Encyclopedia of Crystallographic Prototypes

AFLOW Prototype: A5B5C4_tP28_104_ac_ac_c-001

This structure originally had the label A5B5C4_tP28_104_ac_ac_c. 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)

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https://aflow.org/p/GFEJ
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Ba$_{5}$In$_{4}$Bi$_{5}$ Structure: A5B5C4_tP28_104_ac_ac_c-001

Picture of Structure; Click for Big Picture
Prototype Ba$_{5}$Bi$_{5}$In$_{4}$
AFLOW prototype label A5B5C4_tP28_104_ac_ac_c-001
ICSD 54853
Pearson symbol tP28
Space group number 104
Space group symbol $P4nc$
AFLOW prototype command aflow --proto=A5B5C4_tP28_104_ac_ac_c-001
--params=$a, \allowbreak c/a, \allowbreak z_{1}, \allowbreak z_{2}, \allowbreak x_{3}, \allowbreak y_{3}, \allowbreak z_{3}, \allowbreak x_{4}, \allowbreak y_{4}, \allowbreak z_{4}, \allowbreak x_{5}, \allowbreak y_{5}, \allowbreak z_{5}$

  • Space group $P4nc$ #104 allows an arbitary placement of the origin of the $z$-axis. Here we use this freedom to set $z_{1} = 1/2$ for the Ba-I atoms.

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

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) Ba I
$\mathbf{B_{2}}$ = $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}+\left(z_{1} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{1}{2}a \,\mathbf{\hat{y}}+c \left(z_{1} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (2a) Ba I
$\mathbf{B_{3}}$ = $z_{2} \, \mathbf{a}_{3}$ = $c z_{2} \,\mathbf{\hat{z}}$ (2a) Bi I
$\mathbf{B_{4}}$ = $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}+\left(z_{2} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{1}{2}a \,\mathbf{\hat{y}}+c \left(z_{2} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (2a) Bi I
$\mathbf{B_{5}}$ = $x_{3} \, \mathbf{a}_{1}+y_{3} \, \mathbf{a}_{2}+z_{3} \, \mathbf{a}_{3}$ = $a x_{3} \,\mathbf{\hat{x}}+a y_{3} \,\mathbf{\hat{y}}+c z_{3} \,\mathbf{\hat{z}}$ (8c) Ba II
$\mathbf{B_{6}}$ = $- x_{3} \, \mathbf{a}_{1}- y_{3} \, \mathbf{a}_{2}+z_{3} \, \mathbf{a}_{3}$ = $- a x_{3} \,\mathbf{\hat{x}}- a y_{3} \,\mathbf{\hat{y}}+c z_{3} \,\mathbf{\hat{z}}$ (8c) Ba II
$\mathbf{B_{7}}$ = $- y_{3} \, \mathbf{a}_{1}+x_{3} \, \mathbf{a}_{2}+z_{3} \, \mathbf{a}_{3}$ = $- a y_{3} \,\mathbf{\hat{x}}+a x_{3} \,\mathbf{\hat{y}}+c z_{3} \,\mathbf{\hat{z}}$ (8c) Ba II
$\mathbf{B_{8}}$ = $y_{3} \, \mathbf{a}_{1}- x_{3} \, \mathbf{a}_{2}+z_{3} \, \mathbf{a}_{3}$ = $a y_{3} \,\mathbf{\hat{x}}- a x_{3} \,\mathbf{\hat{y}}+c z_{3} \,\mathbf{\hat{z}}$ (8c) Ba II
$\mathbf{B_{9}}$ = $\left(x_{3} + \frac{1}{2}\right) \, \mathbf{a}_{1}- \left(y_{3} - \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{3} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a \left(x_{3} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- a \left(y_{3} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{3} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (8c) Ba II
$\mathbf{B_{10}}$ = $- \left(x_{3} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\left(y_{3} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{3} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a \left(x_{3} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+a \left(y_{3} + \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{3} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (8c) Ba II
$\mathbf{B_{11}}$ = $- \left(y_{3} - \frac{1}{2}\right) \, \mathbf{a}_{1}- \left(x_{3} - \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{3} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a \left(y_{3} - \frac{1}{2}\right) \,\mathbf{\hat{x}}- a \left(x_{3} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{3} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (8c) Ba II
$\mathbf{B_{12}}$ = $\left(y_{3} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\left(x_{3} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{3} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a \left(y_{3} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+a \left(x_{3} + \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{3} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (8c) Ba II
$\mathbf{B_{13}}$ = $x_{4} \, \mathbf{a}_{1}+y_{4} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ = $a x_{4} \,\mathbf{\hat{x}}+a y_{4} \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ (8c) Bi II
$\mathbf{B_{14}}$ = $- x_{4} \, \mathbf{a}_{1}- y_{4} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ = $- a x_{4} \,\mathbf{\hat{x}}- a y_{4} \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ (8c) Bi II
$\mathbf{B_{15}}$ = $- y_{4} \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ = $- a y_{4} \,\mathbf{\hat{x}}+a x_{4} \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ (8c) Bi II
$\mathbf{B_{16}}$ = $y_{4} \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ = $a y_{4} \,\mathbf{\hat{x}}- a x_{4} \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ (8c) Bi II
$\mathbf{B_{17}}$ = $\left(x_{4} + \frac{1}{2}\right) \, \mathbf{a}_{1}- \left(y_{4} - \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a \left(x_{4} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- a \left(y_{4} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{4} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (8c) Bi II
$\mathbf{B_{18}}$ = $- \left(x_{4} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\left(y_{4} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a \left(x_{4} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+a \left(y_{4} + \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{4} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (8c) Bi II
$\mathbf{B_{19}}$ = $- \left(y_{4} - \frac{1}{2}\right) \, \mathbf{a}_{1}- \left(x_{4} - \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a \left(y_{4} - \frac{1}{2}\right) \,\mathbf{\hat{x}}- a \left(x_{4} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{4} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (8c) Bi II
$\mathbf{B_{20}}$ = $\left(y_{4} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\left(x_{4} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a \left(y_{4} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+a \left(x_{4} + \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{4} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (8c) Bi II
$\mathbf{B_{21}}$ = $x_{5} \, \mathbf{a}_{1}+y_{5} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $a x_{5} \,\mathbf{\hat{x}}+a y_{5} \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ (8c) In I
$\mathbf{B_{22}}$ = $- x_{5} \, \mathbf{a}_{1}- y_{5} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $- a x_{5} \,\mathbf{\hat{x}}- a y_{5} \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ (8c) In I
$\mathbf{B_{23}}$ = $- y_{5} \, \mathbf{a}_{1}+x_{5} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $- a y_{5} \,\mathbf{\hat{x}}+a x_{5} \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ (8c) In I
$\mathbf{B_{24}}$ = $y_{5} \, \mathbf{a}_{1}- x_{5} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $a y_{5} \,\mathbf{\hat{x}}- a x_{5} \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ (8c) In I
$\mathbf{B_{25}}$ = $\left(x_{5} + \frac{1}{2}\right) \, \mathbf{a}_{1}- \left(y_{5} - \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a \left(x_{5} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- a \left(y_{5} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (8c) In I
$\mathbf{B_{26}}$ = $- \left(x_{5} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\left(y_{5} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a \left(x_{5} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+a \left(y_{5} + \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (8c) In I
$\mathbf{B_{27}}$ = $- \left(y_{5} - \frac{1}{2}\right) \, \mathbf{a}_{1}- \left(x_{5} - \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a \left(y_{5} - \frac{1}{2}\right) \,\mathbf{\hat{x}}- a \left(x_{5} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (8c) In I
$\mathbf{B_{28}}$ = $\left(y_{5} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\left(x_{5} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a \left(y_{5} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+a \left(x_{5} + \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (8c) In I

References

  • S. Ponou, T. F. Fässler, G. Tobías, E. Canadell, A. Cho, and S. C. Sevov, Synthesis, Characterization, and Electronic Structure of Ba$_{5}$In$_{4}$Bi$_{5}$: An Acentric and One-Electron Deficient Phase, Chem. Eur. J 10, 3615–3621 (2004), doi:10.1002/chem.200306061.

Found in

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

Prototype Generator

aflow --proto=A5B5C4_tP28_104_ac_ac_c --params=$a,c/a,z_{1},z_{2},x_{3},y_{3},z_{3},x_{4},y_{4},z_{4},x_{5},y_{5},z_{5}$

Species:

Running:

Output: