材料科学基础第2章习题解答_ppt课件

材料科学基础第二章习题解答1. 第 B族元素有多少个价电子？第 B族有几个？第 B族元素 B（ 5）、 Al（ 13）、 Ga（ 31）、 In（ 49）、Tl（ 81）有 3个价电子。第 B族 N （ 7） 、 P （ 15） 、As （ 33） 、 Sb （ 51） 、 Bi（ 83）有 5个价电子。1s/2s/2p/3s/3p/4s/3d/4p/5s/4d/5p/6s/4f/5d/6s/原子结构描述电子运动状态的描述电子运动状态的 量子数量子数 (n、 l、 ml、 ms )p主量子数 n决定了电子在核外出现概率最大区域 (“ 电子层 ” )，离核的远近及其能量的高低 . n决定能量， n越大，电子的能量越高； n也代表电子离核的平均距离， n越大，电子离核越远。 n相同称处于同一电子层。 n是一个非零的整数值，不同的 n 值，对应于不同的电子层 K L M N O轨道角动量量子数描述电子云的不同形状，也与能量有关 l n-1, 取值 0， 1， 2， 3 n-1(亚层）s, p, d, f( 这些符号均来自光谱学 ) p 轨道 角动量量子数 l轨道角动量量子数允许的值 , ln l1234亚层0000s111p22d3f s 轨道球形p 轨道哑铃形d轨道有两种形状 l 值相同的电子 , 具有确定的电子云形状 , 但在磁场中可以有不同的伸展方向 . 磁量子数就是描述电子云在空间的伸展方向 m可取 0， 1, 2 l （ |m| l ） m 值相同的轨道互为等价 (简并 )轨道p磁量子数 ml磁量子数的允许值 , mll ml 轨道数 (2l+1)0(s)1(p)2(d)3(f)0 1 0 1 2 1 0 1 2 3 2 1 0 1 2 31357原子轨道能级的顺序原子轨道能级的顺序鲍林近似能级图按原子轨道能量高低的顺序排列，能级相近的放在一个方框内称为能级组，共七组。第一能级组： 1s第二能级组： 2s, 2p第三能级组： 3s, 3p第四能级组： 4s, 3d, 4p第五能级组： 5s, 4d, 5p第六能级组： 6s, 4f, 5d, 6p第七能级组： 7s, 5f, 6d, 7pl 相同的能级 , n 越大 , E 越高n 相同的能级 , l 越大 , E 越高n和 l 均变动时 , 出现能级交错4s和 3d、 5s和 4d2.在例题 2-1中，我们说明了为什么 Si和 Ge相似，在周期表这一列中的下一个元素是什么？它的电子构型怎样？The next group IVB element in the series is Sn with an electron configuration (from the Periodic Table) of 1s22s22p63s23p63d104s24p64d105s25p2. Note that the valence electron configuration for Sn is also of the form xs2xp2 with x=5. 原子结构3.你预计 Ca和 Zn会表现出类似的性质吗？为什么？Examination of the Periodic Table shows that Ca has the electron configuration1s22s22p63s23p64s2 while Zn has configuration 1s22s22p63s23p63d104s2Since both elements have two valence electrons (4s2) we should expect these elements to display some similar properties.Although Zn and Ca have similar valence electronconfigurations, they have other structural differences that result in differences in properties.原子结构4.如果电子的能量不是量子化的，会有哪些后果？Although there are many possible answers to this question, one of the more important results might be a breakdown in the periodic arrangement of the elements. The Periodic Table owes its existence to the quantization of energy. If quantization of energy did not exist, we would lose the ability to understand and predict properties based on valence electron configuration.原子结构5.Cu有多少个电子、质子和中子？ DATA: From Appendix A, the atomic number of Cu is 29 and the atomic mass is 63.54 g/mole. SOLUTION: Cu has 29 electrons and 29 protons, each proton weighing about 1 g/mole. The balance of the atomic mass is from neutrons. COMMENTS: Elements can have different masses, from having different numbers of neutrons. They are called isotopes. 原子结构6.写出 C的电子结构。 C怎样才能形成 4个相等的键？共价键为高电子密度区域，因此相互排斥，预计在共价键合的 C中 4个间的键合几何（键角）。 sp3杂化：由一个 ns轨道和 3个 np轨道混杂形成 4个能量相同的 sp3杂化轨道，每个轨道含有 1/4s成分和 3/4p成分，各个 sp3杂化轨道之间的夹角为 109.5。原子结构C原子轨道杂化激发4个 sp3轨道7.大多数金属的氧化物在能量上都是比纯金属的能量 “低 ”的。氧化的金属在氧化初期通常增加重量。对于 “金本位制 ”来讲，你需要这种这特性吗？ ASSUMPTIONS: You want the standards critical properties to be invariant with time SOLUTION: Gold is one of the few metals whose pure metallic state is more thermodynamically stable than its oxide. Hence, gold does not oxidize. If it did, then it might gain or lose weight with time of exposure to air. COMMENTS: This is why gold is found in nature as nuggets(天然金块 ), whereas, for example, iron and aluminum are found as oxides or sulfides.热力学与动力学8.在 -1 时能否得到纯的液态水。 Solution: Yes, if the pressure of the system is raised above one atmosphere, water can exist at -1 . Comments: Since most of our daily experiences occur at (or near) atmospheric pressure, we tend to forget about pressure as an important system variable. There are, however, many important engineering processes that occur at either substantially higher or lower pressures.热力学与动力学9.糖浆的流动速率可以描述为激活能约为 50KJ/mol的，当温度从 10 变化到 25 时，流动速率变化多少？ Data: R= 8.314 J/mol.K, K= C +273 Assumptions: Flow rate at temperature T is given byF(T)= Fo exp(-Q/RT) Solution: The ratio of the flow rates at any two temperatures is:F(T1) = Fo exp(-Q/RT1)； F(T2) Fo exp(-Q/RT2) F(25C) = exp (-Q/R(1/T1-1/T2)F(10C)= exp(-50,000J/mol/8.134J/mol-K)(1/298K- 1/283K) = 2.91A temperature increase of 15 C results in almost a factor of three increase in the flow rate of molasses（糖蜜） .热力学与动力学10为了形成聚合物，许多等同的小分子在化学反应中被连接在一起。聚合反应是放热的，或者说是产生热量的，它可描述成阿累尼乌斯过程，激活能约为80KJ/mol。如果温度增加 10 ，反应速率变化多少？l Data: R= 8.314 J/mole-K, K= C +273l Assumption: Polymerization rate at temperature T is givenby P(T)= Po exp(-Q/RT)l Solution: The ratio of the polymerization rates at any twotemperatures is:P(T1) = Po exp(-Q/RT1) = exp(-Q/R(1/T1 - 1/T2)P(T2) Po exp(-Q/RT2) P(T1) = exp-Q/R(T2-T1)/T1T2) = exp-Q/R(T/T1T2)P(T2)This form of the expression shows that the problem can not be solved with the information given. A knowledge of T is not sufficient. We must also know the two temperatures.l Comments: When the temperature increases from 10 C to 20 C, the rate increases by a factor of 3.19. In contrast, a temperature increase from 40 C to 50 C results in a rate increase of 2.59. This example illustrates the general result that a fixed change in temperature has a greater influence on the reaction rate if the average temperature is low.热力学与动力学11为什么高质量的电子导线端点要用金而不是钢、铝或铜制成？ SOLUTION: You do not want the resistance of the connection to increase with time. Oxides are generally good electrical insulators（绝缘体） . Steel points rust and the oxide prevents them from working. COMMENTS: In some electronic devices a slight impedance（电阻，阻抗） increase due to oxide formation can cause a circuit to fail catastrophically, destroying a number of components.热力学与动力学12将 Al2O3还原为 2Al和（ 1.5O2）是一个需要越过能障的过程。由于热力学上有利于氧化物而不是元素的分离，怎样才能将铝矿石还原为纯铝？ Given: Oxide represents a lower energy state than pure Al. Solution: Thermodynamics describes the direction of spontaneous change. That is, balls roll down hill and Al will transform to Al2O3 if the kinetics are favorable and no other factors are acting on the system. We know, however, that a ball can be moved uphill if energy is supplied to the system (i.e. if it is carried uphill). Furthermore, it may remain at a higher elevation if there are activation barriers associated with its return to the lowest energy position. Similar logic applies to the reduction of Al2O3 to 2Al+1.5O2. If man supplies (thermal) energy, the metal can be extracted from its ore and will remain in a metastable state. However, the metal will return to its more stable oxide at a later time if conditions permit.热力学与动力学13对于每种一次键类型，回答下面的问题：1）这种一次键通常包含负电性原子、正电性原子或两种原子吗？2）成键电子是共用的还是转移的？3）如果成键电子是共用的，它们在空间上是定域的还是不定域的？一次键Primary bond typeTypes of atoms usually involved Bonding electrons shared or transferredIf sharing occurs is it localized or delocalizedionic electroneg. and electropositive transferred covalent electronegative shared localizedmetallic electropositive shared delocalized 14.确定下列每种材料中最可能的一次键类型： O2,NaF,InP,Ge,Mg,CaF2,SiC,(CH2)n,MgO,CaO 一次键Element ElectronegativityNo. of valence electronsO 3.44 6Na 0.93 1F 3.98 7In 1.78 3P 2.19 5Ge 2.01 4Mg 1.31 2Ca 1.00 2Si 1.90 4C 2.55 4H 2.20 1 O2, EN=0 so that the bond is not ionic. Since O is an electronegative element and the average number of valence electrons is 6, we predict a covalent bond. NaF, EN = EN(F)-EN(Na) = 3.98-0.93 = 3.05. This EN corresponds to a bond that is approximately 90% ionic. InP, EN = EN(P)-EN(In) = 2.19-1.78 = 0.41. Since this bond is only about 4% ionic, we must examine the average number of electrons, NVE. Since NVE = (3+5)/2 = 4, we predict that the bond will be covalent. Ge, EN=0 so the bond is not ionic. Ge is neither strongly electropositive or electronegative, but it does have NVE=4. Thus we predict covalent bonding. Mg, EN=0 so the bond is not ionic. Mg is an electropositive element with NVE =2. It will have metallic bonds. CaF2, EN = EN(F)-EN(Ca) = 3.98-1.0 = 2.98. This EN corresponds to a bond that is approximately 89% ionic. SiC, EN = EN(C)-EN(Si) = 2.55-1.90 = 0.65. Since this corresponds to a bond that is only 10% ionic, we must consider NVE. The average number of valence electrons is (4+4)/2 = 4 so the bond is predicted to be covalent. (CH2), EN = 2.55-2.20 = 0.35 so the bond is not ionic.Although the average NVE in this compound is (4+1)/2 = 2.5, the bond is predicted to be covalent because H is a recognized exception to the trend and is known to favor the formation of covalent bonds. MgO, EN = EN9O)-EN(Mg) = 3.44-1.31 = 2.13 which corresponds to a bond that is 68% ionic. CaO, EN = EN(O)-EN(Ca) = 3.44-1.0 = 2.44 which corresponds to a bond that is 77% ionic.15.说明 C中 4个未成对电子怎样形成共价键，即说明电子的共用及 “全填满价电层规 ”的应用。 DATA: Carbon forms four equal bonds. SKETCH: Shown is methane（甲烷） , simply as an example: SOLUTION: The 8 dots represent the electrons: 4 come from the central carbon and 1 from each of the 4 hydrogen. The electrons are localized between the atoms sharing the electrons.一次键16：库伦力是怎样随电荷间的距离而变化的？哪种熟知的力也表现出这一特点？在怎样的电荷间隔下 dF(库仑 )/dx最大和最小？ Solution: The Coulombic Force varies inversely with the square of the separation distance between the charge centers. Another familiar example of an inverse square law is the force of gravity. A function of this form has no local maxima or minima because the derivative of equation has a nontrivial values for which it is equal to zero and the maximum value occurs as x approaches infinity.一次键18：为什么共价键只在负电性元素间形成？ Assumptions: During bonding, atoms seek to obtain a filled valence electron shell. Solution: Electronegative elements generally have nearly filled valence shells and are seeking a few additional electrons. If electropositive atoms are nearby, they can transfer electrons to the electronegative atoms and an ionic bond is formed. If, however, only electronegative atoms are present then the only way they can all acquire extra electrons is to share them.This is the definition of a covalent bond. Comments: Hydrogen also forms covalent bonds. 一次键19：若 O与其本身形成共价键， Si也与其本身形成共价键，那么，认为 SiO成共价键是合理的，对吗？ Data: From Appendix B: EN(Si)=1.90 and EN(O)=3.44. Assumptions: The percent ionic character of a bond is given by the table in Appendix A. Solution: It is impossible to form ionic bonds in any pureelement since there will be no difference in EN values for identical atoms. In the case of Si-O, however, the difference in electronegativity is 1.54. This corresponds to a bond that is approximately 45% ionic and 55% covalent. Comments: This type of primary bond is often described as a mixed ionic/covalent bond. 一次键20陶瓷或离子固体有可能成为适当的导电体吗？需要怎样才有可能？ Assumptions: High electrical conductivity