Encyclopedia of Crystallographic Prototypes

AFLOW Prototype: AB9C4_hP28_188_a_kl_ck-001

This structure originally had the label AB9C4_hP28_188_e_kl_ak. 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/2DJ4
or https://aflow.org/p/AB9C4_hP28_188_a_kl_ck-001
or PDF Version

BaSi$_{4}$O$_{9}$ ($S3_{2}$) Structure: AB9C4_hP28_188_a_kl_ck-001

Picture of Structure; Click for Big Picture
Prototype BaSi$_{4}$O$_{9}$
AFLOW prototype label AB9C4_hP28_188_a_kl_ck-001
Strukturbericht designation $S3_{2}$
ICSD 80067
Pearson symbol hP28
Space group number 188
Space group symbol $P\overline{6}c2$
AFLOW prototype command aflow --proto=AB9C4_hP28_188_a_kl_ck-001
--params=$a, \allowbreak c/a, \allowbreak x_{3}, \allowbreak y_{3}, \allowbreak x_{4}, \allowbreak y_{4}, \allowbreak x_{5}, \allowbreak y_{5}, \allowbreak z_{5}$

Other compounds with this structure

Ba(Si$_{x}$Ti$_{1-x}$)Si$_{3}$O$_{9}$ (pabstite),  BaSnGe$_{3}$O$_{9}$,  BaTiSi$_{3}$O$_{9}$ (benitoite),  BaZrSi$_{3}$O$_{9}$ (bazrite),  CaKP$_{3}$O$_{9}$,  MgKP$_{3}$O$_{9}$,  TaKGe$_{3}$O$_{9}$,  TaRbGe$_{3}$O$_{9}$,  TaTlGe$_{3}$O$_{9}$


  • (Hermann, 1937) applied the Strukturbericht label $S3_{2}$ to benitoite, BaTiSi$_{3}$O$_{9}$. The only difference between BaSi$_{4}$O$_{9}$ and benitoite is that titanium replaces silicon at the $(2a)$ Wyckoff position.

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

Basis vectors

Lattice coordinates Cartesian coordinates Wyckoff position Atom type
$\mathbf{B_{1}}$ = $0$ = $0$ (2a) Ba I
$\mathbf{B_{2}}$ = $\frac{1}{2} \, \mathbf{a}_{3}$ = $\frac{1}{2}c \,\mathbf{\hat{z}}$ (2a) Ba I
$\mathbf{B_{3}}$ = $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}$ (2c) Si I
$\mathbf{B_{4}}$ = $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ (2c) Si I
$\mathbf{B_{5}}$ = $x_{3} \, \mathbf{a}_{1}+y_{3} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{3} + y_{3}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \left(x_{3} - y_{3}\right) \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (6k) O I
$\mathbf{B_{6}}$ = $- y_{3} \, \mathbf{a}_{1}+\left(x_{3} - y_{3}\right) \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{3} - 2 y_{3}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{3} \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (6k) O I
$\mathbf{B_{7}}$ = $- \left(x_{3} - y_{3}\right) \, \mathbf{a}_{1}- x_{3} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(2 x_{3} - y_{3}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a y_{3} \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (6k) O I
$\mathbf{B_{8}}$ = $- y_{3} \, \mathbf{a}_{1}- x_{3} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(x_{3} + y_{3}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \left(x_{3} - y_{3}\right) \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (6k) O I
$\mathbf{B_{9}}$ = $- \left(x_{3} - y_{3}\right) \, \mathbf{a}_{1}+y_{3} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(- x_{3} + 2 y_{3}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{3} \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (6k) O I
$\mathbf{B_{10}}$ = $x_{3} \, \mathbf{a}_{1}+\left(x_{3} - y_{3}\right) \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(2 x_{3} - y_{3}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a y_{3} \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (6k) O I
$\mathbf{B_{11}}$ = $x_{4} \, \mathbf{a}_{1}+y_{4} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{4} + y_{4}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \left(x_{4} - y_{4}\right) \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (6k) Si II
$\mathbf{B_{12}}$ = $- y_{4} \, \mathbf{a}_{1}+\left(x_{4} - y_{4}\right) \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{4} - 2 y_{4}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (6k) Si II
$\mathbf{B_{13}}$ = $- \left(x_{4} - y_{4}\right) \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(2 x_{4} - y_{4}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a y_{4} \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (6k) Si II
$\mathbf{B_{14}}$ = $- y_{4} \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(x_{4} + y_{4}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \left(x_{4} - y_{4}\right) \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (6k) Si II
$\mathbf{B_{15}}$ = $- \left(x_{4} - y_{4}\right) \, \mathbf{a}_{1}+y_{4} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(- x_{4} + 2 y_{4}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (6k) Si II
$\mathbf{B_{16}}$ = $x_{4} \, \mathbf{a}_{1}+\left(x_{4} - y_{4}\right) \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(2 x_{4} - y_{4}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a y_{4} \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (6k) Si II
$\mathbf{B_{17}}$ = $x_{5} \, \mathbf{a}_{1}+y_{5} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{5} + y_{5}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \left(x_{5} - y_{5}\right) \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ (12l) O II
$\mathbf{B_{18}}$ = $- y_{5} \, \mathbf{a}_{1}+\left(x_{5} - y_{5}\right) \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{5} - 2 y_{5}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{5} \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ (12l) O II
$\mathbf{B_{19}}$ = $- \left(x_{5} - y_{5}\right) \, \mathbf{a}_{1}- x_{5} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(2 x_{5} - y_{5}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a y_{5} \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ (12l) O II
$\mathbf{B_{20}}$ = $x_{5} \, \mathbf{a}_{1}+y_{5} \, \mathbf{a}_{2}- \left(z_{5} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{5} + y_{5}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \left(x_{5} - y_{5}\right) \,\mathbf{\hat{y}}- c \left(z_{5} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (12l) O II
$\mathbf{B_{21}}$ = $- y_{5} \, \mathbf{a}_{1}+\left(x_{5} - y_{5}\right) \, \mathbf{a}_{2}- \left(z_{5} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{5} - 2 y_{5}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{5} \,\mathbf{\hat{y}}- c \left(z_{5} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (12l) O II
$\mathbf{B_{22}}$ = $- \left(x_{5} - y_{5}\right) \, \mathbf{a}_{1}- x_{5} \, \mathbf{a}_{2}- \left(z_{5} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(2 x_{5} - y_{5}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a y_{5} \,\mathbf{\hat{y}}- c \left(z_{5} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (12l) O II
$\mathbf{B_{23}}$ = $- y_{5} \, \mathbf{a}_{1}- x_{5} \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(x_{5} + y_{5}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \left(x_{5} - y_{5}\right) \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (12l) O II
$\mathbf{B_{24}}$ = $- \left(x_{5} - y_{5}\right) \, \mathbf{a}_{1}+y_{5} \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(- x_{5} + 2 y_{5}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{5} \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (12l) O II
$\mathbf{B_{25}}$ = $x_{5} \, \mathbf{a}_{1}+\left(x_{5} - y_{5}\right) \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(2 x_{5} - y_{5}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a y_{5} \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (12l) O II
$\mathbf{B_{26}}$ = $- y_{5} \, \mathbf{a}_{1}- x_{5} \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(x_{5} + y_{5}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \left(x_{5} - y_{5}\right) \,\mathbf{\hat{y}}- c z_{5} \,\mathbf{\hat{z}}$ (12l) O II
$\mathbf{B_{27}}$ = $- \left(x_{5} - y_{5}\right) \, \mathbf{a}_{1}+y_{5} \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(- x_{5} + 2 y_{5}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{5} \,\mathbf{\hat{y}}- c z_{5} \,\mathbf{\hat{z}}$ (12l) O II
$\mathbf{B_{28}}$ = $x_{5} \, \mathbf{a}_{1}+\left(x_{5} - y_{5}\right) \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(2 x_{5} - y_{5}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a y_{5} \,\mathbf{\hat{y}}- c z_{5} \,\mathbf{\hat{z}}$ (12l) O II

References

  • L. W. Finger, R. M. Hazen, and B. A. Fursenko, Refinement of the crystal structure of BaSi$_{4}$O$_{9}$ in the benitoite form, J. Phys. Chem. Solids 56, 1389–1393 (1995), doi:10.1016/0022-3697(95)00075-5.
  • C. Hermann, O. Lohrmann, and H. Philipp, eds., Strukturbericht Band II 1928-1932 (Akademische Verlagsgesellschaft M. B. H., Leipzig, 1937).

Prototype Generator

aflow --proto=AB9C4_hP28_188_a_kl_ck --params=$a,c/a,x_{3},y_{3},x_{4},y_{4},x_{5},y_{5},z_{5}$

Species:

Running:

Output: