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桥氧+非桥氧的mol数=2×1.33+0.32=2.98mol ∴非桥氧的分数=0.64/2.98=0.215
1
3-22 CaTiO3为标准钙钛矿型结构，简述 其结构特征，分析其中钙离子、钛离子和氧 离子的配位数。
立方晶系：Ca2+占立方体顶角，O2-占立方体面心，Ti4+占立方体体心 配位数：Ca2+为12（12个O2-），Ti4+为6（6个O2-），
O2-负离子数为:242/2 =121 共 121/4 = 30.25 个晶胞
(从FeO(NaCl型)知每晶胞有4个O2-) ∴ ρ= (121×16+114×55.8) / [30.25×6.02×1023(2×0.215×10-7)3 = 5.73 (g/cm3)
5
英文
3.29 On the basis of ionic charge and ionic radii, predict the crystal structures for the following materials: (a) CsI, (b) NiO, (c) KI, and (d) NiS. Justify your selections.
每个晶胞中Cd和S数目相等x个. xM/NA= ρ a3, x=4, So, there are 4 Cd 2+ and S 2- ions in per unit cell.
7
3.41 A hypothetical AX type of ceramic material is known to have a density of 2.65 g/cm 3 and a unit cell of cubic symmetry with a cell edge length of 0.43 nm. The atomic weights of the A and X elements are 86.6 and 40.3 g/mol, respectively. On the basis of this information, which of the following crystal structures is (are) possible for this material: rock salt, cesium chloride, or zinc blende? Justify your choice(s).
第七次作业 中文
3-19 一玻璃含80%(wt)的SiO2和20%(wt)的 Na2O，问非桥氧的分数为多少？
基数 100g=80gSiO2+20gNa2O ∴SiO2: 80g/[(28.1+2×16.0)g/mol]=1.33mol Na2O: 20g/[(2×23.0+16.0)g/mol]=0.32mol ∴非桥氧的mol数=Na的mol数=2×0.32=0.64mol
1, so, it is a cesium chloride crystal structure ,for only this structure includes only one formula unit in per unit.
8
3.63 For each of the following crystal structures, represent the indicated plane in the manner of Figures 3.24 and 3.25, showing both anions and cations: (a) (100) plane for the rock salt crystal structure, (b) (110) plane for the cesium chloride crystal structure, (c) (111) plane for the zinc blende crystal structure, and (d) (110) plane for the perovskite crystal structure.
半径比 CN structures
CsI
0.773 8:8 cesium chloride
NiO
0.493 6:6 Rock salt
KI
ห้องสมุดไป่ตู้
0.627 6:6 Rock salt
NiS
0.375 4:4 Zinc blend (sphalerite)
6
3.38 Cadmium sulfide (CdS) has a cubic unit cell, and from x-ray diffraction data it is known that the cell edge length is 0.582 nm. If the measured density is 4.82 g/cm 3 , how many Cd 2+ and S 2— ions are there per unit cell?
4
2-37如果Fe3+/Fe2+比为0.14，则FeO的密度是多少？ （FeO为NaCl结构；（r0+rFe）平均为0.215nm）
2-37 已知: Fe3+离子数/ Fe2+离子数 = 0.14 = 14/100 即: 114个正离子中有14个Fe3+,100个Fe2+,
正电荷总数为: 2×100+3×14 = 242 保持电荷平衡
3
2-22由X射线衍射数据显示，MgO立方体的单位 晶胞尺寸是0.412nm，其密度3.83Mg/m3，请 问在每单位晶胞中有多少Mg2+离子和O2-离子?
MgO: 40.31 3.83g/cm3 = [40.31×x] / [6.02×1023个
/mol ×(0.412×10-7cm)3] x = 4 (4个Mg2+, 4个O2-)
O2-为（4个Ca2+, 2个Ti4+）
2
3-24 简要说明硅酸盐的几种结构单元 的主要特点。
岛状结构：SiO四面体不共顶，非桥氧比例大， 需形成较多离子键，(4) 环状结构：SiO四面体共二顶成环，非桥氧比例 较大， (2) 链状结构：SiO四面体共二顶成链，非桥氧比例 较大，(2) 层状结构：SiO四面体共三顶成层，非桥氧比例 较小，需形成较少离子键，(1) 架状结构：SiO四面体共四顶（架状），无非桥 氧，不需再成离子键，(0)
1
3-22 CaTiO3为标准钙钛矿型结构，简述 其结构特征，分析其中钙离子、钛离子和氧 离子的配位数。
立方晶系：Ca2+占立方体顶角，O2-占立方体面心，Ti4+占立方体体心 配位数：Ca2+为12（12个O2-），Ti4+为6（6个O2-），
O2-负离子数为:242/2 =121 共 121/4 = 30.25 个晶胞
(从FeO(NaCl型)知每晶胞有4个O2-) ∴ ρ= (121×16+114×55.8) / [30.25×6.02×1023(2×0.215×10-7)3 = 5.73 (g/cm3)
5
英文
3.29 On the basis of ionic charge and ionic radii, predict the crystal structures for the following materials: (a) CsI, (b) NiO, (c) KI, and (d) NiS. Justify your selections.
每个晶胞中Cd和S数目相等x个. xM/NA= ρ a3, x=4, So, there are 4 Cd 2+ and S 2- ions in per unit cell.
7
3.41 A hypothetical AX type of ceramic material is known to have a density of 2.65 g/cm 3 and a unit cell of cubic symmetry with a cell edge length of 0.43 nm. The atomic weights of the A and X elements are 86.6 and 40.3 g/mol, respectively. On the basis of this information, which of the following crystal structures is (are) possible for this material: rock salt, cesium chloride, or zinc blende? Justify your choice(s).
第七次作业 中文
3-19 一玻璃含80%(wt)的SiO2和20%(wt)的 Na2O，问非桥氧的分数为多少？
基数 100g=80gSiO2+20gNa2O ∴SiO2: 80g/[(28.1+2×16.0)g/mol]=1.33mol Na2O: 20g/[(2×23.0+16.0)g/mol]=0.32mol ∴非桥氧的mol数=Na的mol数=2×0.32=0.64mol
1, so, it is a cesium chloride crystal structure ,for only this structure includes only one formula unit in per unit.
8
3.63 For each of the following crystal structures, represent the indicated plane in the manner of Figures 3.24 and 3.25, showing both anions and cations: (a) (100) plane for the rock salt crystal structure, (b) (110) plane for the cesium chloride crystal structure, (c) (111) plane for the zinc blende crystal structure, and (d) (110) plane for the perovskite crystal structure.
半径比 CN structures
CsI
0.773 8:8 cesium chloride
NiO
0.493 6:6 Rock salt
KI
ห้องสมุดไป่ตู้
0.627 6:6 Rock salt
NiS
0.375 4:4 Zinc blend (sphalerite)
6
3.38 Cadmium sulfide (CdS) has a cubic unit cell, and from x-ray diffraction data it is known that the cell edge length is 0.582 nm. If the measured density is 4.82 g/cm 3 , how many Cd 2+ and S 2— ions are there per unit cell?
4
2-37如果Fe3+/Fe2+比为0.14，则FeO的密度是多少？ （FeO为NaCl结构；（r0+rFe）平均为0.215nm）
2-37 已知: Fe3+离子数/ Fe2+离子数 = 0.14 = 14/100 即: 114个正离子中有14个Fe3+,100个Fe2+,
正电荷总数为: 2×100+3×14 = 242 保持电荷平衡
3
2-22由X射线衍射数据显示，MgO立方体的单位 晶胞尺寸是0.412nm，其密度3.83Mg/m3，请 问在每单位晶胞中有多少Mg2+离子和O2-离子?
MgO: 40.31 3.83g/cm3 = [40.31×x] / [6.02×1023个
/mol ×(0.412×10-7cm)3] x = 4 (4个Mg2+, 4个O2-)
O2-为（4个Ca2+, 2个Ti4+）
2
3-24 简要说明硅酸盐的几种结构单元 的主要特点。
岛状结构：SiO四面体不共顶，非桥氧比例大， 需形成较多离子键，(4) 环状结构：SiO四面体共二顶成环，非桥氧比例 较大， (2) 链状结构：SiO四面体共二顶成链，非桥氧比例 较大，(2) 层状结构：SiO四面体共三顶成层，非桥氧比例 较小，需形成较少离子键，(1) 架状结构：SiO四面体共四顶（架状），无非桥 氧，不需再成离子键，(0)