第7-9章Keypoints/concepts1. Lattice of crystal structure: translation symmetrya lattice point = a structure motif -- unit cellCrystal systems (7), Bravais Lattice (14)23Symmetry operations (point &translation)Crystallographicpoint groups(32), space groups (230), miller index of crystalplane, d-spacing etc.X-ray diffraction,Laue equation,Bragg's Law,reciprocallattice,Ewaldsphere,structuralfactor,systemabsencegeneral process of x-ray crystal structure determination.Close-packing of spheres (ccp/A1,hcp/A3,bcp/A2) in metals5andionic compounds,coordination ofcationsCrystal structuresofsometypicalioniccompounds5
第7-9章 Key points/concepts 1. Lattice of crystal structure: translation symmetry a lattice point = a structure motif - unit cell 2. Crystal systems (7), Bravais Lattice (14) 3. Symmetry operations (point & translation) Crystallographic point groups(32), space groups (230), miller index of crystal plane， d-spacing etc. 4. X-ray diffraction, Laue equation, Bragg’s Law, reciprocal lattice, Ewald sphere, structural factor, system absence, general process of x-ray crystal structure determination. 5. Close-packing of spheres (ccp/A1,hcp/A3,bcp/A2) in metals and ionic compounds, coordination of cations. 6. Crystal structures of some typical ionic compounds
Example:p224,7.2conceptoflatticeAstructuremotif (+ occupying space)=a latticepointEach lattice point has identical surroundingsAlatticefulfills translationsymmetryDifferencesbetweena real crystal structureanditslattice.Keypointis tofindthestructure motif(basis)thatfulfillstranslationsymmetry!
Example: p224, 7.2 - concept of lattice • A structure motif （+ occupying space) = a lattice point • Each lattice point has identical surroundings. • A lattice fulfills translation symmetry. • Differences between a real crystal structure and its lattice. Key point is to find the structure motif (basis) that fulfills translation symmetry!
SnFp.227,7.2614/mmmSn: (0,0,0), (1/2,1/2,1/2)F: (0,1/2,0), (1/2,0,0)(0,0,0.237),(0,0,-0.237)1)body-centredtetragonal793pl0.237c0.237c2 lattice point within a unit cell0.237cTheblackdots (2 Sn)and redballs(4F)are definedbythe coordinatesgiven!0.237cOther4Fatomscanbeobtained404pmbytranslationoperation (2LP)FFSnEachSnatomislocatedina2distortedoctahedral hole;R(Sn-F)2=0.5x404=202pmR(Sn-F)1=0.237x793= 187.9pm;2Key: Figure out the atoms within a LP!
p.227, 7.26 SnF4 Sn: (0,0,0), (1/2,1/2,1/2) F: (0,1/2,0), (1/2,0,0) (0,0,0.237), (0,0,-0.237) 1) body-centred tetragonal 2 lattice point within a unit cell. The black dots (2 Sn) and red balls (4 F) are defined by the coordinates given! Other 4 F atoms can be obtained by translation operation (2 LP). 2) Each Sn atom is located in a distorted octahedral hole. R(Sn-F)1 =0.237x 793 = 187.9pm; R(Sn-F)2 = 0.5x404=202 pm Key: Figure out the atoms within a LP!
Simple cubic crystal - example: CsClp.286 9.13Onlyonelatticepointwithinaunitcell.Eachlatticepointcontains2atomsCl (0,0,0), Cs(1/2,1/2,1/2);The structural factor is22元i（hx+kyi+lz,）ZFhklfe=i=l= fc, + fc,e(h+k+I)Therefore, all possible diffractions are observable without systemabsence!However.Fhu=fer+fcsIfh+k+l=2nStrongest diffractionIf h+k+l =2n+1,Fhu = fer- fesWeakest diffraction
Simple cubic crystal – example: CsCl p.286 9.13 • Only one lattice point within a unit cell. • Each lattice point contains 2 atoms, Cl (0,0,0), Cs(1/2,1/2,1/2); • The structural factor is Therefore, all possible diffractions are observable without system absence! However, 2 1 2 i i( h x ky lz ) hkl i i i i F f e π i( h k l ) Cl Cs f f e π If h+k+l =2n, Strongest diffraction hkl Cl Cs If h+k+l =2n+1, F f f Weakest diffraction hkl Cl Cs F f f
p227, 7.21Face-centered cubic crystal-general caseLattice points (LPs): (0,0,0), (1/2,1/2,0), (0,1/2,1/2), (1/2,0,1/2): Suppose each lattice point contains n atoms, ((xj,yj,z,) (j=l,...,n)Each unit cell contains N=4n atoms, e.g., an atom A(xi,yi,z) in oneLP has other three equivalent A atoms within the same unit cell!Then the structure factor isN2元i（hx+hy,+z）LfeSum up over all atomshklilwithin a unit cell!khhK2元i2元i2元i2元i(hx,+hyj+lzj）CO2Fhki =[1+e>f+e+eFeFromtranslationsymmetryoffcc!NowsumupoverallThus, when h,k,I are neither all even nor all oddatoms withinaLP!=0system absence!hkiFurthermore,whenh.k.l areall even orall oddndiffractionobservable!2元i(hx,+hyj+l=,)Fnkl一j=l
Face-centered cubic crystal – general case p227, 7.21 • Lattice points (LPs): (0,0,0), (1/2,1/2,0), (0,1/2,1/2), (1/2,0,1/2) • Suppose each lattice point contains n atoms, {(xj ,yj ,zj )} (j=1,.,n) • Each unit cell contains N=4n atoms, e.g., an atom A(xi ,yi ,zi ) in one LP has other three equivalent A atoms within the same unit cell! • Then the structure factor is • Thus, when h,k,l are neither all even nor all odd, system absence! • Furthermore, when h,k,l are all even or all odd, diffraction observable! N i i h x ky lz hkl i i i i F f e 1 2 ( ) i( h x ky lz ) n j j ) k l ) i( h l ) i( h k i( j j j [ e e e ] f e π π π π 2 1 2 2 2 2 2 2 2 2 2 Fhkl 1 Fhkl 0 i( h x ky lz ) n j j j j j f e 2π 1 Fhkl 4 Sum up over all atoms within a unit cell! Now sum up over all atoms within a LP! From translation symmetry of fcc!