total charge of the [Fe3+O2−4]5− cluster indicates that 5 nearest Fe2+ sites must be occupied by Fe3+. The Kröger–Vink notation of the defects formed are vFe′′, FeFe•, Fei•••.

2.25

The ideal stoichiometry is Fe4O4 per NaCl-type cell, hence per aFeO3. If the cell edge is 2.7×aFeO, the limiting stoichiometric composition is (Fe4O4)×2.73 or Fe78.732O78.732. If one half of these cells contain V13T4, the expected composition is Fe78.732−13/2+4/2O78.732 or Fe74.232O78.732. This corresponds to Fe0.943O. x = 0.057.

2.26

(a) The structure is likely to be similar to that of Sn1.17NbS3.17 described in the text and contain NaCl-like PbS and NbS2 layers with Nb in trigonal prismatic coordination by S. (b) The size of the unit cells is such that there would be approximate alignment after 7 NbS2 (23.19 Å) and 4 PbS (23.37 Å) units suggesting the structure could be described, at least approximately, using a cell with a ≈ 23.3 Å. (c) For a 7/4 ratio, the composition would be (PbS)2×4(NbS2)7 or (PbS)1.14NbS2. Using the exact unit-cell ratio (1.764), the composition would be (PbS)2×a(NbS2)/a(PbS)NbS2 or (PbS)1.134NbS2. (d) This suggests that alternate NbS2 layers are offset from each other in the ab plane. In fact, alternate layers are shifted by b/2. This is analogous to the different c-axes needed to describe the CdCl2 and CdI2 structures as shown in Figure 1.29. (e) A likely structure is to have one layer of PbS followed by two layers of NbS2. This material has been reported with cell parameters of a = 5.829, b = 5.775, c = 35.86 Å for the PbS and a = 3.326, b = 5.775, c = 35.87 Å for the Nb2S4 portion in G.A. Wiegers, Prog. Sol. St. Chem. 24 (1996), 1–139. a(PbS)/a(NbS2) remains at ~1.75. Given that the distance between PbS layers in (PbS)1.14NbS2 is 11.90 Å you’d expect the layer repeat distance for (PbS)1.14Nb2S4 to be around 18 Å (11.90 Å plus the ~6 Å interlayer separation of layered NbS2). Owing to the way in which the layers stack the true c axis is is approximately double this.

2.27

(a) We can calculate the composition by considering the anion packing density in the square- and triangular arrays. In the square grid, we have anion–anion contact along the edge of the square. Call this distance 1, so the square has area 1. Each projected square in Figure 2.16 contains 1 anion and 1 cation and the composition is MX per unit area. Each triangle contains 0.5 anions (3 corners, the anion on each corner is shared by 6 other 12

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

triangles) with anion contact along the edge, which therefore also has distance 1. The triangle area is half the base multiplied by the height = (1/2)(√3/2) = √3/4 ≈ 0.4330. The X packing density is therefore 0.5/0.4330 = X1.155 per unit area. The overall composition is then MX + X1.155 = MX2.155. (b) The height of our equilateral triangle is sin(60°) = √3/2. This does not make a simple ratio with the unity edge of the square, so we would expect an incommensurate structure. There would be an approximate alignment along b of Figure 2.16 after 13 squares and 15 triangles, 15×sin(60°) = 12.99, but this description would require use of a large unit cell. (c) Assume again that the squares have edge 1.0.

Assume that the base of the triangles remains 1.0 but that their height distorts such that 6 triangles have the same length as 5 squares. The height is then 5/6 (0.833 compared to 0.866 for undistorted triangles, a <4% distortion). The area of each triangle is (1/2) × (5/6) = 5/12, and anion packing density is 0.5/(5/12) = 6/5 = 1.2 X per unit area. Overall we have MX + 1.2X = MX1.2. This corresponds to the Y5O4F7 composition of Figure 2.16. For a general system with a squares and b triangles in registry, the packing density is b/a so composition is MX1+(b/a). For a = 6 and b = 7, we have MX2.167. (d) Y7O6F9 can be expressed as YX15/7 = YX1+8/7. We would expect 7 squares to be aligned with 8 triangles. The triangle height is 0.875, a 1% distortion from ideal.

Chapter 3: Defect Chemistry and Nonstoichiometry 3.1

The charge number (ionic charge) of the defect in Kröger–Vink notation is the excess ionic charge in the defect’s immediate neighborhood right after its formation. In a nonionic approach, it is the oxidation state of the defect’s atom (or 0 for empty site) minus the oxidation state of the regular atom. When an oxygen vacancy forms upon O2− removal, ionic excess charge 2+ appears in its immediate neighborhood (and is associated with the vacancy). Each 1+ ionic charge is denoted as a dot, each 1− as a prime; vNi′′, vO••, LiNi′, AlNi•.

3.2

Aluminum acceptor AlTi′, niobium donor NbTi•, fluorine donor FO•, nitrogen acceptor NO′, oxygen interstitial Oi′′ and titanium interstitial Tii••••.

3.3

This structure has dense packing of metal atoms with oxygens in tetrahedral holes. The oxygens will tend to become defects, forming anion-Frenkel intrinsic defect pairs at the point of integer valence/structure.

13

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

3.4

This intrinsic disorder consists of a regular M site forming vM′′ and Mi••. Compared with the Schottky case in the original Table 3.4, reduction is different, as it forms metal interstitials instead of oxygen vacancies: Process

Reaction

Mass-action term

cation Frenkel

MM = vM′′ + Mi••

KcF = [vM′′][Mi••]

Ionization

nil = e′ + h•

Ki = [e′][h•]

Oxidation Reduction

3.5

O2(g) = 2vM′′ + 4h• + 2OO 2O  + 2MM = 2Mi•• + 4e′ + O2(g) O

Kox = [vM′′]2[h•]4ꞏpO2−1 Kred = [Mi••]2[e′]4ꞏpO2

Chemistry: Zr in ZrO2 cannot be further oxidized and it is also hard to reduce (the groupnumber oxidation states dominate for early second- and third-row transition metals).

Physics: The appearance of a cubic-zirconia crystal suggests a wide band gap. The concentration of electronic defects in ZrO2 will therefore be low and almost fixed. The predominant defects will be ionic.

3.6

Fill a table analogous to Table 3.4 with the intrinsic anion-Frenkel reaction taken from Table 3.1, oxidation with O2 forming holes and Oi′′ from Table 3.2 and reduction of OO forming vO••, electrons and O2 from Table 3.3. Write the electroneutrality condition, 2[Oi′′] + [e′] = [h•] + 2[vO••], which can be simplified for the three limiting ranges: Around the point of integer valence into [Oi′′] = [vO••] = constant that, considering the mass-action term in your table, gives [h•] proportional to pO2+1/4 and [e′] to pO2−1/4. In the most oxidized range into 2[Oi′′] = [h•], giving [Oi′′] and [h•] proportional to pO2+1/6. In the most reduced range into [e′] = 2[vO••], giving [e′] and [vO••] proportional to pO2−1/6. The plot is essentially the same as that on the left-hand side of Figure 3.3.

3.7

As shown in Figure 3.1, the intrinsic anti-Schottky disorder means that cation and anion interstitials are the intrinsic defects in PbO (in Schottky disorder, both would be vacancies). Upon reductive nonstoichiometry, Pb interstitials will occur instead of oxygen vacancies in the previous Problem. As both are doubly charged, the limiting slopes will be the same. The reduction equation is 2OO + 2PbPb = 2Pbi•• + 4e′ + O2(g).

3.8

Follow the same procedure as in Problem 3.6. The result only differs around the point of integer valence, because here the electroneutrality equation will simplify into [e′] = [h•] = 14

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

constant that, considering the mass-action term in your table, will give [Oi′′] proportional to pO2+1/2 and [vO••] to pO2−1/2. The plot is similar to that on Figure 3.3 right.

3.9

(a) Oxidation with O2 follows the equation O2(g) = 4vCu′ + 4h• + 2OO , with equilibrium constant Kox = [vCu′]4[h•]4ꞏpO2−1. In the oxidized range, the electroneutrality condition simplifies to [vCu′] = [h•] that plugged into Kox gives [vCu′] proportional to pO2+1/8. (b) Oxidation with O2 is 3O2(g) = 4vCr′′′ + 12h• + 6OO with equilibrium constant Kox = [vCr′′′]4[h•]12ꞏpO2−3. In the oxidized range, the electroneutrality condition simplifies to 3[vCr′′′] = [h•], giving [vCr′′′] proportional to pO2+3/16.

3.10

The Schottky intrinsic defect is a pair of metal and oxygen vacancies. Acceptor (donor) doping will increase the fraction of oxygen (metal) vacancies.

3.11

Acceptor (donor) doping will increase the fraction of holes (electrons); it will promote oxidation (reduction). wide: compensator from the dominant pair

e′ vO••

h•

vO•• +¼

–¼

e′

integer valence

+½

h•

h• vNi′′

e′

–½ integer structure

Dopant is compensated by:

electronic defect

ionic

vNi′′

3.12

vNi′′

vO••

e′ vO••

+½ vNi′′ Dopant is compensated by:

3.13

h•

wide: compensator from dominant pair

e′

vNi′′

–½

integer structure

electronic

–¼

integer valence

vO••

h• v ′′ Ni

+¼

vO••

ionic defect

h•

e′

15

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

3.14

(a) 2ZrO2(s) = 2e′ + 2ZrCr• + 3OO + ½O2(g); the cation and anion sites have a 2:3 ratio.

The solid solubility depends on the oxygen partial pressure since Ksd = [ZrCr•]2[e′]2pO2½. (b) 3ZrO2(s) = vCr′′′ + 3ZrCr• + 6OO; again 2:3 ratio of sites. Solid solubility is independent of pO2 as Ksd = [ZrCr•]3[vCr′′′].

3.15

(a) The variable δ is expressed from Equation (3.5) and evaluated for pO2 = 10−4 bar and 0.21 bar yielding YBa2Cu3O6.31 and YBa2Cu3O6.88, respectively, for Ar gas and air. (b) Expressing pO2 from Equation (3.5) yields an O2 partial pressure of 1.32×10−3 bar for δ = 0.5 equivalent to average CuII.

3.16

The NdBaFe2O5+δ formula suggests NdBaFe2O5 as the point of integer structure.

NdBaFe2O5.5 is the point of integer valence. The point of integer structure is defined via donor doping hypothetical Ba2Fe2O5 by 1NdBa• and is reached by reduction, hence the simplified defect equation follows from subtraction of the Frenkel process OO = vO•• + Oi′′ from the reduction OO = 2e′ + vO•• + ½O2(g). The result is reduction of oxygen interstitials, Oi′′ = 2e′ + ½O2(g). Its mass-action constant Kired is set up by counting interstitials from the point of integer structure, [Oi••] ≈ δ, and electrons from the point of integer valence, [e′] ≈ 2(0.5−δ). This gives Kired = pO2½ × 4(0.5−δ)2/δ valid for δ values well inside the thus defined nonstoichiometry range.

3.17

This type of flux is an amount (mass, volume, mole, number) of something flowing per area per second. Expressing the flux of 6 kg/(0.2 cm2 1 min) in SI units gives 5000 kg/m2s, 5 m/s (volume flux becomes speed), 2.778×105 mol/m2s, or 1.673×1029 molecules m−2s−1.

3.18

The task is to calculate the mass of carbon permeating per unit area per unit time, given the slope of the linear concentration profile inside the wall. The slope is (0.5−1)/0.0005 = −1000 kg/m4. Equation (3.7) gives JC = 10−8 kg/m2s. Verbalized; 0.01 mg of carbon permeates every second through a square meter of the plate.

3.19

Equation (3.7) applies for DO, and we need to evaluate the oxygen-atom flux and the concentration gradient inside the membrane. The O2 flux is 0.00084/24 = 3.5×10−5 mol/m2s, hence JO = 7×10−5 mol/m2s. For steady-state diffusion, the concentration gradient that drives it is expressed in differences. The limiting compositions of the membrane are YBa2Cu3O6.70 at the pressurized-air side and YBa2Cu3O6.63 at the O2 side (obtained as in Problem 3.15). The difference is 0.07 O per formula, and the molar 16

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

volume of 10−4 m3/mol converts it to 700 mol/m3. The concentration slope is (−700)/0.01 = −70000 mol/m4. Equation (3.7) gives DO = 10−9 m2/s at 700 °C. 3.20

(a) The probability concerns making the same choice of probability ½ four times, hence (½)4, multiplied by two because of two directions, hence ⅛. (b) The distribution of probabilities after n steps is: displacement step

−4λ

−3λ

−2λ

−λ

0

2λ

3λ

4λ

1

1

½

0

½

¼

0

½

0

¼

⅛

0

⅜

0

⅜

0

0

¼

0

⅜

0

¼ 0

2 3 4

λ

0

1/16

⅛ 1/16

(c) This is a discrete case with a set of probabilities for each step and a mean equal to n p zero. The mean squared deviation from this mean is  2 =  n

n

(n − 0)2 where pnλ is the

−n

probability at each position after n steps and nλ is the distance in length units, in this case however, λ = 1. From the table above, σ42 = 4 and the root mean square deviation σ4 = 2 from the origin. (d) Evaluation of data in the table above suggests that n 2 = p n

n

(n)2 = n2 ; the root mean square deviation is then σn = λ√n.

−n

3.21

Since the energy is expressed per mol, the Boltzmann constant k must be multiplied with the number of particles in one mol (Appendix J). Then ν times pB of Equation (3.9) gives the jump frequency of 9.59×109 s−1. On average, every 1/pB = 1043th carbon vibration attempt becomes a jump.

3.22

Fe atom has Nn = 8 nearest neighbors. The jump frequency of an Fe atom into a vacant site is ν times pB of Equation (3.9) times the probability pavail = 0.0008 that a vacancy occurs around a selected Fe atom, hence 1.15×109 s−1. The progression rate at a given site is obtained by multiplying this result by the probability pdir that this vacancy will move the jumping atom forwards along the unit-cell edge, pdir = 4/8 = ½, and by the correlation coefficient fc between the jumping atom and the vacancy it left behind, estimated as fc = 1

17

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

− 2/(Nn−1) = 0.714. Thus rprogression = 4.115×108 s−1. On average, there are 1/(pavailpB) = 8679 Fe vibrations between jumps.

3.23

Equation (3.18) suggests multiplying the progression rate with square of the progression distance that was considered for the directional probability pdir. That distance was a/2, and DFe = 8.47×10−12 m2/s.

3.24

The interstitial carbon sites are at the face centers of the bcc cell of Fe. Jump in the flow direction moves the C atom forward by a/2 with probability pdir = 4/12. From Equation (3.10) and Equation (3.18), Δ‡Gm = 61896 J/mol, considering that fc and pavail = 1 for interstitial diffusion with dilute defects.

3.25

The vacancy fraction [vAl] will appear in the site-availability term pavail of Equation (3.10). For vacancy diffusion in fcc elemental structure, where each atom has Nn = 12 nearest neighbors, pavail = 12[vAl]. The probability pdir that this vacancy will forward the jumping atom by λ = a/2 along a is 4/12 = 1/3. The estimated correlation factor fc = 1 − 2/(Nn−1) = 0.82. From Equation (3.10) and Equation (3.18), it follows that [vAl] = 4.17×10−6.

3.26

(a) A single type of vacancy jump can be identified in the unit cell, which forwards an atom by λ = a with probability pdir = 1/6 that it occurs along one direction of the unit cell edge. Since the actual length of the jump ℓ equals a, pdira2 = a2/6 = ℓ2/6 verifies the statement given that Nn = 6. (b) A single type of jump that forwards an atom by λ = a/2 along one direction of the unit-cell edge is found, occurring with probability pdir = 4/8.

Since the actual length of the jump is ℓ = ½ a√3, pdirλ2 = a2/8 = ½ (ℓ/√3)2 = ℓ2/6, to be compared with Nn = 8. (c) A single type of jump that forwards an atom by λ = a/2 along one direction of the unit-cell edge is found, with probability pdir = 4/12, and since the actual length ℓ = ⅓ a√2 of that jump, pdirλ2 = a2/12 = ⅓ (ℓ/√2)2 = ℓ2/6, to be compared with Nn = 12.

3.27

Po has Nn = 6 nearest neighbors, and the probability pavail that a vacancy occurs around a selected Po atom is 0.006. The probability pdir that this vacancy will forward the atom by λ = a in a given direction is 1/6. From Equation (3.18) and Equation (3.10) with fc = 1 − 2/(Nn−1), DPo = 1.17×10−13 m2/s. A simple estimate from jump balance is that the vacancy of fraction 0.001 has diffusivity ~1000 times higher, hence Dv = 1.17×10−10 m2/s.

18

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

3.28

An interstitial Po will reside in the cubic hole, have 6 interstitial sites to jump to, hence pdir = 1/6 that the jump forwards the atom by λ = a along the chosen direction. From Equation (3.18) and Equation (3.10), DPo = 1.247×10−11 m2/s. Three equal jumps forward or backward are obtained by choosing the cell body diagonal as the progression direction.

The pdir increases 3-times, to ½, but the jump-progression length λ in that direction decreases to a/√3, so they yield the same DPo. Visually, the jump-progression lengths λ cut the cube body diagonal a√3 into 3 equal segments, as seen on a 2×2×2 cell oriented such that the unit-cell edge, face diagonal, and body diagonal project in actual lengths.

Or, a mathematically elegant solution: λ is the scalar projection of any of the three jumps, expressed in lattice-vector coordinates as (1,0,0) (0,1,0) (0,0,1), onto their sum vector (1,1,1).

3.29

(a) Since pavail = 1, the answer is given by the inverse of the Boltzmann probability pB, which is expressed from DC with Equation (3.10) and Equation (3.18). Sketching an fcc cell of Fe and placing C into its center (one of the octahedral holes) reveals that the probability of a jump forwarding the carbon by λ = a/2 is pdir = 4/12 (12 octahedral holes at the edges of the sketched cell). Then pB = DC/(pdirνλ2) = 0.0002. (b) The number of jumps n in the given time of one minute is n = 60νpB = 1.2×1011, and the total path of the atom is n(a/2)√2 = 31 m. (c) The root mean square deviation of that atom from its origin one minute earlier is σn = λ√n = 6.317×10−5 m. (d) Equation (3.9) gives ΔGm = −ln(pB)NAkT = 90.16 kJ/mol.

3.30

Right-hand side: μ = σi/qici and from units given in text we write (A/Vm)/(As/m3) = m2/Vs = (m/s)/(V/m) hence speed per electric-field intensity. Left-hand side: μ = Diqi/kT is in (m2/s)(As)/[(J/K)K]. Because J ≡ Ws ≡ VAs, we obtain m2/Vs for the unit of mobility, equivalent to (m/s)/(V/m).

3.31

The unity transference number suggests a pure oxide-anion conductor in an experimental setup where oxygen is transported as oxygen vacancies via zirconia and electrons via the circuit. We have one type of charge carrier, and the Nernst–Einstein equation applies.

Converting σ(vO••) to D(vO••) requires knowing the concentration of the oxygen vacancies.

The fluorite structure (Figure 1.31) has 4×0.15 oxygen vacancies per unit-cell volume, thus the concentration c(vO••) = 4.444×1027 m−3. The diffusivity follows from Equation

19

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

(3.23) as D(v O••) = σ(v O••)kT/(2e)2∙c(v O••) = 2.492×10−10 m2/s and the mobility μ(v O••) = σ(vO••)/2ec(v O••) = 4.21×10−9 m2/Vs. 3.32

The ionic-conductivity transference number is the conductivity fraction due to the ion, hence the total conductivity σ = 104 S/m.

3.33

Calculate the limit of Equation (3.34) for D(vO••) → ∞ and then substitute from the electroneutrality condition of the two opposing fluxes, 2c(vO••) = c(h•).

3.34

As in text, define the enhancement factor as the ratio of the diffusion coefficients of the element A and of its ion. Calculate the limit of Equation (3.34) for D(h•) → ∞. Divide by D(vO••) and generalize the number 4 in the result as a square of the ionic charge to obtain the answer 1+z for the enhancement factor, considering the electroneutrality condition that the fraction of the electronic defect is z times the fraction of the ionic defect of the respective two fluxes.

3.35

(a) The rock-salt structure has 4 metal sites per unit cell, there are 4×10−4 nickel vacancies per unit-cell volume, and this converts to c(vNi′′) = 5.4×1024 m−3 with the concentration of holes twice that. (b) Equation (3.11) suggests that the diffusivity D(vNi′′) of nickel vacancies is (0.9999/0.0001) ∙ DNi ≈ 1×10−11 m2/s. The diffusivity of holes can be calculated after we are sure about the conductivity due to holes in the following point (c). In it, we see that ionic conductivity is very small, hence σ(h•) ≈ 65.5 S/m, from which the Nernst–Einstein formula of Equation (3.23) gives D(h•) = σ(h•)kT/e2c(h•) = 4.48×10−6 m2/s. (c) The electrical conductivity due to nickel vacancies follows from Equation (3.23) as σ(vNi′′) = D(vNi′′)∙(2e)2∙c(vNi′′)/kT = 2.924×10−4 S/m, hence 224000 times smaller than the total. The ionic transference coefficient is the inverse of the latter value, 4.464×10−6, and the electrical conductivity of 65.5 S/m is essentially only due to holes. NiO is an electronic conductor. (d) The mobility of nickel vacancies can be calculated for example from D(vNi′′) as μ(vNi′′) = D(vNi′′)∙2e∙c(vNi′′)/kT = 1.69×10−10 m2/Vs. The mobility of holes can be calculated for example from σ(h•) as μ(h•) = σ(h•)/ec(h•) = 3.79×10−5 m2/Vs.

3.36

The balanced oxidation reaction is 3O2(g) = 12h• + 4vCr′′′ + 6O O, in analogy with Table 3.4. For this equilibrium, Kox = [vCr′′′]4[h•]12/pO23. With holes compensated by chromium vacancies, the electroneutrality condition is 3[vCr′′′] = [h•]. Then [vCr′′′] = Kox1/16  S +16‡ Sm   H ox +16‡ H m   pO 16  exp− . and D =  2  N 2 p  exp ox Cr n dir 4 16R 16RT      3  3

4 3/16

(pO2/3 )

20

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

3.37

Yes, it is dimensionless. The half-length y is in m, D in m2/s, τ in s (all SI units).

3.38

The penetration depth quintuples in 52 of the original time/period, hence in 25 h.

3.39

The maximum relative increase from one side of the blade is 0.0025 = 1 − erf(u), which gives u = 2.1378 (tables, software or the internet). Equation (3.39) then gives minimum thickness values of 0.1622 and 0.8112 mm for 1 and 25 hours of nitridization, resp.

3.40

The 90% change refers to the relative cr variable (Figure 3.11), and Equation (3.40) becomes 0.9 = 1 − erf(u), hence erf(u) = 0.1. An approximate solution considers that erf(u) ≈ u for small u, and u = 0.1 in Equation (3.39) then gives 25000 seconds, or ~7 hours, to reach the 90% change 1 mm below surface. A precise solution yields u = 0.088856, giving 31664 seconds or ~9 hours.

Chapter 4: Phase Diagrams and Phase Transitions 4.1

Definitions are given in Section 4.1.

4.2

Mostly discussed in text. (a) At 100 °C, A, AB, AB2, and B are all thermodynamically stable. (b) Line is called the solidus. (c) Melting points are approximately 460, 460, and 680 °C. (d) AB2 melts incongruently. Between ~480 °C and ~540 °C, a mixture of liquid and B(s) is present. (e) Phases present in the numbered areas are: (1) A(s)+L. (2) A(s)+AB(s). (3) AB(s)+L. (4) AB(s)+L. (5) AB(s)+AB2(s). (6) AB2(s)+L. (7) AB2(s)+B(s). (8) B(s)+L. (9) Homogenous Liquid. (f) No solid solutions are formed. (g) The condensed form of the phase rule is P + F = C + 1, so F = C – P + 1. This is a twocomponent system, C = 2. Point a represents a single liquid phase, so F = 2 – 1 + 1 = 2, and one needs to specify temperature and composition to define the system. Point w is a two-phase mixture of A and liquid, so F = 2 – 2 + 1 = 1, and specifying either temperature or the composition defines the system. Point x is a two-phase mixture of liquid and A, so F = 2 – 2 + 1 = 1, and one variable must be specified to define the system—if one fixes temperature, the composition of the liquid is defined; if one fixes the composition of the liquid, the temperature is defined. Point y again has F = 1, and temperature can be varied without changing the number of phases while the phase composition is fixed (not variable as in the liquid). Point z is the eutectic point. Three phases are present (A, AB and liquid) so F = 2 – 3 + 1 = 0 and no variables have to be specified to fully define the system. (h) Cool down of compositions marked: (a) Below 21

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

~450 °C, solid A crystallizes in equilibrium with remaining liquid. At 250 °C, all liquid solidifies upon forming an A-rich mixture of A and AB. (b) Is a eutectic point. The liquid will solidify to form pure A and AB. (c) Like a but first solid to form will be AB. (d) Like a but with AB2 forming first and then AB+AB2 on further cooling. (e) Below 520 °C, small amounts of B crystallize from the liquid. Right below 500 °C, AB2 and an Arich liquid form via reverse peritectic reaction. At 380 °C, all the liquid has solidified into a mixture of AB and AB2 crystals. (f) Below 620 °C, solid B crystallizes in equilibrium with the liquid until at 500 °C the liquid and B solidify into a mixture of B and AB2. (i) Composition a will be 100% liquid at 500 °C. At 400 °C, lever rule gives xL as 0.33; at 300 °C 0.22. (j) If composition e is cooled rapidly, then it’s likely that you’ll get a nonequilibrium mixture at room temperature. For the peritectic reaction to proceed completely, all of solid B would have to react with liquid to give solid AB2; a significant change upon rapid cooling. (k) At e, all of B reacts to give AB2, at f, only some of B reacts to give AB2.

4.3

From Figure 4.8, ZrW2O8 is thermodynamically stable between 1105 and 1257 °C. One can therefore sinter a polycrystalline powder from component oxides within this temperature interval. Because the product decomposes back to binary oxides on cooling, it needs to be cooled rapidly to a temperature where it is kinetically stable, for example by quenching in liquid N2. Despite considerable technical challenges, single-crystal growth has been performed by Kowach et al., J. Cryst. Growth 212 (2000), 167‒172. To do this, one needs to take into consideration both the incongruent melting of ZrW2O8 and the need to quench the material to room temperature. Starting at a composition a in the figure below, small single crystals of ZrW2O8 form in a solidified eutectic, but it’s hard to separate the crystals from the eutectic. Larger single crystals can actually be grown at composition b by placing a layer of WO3 over a layer of ZrO2+2WO3. It’s thought that as the WO3 melts and percolates through the ZrO2+2WO3 mixture to the bottom of the crucible, crystals grow in a well defined band of the crucible. Gradual evaporation of WO3 is also thought to move the composition slowly towards ZrW2O8. In the published synthesis, the sample (~107 g of stoichiometric oxide covered by 15 g of WO3) was slowcooled from 1300 °C at 0.5 °C per hour to 1230 °C, at which point it was rapidly cooled to room temperature. Crystals up to 1 cm across were grown.

22

1400

Liquid

1300

ZrW 2O8 ZrW 2O8 0.4

1200 1100 0.6

0.8

1.0

WO3

mole fraction WO3

4.4

temperature (ºC)

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

This is based on the 1952 phase diagram of BaO‒Al2O3 in the figure below.

1600

0.0

Al2BaO4

1400

0.2

Al2O3

4.5

Al2Ba3O6

1800

Al12BaO19

temperature (ºC)

2000

0.4

0.6

0.8

mole fraction BaO

1.0

BaO

This is based on a simplified CaTiO3 phase diagram below. Other now-known phases (e.g. Ca4Ti3O10) were added to the phase diagram later and are not shown.

Liquid

CaO + L

1800

L+ CaTiO3

1600

CaO + Ca3Ti2O7 1400 0.0

CaO

4.6

0.2

CaTiO3 + Ca3Ti2O7

temperature (ºC)

2000

CaTiO3 + L

TiO2 + L

CaTiO3 + TiO2

0.4 0.6 mole fraction TiO2

0.8

1.0

TiO2

When Ni metal is exposed to a fixed sub-stoichiometric quantity of O2 gas in a sealed system at high temperature, a portion of Ni will react (2Ni + O2 ↔ 2NiO), forming an equilibrium mixture of Ni, NiO and O2 gas of a very low pressure). The non-condensed 23

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

phase rule P + F = C + 2 applies. We have a two-component system (one must specify the amount of Ni and O to give the composition), C = 2, and three phases, P = 3. There is a single degree of freedom, F = 2+2−3 = 1. When the system is heated to a specified temperature, the partial pressure of oxygen will be uniquely determined and remain constant as long as both Ni and NiO are present in any ratio under equilibrium. ↑ pO2

NiO Ni

temperature → 4.7

A straightforward geometrical proof is based on the areas of triangles. If we draw lines from each corner to point d, then the sum of the areas of triangles ABd + BCd + AdC must equal the total area of the triangle, which is equal to ½ the product of the height and the edge. Let the height of the ABC triangle be h and each edge be e.

Area(ABC) = Area(ABd) + Area(BCd) + Area(AdC) he 2

4.8

=

xC e 2

+

xA e 2

+

xBe

hence h = x +x + x .

A

2

B

C

The points a,b,c have a constant B:C ratio and lie on a straight line; they represent a system getting progressively poorer in A; d,c,e have a constant A:C ratio and lie on a straight line; they represent a system getting progressively poorer in B. A 0.2

a

0.8

0.4

0.6 b

0.6

0.4

e c

0.8

C

0.2

d 0.2

0.4

0.6

mole frac B

24

0.8

B

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

4.9

Points have been set up so they can be read off a single axis using the method in the text.

Note that the read off fractions refer to “normalized” formulae of the triangular diagram’s components Zr, Ti, 2Al (neglecting the oxygen for clarity). Point: a 0.6 ZrO2 and 0.4 Al2O3. b 100% ZrTiO4. c 0.6 ZrO2, 0.2 of ½ Al2TiO5, 0.2 of ½ ZrTiO4. d 0.2 ZrO2, 0.6 Al2O3, 0.2 of ½ Al2TiO5. e 0.4 of ½ ZrTiO4, 0.4 of ½ Al2TiO5, 0.2 TiO2. f 0.1 of ½ ZrTiO4, 0.4 of ½ Al2TiO5, 0.5 TiO2.

4.10

This is intended as a discussion question to highlight different aspects of phase transitions. (a) The transition from -quartz to -quartz at 573 °C is best described as a displacive transition as no chemical bonds are broken. As discussed in the text, it’s driven by a soft phonon that leads to a cooperative tilt of the SiO4/2 corner-shared tetrahedra at the phase transition. It’s unlikely that you could quench the high-temperature structure to low temperature. (b) The transitions from -quartz to -tridymite to -cristobalite at 870 and 1470 °C are reconstructive transitions in which the pattern of bonding between SiO 4 tetrahedra is changed (see also Section 15.2). The transitions are likely to be relatively slow such that the high-temperature forms can be quenched to room temperature where they remain metastable. (c) The transition at 641 °C of the a0b0c+ tilted (P4/mbm) NaNbO3 to non-tilted cubic (Pm3̄ m) perovskite is a displacive transition, for two reasons: No bonds are broken during the tilt of the NbO6/2 octahedron around one of its axes. The P4/mbm space group has a klassengleiche P4/mmm supergroup (Appendix B) that is a translationengleiche subgroup of Pm3̄ m. The cubic structure therefore cannot be quenched through a continuous transition such as this. (d) The high-temperature transition from tetragonal CuAu to cubic CuAu is an order‒disorder transition analogous to FePt described in the text. The high-temperature disordered form can be quenched to room temperature. On heating to moderate temperatures, the metastable disordered form will begin to order.

4.11

At first sight we might conclude that the change from a disordered liquid to the ordered crystal structure increases the symmetry. This would be unusual as symmetry is normally reduced when a phase transition occurs on cooling. However, the time-averaged structure of the liquid phase actually has the higher symmetry as each point in space is “on average” equal, giving liquid water complete translational and rotational symmetry. Solid ice has only the symmetry elements of its space group.

25

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

4.12

The solution is entirely analogous to that presented in Section 4.4.4 except that the order parameter now has symbol M. See Figure 4.15 for a sketch of G vs η ≡ M. The G(M) minima from ∂G/∂M = AM + BM3 = 0 are M = 0 for T > Tc and M = ±[a(Tc – T)/B]½ for T < Tc. In reality, critical experimental exponents are often less than ½ due to the short

M

range nature of magnetic coupling leading to departures from Landau theory.

0.0

0.5

1.0

1.5

T/TC

4.13

For a tricritical transition, B = 0 and a Cη6 term is needed to get positive curvature in the G(η) plot. The expression for G becomes G() = G + a(T − T ) 2 + C6 . Applying 1

0

G 

2

1

0

6

= 0 gives  = (a C )(T − T )1 / 4 . Since  = 1 at T = 0, Tc = C/a. Thus c

 = (T − T ) T 1/4 . c

4.14

c

Subsitution of 2 = 3B/(4C) and T – T0 = 3B2/(16Ca), both valid at T = Tc, into Equation (4.17) yields G = G0 as the other three terms sum to zero. To verify that the Gibbs energy is minimum at Tc, differentiate Equation (4.17) with respect to , divide by  and set = 0 for T = Tc. This gives

dG d

= 0 = a (T − T ) − B2 + C4 , and substitution of 2 = 3B/4C into it 0

confirms that Tc – T0 = 3B2/(16Ca).

4.15

This is just a case of substituting Tc – T0 = 3B2/(16Ca) into the Footnote 26 expression for G(T) at T = Tc. All terms except G0 cancel showing that the value of G at the   0 minimum equals G0 at the critical temperature.

4.16

From Equation 4.17, express S = −dG/dT. Substitution for 2 = 3B/(4C) gives S = −3aB/(8C). Since ΔG = 0 at Tc, use ΔH = TΔS to obtain H = −3aBTc/(8C).

4.17

The data should be straightforward to analyze in a spreadsheet. Ideal-result graphs are shown below. (a) The strain associated with the phase transition is readily apparent when 26

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

compared with extrapolation of the high-temperature data, Diffraction is thus a sensitive technique for following the transition. Although the original work used a more sophisticated expression (as data were analysed over a wider temperature range), a linear extrapolation of high-temperature data is sufficient for this Problem. (b) A plot of et0.5 against T is equivalent to plotting the order parameter. The plot below shows a fit to this using an expression of the type derived in the previous Problem with  = 0.27 and Tc = 1347 K. Note this  is close to the 0.25 expected for a tricritical transition. (c) As in the previous Problem,  should depend on T0.25 for a tricritical transition. One would therefore expect a plot of et2 against T to be linear. This behavior is approximated well.

0.

4.160

0.

4.150

et

0.5

2

4.170

et

cell parameter (Å)

The intercept with the x axis gives Tc = 1342 K.

0.

0.

4.140

4.130 1100

1200

1300

1400

00

1500

temperature (K)

temperature (K)

temperature (K)

Chapter 5: Chemical Bonding 5.1

(a) The successive sums are −2.4853 with 6 cations and 12 anions (2 shells), +2.1335 with 14 cations and 12 anions (3 shells), −0.8665 with 14 cations and 18 anions (4 shells), +9.8666 with 38 cations/18 anions (5 shells), +0.067873 with 38 cations and 42 anions (6 shells), −4.1739 with 38 cations and 54 anions (7 shells). (b) There is no obvious sign of convergence. (c) After three terms, the value of 2.1335 is closest to the true value of the Madelung constant, 1.7476. This also corresponds to the minimum charge of the cluster, which is +1, remembering the central ion is an anion. (d) The series will converge if successive shells are chosen in a way that the net charge of the cluster per atom successively decreases.

5.2

As multiples of the cell-edge length a, the first four terms in the Madelung expansion are 𝑈C = (z1e)(z2e) ( 8 — 6 + 12 — 4πs0a

√3⁄2

1

√2

24

√11⁄2

+ ⋯ ). Given a = (2 3 ) d , the coulombic potential

energy with the bond distance d is: 𝑈C = (z1e)(z2e) (8 — 4πs0d

27

1

6

2⁄√3

+

12

√8⁄√3

—

24

√11⁄√3

+ ⋯) .

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

5.3

(a) Plugging appropriate values into the Born–Mayer equation gives the lattice-formation energy of −3470 kJ/mol. (b) The calculated value is ~60 kJ or 1.5% lower. (c) The repulsive term is 0.144 or 14.4% of the attractive term. (d) Because Sr2+ is larger than Ca2+, the interatomic distance d0 would be larger. All other terms will be the same. Hence the lattice-formation energy will have a smaller absolute value. The experimental SrO lattice-formation energy is −3220 kJ/mol.

5.4

(a) The Born–Mayer equation yields lattice-formation energy of −3850 kJ/mol. (b) Using this energy as an estimate of the lattice-formation enthalpy, a Born–Haber cycle analogous to Figure 5.3 is 0 = −ΔHf (MgO) + ΔHsubl (Mg(s) → Mg(g)) + ΔHat (½O2 → O(g)) + IE1(Mg) + IE2(Mg) + Heg1(O) + Heg2(O) + ΔHL, which gives a value of Heg2(O) = +804 kJ/mol for the 2nd electron gain enthalpy of oxygen. Thus, we see that adding a second electron to oxygen is endothermic.

5.5

CdI2 and CdCl2 form layered structures with cations filling octahedral-hole planes between every other layer of anions. In these structures, anions come into direct contact across the hole planes with no cations. For these structures to be stable, the anions need to be polarizable, and the cation–anion bonds should not be highly ionic. Otherwise the anion–anion repulsions will destabilize the structure in favor of either the rutile or the fluorite structure. Neither of these conditions is met in metal fluorides.

5.6

The London dispersion forces increase upon moving down a column of the periodic table.

This behavior stems from the fact that, as successive electron shells are filled, the outer electrons are less tightly bound to the nucleus, which makes the electron cloud more polarizable.

5.7

(a) False. The s orbitals do not have nodal planes, but ns orbitals with n > 1 have radial nodes. (b) True. (c) False. The number of nodal planes is equal to the orbital angularmomentum quantum number ℓ.

5.8

(a) n = 4, ℓ = 0, 3 radial nodes, 0 nodal planes. (b) n = 5, ℓ = 2, 2 radial nodes, 2 nodal planes. (c) n = 4, ℓ = 3, 0 radial nodes, 3 nodal planes. (d) n = 2, ℓ = 1, 0 radial nodes, 1 nodal plane.

5.9

Peroxide has two more electrons than O2. Therefore, the MO occupation will be σ(2s)2σ*(2s)2σ(2p)2π(2p)4π*(2p)4. The bond order is ½(8−6) = 1; a single bond. The reduction in bond order from 2 in O2 to 1 in O2−2 will lead to a longer bond for peroxide.

28

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

5.10

The HOMO is the doubly degenerate set of orbitals that arise from bonding interactions between the 2px and 2py orbitals of B and the 1s orbitals of the surrounding hydrogen atoms (labeled 1e′ in Figure 5.25). The LUMO is the singly degenerate nonbonding 2pz orbital of B.

5.11

The deviation from the general trend is due to occupation of the eg orbitals that are higher in energy than the t2g orbitals in an octahedral ligand field. While eg is unoccupied for V2+, t2g(π*)3eg(σ*)0, it is singly occupied for Cr2+, t2g(π*)3eg(σ*)1, and doubly occupied for Mn2+, t2g(π*)3eg(σ*)2, as well as for high-spin Fe2+, Co2+ and Ni2+. Because the eg(σ*) orbitals are more strongly antibonding than the t2g(π*) orbitals, adding electrons to the eg(σ*) orbitals weakens the metal–ligand bonding more significantly than adding electrons to the t2g(π*) orbitals. A weaker bond is a longer bond, hence the larger than expected radii of Cr2+ and Mn2+.

5.12

HS and LS states are different for d4, d5, d6 and d7 electron counts. In an octahedral ligand field, there is no distinction between HS and LS for the other electron counts.

5.13

(a) RuCl3 contains LS Ru3+ because Ru is a 4d transition metal, by process of elimination FeCl3 must contains HS Fe3+. (b) Tetrahedrally coordinated first-row transition-metal ions are nearly always HS, so [CoCl4]− will contain HS Co3+ and [Co(NH3)6]3+ will contain LS Co3+.

5.14

(a) Pt2+ is larger than Ni2+ so it is not possible to attribute the lower coordination number of the cooperite structure to the smaller ionic-size of Ni2+. Therefore, covalent interactions must be playing a role. Both Ni2+ and Pt2+ are d8 ions, and in the octahedral field of the NaCl-type structure, the orbital filling will be t2g(π*)6eg(σ*)2. Because the 5d orbitals of Pt are much larger than the 3d orbitals of Ni, they interact more strongly with oxygen. Consequently, occupation of the antibonding eg(σ*) orbitals is much more destabilizing for PtO than for NiO. When the coordination is changed from octahedral to square planar, only the dx2−y2 orbital has a strong destabilizing σ* interaction with the ligands (the d-orbital splitting is similar to a z-elongated octahedron, Figure 5.25, see also the crystal-field splitting diagrams in Figure 7.2). The dx2−y2 orbital is left empty by accommodating all 8 valence electrons in the other four d orbitals. This is what stabilizes square-planar coordination for Pt2+ in PtO. (b) As a general rule, 4d ions have crystalchemical preferences that are closer to 5d ions than 3d ions. So it is reasonable to expect 29

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

PdO to be isostructural with PtO. Magnetic measurements would be informative because a square-planar d8 ion with dx2−y2 orbital empty has all electrons paired, hence cooperitetype PdO would be diamagnetic, whereas NaCl-type PdO would be paramagnetic.

5.15

(a) The HOMO is the doubly degenerate set of nonbonding O 2p orbitals. The LUMO is the singly degenerate MO that arises from the antibonding interaction between the O 2s orbital and the σg H 1s SALC. (b) The O 2s and 2pz orbitals participate in bonding, while the O 2px and 2py orbitals are nonbonding. (c) If we bend the molecule in the xz plane, the O 2py orbital perpendicular to the plane remains nonbonding, but the O 2px orbital can interact with the SALC where the two H 1s orbitals are in-phase with each other. This orbital becomes slightly bonding. (d) In the bent molecule, the O 2s, 2pz and 2px orbitals participate in bonding, while the O 2py orbital remains nonbonding. (e) This is an example of a 2nd order Jahn–Teller distortion.

5.16

(a) Mg is octahedrally coordinated, so the ideal Mg–F bond valence is 2/6 = 1/3. Using 0 this value and RMg−F in Equation (5.18) yields 1.99 Å for the Mg–F bond distance. (b)

Plugging the appropriate values into the Born–Mayer equation yields a lattice-formation energy of −2.78×103 kJ/mol. (c) The calculated value is ~196 kJ or 6.6% higher (less stable) than the experimental lattice-formation energy. As with the alkali metal halides, the calculated energy is higher than the experimental lattice-formation energy, but the percentage is slightly larger. (d) Because Ti4+ and O2− have charges that are twice that of Mg2+ and F−, the lattice-formation energy for TiO2 should be roughly four times lower (have an absolute value that is four times larger), given that the Ti−O and Mg−F distances are similar.

5.17

(a) The bond graph has O(1) bonded to two V, while each O(2) makes one bond to V and one to Zr. (b) The expectation Zr–O(2) bond valence is 4/6 = 2/3, yielding 2 − 2/3 = 4/3 for V–O(2) that yields 5 − 3(4/3) = 1 for V–O(1). (c) Using Equation (5.18) and the expectation bond valences derived in part (b), we can estimate the distances to be 2.08 Å for the Zr–O(2) bonds, 1.70 Å for the V–O(2) bonds, and 1.80 Å for the V–O(1) bonds.

5.18

(a) La[4,4]O[4]F[4]. (b) The expected distances based on ionic radii are 2.47 Å for the La–F bonds and 2.54 Å for the La–O bonds. Although they are not in particularly good agreement with the observed bond distances, they do suggest that the La–F bonds should be shorter than the La–O bonds, hence the 2.42 Å bond is assigned as La–F and the 2.60 30

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

Å bond as La–O. (c) The ideal bond valence is calculated by dividing the absolute value of each anion’s oxidation state by its coordination number, vLa−F = 1/4 and vLa−O = 2/4 =1/2. We can double check our assignment by calculating the bond-valence sum at La = 4(1/2) + 4(1/4) = 3. (d) Using Equation (5.18) and the bond valences from part (c), the predicted bond distances are La–F = 2.57 Å and La–O = 2.43 Å; in a much better agreement with the observed bond distances than our estimates from part (b). Contrary to our radii-based estimate, bond-valence calculations indicate that the 2.42 Å bond is La–O and the 2.60 Å bond is La–F. This is in fact correct. The La–F bond is longer because it has half the bond order of the La–O bond.

5.19

(a) In fluorite, the cation is 8-coordinated, and the ideal M–F bond valence is 2/8 = 1/4.

Plugging this value into Equation (5.18) we get Ca–F = 2.35 Å and Mg–F = 2.09 Å. A geometrical analysis for a cube shows that d(F–F) = (2/√3)d(M–F), which gives F–F distances of 2.71 Å in CaF2 and 2.41 Å MgF2. (b) In rutile, the cation is 6-coordinate, and the ideal bond valence of M–F is 2/6 = 1/3. From Equation (5.18), Ca–F = 2.25 Å and Mg–F = 1.99 Å. For a perfect octahedron, d(F–F) = (√2)d(M–F), which gives F–F distances of 3.18 Å in CaF2 and 2.81 Å in MgF2. (c) The estimated F–F distance for MgF2 of the fluorite type (2.41 Å) is considerably shorter than the 2.62 Å distance where the fluoride ions begin to overlap, whereas the F–F distance in CaF2 (2.71 Å) is not below this critical value. This helps to explain why MgF2 does not adopt the fluorite structure.

In contrast, the F–F distances in both CaF2 and MgF2 with the rutile structure are above the limiting contact distance of 2.60 Å. (d) From Equation (5.18) we get Be–F = 1.69 Å.

For an octahedron, this leads to F–F = 2.39 Å, which is less than the contact distance 2.60 Å, suggesting that BeF2 will not be stable in the rutile structure. Experimentally, BeF2 adopts the quartz structure, Be[4t]F2[2].

5.20

(a) The expectation Mg–O and Si–O bond valences are 1/6 and 2/3, respectively, and correspond via Equation (5.18) to bond lengths of 2.36 Å for Mg–O and 1.77 Å for Si–O. (b) The tolerance factor t = 2.36/(√21.77) = 0.94, which suggests a distorted perovskite. (c) Experimental Si–O distances are in reasonable agreement with the predicted 1.77 Å. (d) The bond-valence sum 1.98 for Mg is close to ideal, so that we can say that this environment is well suited for Mg2+.

31

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

Chapter 6: Electronic Band Structure 6.1

(a) The MOs for H6 are equivalent to the π MOs in benzene in Figure 5.19. The number of nodal planes increases as the MO energy increases; from 0 for the lowest-energy MO in Figure 6.1; to 1 and 2 for the two sets of doubly degenerate MOs; and finally 3 for the highest-energy MO. (b) The number of nodal planes increases from 0 for the lowestenergy MO, to 1, 2, 3 and 4, respectively, for the next four sets of doubly degenerate orbitals, and 5 for the highest-energy MO.

6.2

(a) The MO diagram for a square H4 molecule is shown below on the left. The lowestenergy MO has no nodal planes, the doubly degenerate set of MOs 1 nodal plane, and the highest-energy MO 2 nodal planes. (b) By distorting to a rectangle, one of the half-filled MOs will be lowered in energy and the other will rise in energy as shown below.

6.3

Equation (6.2) gives ψ(r) that we plug into the formula for the electron density, yielding: ψ*(r + Na) ψ(r + Na) = ψ*(r) ψ(r). Next we substitute the mathematical form of the wavefunction, Equation (6.1), and simplify as e−ik(r + Na) u*(r + Na) eik(r + Na) u(r + Na) = e−ik(r) u*(r) eik(r) u(r) by considering that u*(r + Na) u(r + Na) = u*(r) u(r). The latter equation is justified because u(r) must be unaltered upon moving from one unit cell to the same position in any other unit cell, u(r + Na) = u(r) and u*(r + Na) = u*(r), hence that simplifying equality holds regardless of the value of r.

32

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

6.4

(a) −π/a ≤ k ≤ +π/a. (b) Visual inspection shows that the wavefunction repeats every six atoms, therefore, λ = 6a. Given the fact that λ = 2π/k we can deduce that k = π/3a. (c) The momentum p = h/λ, therefore p = h/6a.

6.5

(a) True. (b) False. (c) False, because the first Brillouin zone is the unit cell for the reciprocal-space lattice (see Figure 6.2).

6.6

(a) The band structure is shown below on the left. The 2px and 2py bands are perfectly degenerate, but here they have been slightly offset to show there are two bands. (b) The density-of-states diagram is shown below on the right. (c) Metallic conductivity would be expected because the Fermi level cuts the 2pz band.

-10

2pz band

energy (eV)

-20

2px/2py bands -30

-40

2s band -50 0

6.7

k

π/a

DOS, N(E)

(a) There are two hydrogen atoms (one H2 molecule) per unit cell, so there will be two bands. (b) The crystal orbitals for the band derived from the bonding MO (ψ+) and the band derived from the antibonding MO (ψ−) are shown below on the left. (c) The band structure is shown below in the center. (d) The DOS plot is shown below on the right.

There are two regions in the DOS plot, one for each band. The DOS is higher in those regions where the band curvature is smaller. (e) Because the intramolecular interactions are stronger than the intermolecular interactions, the highest energy point in the ψ+ band will be lower in energy than the lowest energy point in the ψ− band and there will be a gap between the completely filled ψ+ band and the empty ψ− band, hence semiconducting rather than metallic behavior is expected. In other words, only if the intra- and intermolecular interactions were equally strong, would the two bands touch, thereby closing the gap.

33

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

6.8 (a) The H 1s orbitals within a given H2 molecule have the same phase in the bonding ψ+ MO (Figure 5.9). At k = 0, eikr = 1, and the atomic orbitals within the MO keep the same phase on moving from one unit cell to the next, so that only the B 2s orbital has the appropriate symmetry to mix with ψ+, as shown below. At k = π/a, the two AOs that make up the MO reverse phase every unit cell (Figure 6.3), and the B 2pz orbital has the appropriate symmetry to mix with ψ+ MO:

B 2s and ψ+ (bonding) at k = 0

B 2pz and ψ+ (bonding) at k = /a

B 2s and ψ+ (antibonding) at k = 0

B 2pz and ψ+ (antibonding) at k = /a

(b) The individual H 1s orbitals within an H2 molecule have opposite phases in the ψ− MO. At k = 0, these AO phases do not change from one unit cell of the chain to the next, and the B 2px orbital has the proper symmetry to mix with ψ− of the H2 molecule to form a crystal orbital with bonding B–H interactions and one with antibonding interactions, as shown below. At k = π/a no boron orbital has the symmetry needed to mix with the ψ− MO of the H2 molecule, as can be seen by visual inspection:

34

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

B 2px and ψ− (bonding) at k = 0

B 2px and ψ− (antibonding) at k = 0

No boron orbital can mix with ψ – at k = /a

(c) From the answers to parts (a) and (b), we know that of the AOs on boron, only the B 2pz orbital mixes with hydrogen at k = π/a. Of the remaining three boron orbitals, B 2s orbital is lower in energy than the B 2p orbitals, which leaves the B 2py and 2px orbitals to form degenerate crystal orbitals. (d) In order to be both H–H bonding and B–H bonding, the ψ+ MO of H2 must mix in a bonding fashion with a boron orbital. At k = 0, this happens for the crystal orbital that contains B 2s and H2 ψ+, as shown in the answer to part (a). (e) In order to be both H–H antibonding and B–H antibonding, the ψ− MO of H2 must mix in an antibonding fashion with a boron orbital. At k = 0, this happens for the crystal orbital that contains B 2px and H2 ψ−, as shown in the answer to part (b).

6.9

(a) The unit cell contains one titanium and one oxygen so there are in total eight bands related to the Ti 3d and O 2p orbitals. The lowest three bands are “O 2p bands” and the upper five are “Ti 3d bands”. The Ti 3dz2 orbital makes the dominant contribution to the widest band because, while being antibonding at k = π/a with O 2pz orbitals (see figure below), it decreases in energy to nonbonding at k = 0 where mixing with O 2pz orbitals is symmetry forbidden. As shown in the figure below, if we were to try and draw an O 2pz orbital, it would have equal amounts of bonding and antibonding overlap with neighboring Ti 3dz2, orbitals, visually illustrating the symmetry-forbidden nature at k = 0.

The other Ti 3d bands are flatter because they have less change in their bonding/antibonding character on going from k = 0 to π/a.

Ti

O

Ti

O

Ti

O

Ti k = π/a

antibonding Ti 3dz2 − O 2pz

Ti

O

Ti

O

Ti

O

Ti k=0

nonbonding Ti 3dz2

35

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

(b) The bands that arise from the Ti 3dxy and 3dx2−y2 orbitals are essentially nonbonding and very flat because these two orbitals do not have the correct symmetry to mix with any of the O 2p orbitals, regardless of the value of k. Furthermore, since the z axis runs parallel to the chain, they have minimal overlap with each other. (c) The chain contains Ti2+, which is a d2 ion, so there are two electrons to fill the Ti 3d bands. We start by filling the bottom of the 3dz2 band, but, before this band is even ¼ full, we start filling the flat 3dxy and 3dx2−y2 bands. There are not enough electrons to fill these bands (together they can hold four electrons) so the Fermi level must cut through the 3dxy and 3dx2−y2 bands. (d) The chain should behave as a metal because the Fermi level cuts through three bands, one of which is fairly wide.

6.10

(a) The crystal orbitals for the py band at Γ, X, and R are shown below. (b) The crystal orbitals for the three p bands at M are shown below. (c) px = py = pz at Γ, px < py = pz at X, px = py < pz at M, px = py = pz at R.

6.11

py at 

py at X

py at R

px at M

py at M

pz at M

As a first approximation of the orbital energies, we can use the MO diagram for a hypothetical octahedral NbN615− complex that has a central Nb3+ ion (4d 2 configuration) surrounded by six nitride ligands (below on the left). The N 2s orbitals are at the lowest energy and form very weak bonding interactions with Nb. After a substantial gap, we encounter 18 orbitals of N 2p orbital character. In an isolated NbN615− complex ion, these would be a mix of bonding and nonbonding MOs, as discussed in Section 5.3.8, but in NbN (where each N is surrounded by six Nb) all of these electronic states will be bonding crystal orbitals. Above these bonding orbitals, we find the Nb 4d orbitals, which make relatively weak antibonding interactions with nitrogen. The triply degenerate t2g orbitals 36

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

are π* and partially filled, while the doubly degenerate eg orbitals are σ* and empty. The empty niobium 5s and 5p antibonding orbitals lie at still higher energy and are omitted. From the MO diagram, we sketch a DOS plot as shown below on the right, with relative areas referring to the number of these orbitals dictated by the NbN formula: 1 weakly bonding N 2s band, 3 bonding N 2p bands, 3 weakly antibonding Nb 4d π* bands, and 2 Nb 4d σ* bands.

The Fermi level cuts through the partially filled π* bands, which suggests that NbN should be a metallic conductor.

Nb 4d eg (σ*) Nb 4d t2g (π*)

energy

EF N 2p orbitals

N 2s orbitals

DOS, N(E)

6.12

For each compound, the valence-band maximum falls at Γ, as does the conduction-band minimum. Therefore, they are direct-gap semiconductors.

6.13

(a) The orbital interactions involving the 5s orbital are analogous to the 2D square lattice of hydrogen atoms in Figure 6.17, namely σ-bonding at Γ, nonbonding at X, and σantibonding at M. A similar drawing for the 5p-orbitals yields the bonding character of the 5p bands: 5px = σ-antibonding and π-bonding at Γ, σ-bonding and π-bonding at X, σbonding and π-antibonding at M; 5py = σ-antibonding and π-bonding at Γ, σ-antibonding and π-antibonding at X, σ-bonding and π-antibonding at M; 5pz = π-bonding at Γ, nonbonding at X, π-antibonding at M. (b) Considering the relative strengths of bonding and antibonding σ and π interactions above, the band energies at Γ, X and M suggest that band 1 = 5s, band 2 = 5pz, band 3 = 5px, band 4 = 5py. (c) Yes, because the Fermi level cuts through the 5px and 5py bands.

6.14

(a) At Γ, k = 0 and the coefficients in front of all atomic orbitals that make up the crystalorbital wavefunction are equal to 1 (eik∙r = e0 = 1). At X, the coefficients change sign each 37

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

time we translate by one unit cell in the x direction, but they do not change sign when translating in y. At M, the coefficients change sign upon a unit-cell translation in either direction. Armed with this knowledge, we can sketch out the orbital overlap for each type of d-orbital at these three special points in the first Brillouin zone. This is shown here for the dx2−y2 orbitals as an example. y x

 σ-bonding

X nonbonding

M σ-antibonding

From these pictures we see that dx2−y2 is σ-bonding at Γ, nonbonding at X (equal amounts of bonding and antibonding overlap), and σ-antibonding at M. Following the same approach, we can show that dz2 is σ-bonding at Γ, nonbonding at X, σ-antibonding at M. The dxy orbitals are π-antibonding at Γ, nonbonding at X, π-bonding at M. The dxz orbitals are π-antibonding and δ-bonding at Γ, π-bonding and δ-bonding at X, π-bonding and δantibonding at M. The dyz orbitals are π-antibonding and δ-bonding at Γ, π-antibonding and δ-antibonding at X, π-bonding and δ-antibonding at M. (b) Considering the interactions at Γ, X and M discussed above and their expected energy, band 1 = dxy, band 2 = dxz, band 3 = dyz. (c) At Γ, two bands are bonding (dx2−y2 and dz2) and the other three are antibonding, so 4 electrons is the optimal count. At X, there is one bonding band (dxz), one antibonding band (dyz) and three nonbonding bands, so the optimal count is between 2 and 8. At M, three bands are bonding (dxy, dyz, dxz) and two are antibonding (dx2−y2 and dz2) so 6 electrons is the optimal count. Of the three choices, d4 does the best job of filling the bonding bands and leaving the antibonding bands empty.

6.15

(a) The octahedrally coordinated Rh3+ has a d6 configuration that is low-spin because Rh is a 4d transition metal. This is the exact electron count needed to completely fill the three Rh 4d – O 2p π* bands (the M t2g bands in Figure 6.31), which become the valence bands. (b) The orbital character of the conduction bands will be Rh 4d – O 2p σ*. (c) The 38

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

maximum of the Rh t2g bands will fall at the R point, while the conduction-band minimum of the Rh eg bands will be at Γ (see Figure 6.31), so the band gap will be indirect.

6.16

In cubic BaZrO3, the Zr 4d – O 2p π* conduction bands are strictly nonbonding Zr 4d (t2g) at Γ (see Figure 6.30). In CaZrO3, bending of the Zr–O–Zr bonds introduces antibonding character to these crystal orbitals, which raises their energy and narrows the width of the Zr 4d – O 2p π* conduction bands. Raising the energy of the conductionband minimum leads to a larger band gap in CaZrO3.

Chapter 7: Optical Materials 7.1

Rearranging Equation (7.2), λ = hc/E = (6.626×10−34 J∙s)(2.998×108 m/s)/(3.60×10−19 J) = 5.52×10−7 m = 552 nm. The frequency can then be calculated from Equation (7.1) ν = c/λ = (2.998×108 m/s)/(5.52×10−7 m) = 5.43×10−14 s−1. Referring to Figure 7.1 a photon with this wavelength will be green. Finally, we use the relationship 1 eV = 1.602×10−19 J to convert the energy to 2.25 eV.

7.2

(a) Cyan, (b) red, (c) blue, (d) magenta, (e) violet, formed by mixing the not absorbed blue with the remainder of red light, the hue of which lies roughly between blue and magenta.

7.3

(a) Idiochromatic, (b) allochromatic, (c) idiochromatic.

7.4

The dz2 orbital points directly at the ligands and will have the highest energy. The dxz and dyz orbitals will be degenerate and the next highest in energy, while the dxy and dx2−y2 experience the smallest amount of repulsion from the ligands. They will also be degenerate and have the lowest energy.

7.5

(a) Both complexes contain d6 ions, but the ligand-field splitting will be larger for [Co(H2O)6]3+ because Co3+ has a higher oxidation state and will possess shorter metal– ligand bonds. The observed values of Δoct are 1.16 eV for [Fe(H2O)6]2+ and 2.26 eV for [Co(H2O)6]3+. The difference in Δoct is large enough that [Fe(H2O)6]2+ is a high-spin complex while [Co(H2O)6]3+ is low spin. (b) We expect [Rh(H2O)6]3+ to have a larger Δoct than [Co(H2O)6]3+ because the larger 4d orbitals experience more overlap with the ligand orbitals. The observed value of Δoct = 3.35 eV for [Rh(H2O)6]3+ is higher than the 2.26 eV value observed for [Co(H2O)6]3+.

39

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

7.6

(a) For the 1D term L = 2, S = 0 and the total number of microstates is (2L+1)(2S+1) = 5.

For the 3P term L = 1, S = 1 and the total number of microstates is 9. For the 1S term L = 0, S = 0, and the total number of microstates is 1. (b) For the 1D term, the only possible value of J is 2. For the 1P term, the possible values of J are 2, 1 and 0. For the 1S term, the only possible value of J is 0. (c) From part b we see there are five (2S+1)LJ symbols in total. Hund’s first rule (minimize S) predicts that the 3P0 term will be lower in energy than the singlet terms. Hund’s third rule (minimize J because the 2p shell is less than half filled) indicates that the energy of the 3PJ terms will increase as J increases. Hund’s second rule (maximize L) predicts that the energy of the 1D term will be less than the 1S term. Combining these predictions gives the following order 3P0 < 3P1 < 3P2 < 1D2 < 1S0.

7.7

(a) Since the ligand is neutral and the overall charge of the complex ion is +2, the complex must contain a Ni2+ ion with a [Ar]3d8 electron configuration. (b) Using the correlation diagram for a d8 ion given in Figure 7.8 transition (1) can be assigned as the 3

A2g→3T2g transition, the next higher energy transition (2) as the 3A2g→3T1g transition,

and the highest-energy transition as the 3A2g→3T1g(P) transition. (c) The complex does not absorb strongly between 450 and 500 nm, which falls in the blue, nor does it absorb above 650 nm in the red. The combination of reflected blue and red light leads to a magenta color. (d) The corresponding d-to-d transitions in [Ni(H2O)6]2+ fall at 1180 nm, 725 nm and 395 nm, respectively. All three transitions fall at longer wavelengths or lower energy because the spectrochemical series tells us that Δoct for [Ni(H2O)6]2+ is smaller than [Ni(NH2CH2CH2NH2)3]2+.

7.8

Because the Cl− ligands are negatively charged, the Co oxidation state must be 2+, and the electron configuration for a Co2+ ion is [Ar]3d7. The coordination diagram for a tetrahedral dn ion is equivalent to the corresponding octahedral d10−n diagram. In this case n = 7, so the correct diagram will be the correlation diagram for an octahedral d3 ion.

From Figure 7.8, we identify the diagram (a) as the correct one. The term symbols will be almost the same as the octahedral d3 ion, with two exceptions: There will be no g subscript because a tetrahedron has no inversion center, and the spin multiplicity will be a quartet: 2S + 1 = 2(3/2) + 1 = 4. Therefore, W = 4A2, X = 4T2, Y = 4T1(F), and Z = 4T1(P).

7.9

If the coordination polyhedron has an inversion center, the transition is Laporte forbidden, if not, it is Laporte allowed. Spin-allowed transitions will be more intense than 40

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

spin-forbidden transition, but if all five d-orbitals are filled with a single electron (highspin d5), all d-to-d transitions will be spin forbidden. Hence: (a) Laporte and spin allowed. (b) Laporte forbidden, spin allowed. (c) Laporte and spin forbidden. (d) Laporte and spin allowed.

7.10

(a) Figure 7.1 shows that the MoS42− absorbs predominantly blue and cyan. Their complementary colors are yellow and red, so we expect the color to be between red and yellow, which is orange. (b) Using Equation (7.2) and the appropriate conversion from joules to eV, the LMCT transition energy is calculated to be 2.64 eV. This value is smaller than the LMCT in MoO42− (5.3 eV) by roughly a factor of two. Because sulfur is less electronegative than oxygen, less energy is needed to transfer an electron from the ligands to the molybdenum. (c) Because NbV has a lower oxidation state than MoVI, the LMCT transition in NbS43− should be higher in energy and the absorption peak shifted to shorter wavelengths. Experimentally, the lowest-energy LMCT absorption peak in Li3NbS4∙2TMEDA is found at 340 nm, and the compound is pale yellow, see Lee et al., Inorg. Chem. 31 (1992), 4333–4338.

7.11

Semiconductors with Eg > 3.1 eV do not absorb visible light and will be white, while those with Eg < 1.7 eV absorb all wavelengths of visible light and will be black. The compounds with band gaps between these two extremes are: ZnSe (Eg = 2.58 eV, yellow), AlP (Eg = 2.43 eV, yellow), CdS (Eg = 2.42 eV, yellow), GaP (Eg = 2.26 eV, orange), ZnTe (Eg = 2.25 eV, orange), and AlAs (Eg = 2.16 eV, orange).

7.12

If the band gap decreases linearly from CdS to CdSe, the band gap of any composition will be Eg = (2.42 eV)(1−x) + (1.73 eV)∙x. Inserting 2.00 eV for Eg gives x = 0.61 in CdS0.39Se0.61.

7.13

Fluorescence describes luminescence where the electronic transition that leads to emission is spin allowed (ΔS = 0), while phosphorescence describes luminescence involving a spin-forbidden transition (ΔS ≠ 0). Measurements of excited state lifetimes are the best way to distinguish between the two, fluorescence lifetimes (10−11 to 10−8 s) are much shorter than phosphorescence lifetimes (10−6 to 10−2 s).

7.14

(a) [Xe]4f 2. (b) 1D: S = 0, L = 2; 3F: S = 1, L = 3; 1G: S = 0, L = 4; 3H: S = 1, L = 5; 3P: S = 1, L = 1; 1S: S = 0, L = 0. (c) Hund’s first rule (maximize S) predicts that the triplets (3H, 3F, 3P) will be lower in energy than the singlets (1D, 1G, 1S). Applying Hund’s 41

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

second rule (maximize L), we obtain the following order 3H < 3F < 3P < 1G < 1D < 1S. (d) For 3H: J = 4, 5, 6, and according to Hund’s third rule (minimize J for shells that are less than half-filled) the lowest-energy term is 3H4. For 3F: J = 2, 3, 4, and the lowest-energy term is 3F2. For 3P: J = 0, 1, 2, and the lowest energy term is 3P0. For 1G: J = 4, and the lowest-energy term is 1G4. For 1D: J = 2, and the lowest-energy term is 1D2. For 1S: J = 0, and the lowest-energy term is 1S0.

7.15

(a) VO43− charge-transfer transition with strong electron–phonon coupling expected. (b) f-to-f transitions, weak electron–phonon coupling. (c) d-to-f transition, moderate electron–phonon coupling. (d) f-to-f transitions, weak electron–phonon coupling. (e) 6s16p1 → 6s2 transition of the Bi3+ ion, strong electron–phonon coupling.

7.16

(a) Increase, (b) decrease, (c) decrease, (d) increase, (e) increase.

7.17

As the size of the A2+ cation increases, the A–O bonds will become longer and this should allow greater structural reorganization in the excited state where antibonding orbitals of the WO42− group are populated, i.e. larger ΔQ. A larger ΔQ leads to crossing of the ground- and excited-state parabolas at lower energies, which will facilitate nonradiative return to the ground state. This will increase thermal quenching.

7.18

Assuming Vegard’s law, the band gap Eg = (1.43 eV)(1−x) + (2.25 eV)∙x. For x = 0.45, it becomes Eg = 1.80 eV. Converting to joules and using λ = hc/E gives a wavelength of 689 nm in the red region of the visible spectrum.

7.19

(a) Eu2+ emits via a [Xe]4f 65d1 → [Xe]4f 7 transition, and Mn2+ emits via the spinforbidden 4G5/2 → 6S5/2 d-to-d transition. (b) Eu2+ absorbs at an energy E = hc/λ = [(6.626×10−34 Js)(2.998×108 m/s)/(336×10−9 m)](1 eV/1.602×10−19 J) = 3.69 eV, and by a similar calculation a 450 nm photon will have an energy of 2.76 eV. Consequently, the Stokes shift is EStokes = 3.69 – 2.76 = 0.94 eV. (c) Because the optical transitions of Mn2+ are all spin forbidden it will only be weakly absorbing, and energy transfer via a Förster mechanism will be highly inefficient. Therefore, energy transfer must occur through a Dexter mechanism.

7.20

(a) The absorption maximum of 270 nm corresponds to an energy of 4.6 eV. Since the Stokes shift is 0.9 eV, the emission maximum will correspond to 4.6 eV – 0.9 eV = 3.7 eV photons of wavelength 335 nm. (b) The Ce3+ emits in UV, so the green emission must come from Tb3+. Ce3+ is needed because the allowed [Xe]4f 8 → [Xe] 4f 75d1 transitions 42

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

of Tb are too high in energy to be excited by the 254 nm line of the Hg-discharge. (c) Because the accessible f-to-f transitions of Tb3+ are narrow, energy transfer from Ce3+ to Tb3+ must occur through the Dexter mechanism.

7.21

(a) In both hosts, Ce3+ is coordinated by Cl− ions, but the average electronegativity of the cations in the two hosts is different. Using Equation (7.8) and Allred–Rochow electronegativities, the “average” cation electronegativity of LaCl3 is χav = 1.08, while the average cation electronegativity of Cs2NaLaCl6 is χav = (2×1×0.86 + 1×1×1.01 + 1×3×1.08)/6 = 0.99. From this lower value, we expect a larger centroid shift in Cs2NaLaCl6. (b) The ligand-field splitting in an octahedron is the largest of any coordination environment (because of direct repulsions), while the tricapped trigonal prism is generally only 40–50% as large. Hence, the red-shift of the excitation spectrum due to ligand-field effects should be larger in Cs2NaLaCl6. (c) Both effects act in concert and shift the excitation spectrum in Cs2NaLaCl6 to lower energies/longer wavelengths.

7.22

(a) Calculating the energy of the absorbed and emitted photons as E = hc/λ and converting to eV, we obtain Eexc = 2.6 eV and Eem = 2.3 eV, which corresponds to a Stokes shift EStokes = Eexc – Eem = 2.6 eV – 2.3 eV = 0.3 eV. (b) The red shift of the absorption and emission maxima must be either due to an increase in ligand-field splitting or an increase in the centroid shift. In both cases, the ligands surrounding Eu2+ are sulfide anions, but in SrGa2S4 they are also coordinated to the smaller and more electronegative gallium which will reduce their polarizability thereby reducing the centroid shift in SrGa2S4. This would help to explain the longer-wavelength emission observed for SrS:Eu2+. The change in coordination number from 8 in SrGa2S4 to 6 in SrS will increase the crystal-field splitting and further red-shift the emission peak.

7.23

(a) To a first approximation, the anion polarizability should be similar in SrS and CaS, so the centroid shift should also be similar. If anything, the centroid shift should be larger for SrS, because of strontium’s slightly lower electronegativity. On the other hand, Ca2+ is smaller than Sr2+, which will enforce shorter Eu–S distances and larger crystal-field splitting in CaS. This will shift the lowest-energy absorption and emission bands towards red. Dorenbos has estimated very similar centroid shifts of 1.71 eV (SrS) and 1.75 eV (CaS), but noticeably different crystal field splitting 1.49 eV (SrS) and 1.86 eV (CaS). (b) Se is less electronegative and more polarizable than S, which should lead to a larger 43

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

centroid shift and a red shift of the emission maximum in SrSe. On the other hand, the Eu–Se distances are longer than the Eu–S distances, which should lead to a smaller crystal-field splitting and the observed “blue shift” of the emission maximum towards green. We must conclude that the effects of crystal-field splitting are stronger than effects of centroid shift.

Chapter 8: Dielectrics and Nonlinear Optical Materials 8.1

Express P in terms of Eloc, then of E, Equations (8.10) and (8.11). Then express P in terms of χe and E, Equation (8.6), and use the relationship εr = 1 + 4πχe which is valid in CGSes to replace the χe term. Finally equate P from the two expressions and consider it per atom, hence N becomes 1/Va.

8.2

Calculated values are εr = 13 (SnO2), εr = 20 (ZrO2), and εr = 43 (TiO2). They are in good agreement with experiment for SnO2 (εr = 13) and ZrO2 (εr = 22), but not for TiO2 (εr = 94). The poor agreement with TiO2 is due to vibrations that lead to dynamic off centering of the Ti4+ ions through an incipient SOJT-type distortion.

8.3

Because the f-electrons poorly shield the nucleus, there is an increase in the effective nuclear charge as the atomic number increases, which in turn makes the electron cloud less polarizable.

8.4

Calculated values are εr = 21.1 (La), 18.9 (Pr), 17.2 (Nd), 17.7 (Sm) 17.4 (Eu). Although there is a slight bump at Nd, the general trend is for decreasing permitivity with decreasing radius of the rare-earth ion.

8.5

Calculated values are εr = 26 (BaZrO3), 77 (KTaO3), 42 (SrTiO3). The agreement with observed values worsens in the order εr (obs) = 43 (BaZrO3) < 242 (KTaO3) < 330 (SrTiO3) as the tendency for SOJT distortions increases. The Clausius–Mossotti equation cannot be used to estimate εr values for compounds with ions that are prone to SOJT distortions.

8.6

(1) The peak in the temperature dependence of the dielectric permittivity is broad and frequency dependent. (2) Normal ferroelectrics typically undergo a long-range structural phase transition at TC whereas relaxors do not. (3) Relaxors contain polar nanoregions, normal ferroelectrics do not. (4) Relaxors lack the conventional domain structure seen in normal ferroelectrics.

44

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

8.7

A first-order Jahn–Teller distortion usually causes elongation of two bonds “trans” to each other, which does not destroy the local inversion center. Hence, the local coordination polyhedron does not possess a dipole moment.

8.8

(a) Ti4+ because it is a d0 cation, whereas V4+ is a d1 cation, (b) Ta5+ because it is a d0 cation, whereas Mn3+ is a d4 cation (Mn3+ is prone to first-order Jahn–Teller distortions), (c) Mo6+ because it has a higher electronegativity hence the gap between the unfilled metal d orbitals and the nonbonding anion orbitals will be smaller, (d) Te4+ because it is an s2p0 cation, whereas Sn4+ is an s0p0 cation.

8.9

(a) Parsimony rule suggests that all oxygens are equivalent. This is possible with each O making 4 Ba–O bonds, 1 Sc–O bond and 1 Ta–O bond. (b) The bond valences are ⅙ for Ba–O, ½ for Sc–O, and ⅚ for Ta–O. The bond-valence sum at oxygen is 4(1/6) + (1/2) + (5/6) = 2. (c) In the bond graph of Ba3ZnTa2O9 there are two non-equivalent oxygens: The first type of oxygen, O(1), makes 4 Ba–O bonds, 1 Zn–O bond and 1 Ta–O bond, while the second type of oxygen, O(2), makes 4 Ba–O bonds and 2 Ta–O bonds. (d) The bond valences are ⅙ for Ba–O, ⅓ for Zn–O, and ⅚ for Ta–O. The bond-valence sum for O(1) is 4(1/6) + (1/3) + (5/6) = 11/6, while for O(2) it is 4(1/6) + 2(5/6) = 14/6. (e) In Ba2ScTaO6, the oxygen bond valence is 2.00 and no distortions are expected. In Ba3ZnTa2O9, the O(1) atom that links Zn and Ta octahedra is underbonded, while O(2) that links two Ta octahedra is overbonded. To alleviate this bonding instability, Ta shifts away from O(2) and towards O(1).

8.10

From Table 8.3 we see that: (a) Yes, there are three non-centrosymmetric cubic point groups, 23, 4̄ 3 m and 432. (b) Yes, two of the three, 23 and 4̄ 3m, can show piezoelectricity, (c) No, cubic space groups cannot have a polar axis.

8.11

(a) 0.06 m, (b) 0.053 m.

8.12

(a) Replacing non-symmorphic symmetry elements in Pna21 with symmorphic ones, we identify the point symmetry as mm2. According to Table 8.3, this point group is polar, so pyroelectric behavior is expected. (b) While the symmetry also allows for ferroelectricity, it is not clear from the above evidence that the polarity can be switched. Therefore, further experiments are needed to determine if BiInO3 is a ferroelectric, (c) If a crystal meets the symmetry requirements for a pyroelectric, as this one does, piezoelectricity is also permitted. Furthermore, the symmetry requirements for piezoelectricity and SHG are 45

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

the same and SHG is observed. Consequently, there should be some piezoelectric response,

but

its

magnitude

is

uncertain.

(d)

Pyroelectrics,

ferroelectrics,

antiferroelectrics, and relaxor ferroelectrics tend to show a peak in their εr versus temperature plot at the Curie temperature. The available data suggests that TC is greater than 873 K, so it would be reasonable to expect εr to increase up to the decomposition temperature.

8.13

(a) R3c (point group 3m) is a polar space group, while Pcca is a centrosymmetric space group. Therefore, ferroelectricity is possible in BiAlO3 but not in BiGaO3. (b) According to Table 8.3, piezoelectricity is likely in BiAlO3 but not allowed in BiGaO3.

8.14

(a) The bond-valence equation gives d(Pb–O) = 2.63 Å. Twice this distance is the body diagonal of the cubic unit cell, and the relationship between the edge and body diagonal of a cube gives a = [2d(Pb–O)]/3] = 3.04 Å, hence we can estimate the CsCl-type unitcell volume as 28.0 Å3, (b) The actual unit-cell volume per formula unit is a2c/Z = (3.974 Å)2(5.022 Å)/2 = 39.7 Å3, which is a 42% expansion versus the CsCl-type structure. (c) If anion and cation radii are approximated as being equal, then each accounts for 50% of the volume in PbO with the CsCl structure. If the lone pair has the same volume as an oxide ion, we would expect the volume to expand by 50% in the litharge form of PbO. This is not far from the 42% expansion that is observed. The agreement is even better if we account for the fact that Pb2+ is slightly larger than O2−.

8.15

(a) In analogy with the previous Problem, d(Sn–O) = 2.50 Å, a = 2.88 Å and V = 24.0 Å3. (b) V/Z = 35.0 Å3, a 46% expansion versus the CsCl-type structure.

8.16

The MO diagram for a SnH86− ion is shown below. The singly degenerate antibonding Sn 5s – H 1s orbital with a1g symmetry is the HOMO. The triply degenerate set of antibonding Sn 5p – H 1s orbitals with t1u symmetry is the LUMO.

46

energy

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

8.17

The phosphorus displacements are in the ab plane. When averaged over all polyhedra, they cancel out, hence NH4H2PO4 should be centrosymmetric and therefore not SHG active.

8.18

No. P21/n has 2/m point-group symmetry and is centrosymmetric. Hence β- and γ-BiB3O6 should not show any SHG activity.

8.19

(a) tetragonal, point group 4̄ 2m, SHG allowed, (b) cubic, point group m3̄ m, SHG not allowed, (c) orthorhombic, point group mm2, SHG allowed, (d) orthorhombic, point group 222, SHG allowed, (e) monoclinic, point group 2/m, SHG not allowed, (f) monoclinic, point group 2, SHG allowed.

8.20

(a) In β-BaB2O4 each B3O63− polyatomic ion contains three BO3 groups. The simplest way to count the number of BO3 groups is to assume one BO3 group for every boron atom, hence the concentration of BO3 groups for β-BaB2O4, is (182)/(1731 Å3) = 0.020 Å−3. The other two compounds contain isolated BO33− groups, so the concentration for Sr2Be2(BO3)2O is (22)/(290.8 Å3) = 0.014 Å−3 and for KBe2BO3F (31)/(318.0 Å3) = 0.0094 Å−3. (b) β-BaB2O4 has a BO3 concentration that is 2.1 times that of KBe2BO3F and an NLO coefficient that is 2.9 times that of KBe2BO3F. Sr2Be2(BO3)2O has a BO3 concentration that is 1.5 times that of KBe2BO3F and an NLO coefficient that is 1.9 times that of KBe2BO3F. So, the trend appears to be qualitatively valid. (c) CsBe2BO3F2 has a

47

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

concentration of BO3 groups equal to (21)/(243 Å3) = 0.0082. Based on this, its NLO coefficients should be slightly smaller than KBe2BO3F. Chapter 9: Magnetic Materials 9.1

The pole of a bar magnet that in the Earth’s magnetic field turns towards the geographic north is marked N, hence the S pole of the earth’s field is at the Artic. Penguins live in the Antarctic so the answer is no!

9.2

Zero; ℓ = 0.

9.3

56 A/m.

9.4

M is the sum of the atomic magnetic moments μ oriented in the field direction per unit volume. The unit is therefore Am2/m3 = A/m.

9.5

High spin: d3 because the three orbital moments cancel, d4, d8, d9 because the two eg orbitals cannot be unified by rotation about any axis, d5 because L = 0, d10 because S = L = 0. Low spin: d6 because S = L = 0.

9.6

NiII in octahedral coordination has the t2g6eg2 orbital occupation, hence no orbital contribution. In tetrahedral coordination, it has e4t24 that permits a circular motion of the minority t2 spin via any of the two remaining t2 orbitals, and orbital contribution is possible. The spin-only μeff = ge[S(S+1)]0.5 = 2.83 μB via Equation (9.12) where μB is set to unity.

9.7

Summation according to Figure 9.9 gives 2S+1LJ = 4F3/2, 5D0, 6S5/2 for Mn4+, Mn3+ and Mn2+, respectively. The total combined quantum number J is used to calculate μeff with orbital contribution, from Equation (9.14) altered by setting μB to unity and Equation (9.15): μeff = 0.77, 0, 5.92 μB, respectively. The respective spin-only values μeff = 3.88, 4.90, 5.92 μB follow analogously from the total S and Equation (9.12).

9.8

L = 0, J = S, gJ = ge gives Equation (9.12).

9.9

S = 3/2 and Equation (9.12) (μB set to unity) gives μeff = 3.877 μB. Analogously, μsat = 3.003 μB with Equation (9.13).

9.10

For d5, L = 0 and S = 5/2 gives μeff = 5.923 μB with Equation (9.12) and μsat = 5.006 μB with Equation (9.13).

48

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

9.11

(a) Equation (9.20) with the magnetic constant μ0 = 4π×10−7 N/A2 (see Footnote 10 of this Chapter) gives χmol = 6.3002510−6∙S(S+1)/298 = 1.8510−7 m3/mol. (b) Divided by the molar volume VcellNA/4 in m3/mol, χv = 6.39810−4. (c) μr = 1.00064.

9.12

In SI units, dimensionless susceptibility is the susceptibility of 1 m3 of the sample, here of O2 gas. The O2 molecule is magnetic because it has 2 unpaired electrons (Figure 5.13), hence S = 1, and the O2 gas is paramagnetic because ideal gas molecules move independently and do not spontaneously order. We therefore use Equation (9.20) to calculate the molar susceptibility χmol = 46.13×10−9 m3/mol that is then converted to susceptibility per 1 m3, χv = 2.031×10−6, with the standard molar volume of the ideal gas.

9.13

The mass susceptibility of the standard at 299 K is expressed via the Curie–Weiss law.

The diamagnetic mass susceptibility χm in m3/kg of the superconductor is proportional to the voltage per mass ratio: χm = −χm,standard(450/102.5)/(5.73/38.4). Multiplication with density gives χv = −0.25, hence the expelled fraction of the magnetic field is ¼.

9.14

Susceptibility is a function of μeff2, and weighted mean square of the individual μeff values for Mn3+ and Mn4+ must be calculated, whereas μs is a simple average:

eff = 0.5  4.8992 + 0.5  3.8732 = 4.416 μB and μs = (4+3)/2 = 3.5 μB. 9.15

Both Dy and Cr have unpaired electrons. The Dy3+ moment arises from S = 5/2, L = 5, J = L+S = 15/2 determined via the Russell–Saunders scheme in Figure 9.13, gJ = 1.3341 is calculated with Equation (9.15). The Cr3+ moment is spin only, Equation (9.12). Their weighted mean square:  eff =

9.16

0 .5  10 .652 2 + 0 .5  3.877 2 = 8.016 μB.

The oxidation states can be found by determining the combination of unpaired electrons that yields the same effective moment as calculated from C. That is 3kCm/(Nm μ0 μB2) = 4.189 μB, in which Nm = 2NA/0.422114 of Mn and Mo atoms per kilogram and μ0 = 4π×10−7 N/A2. The combination of μeff calculated with Equation (9.12) that satisfies

4.189 = 0.5eff,Mn 2 + 0.5eff,Mo2 is obtained with S = 2.5 and 0 for these two atoms, hence 5 and 0 unpaired electrons, Mn2+ and Mo6+. Note that if you knew that only one atom is magnetic, you could have calculated Nm as NA/0.422114 and then obtain S = 2.5 directly from Equation (9.12).

9.17

Fe2+ adopts either low- or high-spin state. As C = χT, not only susceptibilities, but also Curie constants are additive: The Cr3+ ion of S = 3/2 has μeff = 2[S(S+1)]0.5 = 3.87 μB that 49

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

set equal to 2.828(χCrT)0.5 gives χCrT = 1.875 emu K/mol. High-spin Fe2+ has μeff = 4.90 μB that set equal to 2.828(χFeT)0.5 gives χT = 3.002 emu K/mol. The sum χT = χCrT + χFeT = 4.877 emu K/mol, from which χ = 0.0163 emu/mol at T = 300 K (0.0244 at 200 K).

Low-spin Fe2+ has μeff = 0, hence χFeT = 0. At 200 K, total χT = χCrT = 1.875 emu K/mol, from which χ = 0.0094. The result suggests a low-spin to high-spin crossover somewhere between 200 and 300 K. 9.18

One Bohr magneton of 9.274×10−24 Am2 converts to 9.274×10−21 emu. Dividing Mmol in emu/mol with the magnetic moment of 1 mol of Bohr magnetons in emu (9.274×10−21NA = 5585 emu/mol) gives the number of Bohr magnetons per mol.

9.19

(a) The straight line from 0 K is of a Curie paramagnet. This is expected. In such salts, hydration of magnetic cations, already separated by anions, prevents magnetic interactions. (b) The fit gives θ ≈ 0 K as the intercept, and the Curie constant Cm = 9.638610−5 m3K/kg as the slope of the line. There is one magnetic atom per formula, hence Nm = 1000 NA/392.139 = 1.5361024 kg−1. From Equation (9.19) express eff = 3kCm ( Nm0 = 4.54810−23 Am2 (magnetic constant μ0 = 4π×10−7 N/A2). Equation (9.12) gives S = 2, four unpaired electrons, hence Fe2+ is high spin.

9.20

S = 1.50, obtained as in the preceding Problem, means high-spin Co2+. Weiss constant θ = −330 K suggests CoO to be an antiferromagnet.

9.21

The Brillouin function in Equation (9.21) gives 6.7% of saturation, calculated from the ratio ξ = 0.134 of the magnetic split energy and thermal energy.

9.22

One way is to consider that at TC the energy kT of the thermal chaos equals the magnetic energy μsatBW (when T = TC, practically all induction is due to the Weiss field) in which μsat = geμBS, Equation (9.13). From here, BW = 446 T; about two orders of magnitude larger than fields in a laboratory instrument. Setting BW = μ0HW gives HW = 355000 kA/m. Another approach is the Curie-law approximation and TC ≈ Cλ of Equation (9.24), for which C is expressed from Equation (9.19) and Equation (9.12), and its λ plugged into HW = λMsat = λNμsat. This HW agrees with the first estimate only for S = ½.

9.23

Linear superexchange between octahedrally coordinated cations: a) VO d3–d3 antiferromagnetic, b) LaCrO3 d3–d3 antiferromagnetic, c) La2CrIIIFeIIIO6 d3–d5 ferromagnetic. d) CsNiII[CrIII(CN)6] d8–d3 ferromagnetic.

9.24

Ferromagnetic.

50

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

9.25

The Mn–S bonds are more covalent than the Mn–O bonds. Increased covalence leads to increased superexchange coupling.

9.26

Co has 9 valence electrons. According to Equation (9.13), the moment of 1.751 μB corresponds to 1.614 ordered spins per Co; that is N(up) − N(down) = 1.614 so that N(down) = 5 − 1.614 = 3.386. The number of electrons in the s-band is 9 − 5 − 3.386 = 0.614.

9.27

8 valence electrons means 7.05 electrons in d band. The moment 2.2 μB corresponds to 2.12 spins; hence N(up) − N(down) = 2.12. N(up) + N(down) = 7.05. Solving simultaneous equations gives N(up) = 4.585 and N(down) = 2.465. Thus both bands are partially filled, and d sub-bands in the sketch need to be drawn accordingly.

9.28

VEC = 9.8 suggests yes.

9.29

Magnetization of 1 kg of the product will be due to the ~1 mg of Fe3C formed in it (precisely 1.027 mg). Converting Mremanent from per volume to per mass and multiplying it with the mass fraction of Fe3C in the product gives Mmass = (700000/7000) × 0.000001 = 0.0001 JT−1kg−1.

9.30

Obtained as the point on the demagnetization line where the product of B and H is a maximum. Write the function f(H) = µ0H2 + µ0HMremanent. Differentiating and setting equal to 0, f'(H) = 2µ0H + µ0Mremanent = 0, gives the H coordinate at (BH)max as H = −Mremanent/2.

9.31

Mn2+ is a d5 ion, like Fe3+.

9.32

That with unpaired d electrons, Y3Fe2Fe3O12.

9.33

Msat concerns the uncompensated parallel spins that lend the ferrimagnet its ferromagnetic properties. Of the five d5 high-spin Fe per formula, two and two cancel, and Msat = 5 μB per formula for the iron-garnet ferrimagnet. Owing to half-filled f shell of Gd3+, Russell–Saunders scheme gives L = 0, and, considering the spin-only moment, the gadolinium-garnet ferrimagnet has its saturation moment due to 3Gd3+ minus one Fe3+, hence Msat = 16 μB.

9.34

Such two O links mean that the d9–d9 coupling of two Cu2+ will have about 90° Cu–O– Cu angles, hence be ferromagnetic.

51

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

Chapter 10: Conducting materials 10.1

We first calculate the resistivity, ρ = 1/σ = 1/(5.9×107 S/m) = 1.69×10−8 Ωꞏm. Next we convert to resistance, R = ρL/A = [(1.69×10−8 Ωꞏm)(0.20 m)]/[π(0.0015 m)2] = 4.80×10−4 Ω.

10.2

The electronic charge has units C, the carrier concentration m−3, and the velocity m/s.

The product of these three is C/(m2ꞏs), and because 1 A = 1 C/s this reduces to A/m2.

10.3

The electronic charge has units C, the carrier concentration m−3, and the mobility (m/s)/(V/m). Considering that 1 C = 1 A∙s, the right-hand side of Equation (10.10) has units of A/(Vꞏm). Using Ohm’s law (R = V/I) in unit expression (Ω = V/A), we can reduce the units to 1/(Ωꞏm). Finally, 1 S = 1 Ω−1 hence the units reduce to S/m, the SI unit of conductivity on the left-hand side of Equation (10.10).

10.4

(a) Equation (10.10): μ = σ/ne = (2.1×107 S/m)/[(2.5×1028 m−3)(1.602×10−19 C)] = 5.24×10−3 Sꞏm2/C = 5.24×10−3 m2V−1s−1. (b) The relaxation time is τ = μme/e = (5.24×10−3 m2ꞏV−1ꞏs−1)(9.109×10−31 kg)]/(1.602×10−19 C) = 2.98×10−14 s−1. (c) The root mean square velocity of the electron gas assumed in the Drude model is given by Equation (10.4), v = (3kT/me)1/2 = [3(1.38×10−23 J/K)(300 K)/(9.109×10−31 kg)]1/2 = 1.17×105 m/s. (d) ℓ = τv = (2.98×10−14 s−1)(1.17×105 m/s) = 3.49×10−9 m = 34.9 Å, which is approximately 8 unit cells. The estimates of mean free path given in Section 10.2.7 are about 10 times larger, 300 to 400 Å. Because electrons are fermions and must obey the rules of quantum mechanics, they do not adopt a Maxwell–Boltzmann distribution.

Consequently, only a small fraction of the valence electrons, those near the Fermi level, can move in response to an applied electric field. Since effective carrier concentration n is much smaller than assumed in the Drude model, the original estimate of mobility made in part (a) is far too small.

10.5

(a) We can calculate the valence-electron concentration n by multiplying the number of atoms per unit cell (2 for bcc, 4 for fcc) by the number of valence electrons per atom and dividing by the unit-cell volume: Rb 1.15×1028 m−3, Cs 0.86×1028 m−3, Sr 3.56×1028 m−3, Ba 3.14×1028 m−3. (b) The plot of conductivity versus valence-electron concentration is shown below. (c) Within a periodic-table group, the number of valence electrons per atom is constant but the spacing of the atoms increases as we move down the group, hence the number n of conduction electrons per unit volume will decrease. As shown in 52

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

the figure below, the Drude model qualitatively predicts a decrease in conductivity down each group as observed (Mg is an anomaly because the Fermi level cuts across more than just the bands arising from overlap of s orbitals), but the electron-gas approach cannot describe the combined behavior of both groups.

conductivity (106 S/m)

30

Ca

Na

20

Mg

K 10

Rb Cs

Sr Ba

0 0

2

4

6

8

10

valence-electron concentration (1028 m−3)

10.6

When E = EF, the Fermi–Dirac function reduces to f(E) = 1/[1+exp(0)] = 1/(1 + 1) = 0.5 for any finite temperature.

10.7

We rearrange the Fermi–Dirac function to solve for E − EF = (kBT) ln[f(E)−1 − 1]. Using this equation we obtain (a) 3.9×10−3 eV, (b) 3.9×10−2 eV, (c) 0.12 eV, (d) 0.39 eV above EF.

10.8

(a) Equation (10.4): v = (3kBT/me)1/2 = [3(1.38×10−23 J/K)(300 K)/(9.109×10−31 kg)]1/2 = 1.17×105 m/s. (b) vF = (2EF/m*)1/2 = [2(3.1 eV)(1.602×10−19 J/eV)/(9.109×10−31 kg)]1/2 = 1.0×106 m/s. (c) The Fermi velocity is roughly 10 times larger than the root mean square velocity estimated in the Drude model. (d) In the Drude model, the velocity increases with T1/2, while in the free-electron model, the Fermi energy and thus the Fermi velocity are essentially temperature independent.

10.9

(a) For a metal at very low temperatures, the conductivity is temperature independent and equal to the residual resistivity. Once above the low-temperature regime, the resistivity increases linearly as a function of T, so σ is linearly proportional to the inverse absolute temperature. (b) For an intrinsic semiconductor, the dominant term is the carrier concentration, which means σ increase is proportional to e1/T. Stated another way, ln(σ) is linearly proportional to 1/T. 53

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

10.10 The Fermi level cuts through bands with d-orbital parentage for nickel, but only through the wide band with predominantly s-orbital parentage for copper. Consequently, the density of states at EF and the carrier concentration will be higher for Ni (several bands), but the mobility will be higher for Cu because the band is wide. Table 10.3 shows that the conductivity of Cu is about 4 times larger than Ni. Therefore, the larger mobility of Cu must be more important than the higher carrier density in Ni. 10.11 The conduction-band edge is Eg/2 = 0.35 eV above the Fermi level. Equation (10.21) gives f(E) = 1/[1 + exp{(−0.35 eV)/(8.617×10−5 eV K−1)(300 K)} = 1.3×10−6.

10.12 The ionization energy of the donor electron is calculated from Equation (10.22) as Ed = 13.6(Z/nεr)(m*/me) = 13.6(1/12.1)2(0.07) = 0.0065 eV. Equation (10.24) gives the corresponding Bohr radius rd = εr(me/m*)(0.053 nm) = 12.1(1/0.07)(0.053 nm) = 9.2 nm = 92 Å.

10.13 (a) Increase. If the band gap increases, the crossover to the intrinsic regime will shift to higher temperatures. (b) Decrease. If the effective mass of the carriers increases, the donor/acceptor ionization energy will increase, which means higher temperatures will be needed to crossover from the freeze-out regime to the saturation regime. (c) Increase. If the dielectric permittivity increases, the donor/acceptor ionization energy will decrease and lower the temperature of the crossover from the freeze-out to the saturation regime.

10.14 (a) For isolated Pr3+ ions in the gas phase, U = IE(Pr3+) + Heg(Pr3+) = IE(Pr3+) − IE(Pr2+) = 4th IE(Pr) – 3rd IE(Pr) = 39.0 eV − 21.6 eV = 17.4 eV. A similar calculation for gasphase Gd3+ ions gives U = 44.0 eV − 20.6 eV = 23.4 eV. (b) In solids, U is reduced by two factors; polarization of the surrounding structure and delocalization of the electron wavefunction. The latter effect is minimal for the 4f orbitals, so the ~11 eV reduction in U is due almost entirely to polarization of the surrounding structure. (c) Gd3+ has an electron configuration of [Xe]4f7 where each of the seven 4f orbitals is occupied by a single electron. This configuration leads to an anomalously low absolute value of Heg because Gd2+ will have one doubly occupied 4f orbital whereas Gd3+ does not. This is not the case for praseodymium. (d) The electron configuration for the transition-metal ions in question are [Ar]3d2 for Ti2+ and [Ar]3d5 for Mn2+. The Mn2+ ion has a half-filled 3d shell with one electron in each d-orbital (assuming the high-spin configuration seen for MnO),

54

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

which will lead to an anomalously large value of U. Hence, MnO will have a larger U than TiO. 10.15 (a) [Xe]4f145d3 (b) Due to large crystal-field splitting seen for all 5d ions, the t2g and eg orbitals are well separated in energy. With a 5d3 configuration the t2g orbitals are halffilled which leads to a higher Hubbard U than other perovskites containing 5d ions.

10.16 (a) Fe3+ has the electron configuration [Ar]3d5. Taking into account the crystal-field splitting and high-spin character, we can further specify the electron configuration as t2g3eg2. With partially filled t2g and eg orbitals, Fe2O3 must be a Mott–Hubbard insulator. (b) Cr3+ has the electron configuration [Ar]3d3. In an octahedral crystal field, we can write this as t2g3eg0. With partially filled t2g orbitals, MgCr2O4 must be a Mott–Hubbard insulator. (c) Pt4+ has the low-spin electron configuration [Xe]4f145d6. In an octahedral crystal field, we can write this as t2g6eg0. With a filled t2g set of orbitals and an empty eg set of orbitals, Li2PtO3 is a band insulator.

10.17 (a) A single carbocation–radical pair is a polaron. (b) Oxidation removes an electron from a filled bonding state; the polymer thus obtained is doped p-type.

10.18 (a) The bands touch at six points as shown in Figure 10.30. (b) Dirac points. (c) The bands that touch are the  and * bands; both derived from 2pz orbitals.

10.19 A semiconductor has a nonzero band gap; in graphene the band gap is zero. The carriers in a semiconductor have a velocity that depends on the wave vector k, while the carrier velocity in graphene is independent of k. Near the Fermi level, the bands in a semiconductor have a parabolic shape as a function of k, while in graphene they are linear as a function of k.

10.20 The tubes where n − m  3q, where q is an integer are semiconducting. Statistically this condition is met for 2/3 of the SWCNT in a given synthesis.

10.21 (a) As drawn, this is a (5,1) carbon nanotube. (b) No the T vector shown here is not valid because it connects two carbon atoms that are not equivalent.

10.22 (a) The circumferential vector is C = 4a + 1b, so this is a (4,1) nanotube. (b) n − m = 3, which, being a multiple of 3, makes this a (nearly) metallic nanotube. (c) The diameter is 3.59 Å by Equation (10.28), and the chiral vector is 10.89° by Equation (10.29). (d) The number of atoms in the unit cell is 28 by Equation (10.30), and it matches the number of atoms in the shaded region of the picture.

55

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

Chapter 11: Magnetotransport Materials 11.1

The spin polarization calculated from Equation (11.1) is 0.5. Given that the up and down percentage in the denominator of Equation (11.2) must sum to 100%, the tunneling electrons are 75% spin up and 25% spin down.

11.2

Iron atoms in charge-ordered YBaFe2O5 couple antiferromagnetic in all three directions.

Periodicity of the magnetic cell is then achieved by doubling the unit-cell parameter b of the chemical unit cell. In valence-mixed YBaFe2O5, the coupling along c across the Y layer becomes ferromagnetic. The magnetic cell achieves periodicity upon doubling all three unit-cell parameters of the chemical cell. For clarity, Y and Ba are also drawn: c a

b

–Y−

– Ba −

–Y−

11.3

(a) Given Y3+, Ba2+ and O2−, the average Mn oxidation state is 2.5. This suggests charge ordering of Mn2+ and Mn3+ with respectively 5 and 4 d-electrons. One eg orbital of Mn3+ will be empty. In a square-pyramidal ligand field, the dz2 orbital will be occupied by the “non-bonding” d electron because there is only one oxygen in its direction and that interference is easily minimized by lengthening the Mn3+–O(1) apical bond (if looked upon as a MO orbital, dz2 will hence be slightly antibonding). Orbital dx2–y2, if occupied, would interfere with four oxygen ligands, and it therefore will be empty, keeping the Mn3+–O(2) bonds short. Square pyramid of such a shape corresponds to the Mn(1) site, and Mn(1) is Mn3+. By default, Mn(2) is Mn2+. (b) The orbital ordering involves the dx2–y2 orbitals that are occupied on the Mn(2) sites and empty on the Mn(1) sites.

11.4

(a) Cr3+ is a d3 ion so the t2g orbitals are ½-filled and the eg orbitals are empty. A Jahn– Teller distortion is not expected for this electron configuration. (b) Ti3+ is a d1 ion so there is 1 electron in the t2g orbitals. Adding a single electron into a triply degenerate set 56

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

of orbitals will lead to Jahn–Teller distortion. LaTiO3 does show a subtle cooperative Jahn–Teller distortion. (c) Co3+ is a d6 ion. With a low-spin configuration, the t2g orbitals are fully occupied and the eg orbitals are empty. A Jahn–Teller distortion is not expected for this electron configuration. (d) Mn3+ is a d4 ion so that with a high-spin configuration the t2g orbitals are ½-filled, and there is one electron in the eg orbitals. Because there is a single electron in an originally doubly degenerate set of eg orbitals, a strong Jahn–Teller distortion is expected.

11.5

In O2 of atmospheric pressure, δ ≈ 0.1 would be achieved at ~900 °C. The δ = 0 sample would form at 1300 °C because for this isotherm the point pO2 = 10−4 bar appears at about the center of the (now rather short) integer-structure plateau.

11.6

To identify ferromagnetic couplings, select one spin direction and draw lines of that same spin among nearest neighbors, including same-spin Mn3+. Fill the picture with those zigzag lines. Then draw the lobes of half-occupied dz2 of Mn3+ along the line of each ferromagnetic coupling.

11.7

(a) Ni3+, a d7 ion with electron configuration t2g6eg1. (b) Because LaNiO3 is metallic, the eg electrons are delocalized and we expect symmetric octahedra rather than a Jahn–Teller distortion. (c) The larger ionic radius of La3+ increases the Goldschmidt tolerance factor, allowing less octahedral tilting hence more linear Ni–O–Ni bonds. The Ni–O–Ni bond angles are ~165 in LaNiO3, while ~150 in GdNiO3, and even less straight as the R3+ radius decreases further. The better orbital overlap in LaNiO3 leads to a wider conduction band (Chapter 6.6.5 and 10.4.3), and this favors electron delocalization and metallic conductivity. (d) Higher temperatures are needed to achieve the metallic conductivity as the increased tilting makes the conduction band narrower. (e) There would be a discontinuous increase in the unit-cell volume upon entering the insulating phase. The cooperative Jahn–Teller distortion (Chapter 5.3.10) due to the t2g6eg1 configuration would uniaxially expand the octahedra, making the dz2 orbital occupied (oriented along the longest axis of the distorted octahedron) and dx2–y2 empty. The charge disproportionation of Ni3+ would be manifested by oxygens shifting closer to Ni4+ and away from Ni2+, creating two crystallographically different sites for these two ions distributed in a 3D checkerboard pattern to minimize coulombic repulsion.

57

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

metal

11.8

11.9

superconductor

If searching Heusler alloys by crystal structure, note that while all Heusler-alloy variants can be conveniently shown on the same cell of the F4̄ 3m space group as in Figure 1.34, true symmetry of full-Heusler alloys X2YZ is Fm3̄ m with Wyckoff sites 4a, 4b, 8c.

PdMnSb IrMnSb

TiNiSn

TiCoSb MnCoSb

z

22

21

18

18

21

ferromagnet?

yes

yes

no

no

yes

μsat (in μB)

4

3

0

0

3

halfmetal?

yes

yes

no

no

yes

semiconductor? no

no

yes

yes

no

11.10 Use Figure 11.15 for the orbital scheme and then shift the relevant orbitals on the vertical energy scale so that filling by 29 electrons produces majority spin only at the Fermi level: t2g* eg*

Majority spin

eu

t2g* t1u

eg*

Minority spin eu t1u (t2g)6, (eg)4, s2p6 = 18 of 29 valence electrons

11.11 Equation (11.2) gives the spin polarization P = 0.855. TMR = 0.844 with Equation (11.5) and 5.414 with Equation (11.6).

Chapter 12: Superconductivity 12.1

Discussion question, all answers in text.

12.2

Discussion question covered in text.

58

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

12.3

Use Equation (12.5). Values are 5.54×10−23 J, 0.0035 eV or 2.8 cm−1.

12.4

Use Equation (12.4). Values are 5.48×10−23 J, 0.0034 eV or 2.8 cm−1.

12.5

Using Equation (12.5), we can estimate 20 as 0.0011 and 0.0027 eV or 9 and 18 cm−1 for Sn and Pb respectively. The IR absorption is a measure of the superconducting gap.

Radiation with a corresponding energy smaller than the gap is reflected and radiation with energy above the gap is absorbed.

12.6

The data show that there is a decrease in Tc as the mass M increases. A plot of log(Tc) vs log(M) gives a slope of −0.48 for Sn and −0.44 for Hg. Tc therefore has essentially the same dependence on mass as vibrational frequencies: ν = (kbond/μ)0.5 where μ is the reduced mass of the atoms involved and kbond the force constant (Footnote 8). This can be rationalized using the ideas of virtual phonon exchange for Cooper-pair formation described in the text.

12.7

This problem reviews the material in Chapter 1. The structure is shown in Figure 1.26 or Figure 12.9. Octahedral sites are found half way along each cell edge and at the body center of the cell, hence four per cell and one per C60 molecule. There are eight tetrahedral holes per cell and therefore two per C 60. If the cell is divided into 8 smaller cubes, octants, the tetrahedral sites are at the centers of the octants. The ccp spheres are in contact along the face diagonal and tetrahedral holes have sizes

(

)

−1 r

2a = 4r giving a = 14.14 Å. Octahedral and and

(

)

−1 r

giving 2.07 Å and 1.12 Å,

respectively. Shannon 6-coordinate crystal radii [Acta Crystallogr. Sect. A 32 (1976), 751–767] range from 0.9 to 1.81 for Li+ to Cs+; each could fit in an octahedral hole.

12.8

The data relate to Figure 12.10, where the dependence of Tc on the volume per C60 was presented for a variety of A3C60 materials. For the K3−xRbxC60 series at 0 GPa, there is an increase in Tc with the fcc cell parameter a. As discussed in the text, increasing the radius of A moves C60 molecules further apart, decreases the band width W and increases the density of states N(EF) at the Fermi level. From Equation (12.6), BCS theory would suggest an increase in Tc. A similar effect is seen for Rb3C60 as pressure increases, causing an increase in W, decrease in N(EF), and a fall in Tc. Comparison of the two series suggests that the data can’t be completely explained by just the effects of changing N(EF) predicted by BCS theory. As an example, extrapolation of high-pressure data to the K3C60 cell parameter of 14.25 Å yields Tc = 23 K, significantly higher than observed.

59

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

12.9

The three samples are in the over-, optimal- and under-expanded regimes of Figure 12.10a, respectively, with volumes per C60 of 794, 778 and 754 Å3. For x = 0.35 in the over-expanded regime, non-BCS behavior is expected, and increasing pressure should move the sample “up” the superconducting dome as the volume per C60 decreases. The initial dependence of Tc on pressure, (dTc/dP)0, is +2.2(1) K kbar−1. For under-expanded Rb2CsC60, increased pressure reduces N(EF) and decreases Tc as predicted by BCS theory. (dTc/dP)0 is −0.7(1) K kbar−1. The optimally expanded x = 1.0 sample is at the top of the superconducting dome and dTc/dP)0 is +0.2(1) K kbar−1, close to zero.

12.10 SrBiO3 contains two distinct Bi sites. These can be associated with a Sr2BiIIIBiVO6 oxidation-state assignment. Bond-valence sums for the two sites are 3.35 and 5.02, broadly in line with this. Sr0.4K0.6BiO3 has a single site showing that K doping suppresses charge order. The bond-valence sum for Bi is 5.64, higher than the formal charge of +4.6 suggesting that local disorder in the structure could affect the bond-valence calculation.

The pure-Sr material would be expected to be a semiconductor like BaBiO3 whereas the K doped material, where charge ordering is suppressed, superconducts at 12 K. 12.11 The sketch should be similar to Figure 12.15c. Cu is octahedrally coordinated with four O1 at 1.907 Å and two O2 at 2.46 Å. La is 9 coordinate in a mono-capped square antiprism.

12.12 (a) The structure of YBa2Cu3O7−δ and its relationship to a tripled perovskite is given in the text. It contains planes of CuO5/2 corner-sharing square based pyramids. The apical Cu–O bond is longer than the four bonds in the base such that the electronic properties are similar to the CuO4/2 planes found in most cuprate superconductors. The oxidation state of copper in these sites is slightly higher than Cu2+ such that hole-doped superconductivity occurs. (b/c) Arrow 1 corresponds to a transition from a paramagnetic state to a long-range antiferromagnetic insulator and could be studied by magnetometry or neutron diffraction (to probe magnetic structure, see Box 9.1 Characterizations).

Arrow 2 is a structural transition associated with oxygen ordering as O atoms are added at 0 ½ 0. The change in unit cell from a = b to a ≈ b (Figure 12.17b) could be followed by powder X-ray or neutron diffraction; neutron diffraction would give better structural information on oxygen positions. Electron diffraction/imaging would give information on short-range order. Arrow 3 represents a transition from a poor metal to a superconductor.

60

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

This could be evidenced by an increase in conductivity, measurement of magnetization to show the material becomes a perfect diamagnet, or demonstration of the Meissner effect via levitation of a small permanent magnet above the cooled sample. (d) Methods for preparing a sample with optimum superconducting properties and δ approaching 0 are given in the Box 12.1 Synthetic methods.

12.13 This explores two different ways of hole doping La2CuO4. Sample A is La1.85Ba0.15CuO4.

Under an O2 flow, La2CuO4 gets oxidized to La2CuO4+x (B) but is reduced to La2CuO4 on heating in Ar. Thus 20 mg of B loses 0.059 mg (3.688×10−3 moles) of O giving a molar ratio of 1:0.0749 Cu:O. B is therefore La2CuO4.08. Both materials are optimally hole doped and superconduct below ~35 K. 12.14 The majority of cuprate superconductors are hole doped and have Cu oxidation state around Cu2.15+ for optimum superconductivity. Both materials in this Problem are electron doped. Nd2−xCexCuO4 has a Nd3+/Ce4+ substitution, and Sr1−xNdxCuO2 has Sr2+/Nd3+. The Cu oxidation state in each is Cu(2−x)+.

Chapter 13: Energy Materials: Ionic Conductors, Mixed Conductors, and Intercalation Chemistry 13.1

(a) Cu2+ + 2e− → Cu E° = +0.34 V (higher standard potential defines the oxidizing agent). Sn2+ + 2e− → Sn E° = −0.14 V (lower standard potential defines the reducing agent). Overall reaction: Cu2+ + Sn → Cu + Sn2+; oxidizing agent oxidizes the reducing agent. The plain difference of the two standard reduction potentials is the standard cell potential E°cell = +0.34 − (−0.14) = +0.48 V. (b) Overall reaction 2Co3+ + Sn2+ → 2Co2+ + Sn4+, E°cell = +1.66 V.

13.2

The overall reaction and E°cell can be determined by subtracting the lower equation from the upper: PbO2(s) + 2H3O+(aq) + Pb(s) + 2HSO4−(aq) ↔ 2PbSO4(s) + 4H2O(l); E°cell = 2.041 V. Oxidation occurs at the anode by definition. From the cell equation, the lead electrode is oxidized and PbO2 electrode reduced, making lead the anode.

13.3

Since Zn is the anode, it must be oxidized; for simplicity to ZnO. As there is an approximately 1:2 Zn:MnO2 ratio, a balanced equation suggests MnO2 is reduced to Mn2O3. In an alkaline environment the half equations will involve OH−. Equations are: Overall: Zn(s) + 2MnO2(s) → ZnO(s) + Mn2O3(s) 61

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

Anode: Zn(s) + 2OH−(aq) → ZnO(s) + H2O(l) + 2e− Cathode: 2MnO2(s) + H2O + 2e− → Mn2O3 + 2OH−(aq) 13.4

Since Cd metal must be oxidized during cell operation, it must form the anode. From Section 13.1 we know that it forms Cd(OH)2. NiIII in NiO(OH) is reduced to Ni(OH)2 and electron balance requires a 1:2 ratio. We can write equations as: Anode: Cd(s) + 2OH−(aq) ↔ Cd(OH)2(s) + 2e− Cathode: 2NiO(OH)(s) + 2H2O(l) +2e− ↔ 2Ni(OH)2(s) + 2OH−(aq) Overall: 2NiO(OH)(s) + Cd + 2H2O(l) ↔ 2Ni(OH)2(s) + Cd(OH)2(s)

13.5

We can make use of the Nernst equation as discussed in any physical chemistry textbook.

The cell potential is caused by the different Zn concentrations (or chemical potentials) in the two compartments. This shows how an electrochemical cell could be used as a sensor.

RT  a2  RT  c2  = +0.025 V  ln   ≈ − ln Ecell = Ecell − nF  c  nF a  1  1

13.6

Right-hand electrode: O2(p1) + 4e− → 2O2−. Left-hand: O2(p2) + 4e− → 2O2−. Overall RT  p1  . Voltages of (a) Ecell = 0.91 V; ln reaction (R − L): O2(p1) → O2(p2), and E =   cell nF  p2  (b) Ecell = 0.73 V; (c) Ecell = 0.03 V should be obtained, which show that the sensor output changes significantly under the different conditions.

13.7

The data should give essentially the plot in Figure 13.5. Fitting the extrinsic region (lowest four temperature points) gives a gradient of 8520 K−1 that gives an activation energy in kJ/mol of Em = 8520×R/1000 = 71 kJ/mol using the equations in the text. Note the use of R (not k) to give an answer in kJ/mol. The intrinsic region (highest five temperature points) gives a gradient of 20400 K−1 that is equal to 1000×(Em + ΔHS/2)/R giving ΔHS = 197 kJ/mol.

13.8

We estimated the number of defects in NaCl crystals in Chapter 2 problem 2.3. At 1000 K, we estimate 2×1015 defects in 2×1019 formula units showing that 10−4 carriers per formula unit is the correct order of magnitude. For AgI we can assume 1 carrier per formula unit. Converting to charge carriers per m3 we get 2×1024 and 1.5×1028 respectively (NaCl Z = 4; AgI Z = 2). Putting these values into  = neμ we calculate mobilities of μ ≈ 3×10−9 and 4×10−8 m2V−1s−1 for NaCl and AgI respectively. Thus mobilities are only an order of magnitude different in the two materials even though 62

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

conductivities differ by five orders of magnitude. It’s the high value of n in AgI that gives it such high conductivity.

13.9

(a) Both are fluorite-type ionic conductors as discussed in the text. (b) The data can be used to produce plots similar to Figure 13.6 and Figure 13.18. For PbF2 the conductivity saturates and remains essentially unchanged on melting. (c) For Zr1−xYxO2−x/2 (YSZ) one can plot ln(T) versus 1/T and from the gradient (10900 K−1) extract an activation energy of 90.5 kJ/mol. YSZ can be used in fuel cells and oxygen sensors.

13.10 If we rewrite the formula as LaMoO4.5, the possible relationship to SnWO4 becomes clear. We might therefore expect the high-temperature structure to have metal sites similar to those of SnWO4, but with a higher coordination number for Mo. Neutron diffraction data by Lacorre et al., Nature 404 (2000), 856–858 confirm this picture. The high conductivity suggests that O2− ions are highly mobile. The abrupt fall in conductivity at ~580 °C in Figure 13.18 suggests that O2− ions become frozen into specific sites at the structural phase transition. In fact, the room-temperature structure is remarkably complex with a mixture of four-, five- and six-coordinate Mo sites in a 15:15:18 ratio as reported by Evans et al., Chem. Mater. 17 (2005), 4074–4077. The high-temperature form can then be pictured as a time-averaged dynamic mixture of the local coordinations seen at low temperature such that the variable coordination number of Mo is key to high conductivity.

13.11 The potassium intercalate has a higher conductivity than graphite at both temperatures.

This can be understood from the DOS diagram of Figure 13.21 where K donates electrons to the conduction band, increasing conductivity. Pure graphite shows activated conductivity increasing slightly with T as expected for a semimetal. The intercalate conductivity decreases with T as expected for a metallic conductor.

13.12 The four peaks observed experimentally are 00l reflections l = 2, 4, 6, 8 (the c axis in graphite is actually twice the interlayer spacing). The stage-1 intercalate Rb0.125C has an interlayer spacing of 5.65 Å and the peaks are again 00l reflections that can be indexed with l = 1, 2, 4, 6. The increase in c axis of 2.3 Å is consistent with a Rb+ ionic radius of ~1.5 Å given that the Rb+ will nestle in between C atoms. The 003 reflection is presumably too weak to be observed. Rb0.03C has a much lower Rb content suggesting a staged compound. For a stage-3 intercalate you would expect a c axis of 2  3.35 + 5.65 63

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

= 12.35 Å (2 × graphite + 1 × stage-1 intercalate). The four peaks observed for Rb0.03C are therefore 00l reflections with l = 1, 2, 3 and 4. Original data from Leung et al., Solid State Commun. 33 (1980), 285–288.

13.13 (a) The sketch should be a rectangle enclosing the rhombus shown in the bottom left of Figure 13.21. (b) The layer stacking is AAAA, not ABAB as in hexagonal graphite. This is so that the K environment is the same above and below the guest layer; i.e. the layers shift laterally on intercalation. Only 1 in 4 C6 rings has a K above its center in each layer.

As one goes from one layer to the next, a different ring has K above it to minimize K+ repulsion. This leads to a c axis 4 times the interlayer spacing. (c) The unit cell contains 1 potassium per 8 carbons in a sheet (K on Wyckoff site 8a, 2 × C on 32h), hence the composition K0.125C. The K–K distance in the layers is 4.92 Å. This compares to 4.51 Å in the metal, which can be calculated from the bcc structure (Figure 1.23) as a√3/2. (d) The corresponding distances in metallic Rb and Cs are 4.88 and 5.25 Å such that A+ ions are slightly closer together in the intercalate than in the elemental forms.

13.14 Placing a Li above every graphene hexagon yields LiC6/3. For composition LiC6, you therefore need a Li above every third hexagonal ring in a regular manner. The hexagonal unit cell is the smallest repetitive tile and its unit-cell parameter is the length of 3 C–C bonds or 4.26 Å. The Li–Li separation is larger than sum of metallic radii of ~3.04 Å.

13.15 Capacity is related to the number of electrons that can flow through the external circuit, and the problem is really about changing to units that aren’t commonly used by most chemists. (a) For Li metal, 1 kg is 144.07 moles. Charge is 144.07  NA  1.60210−19 = 1.3899107 C of total electrons available. 1 Ah is 3600 C. The capacity is then 1.3899107/3600 = 3361 Ah/kg. The three intercalates can be calculated in a similar way and give: (b) LiC6 339.3 Ah/kg; (c) LiTiS2 225.3 Ah/kg; and (d) LiCoO2 273.9 Ah/kg.

Note the high capacity of Li metal.

13.16 As rocking-chair batteries are made in the discharged state, it is common to quote capacity per kg of C (i.e. ignoring the Li mass). Analysis as in the previous problem then gives a capacity of 371.9 Ah/kg, a value commonly quoted in the literature.

13.17 The volume of 1 g of the stated mixture is 0.313 cm3. 1 cm3 would therefore contain 2.71 g of LiFePO4, 75% of that for pure material, leading to a 25% reduction in volumetric density.

64

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

13.18 This problem is again about converting from the energy units chemists are used to into those commonly used in the battery field. We can use a cell reaction of Li + TiS 2 → LiTiS2. E = 2V implies G = −193 kJ/mol. The total molar mass of reagents is 6.94 + 112.00 = 118.94 g/mol hence 1 kg of reagents will contain 8.406 moles. The total energy released by the cell in J is given by the product of the number of moles and the free energy per mole. This is then divided by 3600 (because 1 Wh is 3600 J) to obtain the energy density in the required Wh/kg: 1938.408/3600 = 451 Wh/kg. If 50% of the mass is active, the energy density is halved to 225 Wh/kg.

13.19 A simple approach is to use the same method as in the previous Problem, as if the cell reaction were LiC6 + CoO2 → LiCoO2, and then halve it. E = 3.6 V implies G = −347 kJ/mol. Total mass of reagents is 169.9298 g/mol. 1 kg is 5.885 moles. 1 Wh is 3600 J. Energy density is 347  5.885/3600 = 567 Wh/kg. As the actual cell is 50% of the total battery mass and has 50% usable Li capacity, the energy density of the battery is one quarter of this value: 142 Wh/kg. If we use the actual cell equation in the text the corresponding value is 180 Wh/kg. Typical values of 120–150 Wh/kg are achieved in real cells. If Li + CoO2 → LiCoO2 were used as a cell reaction, we’d gain voltage (3.6 + 0.6 = 4.2 V that converts to G = −nFE = −405 kJ/mol), and, without the carbon mass, the capacity increases to 288 Whkg−1. This means that the energy density using a Li anode would be over twice that with LiC6. This battery would “beat” an equivalent LiTiS2 battery (225 Wh/kg), despite only running to half theoretical capacity, due to the larger voltage and lower mass of O versus S. 13.20 The electrochemical data are taken from Ohzuku et al., J. Electrochem. Soc. 137 (1990), 769–775. (a) Spinel has ccp oxygen ions with cations in ⅛ of the tetrahedral and ½ the octahedral holes. LiMn2O4 spinel has MnIII and MnIV stabilized by LFSE on the octahedral sites and Li on tetrahedral sites. Since only a fraction of the ccp interstitial sites are occupied in a spinel, the empty octahedral and tetrahedral sites can potentially accommodate additional Li ions. (b) The first plateau suggests that two phases coexist as domains in the interval 0 < x < 0.5 (compare with Figure 13.28). This occurs as Li0.5Mn2O4 is stabilized by ordered Li at the tetrahedral sites. 0.5 < x < 1 is a solidsolution interval. In the interval 1 < x < 1.8, two phases coexist; the transformation is Li1Mn2O4 to Li2Mn2O4. (c) The change in Ecell at the pure-spinel composition suggests a 65

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

very different chemical potential for the Li deficient spinel and the Li-excess spinel, implying a different coordination environment (in fact, the 4 V process involves Li+ on the regular spinel tetrahedral sites whereas the 3 V process places Li+ on the octahedral interstitital sites).

13.21 Li2Mn2O4 contains d4 MnIIIO6 octahedra, which forces a symmetry reduction due to Jahn–Teller distortion caused by the single eg occupancy. The cell volumes of the cubic and tetragonal phase are 561.4 Å3 and 298.1 Å3, which correspond to volumes per formula unit of 70.2 and 74.5 Å3, respectively. The ~6% difference would eventually lead to loss of capacity upon multiple recharging due to particles cracking away from the electrode.

Chapter 14: Zeolites and Other Porous Materials 14.1

This Problem gives some feel for the high accessible volume of zeolite structures. (a) Cristobalite 2.33 g/cm3; stishovite 4.29 g/cm3; LTA 1.41g/cm3. So the high-pressure stishovite polymorph has a significantly higher density than cristobalite, whereas LTA has a significantly lower density. (b) O radius 1.31 Å; Si radius 0.29 Å. Shannon effective ionic radii are 1.35 and 0.26 Å, respectively [Acta Crystallogr. Sect. A 32 (1976), 751–767]. (c) % volume occupied by hard spheres are: cristobalite 44.3%; stishovite 81.5%; LTA 26.8%. Stishovite exceeds the value for close packed spheres (even if just O volumes are considered), which is unsurprising for a high-pressure phase given our crude hard-sphere approximation. The volume of space filled in LTA is significantly lower than in cristobalite. The accessible pore volume in cristobalite is 0% and in LTA 21.4%.

14.2

(a) From the density, a 1 g spherical single crystal of cristobalite would have a volume of 0.429 cm3 and radius of 0.468 cm. Its surface area would be 2.75 cm2/g or 2.75×10−4 m2/g. (b) 10 m spherical particles would have a surface area of 0.258 m2/g and (c) 10 Å particles 2575 m2/g. Surface area increases rapidly as particle size decreases. The accessible surface area of zeolites is comparable to that of nanoparticles.

14.3

From Figure 14.3 and assuming that oxygen atoms are hard spheres in approximate contact as in Figure 14.1d, a reasonable estimate might be that a 3 Å radius sphere fits into a β cage and a 5.5 Å sphere into an  cage. There will be one of each type of sphere 66

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

per unit cell. The total surface area of these spheres can be calculated as 4.93×10−18 m2 per cell. If we assume a pure SiO2 composition, then the unit-cell content will be Si24O48 and have a mass of 2.39×10−21 g. We would therefore estimate the surface area as ~2000 m2/g. This crude calculation is the same order of magnitude as a more sophisticated one that measures the total surface area visited by the center of a hypothetical 2.8 Å diameter spherical probe molecule when moved throughout all the cavities of the framework. This suggests that the viewpoint of zeolites having a 3D accessible surface is reasonable.

14.4

See table. Note cells are for the framework type, such that T per nm3 values differ slightly from values in Table 14.1.

Name T/nm3

SBU's

Vertex symbol Coordination Sequence

SOD LTA

16.7 14.2

6 8 4-4 6-2 6 1-4-1 4

4ꞏ4ꞏ6ꞏ6ꞏ6ꞏ6 4ꞏ6ꞏ4ꞏ6ꞏ4ꞏ8

4 10 20 4 9 17

FAU

13.3

6-6 6-2 6 4-2 1-4-1 4

4ꞏ4ꞏ4ꞏ6ꞏ6ꞏ12

4 9 16

ANA

19.2

6-2 6 4-[1,1] 1-4-1 4

4ꞏ4ꞏ6ꞏ6ꞏ84ꞏ84

4 10 22

RHO

14.5

8-8 8 6 4

4ꞏ4ꞏ4ꞏ6ꞏ8ꞏ8

4 9 17

14.5

See the table for previous Problem.

14.6

A crude estimate can be made by thinking of a circle of circumference 2rn going through the center of n spheres in contact. The inside of the pore will have a diameter smaller by 2r. Thus the inner radius is given by (2rn/2) − 2r. Values for 6/8/10/12 rings are 2.46/4.2/5.89/7.61 Å; values for the same rings from the zeolite atlas are none/LTA 4.1 Å/ZSM-5 5.1–5.6 Å/FAU 7.4 Å.

14.7

Volumes filled are consistent with H2O filling both the  and  cages but N2 only filling  cages.

14.8

Empirical formula in (a) should lead to a chemical formula (b) of (NPr4)[(AlO2)(SiO2)23] and (c) H[(AlO2)(SiO2)23]. See text for (d) and (e).

14.9

(a/b) Formula (NMe4)[(AlO2)(SiO2)3]H2O. (c) Use of template increases the Si:Al ratio and decreases hydrophilicity.

14.10 Anything above a relative cracking rate of ~1 is commercially interesting. ZSM-5 therefore shows high activity. As discussed in the text, we normally expect acidity to increase as Si content increases due to increasing electronegativity of the framework.

However, when Al:Si is below ≲ 0.01 (Si:Al ≳ 10), all of the second nearest neighbors to 67

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

a given H are likely to be Si, and local electronegativity becomes independent of Si content (more distant sites are less important). In this region, catalytic activity is dominated by the number of Al sites present. The data shown are in this regime and an approximately linear dependence of activity on Al:Si ratio is seen over a wide range.

NMR data show that activity scales with the number of tetrahedral Al sites suggesting that Brønsted sites are responsible for catalytic activity.

14.11 HY is a large-pore zeolite. Similar amounts of isomerisation and disproportionation occur. In the smaller ZSM-5 pores, the bimolecular transition state leading to disproportionation can’t form and isomerisation dominates. This is an example of transition state selectivity. Similarly, in HY one observes an approximately equilibrium distribution of ortho and para isomers in the xylene produced. With ZSM-5, the product is enhanced in the para isomer as this diffuses out of the zeolite most rapidly. This is an example of product selectivity.

14.12 d-spacing of reflection is 40.1 Å. For a 2D hexagonal array, this is the (100) reflection. d100 is at asin(60) suggesting a cell parameter of 46.3 Å. The cell edge is equal to twice the radius of the pore plus the wall thickness. The wall thickness is therefore ~16 Å.

14.13 The contents of the cell shown would be Zn4O(C20O4)3. Occupied volume would be ~1386 Å3. For a cell edge of 21.49 Å, the unit cell volume is 9925 A3, which gives ~86% free volume.

14.14 The PtS structure contains (distorted) square planar PtS4/4 (d8 Pt2+) nodes linked to tetrahedral SPt4/4 nodes. The MOF community calls this the pts net. ZnPt(CN)4 contains square planar Pt(CN)4/4 nodes with N coordinating Zn to form tetrahedral Zn(NC)4/4 nodes. It has therefore the same underlying net. This is an augmented net (vertices are replaced by the corresponding polyhedra). MOF-11 contains square Cu2(OCO)4 nodes joined by ATC linkers that have their carboxylate groups arranged as in a tetrahedron, again leading to the pts net. These examples are discussed in ref. [48] along with a further link to CuO that has the same topology but the structure distorts slightly (Jahn– Teller) allowing Cu to form 2 additional longer bonds to O giving it a 4+2 distorted octahedral geometry.

14.15 A discussion question to highlight some features of confined catalysis. Many of the answers are equally applicable to zeolites, MCM materials and other controlled-pore-size 68

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

frameworks: (a) In general, reaction rates would initially be expected to be slower than in solution as reagents must diffuse to the catalytically active sites. Rates would approach those in solution as pore size increases. (b) If confining the catalytic center increases selectivity, then the ee should be higher in the MOF than in solution. The values might again converge as pore size increases. (c) Confined catalysts should be more stable as degradation pathways involving multiple catalytic centers are prevented. It also ought to be easier to reclaim and recycle a heterogeneous catalyst. (d) One way to test that true heterogeneous catalysis occurs is to filter off the solution above the catalyst. If the filtrate is catalytically inactive, it suggests that catalysis is heterogeneous. (e) This is hard to prove definitively. One method is to see how the catalytic rate depends on the size of the substrate. Catalysis ought to be significantly slowed for molecules larger than the pore size. This conclusion would be strengthened if large molecules were still converted by an equivalent surface-supported catalyst.

Chapter 15: Amorphous and Disordered Materials 15.1

Si is 4-coordinated (Zachariasen’s rule 2), O is 2-coordinated (Zachariasen’s rule 1, SiO4 tetrahedra share corners (Zachariasen’s rule 3 and 4); both Mg and O are 6-coordinated (violates Zachariasen’s rule 1 and 2) and crystal structure reveals edge-sharing polyhedra (violates Zachariasen’s rule 3).

15.2

In Al2O3, Al3+ is in an octahedral coordination (not 3- or 4-coordinated) and O2− is 4coordinated (not 2-coordinated). In aluminosilicates, Al3+ and Si4+ are both 4-coordinated and Al requires charge balance by alkali-metal or alkaline-earth ions. Adding Al2O3 increases the connectivity of an alkali-modified SiO2 glass by replacing non-bridging oxygen (bonded to the alkali-metal cation) with cross-linked Al–O–Si bonds.

15.3

Every M2O turns one –Si–O–Si– bridge into –Si–O− and −O–Si– that bind the two M+ cations, hence xM2O creates 2x non-bridging oxygens. From the given glass composition, their fraction is 2x/(x+2(1−x)) = 2x/(2−x). Their ratio to Si is 2x/(1−x).

15.4

In (Na2O)0.25(SiO2)0.75, the O:Si ratio is 1.75/0.75 = 2.33. The same ratio 1.75/0.75 can be achieved for glasses (Na2O)0.25−x(CaO)x(SiO2)0.75 with x < 0.25 as each Na2O can be exchanged by one CaO without changing the oxygen stoichiometry.

69

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

15.5

(a) The maximum angle that a light ray can make with the wall of the core corresponds to the minimum angle θ1 with the normal that will result in total internal reflection. We find that angle with Snell's Law [n1sin(θ1) = n2sin(θ2)] with θ2 = 90°, and subtracting θ1 from 90°. Thus 1.64 sin(θ1) = 1.50 sin(90°) gives θ1 = 66.2°, hence 23.8° towards the wall of the core. (b) An optical fiber would also work without cladding as nair ≈ 1; cladding is used to protect the core and minimize light leakage and cross talk.

15.6

Follow procedure in previous question with n2 = 1.609 as quoted in the text and n1 1.625.

The angle with the wall is 8.1°.

15.7

Using Equations (15.1) and (7.1), we obtain λFe2O3 = 589.3 nm /3.00 = 196 nm; cFe2O3 = 299792458/3.00 [m/s] = 9.99×107 [m/s]; vFe2O3 = 299792458 [m/s]/196.43 nm = 5.0908×1014 Hz. Only the light frequency does not change upon passing from air into the Fe2O3.

15.8

15.9

Purely ionic bonding implies Zeo = 0. Equation 15.4 can be derived accordingly: 2 — 1)𝑉a = (𝑛2 — 1)𝑉a (𝑛2 — 1)𝑉a (𝑛4π α= = + ( ) (𝑛2 — 1) 4π 2 4π (𝑛 + 2) + ( — 𝑍 ) (𝑛2 — 1) 4π 4π eo 3 3 3 Oxide anions’ contribution of polarizability for 8 oxygen per formula unit follows from Equation (15.5) with 0 = 8×1.79 Å3, N0 = 1.776 Å3 and Van = (720.4/4)/8 Å3 = 22.51 Å3 as αan = 12.990 Å3. Including cation polarizabilities gives the total polarizability αAE = (1.35+0.533+3×0.284 + 12.990) Å3 = 15.725 Å3. Given Va = 720.4 Å3/4 = 180.1 Å3, Equation (15.7) yields n = 1.523 (experiment 1.524).

15.10 KAlSi3O8 is written as (K2O)0.5(Al2O3)0.5(SiO2)3, from which we calculate the mass fractions of the three binary oxides that make up orthoclase: p(K2O) = (0.5×94.20 g/mol)/(278.2 g/mol) = 0.1692, p(Al2O3) = 0.5×101.96/278.2 = 0.1832, and p(SiO2) = 3×60.08/278.2 = 0.6479. The density ρ of orthoclase is calculated as ρ = [4(278.33 g/mol)(1 mol/6.022×1023)]/(720.4×10−24 cm3) = 2.566 g/cm3 from the values in Problem 15.9. The results and the chemical refractivities from Table 15.1 in Equation (15.8) give nGD = 1 + (2.566 g/cm3)[0.1692∙(0.196 cm3/g)] + 0.1832∙(0.207 cm3/g) + 0.6479∙(0.208 cm3/g)] = 1 + (2.566 g/cm3)(0.2058 cm3/g) = 1.528 that is very close to the value obtained in Problem 15.9.

15.11 Volume V = m/ρ. The Va is calculated as the volume in Å3 of the mass in grams of 1 mol of SiO2 divided by NA, divided by the density in g/cm3. Given SiO2 molar mass 60.08 70

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

g/mol, the Va = (60.08 g/mol × 1024 Å3/cm3)/(2.655 g/cm3 × 6.022×1023 mol−1) = 37.58 Å3. Using this value and n = 1.547 in Equation 15.4 gives LL = 2.845 Å3.

15.12 The attenuation is ten times the decadic logarithm of the input/output. The power-out as a percentage of the power-in is 97.73% for 0.1 dB/km and 99.977% for 0.001 dB/km.

71