Encyclopedia of Crystallographic Prototypes

AFLOW Prototype: A5B3C15_oP46_30_a2c_bc_a7c-001

This structure originally had the label A5B3C15_oP46_30_a2c_bc_a7c. Calls to that address will be redirected here.

If you are using this page, please cite:
D. Hicks, M. J. Mehl, E. Gossett, C. Toher, O. Levy, R. M. Hanson, G. L. W. Hart, and S. Curtarolo, The AFLOW Library of Crystallographic Prototypes: Part 2, Comp. Mat. Sci. 161, S1-S1011 (2019). (doi=10.1016/j.commatsci.2018.10.043)

Links to this page

https://aflow.org/p/ABT7
or https://aflow.org/p/A5B3C15_oP46_30_a2c_bc_a7c-001
or PDF Version

Bi$_{5}$Nb$_{3}$O$_{15}$ Structure: A5B3C15_oP46_30_a2c_bc_a7c-001

Picture of Structure; Click for Big Picture
Prototype Bi$_{5}$O$_{15}$Nb$_{3}$
AFLOW prototype label A5B3C15_oP46_30_a2c_bc_a7c-001
ICSD 245707
Pearson symbol oP46
Space group number 30
Space group symbol $Pnc2$
AFLOW prototype command aflow --proto=A5B3C15_oP46_30_a2c_bc_a7c-001
--params=$a, \allowbreak b/a, \allowbreak c/a, \allowbreak z_{1}, \allowbreak z_{2}, \allowbreak z_{3}, \allowbreak x_{4}, \allowbreak y_{4}, \allowbreak z_{4}, \allowbreak x_{5}, \allowbreak y_{5}, \allowbreak z_{5}, \allowbreak x_{6}, \allowbreak y_{6}, \allowbreak z_{6}, \allowbreak x_{7}, \allowbreak y_{7}, \allowbreak z_{7}, \allowbreak x_{8}, \allowbreak y_{8}, \allowbreak z_{8}, \allowbreak x_{9}, \allowbreak y_{9}, \allowbreak z_{9}, \allowbreak x_{10}, \allowbreak y_{10}, \allowbreak z_{10}, \allowbreak x_{11}, \allowbreak y_{11}, \allowbreak z_{11}, \allowbreak x_{12}, \allowbreak y_{12}, \allowbreak z_{12}, \allowbreak x_{13}, \allowbreak y_{13}, \allowbreak z_{13}$

  • Our previous rendition (Hicks, 2019) of this structure had multiple errors in the CIF. We have corrected these here.
  • Space group $Pnc2$ #30 allows an arbitrary choice for the origin of the $z$-axis. We fix it here by setting $z_{1} = 1/2$.

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

Basis vectors

Lattice coordinates Cartesian coordinates Wyckoff position Atom type
$\mathbf{B_{1}}$ = $z_{1} \, \mathbf{a}_{3}$ = $c z_{1} \,\mathbf{\hat{z}}$ (2a) Bi I
$\mathbf{B_{2}}$ = $\frac{1}{2} \, \mathbf{a}_{2}+\left(z_{1} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}b \,\mathbf{\hat{y}}+c \left(z_{1} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (2a) Bi I
$\mathbf{B_{3}}$ = $z_{2} \, \mathbf{a}_{3}$ = $c z_{2} \,\mathbf{\hat{z}}$ (2a) O I
$\mathbf{B_{4}}$ = $\frac{1}{2} \, \mathbf{a}_{2}+\left(z_{2} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}b \,\mathbf{\hat{y}}+c \left(z_{2} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (2a) O I
$\mathbf{B_{5}}$ = $\frac{1}{2} \, \mathbf{a}_{1}+z_{3} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+c z_{3} \,\mathbf{\hat{z}}$ (2b) Nb I
$\mathbf{B_{6}}$ = $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}+\left(z_{3} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{1}{2}b \,\mathbf{\hat{y}}+c \left(z_{3} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (2b) Nb I
$\mathbf{B_{7}}$ = $x_{4} \, \mathbf{a}_{1}+y_{4} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ = $a x_{4} \,\mathbf{\hat{x}}+b y_{4} \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ (4c) Bi II
$\mathbf{B_{8}}$ = $- x_{4} \, \mathbf{a}_{1}- y_{4} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ = $- a x_{4} \,\mathbf{\hat{x}}- b y_{4} \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ (4c) Bi II
$\mathbf{B_{9}}$ = $x_{4} \, \mathbf{a}_{1}- \left(y_{4} - \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a x_{4} \,\mathbf{\hat{x}}- b \left(y_{4} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{4} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) Bi II
$\mathbf{B_{10}}$ = $- x_{4} \, \mathbf{a}_{1}+\left(y_{4} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a x_{4} \,\mathbf{\hat{x}}+b \left(y_{4} + \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{4} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) Bi II
$\mathbf{B_{11}}$ = $x_{5} \, \mathbf{a}_{1}+y_{5} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $a x_{5} \,\mathbf{\hat{x}}+b y_{5} \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ (4c) Bi III
$\mathbf{B_{12}}$ = $- x_{5} \, \mathbf{a}_{1}- y_{5} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $- a x_{5} \,\mathbf{\hat{x}}- b y_{5} \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ (4c) Bi III
$\mathbf{B_{13}}$ = $x_{5} \, \mathbf{a}_{1}- \left(y_{5} - \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a x_{5} \,\mathbf{\hat{x}}- b \left(y_{5} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) Bi III
$\mathbf{B_{14}}$ = $- x_{5} \, \mathbf{a}_{1}+\left(y_{5} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a x_{5} \,\mathbf{\hat{x}}+b \left(y_{5} + \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) Bi III
$\mathbf{B_{15}}$ = $x_{6} \, \mathbf{a}_{1}+y_{6} \, \mathbf{a}_{2}+z_{6} \, \mathbf{a}_{3}$ = $a x_{6} \,\mathbf{\hat{x}}+b y_{6} \,\mathbf{\hat{y}}+c z_{6} \,\mathbf{\hat{z}}$ (4c) Nb II
$\mathbf{B_{16}}$ = $- x_{6} \, \mathbf{a}_{1}- y_{6} \, \mathbf{a}_{2}+z_{6} \, \mathbf{a}_{3}$ = $- a x_{6} \,\mathbf{\hat{x}}- b y_{6} \,\mathbf{\hat{y}}+c z_{6} \,\mathbf{\hat{z}}$ (4c) Nb II
$\mathbf{B_{17}}$ = $x_{6} \, \mathbf{a}_{1}- \left(y_{6} - \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{6} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a x_{6} \,\mathbf{\hat{x}}- b \left(y_{6} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{6} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) Nb II
$\mathbf{B_{18}}$ = $- x_{6} \, \mathbf{a}_{1}+\left(y_{6} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{6} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a x_{6} \,\mathbf{\hat{x}}+b \left(y_{6} + \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{6} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) Nb II
$\mathbf{B_{19}}$ = $x_{7} \, \mathbf{a}_{1}+y_{7} \, \mathbf{a}_{2}+z_{7} \, \mathbf{a}_{3}$ = $a x_{7} \,\mathbf{\hat{x}}+b y_{7} \,\mathbf{\hat{y}}+c z_{7} \,\mathbf{\hat{z}}$ (4c) O II
$\mathbf{B_{20}}$ = $- x_{7} \, \mathbf{a}_{1}- y_{7} \, \mathbf{a}_{2}+z_{7} \, \mathbf{a}_{3}$ = $- a x_{7} \,\mathbf{\hat{x}}- b y_{7} \,\mathbf{\hat{y}}+c z_{7} \,\mathbf{\hat{z}}$ (4c) O II
$\mathbf{B_{21}}$ = $x_{7} \, \mathbf{a}_{1}- \left(y_{7} - \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{7} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a x_{7} \,\mathbf{\hat{x}}- b \left(y_{7} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{7} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) O II
$\mathbf{B_{22}}$ = $- x_{7} \, \mathbf{a}_{1}+\left(y_{7} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{7} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a x_{7} \,\mathbf{\hat{x}}+b \left(y_{7} + \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{7} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) O II
$\mathbf{B_{23}}$ = $x_{8} \, \mathbf{a}_{1}+y_{8} \, \mathbf{a}_{2}+z_{8} \, \mathbf{a}_{3}$ = $a x_{8} \,\mathbf{\hat{x}}+b y_{8} \,\mathbf{\hat{y}}+c z_{8} \,\mathbf{\hat{z}}$ (4c) O III
$\mathbf{B_{24}}$ = $- x_{8} \, \mathbf{a}_{1}- y_{8} \, \mathbf{a}_{2}+z_{8} \, \mathbf{a}_{3}$ = $- a x_{8} \,\mathbf{\hat{x}}- b y_{8} \,\mathbf{\hat{y}}+c z_{8} \,\mathbf{\hat{z}}$ (4c) O III
$\mathbf{B_{25}}$ = $x_{8} \, \mathbf{a}_{1}- \left(y_{8} - \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{8} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a x_{8} \,\mathbf{\hat{x}}- b \left(y_{8} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{8} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) O III
$\mathbf{B_{26}}$ = $- x_{8} \, \mathbf{a}_{1}+\left(y_{8} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{8} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a x_{8} \,\mathbf{\hat{x}}+b \left(y_{8} + \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{8} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) O III
$\mathbf{B_{27}}$ = $x_{9} \, \mathbf{a}_{1}+y_{9} \, \mathbf{a}_{2}+z_{9} \, \mathbf{a}_{3}$ = $a x_{9} \,\mathbf{\hat{x}}+b y_{9} \,\mathbf{\hat{y}}+c z_{9} \,\mathbf{\hat{z}}$ (4c) O IV
$\mathbf{B_{28}}$ = $- x_{9} \, \mathbf{a}_{1}- y_{9} \, \mathbf{a}_{2}+z_{9} \, \mathbf{a}_{3}$ = $- a x_{9} \,\mathbf{\hat{x}}- b y_{9} \,\mathbf{\hat{y}}+c z_{9} \,\mathbf{\hat{z}}$ (4c) O IV
$\mathbf{B_{29}}$ = $x_{9} \, \mathbf{a}_{1}- \left(y_{9} - \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{9} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a x_{9} \,\mathbf{\hat{x}}- b \left(y_{9} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{9} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) O IV
$\mathbf{B_{30}}$ = $- x_{9} \, \mathbf{a}_{1}+\left(y_{9} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{9} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a x_{9} \,\mathbf{\hat{x}}+b \left(y_{9} + \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{9} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) O IV
$\mathbf{B_{31}}$ = $x_{10} \, \mathbf{a}_{1}+y_{10} \, \mathbf{a}_{2}+z_{10} \, \mathbf{a}_{3}$ = $a x_{10} \,\mathbf{\hat{x}}+b y_{10} \,\mathbf{\hat{y}}+c z_{10} \,\mathbf{\hat{z}}$ (4c) O V
$\mathbf{B_{32}}$ = $- x_{10} \, \mathbf{a}_{1}- y_{10} \, \mathbf{a}_{2}+z_{10} \, \mathbf{a}_{3}$ = $- a x_{10} \,\mathbf{\hat{x}}- b y_{10} \,\mathbf{\hat{y}}+c z_{10} \,\mathbf{\hat{z}}$ (4c) O V
$\mathbf{B_{33}}$ = $x_{10} \, \mathbf{a}_{1}- \left(y_{10} - \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{10} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a x_{10} \,\mathbf{\hat{x}}- b \left(y_{10} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{10} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) O V
$\mathbf{B_{34}}$ = $- x_{10} \, \mathbf{a}_{1}+\left(y_{10} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{10} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a x_{10} \,\mathbf{\hat{x}}+b \left(y_{10} + \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{10} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) O V
$\mathbf{B_{35}}$ = $x_{11} \, \mathbf{a}_{1}+y_{11} \, \mathbf{a}_{2}+z_{11} \, \mathbf{a}_{3}$ = $a x_{11} \,\mathbf{\hat{x}}+b y_{11} \,\mathbf{\hat{y}}+c z_{11} \,\mathbf{\hat{z}}$ (4c) O VI
$\mathbf{B_{36}}$ = $- x_{11} \, \mathbf{a}_{1}- y_{11} \, \mathbf{a}_{2}+z_{11} \, \mathbf{a}_{3}$ = $- a x_{11} \,\mathbf{\hat{x}}- b y_{11} \,\mathbf{\hat{y}}+c z_{11} \,\mathbf{\hat{z}}$ (4c) O VI
$\mathbf{B_{37}}$ = $x_{11} \, \mathbf{a}_{1}- \left(y_{11} - \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{11} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a x_{11} \,\mathbf{\hat{x}}- b \left(y_{11} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{11} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) O VI
$\mathbf{B_{38}}$ = $- x_{11} \, \mathbf{a}_{1}+\left(y_{11} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{11} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a x_{11} \,\mathbf{\hat{x}}+b \left(y_{11} + \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{11} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) O VI
$\mathbf{B_{39}}$ = $x_{12} \, \mathbf{a}_{1}+y_{12} \, \mathbf{a}_{2}+z_{12} \, \mathbf{a}_{3}$ = $a x_{12} \,\mathbf{\hat{x}}+b y_{12} \,\mathbf{\hat{y}}+c z_{12} \,\mathbf{\hat{z}}$ (4c) O VII
$\mathbf{B_{40}}$ = $- x_{12} \, \mathbf{a}_{1}- y_{12} \, \mathbf{a}_{2}+z_{12} \, \mathbf{a}_{3}$ = $- a x_{12} \,\mathbf{\hat{x}}- b y_{12} \,\mathbf{\hat{y}}+c z_{12} \,\mathbf{\hat{z}}$ (4c) O VII
$\mathbf{B_{41}}$ = $x_{12} \, \mathbf{a}_{1}- \left(y_{12} - \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{12} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a x_{12} \,\mathbf{\hat{x}}- b \left(y_{12} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{12} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) O VII
$\mathbf{B_{42}}$ = $- x_{12} \, \mathbf{a}_{1}+\left(y_{12} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{12} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a x_{12} \,\mathbf{\hat{x}}+b \left(y_{12} + \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{12} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) O VII
$\mathbf{B_{43}}$ = $x_{13} \, \mathbf{a}_{1}+y_{13} \, \mathbf{a}_{2}+z_{13} \, \mathbf{a}_{3}$ = $a x_{13} \,\mathbf{\hat{x}}+b y_{13} \,\mathbf{\hat{y}}+c z_{13} \,\mathbf{\hat{z}}$ (4c) O VIII
$\mathbf{B_{44}}$ = $- x_{13} \, \mathbf{a}_{1}- y_{13} \, \mathbf{a}_{2}+z_{13} \, \mathbf{a}_{3}$ = $- a x_{13} \,\mathbf{\hat{x}}- b y_{13} \,\mathbf{\hat{y}}+c z_{13} \,\mathbf{\hat{z}}$ (4c) O VIII
$\mathbf{B_{45}}$ = $x_{13} \, \mathbf{a}_{1}- \left(y_{13} - \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{13} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a x_{13} \,\mathbf{\hat{x}}- b \left(y_{13} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{13} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) O VIII
$\mathbf{B_{46}}$ = $- x_{13} \, \mathbf{a}_{1}+\left(y_{13} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{13} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a x_{13} \,\mathbf{\hat{x}}+b \left(y_{13} + \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{13} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) O VIII

References

  • S. Tahara, A. Shimada, N. Kumada, and Y. Sugahara, Characterization of Bi5Nb3O15 by refinement of neutron diffraction pattern, acid treatment and reaction of the acid-treated product with {\em n}-alkylamines, J. Solid State Chem. 180, 2517–2524 (2007), doi:10.1016/j.jssc.2007.05.017.
  • D. Hicks, M. J. Mehl, E. Gossett, C. Toher, O. Levy, R. M. Hanson, G. Hart, and S. Curtarolo, The AFLOW Library of Crystallographic Prototypes: Part 2, Comput. Mater. Sci. 161, S1–S1011 (2019), doi:10.1016/j.commatsci.2018.10.043.

Found in

  • P. Villars and K. Cenzual, Pearson's Crystal Data – Crystal Structure Database for Inorganic Compounds (2013). ASM International.

Prototype Generator

aflow --proto=A5B3C15_oP46_30_a2c_bc_a7c --params=$a,b/a,c/a,z_{1},z_{2},z_{3},x_{4},y_{4},z_{4},x_{5},y_{5},z_{5},x_{6},y_{6},z_{6},x_{7},y_{7},z_{7},x_{8},y_{8},z_{8},x_{9},y_{9},z_{9},x_{10},y_{10},z_{10},x_{11},y_{11},z_{11},x_{12},y_{12},z_{12},x_{13},y_{13},z_{13}$

Species:

Running:

Output: