《结构化学 Structural Chemistry》课程教学资源（课件讲稿）Selected Problems and Solutions for Chapters 5-9

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!

Face-centered cubic crystal -p227, 7.23Special case: NaClUnit cell contains 4 lattice points, or 4NaClEach lattice point (LP) corresponds to a NaClPut Cl at (0,0,0), then a neighboring Na at (1/2,0,0)Onlywhenh.k.l are all evenorall odd candiffractions be observed!2i(hx,+hy,+t,) = 4[ fe, + fnehi ]Fhkl = 4j-lNowsumupoverallatomswithinaLP!Case 1: if h =2n (note we also have l-2n and k-2n)Strong diffraction!Fhk1 =4[ fc, + fNa ]Case2:ifh=2n+1(notewealsohave[=2n+1,k=2n+1)Weakdiffraction!Fhki =4[ fc, - fna]

Face-centered cubic crystal – Special case: NaCl p227, 7.23 • Unit cell contains 4 lattice points, or 4NaCl • Each lattice point (LP) corresponds to a NaCl. • Put Cl at (0,0,0), then a neighboring Na at (1/2,0,0). • Only when h,k,l are all even or all odd can diffractions be observed! Case 1: if h = 2n (note we also have l=2n and k=2n) Strong diffraction! Case 2: if h = 2n+1 (note we also have l=2n+1, k=2n+1) Weak diffraction! f e [ f f e ] h i C l N a i( h x ky lz ) j j π j j j π       F 4 4 2 2 1 hkl [ f f ]  4 Cl  Na F hkl [ f f ]  4 Cl  Na F hkl Now sum up over all atoms within a LP!

Face-centered cubic crystal -Special case: ZnS (sphalerite)Unit cell contains 4 lattice points, or 4ZnSEach lattice point corresponds to a ZnSS(0,0,0), Zn(1/4,1/4,1/4)(different elements!)When h,k.l are all even or all odd, diffractions observable2e2mi(h+)+,)=4[s+fe(h++)/2 ]Fhkl=4.ej-1> (111),(200),(220),(311),(222),(400),(331),(420),(422),Case 1: if h+k+l = 4n, e.g., (220),(400),(440).Strongestdiffraction!Fhk1 = 4[ fs + fzmn ]Case 2: if h+k+l = 4n+2, e.g., (200),(222),(420),(442)FWeakestdiffraction!=4[fs- fzn]hkl

Face-centered cubic crystal – Special case: ZnS (sphalerite) • Unit cell contains 4 lattice points, or 4ZnS • Each lattice point corresponds to a ZnS. S(0,0,0), Zn(1/4,1/4,1/4) (different elements!) • When h,k,l are all even or all odd, diffractions observable,  (111),(200),(220),(311),(222),(400),(331),(420),(422),. Case 1: if h+k+l = 4n, e.g., (220),(400),(440). Strongest diffraction! Case 2: if h+k+l = 4n+2, e.g., (200),(222),(420),(442). Weakest diffraction! f e [ f f e ] i( h k l )/ S Zn i( h x ky lz ) j j 2 j j j 2 2 1 Fhkl 4 4          π π [ f f ]  4 S  Zn F hkl [ f f ]  4 S  Zn F hkl

Face-centred cubiccrystal--Derivethesystemabsenceofdiamond!Specialcase:DiamondO,7-Fd3mLattice points: (0,0,0)+, (1/2,1/2,0)+, (0,1/2,1/2)+, (1/2,0,1/2)+Each LP contains two C atoms (i.e., structure motif=2C)C1-- (0,0,0), C2-- (1/4,1/4,1/4) (the same element)The other six C atoms within a unit cell can be derived as(1/2,1/2,0),(3/4,3/4,1/4) ; (0,1/2,1/2),(1/4,3/4,3/4);(1/2,0,1/2),(3/4,1/4,3/4)Such an arrangement of C atoms produces new translationsymmetry elements, i.e., screw axes and d glide planes, which inturn introduce special system absence of diffractions (in additionto the system absence from normal FCC lattice!!!!

• Lattice points: (0,0,0)+, (1/2,1/2,0)+, (0,1/2,1/2)+, (1/2,0,1/2)+ • Each LP contains two C atoms (i.e., structure motif =2C) C1- (0,0,0), C2- (1/4,1/4,1/4) (the same element) • The other six C atoms within a unit cell can be derived as (1/2,1/2,0), (3/4,3/4,1/4) ; (0,1/2,1/2), (1/4,3/4,3/4) ; (1/2,0,1/2), (3/4,1/4,3/4) • Such an arrangement of C atoms produces new translation symmetry elements, i.e., screw axes and d glide planes, which in turn introduce special system absence of diffractions (in addition to the system absence from normal FCC lattice !!!!!!! Special case: Diamond Oh 7 - Fd3m Face-centred cubic crystal - Derive the system absence of diamond!

Face-centred cubiccrystal--Derivethesystemabsence ofdiamondSpecialcase:DiamondO,7-Fd3mSuch an arrangement of C atoms produces new translationsymmetry elements, i.e., screw axes and d glide planes, which inturn introduce special system absence of diffractions (in additiontothe system absencefrom normalFCC lattice!!!d glideplane1/8,3/85/8,7/8Let's derivethe structural factor of diamond tounravelitssystemabsence

• Such an arrangement of C atoms produces new translation symmetry elements, i.e., screw axes and d glide planes, which in turn introduce special system absence of diffractions (in addition to the system absence from normal FCC lattice !!!!!!! Special case: Diamond Oh 7 - Fd3m Sideview topview 41 d glide plane 1/8,3/8, 5/8,7/8 Face-centred cubic crystal - Derive the system absence of diamond! • Let’s derive the structural factor of diamond to unravel its system absence

Face-centered cubic crystal -Special case: DiamondNow sum up over all atoms within a unit cell!82mi(hx+ky,+lz)Fnk!fceNow sum up over all atoms within a LP!i=l1125I七2元2元i2元i)2mi（hx,+hy,+z）1222"x2= fc[l+e+e+eej-1（h+)K-212元i02元i02元i022222J(1 + e(h+k+1)/2)=fc[l+e+e+e(Note: two carbon atoms within a lattice point: (0,0,0), (1/4,1/4,1/4)>Systemabsencehkh1k12元i（2元i2元222222a)[l+e=0or+e+eb) (1+emi(h+k+1)/2=0i.e., a) h,k,l are neither all even nor all odd! & b) h+k+l= 4n+2> Observable diffractions: (111), (220),(311),(400),(331),(422) &If h+k+l=4n, Fhki = 8fc, (220),(400)... strongest diffraction!

Face-centered cubic crystal – Special case: Diamond (Note: two carbon atoms within a lattice point: (0,0,0), (1/4,1/4,1/4)) System absence: i.e., a) h,k,l are neither all even nor all odd! & b) h+k+l = 4n+2  Observable diffractions: (111), (220),(311),(400),(331),(422) & If h+k+l=4n, Fhkl = 8fC , (220),(400). strongest diffraction! [1 ](1 ) [1 ] F ( )/ 2 ) 2 2 ) 2 ( 2 2 ) 2 ( 2 2 2 ( 2 ( ) 2 1 ) 2 2 ) 2 ( 2 2 ) 2 ( 2 2 2 ( 2 ( ) 8 1 hkl i h k l k l i h l i h k i C i h x ky lz j k l i h l i h k i C i h x ky lz i C f e e e e f e e e e f e j j j i i i                                     Now sum up over all atoms within a unit cell! Now sum up over all atoms within a LP! b) (1 ) 0 ) [1 ] 0 or ( )/ 2 ) 2 2 ) 2 ( 2 2 ) 2 ( 2 2 2 (            i h k l k l i h l i h k i e a e e e    