Encyclopedia of Crystallographic Prototypes

AFLOW Prototype: A2BC6_oP36_29_2a_a_6a-001

If you are using this page, please cite:
H. Eckert, S. Divilov, M. J. Mehl, D. Hicks, A. C. Zettel, M. Esters. X. Campilongo and S. Curtarolo, The AFLOW Library of Crystallographic Prototypes: Part 4. Submitted to Computational Materials Science.

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Koechlinite (Room temperature Bi$_{2}$MoO$_{6}$) Structure: A2BC6_oP36_29_2a_a_6a-001

Picture of Structure; Click for Big Picture
Prototype Bi$_{2}$MoO$_{6}$
AFLOW prototype label A2BC6_oP36_29_2a_a_6a-001
Mineral name koechlinite
ICSD 47139
Pearson symbol oP36
Space group number 29
Space group symbol $Pca2_1$
AFLOW prototype command aflow --proto=A2BC6_oP36_29_2a_a_6a-001
--params=$a, \allowbreak b/a, \allowbreak c/a, \allowbreak x_{1}, \allowbreak y_{1}, \allowbreak z_{1}, \allowbreak x_{2}, \allowbreak y_{2}, \allowbreak z_{2}, \allowbreak x_{3}, \allowbreak y_{3}, \allowbreak z_{3}, \allowbreak x_{4}, \allowbreak y_{4}, \allowbreak z_{4}, \allowbreak x_{5}, \allowbreak y_{5}, \allowbreak z_{5}, \allowbreak x_{6}, \allowbreak y_{6}, \allowbreak z_{6}, \allowbreak x_{7}, \allowbreak y_{7}, \allowbreak z_{7}, \allowbreak x_{8}, \allowbreak y_{8}, \allowbreak z_{8}, \allowbreak x_{9}, \allowbreak y_{9}, \allowbreak z_{9}$

Other compounds with this structure

Bi$_{2}$WO$_{6}$ (russelite)


  • This is the room temperature structure of Bi$_{2}$MoO$_{6}$. Above 896°C it transforms to the high-temperature monoclinic Bi$_{2}$MoO$_{6}$ structure (Villars, 2018).
  • (Teller, 1984) state that $\gamma$–Bi$_{2}$MoO$_{6}$ is in the $Pna2_{1}$ #33 space group, but the symmetry operations they give are for $Pca2_{1}$ #29, and interatomic distances they give are consistent with this space group. The ICSD, (Villars, 2018), and others are in agreement with this assessment.
  • Space group $Pca2_{1}$ does not specify the origin of the $z$-axis. We follow (Teller, 1984) and chose it so that the molybdenum atom has $z_{3} = 0$.

\[ \begin{array}{ccc} \mathbf{a_{1}}&=&a \,\mathbf{\hat{x}}\\\mathbf{a_{2}}&=&b \,\mathbf{\hat{y}}\\\mathbf{a_{3}}&=&c \,\mathbf{\hat{z}} \end{array}\]

Basis vectors

Lattice coordinates Cartesian coordinates Wyckoff position Atom type
$\mathbf{B_{1}}$ = $x_{1} \, \mathbf{a}_{1}+y_{1} \, \mathbf{a}_{2}+z_{1} \, \mathbf{a}_{3}$ = $a x_{1} \,\mathbf{\hat{x}}+b y_{1} \,\mathbf{\hat{y}}+c z_{1} \,\mathbf{\hat{z}}$ (4a) Bi I
$\mathbf{B_{2}}$ = $- x_{1} \, \mathbf{a}_{1}- y_{1} \, \mathbf{a}_{2}+\left(z_{1} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a x_{1} \,\mathbf{\hat{x}}- b y_{1} \,\mathbf{\hat{y}}+c \left(z_{1} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4a) Bi I
$\mathbf{B_{3}}$ = $\left(x_{1} + \frac{1}{2}\right) \, \mathbf{a}_{1}- y_{1} \, \mathbf{a}_{2}+z_{1} \, \mathbf{a}_{3}$ = $a \left(x_{1} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- b y_{1} \,\mathbf{\hat{y}}+c z_{1} \,\mathbf{\hat{z}}$ (4a) Bi I
$\mathbf{B_{4}}$ = $- \left(x_{1} - \frac{1}{2}\right) \, \mathbf{a}_{1}+y_{1} \, \mathbf{a}_{2}+\left(z_{1} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a \left(x_{1} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+b y_{1} \,\mathbf{\hat{y}}+c \left(z_{1} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4a) Bi I
$\mathbf{B_{5}}$ = $x_{2} \, \mathbf{a}_{1}+y_{2} \, \mathbf{a}_{2}+z_{2} \, \mathbf{a}_{3}$ = $a x_{2} \,\mathbf{\hat{x}}+b y_{2} \,\mathbf{\hat{y}}+c z_{2} \,\mathbf{\hat{z}}$ (4a) Bi II
$\mathbf{B_{6}}$ = $- x_{2} \, \mathbf{a}_{1}- y_{2} \, \mathbf{a}_{2}+\left(z_{2} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a x_{2} \,\mathbf{\hat{x}}- b y_{2} \,\mathbf{\hat{y}}+c \left(z_{2} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4a) Bi II
$\mathbf{B_{7}}$ = $\left(x_{2} + \frac{1}{2}\right) \, \mathbf{a}_{1}- y_{2} \, \mathbf{a}_{2}+z_{2} \, \mathbf{a}_{3}$ = $a \left(x_{2} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- b y_{2} \,\mathbf{\hat{y}}+c z_{2} \,\mathbf{\hat{z}}$ (4a) Bi II
$\mathbf{B_{8}}$ = $- \left(x_{2} - \frac{1}{2}\right) \, \mathbf{a}_{1}+y_{2} \, \mathbf{a}_{2}+\left(z_{2} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a \left(x_{2} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+b y_{2} \,\mathbf{\hat{y}}+c \left(z_{2} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4a) Bi II
$\mathbf{B_{9}}$ = $x_{3} \, \mathbf{a}_{1}+y_{3} \, \mathbf{a}_{2}+z_{3} \, \mathbf{a}_{3}$ = $a x_{3} \,\mathbf{\hat{x}}+b y_{3} \,\mathbf{\hat{y}}+c z_{3} \,\mathbf{\hat{z}}$ (4a) Mo I
$\mathbf{B_{10}}$ = $- x_{3} \, \mathbf{a}_{1}- y_{3} \, \mathbf{a}_{2}+\left(z_{3} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a x_{3} \,\mathbf{\hat{x}}- b y_{3} \,\mathbf{\hat{y}}+c \left(z_{3} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4a) Mo I
$\mathbf{B_{11}}$ = $\left(x_{3} + \frac{1}{2}\right) \, \mathbf{a}_{1}- y_{3} \, \mathbf{a}_{2}+z_{3} \, \mathbf{a}_{3}$ = $a \left(x_{3} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- b y_{3} \,\mathbf{\hat{y}}+c z_{3} \,\mathbf{\hat{z}}$ (4a) Mo I
$\mathbf{B_{12}}$ = $- \left(x_{3} - \frac{1}{2}\right) \, \mathbf{a}_{1}+y_{3} \, \mathbf{a}_{2}+\left(z_{3} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a \left(x_{3} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+b y_{3} \,\mathbf{\hat{y}}+c \left(z_{3} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4a) Mo I
$\mathbf{B_{13}}$ = $x_{4} \, \mathbf{a}_{1}+y_{4} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ = $a x_{4} \,\mathbf{\hat{x}}+b y_{4} \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ (4a) O I
$\mathbf{B_{14}}$ = $- x_{4} \, \mathbf{a}_{1}- y_{4} \, \mathbf{a}_{2}+\left(z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a x_{4} \,\mathbf{\hat{x}}- b y_{4} \,\mathbf{\hat{y}}+c \left(z_{4} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4a) O I
$\mathbf{B_{15}}$ = $\left(x_{4} + \frac{1}{2}\right) \, \mathbf{a}_{1}- y_{4} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ = $a \left(x_{4} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- b y_{4} \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ (4a) O I
$\mathbf{B_{16}}$ = $- \left(x_{4} - \frac{1}{2}\right) \, \mathbf{a}_{1}+y_{4} \, \mathbf{a}_{2}+\left(z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a \left(x_{4} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+b y_{4} \,\mathbf{\hat{y}}+c \left(z_{4} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4a) O I
$\mathbf{B_{17}}$ = $x_{5} \, \mathbf{a}_{1}+y_{5} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $a x_{5} \,\mathbf{\hat{x}}+b y_{5} \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ (4a) O II
$\mathbf{B_{18}}$ = $- x_{5} \, \mathbf{a}_{1}- y_{5} \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a x_{5} \,\mathbf{\hat{x}}- b y_{5} \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4a) O II
$\mathbf{B_{19}}$ = $\left(x_{5} + \frac{1}{2}\right) \, \mathbf{a}_{1}- y_{5} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $a \left(x_{5} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- b y_{5} \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ (4a) O II
$\mathbf{B_{20}}$ = $- \left(x_{5} - \frac{1}{2}\right) \, \mathbf{a}_{1}+y_{5} \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a \left(x_{5} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+b y_{5} \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4a) O II
$\mathbf{B_{21}}$ = $x_{6} \, \mathbf{a}_{1}+y_{6} \, \mathbf{a}_{2}+z_{6} \, \mathbf{a}_{3}$ = $a x_{6} \,\mathbf{\hat{x}}+b y_{6} \,\mathbf{\hat{y}}+c z_{6} \,\mathbf{\hat{z}}$ (4a) O III
$\mathbf{B_{22}}$ = $- x_{6} \, \mathbf{a}_{1}- y_{6} \, \mathbf{a}_{2}+\left(z_{6} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a x_{6} \,\mathbf{\hat{x}}- b y_{6} \,\mathbf{\hat{y}}+c \left(z_{6} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4a) O III
$\mathbf{B_{23}}$ = $\left(x_{6} + \frac{1}{2}\right) \, \mathbf{a}_{1}- y_{6} \, \mathbf{a}_{2}+z_{6} \, \mathbf{a}_{3}$ = $a \left(x_{6} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- b y_{6} \,\mathbf{\hat{y}}+c z_{6} \,\mathbf{\hat{z}}$ (4a) O III
$\mathbf{B_{24}}$ = $- \left(x_{6} - \frac{1}{2}\right) \, \mathbf{a}_{1}+y_{6} \, \mathbf{a}_{2}+\left(z_{6} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a \left(x_{6} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+b y_{6} \,\mathbf{\hat{y}}+c \left(z_{6} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4a) O III
$\mathbf{B_{25}}$ = $x_{7} \, \mathbf{a}_{1}+y_{7} \, \mathbf{a}_{2}+z_{7} \, \mathbf{a}_{3}$ = $a x_{7} \,\mathbf{\hat{x}}+b y_{7} \,\mathbf{\hat{y}}+c z_{7} \,\mathbf{\hat{z}}$ (4a) O IV
$\mathbf{B_{26}}$ = $- x_{7} \, \mathbf{a}_{1}- y_{7} \, \mathbf{a}_{2}+\left(z_{7} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a x_{7} \,\mathbf{\hat{x}}- b y_{7} \,\mathbf{\hat{y}}+c \left(z_{7} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4a) O IV
$\mathbf{B_{27}}$ = $\left(x_{7} + \frac{1}{2}\right) \, \mathbf{a}_{1}- y_{7} \, \mathbf{a}_{2}+z_{7} \, \mathbf{a}_{3}$ = $a \left(x_{7} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- b y_{7} \,\mathbf{\hat{y}}+c z_{7} \,\mathbf{\hat{z}}$ (4a) O IV
$\mathbf{B_{28}}$ = $- \left(x_{7} - \frac{1}{2}\right) \, \mathbf{a}_{1}+y_{7} \, \mathbf{a}_{2}+\left(z_{7} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a \left(x_{7} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+b y_{7} \,\mathbf{\hat{y}}+c \left(z_{7} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4a) O IV
$\mathbf{B_{29}}$ = $x_{8} \, \mathbf{a}_{1}+y_{8} \, \mathbf{a}_{2}+z_{8} \, \mathbf{a}_{3}$ = $a x_{8} \,\mathbf{\hat{x}}+b y_{8} \,\mathbf{\hat{y}}+c z_{8} \,\mathbf{\hat{z}}$ (4a) O V
$\mathbf{B_{30}}$ = $- x_{8} \, \mathbf{a}_{1}- y_{8} \, \mathbf{a}_{2}+\left(z_{8} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a x_{8} \,\mathbf{\hat{x}}- b y_{8} \,\mathbf{\hat{y}}+c \left(z_{8} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4a) O V
$\mathbf{B_{31}}$ = $\left(x_{8} + \frac{1}{2}\right) \, \mathbf{a}_{1}- y_{8} \, \mathbf{a}_{2}+z_{8} \, \mathbf{a}_{3}$ = $a \left(x_{8} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- b y_{8} \,\mathbf{\hat{y}}+c z_{8} \,\mathbf{\hat{z}}$ (4a) O V
$\mathbf{B_{32}}$ = $- \left(x_{8} - \frac{1}{2}\right) \, \mathbf{a}_{1}+y_{8} \, \mathbf{a}_{2}+\left(z_{8} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a \left(x_{8} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+b y_{8} \,\mathbf{\hat{y}}+c \left(z_{8} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4a) O V
$\mathbf{B_{33}}$ = $x_{9} \, \mathbf{a}_{1}+y_{9} \, \mathbf{a}_{2}+z_{9} \, \mathbf{a}_{3}$ = $a x_{9} \,\mathbf{\hat{x}}+b y_{9} \,\mathbf{\hat{y}}+c z_{9} \,\mathbf{\hat{z}}$ (4a) O VI
$\mathbf{B_{34}}$ = $- x_{9} \, \mathbf{a}_{1}- y_{9} \, \mathbf{a}_{2}+\left(z_{9} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a x_{9} \,\mathbf{\hat{x}}- b y_{9} \,\mathbf{\hat{y}}+c \left(z_{9} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4a) O VI
$\mathbf{B_{35}}$ = $\left(x_{9} + \frac{1}{2}\right) \, \mathbf{a}_{1}- y_{9} \, \mathbf{a}_{2}+z_{9} \, \mathbf{a}_{3}$ = $a \left(x_{9} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- b y_{9} \,\mathbf{\hat{y}}+c z_{9} \,\mathbf{\hat{z}}$ (4a) O VI
$\mathbf{B_{36}}$ = $- \left(x_{9} - \frac{1}{2}\right) \, \mathbf{a}_{1}+y_{9} \, \mathbf{a}_{2}+\left(z_{9} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a \left(x_{9} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+b y_{9} \,\mathbf{\hat{y}}+c \left(z_{9} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4a) O VI

References

  • R. G. Teller, J. F. Brazdil, R. K. Grasselli, and J. D. Jorgensen, The structure of γ-bismuth molybdate, Bi$_{2}$MoO$_{6}$, by powder neutron diffraction, Acta Crystallogr. Sect. C 40, 2001–2005 (1984), doi:10.1107/S0108270184010398.

Found in

  • P. Villars, H. Okamoto, and K. Cenzual, eds., ASM Alloy Phase Diagram Database (ASM International, 2018), chap. Bismuth-Molybdenum-Oxygen Ternary, Vertical Section (1971 Erman L.Y.). Copyright © 2006-2018 ASM International.

Prototype Generator

aflow --proto=A2BC6_oP36_29_2a_a_6a --params=$a,b/a,c/a,x_{1},y_{1},z_{1},x_{2},y_{2},z_{2},x_{3},y_{3},z_{3},x_{4},y_{4},z_{4},x_{5},y_{5},z_{5},x_{6},y_{6},z_{6},x_{7},y_{7},z_{7},x_{8},y_{8},z_{8},x_{9},y_{9},z_{9}$

Species:

Running:

Output: