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

p227, 7.22Body-center crystal -General case: each lattice point contains n atomsThe total number of atoms within a unit cell is 2n: For jth atom in a structure motif (a lattice point): (xj,yj,z): Its body-center equivalent is: (0.5+x, 0.5+yj, 0.5+z)2nSum up over all atoms2元i(hx+hy,+lz）ZFhki =fePwithin a unit cell!i=1n2m[h(→+x;)+k(+y,)+(+z,)]2mi(hx,+hy,+lz,)ZufeLLj=lFrom translation-symmetryn2mi(hx,+y,+lz,)[1 +em(h+k+l)ZJf,eSum up over all atoms in a LP!j=lWhile h+k+l =2n+1Systemeri(h+k+l)（2n+1)）元i=-1→Fhu =0absence

Body-center crystal – p227, 7.22 General case: each lattice point contains n atoms • The total number of atoms within a unit cell is 2n; • For jth atom in a structure motif (a lattice point): (xj ,yj ,zj ) • Its body-center equivalent is: (0.5+xj , 0.5+yj , 0.5+zj )     n i i( h x ky lz ) hkl i i i i F f e 2 1 2π                    n j i h x ky lz j i h k l i h x k y l z j i h x ky lz j n j j j j j j j j j j e f e f e f e 1 ( ) 2 ( ) )] 2 1 ) ( 2 1 ) ( 2 1 2 [ ( 2 ( ) 1 [1 ] { }     While h+k+l =2n+1, 1 0 2 1         hkl i( h k l ) ( n ) i e e F π π System absence Sum up over all atoms within a unit cell! Sum up over all atoms in a LP! From translation-symmetry

Body-center crystal - special case(p.227, 7.24)Each lattice point (LP) contains 2 atoms, A (O,O,O), B(x,y,z):Thus in another LP, A (1/2,1/2,1/2), B (x+1/2,y+1/2,z+1/2)4The structural factor isfe2mi(hg+hy+E,)ZEhkli=l22元i(hx,+kyj+lz,)=[1+em(h)]Zf,e?j=l=[1+e(h++1)]×[fa+ fre2?m(hx+h+1)]Sum up over all atoms in a LP!Translation-symmetryterm!While h+k+l =2n+1System absenceexi(h+k+l) =e(2n+1)i =-1=→FChkl

Body-center crystal – special case (p.227, 7.24) • Each lattice point (LP) contains 2 atoms, A (0,0,0), B(x,y,z); • Thus in another LP, A (1/2,1/2,1/2), B (x+1/2,y+1/2,z+1/2) • The structural factor is     4 1 2 i i( h x ky lz ) hkl i i i i F f e π [1 ] [ ] [1 ] ( ) 2 ( ) 2 1 ( ) 2 ( ) i h x ky lz A B i h k l j i h x ky lz j i h k l e f f e e f e j j j                     While h+k+l =2n+1, 1 0 2 1         hkl i( h k l ) ( n ) i e e F π π System absence Sum up over all atoms in a LP! Translation-symmetry term!

Example:diffraction data → indexing → cell parameter!，钨为立方晶系，其粉末衍射线指标为：110,200.211.220.310222.321,400.....属何种点阵类型？若X射线2=154.4pm，220衍射角43.6°，试计算晶胞参数。p.226, 7.18Answer:h+k+l=2ndiffractions observed!=2n+1 systemabsence!or (h2+k2+12): 2:4:6:8:10:12:14:16..... > bcc lattice!Bragg law: 2dhk sinθ= α → dhk= V(2sin?)since 0220 = 43.6°cell parameter:a=dh×/h?+?+[? =h?+k?+P? /2sinOh=154.4×/8/2sin(43.6)=316.6pmThe answer given in p. 317 is wrong!Atomic radius of W atom: (bcc):4R=/3a=R=V3a/4=316.6x/3/4=137.1ppm

Example: diffraction data  indexing  cell parameter! • 钨为立方晶系，其粉末衍射线指标为: 110,200,211,220, 310, 222,321,400,. 属何种点阵类型？若X射线=154.4 pm，220衍 射角43.6°，试计算晶胞参数。 • Answer: h+k+l =2n diffractions observed! =2n+1 system absence! or (h2+k2+l2 ): 2:4:6:8:10:12:14:16.  bcc lattice! Bragg law: 2dhkl sin =   dhkl = /(2sin) since 220 = 43.6° cell parameter: pm a d h k l h k l hkl hkl 154.4 8 / 2sin(43.6) 316.6 / 2sin 2 2 2 2 2 2             The answer given in p. 317 is wrong! Atomic radius of W atom: (bcc) 4R  3a  R  3a / 4  316.6x 3 / 4 137.1ppm p.226, 7.18

P227, 7.27Bragg's law, d-spacing. Number of S.in a unit cell of orthorhombic crystal:nss = N.Vp/ Mss = N(abc)p/ Ms8=6.022×1023×(1048×1292×2455)×10-30×2.07/(32×8)=16p.200For orthohombic crystal, 1/dnk=[(h/a)2 + (k/b)2 + (/c)*j1/2According to Bragg's Law, we havesin0 = ~/2dhk = [(h/a)2 + (k/b)2 + (V/c)"j1/2 × N/2=[(2/1048)2 + (2/1292)2 + (4/2455)^j1/2X×154.18/2=[(1/1048)2 + (1/1292)2 + (2/2455)2j1/2×154.18 = 0.2273> 0= 13.10(Cu Kα, x-ray, a= 154.18 pm)Note: the value of a is not given in question 7.27! !!

P227, 7.27 Bragg’s law, d-spacing • Number of S8 in a unit cell of orthorhombic crystal: For orthohombic crystal, 1/dhkl=[(h/a)2 + (k/b)2 + (l/c)2 ] 1/2 p.200 According to Bragg’s Law, we have sin = /2dhkl = [(h/a)2 + (k/b)2 + (l/c)2 ] 1/2×/2 = [(2/1048)2 + (2/1292)2 + (4/2455)2 ] 1/2×154.18/2 = [(1/1048)2 + (1/1292)2 + (2/2455)2 ] 1/2×154.18 = 0.2273   = 13.1º (Cu K1 x-ray,  = 154.18 pm) 6.022 10 (1048 1292 2455) 10 2.07 /(32 8) 16 ( ) / ~ / ~ 2 3 3 0 8 0 8 0 8             nS N V MS N abc  MS Note: the value of  is not given in question 7.27! !!!

p.257, 8.8indexing of diffraction data!Ta metal's x-ray diffraction data, (sin?o) is known1)Indexing:sin?0,:sin?02:sin?20,:sin?04:sin?0,:sin?06:sin?0,:sin20g:sin?0g...= 1:2:3:4:5:6:7:8:9:... = 2:4:6:8:10:12:14:16:18:body-centered cubic lattice, h+k+l=2n+1 system absence!0bserved hkl: (110)(200)(211)(220)(310)(222)(321)(400)(330)..2) Cell parameter: dhkr= a/(h2+k2+P)1/2 a = dhk (h?+k2+P)1/2dhk=2sineAccording to Bragg's Law, we have> a = (h2+k2+P)1/2 /2sineCho0sing the (330) diffraction with sin?0-0.97826 (2=154.1 pm),,wehavea =a (h2+k2+12)1/2/2sine= 154.1 x (18)1/2 /(2x0.98907) = 330.5 pm

p.257, 8.8 indexing of diffraction data! Ta metal’s x-ray diffraction data, (sin2) is known. 1)Indexing: sin21 :sin22 :sin23 :sin24 :sin25 :sin26 :sin27 :sin28 :sin29. = 1:2:3:4:5:6:7:8:9:. = 2:4:6:8:10:12:14:16:18:. body-centered cubic lattice, h+k+l =2n+1 system absence! Observed hkl: (110)(200)(211)(220)(310)(222)(321)(400)(330). 2) Cell parameter: dhkl= a/(h2+k2+l2 ) 1/2  a = dhkl(h2+k2+l2 ) 1/2 According to Bragg’s Law, we have dhkl = /2sin  a =  (h2+k2+l2 ) 1/2 /2sin Choosing the (330) diffraction with sin20.97826 (=154.1 pm), we have a =  (h2+k2+l2 ) 1/2 /2sin = 154.1 x (18)1/2 /(2x0.98907) = 330.5 pm