Chapter 18, Problem 18.4 The origin of the defect is due to that a small portion of the iron is Fe(III) instead of Fe(II). For a pair of Fe(III) ions to appear, an O2- would be missed from its lattice site to compensate the charge difference.

Ref. West, A. R. Basic Solid State Chemistry 2nd, p223.

18.6 Ref. Textbook, p618 and p626.

The introduction of Li2O into NiO places Li+ in Ni2+ sites, the system balances the charge by oxidation of one Ni2+ to Ni3+ for each Li+ ion present. This process introduces holes near the valance band of NiO and the conductivity increases greatly as a p-type semiconductor. 18.7 Ref. textbook, Figure 13.10, p316-317 and p453-454.

For Cr+3 in the Oh site, the three d-electrons are filled in t2g orbitals. The dark green color of Cr2O3 is due to the absorption of red light of incident white light by its 3d electron.

When small amount of Cr3+ doped into the alumina matrix, the Cr3+ ion replaced Al3+ in Oh site. The Cr3+ was compressed by the O2- liqands because the diameter of Al3+ is smaller than that of Cr3+. The compression increases the liqand-field splitting parameter and shifts the first-allowed d-d band to higher energy. Therefore, the red color of ruby is due to the adsorption of violet and green light of incident white light by the 3d electron of Cr3+ doped in the alumina matrix.

The mechanism of phosphorescence was described in the textbook.

18.9 For normal spinel, the formula is AII[BIII]2O4, where AII and BIII are occupied in

the Td and Oh sites, respectively. The occupation factor, ?, of a spinel is the faction of BIII cations in the tetrahedral sites. Thus, ? = (BIII)T d/ (BIII)Total. ? = 0 for normal spinel, and ? = 0.5 for an inverse spinel BIII [AIIBIII]O4, where AII is in the Oh sites and a half of the BIII ions are occupied in the Td and another half in Oh sites. (a) Fe3O4 In octahedral (Oh) symmetry, the liqand field stabilized energy (LFSE) of Fe3+ in high spin state (or in weak ligand field) is 0 ∆o. In tetrahedral (Td) symmetry, the LFSE of Fe3+ is also 0 ∆o. Define ∆LFSE = (LFSE)Oh - (LFSE)T d. If ∆LFSE < 0, Oh site is more stable than Td site. Since the LFSEs of Fe3+ in Td and Oh sites are the same, ∆LFSE = 0, the Fe3+ ion can be in either sites. For Fe2+, electron configuration is 3d6. ∆LFSE = [4(-0.4 ∆o) + 2(0.6) ∆o] – [3(-0.6) + 3(0.4) ∆t] = (-0.4 ∆o) – (-0.6 ∆t) = (-0.4 ∆o) – (-0.6 x 4/9 ∆o) = -0.13 ∆o Fe2+ in Oh site is more stable than in Td site. Thus, Fe3O4 is inverse spinel structure and its ? is 0.5.

(b) Cr3O4. For Cr3+, electron configuration is 3d3. ∆LFSE = [3(-0.4 ∆o)] – [2(-0.6)+(0.4) ∆t] = [-1.2 ∆o- (-0.8) ∆t] = [-1.2 ∆o- 4/9 x (-0.8) ∆o] = -0.84 ∆o. For Cr2+, electron configuration is 3d4. ∆LFSE = [3(-0.4 ∆o) +(0.6) ∆o] – [2(-0.6)+2(0.4) ∆t] = [-0.6 ∆o- (-0.4) ∆t] = [-0.6 ∆o- 4/9 x (-0.4) ∆o] = -0.42 ∆o. Cr3+ prefers in Oh site more than Cr2+. Thus, Cr3O4 is normal spinel structure. (c) Mn3O4 For Mn3+, electron configuration is 3d4. ∆LFSE = [3(-0.4 ∆o) + (0.6) ∆o] – [2(-0.6) + 2(0.4) ∆t] = [-0.6 ∆o- (-0.4) ∆t] = [-0.6 ∆o- 4/9 x (-0.4) ∆o] = -0.42 ∆o. For Mn2+, electron configuration is 3d5. ∆LFSE = [3(-0.4 ∆o) + 2(0.6) ∆o] – [2(-0.6)+3(0.4) ∆t] = 0. Mn3+ prefers in Oh site and no difference for Mn2+. Thus, Mn3O4 is normal spinel structure. Additional problems 1. The inter-facial angles of the unit cell are defined to be: α: angle between edges b and c. β: angle between edges a and c. γ: angle between edges a and b. In 3D, non-primitive cells are of three kinds: End-centered : an extra lattice point is centered in each of two opposing faces of the cell. Face-centered : an extra lattice point is centered in every face of the cell. Body-centered : an extra lattice point is centered in the exact middle of the cell. Correction for question 1(a): There is no unit cell as “end-centered tetragonal”. It can be simplified to “primary tetragonal”. The end-centered unit cells only occur in “orthorhombic” and in “monoclinic” as shown below.

2.

(b) 2 lattice

points

(c) 1 lattice points

(a) 2 lattice

points

Angle = 90 degree

(a) 2 lattice

points

3. The space group P21/c means

(a) P: primitive cell. 21: translate 1/2, and then rotate 180o. /: c-glide plane is perpendicular to 21 screw axis C: axial glide plane, translate c/2 alone z-axis and then reflect through ac-plane or bc-plane. (b) 4 lattice points.