β-Berlinite (AlPO$_{4}$) Structure: AB4C_hP18_181_c_k_d-001

Picture of Structure; Click for Big Picture
Prototype AlO$_{4}$P
AFLOW prototype label AB4C_hP18_181_c_k_d-001
Mineral name berlinite
ICSD 9643
CCDC 1594615
Pearson symbol hP18
Space group number 181
Space group symbol $P6_422$
AFLOW prototype command aflow --proto=AB4C_hP18_181_c_k_d-001
--params=$a, \allowbreak c/a, \allowbreak x_{3}, \allowbreak y_{3}, \allowbreak z_{3}$

  • This is the high-temperature form of berlinite, the analog of $\beta$-quartz ($C8$).
  • Below 853K (Ng, 1976) it transforms into $\alpha$-berlinite in the alarsite (AlAsO$_{4}$) structure.
  • Our silica and aluminum phosphate page has much more about the relationship between these structures.
  • We originally found this structure described in (Muraoka, 1997), but their aluminum and phosphorous coordinates are inconsistent with the Al-O and P-O distances and bond angles, so we use this earlier work.
  • Like $\beta$-quartz, this structure is chiral, Sohncke Class II, and so may also be found in the enantiomorphic space group $P6_{2}22$ #180.
  • (Ng, 1976) put the phosphorous atoms on the (3c) Wyckoff postion, but the AFLOW Wyckoff ordering protocol shifts the origin of the $z$-axis by c/2, moving the aluminum atoms to the (3c) site.

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

Basis vectors

Lattice coordinates Cartesian coordinates Wyckoff position Atom type
$\mathbf{B_{1}}$ = $\frac{1}{2} \, \mathbf{a}_{1}$ = $\frac{1}{4}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{4}a \,\mathbf{\hat{y}}$ (3c) Al I
$\mathbf{B_{2}}$ = $\frac{1}{2} \, \mathbf{a}_{2}+\frac{1}{3} \, \mathbf{a}_{3}$ = $\frac{1}{4}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{4}a \,\mathbf{\hat{y}}+\frac{1}{3}c \,\mathbf{\hat{z}}$ (3c) Al I
$\mathbf{B_{3}}$ = $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}+\frac{2}{3} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{2}{3}c \,\mathbf{\hat{z}}$ (3c) Al I
$\mathbf{B_{4}}$ = $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{3}$ = $\frac{1}{4}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{4}a \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ (3d) P I
$\mathbf{B_{5}}$ = $\frac{1}{2} \, \mathbf{a}_{2}+\frac{5}{6} \, \mathbf{a}_{3}$ = $\frac{1}{4}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{4}a \,\mathbf{\hat{y}}+\frac{5}{6}c \,\mathbf{\hat{z}}$ (3d) P I
$\mathbf{B_{6}}$ = $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}+\frac{1}{6} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{1}{6}c \,\mathbf{\hat{z}}$ (3d) P I
$\mathbf{B_{7}}$ = $x_{3} \, \mathbf{a}_{1}+y_{3} \, \mathbf{a}_{2}+z_{3} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{3} + y_{3}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \left(x_{3} - y_{3}\right) \,\mathbf{\hat{y}}+c z_{3} \,\mathbf{\hat{z}}$ (12k) O I
$\mathbf{B_{8}}$ = $- y_{3} \, \mathbf{a}_{1}+\left(x_{3} - y_{3}\right) \, \mathbf{a}_{2}+\left(z_{3} + \frac{1}{3}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{3} - 2 y_{3}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{3} \,\mathbf{\hat{y}}+c \left(z_{3} + \frac{1}{3}\right) \,\mathbf{\hat{z}}$ (12k) O I
$\mathbf{B_{9}}$ = $- \left(x_{3} - y_{3}\right) \, \mathbf{a}_{1}- x_{3} \, \mathbf{a}_{2}+\left(z_{3} + \frac{2}{3}\right) \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(2 x_{3} - y_{3}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a y_{3} \,\mathbf{\hat{y}}+\frac{1}{3}c \left(3 z_{3} + 2\right) \,\mathbf{\hat{z}}$ (12k) O I
$\mathbf{B_{10}}$ = $- x_{3} \, \mathbf{a}_{1}- y_{3} \, \mathbf{a}_{2}+z_{3} \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(x_{3} + y_{3}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a \left(x_{3} - y_{3}\right) \,\mathbf{\hat{y}}+c z_{3} \,\mathbf{\hat{z}}$ (12k) O I
$\mathbf{B_{11}}$ = $y_{3} \, \mathbf{a}_{1}- \left(x_{3} - y_{3}\right) \, \mathbf{a}_{2}+\left(z_{3} + \frac{1}{3}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(- x_{3} + 2 y_{3}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{3} \,\mathbf{\hat{y}}+c \left(z_{3} + \frac{1}{3}\right) \,\mathbf{\hat{z}}$ (12k) O I
$\mathbf{B_{12}}$ = $\left(x_{3} - y_{3}\right) \, \mathbf{a}_{1}+x_{3} \, \mathbf{a}_{2}+\left(z_{3} + \frac{2}{3}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(2 x_{3} - y_{3}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a y_{3} \,\mathbf{\hat{y}}+\frac{1}{3}c \left(3 z_{3} + 2\right) \,\mathbf{\hat{z}}$ (12k) O I
$\mathbf{B_{13}}$ = $y_{3} \, \mathbf{a}_{1}+x_{3} \, \mathbf{a}_{2}- \left(z_{3} - \frac{1}{3}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{3} + y_{3}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a \left(x_{3} - y_{3}\right) \,\mathbf{\hat{y}}- c \left(z_{3} - \frac{1}{3}\right) \,\mathbf{\hat{z}}$ (12k) O I
$\mathbf{B_{14}}$ = $\left(x_{3} - y_{3}\right) \, \mathbf{a}_{1}- y_{3} \, \mathbf{a}_{2}- z_{3} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{3} - 2 y_{3}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{3} \,\mathbf{\hat{y}}- c z_{3} \,\mathbf{\hat{z}}$ (12k) O I
$\mathbf{B_{15}}$ = $- x_{3} \, \mathbf{a}_{1}- \left(x_{3} - y_{3}\right) \, \mathbf{a}_{2}- \left(z_{3} - \frac{2}{3}\right) \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(2 x_{3} - y_{3}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a y_{3} \,\mathbf{\hat{y}}- \frac{1}{3}c \left(3 z_{3} - 2\right) \,\mathbf{\hat{z}}$ (12k) O I
$\mathbf{B_{16}}$ = $- y_{3} \, \mathbf{a}_{1}- x_{3} \, \mathbf{a}_{2}- \left(z_{3} - \frac{1}{3}\right) \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(x_{3} + y_{3}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \left(x_{3} - y_{3}\right) \,\mathbf{\hat{y}}- c \left(z_{3} - \frac{1}{3}\right) \,\mathbf{\hat{z}}$ (12k) O I
$\mathbf{B_{17}}$ = $- \left(x_{3} - y_{3}\right) \, \mathbf{a}_{1}+y_{3} \, \mathbf{a}_{2}- z_{3} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(- x_{3} + 2 y_{3}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{3} \,\mathbf{\hat{y}}- c z_{3} \,\mathbf{\hat{z}}$ (12k) O I
$\mathbf{B_{18}}$ = $x_{3} \, \mathbf{a}_{1}+\left(x_{3} - y_{3}\right) \, \mathbf{a}_{2}- \left(z_{3} - \frac{2}{3}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(2 x_{3} - y_{3}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a y_{3} \,\mathbf{\hat{y}}- \frac{1}{3}c \left(3 z_{3} - 2\right) \,\mathbf{\hat{z}}$ (12k) O I

References

  • H. N. Ng and C. Calvo, X-ray study of the α-β transformation of berlinite (AlPO$_{4}$), Can. J. Physics 54, 638–647 (1976), doi:10.1139/p76-070.

Found in

  • Y. Muraoka and K. Kihara, The temperature dependence of the crystal structure of berlinite, a quartz-type form of AlPO$_{4}$, Phys. Chem. of Miner. 24, 243–253 (1997), doi:10.1007/s002690050036.

First cited in

  • N. Anderson, M. J. Mehl, H. Eckert, S. Divilov, X. Campilongo, S. Curtarolo, The AFLOW Library of Crystallographic Prototypes: Part 5. Submitted to Computational Materials Science (2026).

Geometry files


Prototype Generator

aflow --proto=AB4C_hP18_181_c_k_d --params=$a,c/a,x_{3},y_{3},z_{3}$

Species:

Running:

Output: