1)List three types of intrinsic point defects for crystalline materials and briefly describe the features of the defects.(列出三种晶体材料的本征点缺陷，并简要描述这些缺陷的特征)(9p)

1.A Schottky defect is the absence of ions on normally occupied sites; for charge neutrality there must be equal numbers of cation and anion vacancies in a 1:1 compound.(肖特基缺陷是离子在原本被占据的晶格位置上缺失；为了保持电中性，在 1:1 型化合物中，阳离子空位和阴离子空位的数量必须相等。)

2.A Frenkel defect forms when an ion moves to an interstitial site.(当一个离子移动到间隙位置时，就会形成弗伦克尔缺陷。)

3.Atom exchange can also give rise to a point defect as in CuAu.(原子交换也会产生点缺陷，例如在铜金合金（CuAu）中。)

2) a) Prepare a molecular orbital (MO) energy-level diagram for the cyanide ion (CN-). Use sketches to show clearly how the atomic orbitals interact to form MOs. Please calculate the bond order and unpaired electrons. （绘制氰根离子（CN-）的分子轨道（MO）能级图。请用轨道示意图清晰说明原子轨道如何相互作用形成分子轨道，并计算其键级和未成对电子数。(4p)

Bonding electrons: 2 (in σ(2s)) +4 (in π(2p)) +2 (in σ(2p)) =8 ；Antibonding electrons: 2 (in σ∗(2s)) =2

BO = 1/2 x ( 8-2) = 3.All electrons in the MO configuration are paired.

So,the cyanide ion (CN-) has a bond order of 3 and 0 unpaired electrons.

2)b)) Prepare a molecular orbital energy-level diagram for linear ion,H3+.Use sketches to show clearly how the atomic orbitals interact to form MOs. Please calculate the bond order and unpaired electrons. (绘制线性离子H3+的分子轨道能级图。请用轨道示意图清晰说明原子轨道如何相互作用形成分子轨道，并计算其键级和未成对电子数)(6p)

BO = 1/2 x ( 2-0) = 1.All electrons in the MO configuration are paired.

So,the cyanide ion (CN-) has a bond order of 1 and 0 unpaired electrons.

3) Determine the point group for each of the following molecules or molecular ions. Neglect all hydrogen atoms and assume conjugated systems are fully delocalized. (确定下列每个分子或分子离子的点群。忽略所有氢原子，并假设共轭体系完全离域)(12p)

a.

• 低对称群？→ No
• 高对称群？→ No（非 T/O/∞对称）
• 找最高阶旋转轴：C₄（垂直于分子平面）
• 存在 4 个垂直于 C₄的 C₂轴？→ Yes（沿 F-Xe-F 键和键角平分线）→ 进入D 型群
• 存在 σₕ？→ Yes（分子平面本身就是垂直于 C₄的镜面）→ 结论：D₄ₕ

b.

• 低对称群？→ No
• 高对称群？→ No
• 找最高阶旋转轴：C₂（穿过中心 O，平分键角）
• 存在垂直于 C₂的 C₂轴？→ No→ 进入C/S 型群
• 存在 σₕ？→ No（分子平面不垂直于 C₂）
• 存在 σᵥ？→ Yes（分子平面和垂直于分子平面的两个镜面，都包含 C₂轴）→ 结论：C₂ᵥ

c.

• 低对称群？→ No
• 高对称群？→ No
• 找最高阶旋转轴：C₂（沿 Br-Cr-Br 轴，同理 NH₃和 H₂O 轴也是 C₂）
• 存在 2 个垂直于主 C₂轴的 C₂轴？→ Yes（三个互相垂直的 C₂轴）→ 进入D 型群
• 存在 σₕ？→ Yes（垂直于主 C₂轴的镜面）→ 结论：D₂ₕ

d.

• 低对称群？→ No
• 高对称群？→ No
• 找最高阶旋转轴：C₃（穿过 V 原子，垂直于 S 原子平面）
• 存在垂直于 C₃的 C₂轴？→ No→ 进入C/S 型群
• 存在 σₕ？→ No
• 存在 σᵥ？→ Yes（3 个包含 C₃轴、平分 S-V-S 夹角的镜面）→ 结论：C₃ᵥ

e.

• 低对称群？→ No
• 高对称群？→ No
• 找最高阶旋转轴：C₅（穿过 Cp 环中心、Ni、NO 键轴）
• 存在垂直于 C₅的 C₂轴？→ No→ 进入C/S 型群
• 存在 σₕ？→ No
• 存在 σᵥ？→ Yes（5 个包含 C₅轴、穿过 Cp 环顶点和 NO 的镜面）→ 结论：C₅ᵥ

f.

• 低对称群？→ No
• 高对称群？→ No
• 找最高阶旋转轴：C₂（穿过环中心，平分对边）
• 存在垂直于 C₂的 C₂轴？→ No→ 进入C/S 型群
• 存在 σₕ？→ Yes（垂直于 C₂轴的镜面，与 C₂轴一起产生对称中心）→ 结论：C₂ₕ

4) The mineral pyrargyrite,Ag₃SbS₃, utilizes the hexagonal lattice system. The sulfides are space-filling and the Ag+ ions occupy one sort of hole (either octahedral or tetrahedral) while the Sb3+ ions occupy the other sort of hole. (矿物硫锑银矿采用六方晶系晶格。硫离子做密堆积，Ag⁺离子占据一种空隙（八面体或四面体空隙），Sb³⁺离子占据另一种空隙)(11p)

5)One of the first materials used in solid state electronics was red copper(I) oxide. Interest is renewed nowadays because it could be a non-toxic and cheap component of solar cells. (固态电子学中最早使用的材料之一是氧化亚铜（Cu₂O）。如今人们对它的兴趣重新高涨，因为它有望成为太阳能电池中一种无毒且廉价的组件)(19p)

The two figures above depict the cubic unit cell of the Cu₂O crystal. The lattice constant of the structure is 427.0 pm.(图中所示为 Cu₂O 晶体的立方晶胞，该结构的晶格常数为427.0 pm)

There are 2A atoms and 4B atoms in the cell. Cu:B
B:fcc A:bcc
B:2 A:4

b) Calculate the smallest O-O, Cu-O and Cu-Cu distances in the structure.(计算该结构中 O-O、Cu-O 和 Cu-Cu 的最小原子间距) (6p)

d_O-O :half of the cell body diagonal 1/2*427.0pm*√3 = 369.8pm
d_Cu-O:1/4^th of the cell body diagonal 1/4*427.0pm*√3 =184.0pm
d_Cu-Cu:half of the face diagonal:1/2*427.0pm*√2 =301.9pm
1p each, no penalty for Cu and O switched

c) What is the density of pure copper(I) oxide? (纯氧化亚铜的密度是多少)(2p)

The volume of the unit cell is (427.0pm)³
The mass of a unit cell is (4M_Cu+2M_O)/N_A.The density is 6.106g/cm³

d)A common defect in this crystal is some copper atoms missing with the oxygen lattice unchanged. The composition of one such crystal sample was studied, and 0.2% of all copper atoms were found to be in oxidation state +2. What percentage of normal copper sites are empty in the crystal sample? What is x in the empirical formula Cu₂₋ₓO of the crystal?(该晶体中一种常见缺陷是部分铜原子缺失，而氧原子晶格保持不变。对某一样品进行成分分析后发现，0.2% 的铜原子处于 + 2 氧化态。请问该晶体样品中，正常铜原子位置的空位占比是多少？经验式Cu_2-xO中的x值是多少(4p)

6)Zinc blende (sphalerite, ZnS) crystallizes in a different crystal structure type, which is closely related to the structure of diamond. Both types of structures are shown in Figure below(闪锌矿（立方硫化锌，ZnS）的晶体结构与金刚石结构密切相关，两种结构如下图所示) (13p):

a) How many formula units (ZnS) are there in the unit cell of sphalerite? (闪锌矿晶胞中包含多少个 ZnS 结构单元)(2p)

b) Heavier elements of group IV (i.e. group 14), silicon and germanium, also adopt the structure of diamond. The radius of elemental germanium is r(Ge)=1.23 A˚. Calculate the density of solid germanium. (第 14 族（碳族）的较重元素（如硅、锗）也会形成金刚石型结构。已知金属锗的原子半径r(Ge)=1.23 A˚，计算固态锗的密度)(4p)

c) Germanium is a semiconductor similar to silicon. It is used in electro-technology and similarly to silicon, it is also very fragile. Therefore, more flexible isoelectronic gallium arsenide, GaAs, is used in some practical applications. This compound belongs to semiconductors of III-V type (compounds of elements from groups III and V, i.e. groups 13 and 15, respectively) and adopts the structure of sphalerite. The lattice constants of Ge and GaAs are very similar, and a(GaAs)=5.65 A˚. An analogous compound GaP also adopts the structure of sphalerite, but has a smaller unit cell with a(GaP)=5.45 A˚. Calculate the difference between the radii of P and As in the respective compounds with gallium (GaP versus GaAs).(锗是一种和硅类似的半导体，在电子技术中应用广泛，但和硅一样质地很脆。因此，在一些实际应用中会使用更柔韧的 III-V 族半导体砷化镓（GaAs），这类半导体由第 13 族和第 15 族元素组成，同样采用闪锌矿结构。锗和砷化镓的晶格常数非常接近，a(GaAs)=5.65 A˚。类似地，磷化镓（GaP）也采用闪锌矿结构，但晶胞更小，晶格常数a(GaP)=5.45 A˚。计算在这两种化合物中，磷（P）和砷（As）的离子半径差（GaP 与 GaAs 对比）) (7p)

7)The bond lengths inO₂,O₂⁺andO₂^2-are 121, 112, and 149 pm,respectively. Please rationalize the differences in bond lengths. (6p)