answers to question set

Answers

1.1 a) CH3Cl has a single 3-fold axis running through the C-Cl bond Fig. S1. There are three mirror planeslabelled σ 1, σ 2 and σ 3 for each of which, the C-Cl bond and one of the C-H bonds lies on the mirrorwith the other two C-H bonds to either side.

b) CH2Cl2 has one 2-fold axis passing through C and bisecting both the Cl-C-Cl and H-C-H bondangles Fig. S1 (c,d). It also has two mirrors, at 90◦ to each other; in one, σ 1, C and the two H atomslie on the mirror with the two Cl atoms to either side; in the other, σ 2, C and the two Cl atoms lieon the mirror with the two H atoms to either side. A diagram of a 2-fold rotation axis similar to thatin CH2Cl2 is shown for the SiO4 tetrahedron in Fig. 1.4(b); a mirror plane in an SiO4 tetrahedronis shown in Fig. 1.7(a).

c) The 2-fold axis is now lost because atoms Cl and Br are not identical and therefore are notinterchangeable on rotation. Also one of the mirrors is lost leaving a single mirror with C-Cl andC-Br bonds on the plane and the two H atoms to either side.

d) On replacing Cl by H in part a), it should be apparent that there are now four 3-fold axes alongeach C-H bond. To see the mirror planes (six) and 2-fold axes (three) it is useful to draw the CH4

molecule as occupying a cube with H at alternate corners only, Fig. 1.25 and Fig. S1(e).

1.2 A mirror plane.

1.3 Four 3-fold axes (parallel to each of the four cube body diagonals), three 4-fold axes (passing throughpairs of opposite cube faces), six 2-fold axes (passing through pairs of opposite cube edges), ninemirror planes, one centre of symmetry. A selection of these is shown in Fig. S2.

1.4 Comparing Figs. S1 and S2, all 3-fold axes are lost in S2, as are two of the 4-fold axes which insteadbecome 2-fold axes; four of the 2-fold axes and four of the mirror planes are also lost, leaving: One4-fold axis, four 2-fold axes, five mirrors and one centre of symmetry, Fig. S3.

1.5 One 2-fold axis, two mirrors.

1.6 See Fig. S4: In (a), with atoms X, 5 and 6, in the trans orientation, there is a centre of symmetry atatom M; in (b) with X in cis positions, there is no centre of symmetry because, for example, atoms2 and 4 are not interchangeable. Other symmetries in (a) are one 4-fold axis, four 2-fold axes, fivemirrors and one centre of symmetry; in (b) there is one 2-fold axis and two mirrors; to see this moreclearly, imagine the molecule inside a cube with X, Y atoms at cube face centers.

1.7 a) 1: there are lattice points at each corner of the unit cell, but each counts as only 1/8.b) 2: one from the corner positions (i.e. 8 × 1/8) and one from the body centre.

Solid State Chemistry and its Applications, Second Edition, Student Edition. Anthony R. West.© 2014 John Wiley & Sons, Ltd. Published 2014 by John Wiley & Sons, Ltd.Companion Website: http://www.wiley.com/go/west/solidstatechemistrystudent

Solid State Chemistry and its Applications, Second Edition, Student Edition 2

(a)

H3

H2

σ3

σ2

σ1

H1

Cl

(b)

H3

n = 3

H1 H2

Cl

H2

σ2

σ1H1

Cl2

Cl1

(c)

H2H1

Cl2

n = 2

Cl1

(d)

H2

H3H4

(e)

H1n = 2

n = 2

Figure S1

mirror n = 2

n = 3n = 4

Figure S2

3 Answers

n = 2

n = 2

n = 4

Figure S3

Y

X

6

(a)

4

3

52

1

(b)

6

4

3

52

1

Figure S4

c) 2: one from the corner positions and one from the C-centre positions (in the middle of opposite a bfaces) which count as 1/2 each.

d) 4: 1 from corners, 3 from face centres (6 × 1/2).

1.8 Body centred tetragonal.

1.9 a) A C-centred lattice has no 3-fold rotation axes; if the unit cell is geometrically cubic, the truesymmetry would be tetragonal.

b) It is always possible to draw a I-centred tetragonal cell with half the volume of the F-centred celland for which a(I-centred) = √

2a/2(F-centred): see Fig. 1.7(a).c) It is always possible to draw a P tetragonal cell with half the volume of the C-centred tetragonal

cell. Again, a(P) = √2a/2(C-centred).

See Fig. S5 showing how a C-centred tetragonal cell can be redefined as a smaller primitive cell orhow an F-centred tetragonal cell can be redefined as a smaller I-centred cell.

ac

c

ac

ap

ap

Figure S5


1.10 a) (001) b) (200) c) (222)

1.11 a) [010] b) [010]

1.12 1d2 = h2+k2+l2

a2 . Since h, k, l are integers, the sum (h2 + k2 + l2) can take only the values: 1, 2, 3, 4, 5,6, 8, 9, 10 etc. (note absence of value 7; 15 is also absent). As d = a (h2 + k2 + l2)−1/2, lines can belisted in order of decreasing d-spacing, corresponding to increasing (h2 + k2 + l2) i.e.

h2 + k2 + l2 hkl d/A

1 100 5.002 110 3.533 111 2.894 200 2.505 210 2.24

1.13 D = F wt × cell contents Z

cell volume V × Avogadro′s No.

D = 74.6 × 4

(6.2931 × 10−24)3 × 6.023 × 1023= 2.03 g cm−3

1.14 Use the formula for D (Answer to 1.13). F wt = 284.3, values for a and D are given; therefore Z canbe calculated. Z must be an integer, by definition. Z may not calculate to be exactly an integer if thereare errors in e.g. D or a, but should be very close to an integral value. Here Z = 0.99, i.e. Z = 1. Thestructure is CsCl.

1.15 a) Close packed layers occur in four orientations (111), (111), (111), (111). Each of these fourorientations can be redefined using negative directions of the unit cell, giving (111), (111), (111)and (111), respectively, i.e. these are the same four sets of orientations and layers, but viewed fromthe opposite direction. Collectively they form the set {111}. See Fig. S6 showing the (111) planes.

b) Close packed directions belong to the set <101> (i.e. the cell face diagonals); two such directions,[101] and [101], are shown in Fig. S6(b).

c) Using Pythagoras, atoms are in contact along cube face diagonals, Fig. S6, i.e. a2 + a2 = (4r)2,r = a/

√8

1.16 1.44, 1.38 A

(a) (b)

(111)

z

yx

(111)

(111)

r

az

x

[101] [101]

Figure S6

5 Answers

1.17 a) 8; the atom at the body centre is equidistant from the 8 corners.b) No; in a cp structure, the sphere coordination number is 12.c) <111>; atoms are in contact along the cube body diagonals.d) body diagonal = a

√3 = 4r; ∴ r = a

√3/4

1.18 a) 7.88, 7.59, 7.30 g cm−3

b) 1.24, 1.29, 1.27 A

1.19 From 1.17d, r = a√

3/4; also Z = 2 ∴ D = sphere volume12 (unit cell volume)

= 1.33π r3

0.5a3= 1.33π r3

0.5(

4r√3

)3 = 0.6802

1.20 From 1.15c, r = a√

2/4; also Z = 4

∴ sphere volume

0.25 (unit cell volume)= 1.33π r3

0.25a3= 1.33π r3

0.25(

4r√2

)3 = 0.740

1.21 a) antifluorite, b) zinc blende (sphalerite), c) rock salt, d) CdCl2.

1.22 a) none, b) wurtzite, c) NiAs, d) CdI2. In a), because the cation–cation distance in pairs of face-sharingtetrahedra would be too short.

1.23 i) rock salt, ii) zinc blende, iii) CsCl, iv) antifluorite, v) ReO3, vi) perovskite.

1.24 i) ccp, ii) CdCl2, iii) zinc blende.

1.25 Because, in the z direction, perpendicular to the cp layers, Ni atoms are sufficiently close for overlapof dz2 orbitals and metallic bonding. Such interactions do not occur in ionic structures; instead cation–cation distances are greater and the rock salt structure is adopted.

1.26 Mg-O = a/2 = 2.107 ArMg = 2.107 − rO = 0.847 AO-O = 1/2(a2 + a2)1/2 = 2.979 A2rO = 2 × 1.26 A = 2.52 A

Therefore, the oxide ions are not in contact and the structure is eutectic cp.

1.27 Zn-S = 1/2(a2 + a2 + a2)1/2 = 2.341 A

1.28 a) Li-O = 1/4a√

3 = 1.997 AO-O = 1/2a

√2 = 3.260 A

Li-Li = 1/2a = 2.305 AO-O = 3.26 A; 2rO = 2.52 ATherefore, the oxide ions are not in contact.

1.29 a) See Fig. 1.36.b) i: 12, ii: 6, iii: Two Ti at 1.953 A, four Sr at 2.761 A and eight O also at 2.761 Ac) Sr-O = 1/2

√2a = 2.761 A, Ti-O = a/2 = 1.953 A, D = FW × Z

V × N = (87.6 + 47.9 + 48) × 13.9053 × 10−24 × 6.023 × 1023 =

5.116 g cm−3

d) Sr and O together form a ccp array; primitive; it must be since Z = 1.e) i) for ferroelectricity, a polar structure is required with ‘stretched’ Ti-O bonds: Sr should be

substituted, perhaps partially, by a larger divalent cation such as Pb2+ or Ba2+.


ii) it has been suggested that a cluster of superconducting phases may occur for phases containingd1 ions, as a ‘mirror image’ of the cuprates based on d9 Cu2+. SrTiO3 is an insulator andwould need to be reduced either by oxygen loss, to give SrTiO3-δ or by substitution of highervalent ions (e.g. La for Sr; Nb for Ti).

iii) ionic conductivity depends on the presence of vacancies in one of the sublattices. These couldbe induced by aliovalent substitution (e.g. 3Sr = 2La; 5Ti = 4Nb) or by oxygen loss; the latterwould, however, be likely to show electronic conduction (mixed valence Ti) as well as possibleoxide ion conduction. Of course, there is no guarantee that the vacancies, if they can be created,will be ‘mobile’.

1.30 rSi = 12 (Si − Si) = 1

2 · 14 a

√3 = 1.176A.

1.31 To derive the new coordinates, subtract 1/41/4

1/4 from each since Ag at 1/41/4

1/4 then becomes Agat the origin i.e. 000; if any of the resulting coordinates are negative, add 1 to translate the atominto the unit cell; this applies to O at 000 which becomes −1/4−1/4−1/4 and therefore, 3/4

3/43/4. New

coordinates are:- Ag: 000, 1/21/20, 1/201/2, 01/2

1/2; O : 3/43/4

3/4. 1/41/4

1/4, see Fig. S7: The structure hasa ccp arrangement of Ag but only two of the tetrahedral sites are occupied by O. The structure has abcc arrangement of O but overall, the lattice type is primitive. The Ag is linear, 2-coordinate to O. Ois tetrahedrally coordinated by Ag. Ag-O = 1/4a

√3 = 2.046 A. Yes, located at Ag.

Ag

0

0

½

½

Oxygen

¼

-¼

¾

Figure S7

1.32 rM + rX = 1/2a; rX = 1/4(a√

2)∴ rM = 1/2a-rX = 4rX

2√

2− rX = rX(

√2 − 1)

∴ rM/rX = 0.414

1.33 rM + rX = 1/4a√

3; rX = 1/4a√

2rM = 1/4a

√3 − rX = 4rX√

2· 1

4

√3 − rX

= rX

(√3√2

− 1)

∴ rMrX

= 0.225

1.34 rM + rX = 1/2a√

3; rX = a2

rM = 1/2a√

3 − rX = √3rX − rX

rMrX

= √3 − 1 = 0.732

1.35 a) this value is greater than the minimum value for the zinc blende structure with 4:4 coordination butsmaller than that for the rock salt structure with 6:6 coordination. Hence the zinc blende structure isexpected but without anion–anion contacts. Using a similar argument:- b) rock salt, c) CsCl.

7 Answers

1.36 NaCl:- anions touching

rX = a√

2

4

(rM + rX) = a

2

V4X

Vcell=

4

3π r3

X · 4

a3=

4

3π

2√

2a34

64a3

= 0.740

rM = a

2− rX = 2rX√

2− rX = rX

(√2 − 1

)= 0.414rX

V4M =4

3π (0.414rX)3 · 4

23264r3X

2√

2

= π (0.414)3√

2

3 × 2= 0.053

∴ total packing density = 0.740 + 0.053 = 0.793.

An alternative and simpler method is as follows:-For the ideal rock salt structure,

rM

rX= 0.414

∴ VM

VX= (0.414)3 = 0.071

DensityM + DensityX = 0.74 +(

0.071

0.74

)= 0.79

CsCl:- anions touching

rX = a

2

VX

Vcell=

4

3π

a3

8a3

= 0.524

2 (rM + rX) =√

3a

VM

Vcell=

√3a

2− a

2

=4

3π

(0.732)3a3

8a3

= 0.205

∴ total packing density = 0.524 + 0.205 = 0.729


1.37 It is assumed that silicate structures are built of SiO4 tetrahedra which may/may not share corneroxygens with other SiO4 tetrahedra. In many aluminosilicates, AlO4 tetrahedra are present and actsimilarly to SiO4 tetrahedra. From the overall (Si, Al):O ratio it is possible to calculate whether thetetrahedra are isolated or share corners (bridging oxygens) with other tetrahedra. Therefore, for eachframework element (tetrahedral Si, Al), non-bridging oxygens, nbo’s, count as 1 and bridging oxygens,bo’s count as 0.5. Therefore,a) Si:O = 1:4, i.e. 4 nbo’s and the silicate anion is isolated SiO4−

4 .b) (Al + Si):O = 1:2, i.e. 4 bo’s, therefore the structure is a 3D framework of corner-sharing AlO4

and SiO4 tetrahedra.c) Si:O = 1:3, 2 nbo’s, 2 bo’s (Si3O9)6− 3-membered rings [from the Si:O ratio of 3, it is not possible

to determine the anion consists of closed rings or infinite chains since for both, two of the corneroxygens are shared with other tetrahedra].

d) Si:O = 1.35, 3 nbo’s, 1 bo, Si2O6−7 dimers.

e) Si:O = 1:3, 2 nbo’s, 2 bo’s (SiO3)2− infinite chains.f) Si:O = 2.75, giving on average 1.5 nbo’s, 2.5 bo’s, anion expected to be between a chain and a

sheet; in fact, it is a double chain.g) (Si + Al):O = 1:2.5, 1 nbo, 3 bo’s, infinite sheets of SiO4, AlO4 tetrahedra sharing three corners.h) Si:O = 1:2.5, 1 nbo, 3 bo’s, infinite sheets of SiO4 tetrahedra sharing corners.

1.38 a) the distance between the centres of adjacent C60 molecules is 0.707a; therefore, the Van der Waalsradius is 0.707a/2 = 5.01 A.

b) for octahedral sites, consider the rock salt structure:

a = 2(rC60 + rOCT

); ∴ rOCT = 2.08A

for tetrahedral sites, consider the zinc blende structure:

(rTET + rC60

) = 0.433a; ∴ rTET = 1.13A

If inert gas fullerides form, the inert gas atoms are likely to be cationic and therefore have radii lessthan the Van der Waals radii. All three could easily enter octahedral sites, giving formulae AC60 witha rock salt-related structure; the smaller ones may enter tetrahedral sites instead, giving either AC60 orA2C60 with zinc blende and fluorite related structures, respectively; or if all octahedral and tetrahedralsites are occupied, the formula A3C60 is feasible.

For further details, see G. E. Gadd et al. (1998) J. Phys. Chem. Solids, 59, 937–944; although thisdeals primarily with C70 compounds, details on C60 compounds are given in the Introduction.

1.39 I shall not attempt a model answer for this!

2.1 Because of the effect of T in the equation:

�G = �H − T�S

To create defects, the enthalpy of formation must be provided; this is likely to be large and positive.Associated with the defect will be a large, positive increase in entropy. At high temperatures, if theterm T�S > �H, then �G < O and defects may form at thermodynamic equilibrium.

2.2 The line is slightly curved. Ignoring the curvature, slope ≈ 1.80.0005

= −E log10 e

2R∴ E = 138 kJ mol−1

9 Answers

2.3 a) cation vacancies 2NaXNa ⇒ Mn•

Na+V′Na

b) oxide ion vacancies Zr XZr + O X

O ⇒ Y ′Zr + 1

2 V ••O

c) interstitial F CaXCa ⇒ Y •

Ca + F ′i

d) electrons (n-type doping) SiSi ⇒ As•Si + e

e) dislocationsf) oxygen vacancies with mixed valence W, i.e. WO3-δ or W 6+

1−2δW 5+2δ O3−δ

2.4 Cu is face centred cubic and therefore, a piece of metal has {111} slip planes in many orientations,permitting easy dislocation motion. W is body centred cubic and slip is a much more difficult process;there are no cp layers which would favour easy dislocation motion.

2.5 Extra lines should appear in the XRD pattern of β ′ brass. This is because β brass is body centred cubicwhereas β ′ is primitive cubic. Hence reflections which violate the condition (h + k + l) = 2n shouldappear in the β ′ pattern but be absent from the β pattern.

2.6 The law of mass action applies to the creation of defects under conditions of thermodynamic equi-librium. It is assumed that, in the limit of low defect concentrations, the equilibrium constant isindependent of defect concentration. At high defect concentrations, defect interactions are very likelyto occur, modifying K and making the predictions at lower concentrations invalid.

2.7 a) edge dislocation b) screw dislocation.

2.8 In each case, parallel to close packed directions in the crystal structure i.e. a) in the basal plane of hcpZn, [100], [010], [110], b) parallel to the face diagonals in fcc Cu i.e. <110>, c) parallel to the cubicbody diagonal in αFe i.e. <111> d) as in b), <110>.

2.9 Using Nv/N = A exp (−E/2RT)E = 2.3 × 105 J mole−1

R = 8.314T = 1023 KNv/N = 1 × 10−5

Substituting, A = 7.450Then, at 573 K, Nv/N = 2.44 × 10−8

At 298 K, Nv/N = 5.17 × 10−20

2.10 a) Ca1-3xY2xF2 : 0 < x < 0.33F wt = 40.08 (1 − 3x) + (88.91 x 2x) + 38

D = F wt × Z

Vuc × N= (78.08 + 57.58x) 4

(5.46)3 × 10−24 × 6.023 × 1023

= 3.19 + 2.35x g cm−3 : 0 < x < 0.33b) Ca1-yYF2+y : 0 < y < 1

F wt = 40.08 (1 − y) + 88.91y + 19 (2 + y)D = 3.19 + 2.77y g cm−3 : 0 < y < 1

Compare the results with Fig. 2.15; both mechanisms lead to an increase in density, but this is greaterfor b).

3.1 � ebs(Si + Al + 2Ca) = 44 + 3

6 + 2×28 = 2.

3.2 If three SiO4 tetrahedra shared a common corner, � ebs (3 Si) = 3 × 44 = 3. For electroneutrality,

� ebs = 2.

3.3 From Fig. 2.3

rBe(tet) = 0.40 A

rMg(oct) = 0.85 A


rCa(cubic) = 1.25 A

rF− = 1.19 A

r+r−

= 0.40

1.19= 0.34 for Be

0.85

1.19= 0.71 for Mg

1.25

1.19= 1.05 for Ca

From the radius ratio rules, Table 2.3, the values for BeF2 and MgF2 are consistent with tetrahedraland octahedral coordination of Be, Mg, respectively. For CaF2, the value is outside the range for 8coordination but is qualitatively reasonable. If the inverse radius ratio, r-/r+, is used instead with a valueof 0.95, this value is far from the range predicted for tetrahedral coordination of F. Hence, althoughcation coordination numbers increase with cation size, quantitative predictions using the radius rulesare not always successful.

3.4 In the rock salt structure, a = 2(rM + rO)

Compound a/A rM if rO = 1.26 A rM/rO rM if rO = 1.40 A rM/rO

MgO 4.123 0.847 0.67 0.707 0.505CaO 4.811 1.146 0.91 1.006 0.718SrO 5.160 1.32 1.05 1.18 0.84BaO 5.539 1.51 1.20 1.37 0.98TiO 4.177 0.83 0.66 0.689 0.49MnO 4.445 0.963 0.76 0.823 0.589FeO 4.307 0.894 0.71 0.754 0.538CoO 4.260 0.87 0.69 0.73 0.521NiO 4.177 0.829 0.66 0.689 0.492CdO 4.695 1.088 0.86 0.948 0.677

Many of the calculated rM/rO values are greater than the value of 0.732 at which a change to an8-coordinated crystal structure would be expected.

a/A c/A rM rM/rO rM rM/rO

ZnO 3.2495 5.2069 0.693 0.55 0.55 0.395BeO 2.698 4.38 0.38 0.30 0.24 0.173

The simplified formula, M-O = 0.375c is used. rM/rO values do not stay within the range 0.225 <

rM/rO < 0.414 expected for tetrahedral coordination.

11 Answers

a/A rM/A rM/rO rM/A rM/rO

PbO2 5.349 1.056 0.84 0.955 0.68CeO2 5.411 1.083 0.86 0.943 0.67PrO2 5.392 1.075 0.85 0.935 0.67ThO2 5.600 1.165 0.92 1.025 0.73UO2 5.372 1.066 0.85 0.926 0.66NpO2 5.4334 1.093 0.87 0.953 0.68Li2O 4.6114 0.737 0.58 0.597 0.43Na2O 5.55 1.143 0.91 1.003 0.72K2O 6.449 1.532 1.22 1.392 0.99Rb2O 6.74 1.658 1.32 1.518 1.08

For the oxides MO2, the rM/rO values are greater than the minimum value of 0.732 for 8-coordinationusing rO2− = 1.26 A but are close to this value for rO2− = 1.40 A.

For the oxides M2O, the rM/rO values are all greater than the value of 0.414 expected at the changeoverfrom 4- to 6-coordination.

Again, therefore, the radius ratio rules are not quantitatively applicable.

3.5 Substituting V = 2 (two ions/formula unit) and (Z+Z-) = 4 (product of ion charges) into Kapustinkii’sequation, with data for ra + rc = a/2:

oxide ra + rc/A U/kJ/mole U/lit. data

MgO 2.1065 3,813 3,938CaO 2.4053 3,420 3,566SrO 2.58 3,225 3,369BaO 2.7695 3,036 3,202

There are small differences in the values of U calculated from the simple Kapustinskii’s, column 3,equation and lattice energy data reported in the the literature. Nevertheless, the data are reasonablyclose and the trends are clear.

3.6 This requires a discussion of the relative importance of the different parameters that influence �Hf.For Cu, the key parameters are the second ionisation potential, for Cu+ → Cu2+, which is large andpositive and the increase in lattice energy for Cu dihalides compared to Cu monohalides, which isnegative. For F, both CuF and CuF2 have an overall negative �Hf but that of CuF2 is much largerand so CuF2 forms preferentially. For I, the reverse holds; the large size of I− leads to an increase ininteratomic distances and a decrease in lattice energy (it is arguable whether the ionic bonding modelis valid for copper iodides, but we assume so for the present purposes). Consequently, �Hf for CuI ismuch greater (more −ve) than for CuI2.

3.7 There is a general tendency for ionic structures to be most stable when anions and cations are ofcomparable size. Thus, for both alkali and alkaline earth oxides, there is an increasing trend for theperoxides (O2

2−) and superoxides (O2−) to be more stable with the larger cations and the simple oxides

(O2−) to be more stable with smaller cations. This is a size effect; an additional parameter that is likelyto be important in these cases is that the second electron affinity of oxygen (i.e. for O− → O2−) ispositive and this will act to reduce the enthalpy of formation of simple oxides.

3.8 Using data given for S, Sb may be calculated using Sb = √SNaSX for each NaX; δ may be calculated

from δ = (S −Sb)/�Sc, using tabulated �Sc data.


This gives:

NaF : δ = 0.75NaCl : δ = 0.67NaBr : δ = 0.62

NaI : δ = 0.54

where δ indicates the fractional ionic character in the NaX bond.r may be calculated for Na and X in each case from r = rc = Bδ using tabulated rc, B values. Unit

cell dimensions, a, may be calculated from a = 2(rNa + rX):-

Calculated Experimental

NaF 4.68 4.64NaCl 5.52 5.64NaBr 5.76 5.98NaI 6.10 6.47

Values of calculated and experimental data agree to within a few percentage points.

3.9 From purely size considerations, oxides MO of the 3d transition metals have the rock salt structure, asfound, with 6:6 coordination. CuO contains Cu2+ (d9); in the undistorted octahedral coordination, thetwo eg orbitals d

x2−y2 and dz2, cannot be degenerate since one contains a single electron and one contains

two electrons. Instead a structural distortion occurs, the Jahn–Teller effect, with an elongation of theoctahedron along the apical bonds. If the apical bonds are defined as being parallel to the z direction,the d

z2 orbital contains a pair of electrons and the Cu-O bond in this direction becomes elongated dueto repulsion between d

z2 pair and the Cu-O bonding electrons. The dx2−y2 orbital contains only a single

electron and so the repulsions between it, and the Cu -O bonding electrons in the x y plane are smaller.

3.10 a) As shown in the previous example, the non-bonding eg electrons on Cu2+ lead to structural dis-tortions associated with the Jahn-Teller effect. Other d9 and d4 ions may also show Jahn-Tellerdistortions.

b) Several ions show a preference for certain coordination environments associated with crystal fieldstabilisation energies which are a direct consequence of their d-electron configurations. Thus, Cr3+

and Ni2+ usually show a strong preference for octahedral coordination.

3.11 Atomic Mg has the configuration (Ne core) 3 s2. In the metallic state, the 3 s orbitals overlap to forma band of levels, as do the 3p orbitals. There is also an overlap in the energies of the 3s and 3p levels.Consequently, both 3s and 3p bands are partially occupied, allowing electrons near the Fermi energyto move freely and exhibit metallic conduction.

3.12 TiO is metallic because the t2g orbitals on adjacent Ti are able to overlap to form a t2g band which ispartially occupied. On moving across the 3d series, the 3d orbitals become more tightly located near tothe nucleus and so orbital overlap gradually reduces (so for instance, MnO is a semiconductor). WithNiO, the t2g levels are full and the eg levels do not overlap to form an eg band since the eg orbitals arelocated in the vicinity of the Ni-O bonds. Consequently, easy metallic conduction does not occur.

NiO exhibits semiconductivity only when some Ni2+ is oxidised to Ni3+, and then electrons areable to hop between adjacent Ni2+ and Ni3+ ions.

3.13 The answer to this is, in fact, not clear. One possible scenario is shown in Fig. S8; the diagram showsthe unit cell face of the ReO3/perovskite structure with W at corner and O at edge centre positions. The

13 Answers

a

aW W

WW

Figure S8

t2g orbitals (W 5d) point towards each other across the face diagonals and are able to overlap forminga 5d band which contains 0.5 electrons/W in Na0.5WO3.

It is perhaps, more likely that the W-W distances across the face diagonals are too great to permitgood overlap of the t2g orbitals and that, instead, t2g orbitals on W overlap with 2px,y,z orbitals on Oand this provides the mechanism for electrical conductivity.

3.14 Points to bring out:- mainly ionic bonding in LiF; increasing covalent character with increasing formalvalence, especially in LaN and TiC; additional metallic bonding in TiO (as shown by its good electricalconductivity); unlike the example in 3.13, the Ti-Ti distance in TiO should be short enough for t2g

orbitals on adjacent Ti to overlap.

3.15 U = 1200.5 × 2 × 4

2.12

(1 − 0.345

2.12

)

= − 3793 kJ/mole

�Hf = −602 = 148 + 498

2+ 2188 + EA − 3793

EA = 606kJ

mole

Note, this value is positive since the formation of O2− is endothermic relative to O.

4.1 Methods are described in Chapter 4. Advantages, briefly are i) less energy intensive, ii) better producthomogeneity when conditions are optimised, iii) versatile method, especially for synthesis of phasesthat are not thermodynamically stable. Disadvantages are i) often, high cost of chemicals, ii) oftenmuch work needed to optimise synthesis conditions, iii) methods developed for one material may notbe immediately transferable to the synthesis of other materials, iv) reactions are often small-scale andscale-up can be difficult.

4.2 LiTiS2 : reaction of TiS2 with n-butyl Li.Li0.5CoO2 : electrochemical extraction of Li from LiCoO2.BaTiO3 : several routes, including sol–gel, preparation of catecholate intermediates and formation of

mixed alkoxide intermediates.CoFe2O4 : several routes, including sol–gel, heating of aqueous nitrate mixtures and preparation of

pyridinate intermediates.

4.3 Briefly, the key stages are nucleation and growth. Nucleation requires all the reacting species to bepresent at the nucleation site; this is facilitated by fine-scale mixing and frequent regrinding in solid statesyntheses. Growth can be very slow, especially as counter-diffusion of reactants through a thickeningproduct layer is needed. Growth rates increase with temperature and with presence of crystal defects.

4.4 See Chapter 4.


4.5 According to the phase diagram for carbon, diamond is thermodynamically metastable under ambientconditions. It is, however, kinetically stable once formed. Traditional HPHT routes follow the con-straints imposed by the phase diagram. In low temperature gas-phase routes, it is not necessary to trans-form graphite into diamond since the starting materials are carbon-containing molecules and radicals.

4.6 Whilst pure crystalline Si becomes either a) p-type or b) n-type on doping, pure amorphous Si containsa number of broken Si-Si bonds, leading to unsatisfied bonding on Si and the dopants merely mop upthese unpaired electrons. Hydrogenated amorphous Si, by contrast, has numerous Si-H bonds insteadof unpaired electrons and such material can be doped to make it p-type or n-type.

4.7 TiS2 has a layered, or sandwich structure, similar to that of CdI2 and CdCl2. Sandwiches of S-Ti-Slayers are separated from adjacent sandwiches by van der Waals bonding and Li+ ions are able tointercalate into this space, leading to LiTiS2. The Li can be added and extracted reversibly, hencegiving rise to its behaviour as a cathode in rechargeable Li batteries.

Unlikely; NiS does not have a layered structure (it is likely to have either a rock salt or a NiAs struc-ture) and the interstitial sites are probably too small to accommodate Li ions. In addition, intercalationof Li would require reduction of Ni2+ to Ni+, which is not a common oxidation state of Ni.

4.8 A range of liquid-based low temperature routes. Alternatively, gas-phase routes provided suitableprecursors can be identified.

4.9 The silicate and silicate-derivative anions in the reaction mixture crystallise around the templatemolecules or ions. Hence the template influences greatly the size and shape of the cavities in thezeolite structure.

5.1 To verify Moseley’s law, a plot of λ−1/2 vs Z should be linear, Fig. S9.

20

0.6

0.8

1.0

1.2

1.4

30

Z

λ-½

40 50

Figure S9

Element λ/A λ−1/2 Z

Cr 2.2896 0.66 24Fe 1.936 0.72 26Cu 1.5405 0.80 29Mo 0.7093 1.19 42Ag 0.5594 1.34 47

By interpolation, Co (Z = 27) should have λKα1 of ∼1.78 A.

15 Answers

5.2 a) body centred, I (h + k + l = 2n).b) face centred, F (h k l all odd or all even).c) primitive, P (no conditions).d) C-centred (h + k = 2n).

5.3 Using 1d2 = (h2 + k2 + l2)/a2, λ = 2d sin θ and λ = 1.54 A

h k l d/A 2θ /◦

1 1 1 2.887 30.942 0 0 2.500 35.88

5.4 These may be observed but the lattice planes for diffraction are relabelled. In general the nth orderreflection from a set of planes (hkl) is treated as 1st order (n = 1) from the set of planes (nh nk nl).Thus, the number of planes in the set is increased n-fold and their d-spacing is reduced according to1/n. For example, the second order diffraction from (110) is treated as the first order diffraction from(220).

5.5 Using 1d2 = (h2 + k2 + l2)/a2, we look for a pattern in the values of 1

d2

d/A 1/d2 (h2 + k2 + l2) h k l

4.08 0.060 3 1 1 13.53 0.080 4 2 0 02.50 0.160 8 2 2 02.13 0.220 11 3 1 12.04 0.240 12 2 2 21.77 0.319 16 4 0 0

Since (h2 + k2 + l2) is an integer and 1/a2 is a constant, then (h2 + k2 + l2)/a2 ≡ 1/d2 must increasein steps. The smallest increment between adjacent 1/d2 values is 0.02 which should correspond to(h2 + k2 + l2) = 1 e.g. between 1 1 1 and 2 0 0 (i.e. (h2 + k2 + l2) = 3 and 4 respectively). We,therefore, tentatively assign the first line to (h2 + k2 + l2)/a2 = 3/a2. The whole pattern can then beindexed, as shown in columns 3 and 4.

From d200 = 3.53 A, a = 7.06 A.The indices indicate a face centred structure, i.e. Z = 4.Density is given by D = FW × Z

Vcell × N

Substituting:- D = 3.126 g cm−3, Z = 4, V = (7.06 × 10−8)3 A3 and N = 6.023 × 1023 givesFW = 165.6. Considering possible formula weights for the various alkali halide combinations leads toeither KI (FW = 166) or RbBr (FW = 165.36). Further information would be required to distinguishbetween these two possibilities, such as checking the value of a against literature values. As a matterof interest, the answer is K1 (from Table 1.8).

5.6 The origin coincides with a centre of symmetry and, therefore, we may use the simplified equation forF (i.e. no sine terms)

F = �

jfj cos 2π (hxj + kyj + lzj)


Substituting for the two atoms at 000 and 1/21/20 gives

F = fj

[cos 2π.0 + cos 2π

(h

2+ k

2+ 0

)]

= fj [1 + cos π (h + k)]

If (h + k) = 2n, then F = 2fj OR if(h + k) = 2n + 1, then F = 0.

Thus a reflection can be observed (F = 0) only if (h + k) = 2n. This same condition applies to anystructure that is C-centred since, for every atom with coordinates xyz there is an identical atom at x +1/2 y + 1/2 z.

5.7 The structure has a centre of symmetry centred at Ti. We may therefore use the coordinates given andthe cosine-only formula for F.

F = fTi cos π (h.0 + k.0 + l.0) + fSr cos 2π

(h

2+ k

2+ l

2

)

+ f0

[cos 2π

(h

2+ k.0 + l.0

)+ cos 2π

(h.0 + k

2+ l.0

)+ cos 2π

(h.0 + k.0 + l

2

)]

= fTi + fSr cos π (h + k + l) + f0 [cos hπ + cos kπ + cos lπ]

5.8 In the rock salt crystal structure of both KF and KCl, the (111) planes coincide with cp layers of anions(choosing the origin to coincide with an anion). The cations occupy octahedral sites midway betweenthe (111) planes. For diffraction from (111), anions and cations are out of phase. Since K+ and Cl− areisoelectronic, the resultant intensity is zero for KCl. In general, for the rock salt structure, the intensityof (111) depends on the different in form factor of the two ions in the structure.

5.9 There are two possible consequences. (1) If the unit cell of the ordered form is smaller, then all lineswill shift to lower d-spacing/higher 2θ . (2) Although both structures are cubic, the systematic absenceswill differ.a) Those for the disordered structure are: h, k, l should be all odd or all even for observed

relections.b) It is necessary to determine the lattice type for the ordered structure. It is centred but on one pair

of cell faces only; e.g. it is C-centred if Cu is at 000, 1/21/20. The Au atoms also form a C-centred

array since atoms at 1/201/2 are related by the C-centring transformation: xyz → x + 1/2 y + 1/2 z.Hence, extra reflections will appear in the ordered form that violate the F condition, but obey the Ccondition, i.e. reflections for which k + l = 2n + 1 and h + l = 2n +1 may be observed in the orderedform but not in the disordered form.

5.10 d002 = 4.68 A, ∴ c = 9.36 A

for 101, 1d2 = 1

a2 + 1c2 ∴ a = 3.74 A

for 112, 1d2 = 1

a2 + 1b2 + 4

c2 ∴ b = 2.97 A

D = 4.87 = 152.3 × Z

(9.36 × 3.74 × 2.97) × 10−24 × 6.023 × 1023

∴ Z = 2.00

17 Answers

5.11 a) Z = 1, V = (4.059 × 10−8)3 cm3, D = 2.431 g cm−3

∴ FW = 2.431 × (4.059 × 10−8)3 × 1023

= 97.9∴ the halide has atomic weight 97.9 −18 = 79.9 i.e. Br

b) i) Primitive lattice since Z = 1; therefore:d100 = 4.059 Ad110 = 2.870 Ad111 = 2.343 Ad200 = 2.030 A

ii) Face centred lattice for the rock salt polymorph with Z = 4; therefore:d111 = 3.965 Ad200 = 3.434 Ad220 = 2.428 Ad311 = 2.070 A

c) volumes per formula unit:-primitive = (4.059)3 = 66.87 Af cc = (6.867)3/4 = 80.95hence the primitive form has 82.6% the molar volume of the face centred form.

d) i) primitive form: ions in contact along cube body diagonals.

(rNH4 + rBr) = 12 a2

√3 = 3.515 A ∴ rBr = 2.015 A.

Since the cell edge, a = 4.059 A, is only slightly greater than 2rBr (4.030 A), anions are almostin contact.

ii) face centred form: ions in contact along cell edge.

(rNH4 + rBr) = 12 a = 3.434 A ∴ rBr− = 1.934 A.

The unit cell face diagonal√

2 a = 9.17 A, is much greater than the radii of four Br− ions, 7.74 A,and therefore Br− ions are not in contact.

5.12 Key points:- Powder patterns are characterised by positions and intensities of lines. Positions aregoverned by unit cell dimensions; therefore if isostructural substances have different cell dimensions,their patterns must differ. Intensities are governed by atomic coordinates and atomic scattering factors.Therefore, if two substances have the same structure, their intensities will differ if the constituentatoms differ significantly in atomic number.

5.13 First, although all three structures are face centred cubic, their cell dimension will differ and hencepeaks occur at different d-spacings. For NaCl, X-ray and neutron patterns will give peaks at similarpositions but with very different intensities as the scattering powers of the atoms are different towardsneutrons and X-rays.

Similar comments apply to Fe.NiO is, however, antiferromagnetic (afm) giving an ordered superstructure which is detected by

neutrons but not by X-rays. Hence, an extra set of lines will appear in the neutron diffraction pattern.If we ignore the small rhombohedral distortion in afm NiO, the subcell and supercell are related:

asupercell = 2asubcell. The supercell is fcc (as is the subcell). Subcell lines need to be reindexed bydoubling their indices e.g. (111)subcell → (222)supercell. New peaks associated exclusively with thesupercell appear e.g. (111)supercell, (200)supercell.


5.14 Note: intensities are arbitrary, Fig. S10. In a) separate peaks for NaCl and AgCl are expected sincethey have the same rock salt structure but different cell dimensions. In b) a homogeneous solid solutionforms whose unit cell, and hence d-spacings, is intermediate between those of NaCl and AgCl.

(200)

Mixture

2θ

Solidsolution

(220)

(111)

(311)

(222)

Figure S10

5.15 a) Separate peaks associated with Fe(OH)3 should be seen, although of much smaller intensity thanthose for the Al(OH)3 main phase.

b) If a solid solution forms with partial replacement of Al by Fe, then a small change in unit celldimensions and hence in position of the lines of the Al(OH)3 phase in its powder XRD patternwould be expected due to the different sizes of Al3+ and Fe3+.

6.1 i) For optical microscopy, the impurities must be present as separate phases and with particle sizes ofat least 1 μm. There is no lower number/concentration detection limit:- one impurity particle can, inprinciple, be detected. Impurities present as dopants/solid solutions would be detectable by opticalmicroscopy only if the optical properties (e.g. refractive index) changed as a consequence.

ii) Impurity phases can be detected by XRD but a concentration of 3−5% is usually needed for themto be readily detected. As for optical microscopy, dopant/solid solution impurities are detectableonly if there is a change in properties, i.e. unit cell dimensions.

6.2 Both substances are crystalline and could be identified by powder X-ray diffraction, assuming thattheir data are included in the ICDD File of reference powder patterns. Since CHI3 is a molecularsubstance, the various spectroscopic techniques – NMR, MS and IR – could be used for a fullstructural characterisation, whereas these would be of limited use for CdI2.

6.3 a) Isotropic; dark in crossed polars; irregular-shaped fragments.b) Hexagonal symmetry, therefore bright in crossed polars unless crystals are oriented with c axis

parallel to the transmitted light beam [i.e. the crystal optic axis is parallel to c]. Crystals shouldhave well-defined shapes; refractive index may vary with crystal orientation.

c) Isotropic; dark in crossed polars; well-defined shapes, probably cubes but could be octahedra.d) Opaque to transmitted light.

6.4 Substituting into λ = h(2 meV)−1/2

h = 6.63 × 10−34 Jsm = 9.11 × 10−31 kge = 1.602 × 10−19 C

19 Answers

V = 5 × 104 Vgives λ = 5.49 × 10−12 m

= 0.055 A

6.5 Key points are as follows:a) Extended X-ray Absorption Fine Structure.

Refers to the absorption of X-rays by a sample in the region of, but to the high energy side of,an absorption edge. For instance, in Cu-containing samples, the CuKα edge (for ionisation of 1selectrons) occurs at ∼8.9 keV or 1.38 A. The EXAFS region for CuKα covers the range ∼9 to9.5 keV. From analysis of EXAFS, information on local coordination environments (in this casearound Cu) may be obtained.

b) X-ray Absorption Near Edge Structure.Refers to the absorption of X-rays by a sample in the region of, and close to, an absorption edge.

Thus, for CuKα, the absorption threshold (or onset) and the detailed shape of the absorption edgeare sensitive to the oxidation state of Cu and also to its environment.

c) Electron Energy Loss Spectroscopy.Refers to the residual energy of incident electrons after they have bombarded a sample and caused

inner shell ionisations, i.e. EEELS = hν − Eb, where hν is the energy of the incident electrons andEb is the binding energy of the ionised electrons. EELS is a complementary technique to electronmicroscopy and can be used to detect light elements in particular and provide a mapping of theirdistribution throughout a sample.

d) X-Ray Fluorescence Spectroscopy.Refers to the spectrum of X-rays emitted by a sample on bombardment by a high energy electron

beam [as in an X-ray tube used to generate X-rays]. The spectra, caused by transitions such as2p → 1s (Kα) and 3p → 1s (Kβ), have wavelengths characteristic of the elements concernedand, therefore, form the basis of an analytical technique for determining the concentration of suchelements present in a sample.

e) Electron Spectroscopy for Chemical Analysis.Another technique associated with the bombardment of a sample by high energy radia-

tion/electrons. If all the incident radiation is used to a) ionise an electron from a particular shell andb) give it some kinetic energy, then the following relation holds

Ehv = Eb + Ekin

Clearly, the incident radiation must have monochromatic energy and the energy of the ionisedelectrons may then be used for a range of analytical purposes. If the incident radiation is highenergy X-rays, the technique is also known as:

f) X-ray Photoelectron Spectroscopy and is used to study inner-shell electrons; if the radiation is lowerenergy UV light, it is also known as:

g) Ultraviolet Photoelectron Spectroscopy and is used to study bonding or outer shell electrons.h) Magic Angle Spinning Nuclear Magnetic Resonance Spectroscopy.

Refers to the application of the NMR technique to solid samples. Conventional NMR on solidsamples gives broad, featureless absorption peaks. In the MAS modification, the sample is spun athigh velocity at a critical angle to the applied magnetic field. This causes the broad absorption bandto collapse and reveal a number of sharp peaks, as in liquid state NMR, from which much usefulstructural information may be obtained.

i) Scanning Electron Microscopy, andj) Transmission Electron Microscopy.

The key difference is that SEM uses a reflection geometry technique for looking at the surfacesof thick samples, whereas TEM is a transmission technique requiring very thin samples.


k) Analytical Electron Microscopy,l) Energy Dispersive analysis of X-rays, and

m) Electron Probe MicroAnalysis.These operate on the same principles as XRF and refer to the X-ray spectra generated on examin-

ation of a sample in an electron microscope. They may be used for both qualitative, semiquantitative(EDX) and quantitative (EPMA) analysis.

6.6 a) UV/ vis spectroscopy should give characteristic absorption bands; alternatively XANES for oxida-tion state and EXAFS for coordination number (and bond distances).

b) various methods, as in a); esr may be useful.c) surface-sensitive techniques required, such as XPS/UPS for composition, SEM for texture.d) SEM of fracture surfaces/polished sections; TEM of thin sections for diffraction studies;

AEM/EDX/EPMA for elemental analysis.e) AEM/EDX if particle size < 1 μm.f) powder neutron diffraction.g) powder neutron diffraction.

6.7 (i)a 4 × 10−5 eV, (i)b 0.004 kJmol−1, (i)c 0.3 cm−1, (i)d 1010 Hz(ii)a 1 eV, (ii)b 100 kJmol−1, (ii)c 104 cm−1, (ii)d 3 × 1014 Hz(iii)a 9000 eV, (iii)b 9 × 105 kJmol−1, (iii)c 7.5 × 107 cm−1, (iii)d 2 × 1018 Hz(iv)a 0.01 eV, (iv)b 1 kJmol−1, (iv)c 80 cm−1, (iv)d 2.5 × 1012 Hz

6.8 a) SEM or reflected light microscopy (depending on scale).b) Transmission optical microscopy.c) TEM/HREM.d) as in b).

6.9 The spectroscopic techniques IR, NMR and MS together give functional groups, numbers of atomicneighbours and molecular weight/mass fragments. Together these usually yield formulae and structuresand are applicable to most molecular substances.

These techniques are of limited used for non-molecular solids. For the latter, a wide range ofspectroscopic, diffraction and microscopic techniques are required to fully characterise them.

6.10 Even though the substance is new, its powder XRD pattern may be very similar to that of other knownperovskites. Hence, treat the substance as an unknown, record its powder XRD pattern and carry outa search using the ICDD (or JCPDS) Powder Diffraction File.

To confirm that it has a perovskite structure, calculation of the intensities of an XRD pattern wouldbe needed. This could be done with a calculator for a simple structure such as perovskite. For a fulleranalysis, Rietveld refinement of XRD (or neutron diffraction) data should be used.

For compositional analysis, a bulk chemical analysis could be used only if the sample were phase-pure. Various methods could be used: XRF on solids or AA/ICPMS on solutions of the sample. Foranalysis of individual grains in a mixture, EPMA can be used, provided the grain size is ≥ 1 μm. Forsmaller grains EDX is necessary.

To estimate purity, microscopic techniques – SEM/TEM/EPMA/optical microscopy – are needed.XRD can be used if ≥ 3% of the impurity phase(s) is present.

6.11 a) DTA: endotherm at MPt; TG: no changes.b) DTA: small heat anomaly at Tg; TG: no changes.c) DTA: endotherm at MPt, 800 ◦C; TG: no changes.d) DTA: heat effects associated with decomposition; endotherm at MPt of Na2CO3; TG: weight loss

due to loss of CO2/H2O.e) MgSO4 × H2O: DTA: endotherm due to loss of H2O, second endotherm on melting of MgSO4;

TG: weight loss on dehydration.

21 Answers

f) DTA: endotherm on melting; TG: no changes unless some oxidation occurred on heating.g) DTA: endotherm at 120 ◦C at tetragonal to cubic phase transition; second endotherm at ∼1400 ◦C on

transformation to hexagonal structure; third endotherm on melting at > 1500 ◦C; TG: no changes.h) First, DTA: endotherm and TG: weight loss both attributable to dehydration; no further changes by

TG, but DTA: possible exotherm on recrystallization; endotherm(s) on melting at higher tempera-ture.

6.12 a) Rapid reversible melting/crystallisationb) irreversible since carbonation of CaO on cooling would proceed only slowlyc) may be reversible unless i) the sequence of changes on melting is complex or ii) the resulting liquid

formed a glass on coolingd) irreversible; to convert MgO back to Mg, a highly reducing atmosphere would be needede) could be reversible as CaO, formed by dehydration of Ca(OH)2, is hygroscopic

6.13 Pure ice should give a DTA endotherm on heating at 0 ◦C. With added salt, the endotherm should bedisplaced to lower temperatures: there is a eutectic in the H2O-NaCl system at ∼−20 ◦C.

6.14 See Fig. S11.

(a)

Liquidα γ δ

(b)

α + Fe3C γ ss

Liquid

Figure S11

a) Should show the transition sequence:

α → γ → δ → liquid

b) should show transformation of the (α + Fe3C) mixture to γ ss at the eutectoid temperature, followedby melting at some higher temperature.

6.15 A very open-ended question; I shall not attempt a model answer!

7.1 a) Al6Si2O13 can be rewritten as an oxide ratio, 3Al2O3.2SiO2

The mole % Al2O3 is therefore 33+2 × 100 = 60%

The formula weights are Al2O3: 101.96, SiO2: 60.09Therefore, the wt% Al2O3 = 3 × 101.96

(3 × 101.96) + (2 × 60.09) × 100 = 71.79%b) Na2Ca3Si6O16 ≡ Na2O • 3CaO • 6SiO2

mole % Na2O = 1

(1 + 3 + 6)× 100 = 10%

wt% Na2O = (1 × 61.98) × 100

(1 × 61.98) + (3 × 56.08) + (6 × 60.09)= 5.21%


c) Y3Fe5O12 ≡ 11/2 Y2O3 • 21/2 Fe2O3

mole % Y2O3 = 1 12 × 100

1 12 + 2 1

2

= 37.5%

wt% Y2O3 =(1 1

2 × 225.81) × 100(

1 12 × 225.81

) + (2 12 × 159.69)

= 45.90%

7.2 wt% B = (mole % B × F wt B) × 100

(mole % B × F wt B) + (mole % A × F wt A)

mole % B = (wt% B/F wt B) × 100

(wt% B/F wt B) + (wt% A/F wt A)

7.3 Ca3SiO5 = 3CaO • SiO2 = 25 mole % SiO2

Ca2SiO4 = 2CaO • SiO2 = 33.3 mole % SiO2

Ca3Si2O7 = 3CaO • 2SiO2 = 40 mole % SiO2

CaSiO3 = CaO • SiO2 = 50 mole % SiO2

7.4 Na3PO4 = 3Na2O • P2O5 = 25 mole % P2O5

Na4P2O7 = 2Na2O • P2O5 = 33.3 mole % P2O5

Na5P3O10 = 5Na2O • 3P2O5 = 37.5 mole % P2O5

NaPO3 = Na2O • P2O5 = 50 mole % P2O

7.5 See diagram, S12.

A + A2BAB +AB2

A2B +AB

A2B +Liquid

Liquid

AB2+ liq

B + liq

AB+ liq

AB2 + B

A2B + AB2

A +liq

A2B AB2 BABA

Figure S12




23 Answers

Liquid

Al2O3+

Liquid

Al2O3+

Al6Si2O13

Al6Si2O13 + SiO2

SiO2 +Liquid

Al6Si2O13+ Liquid

1595

1500

1600

1700

1800

1900

2000

°C

20 60 80 SiO2Al6Si2O13Al2O3

Figure S13

400

600

800

1000

1200

1400Liquid

Na2O

Key 3:1 = Na3NbO4 1:1 = NaNbO3

1:4 = Na2Nb8O211:10 = Na2Nb20O51

Na2O+ liq

°C

3:1 + liq γ1:1 + liq

γ1:1 + 1:4

β1:1 + 1:4

α1:1 + 1:4

1:4+

1:10

3:1+

γ1:1

3:1+

β1:1

3:1 +α1:1

1:4 + liq

1:10 + liq

Nb2O5 + liq

1:10+

Nb2O5Na2O

+3:1

3:1 1:1

The three polymorphs of 1:1 are labelled as α, β, γ

1:4 1:10 Nb2O5

Figure S14


αss + Fe3C

723

910

e

γss + Fe3C

γss

αss

γss +

αss

Fe % C

Figure S15

You are given no information on solid solution limits or on other phases that form. Nevertheless, youshould know that a eutectoid, e, forms at 723 ◦C.You are also told nothing both the high temperature δ polymorph.Compare your diagram with Fig. 7.22.General features should be similar.


A B

Liquid

Figure S16

Since you are given no information on melting temperatures or solid solution limits, the diagram isonly schematic. It will probably be very different to that for Mg2SiO4-Zn2SiO4 as regards the shapesof the different regions, but the general features should be similar.a) Heat mixtures of different composition under different conditions (temperature, time) until equi-

librium is reached (this is difficult to ensure, for certain; at least, heat the samples until no furtherchanges occur on prolonged heating). Quench samples to room temperature (this assumes that nochanges occur during cooling and you may determine by analysis at room temperature the phases

25 Answers

that were in equilibrium at the temperature of heating). Analyse the products by X-ray diffrac-tion/microscopy. In particular, identify phase(s) present using powder XRD. If solid solutions areevident, measure lattice parameters (unit cell dimensions) as a function of composition; these usu-ally change linearly or not at all with composition. If the former, a calibration graph of latticeparameter against composition may be constructed and used to determine the composition of othersolid solutions, or to estimate solid solution limits. By microscopy (optical, electron), determinewhether products are single phase or mixtures. If EDX/EPMA is available, compositional analysisof the individual phases may be performed.

b) By density measurements: for different plausible solid solution mechanisms, the average unit cellcontents and hence the F wt will change; density measurements may therefore distinguish these. Bycrystallographic/spectroscopic studies, especially Rietveld refinement of powder XRD or neutrondiffraction data to determine site occupancies and infer defect structures.

c) Thermal analysis, especially DTA or DSC; should give endotherm on heating and reversibleexotherm on cooling.

7.10 A phase is a physically distinct, homogeneous compound or structure or material. For example, themagnesium silicates Mg2SiO4 and MgSiO3 are separate phases since they have fixed compositions anddifferent structures. MgO and SiO2 are also separate phases. Phases can have variable composition:on most phase diagrams, a complete range of compositions forms a single, homogeneous liquid, orsolution phase. In some cases, however, liquid immiscibility occurs; inside an immiscibility dome, amixture of two liquid phases exists. For example, oil and water are immiscible liquids, and thereforeseparate phases, at room temperature. Polymorphs of the same composition are also separate phases,e.g. the different polymorphs of SiO2: i.e. quartz, cristobalite, etc.

A component is an end-member composition of a phase diagram. To describe the composition of aphase, it is necessary to specify the components and their amounts. Thus, both MgSiO3 and Mg2SiO4

are made from two components MgO and SiO2.In the MgO-SiO2 system, MgO and SiO2 are the components, but they are also phases that appear

over certain composition ranges.In phase diagrams, one usually looks for the smallest number of components to describe a diagram

and the phases that form. Thus, magnesium silicates could be described in terms of three components,Mg, Si and O but it is simpler to use the two components MgO and SiO2: all the phases that formcan be represented by positive quantities of MgO and SiO2. The same would not necessarily be trueof iron silicates; although Fe2SiO4 contains Fe in only the 2+ state, both Fe2+ and Fe3+ are equallycommon and the oxide FeO is, in fact, a complex material of variable Fe:O ratio. To adequately describeiron silicates and their equilibria, it would be necessary to use the 3-component or ternary system,Fe-Si-O.

7.11 The liquidus temperature for this composition is ∼2120 ◦C. On cooling, spinel solid solutionof composition ∼55% Al2O3 starts to crystallise. On cooling slowly to the solidus temperature,∼2080 ◦C, more spinel crystallises but its composition becomes gradually richer in Al2O3 until,at ∼2080 ◦C, the spinel solid solution has composition 60% Al2O3 and all the remaining liquiddisappears. This solid solution is then stable as a single phase until the temperature drops below∼1350 ◦C, at which point, corundum should start to precipitate. With further cooling, more corundumprecipitates and the spinel solid solution gradually becomes more MgO-rich. At 1000 ◦C, the spinelsolid solution has composition ∼53% Al2O3.

On rapid cooling, fractional crystallisation of the liquid is likely. Thus, the first spinel nuclei to format ∼2120 ◦C have composition ∼55% Al2O3. These do not re-equilibriate on rapid cooling, howeverand instead, a range of increasingly Al2O3-rich solid solutions crystallise as the temperature drops.


8.1 The purpose of this question is to clarify nomenclature. Although ethylene has a carbon-carbon doublebond, polyethylene does not. This is a saturated molecule CH3-(CH2)n-CH3. Acetylene by contrast hasa triple bond and in polyacetylene the structure has an alternating single bond–double bond sequencein the carbon chain; it is this sequence of alternate single and double bonds that gives the possibilityof electrical conductivity in polyacetylene.

8.2 a) To make polypyrrole n-type, it is necessary to add a species which is likely to behave as an electrondonor, and therefore, a good possibility would be an alkal metal giving nominally Na+ polypyrrole−.b) For this, an electron acceptor is needed, for example, a halogen, giving polypyrrole+ Br−.

8.3 For simplicity, chose a scale of a = b= 4 cm, c = 12 cm. This should give fairly accurate drawings.See Fig. S17.

Ba

Y

Ba

0,1

0,1

0,1

0,1

1/2

1/2

1/2

1/2

1/2

a

(a)

1/2

Ba

Ycc

Ba

0,1

0,1

0,1

0,1

0,1

1/2

1/2

b

(b)

Figure S17

Cu(1)–O(1) = 0.16c = 1.87 ACu(1)–O(4) = 0.5b = 1.94 ACu(2)–O(1) = 0.19c = 2.22 ACu(2)–O(2) = [(0.03c)2 + (0.5a)2]1/2 = 1.94 ACu(2)–O(3) = [(0.03c)2 + (0.5b)2]1/2 = 1.97 ABa–O(1) = [(0.5a)2 + (0.5b)2 + (0.02c)2]1/2 = 2.73 ABa–O(2) = [(0.5b)2 + (0.02c)2] = 3.04 ABa–O(3) = [(0.5a)2 + (0.2c)2] = 3.02 ABa–O(4) = [(0.5a)2 + (0.18c)2] = 2.84 AY–O(2) = [(0.5b)2 + (0.12c)2] = 2.39 AY–O(3) = [(0.5a)2 + (0.12c)2] = 2.37 A

8.4 Since silicon is tetravalent, additives to make it p-type would be ideally trivalent. For example, Alor Ga. In this case it is possible to regard the bonding as donation of an electron from silicon to thetrivalent element leaving a positive hold on silicon. See Fig. S18.Probability = exp(−E/RT)E = 0.01 eV = 1000 J/moleProbability = 0.68

27 Answers

C.B.

V.B.

C.B.

V.B.0.01eV

1.1eV

Figure S18

[Impurity] = 1 × 10−4 atom %= 1 × 10−4 × 10−2 × 6.023 × 1023 mole−1

= 6.023 × 1017 mole−1

∴ Carrier density = 0.68 × 6.023 × 1017 = 4 × 1017 mole−1

Assume Nintrinsic/N = exp(−1.1 × 105/RT) = 5.22 × 10−20

∴ Intrinsic carrier density = 3.14 × 104 mole−1 at 298 KT at which carrier density = 4 × 1017 = 930 K

8.5 a) 0.68 b) 4 × 1017 mole−1 c) 3.23 × 1011 mole−1 d) 592 K

8.6 This is to ensure that the material under operating conditions is in the extrinsic region in which carrierconcentration is temperature-independent. At lower temperatures, the exhaustion region is found atwhich the mobile carriers become increasingly trapped; therefore, a very shallow trapping energy willdisplace the exhaustion region to lower temperatures. By contrast, at higher temperatures the intrinsicregion involves promotion of electrons across the band gap and in large band gap materials this will notbe a significant process until very high temperatures; hence, using those two criteria, the temperaturerange over which the extrinsic region is observed is as wide as possible.

8.7 a) The intrinsic defect concentration, n, is equal to both [e] and [h].Therefore [e] = [h] = 1 × 1010 cm−3 in undoped Si.Here, Si is B-doped and is p-type with [h] = 5 × 1018 cm−3

From K = [e][h] = [n]2 = 1 × 1020

[e] = 1×1020

5×1018 = 20 cm−3

b) Conductivity is given by : σ = neeμe + nheμh

Since ne �� nh, the equation for σ reduces to: σ = nheμh

nh = 5 × 1018 cm−3; μh = 0.05 m2 V−1 sec−1

therefore σ = 5 × 1018 × 1.602 × 10−19 × 0.05= 4 × 10−2 ohm−1 cm−1

8.8 a) NaCl is an ionic conductor, primarily a sodium ion conductor, in which the conductivity dependsupon the number of sodium vacancies and secondly, a chloride ion conductor associated withchloride vacancies.i) No change, ii) no change, iii) assuming that CaCl2 dissolves in sodium chloride, the substitutionmechanism would be two sodiums replaced by one calcium and one sodium vacancy. Hence thesodium ion conductivity should increase as a consequence. iv) No change, v) if a solid solutionforms, it is unlikely that interstitial sodium ions are created and instead, the mechanism is likelyto involve anion vacancies, i.e. 2Cl− → O2− + VCl. Hence the conductivity due to chloride ionsshould increase. Since the conductivity of sodium chloride is dominated by sodium ion conductivity,little effect on the overall conductivity will be seen until the chloride ion conductivity becomescomparable to that of the sodium ion conductivity.

b) The conductivity of silver chloride is predominantly attributed to silver ions in interstitial sites andsecondly to silver ion vacancies. Therefore, i) no change, ii) the silver ion conductivity will increase


due to the replacement mechanism: 2Ag+ → Zn2+ + VAg, iii) the conductivity will increase dueto an increase in the interstitial silver concentration. The replacement mechanism is ClCl → OCl +Agi.

8.9 Region II:Slope = −2.9/(0.8 × 10−3) = −3.63 × 103

= Em log10 e/REm = 0.73 eV

Region I:Slope = −0.97 × 104

= (Em + Ef/2) log10 e/REf = 2.46 eV

Notes: 1. There is considerable uncertainty since the data in Fig. 8.18 do not fall on straight lines. 2.Although the data are presented as log σ T vs T−1, similar slopes are obtained if the data are plottedas log σ vs T−1.

8.10 There are no simple, generally applicable methods. The two main approaches involve dc measurementswith different electrode combinations and emf measurements with different activities of a particularspecies at the two electrodes.a) With metal electrodes, it should be possible to pass a continuous dc current. Sometimes, however,

Schottky barriers form at the electrode-sample interface due to the difference in Fermi level in theelectrode and the sample and these lead to an increase in effective resistance of the electrode-sampleassembly. The measurement of a dc current does not prove that the carriers are electrons (or holes);if the carriers are ions and the electrodes are ‘non-blocking’ then a dc current may pass. This isparticularly important if the charge carrier is the oxide ion.

b) Use of dc measurement with Na metal electrodes will not distinguish between conduction ofelectrons and Na+ ions, but will eliminate the possibility of conduction by other ions. Subsequentmeasurements with other metal electrodes should distinguish between the two carriers.

c) The definitive test would involve measuring the open circuit emf of a concentration cell, in whichthe electrode compartments contain oxygen gas at different oxygen partial pressures. Comparisonbetween the observed emf and that expected for oxide ion conduction from the Nernst equationgives a measure of the transport number of oxide ions.

8.11

Ion Slope E/eV log AT

Na −8.6 × 102 0.17 3.6Ag −8.6 × 102 0.17 3.4K −1.55 × 103 0.31 3.7Tl −2 × 103 0.40 3.0

8.12 a) At 200 ◦C, the S-containing cathode compartment would remain partially liquid over the range 0to ∼25% Na, but would increasingly solidify during further discharge.

b) At 300 ◦C, the cathode compartment would be completely liquid up to ∼40% Na and would thengradually solidify over the range 40–50% Na. At 300 ◦C, therefore, the cell would permit a muchgreater depth of discharge and the discharge rate would be much greater.

8.13 Higher energy/weight capacity favours Na/S. Disadvantages are the high temperature of operation,giving a need for suitable thermal insulation, potential hazards in the case of accident with liquid Na(although not as hazardous as a tank full of petrol!), and questions over the reliability of Na/S oncharging and discharging for many hundreds of cycles.

29 Answers

8.14 A fluorine concentration cell in which one electrode compartment contains fluorine at a standard activityand the other contains the fluorine to be measured. Note: there are likely to be severe design/safetyproblems with containment of F2 gas.

8.15 Paraelectric materials are polarisable to some extent but do not exhibit a permanent electrical dipolemoment on poling. Ferro- and ferri-materials do exhibit an overall dipole moment which can bereversed/modified in an applied electric field, exhibiting hysteresis effects. In ferri-materials somedipoles are aligned antiparallel, reducing the magnitude of the net dipole moment. In antiferro-materialsthere are equal numbers of parallel and antiparallel dipoles, giving no net overall dipole moment.

8.16 a) 1 (similar to a vacuum); b) 81; c) less than for water, but if any surface conduction of H+ or OH− ionsoccurs, anomalously high capacitances associates with electrode polarisation, may occur; d) small;e) 5–10; f) greater than for pure KCl due to K+ ion mobility, associated with K+ vacancies, giving riseto grain boundary and/or electrode polarisation capacitances; g) 106–107, due to formation of largedouble layer capacitances at electrodes, associated with high concentration of mobile Na+ ions; h) 104.

8.17 Minimum requirement for piezoelectricity is a non-centric crystal structure: a) No, b) No, c) No,d) Yes, e) No, f) No.

8.18 Not strictly true. Pyroelectric substances always display a spontaneous polarisation but the amountof polarisation may be temperature dependent. Pyroelectric substances containing absorbed surfacelayers may have their charges effectively neutralised and, if these absorbed materials are removed onheating, then the polarisation appears to develop.

8.19 Dielectric losses are associated with leakage conductivity in insulators. It is very difficult to eliminatethis completely since defects, impurities and absorbed surface species can all give rise to residualelectron hopping. High permittivity solids, involving some ionic/electronic charge displacement arealso likely to show significant dielectric losses and it is an interesting question as to whether dielectricloss is unavoidable, especially at microwave frequencies, in high permittivity materials.

8.20 Key requirements are: partial occupancy of those sites occupied by the mobile ions, easy transfer ofions between sites (i.e. bottlenecks are wide), shallow potential wells associated with the mobile ionsin their lattice sites, long range conduction pathway through a sequence of empty/occupied sites.

8.21 a) Li vacancies associated with the substitution mechanism 2 Li = Mg, b) mixed valence of Niassociated with the mechanism 2Ni2+ = Ni3+ + Li.a) Should show an increase in ionic conductivity, b) should show a large increase in electronicconductivity as pure NiO is an electronic insulator.

8.22 a) The valence states of Li and F are fixed as +1 and −1 respectively. No additional electrons or holesare present in LiF. The band gap between the VB and CB is likely to be very large >5 eV and henceno significant concentration of electrons in the CB is expected.

b) Pure NiO contains Ni2+ ions; the d8 electrons are localised on the individual ions and the electronicconductivity is zero; pale green colour is due to d-d transitions. On heating in air, oxidiation occurs,with a small amount of oxygen uptake, giving Ni1-δO containing a mixture of Ni2+ and Ni3+ andelectronic conductivity associated with the mixed valence of Ni.

c) In TiO, there is a good overlap of the t2g orbitals on adjacent Ti2+ ions, giving rise to a partiallyoccupied t2g band and metallic conductivity. In MnO, the t2g orbital overlap is very limited anddiscrete Mn2+ ions occur; electron hopping between adjacent Mn2+ leads to semiconductivity. InNiO, the t2g levels are full; the half-occupied eg levels cannot overlap to form an eg band because ofthe intervening oxide ions; consequently the Ni2+ ions are discrete and the conductivity is zero. Thekey to the difference in properties is the increased nuclear charge associated with the later transitionmetal ions which means that the 3d electrons are bound more tightly to the individual nuclei (andtheir ionic radii are also reduced).


8.23 A plot of log n/N vs T−1 is linear, Fig. S19 and hence shows Arrhenius behaviour.

0 1 2

1000 K / T

Log

n / N

3 416

14

12

10

8

6

4

Figure S19

Slope = −102.2×10−3 = −E f

2R log10 e = 176 kJ mole−1 (b) = 2.9 × 10−19 J defect−1 (a)

8.24 Key points are that Cu exhibits three oxidation states +I, +II and +III. Many examples may be identifiedof two-coordinate Cu, corresponding to Cu(I), five coordinate (pyramidal) and distorted octahedralcoordinate (Cu II) and square planar (Cu III). Cuprate perovskites would have general formula MCuO3

but most of the high Tc phases are oxygen deficient, MCuO3-x and the oxygen-deficiency gives rise tocoordination numbers less than 6.

8.25 Several methods may be used:i) Monitoring of weight loss as a function of temperature, using a thermoblance. For this, the oxygen

content of the starting material must be known.ii) Hydrogen-reduction thermogravimetry; the sample is heated in e.g. 5% H2 95% N2 to decompose the

sample, yielding La2O3, Cu and BaO. The value of δ in the starting material can then be evaluated.iii) Chemical titration. The material is dissolved in a suitable solvent and the amount of Cu3+,

determined by titration, e.g. by iodometry. Care is required at the dissolution stage, to avoidoxidation of the solvent.

8.26 Using E = RT4F loge

P2P1

E = +0.2 VT = 1500 KP1 = 10−6

R = 8.314 J/moleF = 96, 500 C

Substituting, P2 = 4.88 × 10−4 atm.

31 Answers

8.27 At very low pO2, YSZ is an electronic conductor; loss of oxygen generates electrons and the materialis n-type. Conversely, at high pO2, YSZ becomes p-type due to absorption of oxygen and extractionof electrons from the sample. For sensor applications, the YSZ must be in the ‘electrolytic domain’in which it is primarily an oxide ion conductor; outside the electrolytic domain it becomes a mixedconductor as the electronic conduction becomes increasingly important.

9.1 Paramagnetic substances show a weak attraction towards a magnetic field and can therefore be char-acterised by their mass in the presence/absence of a magnetic field. Ferromagnetic substances are verystrongly attracted to a magnetic field and exhibit a large increase in weight on switching on the elec-tromagnet. Antiferromagnetic substances have zero attraction to a magnetic field in the fully orderedAFM state but become increasingly paramagnetic as disorder sets in with increasing temperature.

9.2 VO is paramagnetic due to t32g state.In NiO, the magnetic e2

g ions order antiferromagnetically to give alternate layers ‘spin up’ and ‘spindown’ in the rock salt structure. Above the Neel temperature, TN, NiO is paramagnetic. For electricalproperties, see 8.22c.

9.3 a) ZnFe2O4 is an inverse spinel: Fe(Zn, Fe)O4. There are equal numbers of Fe3+ on tetrahedral andoctahedral sites, with opposed spins, giving antiferromagnetic behaviour.

b) MgFe2O4 is only partially inverse, giving ferromagnetic behaviour; with increasing temperature,the structure goes increasingly normal as Mg occupies preferentially the tetrahedral sites and so themagnetic moment rises.

c) MnFe2O4 is also partially inverse but contains two magnetically active d5 ions, Mn2+ and Fe3+ andhence the overall magnetic moment is insensitive to any normal-inverse changes.

9.4 To permit a clear, sharp switch in magnetisation direction with changing magnetic field.

9.5 Pure Fe has a low resistivity giving rise to eddy current losses, especially at high frequency; high lossesmust be avoided in transformer cores.

9.6 Plot X−1 vs T; see Fig. 9.4b. Since data ≥ 800K only are provided, it is not possible to separate Tc andθ from the data given.

9.7 Zinc is not a magnetically active ion. NiFe2O4 is an inverse spinel; the magnetic moments on tetrahedraland octahedral Fe3+ are opposed and cancel, leaving residual ferrimagnetism due to Ni2+. On partialreplacement of Ni by Zn, the structure becomes increasingly normal as Zn occupies tetrahedral sites.The increasing proportion of Fe3+ on octahedral sites causes the magnetisation to increase.

10.1 Use equation (6.3):

Halide Band Gap/eV λ

KF 11 7.8 AKCl 8.5 10.2 AKBr 7.5 11.5 AKl 5.8 14.9 A

10.2 An excited state with a reasonably long lifetime that permits population inversion to build up onexcitation.

10.3 Excitation takes place in two (or more) stages, permitting the emission to have a shorter wavelength(and higher energy) than the individual excitation steps.

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