Encyclopedia of Crystallographic Prototypes

AFLOW Prototype: A3B_tI64_142_def_d-001

If you are using this page, please cite:
H. Eckert, S. Divilov, M. J. Mehl, D. Hicks, A. C. Zettel, M. Esters. X. Campilongo and S. Curtarolo, The AFLOW Library of Crystallographic Prototypes: Part 4. Submitted to Computational Materials Science.

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R$_{1}$ Au$_{3}$Zn Structure: A3B_tI64_142_def_d-001

Picture of Structure; Click for Big Picture
Prototype Au$_{3}$Zn
AFLOW prototype label A3B_tI64_142_def_d-001
ICSD 58628
Pearson symbol tI64
Space group number 142
Space group symbol $I4_1/acd$
AFLOW prototype command aflow --proto=A3B_tI64_142_def_d-001
--params=$a, \allowbreak c/a, \allowbreak z_{1}, \allowbreak z_{2}, \allowbreak x_{3}, \allowbreak x_{4}$

  • Au$_{3}$Zn is known to exist in three forms, depending upon the exact composition and temperature (Hisatsune, 1998):
    • The tetragonal $R_{1}$ phase (this structure) is stable below $≈ 475$K with a composition very nearly stoichiometric.
    • The orthorhombic $R_{2}$ phase is stable below $≈ 550$K with a composition range somewhat wider than the $R_{1}$ phase.
    • The tetragonal $H$ phase has the $D0_{23}$ structure and is stable at temperatures up to $≈ 700$K over a considerably wider range of stoichiometries than either the $R_{1}$ or $R_{2}$ phases.
  • (Iwaskai, 1962) gave structure of the $R_{1}$ phase in setting 1 of space group $I4_{1}/acd$ #142. We used FINDSYM to transform this to the standard setting 2.
  • (Iwaskai, 1962) gave the lattice constants in kX units. We used the conversion factor 1 kX = 1.00202Å. (Wood, 1947)

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

Basis vectors

Lattice coordinates Cartesian coordinates Wyckoff position Atom type
$\mathbf{B_{1}}$ = $\left(z_{1} + \frac{1}{4}\right) \, \mathbf{a}_{1}+z_{1} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $\frac{1}{4}a \,\mathbf{\hat{y}}+c z_{1} \,\mathbf{\hat{z}}$ (16d) Au I
$\mathbf{B_{2}}$ = $z_{1} \, \mathbf{a}_{1}+\left(z_{1} + \frac{1}{4}\right) \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{1}{4}a \,\mathbf{\hat{y}}+c \left(z_{1} - \frac{1}{4}\right) \,\mathbf{\hat{z}}$ (16d) Au I
$\mathbf{B_{3}}$ = $- \left(z_{1} - \frac{1}{4}\right) \, \mathbf{a}_{1}- \left(z_{1} - \frac{1}{2}\right) \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{1}{4}a \,\mathbf{\hat{y}}- c z_{1} \,\mathbf{\hat{z}}$ (16d) Au I
$\mathbf{B_{4}}$ = $- \left(z_{1} - \frac{1}{2}\right) \, \mathbf{a}_{1}- \left(z_{1} - \frac{1}{4}\right) \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $\frac{1}{4}a \,\mathbf{\hat{y}}- c \left(z_{1} - \frac{1}{4}\right) \,\mathbf{\hat{z}}$ (16d) Au I
$\mathbf{B_{5}}$ = $- \left(z_{1} - \frac{3}{4}\right) \, \mathbf{a}_{1}- z_{1} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $\frac{3}{4}a \,\mathbf{\hat{y}}- c z_{1} \,\mathbf{\hat{z}}$ (16d) Au I
$\mathbf{B_{6}}$ = $- z_{1} \, \mathbf{a}_{1}- \left(z_{1} - \frac{3}{4}\right) \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{1}{4}a \,\mathbf{\hat{y}}- c \left(z_{1} - \frac{1}{4}\right) \,\mathbf{\hat{z}}$ (16d) Au I
$\mathbf{B_{7}}$ = $\left(z_{1} + \frac{3}{4}\right) \, \mathbf{a}_{1}+\left(z_{1} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $\frac{1}{4}a \,\mathbf{\hat{y}}+c \left(z_{1} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (16d) Au I
$\mathbf{B_{8}}$ = $\left(z_{1} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\left(z_{1} + \frac{3}{4}\right) \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{1}{4}a \,\mathbf{\hat{y}}+c \left(z_{1} + \frac{1}{4}\right) \,\mathbf{\hat{z}}$ (16d) Au I
$\mathbf{B_{9}}$ = $\left(z_{2} + \frac{1}{4}\right) \, \mathbf{a}_{1}+z_{2} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $\frac{1}{4}a \,\mathbf{\hat{y}}+c z_{2} \,\mathbf{\hat{z}}$ (16d) Zn I
$\mathbf{B_{10}}$ = $z_{2} \, \mathbf{a}_{1}+\left(z_{2} + \frac{1}{4}\right) \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{1}{4}a \,\mathbf{\hat{y}}+c \left(z_{2} - \frac{1}{4}\right) \,\mathbf{\hat{z}}$ (16d) Zn I
$\mathbf{B_{11}}$ = $- \left(z_{2} - \frac{1}{4}\right) \, \mathbf{a}_{1}- \left(z_{2} - \frac{1}{2}\right) \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{1}{4}a \,\mathbf{\hat{y}}- c z_{2} \,\mathbf{\hat{z}}$ (16d) Zn I
$\mathbf{B_{12}}$ = $- \left(z_{2} - \frac{1}{2}\right) \, \mathbf{a}_{1}- \left(z_{2} - \frac{1}{4}\right) \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $\frac{1}{4}a \,\mathbf{\hat{y}}- c \left(z_{2} - \frac{1}{4}\right) \,\mathbf{\hat{z}}$ (16d) Zn I
$\mathbf{B_{13}}$ = $- \left(z_{2} - \frac{3}{4}\right) \, \mathbf{a}_{1}- z_{2} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $\frac{3}{4}a \,\mathbf{\hat{y}}- c z_{2} \,\mathbf{\hat{z}}$ (16d) Zn I
$\mathbf{B_{14}}$ = $- z_{2} \, \mathbf{a}_{1}- \left(z_{2} - \frac{3}{4}\right) \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{1}{4}a \,\mathbf{\hat{y}}- c \left(z_{2} - \frac{1}{4}\right) \,\mathbf{\hat{z}}$ (16d) Zn I
$\mathbf{B_{15}}$ = $\left(z_{2} + \frac{3}{4}\right) \, \mathbf{a}_{1}+\left(z_{2} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $\frac{1}{4}a \,\mathbf{\hat{y}}+c \left(z_{2} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (16d) Zn I
$\mathbf{B_{16}}$ = $\left(z_{2} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\left(z_{2} + \frac{3}{4}\right) \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{1}{4}a \,\mathbf{\hat{y}}+c \left(z_{2} + \frac{1}{4}\right) \,\mathbf{\hat{z}}$ (16d) Zn I
$\mathbf{B_{17}}$ = $\frac{1}{4} \, \mathbf{a}_{1}+\left(x_{3} + \frac{1}{4}\right) \, \mathbf{a}_{2}+x_{3} \, \mathbf{a}_{3}$ = $a x_{3} \,\mathbf{\hat{x}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (16e) Au II
$\mathbf{B_{18}}$ = $\frac{3}{4} \, \mathbf{a}_{1}- \left(x_{3} - \frac{1}{4}\right) \, \mathbf{a}_{2}- \left(x_{3} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a x_{3} \,\mathbf{\hat{x}}+\frac{1}{2}a \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (16e) Au II
$\mathbf{B_{19}}$ = $\left(x_{3} + \frac{1}{4}\right) \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}+x_{3} \, \mathbf{a}_{3}$ = $\frac{1}{4}a \,\mathbf{\hat{x}}+a \left(x_{3} - \frac{1}{4}\right) \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ (16e) Au II
$\mathbf{B_{20}}$ = $- \left(x_{3} - \frac{1}{4}\right) \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}- \left(x_{3} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{4}a \,\mathbf{\hat{x}}- a \left(x_{3} - \frac{1}{4}\right) \,\mathbf{\hat{y}}$ (16e) Au II
$\mathbf{B_{21}}$ = $\frac{3}{4} \, \mathbf{a}_{1}- \left(x_{3} - \frac{3}{4}\right) \, \mathbf{a}_{2}- x_{3} \, \mathbf{a}_{3}$ = $- a x_{3} \,\mathbf{\hat{x}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (16e) Au II
$\mathbf{B_{22}}$ = $\frac{1}{4} \, \mathbf{a}_{1}+\left(x_{3} + \frac{3}{4}\right) \, \mathbf{a}_{2}+\left(x_{3} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a \left(x_{3} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (16e) Au II
$\mathbf{B_{23}}$ = $- \left(x_{3} - \frac{3}{4}\right) \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}- x_{3} \, \mathbf{a}_{3}$ = $- \frac{1}{4}a \,\mathbf{\hat{x}}- a \left(x_{3} - \frac{1}{4}\right) \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ (16e) Au II
$\mathbf{B_{24}}$ = $\left(x_{3} + \frac{3}{4}\right) \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}+\left(x_{3} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{4}a \,\mathbf{\hat{x}}+a \left(x_{3} + \frac{1}{4}\right) \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ (16e) Au II
$\mathbf{B_{25}}$ = $\left(x_{4} + \frac{3}{8}\right) \, \mathbf{a}_{1}+\left(x_{4} + \frac{1}{8}\right) \, \mathbf{a}_{2}+\left(2 x_{4} + \frac{1}{4}\right) \, \mathbf{a}_{3}$ = $a x_{4} \,\mathbf{\hat{x}}+a \left(x_{4} + \frac{1}{4}\right) \,\mathbf{\hat{y}}+\frac{1}{8}c \,\mathbf{\hat{z}}$ (16f) Au III
$\mathbf{B_{26}}$ = $- \left(x_{4} - \frac{3}{8}\right) \, \mathbf{a}_{1}- \left(x_{4} - \frac{1}{8}\right) \, \mathbf{a}_{2}- \left(2 x_{4} - \frac{1}{4}\right) \, \mathbf{a}_{3}$ = $- a x_{4} \,\mathbf{\hat{x}}- a \left(x_{4} - \frac{1}{4}\right) \,\mathbf{\hat{y}}+\frac{1}{8}c \,\mathbf{\hat{z}}$ (16f) Au III
$\mathbf{B_{27}}$ = $\left(x_{4} + \frac{1}{8}\right) \, \mathbf{a}_{1}- \left(x_{4} - \frac{3}{8}\right) \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $- a \left(x_{4} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+a \left(x_{4} + \frac{1}{4}\right) \,\mathbf{\hat{y}}- \frac{1}{8}c \,\mathbf{\hat{z}}$ (16f) Au III
$\mathbf{B_{28}}$ = $- \left(x_{4} - \frac{1}{8}\right) \, \mathbf{a}_{1}+\left(x_{4} + \frac{3}{8}\right) \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $a \left(x_{4} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- a \left(x_{4} - \frac{1}{4}\right) \,\mathbf{\hat{y}}- \frac{1}{8}c \,\mathbf{\hat{z}}$ (16f) Au III
$\mathbf{B_{29}}$ = $- \left(x_{4} - \frac{5}{8}\right) \, \mathbf{a}_{1}- \left(x_{4} - \frac{7}{8}\right) \, \mathbf{a}_{2}- \left(2 x_{4} - \frac{3}{4}\right) \, \mathbf{a}_{3}$ = $- a \left(x_{4} - \frac{1}{2}\right) \,\mathbf{\hat{x}}- a \left(x_{4} - \frac{1}{4}\right) \,\mathbf{\hat{y}}+\frac{3}{8}c \,\mathbf{\hat{z}}$ (16f) Au III
$\mathbf{B_{30}}$ = $\left(x_{4} + \frac{5}{8}\right) \, \mathbf{a}_{1}+\left(x_{4} + \frac{7}{8}\right) \, \mathbf{a}_{2}+\left(2 x_{4} + \frac{3}{4}\right) \, \mathbf{a}_{3}$ = $a \left(x_{4} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+a \left(x_{4} + \frac{1}{4}\right) \,\mathbf{\hat{y}}+\frac{3}{8}c \,\mathbf{\hat{z}}$ (16f) Au III
$\mathbf{B_{31}}$ = $- \left(x_{4} - \frac{7}{8}\right) \, \mathbf{a}_{1}+\left(x_{4} + \frac{5}{8}\right) \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $a x_{4} \,\mathbf{\hat{x}}- a \left(x_{4} - \frac{1}{4}\right) \,\mathbf{\hat{y}}+\frac{5}{8}c \,\mathbf{\hat{z}}$ (16f) Au III
$\mathbf{B_{32}}$ = $\left(x_{4} + \frac{7}{8}\right) \, \mathbf{a}_{1}- \left(x_{4} - \frac{5}{8}\right) \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $- a x_{4} \,\mathbf{\hat{x}}+a \left(x_{4} + \frac{1}{4}\right) \,\mathbf{\hat{y}}+\frac{5}{8}c \,\mathbf{\hat{z}}$ (16f) Au III

References

  • H. Iwaskai, Study on the Ordered Phases with Long Period in the Gold-Zinc Alloy System II. Structure Analysis of Au$_{3}$Zn[R$_{1}$], Au$_{3}$Zn[R$_{2}$] and Au$_{3}+$Zn, J. Phys. Soc. Jpn. 17, 1620–1633 (1962), doi:10.1143/JPSJ.17.1620.
  • E. A. Wood, The Conversion Factor for kX Units to Angström Units, J. Appl. Phys. 18, 929–930 (1947), doi:10.1063/1.1697570.

Found in

  • K. Hisatsune, Y. Takuma, Y. Tanaka, K. Udoh, T. Morimura, and M. Hasaka, Martensite transformation in Au$_{3}$Zn alloy, Solid State Commun. 106, 509–512 (1998), doi:10.1016/S0038-1098(98)00077-5.

Prototype Generator

aflow --proto=A3B_tI64_142_def_d --params=$a,c/a,z_{1},z_{2},x_{3},x_{4}$

Species:

Running:

Output: