Encyclopedia of Crystallographic Prototypes

AFLOW Prototype: A3B7C_hR11_160_b_a2b_a-001

This structure originally had the label A3B7C_hR11_160_b_a2b_a. Calls to that address will be redirected here.

If you are using this page, please cite:
D. Hicks, M.J. Mehl, M. Esters, C. Oses, O. Levy, G.L.W. Hart, C. Toher, and S. Curtarolo, The AFLOW Library of Crystallographic Prototypes: Part 3, Comp. Mat. Sci. 199, 110450 (2021). (doi=10.1016/j.commatsci.2021.110450)

Links to this page

https://aflow.org/p/JN4A
or https://aflow.org/p/A3B7C_hR11_160_b_a2b_a-001
or PDF Version

Fe$_{3}$PO$_{7}$ Structure: A3B7C_hR11_160_b_a2b_a-001

Picture of Structure; Click for Big Picture
Prototype Fe$_{3}$O$_{7}$P
AFLOW prototype label A3B7C_hR11_160_b_a2b_a-001
ICSD 36207
Pearson symbol hR11
Space group number 160
Space group symbol $R3m$
AFLOW prototype command aflow --proto=A3B7C_hR11_160_b_a2b_a-001
--params=$a, \allowbreak c/a, \allowbreak x_{1}, \allowbreak x_{2}, \allowbreak x_{3}, \allowbreak z_{3}, \allowbreak x_{4}, \allowbreak z_{4}, \allowbreak x_{5}, \allowbreak z_{5}$

  • Space group $R3m$ #160 allows an arbitary definition of the zero of the $z$-axis. Here we select $z_{1} = 0$, putting the phosphorous atom at the origin.

\[ \begin{array}{ccc} \mathbf{a_{1}}&=&\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+\frac{1}{3}c \,\mathbf{\hat{z}}\\\mathbf{a_{2}}&=&\frac{1}{\sqrt{3}}a \,\mathbf{\hat{y}}+\frac{1}{3}c \,\mathbf{\hat{z}}\\\mathbf{a_{3}}&=&- \frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+\frac{1}{3}c \,\mathbf{\hat{z}} \end{array}\]

Basis vectors

Lattice coordinates Cartesian coordinates Wyckoff position Atom type
$\mathbf{B_{1}}$ = $x_{1} \, \mathbf{a}_{1}+x_{1} \, \mathbf{a}_{2}+x_{1} \, \mathbf{a}_{3}$ = $c x_{1} \,\mathbf{\hat{z}}$ (1a) O I
$\mathbf{B_{2}}$ = $x_{2} \, \mathbf{a}_{1}+x_{2} \, \mathbf{a}_{2}+x_{2} \, \mathbf{a}_{3}$ = $c x_{2} \,\mathbf{\hat{z}}$ (1a) P I
$\mathbf{B_{3}}$ = $x_{3} \, \mathbf{a}_{1}+x_{3} \, \mathbf{a}_{2}+z_{3} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{3} - z_{3}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \left(x_{3} - z_{3}\right) \,\mathbf{\hat{y}}+\frac{1}{3}c \left(2 x_{3} + z_{3}\right) \,\mathbf{\hat{z}}$ (3b) Fe I
$\mathbf{B_{4}}$ = $z_{3} \, \mathbf{a}_{1}+x_{3} \, \mathbf{a}_{2}+x_{3} \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(x_{3} - z_{3}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \left(x_{3} - z_{3}\right) \,\mathbf{\hat{y}}+\frac{1}{3}c \left(2 x_{3} + z_{3}\right) \,\mathbf{\hat{z}}$ (3b) Fe I
$\mathbf{B_{5}}$ = $x_{3} \, \mathbf{a}_{1}+z_{3} \, \mathbf{a}_{2}+x_{3} \, \mathbf{a}_{3}$ = $- \frac{1}{\sqrt{3}}a \left(x_{3} - z_{3}\right) \,\mathbf{\hat{y}}+\frac{1}{3}c \left(2 x_{3} + z_{3}\right) \,\mathbf{\hat{z}}$ (3b) Fe I
$\mathbf{B_{6}}$ = $x_{4} \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{4} - z_{4}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \left(x_{4} - z_{4}\right) \,\mathbf{\hat{y}}+\frac{1}{3}c \left(2 x_{4} + z_{4}\right) \,\mathbf{\hat{z}}$ (3b) O II
$\mathbf{B_{7}}$ = $z_{4} \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}+x_{4} \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(x_{4} - z_{4}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \left(x_{4} - z_{4}\right) \,\mathbf{\hat{y}}+\frac{1}{3}c \left(2 x_{4} + z_{4}\right) \,\mathbf{\hat{z}}$ (3b) O II
$\mathbf{B_{8}}$ = $x_{4} \, \mathbf{a}_{1}+z_{4} \, \mathbf{a}_{2}+x_{4} \, \mathbf{a}_{3}$ = $- \frac{1}{\sqrt{3}}a \left(x_{4} - z_{4}\right) \,\mathbf{\hat{y}}+\frac{1}{3}c \left(2 x_{4} + z_{4}\right) \,\mathbf{\hat{z}}$ (3b) O II
$\mathbf{B_{9}}$ = $x_{5} \, \mathbf{a}_{1}+x_{5} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{5} - z_{5}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \left(x_{5} - z_{5}\right) \,\mathbf{\hat{y}}+\frac{1}{3}c \left(2 x_{5} + z_{5}\right) \,\mathbf{\hat{z}}$ (3b) O III
$\mathbf{B_{10}}$ = $z_{5} \, \mathbf{a}_{1}+x_{5} \, \mathbf{a}_{2}+x_{5} \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(x_{5} - z_{5}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \left(x_{5} - z_{5}\right) \,\mathbf{\hat{y}}+\frac{1}{3}c \left(2 x_{5} + z_{5}\right) \,\mathbf{\hat{z}}$ (3b) O III
$\mathbf{B_{11}}$ = $x_{5} \, \mathbf{a}_{1}+z_{5} \, \mathbf{a}_{2}+x_{5} \, \mathbf{a}_{3}$ = $- \frac{1}{\sqrt{3}}a \left(x_{5} - z_{5}\right) \,\mathbf{\hat{y}}+\frac{1}{3}c \left(2 x_{5} + z_{5}\right) \,\mathbf{\hat{z}}$ (3b) O III

References

  • A. Modaressi, A. Courtois, R. Gerardin, B. Malaman, and C. Gleitzer, Fe$_{3}$PO$_{7}$, Un cas de coordinence 5 du fer trivalent, etude structurale et magnetique, J. Solid State Chem. 47, 245–255 (1983), doi:10.1016/0022-4596(83)90016-6.

Found in

  • C. L. Sarkis, M. J. Tarne, J. R. Neilson, H. B. Cao, E. Coldren, M. P. Gelfand, and K. A. Ross, Partial Antiferromagnetic Helical Order in Single Crystal Fe$_{3}$PO$_{4}$O$_{3}$, Phys. Rev. B 101, 184417 (2020), doi:10.1103/PhysRevB.101.184417.

Prototype Generator

aflow --proto=A3B7C_hR11_160_b_a2b_a --params=$a,c/a,x_{1},x_{2},x_{3},z_{3},x_{4},z_{4},x_{5},z_{5}$

Species:

Running:

Output: