1、天津理工大学作业答题纸固体化学期末题库 课程代码： 任课教师： Lect-1:1.What are some differences between “Materials Chemistry” and “Solid-State Chemistry”?（“材料化学”和“固体化学”有什么区别”？）Solution：The broad defined discipline of materials chemistry is focused on understanding the relationships between the arrangement of atoms, ions, or mol
2、ecules comprising a material, and its overall bulk structural and physical properties. By this designation, common disciplines such as polymer, solid-state, and surface chemistry would all be placed within the scope of materials chemistry. Solid-State Chemistry is a branch of chemistry that studies
3、the preparation, composition, structure and properties of solid substances.2. What is meant by “top-down” and “bottom-up” synthetic approaches? Provide applications of each (both man-made or natural materials).（ 什么是“自上而下”和“自下而上”的合成方法？ 提供每种（人造或天然材料)的应用）Solution:The top-down route：The top-down route i
4、s often used to transform naturally occurring products into useful materials. Representations include the conversion of wood into paper products, as well as certain golf ball covers.The bottom-up route：The representation is the fabrication of plastics and vinyl found in common household products and
5、 automotive interiors, through polymerization processes starting from simple monomeric compounds。3.Are complex liquids such as crude oil or detergents considered materials? Explain your reasoning.（ 复杂的液体，如原油或洗涤剂，是否被认为是“材料？ 解释你的推理）Solution：Both of them are not materials.The term material may be broad
6、ly defined as any solid-state substance or device that may be used to address a current or future societal need. Crude oil is liquid substance that be considered as precursors for materials. Detergents are substances made from materials.4.Evaluate the overall sustainability of the alternative energi
7、es listed below, considering the environmental impacts during their fabrication, installation and use, as well as end-of-use scenarios.（评估以下所列替代能源的总体可持续性，同时考虑到其制造、安装和使用过程中的环境影响，以及最终使用情况。）(a) wind farms（风电场） (b) solar parks（太阳能公园） (c) hydroelectric plants（水力发电厂） Solution：（a）Wind farms can reduce the
8、use of coal and gas, thus having a beneficial impact on the environment in the future.But wind farms need to be built in large wide areas.Wind farms can also cause environmental problems, such as soil erosion. Besides, the use of the process will affect the habitat and migration of birds, and produc
9、e noise pollution. (b) Solar energy is one of the most promising renewable energy and widely distributed.Solar parks will not pollute the environment when they are used, but during their fabrication, the pollution to the environment is great. there are lots of GHGs were released during the energy-in
10、tensive processing steps required to convert sand to silicon. In addition, there is lots of energy be expended and environmental impacts generated by recycling solar panels after reaching their lifetime of use(c) Hydroelectric power accounts for a large proportion of clean energy in our country. In
11、addition to providing cheap electricity,hydroelectric plants can also control flooding, provide irrigation water, improve river navigation, and improve the traffic, power supply and economy of the area, especially the development of tourism and aquaculture.But Hydroelectric plants may cause surface
12、activity and even induce earthquakes during their fabrication.Lect-2:1. Calculate ionic character of Al2O3 and SiO2.(计算Al2O3和SiO2的离子性质。)%ionic character(Al2O3)=(1- e-0.25（XA-XB）2)100%=(1- e-0.25（1.61-3.44）2)=56.71%ionic character(SiO2)=(1- e-0.25（XA-XB）2)100%=(1- e-0.25（1.90-3.44）2)=44.72%2. Describ
13、e free electron model of metallic solids.(描述金属固体的自由电子模型。)Free electron model of metallic solids considers the solid as a close-packed array of atoms, with valence electrons completely delocalized throughout the extended structure.3. Explain why Hg exists in liquid at STP by using its electronic conf
14、iguration. (解释为什么汞在STP液体中存在，使用其电子构型。)Solution:The liquid state of Hg is a consequence of the electronic congurations of its individual atoms. The 6s valence electrons are shielded from the nuclear charge by a lled shell of 4f electrons. This shielding causes the effective nuclear charge (Zeff) to be
15、 higher for these electrons, resulting in less sharing/delocalization of valence electrons relative to other metals. Further,relativistic contraction of the 6s orbital causes these electrons to be situated closer to the nucleus, making them less available to share with neighboring Hg atoms. In fact,
16、 mercury is the only metal that does not form diatomic molecules in the gas phase. Energetically, the individual atomsdonotpackintoasolidlatticesincethelatticeenergydoesnotcompensatefor the energy required to remove electrons from the valence shell.Lect-3: 1. A metal alloy consists of 5 at% Au and 9
17、5% at% Pt; calculate the composition in terms of wt%. How many atoms of gold will be present per cubic meter of the alloy?(金属合金由5at%Au和95%at%Pt组成，用wt%计算成分%。 每立方米合金将有多少个金原子？)18.8wt%(Au)=wt%(Pt) wt%(Au)=1（1+18.8）5%wt%(Pt) =18.8（1+18.8）95%=3.2610262.Calculate the volume, atomic packing factor , theoret
18、ical density of unit cell for aluminum.aluminum:volume=a3=（22r）3=6.6310-23cm2 (计算铝的体积、原子堆积因子、单位电池的理论密度。 铝：体积=A3=(22r)3=6.6310-23cm2)Lect-4:1. For a ceramic compound, what are the two characteristics of the component ions that determine the crystal structure?对于陶瓷化合物，决定晶体结构的成分离子有哪两个特点？Solution:.-The c
19、oordination Number (CN): number of anion nearest neighbors for a cation.配位数(CN)：阳离子的阴离子最近邻数。-.cationanion radius ratio: rC/rA.阳离子-阴离子半径比：rC/rA .-Effective ionic radii.(有效的离子半径)2.Show that the minimum cation-to-anion radius ratio for a coordination number of 4 is 0.225.说明配位数4的最小阳离子与阴离子半径之比为0.225。3.Sh
20、ow that the minimum cation-to-anion radius ratio for a coordination number of 6 is 0.414. Hint: Use the NaCl crystal structure, and assume that anions and cations are just touching along cube edges and across face diagonals.表明配位数6的最小阳离子与阴离子半径比为0.414。 提示：使用NaCl晶体结构，并假设阴离子和阳离子只是沿着立方体边缘接触 穿过面部对角线。Solut
21、ion:For this coordination, the small cation is surrounded by six anions, and four of them form a square as shown below ABCD; the centers of all five ions are coplanar.AO=rc+raAB=2raAO/AB=cos45=2/2 (rc+ra)/2ra=2/2rc/ra=0.4144. Demonstrate that the minimum cation-to anion radius ratio for a coordinati
22、on number of 8 is 0.732. 证明配位数8的最小阳离子-阴离子半径比为0.732。Solution:AC=23rA，AO=3rA3rA-rA=rCrC/rA=3-1=0.7325.Calculate the density of FeO, given that it has the rock salt crystal structure.计算FeO的密度，因为它具有岩盐晶体结构。AC=AFe=55.85g/molAA=AO=16.00g/mola=2rFe2+2rO2-Vc=a3=(2rFe2+2rO2-)3=n（ AFe +AO）/（2rFe2+2rO2-）3NA=4（5
23、5.85+16.00）/2(0.07710-7+2（0.14010-7）3（6.0231023）=5.83g/cm36.Magnesium oxide has the rock salt crystal structure and a density of 3.58 g/cm3.Determine the unit cell edge length.氧化镁具有岩盐晶体结构,密度为3.58g/cm3。确定单位细胞边缘长度。Solution:=（4AMg+4Ao）/(a3NA）=（424+416）/6.021023a3=3.581021 a=0.42nm7.Compute the theoreti
24、cal density of diamond given that the C-C distance and bond angle are 0.154 nm and 109.5 o, respectively.计算金刚石的理论密度，考虑到C-C距离和键角分别为0.154nm和109.5o。a=（0.1544）/3=0.356=（8M）/（NAa3） =（812)/（6.02310230.356310-21） =3.53g/cm38. Compute the theoretical density of ZnS given that the Zn-S distance and bond angl
25、e are 0.234 nm and 109.5 o, respectively.计算了ZnS的理论密度，给出了ZnS距离和键角分别为0.234nm和109.5o。Solution:a=40.234/3=0.54=（4M）/（NAa3） =4（65.41+32.07）/（6.02310230.5410-21）=4.102g/cm39. A hypothetical AX type of ceramic material is known to have a density of 2.65 g/cm3 and a unit cell of cubic symmetry with a cell e
26、dge 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? (已知一种假设的AX型陶瓷材料的密度为2.65g/cm3，单位细胞为立方对称
27、细胞边缘长度为0.43nm。 的原子量A和X元素 分别为86.6和40.3g/mol。 根据这些信息，下列哪种晶体结构是这种材料可能的：岩盐、氯化铯或共混锌？)Solution:Let the number of atoms of A and X in the cell be n=n(86.6+40.3)/（0.4310-7）36.021023=2.65 n=1Because a cesium chloride cell contains 1 Cs+ and 1 Cl-, the cesium chloride structure is possible for this material.
28、10. The unit cell for Al2O3 has hexagonal symmetry with lattice parameters a = 0.4759 nm and c= 1.2989 nm. If the density of this material is 3.99 g/cm3, calculate its atomic packing factor.Al2O3单元具有六方对称，晶格参数a=0.4759nm，c=1.2989nm。 如果这种材料的密度为3.99g/cm3，则计算其原子堆积因子。Solution:APF=Vs/Vc=2（4/3）（3a/5）3/6（3）a
29、2/4c=2（4/3）（30.4759/5）3/6（3）0.47592/41.2989=0.7411. In terms of bonding, explain why silicate materials have relatively low densities.在键合方面，解释为什么硅酸盐材料的密度相对较低。Solution:Silicates are composed of various silicon-oxygen groups with the silicon-oxygen tetrahedron SiO4 as the basic structural unit. In add
30、ition, silicates do not have higher densities because their spatial structure is not as closely aligned as that of metals; O,si bonds have spatial bonds and theyre longer, but unlike metals, they have very short bonds and they have a lot of atomic weight, which is why they have relatively low densit
31、ies.12. Determine the angle between covalent bonds in an SiO44- tetrahedron. 确定SiO44-四面体中共价键之间的夹角109.5Four silicon atoms can form a regular tetrahedron,let the angle be AThe radius of circumference of the regular tetrahedron is RR=a(6(1/2)/4cosA=(R2+R2-a)/(2RR)=-1/3 A=109.5INDEPTH：Crystal Structure
32、and Properties of CsPbX3（X=Cl、Br、I）Anion packing；FCC；Cation：12(A);Anion:6Fully inorganic perovskite quantum dots (CsPbX3, X = Cl, Br, I) have excellent photoelectric performance and have shown great application prospects in photovoltaic devices such as solar cells, light-emitting diodes and lasers.L
33、ect-5678:1. Calculate the fraction of atom sites that are vacant for lead at its melting temperature of 327oC. Assume an energy for vacancy formation of 0.55 eV/atom.计算铅在327C熔化温度下空位的原子位的分数。 假设空位形成的能量为0.55eV/原子。Solution:According to the formula ,Thus, the proportion of vacancies is 2. Calculate the n
34、umber of vacancies per cubic meter in iron at 850oC. The energy for vacancy formation is 1.08 eV/atom. Furthermore, the density and atomic weight for Fe are 7.65 g/cm3 and 55.85 g/mol, respectively.计算850C时每立方米铁中的空位数。 空位形成的能量为1.08eV/原子。 此外，Fe的密度和原子量分别为7.65g/cm3和55.85g/mol，Solution:to determine the va
35、lue of N, the number of atomic sites per cubic meter for iron N, from its atomic weight 𝐴𝐹𝑒Its density and Avogadros number according to: Thus, the number of vacancies at 850oC (1123 K) is equal to: 3. Calculate the energy for vacancy formation in silver, given that the equil
36、ibrium number of vacancies at 800oC (1073 K) is 3.61023 vacancies/m3. The atomic weight and density (at 800oC) for silver are, respectively, 107.9 g/mol and 9.5 g/cm3.计算银中空位形成的能量，因为800C(1073K)的空位平衡数为3.6(1023空位/m3。 银Ar的原子重量和密度(在800C 分别为107.9g/mol和9.5g/cm3。Solution: 4. What is the composition, in atom
37、 percent, of an alloy that consists of 30 wt% Zn and 70 wt% Cu? 由30wt%Zn和70wt%Cu组成的合金的原子百分比是多少？Solution:If we denote the respective atom percent compositions as and, substitution into Equations yields: 5. What is the composition, in weight percent, of an alloy that consists of 6 at% Pb and 94 at% Sn
38、 由6at%Pb和94at%Sn组成的合金的重量百分比是多少？Solution:If we denote the respective weight percent compositions as and, substitution into Equations yields: 6. Calculate the composition, in weight percent, of an alloy that contains 218.0 Kg titanium,14.6 kg of aluminum, and 9.7 kg of vanadium.计算含有218.0公斤钛、14.6公斤铝和9
39、7公斤钒的合金的成分，重量百分比。Solution:If we denote the respective weight percent compositions as C: 7. What is the composition, in atom percent, of an alloy that contains 98 g tin and 65 g of lead?含98克锡和65克铅的合金的原子百分比是多少？Solution:If we use C for the composition of the weight percentages, C for the atomic percen
40、tages 8. What is the composition, in atom percent, of an alloy that consists of 97wt% Fe and 3wt% Si? 由97wt%Fe和3wt%Si组成的合金的原子百分比是什么？Solution:If we denote the respective weight percent compositions as =97 and =3, substitution into Equations yields. 9. The concentration of carbon in an iron-carbon all
41、oy is 0.15 wt%. What is the concentration in kilograms of carbon per cubic meter of alloy?铁碳合金中碳的浓度为0.15wt%。 每立方米合金的碳浓度是多少千克？10. Determine the approximate density of a high-leaded brass that has a composition of 64.5 wt% Cu, 33.5 wt% Zn, and 2 wt% Pb. 确定高铅黄铜的近似密度，其组成为64.5wt%Cu、33.5wt%Zn和2wt%Pb。11. G
42、old forms a substitutional solid solution with silver. Compute the number of gold atoms per cubic centimeter for a silver-gold alloy that contains 10 wt% Au and 90 wt% Ag. The densities of pure gold and silver are 19.32 and 10.49 g/cm3, respectively. 金与银形成替代固溶体。 计算含10wt%Au和90wt%Ag的银金合金的每立方厘米金原子数。 纯金
43、的密度 银分别为19.32克和10.49克/厘米。12. Germanium forms a substitutional solid solution with silicon. Compute the number of germanium atoms per cubic centimeter for a germanium-silicon alloy that contains 15 wt% Ge and 85 wt% Si. The densities of pure germanium and silicon are 5.32 and 2.33 g/cm3, respectively
44、 锗与硅形成替代固溶体。 计算含有15wt%Ge和85wt%Si的锗硅合金的每立方厘米锗原子数。 丹西 纯锗和硅的结合分别为5.32和2.33g/cm3。13. Copper and platinum both have the FCC crystal structure and Cu forms a substitutional solid solution for concentrations up to approximately 6 wt% Cu at room temperature. Compute the unit cell edge length for a 95 wt% P
45、t-5 wt% Cu alloy.铜和铂都具有FCC晶体结构，Cu在室温下形成浓度约为6wt%Cu的替代固溶体。 计算单元格边le 对于95wt%Pt-5wt%Cu合金。14. Would you expect Frenkel defects for anions to exist in ionic ceramics in relatively large concentrations? Why or why not?你会期望阴离子在相对较大浓度的离子陶瓷中存在Frenkel缺陷吗？ 为什么或者为什么不？Solution: Frinkel defect is characterized by
46、the pairing of vacancy and gap. For the freckler defect of negative ions in ionic ceramics, it is composed of a interstitial anion and an anion vacancy. The concentration of frinkel defect is related to temperature. The higher the temperature is, the higher the defect concentration is. So I think it
47、 might exist in a higher concentration.15. In your own words, briefly define the term stoichiometric.用你自己的话来说，简单地定义一下“化学计量。Solution：Calculationofthequantitiesofreactantsandproductsinachemicalreaction.16. If cupric oxide (CuO) is exposed to reducing atmospheres at elevated temperatures, some of the Cu2+ ions will become Cu+. 如果铜氧化物(CuO)在高温下暴露于还原气氛中，一些Cu2离子将变成Cu。(a) Under these conditions, name one crystalline defect that you would expect to form in order to maintain charge neutrality.在这些条件下，命名一个晶体缺陷，你将期望形成，以保持电荷中立。
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