wyckoff position 各种晶格群的表格 http://www.cryst.ehu.es/cgi-bin/cryst/programs/nph-wp-list 下面是从其他网站转的，读过后，对wyckoff position 可以有大致的了解了。 Space Groups The International Tables for Crystallography From examination of a space group in “The International Tables for Crystallography” Vol. A, you should be able to ascertain the following information: · Herman-Mauguin (HM) Symbol (Long, Short) · Point Group (HM, Schoenflies) · Locate and identify symmetry elements · Understand Wyckoff site multiplicity and symmetry · Distinguish general and special positions · Extinction conditions · Identify possible subgroups and supergroups Understanding the Herman-Mauguin Space Group Symbol Space groups are typically identified by their short Herman-Mauguin symbol (i.e. Pnma, I4/mmm, etc.). The symmetry elements contained in the short symbol are the minimum number needed to generate all of the remaining symmetry elements. This symbolism is very efficient, condensed form of noting all of the symmetry present in a given space group. We won’t go into all of the details of the space group symbol, but I will expect you to be able to determine the Crystal system, Bravais Lattice and Point group from the short H-M symbol. You should also be able to determine the presence and orientation of certain symmetry elements from the short H-M symbol and vice versa. The HM space group symbol can be derived from the symmetry elements present using the following logic. The first letter identifies the centering of the lattice, I will hereafter refer to this as the lattice descriptor : · P ® Primitive · I ® Body centered · F ® Face centered · C ® C-centered · B ® B-centered · A ® A-centered The next three symbols denote symmetry elements present in certain directions, those directions are as follows: Crystal System Symmetry Direction Primary Secondary Tertiary Triclinic None Monoclinic Orthorhombic Tetragonal / Hexagonal/ Trigonal / / Cubic / / – Axis parallel or plane perpendicular to the x-axis. – Axis parallel or plane perpendicular to the y-axis. – Axis parallel or plane perpendicular to the z-axis. – Axis parallel or plane perpendicular to the line running at 45° to the x and y axes. – Axis parallel or plane perpendicular to the long face diagonal of the ab face of a hexagonal cell. – Axis parallel or plane perpendicular to the body diagonal. For a better understanding see specific examples from class notes. However, with no knowledge of the symmetry diagram we can identify the crystal system from the space group symbol. · Cubic – The secondary symmetry symbol will always be either 3 or –3 (i.e. Ia3, Pm3m, Fd3m) · Tetragonal – The primary symmetry symbol will always be either 4, (-4), 4 1 , 4 2 or 4 3 (i.e. P4 1 2 1 2, I4/m, P4/mcc) · Hexagonal – The primary symmetry symbol will always be a 6, (-6), 6 1 , 6 2 , 6 3 , 6 4 or 6 5 (i.e. P6mm, P6 3 /mcm) · Trigonal – The primary symmetry symbol will always be a 3, (-3) 3 1 or 3 2 (i.e P31m, R3, R3c, P312) · Orthorhombic – All three symbols following the lattice descriptor will be either mirror planes, glide planes, 2-fold rotation or screw axes (i.e. Pnma, Cmc2 1 , Pnc2) · Monoclinic – The lattice descriptor will be followed by either a single mirror plane, glide plane, 2-fold rotation or screw axis or an axis/plane symbol (i.e. Cc, P2, P2 1 /n) · Triclinic – The lattice descriptor will be followed by either a 1 or a (-1). The point group can be determined from the short H-M symbol by converting glide planes to mirror planes and screw axes to rotation axes. For example: Space Group = Pnma ® Point Group = mmm Space Group = I`4c2 ® Point Group =`4m2 Space Group = P4 2 /n ® Point Group = 4/m Wyckoff Sites One of the most useful pieces of information contained in the International Tables are the Wyckoff positions. The Wyckoff positions tell us where the atoms in a crystal can be found. To understand how they work consider the monoclinic space group Pm. This space group has only two symmetry elements, both mirror planes perpendicular to the b-axis. One at y = 0 and one at y = ½ (halfway up the unit cell in the b direction). Now let’s place an atom in the unit cell at an arbitrary position, x,y,z. If we now carry out the symmetry operation associated with this space group a second atom will be generated by the mirror plane at x,-y,z. However, if we were to place the atom on one of the mirror planes (its y coordinate would have to be either 0 or ½) then the reflection operation would not create a second atom.All of the information in the proceeding paragraph is contained in Wyckoff positions section of the International Tables. Pm has three Wyckoff sites as shown in the table below: Multiplicity Wyckoff Letter Site Symmetry Coordinates 2 c 1 (1) x,y,z (2) x,-y,z 1 b m x,½,z 1 a m x,0,z The multiplicity tells us how many atoms are generated by symmetry if we place a single atom at that position. In this case for every atom we insert at an arbitrary position (x,y,z) in the unit cell a second atom will be generated by the mirror plane at x,-y,z. This corresponds to the uppermost Wyckoff position 2c. The letter is simply a label and has no physical meaning. They are assigned alphabetically from the bottom up. The symmetry tells us what symmetry elements the atom resides upon. The uppermost Wyckoff position, corresponding to an atom at an arbitrary position never resides upon any symmetry elements. This Wyckoff position is called the general position. The coordinates column tells us the coordinates of all of the symmetry related atoms (two in this case). All of the remaining Wyckoff positions are called special positions. They correspond to atoms which lie upon one of more symmetry elements, because of this they always have a smaller multiplicity than the general position. Furthermore, one or more of their fractional coordinates must be fixed. In this case the y value must be either 0 or ½ or the atom would no longer lie on the mirror plane. Generating a Crystal Structure from its Crystallographic Description Using the space group information contained in the International Tables we can do many things. One powerful use is to generate an entire crystal structure from a brief description.Consider the following description of the crystal structure of Sr 2 AlTaO 6 .Space Group = Fm`3ma = 7.80 ÅAtomic PositionsAtom X Y ZSr 0.25 0.25 0.25Al 0.0 0.0 0.0Ta 0.5 0.5 0.5O 0.24 0.0 0.0 From the space group tables we see that the atoms are located on the following Wyckoff sites Sr ® 8c Al ® 4a Ta ® 4b O ® 24e The number associated with the Wyckoff sites tells us how many atoms of that type there are in the unit cell. In this So there are 40 atoms in the unit cell, with stoichiometry Sr 8 Al 4 Ta 4 O 24 which reduces to the empirical formula Sr 2 AlTaO 6 . Since the number of atoms in the unit cell is four times the number of atoms in the formula unit, we say that Z = 4. Using the face centering generators (0,0,0), (½,½,0), (½,0,½), (0,½,½) together with the coordinates of each Wyckoff site we can generate the fractional coordinates of all atoms in the unit cell: Sr 1:(0.25,0.25,0.25), 2:(0.75,0.75,0.25), 3:(0.75,0.25,0.75), 4:(0.25,0.75,0.75) 5:(0.25,0.25,0.75), 6:(0.75,0.75,0.75), 7:(0.75,0.25,0.25), 8:(0.25,0.75,0.25) Al 1:(0.0,0.0,0.0), 2:(0.5,0.5,0.0), 3:(0.5,0.0,0.5), 4:(0.0,0.5,0.5) Ta 1:(0.5,0.5,0.5), 2:(0.0,0.0,0.5), 3:(0.0,0.5,0.0), 4:(0.5,0.0,0.0) O 1:(0.24,0.0,0.0), 2:(0.74,0.5,0.0), 3:(0.74,0.0,0.5), 4:(0.24,0.5,0.5) 5:(0.76,0.0,0.0), 6:(0.26,0.5,0.0), 7:(0.26,0.0,0.5), 8:(0.76,0.5,0.5) 9:(0.0,0.24,0.0), 10:(0.5,0.74,0.0), 11:(0.5,0.24,0.5), 12:(0.0,0.74,0.5) 13:(0.0,0.76,0.0), 14:(0.5,0.26,0.0), 15:(0.5,0.76,0.5), 16:(0.0,0.26,0.5) 17:(0.0,0.0,0.24), 18:(0.5,0.5,0.24), 19:(0.5,0.0,0.74), 20:(0.0,0.5,0.74) 21:(0.0,0.0,0.76), 22:(0.5,0.5,0.76), 23:(0.5,0.0,0.26), 24:(0.0,0.5,0.26) From these fractional coordinates you can sketch out the structure of Sr 2 AlTaO 6 . With some luck I will provide a link to a picture of the structure here, at some point in the future. We can also work out bond distances from this information. The first Al ion is octahedrally coordinated by six oxygens (1,5,9,13,17,21) and the Al-O distance is : d = 7.80′ 1/2 = 1.87Å while the first Ta ion is also surrounded by 6 oxygens (4,8,11,15,18,22) at a distance of d = 7.80′ 1/2 = 2.03Å and Sr is surrounded by 12 oxygens (1,4,6,7,9,11,14,16,17,18,23,24) at a distance of d = 7.80′ 1/2 = 2.76Å Determining a Crystal Structure from Symmetry Composition Another use is that given the stoichiometry, space group and unit cell size (which can typically be determined from diffraction techniques) and the density of a compound we can often deduce the crystal structure of relatively simple compounds. As an example consider the following information: Stoichiometry = SrTiO 3 Space Group = Pm3m a = 3.90 Å Density = 5.1 g/cm 3 To derive the crystal structure from this information the first step is to calculate the number of formula units per unit cell : Formula Weight SrTiO 3 = 87.62 + 47.87 + 3′(16.00) = 183.49 g/mol Unit Cell Volume = (3.90′10 -8 cm) 3 = 5.93′10 -23 cm 3 (5.1 g/cm 3 )′(5.93′10 -23 cm 3 )′(mol/183.49 g)′(6.022′10 23 /mol) = 0.99 Thus there is one formula unit per unit cell (Z=1), and the number of atoms per unit cell is : 1 Sr, 1 Ti and 3 O. Next we compare the number of atoms in the unit cell with the multiplicities of the Wyckoff sites. · From the multiplicities of the special positions in space group Pm3m we see that Sr must occupy either the 1a or 1b positions (otherwise there would be more than one Sr in the unit cell) · By the same reasoning Ti must also reside in either the 1a or 1b position, and, since there are no free positional parameters (x,y or z) in either 1a or 1b, the two ions cannot occupy the same site. · To maintain 3 oxygen ions in the unit cell it must reside at either site 3c or 3d. If we arbitrarily put Ti at the origin (1a), then by default Sr must go to 1b. To evaluate the prospects of putting O at either 3c or 3d we calculate the Ti-O bond distances: D (O @ 3c) = 3.90′ 1/2 = 2.76Å D (O @ 3d) = 3.90′ 1/2 = 1.95Å Of these two the latter (3d) is obviously more appropriate for a Ti-O bond (consult tables of ionic radii to convince yourself of this statement). Thus we obtain the structure of SrTiO 3 to be Space Group = Pm3ma = 3.90 ÅAtomic PositionsAtom Site X Y ZSr 1b 0.5 0.5 0.5Ti 1a 0.0 0.0 0.0O 3d 0.5 0.0 0.0
答网友晶体学问题：单斜晶系的空间群符号 问题在： http://blog.sciencenet.cn/home.php?mod=spaceuid=2321do=wall 回答如下： 根据你提供的信息（ ICSD #33243 ：单斜钒酸铋 ）。 ICSD 33243 数据（摘要）如下： Structured Bi (V O4) Unit Cell 5.197 5.096 11.702 90. 90. 90.4 Space Group I 1 1 2/b SG Number 15 Atom # OX SITE x y z SOF Bi 1 +3 4 e 0 0.25 0.6337(1) 1. V 1 +5 4 e 0 0.25 0.1352(5) 1. O 1 -2 8 f 0.149(3) 0.506(8) 0.210(2) 1. O 2 -2 8 f 0.258(3) 0.379(8) 0.451(2) 1. *end for ICSD #33243 在国际空间群表里， 15 号空间群有 6 种表达方式： Uniq-b 轴有三种晶胞， Uniq-c 轴有三种晶胞，共有 6 种。 注意， 6 种晶胞对应一个空间群编号，但是 6 种晶胞的 Patterson symmetry 符号不同。例如，你的数据（ ICSD #33243 ）对应 I 1 1 2/b ， 根据这个符号确定空间群表里的具体晶胞选择。根据 Unit Cell 的（ 5.197 5.096 11.702 90. 90. 90.4 ） 信息，可以确定属于 Uniq-c 轴，对应的晶胞如下： 注意，原子坐标的数据是和晶胞数据对应的。 BiVO4 的原子环境数据如下： a=,5.197,b=,5.096,c=,11.702,Delta=,.005 1 4e-Bi, is a reference atom Cartesian = ,-2.5896,-1.274,1.5681 SP coordinates=,0,.25,.634 ,1,1,The shell structure as a cluster:,shell,No,1 d of neighbors from the reference atom =,2.3434 The number of neighbor atoms, N=,2,and the number of shortest bonds formed by these atoms=,1 and the shortest bond length=,2.94 Neighbor atom:,8f-1-O Cartesian = ,-3.3726,-.0306,3.3936 SP coordinates= ,-.149,.494,.79 Neighbor atom:,8f-1-O Cartesian = ,-1.8066,-2.5174,3.3936 SP coordinates= ,.149,.006,.79 ,1,2,The shell structure as a cluster:,shell,No,2 d of neighbors from the reference atom =,2.3734 The number of neighbor atoms, N=,2,and the number of shortest bonds formed by these atoms=,1 and the shortest bond length=,4.4 Neighbor atom:,8f-1-O Cartesian = ,-4.4051,-2.5174,2.4574 SP coordinates= ,-.351,.006,.71 Neighbor atom:,8f-1-O Cartesian = ,-.7741,-.0306,2.4574 SP coordinates= ,.351,.494,.71 ,1,3,The shell structure as a cluster:,shell,No,3 d of neighbors from the reference atom =,2.5292 The number of neighbor atoms, N=,2,and the number of shortest bonds formed by these atoms=,1 and the shortest bond length=,4.65 Neighbor atom:,8f-2-O Cartesian = ,-3.9436,.6166,.5734 SP coordinates= ,-.258,.621,.549 Neighbor atom:,8f-2-O Cartesian = ,-1.2356,-3.1645,.5734 SP coordinates= ,.258,-.121,.549 ,1,4,The shell structure as a cluster:,shell,No,4 d of neighbors from the reference atom =,2.6084 The number of neighbor atoms, N=,2,and the number of shortest bonds formed by these atoms=,1 and the shortest bond length=,2.98 Neighbor atom:,8f-2-O Cartesian = ,-3.9258,-1.9313,-.5734 SP coordinates= ,-.258,.121,.451 Neighbor atom:,8f-2-O Cartesian = ,-1.2534,-.6166,-.5734 SP coordinates= ,.258,.379,.451 ,1,5,The shell structure as a cluster:,shell,No,5 d of neighbors from the reference atom =,3.6058 The number of neighbor atoms, N=,2,and the number of shortest bonds formed by these atoms=,1 and the shortest bond length=,6.93 Neighbor atom:,8f-2-O Cartesian = ,-3.9081,-4.4793,.5734 SP coordinates= ,-.258,-.379,.549 Neighbor atom:,8f-2-O Cartesian = ,-1.2712,1.9313,.5734 SP coordinates= ,.258,.879,.549 ,1,6,The shell structure as a cluster:,shell,No,6 d of neighbors from the reference atom =,3.6266 The number of neighbor atoms, N=,2,and the number of shortest bonds formed by these atoms=,1 and the shortest bond length=,7.25 Neighbor atom:,4e-V Cartesian = ,-5.1703,-3.8219,1.5798 SP coordinates= ,-.5,-.25,.635 Neighbor atom:,4e-V Cartesian = ,-.0089,1.274,1.5798 SP coordinates= ,.5,.75,.635 2 4e-V, is a reference atom Cartesian = ,-2.5896,-1.274,-4.2712 SP coordinates=,0,.25,.135 ,2,1,The shell structure as a cluster:,shell,No,1 d of neighbors from the reference atom =,1.7431 The number of neighbor atoms, N=,2,and the number of shortest bonds formed by these atoms=,1 and the shortest bond length=,2.85 Neighbor atom:,8f-2-O Cartesian = ,-3.8519,-.6166,-5.2776 SP coordinates= ,-.242,.379,.049 Neighbor atom:,8f-2-O Cartesian = ,-1.3273,-1.9313,-5.2776 SP coordinates= ,.242,.121,.049 ,2,2,The shell structure as a cluster:,shell,No,2 d of neighbors from the reference atom =,1.7486 The number of neighbor atoms, N=,2,and the number of shortest bonds formed by these atoms=,1 and the shortest bond length=,3.02 Neighbor atom:,8f-1-O Cartesian = ,-3.3549,-2.5785,-3.3936 SP coordinates= ,-.149,-.006,.21 Neighbor atom:,8f-1-O Cartesian = ,-1.8244,.0306,-3.3936 SP coordinates= ,.149,.506,.21 ,2,3,The shell structure as a cluster:,shell,No,3 d of neighbors from the reference atom =,2.89 The number of neighbor atoms, N=,2,and the number of shortest bonds formed by these atoms=,1 and the shortest bond length=,4.5 Neighbor atom:,8f-1-O Cartesian = ,-4.4229,.0306,-2.4574 SP coordinates= ,-.351,.506,.29 Neighbor atom:,8f-1-O Cartesian = ,-.7564,-2.5785,-2.4574 SP coordinates= ,.351,-.006,.29 ,2,4,The shell structure as a cluster:,shell,No,4 d of neighbors from the reference atom =,3.1239 The number of neighbor atoms, N=,2,and the number of shortest bonds formed by these atoms=,1 and the shortest bond length=,4.53 Neighbor atom:,8f-2-O Cartesian = ,-3.8341,-3.1645,-6.4244 SP coordinates= ,-.242,-.121,-.049 Neighbor atom:,8f-2-O Cartesian = ,-1.3451,.6166,-6.4244 SP coordinates= ,.242,.621,-.049 ,2,5,The shell structure as a cluster:,shell,No,5 d of neighbors from the reference atom =,3.6266 The number of neighbor atoms, N=,2,and the number of shortest bonds formed by these atoms=,1 and the shortest bond length=,7.25 Neighbor atom:,4e-Bi Cartesian = ,-5.1703,-3.8219,-4.2829 SP coordinates= ,-.5,-.25,.134 Neighbor atom:,4e-Bi Cartesian = ,-.0089,1.274,-4.2829 SP coordinates= ,.5,.75,.134 ,2,6,The shell structure as a cluster:,shell,No,6 d of neighbors from the reference atom =,3.652 The number of neighbor atoms, N=,2,and the number of shortest bonds formed by these atoms=,1 and the shortest bond length=,7.3 Neighbor atom:,4e-Bi Cartesian = ,-5.2059,1.274,-4.2829 SP coordinates= ,-.5,.75,.134 Neighbor atom:,4e-Bi Cartesian = ,.0267,-3.8219,-4.2829 SP coordinates= ,.5,-.25,.134 3 8f-1-O, is a reference atom Cartesian = ,-1.8066,-2.5174,3.3936 SP coordinates=,.149,.006,.79 ,3,1,The shell structure as a cluster:,shell,No,1 d of neighbors from the reference atom =,1.7486 The number of neighbor atoms, N=,1,and the number of shortest bonds formed by these atoms=,0 and the shortest bond length=,0 Neighbor atom:,4e-V Cartesian = ,-2.5718,-3.8219,4.2712 SP coordinates= ,0,-.25,.865 ,3,2,The shell structure as a cluster:,shell,No,2 d of neighbors from the reference atom =,2.3434 The number of neighbor atoms, N=,1,and the number of shortest bonds formed by these atoms=,0 and the shortest bond length=,0 Neighbor atom:,4e-Bi Cartesian = ,-2.5896,-1.274,1.5681 SP coordinates= ,0,.25,.634 ,3,3,The shell structure as a cluster:,shell,No,3 d of neighbors from the reference atom =,2.3734 The number of neighbor atoms, N=,1,and the number of shortest bonds formed by these atoms=,0 and the shortest bond length=,0 Neighbor atom:,4e-Bi Cartesian = ,.0089,-1.274,4.2829 SP coordinates= ,.5,.25,.866 4 8f-2-O, is a reference atom Cartesian = ,-1.3273,-1.9313,-5.2776 SP coordinates=,.242,.121,.049 ,4,1,The shell structure as a cluster:,shell,No,1 d of neighbors from the reference atom =,1.7431 The number of neighbor atoms, N=,1,and the number of shortest bonds formed by these atoms=,0 and the shortest bond length=,0 Neighbor atom:,4e-V Cartesian = ,-2.5896,-1.274,-4.2712 SP coordinates= ,0,.25,.135 ,4,2,The shell structure as a cluster:,shell,No,2 d of neighbors from the reference atom =,2.5293 The number of neighbor atoms, N=,1,and the number of shortest bonds formed by these atoms=,0 and the shortest bond length=,0 Neighbor atom:,4e-Bi Cartesian = ,.0267,-3.8219,-4.2829 SP coordinates= ,.5,-.25,.134 ,4,3,The shell structure as a cluster:,shell,No,3 d of neighbors from the reference atom =,2.6084 The number of neighbor atoms, N=,1,and the number of shortest bonds formed by these atoms=,0 and the shortest bond length=,0 Neighbor atom:,4e-Bi Cartesian = ,.0089,-1.274,-7.4191 SP coordinates= ,.5,.25,-.134 更多的数据可下载： BiVO4-15-Mono-Uniq-c-AEC-1-原子输出- .005.rar BiVO4-15-Mono-Uniq-c-AEC-1-配位数-简化版-修改- .005.rar
标签: 空间群
