Encyclopedia of Crystallographic Prototypes

AFLOW Prototype: A2B_mP18_7_6a_3a-001

This structure originally had the label A2B_mP18_7_6a_3a. 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/34FL
or https://aflow.org/p/A2B_mP18_7_6a_3a-001
or PDF Version

BaAs$_{2}$ Structure: A2B_mP18_7_6a_3a-001

Picture of Structure; Click for Big Picture
Prototype As$_{2}$Ba
AFLOW prototype label A2B_mP18_7_6a_3a-001
ICSD 414139
Pearson symbol mP18
Space group number 7
Space group symbol $Pc$
AFLOW prototype command aflow --proto=A2B_mP18_7_6a_3a-001
--params=$a, \allowbreak b/a, \allowbreak c/a, \allowbreak \beta, \allowbreak x_{1}, \allowbreak y_{1}, \allowbreak z_{1}, \allowbreak x_{2}, \allowbreak y_{2}, \allowbreak z_{2}, \allowbreak x_{3}, \allowbreak y_{3}, \allowbreak z_{3}, \allowbreak x_{4}, \allowbreak y_{4}, \allowbreak z_{4}, \allowbreak x_{5}, \allowbreak y_{5}, \allowbreak z_{5}, \allowbreak x_{6}, \allowbreak y_{6}, \allowbreak z_{6}, \allowbreak x_{7}, \allowbreak y_{7}, \allowbreak z_{7}, \allowbreak x_{8}, \allowbreak y_{8}, \allowbreak z_{8}, \allowbreak x_{9}, \allowbreak y_{9}, \allowbreak z_{9}$

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

Basis vectors

Lattice coordinates Cartesian coordinates Wyckoff position Atom type
$\mathbf{B_{1}}$ = $x_{1} \, \mathbf{a}_{1}+y_{1} \, \mathbf{a}_{2}+z_{1} \, \mathbf{a}_{3}$ = $\left(a x_{1} + c z_{1} \cos{\beta}\right) \,\mathbf{\hat{x}}+b y_{1} \,\mathbf{\hat{y}}+c z_{1} \sin{\beta} \,\mathbf{\hat{z}}$ (2a) As I
$\mathbf{B_{2}}$ = $x_{1} \, \mathbf{a}_{1}- y_{1} \, \mathbf{a}_{2}+\left(z_{1} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\left(a x_{1} + c \left(z_{1} + \frac{1}{2}\right) \cos{\beta}\right) \,\mathbf{\hat{x}}- b y_{1} \,\mathbf{\hat{y}}+c \left(z_{1} + \frac{1}{2}\right) \sin{\beta} \,\mathbf{\hat{z}}$ (2a) As I
$\mathbf{B_{3}}$ = $x_{2} \, \mathbf{a}_{1}+y_{2} \, \mathbf{a}_{2}+z_{2} \, \mathbf{a}_{3}$ = $\left(a x_{2} + c z_{2} \cos{\beta}\right) \,\mathbf{\hat{x}}+b y_{2} \,\mathbf{\hat{y}}+c z_{2} \sin{\beta} \,\mathbf{\hat{z}}$ (2a) As II
$\mathbf{B_{4}}$ = $x_{2} \, \mathbf{a}_{1}- y_{2} \, \mathbf{a}_{2}+\left(z_{2} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\left(a x_{2} + c \left(z_{2} + \frac{1}{2}\right) \cos{\beta}\right) \,\mathbf{\hat{x}}- b y_{2} \,\mathbf{\hat{y}}+c \left(z_{2} + \frac{1}{2}\right) \sin{\beta} \,\mathbf{\hat{z}}$ (2a) As II
$\mathbf{B_{5}}$ = $x_{3} \, \mathbf{a}_{1}+y_{3} \, \mathbf{a}_{2}+z_{3} \, \mathbf{a}_{3}$ = $\left(a x_{3} + c z_{3} \cos{\beta}\right) \,\mathbf{\hat{x}}+b y_{3} \,\mathbf{\hat{y}}+c z_{3} \sin{\beta} \,\mathbf{\hat{z}}$ (2a) As III
$\mathbf{B_{6}}$ = $x_{3} \, \mathbf{a}_{1}- y_{3} \, \mathbf{a}_{2}+\left(z_{3} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\left(a x_{3} + c \left(z_{3} + \frac{1}{2}\right) \cos{\beta}\right) \,\mathbf{\hat{x}}- b y_{3} \,\mathbf{\hat{y}}+c \left(z_{3} + \frac{1}{2}\right) \sin{\beta} \,\mathbf{\hat{z}}$ (2a) As III
$\mathbf{B_{7}}$ = $x_{4} \, \mathbf{a}_{1}+y_{4} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ = $\left(a x_{4} + c z_{4} \cos{\beta}\right) \,\mathbf{\hat{x}}+b y_{4} \,\mathbf{\hat{y}}+c z_{4} \sin{\beta} \,\mathbf{\hat{z}}$ (2a) As IV
$\mathbf{B_{8}}$ = $x_{4} \, \mathbf{a}_{1}- y_{4} \, \mathbf{a}_{2}+\left(z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\left(a x_{4} + c \left(z_{4} + \frac{1}{2}\right) \cos{\beta}\right) \,\mathbf{\hat{x}}- b y_{4} \,\mathbf{\hat{y}}+c \left(z_{4} + \frac{1}{2}\right) \sin{\beta} \,\mathbf{\hat{z}}$ (2a) As IV
$\mathbf{B_{9}}$ = $x_{5} \, \mathbf{a}_{1}+y_{5} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $\left(a x_{5} + c z_{5} \cos{\beta}\right) \,\mathbf{\hat{x}}+b y_{5} \,\mathbf{\hat{y}}+c z_{5} \sin{\beta} \,\mathbf{\hat{z}}$ (2a) As V
$\mathbf{B_{10}}$ = $x_{5} \, \mathbf{a}_{1}- y_{5} \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\left(a x_{5} + c \left(z_{5} + \frac{1}{2}\right) \cos{\beta}\right) \,\mathbf{\hat{x}}- b y_{5} \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{2}\right) \sin{\beta} \,\mathbf{\hat{z}}$ (2a) As V
$\mathbf{B_{11}}$ = $x_{6} \, \mathbf{a}_{1}+y_{6} \, \mathbf{a}_{2}+z_{6} \, \mathbf{a}_{3}$ = $\left(a x_{6} + c z_{6} \cos{\beta}\right) \,\mathbf{\hat{x}}+b y_{6} \,\mathbf{\hat{y}}+c z_{6} \sin{\beta} \,\mathbf{\hat{z}}$ (2a) As VI
$\mathbf{B_{12}}$ = $x_{6} \, \mathbf{a}_{1}- y_{6} \, \mathbf{a}_{2}+\left(z_{6} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\left(a x_{6} + c \left(z_{6} + \frac{1}{2}\right) \cos{\beta}\right) \,\mathbf{\hat{x}}- b y_{6} \,\mathbf{\hat{y}}+c \left(z_{6} + \frac{1}{2}\right) \sin{\beta} \,\mathbf{\hat{z}}$ (2a) As VI
$\mathbf{B_{13}}$ = $x_{7} \, \mathbf{a}_{1}+y_{7} \, \mathbf{a}_{2}+z_{7} \, \mathbf{a}_{3}$ = $\left(a x_{7} + c z_{7} \cos{\beta}\right) \,\mathbf{\hat{x}}+b y_{7} \,\mathbf{\hat{y}}+c z_{7} \sin{\beta} \,\mathbf{\hat{z}}$ (2a) Ba I
$\mathbf{B_{14}}$ = $x_{7} \, \mathbf{a}_{1}- y_{7} \, \mathbf{a}_{2}+\left(z_{7} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\left(a x_{7} + c \left(z_{7} + \frac{1}{2}\right) \cos{\beta}\right) \,\mathbf{\hat{x}}- b y_{7} \,\mathbf{\hat{y}}+c \left(z_{7} + \frac{1}{2}\right) \sin{\beta} \,\mathbf{\hat{z}}$ (2a) Ba I
$\mathbf{B_{15}}$ = $x_{8} \, \mathbf{a}_{1}+y_{8} \, \mathbf{a}_{2}+z_{8} \, \mathbf{a}_{3}$ = $\left(a x_{8} + c z_{8} \cos{\beta}\right) \,\mathbf{\hat{x}}+b y_{8} \,\mathbf{\hat{y}}+c z_{8} \sin{\beta} \,\mathbf{\hat{z}}$ (2a) Ba II
$\mathbf{B_{16}}$ = $x_{8} \, \mathbf{a}_{1}- y_{8} \, \mathbf{a}_{2}+\left(z_{8} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\left(a x_{8} + c \left(z_{8} + \frac{1}{2}\right) \cos{\beta}\right) \,\mathbf{\hat{x}}- b y_{8} \,\mathbf{\hat{y}}+c \left(z_{8} + \frac{1}{2}\right) \sin{\beta} \,\mathbf{\hat{z}}$ (2a) Ba II
$\mathbf{B_{17}}$ = $x_{9} \, \mathbf{a}_{1}+y_{9} \, \mathbf{a}_{2}+z_{9} \, \mathbf{a}_{3}$ = $\left(a x_{9} + c z_{9} \cos{\beta}\right) \,\mathbf{\hat{x}}+b y_{9} \,\mathbf{\hat{y}}+c z_{9} \sin{\beta} \,\mathbf{\hat{z}}$ (2a) Ba III
$\mathbf{B_{18}}$ = $x_{9} \, \mathbf{a}_{1}- y_{9} \, \mathbf{a}_{2}+\left(z_{9} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\left(a x_{9} + c \left(z_{9} + \frac{1}{2}\right) \cos{\beta}\right) \,\mathbf{\hat{x}}- b y_{9} \,\mathbf{\hat{y}}+c \left(z_{9} + \frac{1}{2}\right) \sin{\beta} \,\mathbf{\hat{z}}$ (2a) Ba III

References

  • F. Emmerling, D. Petri, and C. Röhr, Neue Arsenide mit As$^{-}$-Ketten und -Ringen: BaAs$_{2}$ und A$^{I}$Ba$_{2}$As$_{5}$ (A$^{I}$ = K, Rb), Z. Anorganische und Allgemeine Chemie 630, 2490–2501 (2004), doi:10.1002/zaac.200400257.

Found in

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

Prototype Generator

aflow --proto=A2B_mP18_7_6a_3a --params=$a,b/a,c/a,\beta,x_{1},y_{1},z_{1},x_{2},y_{2},z_{2},x_{3},y_{3},z_{3},x_{4},y_{4},z_{4},x_{5},y_{5},z_{5},x_{6},y_{6},z_{6},x_{7},y_{7},z_{7},x_{8},y_{8},z_{8},x_{9},y_{9},z_{9}$

Species:

Running:

Output: