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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 corresp
2、onds 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 subshel
3、ls,K: 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(-
4、1/2); 322(1/2); 322(-1/2),2.7 Give the electron configurations for the following ions: Fe2+, Fe3+, Cu+, Ba2+, Br-, and S2-. SOLUTION Fe2+ : 1s22s22p63s23p63d6 Fe3+ : 1s22s22p63s23p63d5 Cu+ : 1s22s22p63s23p63d10 Ba2+ : 1s22s22p63s23p63d104s24p64d105s25p6 Br- : 1s22s22p63s23p63d104s24p6 S 2- : 1s22s22
5、p63s23p6,2.17 (a) 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
6、 character of the interatomic bonds for the following compounds: TiO2, ZnTe, CsCl, InSb, and MgCl2 . SOLUTION,TiO2, XTi = 1.5 and XO = 3.5,ZnTe, %IC=6.05% CsCl, %IC=73.4% InSb, %IC=1.0% MgCl2, %IC=55.5%,2.24，On the basis of the hydrogen bond, explain the anomalous behavior of water when it freezes.
7、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 explanation as to why covalently bonded materials are generally less dense than ionically or metallicall
8、y 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 chlorine. SOLUTION Ge : 4 P : 3 Se : 2 Cl : 1,2-6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
9、 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=0.133/0.414 = 0.321 nm CN=4 R = r/0.225=0.133/0.225 = 0.591 nm CN=8 R = r+/0.732=0.133/0.732
10、= 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 by the two vectors in sketch (a)? b What are the indices for the two planes drawn i
11、n 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 012 direction 2, x y z Projections a/2 b/2 -c Projections in terms of a, b, and c 1/2 1/2 -1 Reduction to integers 1 1 -2 Enclosure 11 2 （b）Plane 1， ：1/2 ： ； 0：2
12、：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 b,3.53 Determine the indices for the directions shown 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 0,3.57 Determine the Miller indices for
13、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 2/1 -2/1 (3 2 2),3.58 Determine the Miller indices for the planes shown in the following unit cell:,plane A 以（0，1，0）为新原点 x y z 2/3a -b c/2 2/3 -1 1/2 3/2 -1/1 2/1 3/2 -2/2 4/2 (3 2 4),3.61* Sketch within a cubi
14、c unit cell the following planes:,a,3.62 Sketch the atomic packing of (a) the (100) plane for the FCC crystal structure, and (b) the (111) plane for the BCC crystal structure (similar to Figures 3.24b and 3.25b).,(a) FCC: (100) plane (b) BCC: (111) plane,3.81The metal iridium has an FCC crystal stru
15、cture. If the angle of diffraction for the (220) set of planes occurs at 69.22(first-order reflection) when monochromatic x-radiation having a wavelength of 0.1542 nm is used, compute (a) the interplanar spacing for this set of planes, and (b) the atomic radius for an iridium atom. SOLUTION n= 2 d sin d=n/(2sin) =0.1542nm/2sin(69.22/2)=0.136nm d = a / h2 + k2 + l2 a=d h2 + k2 + l2 =0.384nm FCC, a = 4R /2 则R=a 2 /4=0.136nm,2-14计算（a）面心立方金属的原子致密度；（ b）面心立方化合物NaCl的离子致密度（离子半径r(Na+)=0.097，r(Cl-) =0.181）；（C
