（a）directctions in terms of a, b, and c Reduction to integers Enclosure
xyz 0a b/2 c 0 1/2 1
01 2 [012]
direction 2,
Projections Projections in terms of a, b, and c Reduction to integers Enclosure
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(-1/2); 322(1/2); 322(-1/2)
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 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
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
interatomic bonds for the following compounds: TiO2, ZnTe, CsCl, InSb, and MgCl2 .
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
xyz a/2 b/2 -c 1/2 1/2 -1 1 1 -2
[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:
SOLUTION：
已知：铱FCC的（220） 晶面,2θ= 69.22°;λ= 0.1542 nm； n = 1
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=55.5%
plane A 以（0，1， 0）为新原点 xyz
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)
plane B ( 2 2 1)
3.61* Sketch within a cubic unit cell the following planes:
a
b
（c）[0 1 2]
（d）[1 3 3]
（e）[1 1 1]
（f）[1 2 2]
（g）[1 2 3]
（h）[1 0 3]
3.53 Determine the indices for the directions shown in the following cubic unit cell:
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- : 1s22s22p63s23p6
CN=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 by the two vectors in sketch (a)? b What are the indices for the two planes drawn in sketch (b)?
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
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 metallically bonded ones.
Direction C xyz
1/3a -b -c 1/3 -1 -1 1 -3 -3
[1 3 3]
Direction D xyz a/6 b/2 -c 1/6 1/2 -1 1 3 -6
[1 3 6]
3.57 Determine the Miller indices for the planes shown in the following unit cell:
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?
(a) FCC: (100) plane (b) BCC: (111) plane
3.81The metal iridium has an FCC crystal structure. If the angle of diffraction for the (220) set of planes occurs at 69.22°(first-order
plane A xyz a /3 b/2 -c/2 1/3 1/2 -1/2 3/1 2/1 -2/1 (3 2 2)
plane B (1 0 1)
3.58 Determine the Miller indices for the planes shown in the following unit cell:
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.
Direction A: xy z
-2/3a b/2 0c -2/3 1/2 0
-4 3 0 [4 3 0]
4
Direction A: xyz
-2/3 a b/2 0c -2/3 1/2 0 -4 3 0 [4 3 0]
Direction B: xyz
2/3 a -b 2/3 c 2/3 -1 2/3 2 -3 2 [2 3 2]
习题讲解
第一次作业
英文 2.6 Allowed values for the quantum numbers of electrons are as follows: