第二章 材料的结构(答案更正)

第二章材料的结构Chapter 2 Structure of materials1. 原子间的结合键共有几种？各自特点如何？How many kinds of binding bonds among atoms? What are their characteristics?原子间的结合键共有5种，分别是金属键、离子键、共价键、氢键和范德华键。

There are five kinds of binding bonds among atoms, namely, metallic bond, ionic bond, covalent bond, hydrogen bonding and van der Waals bond.1）金属键是金属中的正离子和自由电子之间形成的键合。

其特点是电子共有化，可以自由流动。

金属键无方向性和饱和性。

金属键合力较强，键能为几百kJ/mol。

Metallic bond is formed between the positive ions and free electrons. Metal bond is characterized by many sharable electrons and free mobile electrons. It is nondirectional and unsaturated. Metal Binding force in metal bond is strong and the bond energy may up to hundreds of kJ / mol. 2）离子键是正负离子之间由于静电吸引而形成的键合，离子键无方向性也无饱和性，配位数高。

离子键具有较强的键合力，键能为几百到几千kJ/mol。

Ionic bond is formed through electrostatic attraction between oppositely charged ions. It is nondirecitonal and unsaturated. It has high coordination number, and the bond force may up to several hundreds to thousands of kJ/mol.3）共价键是原子间共用电子对而形成的键合，有方向性和饱和性，配位数低。

共价键的键强度较高，键能通常为几百kJ/mol。

Covalent bond is formed by sharing of pairs of electrons between atoms. It is directional and saturated. It has small coordination number and the bond energy is several hundreds of kJ/mol.4）氢键是由氢原子与电负性较大的原子之间形成的X—H…Y的键合。

具有方向性和饱和性。

氢键的键合较弱，一般为几十kJ/mol，但是对材料性能的影响较大。

Hydrogen bond is formed between hydrogen atom and two electronegative atoms such as N—H…O. It is directional and saturated. The hydrogen bond is weak, and it is only dozens of kJ/mol，but is has much influence on the materials’ proper ties.5）范德华键是分子间形成的一种作用力，其键能很弱，为几到几十kJ/mol。

不具有方向性和饱和性。

作用范围在几百个皮米之间。

它对材料的沸点、熔点、汽化热、熔化热、溶解度、表面张力、粘度等物理化学性质有决定性影响。

The van der Waals force is the sum ofattractive or repulsive forces between molecules. It is weak and the bond energy is several to dozens of kJ/mol. The van der Waals force is nondirectional and unsaturated, and the range of the force is only several hundreds of picometers. It has much influence on material’s boiling point, melting point, heat of vaporization and fusion, solubility, surface tension, viscosity and other physical and chemical properties.2.为什么可将金属单质的结构问题归结为等径圆球的密堆积问题？Why the structure of metallic single matter can be considered as the close-packed accumulation of balls with equal radius?答：金属键是由金属阳离子和自由电子形成的结合键，它既无方向性，也无饱和性。

因此，金属晶体中的原子不存在受临近质点的异号电荷限制和化学量比的限制，在一个金属原子的周围可以围绕着尽可能多的又符合几何图形的临近原子。

这样的结构由于充分利用了空间，从而使体系的势能尽可能降低，使体系稳定，故金属晶体具有较高的配位数。

综合以上原因，可将金属单质的结构问题归结为等径圆球的密堆积问题。

Metallic bond is formed between the positive metallic ions and free electrons. It is nondirectional and unsaturated. Therefore, the atoms in metal are not limited by the opposite charges on neighboring particles or by stoichiometric ratio. Atoms can be around as many as possible around a metal atom. In such a structure, the space is fully utilized, so as to minimize the potential energy of the system, so the system is stable, and the metal crystals always have high coordination number. For these reasons, the structure of metallic single matter can be considered as the close-packed accumulation of balls with equal radius?3. 计算体心立方结构和六方密堆结构的堆积系数。

Please calculate the accumulation factor of body-centered cubic and hexagonal close-packed structure.1）体心立方结构2）六方密堆结构1) body-centered cubic 2) hexagonal close-packed structure 4. 试确定简单立方、体心立方和面心立方结构中原子半径和点阵参数之间的关系。

Please determine the relationship between the atomic radius and lattice parameter in simple cubic, body-centered cubic and face-centered cubic structure.1）简单立方结构a=b=c=2R a =β =γ = 90°2）体心立方结构如题3的图，则a=b=c=2.309 a =β =γ = 90°3）面心立方结构点阵参数为a=b=c=2.829 a =β =γ = 90°5.金属铷（Rb）为A2型结构，Rb的原子半径为0.2468 nm，密度为1.53g/cm3，试求晶格参数a和Rb的相对原子质量。

Structure of rubidium (Rb) belongs to A2-type, the atomic radius of Rb is 0.2468 nm, the density is 1.53g/cm3, please calculated the lattice parameter a and the relative atomic mass of Rb.答：A2型结构为体心立方结构，则a=(4/1.732)×R＝0.57 nm 镍的晶格参数为a=b=c=0.57 nm a =β =γ = 90°单位晶胞原子数为2设其相对原子质量为M g/mol，则ρ＝m/V=(2/N A)×M/V N A=6.02×1023个/molV=a3=(0.57 nm)3已知ρ＝1.53 g/cm3代入得M=85.3 g/molAnswer: A2 type structure is body-centered cubic, as shown in the figure. From the figure, we can get that a=(4/1.732)×R＝0.57 nm. The number of atoms in unit cell is 1+8×(1/8)=2. According to the formulaρ＝m/V=(2/N A)×M/V, and as is known, N A=6.02×1023个/mol, V=a3=(0.57 nm)3,ρ＝1.53 g/cm3, so M=85.3 g/mol, that is to say, the relative atomic mass of Rb is 85.3 g/mol.6. FCC结构的镍原子半径为0.1243 nm，试求镍的晶格参数和密度。