Encyclopedia of Crystallographic Prototypes

AFLOW Prototype: A2B_tP30_85_ab2g_cg-001

This structure originally had the label A2B_tP30_85_ab2g_cg. 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/G8GD
or https://aflow.org/p/A2B_tP30_85_ab2g_cg-001
or PDF Version

α-SrBr$_{2}$ Structure: A2B_tP30_85_ab2g_cg-001

Picture of Structure; Click for Big Picture
Prototype Br$_{2}$Sr
AFLOW prototype label A2B_tP30_85_ab2g_cg-001
ICSD 262673
Pearson symbol tP30
Space group number 85
Space group symbol $P4/n$
AFLOW prototype command aflow --proto=A2B_tP30_85_ab2g_cg-001
--params=$a, \allowbreak c/a, \allowbreak z_{3}, \allowbreak x_{4}, \allowbreak y_{4}, \allowbreak z_{4}, \allowbreak x_{5}, \allowbreak y_{5}, \allowbreak z_{5}, \allowbreak x_{6}, \allowbreak y_{6}, \allowbreak z_{6}$

Other compounds with this structure

EuBr${2}$,  $\alpha$-Se$_{2}$U


  • (Hermann, 1943) originally assigned SrBr$_{2}$ to space group $Pnma$ #62, and gave it the Strukturbericht designation $C53$. Subsequent investigation showed that the structure should be in space group $P4/n$ #85, but this structure was never given a Strukturbericht symbol, although (Parthé) does list it as $C53$.
  • This compound can also be found as $\beta$–SrBr$_{2}$ in the fluorite ($C1$) structure.

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

Basis vectors

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

References

  • S. Hull, S. T. Norberg, I. Ahmed, S. G. Eriksson, and C. E. Mohn, High temperature crystal structures and superionic properties of SrCl$_{2}$, SrBr$_{2}$, BaCl$_{2}$ and BaBr$_{2}$, J. Solid State Chem. 184, 2925–2935 (2011), doi:10.1016/j.jssc.2011.09.004.
  • K. Herrmann, ed., Strukturbericht Band VII 1939} (Akademische Verlagsgesellschaft M. B. H., Leipzig, 1943).
  • \bibitem{parthe93:TYPIXE. Parthé, L. Gelato, B. Chabot, M. Penso, K. Cenzula, and R. Gladyshevskii, Standardized Data and Crystal Chemical Characterization of Inorganic Structure Types, Gmelin Handbook of Inorganic and Organometallic Chemistry}, vol. 2 (Springer-Verlag, Berlin, Heidelberg, 1993), 8 edn., doi:10.1007/978-3-662-02909-1_3.\bibAnnoteFile{parthe93:TYPIX

Prototype Generator

aflow --proto=A2B_tP30_85_ab2g_cg --params=$a,c/a,z_{3},x_{4},y_{4},z_{4},x_{5},y_{5},z_{5},x_{6},y_{6},z_{6}$

Species:

Running:

Output: