α-Li$_{3}$PS$_{4}$ Structure: A2BC4_oP28_60_d_c_2d-002

Picture of Structure; Click for Big Picture
Prototype Li$_{3}$PS$_{4}$
AFLOW prototype label A2BC4_oP28_60_d_c_2d-002
Pearson symbol oP28
Space group number 60
Space group symbol $Pbcn$
AFLOW prototype command aflow --proto=A2BC4_oP28_60_d_c_2d-002
--params=$a, \allowbreak b/a, \allowbreak c/a, \allowbreak y_{1}, \allowbreak x_{2}, \allowbreak y_{2}, \allowbreak z_{2}, \allowbreak x_{3}, \allowbreak y_{3}, \allowbreak z_{3}, \allowbreak x_{4}, \allowbreak y_{4}, \allowbreak z_{4}$

  • Li$_{3}$PS$_{4}$ is known to exist in four forms (Homma, 2011):
  • The astute reader will note that we have left 1/3 of the lithium atoms unaccounted for. (Homma, 2010) could not find the positions of these atoms. As Li$_{3}$PS$_{4}$ is a superionic conductor it is possible that these atoms move freely through the structure.
  • Data for this structure was taken at 906K.
  • CaB$_{2}$O$_{4}$ I and $\alpha$–Li$_{3}$PO$_{4}$ have the same AFLOW prototype label, A2BC4_oP28_60_d_c_2d. They are generated by the same symmetry operations with different sets of parameters (--params) specified in their corresponding CIF files.

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

Basis vectors

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

References

  • K. Homma, M. Yonemura, M. Nagao, M. Hirayama, and R. Kanno, Crystal Structure of High-Temperature Phase of Lithium Ionic Conductor, Li$_{3}$PS$_{4}$, J. Phys. Soc. Jpn. 79, 90–93 (2010), doi:10.1143/JPSJS.79SA.90.

Found in

  • K. Homma, M. Yonemura, T. Kobayashi, M. Nagao, M. Hirayama, and R. Kanno, Crystal structure and phase transitions of the lithium ionic conductor Li$_{3}$PS$_{4}$, Solid State Ionics 182, 53–58 (2011), doi:10.1016/j.ssi.2010.10.001.

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=A2BC4_oP28_60_d_c_2d --params=$a,b/a,c/a,y_{1},x_{2},y_{2},z_{2},x_{3},y_{3},z_{3},x_{4},y_{4},z_{4}$

Species:

Running:

Output: