《材料科学与工程基础习题评讲-1》由会员分享，可在线阅读，更多相关《材料科学与工程基础习题评讲-1（59页珍藏版）》请在金锄头文库上搜索。
1、习题讲解第一次作业英文2.6 Allowed values for the quantum numbers of electrons are as follows: The relationships between n and the shell designations are noted in Table 2.1. Relative to the subshells, l 0 corresponds to an s subshell l 1 corresponds to a p subshell l 2 corresponds to a d subshell l 3 correspond
2、s to an f subshell For the K shell, the four quantum numbers for each of the two electrons in the 1s state, in the order of n l ml ms , are 100(1/2) and 100( -1/2 ). Write the four quantum numbers for all of the electrons in the L and M shells, and note which correspond to the s, p, and d subshellsK
3、: s: 100(1/2); 100(-1/2)L: s: 200(1/2); 200(-1/2)p: 210(1/2); 210(-1/2); 21-1(1/2); 21-1(-1/2); 211(1/2); 211(-1/2)M: s: 300(1/2); 300(-1/2)p: 310(1/2); 310(-1/2); 31-1(1/2); 31-1(-1/2);311(1/2); 311(-1/2)d: 320(1/2); 320(-1/2); 32-1(1/2); 32-1(-1/2); 321(1/2); 321(-1/2);32-2(1/2); 32-2(-1/2);322(1/
4、2); 322(-1/2)2.7 Give the electron configurations for the following ions: Fe2+, Fe3+, Cu+, Ba2+, Br-, and S2-. SOLUTION Fe2+ : 1s22s22p63s23p63d6Fe3+ : 1s22s22p63s23p63d5Cu+ : 1s22s22p63s23p63d10Ba2+ : 1s22s22p63s23p63d104s24p64d105s25p6 Br- : 1s22s22p63s23p63d104s24p6S 2- : 1s22s22p63s23p6 2.17 (a)
5、 Briefly cite the main differences between ionic, covalent, and metallic bonding. (b) State the Pauli exclusion principle. SOLUTION (a) 离子键： 无方向性 球形正、负离子堆垛取决 电荷数电荷平衡体积（离子半径）金属键： 无方向性 球形正离子较紧密堆垛共价键： 有方向性、饱和性，电子云最大重叠 (b)原子中的每个电子不可能有完全相同的四个量子 数（或运动状态）2.19 Compute the percents ionic character of the int
6、eratomic bonds for the following compounds: TiO2, ZnTe, CsCl, InSb, and MgCl2 . SOLUTION由公式：已知：TiO2, XTi = 1.5 and XO = 3.5 ZnTe,已知：XZn = 1.6 and XTe = 2.1 ，故，%IC=6.05%CsCl,已知： XCs = 0.7 and XCl = 3.0 ， 故： %IC=73.4%InSb,已知： XIn = 1.7 and XSb = 1.9， 故： %IC=1.0%MgCl2,已知：XMg = 1.2 and XCl = 3.0故： %IC=5
7、5.5%2.24，On the basis of the hydrogen bond, explain the anomalous behavior of water when it freezes. That is, why is there volume expansion upon solidification? 水冻结时结晶，非球形的水分子规整排列 时受氢键方向性和饱和性的更强限制， 不能更紧密地堆积，故密度变小，体积 增大。2-7影响离子化合物和共价化合物配位数的 因素有那些？离子化合物： 体积电荷共价化合物： 价电子数电子云最大重叠第二次作业2.18 Offer an explan
8、ation as to why covalently bonded materials are generally less dense than ionically or metallically bonded ones.共价键需按键长、键角要求堆垛， 相 对离子键和金属键较疏松 2.21Using Table 2.2, determine the number of covalent bonds that are possible for atoms of the following elements: germanium, phosphorus, selenium, and chlori
9、ne. SOLUTIONGe : 4P : 3Se : 2Cl : 12-6按照杂化轨道理论，说明下列的键合形式：（1）CO2的分子键合 C sp 杂化 （2）甲烷CH4的分子键合 C sp3杂化 （3）乙烯C2H4的分子键合 C sp2杂化 （4）水H2O的分子键合 O sp3杂化 （5）苯环的分子键合C sp2杂化 （6）羰基中C、O间的原子键合 C sp2杂化 2-10 当CN=6时，K+离子的半径为0.133nm (a) 当CN=4时，对应负离子半径是多少？(b) 当CN=8时，对应负离子半径是多少？ 若(按K+半径不变) 求负离子半径, 则：CN=6 R = r/0.414=0.13
10、3/0.414 = 0.321 nmCN=4 R = r/0.225=0.133/0.225 = 0.591 nmCN=8 R = r+/0.732=0.133/0.732 = 0.182 nm第三次作业3.48 Draw an orthorhombic unit cell, and within that cell a 121 direction and a (210) plane.3.50 Here are unit cells for two hypothetical metals:a. What are the indices for the directions indicated
11、by the two vectors in sketch (a)? b What are the indices for the two planes drawn in sketch (b)?（a）direction 1,x y z Projections 0a b/2 c Projections in terms of a, b, and c 0 1/2 1 Reduction to integers 0 1 2 Enclosure 012direction 2,x y z Projections a/2 b/2 -c Projections in terms of a, b, and c
12、1/2 1/2 -1 Reduction to integers 1 1 -2 Enclosure 11 2（b）Plane 1， ：1/2 ： ； 0：2：0 ； (020)Plane 2， 1/2：-1/2 ： 1 ； 2：-2：1； (2 2 1)3.51* Within a cubic unit cell, sketch the following directions:a bn（c）0 1 2 （d）1 3 3n（e）1 1 1 （f）1 2 2n（g）1 2 3 （h）1 0 33.53 Determine the indices for the directions shown
13、in the following cubic unit cell:Direction A:x y z-2/3a b/2 0c-2/3 1/2 0-4 3 0 4 3 0Direction A: Direction B:x y z x y z-2/3 a b/2 0c 2/3 a -b 2/3 c -2/3 1/2 0 2/3 -1 2/3-4 3 0 2 -3 24 3 0 2 3 2Direction C Direction Dx y z x y z1/3a -b -c a/6 b/2 -c1/3 -1 -1 1/6 1/2 -11 -3 -3 1 3 -61 3 3 1 3 6 3.57 Determine the Miller indices for the planes shown in the following unit cell:plane A x y z a /3 b/2 -c/2 1/3 1/2 -1/2 3/1 2/
