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清华大学:《材料科学基础》课程教学资源(PPT课件讲稿)Chapter 1 .2 Atomic bonding

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1.2 Atomic bonding Bonding forces and energies FF net force attractive force repulsive force
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31.2 Atomic bond ing Bond ing forces and energies FN=FA+FR F入 N: net force F: attractive force FR: repulsive force

§1.2 Atomic bonding ◆Ⅰ.Bonding forces and energies FN : net force FA : attractive force FR : repulsive force FN = FA + FR

Atrrctve Force Net 号PEgL Force Repulsive Force Distance between atoms

Bonding energies E=「Far For atomic systems Ex=E+E N R

 E = Fdr For atomic systems: EN = EA + ER Bonding energies:

Low Temperature (Equilibrium Spacing) Atom Distance

l Primary interatomic bonds 1. lonic bonding lonic bonding is always found in compounds that are composed of both metallic and nonmetallic elements, elements that are situated at the horizontal extremities of the periodic table

◆ Ⅱ.Primary interatomic bonds 1. Ionic bonding Ionic bonding is always found in compounds that are composed of both metallic and nonmetallic elements, elements that are situated at the horizontal extremities of the periodic table

( Na

◇ Interionic Forces nb attractive 4兀na repulsive n+1 nb →F 45a2 n+1 Z1, Z2=number of electrons removed or added from the atoms during the ion formation e=electron charge a =interionic separation distance Co- permittivity of free space=8.85X10-2C2/(N m2)

2 0 2 1 2 attractive 4 a Z Z e F   = − repulsive +1 = − n a nb F 2 1 0 2 1 2 net 4 +  = − − n a nb a Z Z e F   Interionic Forces Z1 , Z2 =number of electrons removed or added from the atoms during the ion formation e =electron charge a =interionic separation distance ε0 =permittivity of free space=8.85×10-12C2 /(N·m2 )

o Interionic Energies b E net 4丌Eaa Attractive Repul energy energy

Interionic Energies n a b a Z Z e E = + + 0 2 1 2 net 4  Attractive energy Repulsive energy

Example problem 2.1 If the attractive force between a pair of Mg2tand S2- is 1.49X10-N and if the S2- ion has a radius of 0.184nm, calculate a value for the ionic radius of the Mg+ ion in nanometers Solution attractive 460a (2)(-2(1.60×10C a0-14 7 So Attractive V48.85×10C2/(N,m2)](.49×10N) =0.249mm 又:a Mg2+=a0-rs2=0249m-0.184m=0.065m

Example problem 2.1 If the attractive force between a pair of Mg2+and S2- is 1.49×10-8N and if the S2- ion has a radius of 0.184nm, calculate a value for the ionic radius of the Mg2+ ion in nanometers. Solution 2 0 2 1 2 attractive 4 a Z Z e F    = − 0.249nm 4 [8.85 10 C (N m ](1.49 10 N) (2)( 2)(1.60 10 C) 4 1 2 2 2 8 1 9 2 0 attractive 2 1 2 0 =    − −  = −  = − − −   F  ) Z Z e a 2 2 0.249nm 0.184nm 0.065nm 2 2 M g 0 M g 0  = − = − = = + + − + − S S r a r 又a r r

2. Covalent bonding Materials with covalent bonding are characterized by bonds that are formed by sharing of valence electrons among two or more atoms o:::o o:o:o:o :o:0:o

2. Covalent bonding Materials with covalent bonding are characterized by bonds that are formed by sharing of valence electrons among two or more atoms

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