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1、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量子数(n、l、ml、ms ),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
2、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 0，1，2，3n-1(亚层）s, p, d, f(这些符号均来自光谱学),轨道角动量量子数 l, 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, 2l （|m| l ） m 值相同的轨道互为等价(简并)轨道,磁量子数 ml,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：1
3、s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, 7p,l 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4d,2.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 (f
4、rom 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 config
5、uration 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,
6、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 woul
7、d 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 bal
8、ance 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
9、的夹角为109.5。,原子结构,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 n
10、ot 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 rai
11、sed 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
12、 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 +273Assumptions: 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)
13、 = 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？,Da
14、ta: R= 8.314 J/mole-K, K= C +273 Assumption: Polymerization rate at temperature T is givenby P(T)= Po exp(-Q/RT) 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)
15、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.Comments: When the temperature increases from 10 C to 20 C, the rate increases by a factor of 3.19. In contrast, a temperat
16、ure 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 resist
17、ance 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.,
